STATIC AND DYNAMIC ELECTRICITY
ELECTROMAGNETICS LIBRARY C. E. BAUM, Editor
Baum and Kritikos Electromagnetic Symmetry Baum and Stone Transient Lens Synthesis: Differential Geometry in Electromagnetic Theory Bowman, Senior, and Uslenghi Electromagnetic and Acoustic Scattering by Simple Shapes
Gardner Lightning Electromagnetics Hoppe and Rahmat Samii Impedance Boundary Conditions in Electromagnetics -
Lee EMP Interaction: Principles, Techniques, and Reference Data
Mittra Computer Techniques for Electromagnetics Smythe Static and Dynamic Electricity
Taylor and Giri High-Power Microwave Systems and Effects Van Bladel Electromagnetic Fields
STATIC AND DYNAMIC ELECTRICITY Third Edition, Revised Printing
WILLIAM R. SMYTHE Professor Emeritus of Physics California Institute of Technology
A SUMMA BOOK
Taylor &Francis Publishers since 1798
STATIC AND DYNAMIC ELECTRICITY: Third Edition, Revised Printing Copyright © 1989 by Hemisphere Publishing Corporation. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 234567890 BRBR 898765 Library of Congress Cataloging in Publication Data -
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Smythe, William Ralph, date. Static and dynamic electricity. (A SUMMA book) Includes bibliographies and index. 1. Electrostatics. 2. Electrodynamics. I. Title. QC571.S59 1989 88-24722 537'.2 ISBN 0-89116-916-4 (hard) ISBN 0-89116-917-2 (soft)
CONTENTS
TABLE OF SYMBOLS
xi
PREFACE ID THE THIRD EDITION, REVISED PRINTING
xvii
PREFACE TO THE THIRD EDITION
xix
PREFACE TO THE SECOND EDITION
xxi
PREFACE ID THE FIRST EDITION
xxiii
CHAPTER I BASIC IDEAS OF ELECTROSTATICS
1
Electrification, conductors, and insulators—Positive and negative electricity — Coulomb's law, unit charge, dielectrics—Limitations of the inverse-square law—Electrical induction—The elementary electric charges—Electric field intensity—Electrostatic potential—Electric dipoles and multipoles—Interaction of dipoles—Lines of force—Equipotential surfaces—Gauss's electric flux theorem — Lines of force from collinear charges—Lines of force at infinity—Potential maxima and minima. Earnshaw's theorem—Potential of electric double layer — Electric displacement and tubes of force—Stresses in an electric field— Gauss's electric flux theorem for nonhomogeneous mediums—Boundary conditions and stresses on the surface of conductors—Boundary conditions and stresses on the surface of a dielectric—Displacement and intensity in solid dielectric—Crystalline dielectrics—Problems—References.
CHAPTER II CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
25
Uniqueness theorem—Capacitance—Capacitors in series and parallel— Spherical capacitors—Cylindrical capacitors—Parallel-plate capacitors— Guard rings—Energy of a charged capacitor—Energy in an electric fieldParallel-plate capacitor with crystalline dielectric—Stresses when the capacitivity is a function of density—Electrostriction in liquid dielectrics—Force on conductor in dielectric—Green's reciprocation theorem—Superposition of fields—Induced charges on earthed conductors—Self- and mutual elastanceSelf- and mutual capacitance—Electric screening—Elastances and capacitances for two distance conductors—Energy of a charged system—Forces and torques on charged conductors—Problems—References.
CHAPTER III GENERAL THEOREMS
Gauss's theorem—Stokes' s theorem—Equations of Poisson and Laplace— Orthogonal curvilinear coordinates—Curl in orthogonal curvilinear coordinates v
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CONTENTS — 4 (eVV) in other coordinate systems—Green's theorems—Green's reciprocation theorem for dielectrics—Green's function—Solution of Poisson's equation —Uniqueness theorem with dielectrics present—Introduction of new conductor —Green's equivalent stratum—Energy of a dielectric body in an electric field— Effect of an increase of capacitivity—Potential of axially symmetrical fieldProblems—References.
CHAPTER IV TWO—DIMENSIONAL POTENTIAL DISTRIBUTIONS
63
Field and potential in two dimensions—Circular harmonics—Harmonic expansion of line charge potential—Conducting or dielectric cylinder in uniform field—Dielectric cylinder. Method of images—Image in conducting cylinder—Image in plane face of dielectric or conductor. Intersecting conducting planes—Dielectric wedge—Complex quantities—Conjugate functions — The stream function—Electric field intensity. Electric flux—Functions for a line charge—Capacitance between two circular cylinders—Capacitance between cylinder and plane and between two similar cylinders—Conformal transformations—Given equations of boundary in parametric form—Determination of required conjugate functions—The Schwarz transformation—Polygons with one positive angle—Polygon with angle zero—Polygon with one negative angle. Doublet. Inversion—Images by two-dimensional inversion—Polygon with two angles—Slotted plane—Riemann surfaces—Circular cylinder into elliptic cylinder—Dielectric boundary conditions—Elliptic dielectric cylinder—Torque on dielectric cylinder—Polygon with rounded corner—Plane grating of large cylindrical wires—Elliptic function transformations. Two coplanar strips— Unequal coplanar strips by inversion—Charged circular cylinder between plates —Problems—References.
CHAPTER
V
THREE—DIMENSIONAL POTENTIAL DISTRIBUTIONS
When can a set of surfaces be equipotentials2—Potentials for confocal conicoids —Charged conducting ellipsoid—Elliptic and circular disks—Method of images. Conducting planes—Plane boundary between dielectrics—Image in spherical conductor—Example of images of point charge—Infinite set of images. Two spheres—Difference equations. Two spheres—Sphere and plane and two equal spheres—Inversion in three dimensions. Geometrical properties—Inverse of potential and image systems—Example of inversion of images—Inversion of charged conducting surface—Capacitance by inversion— Three-dimensional harmonics—Surface of revolution and orthogonal wedge-Spherical harmonics—General property of surface harmonics—Potential of harmonic charge distribution—Differential equations for surface harmonics- Surface zonal harmonics. Legendre's equation—Series solution of Legendre's equation—Legendre polynomidls. Rodrigues's formula—Legendre coefficients. Inverse distance—Recurrence formulas for Legendre polynomials— Integral of product of Legendre polynomials—Expansion of function in Legendre polynomials—Table of Legendre polynomials—Legendre polynomial with imaginary variable—Potential of charged ring—Charged ring in conducting sphere—Dielectric shell in uniform field—Off-center spherical capacitor-
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CONTENTS
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Simple conical boundary—Zonal harmonics of the second kind—Recurrence formulas for Legendre functions of the second kind—Legendre functions of the second kind in terms of Legendre polynomials—Special values of Legendre functions of the second kind—Legendre function of the second kind with imaginary variable—Use of Legendre function of the second kind in potential problems—Nonintegral zonal harmonics—Associated Legendre functions—Integrals of products of associated functions—Associated functions with imaginary argument—Recurrence formulas for Legendre associated functions—Special values of associated Legendre functions—Neutral points and lines—Biaxial harmonics—Conical boundaries—Nonintegral associated Legendre functions-Green's function for a cone—Green's function for a conical box—Oblate spheroidal coordinates—Oblate spheroidal harmonics—Conducting sheet with circular hole—Torque on disk in uniform field—Potential of charge distribution on spheroid—Potential of point charge in oblate spheroidal harmonics— Prolate spheroidal harmonics—Prolate spheroid in uniform field—Laplace's equation in cylindrical coordinates—Bessel's equation and Bessel functions —Modified Bessel equation and functions—Solution of Bessel's equation— Recurrence formulas for Bessel functions—Values of Bessel functions at infinity—Integrals of Bessel functions—Expansion in series of Bessel functions—Green's function for cylinder. Inverse distance—Green's function for cylindrical box—Bessel functions of zero order—Roots and numerical values of Bessel functions of zero order—Derivatives and integrals of Bessel functions of zero order—Point charge and dielectric plate—Potential inside hollow cylindrical ring—Nonintegral order and spherical Bessel functions—Modified Bessel functions—Recurrence formulas for modified Bessel functions—Values of modified Bessel functions at infinity—Integral of a product of complex modified Bessel functions—Green's function for a hollow cylindrical ring—Modified Bessel functions of zero order—Definite integrals for the modified Bessel function of the second kind. Value at infinity—Definite integrals for Bessel functions of zero order—Inverse distance in terms of modified Bessel functionsCylinarical dielectric boundaries—Potential inside hollow cylindrical ring— Modified Bessel functions of nonintegral order—Wedge functions—Mixed boundaries. Charged right circular cylinder—Mixed spherical, spheroidal, and cylindrical boundaries—Sphere in concentric, coaxial, earthed spheroid — Charged sphere in cylinder—Cylinder in hole in sheet—Off-axis charged sphere in cylinder—Problems—References. ,
CHAPTER VI ELECTRIC CURRENT
Electric current density. Equation of continuity—Electromotance—Ohm's law. Resistivity—Heating effect of electric current—Steady currents in extended mediums—General theorems—Current flow in two dimensions—Long strip with abrupt change in width—Current flow in three dimensions—Systems of electrodes. Two spheres. Distant electrodes—Source and sink in solid sphere. Spherical bubble—Solid conducting cylinder—Earth resistance — Currents in thin curved sheets—Current distribution on spherical shell— Surface of revolution—Limits of resistance—Currents in nonisotropic mediums. Earth strata—Vector potential. Flow around sphere in tube—Spacecharge current. Child's equation—Problems—References.
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CONTENTS
CHAPTER VII MAGNETIC INTERACTION OF CURRENTS
280
Definition of the ampere in terms of the magnetic moment—Magnetic induction and permeability—Magnetic vector potential. Uniform field—Uniqueness theorems for magnetostatics—Orthogonal expansions for vector potential— Vector potential in cylindrical coordinates—Vector potential in spherical coordinates—Vector potential in terms of magnetic induction on axis—Equation of axially symmetrical tubes of induction—Vector potential and field of bifilar circuit—Vector potential and field of circular loop—Field of currents in spherical shell—Zonal currents in spherical shell—Field of circular loop in spherical harmonics—Biot and Savart's law. Field of straight wire—Field of helical solenoid—Field in cylindrical hole in conducting rod—Field of rectilinear currents in cylindrical conducting shell—Force on electric circuit in magnetic field—Examples of forces between electric circuits—Vector potential and magnetization—Magnetic boundary conditions—Example of the use of A and A— Current images in plane face—Magnetic induction and permeability in crystals —Two-dimensional magnetic fields—Magnetic shielding of bifilar circuit— Current images in two dimensions—Magnetomotance and magnetic intensity— The magnetic circuit. Anchor ring—Air gaps in magnetic circuits—Field in shell-type transformer—Slotted pole piece. Effective air gap—ProblemsReferences.
CHAPTER VIII ELECTROMAGNETIC INDUCTION
329
Faraday's law of induction—Mutual energy of two circuits—Energy in a magnetic field—Mutual inductance—Boundary conditions on A—Mutual inductance of simple circuits—Mutual inductance of circular loops—Variable mutual inductance—Self-inductance—Computation of self-inductance. Thin wire—Self-inductance of circular loop—Self-inductance of solenoid—Selfinductance of bifilar lead—Energy of n circuits—Stresses in a magnetic field— Energy of a permeable body in a magnetostatic field—Problems—References.
CHAPTER IX
349
MAGNETISM
Paramagnetism and diamagnetism—Magnetic susceptibility—Magnetic properties of crystals—Crystalline sphere in uniform magnetic field—Ferromagnetism—Hysteresis. Permanent magnetism—The nature of permanent magnetism—Uniform magnetization. Equivalent current shell—Magnetized sphere and cylinder. Magnetic poles—Boundary conditions on permanent magnets—Spherical permanent magnet in uniform field—Lifting power of horseshoe magnet—Field of cylindrical magnet—Magnetic needles—Problems —References.
CHAPTER X EDDY CURRENTS
Induced currents in extended conductors—Solution for vector potential of eddy currents—Steady-state skin e f fect—Skin effect on tubular conductor-
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CONTENTS
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Skin ef fect on solid cylindrical conductor—Solution in spherical coordinates for axial symmetry—Conducting sphere in alternating field—Power absorbed by sphere in alternating magnetic field—Transients in conducting sphere— Eddy currents in plane sheets—Eddy currents in infinite plane sheet by image method—Torque on small rotating current loop or magnetic dipole—Eddy currents from rotating dipole—Shielding of circular coil by thin conducting sheet— Rotating sheet in magnetic gap—Rotating disk in magnetic gap—Zonal eddy currents in spherical shell—Spherical shell in alternating field solenoid gap— General eddy currents in spherical shell—Torque on spinning spherical shell between magnet poles—Eddy currents in thin cylindrical shell—Eddy currents in rotating finite cylindrical shell—Eddy current damping—Transient shielding by a thick cylindrical shell—Problems—References.
CHAPTER XI PLANE ELECTROMAGNETIC WAVES
415
Maxwell's field equations—Propagation equation. Dynamic potentials. Gauges. Hertz vector—Poynting's vector—Plane waves in homogeneous uncharged dielectric insulator—Plane wave velocity in anisotropic mediums— Ray surface and polarization in anisotropic mediums—Energy, pressure, and momentum of a plane wave—Refraction and reflection of a plane wave— Intensity of reflected and refracted waves—Frequency, wave length, elliptic polarization—Total reflection—Electromagnetic waves in homogeneous conductors—Plane waves in homogeneous isotropic conductors—Reflection from conducting surface—Plane waves on cylindrical perfect conductors—Intrinsic impedance of a medium—Reflection at a discontinuity. Matching section— Complex Poynting vector—Nearly plane waves on imperfect conductors. Lecher wires—Group velocity—Problems—References.
CHAPTER XII ELECTROMAGNETIC RADIATION
448
The radiation problem—Spherical electromagnetic waves. Dipole and quadrupole radiation—Retarded Potentials—Radiation from linear antenna— Distant radiation from linear antenna—Radiation from progressive waves— Conical transmission lines—The biconical antenna—Antenna arrays—Earth effects—Uniqueness of solution—Solutions of the wave equation in spherical coordinates—Polynomial expansion for a plane wave—Radiation from uniform current loop. Magnetic dipole—Free oscillations of a conducting sphere —Forced oscillations of dielectric or conducting sphere—Solution of propagation equation in cylindrical coordinates—Expansion in cylindrical harmonics for a plane wave—Radiation from apertures in plane conducting screens— Diffraction from rectangular aperture in conducting plane—Orthogonal functions in diffraction problems. Coaxial line—Problems—References.
CHAPTER XIII WAVE GUIDES AND CAVITY RESONATORS
Waves in hollow cylindrical tubes—Attenuation in hollow wave guides—The rectangular wave guide—Green's function for rectangular guide—Aperture excitation of wave guides. Coaxial opening—Rectangular guide filled by two
493
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CONTENTS mediums—Thin iris in rectangular guide—A variational method for improving approximate solutions—Inductive iris susceptance by variational method— The circular wave guide=-Green's function for a circular guide—Loop coupling with circular guide—Orifice coupling with circular guide—The coaxial wave guide—Plane discontinuities in coaxial lines—Guides of arbitrary section. Elliptic guides—Cavity resonators. Normal modes—Independent oscillation modes of a cavity—Energy and damping of normal modes. Quality—Normal modes of a cylindrical cavity—Properties of a rectangular cavity—Properties of a right circular cylindrical cavity—Multiply connected cylindrical cavities— Coaxial cable resonators—Normal modes of a spherical cavity—Normal modes of an imperfect cavity—Effect of small foreign body on cavity mode frequency— Wall deformation effect on mode frequencies—Stationary formulas for cavity resonance frequency—Complex cavities—Excitation of cavities. Inductive coupling—Inductive coupling to a circular cylindrical cavity—Excitation of cavity by internal electrode—Cavity excitation through orifice—ProblemsReferences. CHAPTER XIV
SPECIAL RELATIVITY AND THE MOTION OF CHARGED PARTICLES
560
The postulates of special relativity—The Lorentz transformation equations— Transformation equations for velocity and acceleration—Variation of mass with velocity—The transformation equations for force—Force on charge moving in magnetic field—Motion of charge in uniform magnetic field—Energy of a charged moving particle—Magnetic cutoff of thermionic rectifier—Path of cosmic particle in uniform field—Magnetic field of moving charge—Retarded fields and potentials of moving charge—Radiation from an accelerated charge— Charge radiation when acceleration parallels velocity—Charge radiation with acceleration at right angles to velocity—Cherenkov radiation—Transformation of Maxwell's equations—Ground speed of an airplane—Motion of charged particle in crossed electric and magnetic fields—Aberration and Doppler effectProblems—References.
APPENDIX SYSTEMS OF ELECTRICAL UNITS
591
TABLE I. RELATIONS BETWEEN CGS AND MKS MECHANICAL UNITS
592
TABLE II. TRANSPOSITION OF MKS FORMULAS INTO CGS ESU
592
TABLE III. TRANSPOSITION OF MKS FORMULAS INTO CGS EMU
593
TABLE IV. TRANSPOSITION OF CGS ESU OR CGS EMU FORMULAS INTO MKS
594
TABLE V. DIMENSIONS OF ELECTRIC AND MAGNETIC QUANTITIES
596
TABLE VI. NUMERICAL VALUES
597
INDEX
599
TABLE OF SYMBOLS
Note: In this table bold-face symbols (v, u, 4), . . .) are space vectors. Phasors (I, E, . . .), phasor space vectors (E, B, II, . . .), conjugate phasors (I, E, . . .) and conjugate phasor space vectors (E, B, II, . . .) are shown by an erect ( or an inverted (^) flat vee above the symbol. Magnitudes of vectors and scalars, whether time-dependent or not, are written without designation.
A, Ao, Az, etc. Vector potential. A° Normalized vector potential. A, A., etc. Quasi-vector potential B, Bz, etc. Magnetic induction B Susceptance. B° Normalized or relative susceptance, BZk . C Capacitance. A constant. C° Normalized or relative capacitance, CZk. Velocity of light. A length. Self-capacitance. Cnn Cmn Mutual capacitance. D, Dz, etc. Electric displacement. Dw Dwight integral tables. ds Differential element along s. dr Differential change in r. E, E, E, E etc. Electric field intensity. E(k) Complete elliptic integral. e Electronic charge. 2.71828. e, 6, E, E, etc. Electromotance. se Effective or rms electromotance. F, Fz Force G Conductance, Y = G jB. Acceleration of gravity. H, H, H, H, etc. Magnetic field intensity. H(1), 11(:) (v), H.;,!)(v) Hankel functions. h Planck's constant. 4112, h3 In orthogonal curvilinear coordinates. Length elements are hidui, h2du2, hadu3. :), !IT )(v Spherical Hankel functions. o , h;," (v), 11( xi ,
xii
TABLE OF SYMBOLS 1,1,1,1, etc. Electric current.
/e Effective or rms current. ie Effective or rms current density. 1, i, ix, etc. Current density. Current. k Unit x, y, z vectors. Jn, J, (v) Bessel functions.
j (-1)'. Spherical Bessel functions. K Relative capacitivity, a/ e,,. K(k) Complete elliptic integral. Km Relative permeability, K., K.(v) Modified Bessel functions. k„, k„(v) Modified spherical Bessel functions. k Boltzman constant. L, Lnn, Ln Inductance. Lm. Mutual inductance. L° Normalized or relative inductance, L/Zk. Direction cosines with x, y, z axis. M, M Magnetization. M Mutual inductance. M, M, etc. Dipole or loop moment. M', M' Classical magnetic dipole moment (IX). m Mass. A number (usually integer). N Electric or magnetic flux. n Unit normal vector. n Index of refraction. A number. nn(y) Spherical Bessel function. nn, 2n 2n!! 2 .-4 . 6 (2n + 1)!! 1 3 . 5 • • • (2n + 1) P Polarization P, P Power. P Average power. P7(1) Associated Legendre function. Pc Peirce integral tables. p, p Momentum. p A number. wiry. w. Q Electric charge. Quality of cavity. Q Quadrupole moment. (27, Q7(.1.) Associated Legendre function. q Point or variable charge. R, R, R.., R.,. Resistance. R, R(r) Function of r only. R., R.(v) Solution of Bessel's equation. j n, jn(v)
TABLE OF SYMBOLS
R;)„ nu) Solution of Bessel's modified equation. R, R Distance between two points. r, r Distance from origin. S Area or surface. So, So Cavity cross-section areas. S, S',7 Surface harmonic. S, Sn, Smn, Snn Elastance. Self-elastance. s71n s„,,, Mutual elastance. s Distance along curve. An integer. T, T Torque. T Absolute temperature. Period. t Time. TE Transverse electric. TM Transverse magnetic. to Subscript for TE wave quantities. tm Subscript for TM wave quantities. U Stream or potential function. [U] dU around constant V curve. u, u Velocity. u cos O. u2, 2/3 Orthogonal curvilinear coordinates. V Potential or steam function. dV around constant U curve. [V] v Volume. v, v Velocity. W, W, W, W Solutions of scalar wave equation. W Energy. U jV. Wte Solutions yielding TE waves. Wt„, Solutions yielding TM waves. X Reactance. x, y, z Rectangular coordinates. Yn, Y.,„(v) Bessel functions. Y, Y Admittance, G jB. Y° Normalized or relative admittance, YZk . Z, Z Hertz vector. 2, Z, Z, Z.., Z. Impedance. 2k Characteristic impedance. Z° Normalized or relative impedance, 2/2k. Z, Z(z) Function of z only. z Complex variable, x jy. a, 0, y, 8, el
Often used for angles.
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TABLE OF SYMBOLS
p„). 0 Ratio v/c. Ratio (p. — p,)/(12 0 Free space wave number, (0(1.40 1. „ 0:,„ Wave-guide wave number, (02 — 02.7,)1. Cutoff wave number. 0..p Cavity resonance wave number. F, P Phasor propagation constant. y Electrical conductivity. A, Ara Determinant. A small part of. (5 Skin depth. Phase difference. 5 A small quantity. A small part of. 67,7 Kronecker delta, zero if m n, one if m = n. • Capacitivity. A small quantity. E„ Free space capacitivity. E, En Phase angle. n Intrinsic impedance. 0, 0(0) Function of 0 only. Colatitude angle. Unit vector in 0 direction. 0, 0', 0" Angles of incidence, reflection, and refraction. • Magnetic susceptibility. (1 — /32)-1. X Wave length. Cutoff wave length. Xg, Wave-guide wave length. X.„„ Cavity resonance wave length. Permeability. cos 0. 11„ Free space permeability, 47 X 10-7 . • Frequency in cycles per second. limn Cutoff frequency. P mn p Cavity resonance frequency. :EZ(E) Function of E only. E, 0 Oblate spheroidal coordinates. n, 4 Prolate spheroidal coordinates. H, H, lI Poynting vector. lI Rms Poynting vector. p Distance from z- or 0-axis. Charge density. Ni Unit vector in p-direction. a Surface electric charge density. s Area or surface resistivity. • Volume resistivity. Density. Time. 4' Function of 0 only. + Unit vector in 0-direction. 0 Longitude angle. Phase angle. 1' Scalar potential.
TABLE OF SYMBOLS
St Magnetomotance. Solid angle. 0.1 Frequency in radians per second. V Vector operator, is/ax is/ay ka/az. V2 Two-dimensional vector operator. a x b Vector product of a and b. a • b Scalar product of a and b. V2 Laplace operator. [v] Retarded v.
xv
PREFACE TO THE THIRD EDITION, REVISED PRINTING
It is a source of great satisfaction to my father on his 95th birthday that a revision of this book is being published. In the past 50 years it has been used as a textbook by countless students who have, sometimes painfully, learned the value of a rigorous problem course. Others have found it valuable as a reference that contains solutions to hundreds of difficult electromagnetic problems with widespread applicability. WILLIAM RODMAN SMYTHE
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PREFACE TO THE THIRD EDITION
The arrival of the digital computer since the appearance of the second edition has considerably altered the electromagnetic computation problem and made methods feasible that were probably discarded years ago as impractical. These devices iterate so efficiently that the summation of a simple, slowly convergent series is often faster than closed-form evaluations requiring interpolation in an inadequate function table. Solution of linear simultaneous algebraic equations is now easy and very useful for obtaining coefficients in the series solutions of mixed boundary and mixed coordinate problems such as those added to the text and problems of Chap. V. Many of these appear here for the first time. About 25 useful elliptic function conformal transformations are added to text and problems of Chap. IV as well as minor alterations and additions to Chaps. VI, VII, and XI (former Chap. XIII). Old Chaps. IX and X have been deleted because for about 10 years all physics and electrical engineering graduate students at the California Institute of Technology, and probably elsewhere, have seemed to be already familiar with linear circuits. The coverage of eddy currents in Chap. X (former Chap. XI) is greatly extended. Little or none of this added material appears elsewhere. Chapter XII (former Chap. XIV) on wave guides and cavities is entirely rewritten, with the addition of new matter and use of the variational method. The treatment of radiation from accelerated particles in Chap. XIII (former Chap. XV) has been expanded. The new material herein, unlike that in former editions, is unchecked by students. As much as possible of the almost error-free last edition has been left undisturbed. The author has checked the new material as well as he can but knows from past experience that some errors may survive. It is hoped that those finding them will inform him so that they may be corrected in future printings.
WILLIAM R. SMYTHE
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PREFACE TO THE SECOND EDITION
The wide use of rationalized mks units and the increased importance of microwaves made this radical revision of the first edition imperative. The units are changed throughout. The resultant extensive resetting of the text permits a modernization of nomenclature through such changes as "capacitor" for "condenser" and "electromotance" for "electromotive force," The original wording has been preserved only in the Cambridge problems. In static-field chapters, 40 problems of aboveaverage difficulty have been added, usually covering boundary conditions omitted in the first edition. The expanded treatment of electromagnetic waves made necessary the rewriting of the parts of Chap. V dealing with Bessel functions and led to the introduction of vector surface harmonics, which greatly simplify some calculations. Much of Chap. XI on eddy currents has been rewritten, and two of the three electromagnetic-wave chapters are entirely new. Both the text and the 150 problems include methods and results not found in the literature. Two groups of advanced Ph.D. students worked over this material to get practice in attacking every type of wave-field problem. Many are too difficult for first-year graduate students, but every problem was solved by at least one of the advanced students. They can be worked either directly from the text or by fairly obvious extensions of it. Some useful results appear in the problems and are listed in the Index, which should be consulted by engineers with boundary value problems to solve. Chapter XV of the first edition is omitted because none of the remaining theory is based on it and because to bring it up to date would require an excessive amount of space. None of the new topics appears to lie outside the scope of the mathematical preparation assumed for readers of the first edition. That the successful solution of electrical problems depends on physical rather than mathematical insight is borne out by the author's experience with the first edition, which shows that graduate students in electrical engineering and physics greatly excel those in mathematics. It is believed that very few of the errors and obscure or ambiguous statements in the first edition escaped the scrutiny of the 375 students at the California Institute of Technology who worked it through. No infallible system for locating errors caused by the transposition of units has been found, and the author will appreciate letters from readers pointing them out.
WILLIAM R. SMYTHE July, 1950
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PREFACE TO THE FIRST EDITION
It has been found that, in most cases, the average graduate student, even though he seems to be thoroughly familiar with advanced electrical theory, is unable to solve the electrical problems he encounters in research if they fall outside the routine types and so must be worked out from first principles. The present book is the result of having taught, for the last twelve years, a course designed to train first-year graduate students in physics, electrical engineering, geophysics, and mathematics to apply the principles of electricity and magnetism. Students in these fields must show a proficiency equivalent to an average grade in this course to be admitted to candidacy for the Ph.D. degree at this institute. It is hoped that this book will also provide a reference where workers in these fields may find the methods of attack on those problems, common in research, for which the formulas in the handbooks are inadequate. It is assumed that the reader has the mathematical preparation usually acquired in a good undergraduate course in introduction to mathematical physics. This implies a reasonable familiarity with vector analysis, the calculus, and elementary differential equations. All the mathematical development beyond this point is done herein by methods that a reader with such training can follow. The author has succeeded, with some difficulty, in avoiding the use of contour integration but believes that anyone going further into the subject should acquire a working knowledge of this powerful mathematical tool. As has already been implied, this book is written for the experimental research physicist and engineer rather than for the theoretical man. For this reason, only that theory which has applications is included, and this is developed by the most concise method compatible with the assumed preparation of the reader. No subject is included for its historical interest alone. More than the usual number of problems have been worked out in the text, and these have been selected for one or more of the following reasons: the result has important applications, it clarifies the theory, it illustrates some useful mathematical device, it proves the utility of some concept in the theory. At the end of each chapter is an extensive collection of problems involving nearly all the theory in the text. Many of these are taken from the Cambridge University examination questions as printed in Jeans's "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925, and are included with the permission of the Cambridge University Press to whom we express our thanks. Although the best students can work all the problems, the average student cannot. They provide an opporxxiii
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PREFACE TO THE FIRST EDITION
tunity for the reader who is working up the subject by himself to test his proficiency. In many cases important results, which lack of -space prevented working out in the text, are included as problems and are listed in the index. The treatment and arrangement of the subject matter depart from the most common practice in several respects. In the first place, it will be noticed that all developments are based directly on the macroscopic experimental facts rather than on the hypothetical microscopic structure of conductors and dielectrics. There are two reasons for this. The first is that, although the microscopic theory meets the crucial test, namely, gives the observed macroscopic laws within the precision of observation, this does not imply that the theory is unique or that other deductions from it are correct. The second is that the development of the most satisfactory theories, those based on wave mechanics, require mathematical technique with which we have assumed the reader to be unfamiliar when starting this book, but with which he will be quite familiar after finishing two-thirds of it. The development of these theories, necessarily brief, is therefore postponed until the last chapter. The second departure from the most common practice consists in the complete omission of the concept of the isolated magnetic pole. All magnetic theory is therefore based on the interactions of electric currents or moving charges. This logically leads to the use of the magnetic vector potential rather than the scalar potential, so that the former is used extensively, although not exclusively, in all magnetic and electromagnetic theory. It will surprise many to find that this leads to considerable simplification in certain places, notably in the calculation of inductance and the treatment of eddy currents and electromagnetic radiation. Other minor departures from the usual practice include the more extensive use of Bessel functions and conformal transformations and the treatment of forces between moving charges exclusively by the methods of special relativity. The latter procedure, besides having a very firm experimental basis, makes hypotheses as to the shape or size of electric charges unnecessary and gives a clearer idea of the limits within which the standard formulas can be applied without using such hypotheses. Several subjects usually included in books on electricity and magnetism are omitted here. The theories of electrolytic conduction, thermionic emission, photoelectric phenomena, thermoelectric effects, etc., of which the reader is assumed to have an elementary knowledge, are entirely omitted because it is believed that a treatment of them, on the same level as the remainder of the book, would require a background of physical chemistry, thermodynamics, and quantum theory not possessed by the average reader. The theory of electrical machines and instruments, including vacuum tubes, is also omitted because it is
PREFACE TO THE FIRST EDITION
xxv
believed that such subjects are best treated in connection with laboratory courses. Certain alternative methods such, for example, as the Heaviside operational method and the dynamic method of circuit analysis have been omitted for lack of space. Before starting this book the average reader will have become familiar with all the common systems of electrical units and will have his own preference. The actual system used will make little difference provided it is clearly indicated. For each section of the subject, the author has used that system which seemed simplest to work with. Thus the c.g.s. electrostatic units are used in Chaps. I to V, the c.g.s. electromagnetic units in Chaps. VII to XII, and the Gaussian system in Chaps. XIII, XIV, and XV. To avoid confusion, the units used are noted at the bottom of each page and a very complete system of conversion tables is given in the Appendix, enabling the results of any calculation to be expressed in any units. To see whether the use of rational units would have simplified calculations, all the numbered formulas occurring in the preliminary lithoprinted edition were scrutinized. It was found that the complexity of 1196 formulas would be unchanged, that of 169 would be decreased, and that of 123 would be increased. Thus in the theory there is little to choose between rationalized and unrationalized units. The results in problems were not investigated. It has been the practice throughout, whenever an integration is performed or a mathematical transformation made, to refer, by number, to the appropriate formula in both Peirce's "A Short Table of Integrals," Ginn, 1929, and Dwight's "Tables of Integrals and Other Mathematical Data," Macmillan, 1934 [4th ed., 1961]. It is therefore advisable for the prospective reader to procure a copy of one of these inexpensive tables. The bibliographies at the ends of the chapters are not by any means complete but include merely those books that have come to the author's attention which contain useful additional material or instructive alternative treatments. The author has used every device he can think of to eliminate mistakes but is perfectly certain that they still persist and will be grateful to anyone pointing them out. The author wishes to express his gratitude to the hundred graduate students, especially Dr. Pasternack, who worked through the preliminary lithoprinted edition, zealously detecting errors and obscure spots, and checking the answers to problems. He also wishes to thank Professors Ira S. Bowen and William V. Houston for reading sections of the manuscript and for valuable discussions. In particular, he wishes to thank Dr. Charles H. Townes for checking all the derivations in the final manuscript and the answers to those problems not appearing in the preliminary edition. WILLIAM R. SMYTHE August, 1939
CHAPTER I BASIC IDEAS OF ELECTROSTATICS 1.00. Electrification, Conductors, and Insulators.—The word "electricity" is derived from the Greek word for amber. Apparently Thales of Miletus first discovered, about 600 B.C. that amber, when rubbed, attracts light objects to itself. It is now known that most substances possess this property to some extent. If we rub with a piece of silk a glass rod or a metal rod supported by a glass handle, we find that both will attract small bits of paper, and we say they are electrified. We see that only in the case of the metal rod can we destroy the electrification by touching with the finger. Furthermore, we find when we touch the electrified metal rod with pieces of various materials held in the hand that metals and damp objects destroy the electrification whereas substances like glass and silk do not. We call substances that remove the electrification conductors and those which do not insulators. We can find materials that remove the electrification very slowly and that we might call poor conductors or poor insulators. Thus there is no definite division between the two classes. 1.01. Positive and Negative Electricity.—If we rub a glass rod with silk and touch some light conducting body, such as a gilded pith ball suspended by a silk string, with either the rod or the silk, we find it is electrified. If two balls are electrified by the glass or by the silk, they repel each other; but if one is electrified by the rod and one by the silk, they attract. Thus we conclude that there are two kinds of electrification and that bodies similarly electrified repel each other and those oppositely electrified attract each other. By experimenting with many substances, we conclude that there are only two kinds of electricity. Arbitrarily we define the electricity on the glass rod to be positive and that on the silk negative. 1.02. Coulomb's Law, Unit Charge, Dielectrics.—We next observe that the force between the balls decreases rapidly as they are separated. Coulomb investigated these forces with a torsion balance and found that the force between two small electrified bodies is directed along the line joining them, is proportional to the product of the charges, and is inversely proportional to the square of the distance between them. This is known as Coulomb's or Priestley's law. The statement of this law implies the definition of a quantity of electrification or of electric 1
2
BASIC IDEAS OF ELECTROSTATICS
§1.03
charge. We formulate this as follows: An electrostatic unit of charge is that charge which, when placed one centimeter away from a like charge in a vacuum, repels it with a force of one dyne. The practical unit of charge is the coulomb which is 3.0 X 109 electrostatic units. In many homogeneous, isotropic, nonconducting mediums, the inversesquare law is found to hold, but the force between the same charges is reduced, so that for such a medium we may write this law, in mks units, 4
F =
21
4/rer2
(1)
7.1
where F is the force on the charge of q' coulombs due to the charge q, r the vector from q to q' whose magnitude is r meters, r1a unit vector along r, and e the capacitivity which is a constant characteristic of the medium. In vacuo the numerical value of e is 8.85 X 10-12 farad per meter and is written Ev. The specific inductive capacity or relative capacitivity K is the ratio E/E,, which is dimensionless and has the same numerical value for all systems of units. 1.021. Limitations of the Inverse square Law. The precision of Coulomb's measurement has been greatly exceeded by modern methods, and we may now state that the exponent of r in 1.02 (1) has been verified, recently, to about 1 part in 109 for measurable values of r. It should be borne in mind that we can apply Coulomb's law with certainty only to dimensions for which it has been verified. We shall endeavor throughout this book to avoid basing any macroscopic theory on the assumption of the validity of this law at atomic distances. The law applies strictly only to charged bodies whose dimensions are small compared with the distances between them. Their shape and composition are immaterial. 1.03. Electrical Induction. The fact that the electrical charge on a conductor is mobile indicates that when an electric charge is brought near an uncharged conductor electricity of the opposite sign will move to the parts nearer the charge and that of the same sign to the parts more remote from it. From Coulomb's law, the force between nearer charges will be greater so that the charge will attract the conductor. These charges on the conductor are known as induced charges. Unless the more remote charge is removed, for example, by touching with the finger, the conductor will return to its neutral state when the inducing charge is removed. If, with the inducing charge in place, we separate the near and far parts of the conductor, being careful to keep them insulated, we find that the two parts are oppositely charged, as we would expect. Many "static machines" have been devised that repeat this operation automatically and accumulate the separated charges. When sufficiently sensitive methods are used, it is found that a charge -
—
—
§1.05
ELECTRIC FIELD INTENSITY
3
also exerts a small attractive force on uncharged insulators. This seems to indicate that even in insulators charges are present and are not absolutely fixed but may suffer some displacement. The hypotheses concerning the actual behavior of the electric charges in conductors and insulators will not be discussed here. The theories are still imperfect but have greatly improved since 1930. 1.04. The Elementary Electric Charges.—It has been found that an electric charge cannot be subdivided indefinitely. The smallest known negative charge is that of the negative electron and mesotron, first determined with considerable precision by Millikan. The accepted value is now 1.60 X 10-" coulomb. The smallest known positive charge is that on the positron, or positive electron, on the mesotron, and on the proton. To a high order of precision, all these elementary charges are the same in magnitude. The mass of the negative electron and also, probably, that of the positron is 9.1 x 10 -31kg. The mass of the proton is about 1837 times that of the electron. In handling electrical problems, we shall treat electrical charges as infinitely divisible, using charge density as if this were so. Clearly such a procedure is justifiable only when we are dealing with quantities much greater than 1.60 X 10-'9 coulomb and certainly is useless if we go to atomic dimensions. We have seen that the electricity in conductors is free to move, so that if it possess inertia we might expect it to lag behind when the body is accelerated, thus producing an electric current which could be detected by its magnetic field. Maxwell predicted such phenomena but their magnitude is so small that they were detected and measured by Tolman, Barnett, and others not until long after his death. The results show that the mobile electricity in conductors is negative and that the ratio of its charge to its mass equals, within experimental error, the ratio for the negative electron. It appears that almost all the phenomena with which we deal in this book involve the distribution or motion of such electrons, a positive charge appearing when there is a dearth of electrons. As far as the mathematics is concerned, it is immaterial whether the positive, the negative, or both take part in the transfer of electric charge. 1.05. Electric Field Intensity.—When an infinitesimal electric charge, placed in a region, experiences a force, we say an electric field exists there. We define the electric intensity to be a vector equal to the force per unit charge acting on a positive charge placed at that point, the charge being so minute that no redistribution of charge takes place due to its presence. The last qualification is necessary because of the phenomenon of electric induction. Just as a number of mechanical forces acting on a body can be resolved into a single resultant force by taking their vector sum, so the resultant
§1.06
BASIC IDEAS OF ELECTROSTATICS
4
electric intensity due to any charge distribution can be obtained by taking the vector sum of the electric intensities due to each element of the distribution. Thus the electric intensity at point P due to n charges is
E=
1
qi
4re
r2
(1)
i=i
where E„ is the electric intensity in volts per meter, ri is the vector of magnitude ri directed from P to qi, and e is the capacitivity of the infinite homogeneous medium in which the system is immersed. 1.06. Electrostatic Potential.—Work is done against electric forces when a charge is moved in an electric field. The potential in volts at a point P in an electrostatic field is the work in joules per coulomb to bring a charge from the point of zero potential to the point 1‘,P, Vie charge being so 0 0 small that there is no redistribution FIG. 1.06. of electricity due to its presence. The choice of the point of zero potential is a matter of convenience. It is frequently, but not always, chosen at infinity. To avoid other than electrostatic effects, it will be necessary to move the charge very slowly. Let us compute the potential due to a point charge q. The work dV required to move a unit charge a distance ds in a field of intensity E is —E • ds or —E ds cos 0 where 0 is the angle between E and ds. In the case of the point charge, this becomes
dV = _q cos 0 ds 4irer2
where r is the vector from the charge q to the element of path ds and 0 is the angle between r and ds as shown in the figure. Evidently dr = ds cos 0, so that we have for the potential in volts
1.
VP
q dV = — —
'
dr
4ire fr 2
Or
VP =
471.6 \r,,
—
)
ro /
(1)
If r0is chosen at infinity, this becomes
V, =
(2)
q
47i- Erp
the electrostatic potential is a scalar point function and is independent of the path by which we bring our charge to the point. The potential at any point in an electrostatic field can be obtained by adding up the potentials due to the individual charges producing the field; thus qi
Vp = 1
LlireLJri 1=1
where ri is the distance from P to qi in meters.
(3)
ELECTRIC DIPOLES AND MULTIPOLES
§1.07
5
Since it is usually much easier to take a scalar sum than a vector sum, it is evident why (3) is to be preferred to 1.05 (1) for purposes of computation. The field intensity at P can be obtained from (3) by taking the gradient; thus E = —grad V = —17V (4) In rectangular coordinates, the components are
av Ex = ax
ay E, = --, ay
E, = —
aV az
(5)
The components of the gradient in any other fixed coordinate system can be obtained by expressing V and the components of the gradient in terms of that system. A method of doing this is given in 3.03 and 3.05. If the elementary charges are very close together compared with the dimensions involved, as is always the case in ordinary practice, we may treat the distribution as continuous and speak of the charge per unit volume as volume charge density p and the charge per unit area as the surface charge density o. The summation in (3) now becomes an integral V, =
1 fpdv zire v r
1 radS zirEJ s r
(6)
where dv is the element of volume and dS is the element of surface. It should be noted that this formula applies only when all space, including the interior of any material bodies present, has the capacitivity e. When this is not the case, we must use the methods of Chaps. IV and V. 1.07. Electric Dipoles and Multipoles. Consider the potential at a point P at x, y, z due to charges — q at xo, Ye, zo and +q at xo h,yo, zo for small h. If the distance r_ = f(xo, yo, zo, x, y, z) from — q to the field point P is [(x — xo) 2 (y — yo)2 (z — zo)91, then from +q to P it will be r + = f(xo h, yo, zo, x, y, z). Write down this potential from 1.06 (2) and expand r;1in a Taylor series by Dw 39, which gives —
V
4ire\r+
4reLr_
a2 1 \ axAr_) 2! axo2V_)
If h 0 and q 00 in such a way that hq = M remains a finite constant, the result is an electric dipole of moment M in the x-direction. Only the h term in the expression above survives, so that, if 0 is the angle between 11 and r, which replaces r_ as h 0, there results V—
M a (1) — m a (1) m cos o m • r 4re axo r 4ire ax r 4rer2 Lirer3
(1)
Evidently this can be extended so that if the potential V, at the point P due to a set of n charges, the radius vector from q, to P being r„ is given by 1.06 (3), then the potential P, due to a set of n dipoles of the same strength and sign located at the same points with axes parallel to x is
BASIC IDEAS OF ELECTROSTATICS
6
vt
§1.071
n
p aV
ax,
(xp —
(2)
41-Er2 := 1
By differentiating the expression for the potential of a unit electric dipole with respect to any of the rectangular coordinates, we get the potential due to a unit quadrupole whose dimensions are QL2; thus 1 a2(1/r)
1 a2(1/r) 47E axe
etc.
ax ay
represent the potentials due to a linear quadrupole (Fig. 1.07a) and a square quadrupole (Fig. 1.07b), etc. Further differentiations give us the potentials due to more complicated multipole arrangements but always ones in which the total charge is zero. For other cases see 1.12. y .dxrdx 4-
•
4 -2q +4
x
dy +q • dx •-q
(b)
(a) FIG. 1.07.
The translational force on a dipole M in a field E will be the vector sum of the forces on each charge. Since the charges are equal and opposite, this means the vector difference in the field strength (ds • V )E, at the two ends, multiplied by q; thus F = q(ds • V)E = (m • V)E
(3) In a uniform field, this force is zero. In a uniform field, the charges are subject to forces -FqE and — qE applied at a distance ds sin 0 apart where 0 is the angle between ds and E. Thus there is a torque acting on the dipole of amount (4) T = tEq ds sin 0 = tME sin 0 = M x E where t is a unit vector normal to M and E. 1.071. Interaction of Dipoles. The potential energy of a dipole, in any field whose potential is V, equals the total work done, against this field, in bringing each charge separately into place. If the potential of the field at P1where +q is placed is Vi and at P2 where — q is placed it is V2, then av iv = q(V — V 2) = qPir- 2av (1) as = m as where s is in the direction of the dipole axis and M is its moment. In vector notation, this is IV = (m • V) V (2) —
LINES OF FORCE
*1.08
7
If miand M2 are the vector moments of two dipoles A and B and if r is the vector from A to B, then the potential at B due to A is, from 1.07 (1), i cos 0 mm• i
47re V -
r2
r = -mi • V( 1)
(3)
Substituting this value for V in (2) gives 471-eW = ±m2 • V(mi • r7-3) = m2 • (7-3Vmi • r Ml •T or-3) M2 • (m1r-3 - M1 • r3rr-5) = M1 • m27-3- 3M1 • TM2 • rr-5 If M1 and M2 make angles 0 and 0' with r and ¢ with each other, we have W=
mi m
2 (cos
4rer 3
47 - 3 cos 0 cos 0')
(4)
If ik is the angle between the planes intersecting in r which contain M1 and M2, then, taking r in the direction of x and M1in the xy-plane, the direction cosines are i1 = cos 0,12 = cos 0', m l =sin 0, m2 = sin 0' cos and n1 = 0, so that cos 4,and W become, respectively, cos 47 = 1112 + mim2 W=
mim2
4rer 3
n1n2 = cos 0 cos 0' + sin 0 sin 0' cos
(si n0 sin 0' cos ¢ - 2 cos
cos 0')
(5)
This gives the radial force between two dipoles, by differentiation, to be
F=
aw = 3mim2(sin 0 sin 0' cos ¢- 2 cos 0 cos 0') car
4rfr4
(6)
This has its maximum value when IP = 0, 0 = 0' = 0. The torque trying to rotate the dipole in the direction a is obtained similarly, giving
aw
T = -as
(7)
1.08. Lines of Force.—A most useful method of visualizing an electric field is by drawing the "lines of force" and " equipotentials." A line of force is a directed curve in an electric field such that the forward drawn tangent at any point has the direction of the electric intensity there. It follows that if ds is an element of this curve, ds = AE (1) where X is a scalar factor. Writing out the components in rectangular coordinates and equating values of X, we have the differential equation of the lines of force
dx _ dy dz (2) E. E„ E L Analogous equations can be written in terms of other coordinate systems by using 3.03 and 3.05. Usually much easier methods of getting the equations of the lines of force exist than by integrating these equations.
8
BASIC IDEAS OF ELECTROSTATICS
§1.08
We shall, however, give one example of the integration of this equation. Consider the field due to two charges +q at x = a and ±q at x = —a. Since, from symmetry, any section of the field by a plane including the x-axis will look the same, we may take the section made by the xy-plane. The sum of the x-components of the intensity at any point due to the two charges is E, where
ex(x——a)a)2]i — [y2 ex(x+ a)a)21i 4reEl — [y2 + If we substitute, = x + a and v — x — a y y this becomes qv 47EL y2(1 + v2)1 y2(1 + u2)1 Similarly, 47rEE'r —
y2(1 + v2)1
(3)
+
y2(1 + u2)1 Equation (2) becomes dy _ E, _ (1+ v2)i ± (1 + u2)1 u(1 v2)1 ± v(1 + u2)1 dx Solving (3) for y and x and taking the ratio of the differentials give
dv—du dx u dv — v du
dy
Comparing these two expressions for dy/dx, we see that
du_ (1 ± WV' + ) dv = Separating variables and integrating, we have u(1 + u2)-1 ± v(1 + v2)-1 = C In terms of x and y, this becomes
(x
a)[(x
a)2 + y2]-1 ± (x — a)[(x — a)2 + y2]-1 = C
(4)
This is the equation giving the lines of force shown in Figs. 1.08a and 1.08b in which the value of C for each line is marked. An easier method for getting this equation, by Gauss's electric flux theorem, is given in 1.101. We may write the left side of (4) in the form
(x
a)r-1(1 + 2axr-2 + a2r-2)-1— (x — a)r-1(1 — 2axr-2 + a2r-2)-1
where r2 = x2 + y2.If we let a —> 0, expand the radicals by Pc 750 or DID 9.03, neglect the square and higher powers of a, and write C' for C/(2a), we obtain
EQUIPOTENTIAL SURFACES
§1.09
Y2
— 3
= CI =
sin2
9 (5)
which is the equation of the lines of force for an electric dipole, shown in Fig. 1.08c.
FIG. 1.08a.—Field about equal charges of opposite sign. Lines of force and equipotential lines are shown by solid and dotted lines, respectively.
1.09. Equipotential Surfaces.—An equipotential surface in an electric field is one at all points of which the potential is the same. The equation of an equipotential surface is therefore
V= C
(1)
where C is a constant. Numerous maps of electric fields showing equipotentials and lines of force will be found in subsequent chapters. It should be noted that since no work is done in moving charges over an equipotential surface, the lines of force and equipotentials must intersect orthogonally. As an example of the use of this equation, we shall take the case just considered. The potential at any point P will be C, where q[(x — a)2
Y2]-1-T q[(x
a)2
y2]-1= 4rreC
(2)
10
BASIC IDEAS OF ELECTROSTATICS
§1.10
This is the equation of the equipotential surfaces shown in section in Figs. 1.08a and 1.08b by broken lines. C values are for q = 4re. There are often points or lines in an electrostatic field where an equipotential surface crosses itself at least twice so that V V vanishes there.
1.08b.—Field about equal charges of the same sign. Lines of force and equipotential lines are shown by solid and dotted lines, respectively.
These are called neutral or equilibrium or singular points or lines. The origin in Fig. 1.08b is such a point. Some of the properties of such points are considered in 5.235. 1.10. Gauss's Electric Flux Theorem.—We shall derive this flux theorem from the inverse-square law on the assumption, to be modified later, that all space is filled with a uniform dielectric.
§1.10
GAUSS'S ELECTRIC FLUX THEOREM
11
Consider a small element dS of a closed surface (Fig. 1.10), whose outward normal makes an angle a with the radius vector from a point charge q at P. Now draw a line from every point of the boundary of dS to the point P so that the small cone so formed cuts out an area dI from the spherical surface, having P as a center and passing through Q. Then
= dS cos a. The normal component of the intensity at Q due to a charge q at P is —
qr • n _ q cos a 471-Er3 4T-Er2
The normal component of the flux through dS is defined as
dAr= eEn dS =
q cos a dS _ q dZ 471-r2 tirr 2
But the solid angle subtended by dS at P is dIr-2= d2, so that 4r dN = q d12 We notice that if the point is inside the surface the cone cuts the surface n times where n is an odd integer, and the angle a will be acute *(n + 1) times and obtuse i(n — 1) times so that the net value of the flux through the cone is (q/47) da But if the point is outside the surface, n is even and we have the same number of positive and negative values of (12 so that the net contribution is zero. To obtain the total flux through the surface surrounding the charge, we integrate the normal component and obtain or f dN = qJo N= q Js By adding up the flux due to all the charges inside, we arrive at Gauss's electric flux theorem which is: If any closed surface is taken in an electric field and if E is the electric intensity at any point on the surface and n the 4r
12
BASIC IDEAS OF ELECTROSTATICS
§1101
unit outward normal vector to the surface, then
efsE • n dS = q
(1)
where the integration extends over the whole surface which includes the charge q. If the space outside this surface is not homogeneous but contains various dielectric and conducting bodies, it is necessary to make certain hypotheses as to the electrical behavior of matter in electrostatic fields. It is assumed therefore that, as regards such fields, matter is entirely electrical in nature consisting of positive and negative charges whose fields obey the inverse-square law. With this hypothesis the electrostatic effect of any material body is obtained by superimposing the fields of its constituent charges as calculated from this law. Equation (1) is valid, therefore, no matter what the nature of the dielectric or conducting material outside the surface considered may be because the fields of external charges were considered in its derivation. The above hypothesis is used, explicitly or implicitly, in most treatments of electrostatics. 1.101. Lines of Force from Collinear Charges. To illustrate the application of this theorem, we shall use it to find the equation of the —
lines of force about any set of collinear electric charges qi, q2, q3 . . . situated at x,, x2, x3. . . along the x-axis. From symmetry, no lines of force will pass through the surfaces of revolution generated by rotating the lines of force lying in the xy-plane about the x-axis. Gauss's electric flux theorem, applied to the charge free space inside such a surface between the planes x = A and x = B (Fig. 1.101), tells us that the total normal flux N entering through section A equals that leaving through section B since none passes through the surface. The equation of the surface is therefore obtained by setting N equal to a constant. From 1.05 (1), N equals the sum of the normal fluxes due to each charge and, as we have just found in proving Gauss's electric flux theorem, this is 47N =
q202 q30 3 ± • ' •
13
POTENTIAL MAXIMA AND MINIMA
§1.11
where 03, 22, 23 . . are the solid angles which section A subtends at Xi, x2, x8 . . . In terms of the angles al, a2, a3 . . . in Fig. 1.101 this is
N = Ziq,(1 — cos «,) = C' — -}Zqi cos ai i=i i-1 Collecting the constants on one side of the equation and substituting for the cosines in terms of the coordinates x, y of a line of force in the xy-plane, we have C = Zq,(x — xi)[(x — x,.)2 +y2]-1
(1)
for the equation of a line of force. This agrees with 1.08 (4). 1.102. Lines of Force at Infinity.—If f = [( x — X)2 y9 1we neglect 2 and write f and n [(2 — /f]n when x —
c
=x
211qi
f-
Fi Lf
(
x — X)2]qi(2 f3
i=i
=
z —
qi (1) iml
where 2 is the coordinate of the "center of gravity" of the charges. Thus at infinity, the field is the same as if the algebraic sum of the charges were concentrated at their center of gravity. We may extend this rule to noncollinear charges by pairing off members of a group of charges and taking the centers of gravity of each pair by (1), then pairing off these centers of gravity, etc., until we arrive at the center of gravity of the group.
1.11. Potential Maxima and Minima. Earnshaw's Theorem.—Consider a small spherical surface enclosing a point P in an electric field. The average value of the potential over this sphere is
rr
1 2' V dS = — f irf V sin 0 d0 47r2s 47 o Taking derivatives and applying Gauss's electric flux theorem gives V =
1
cif _ 1 far 21. v sin n 0 de 45 dr 47rjo Jo ar f dV — dS = — q 4irr2 47r-r2 s dr where q is the charge inside the sphere, giving, on integration, the result
=4irer + c If q = 0, the average value of the potential over the small sphere enclosing P is the same as at P. Hence the theorem that the potential cannot have a maximum or a minimum value at any point in space not occupied
14
BASIC IDEAS OF ELECTROSTATICS
§1.13
by an electric charge. It follows from the definition of potential that to be in stable equilibrium a positive point charge must be at a point of minimum potential and a negative point charge must be at a point of maximum potential, the potential due to the charge itself being excluded. From the theorem just proved, this is impossible, giving us Earnshaw's theorem which states that a charge, acted on by electric forces only, cannot rest in stable equilibrium in an electric field. This shows that if we are to consider matter as purely electrical in nature and as made up of positive and negative charges controlled by electrical forces only, these forces must be different from those operating on a laboratory scale in electrostatics. 1.12. Potential of Electric Double Layer.—We saw in 1.07 that the potential due to a dipole may be obtained from that due to a single charge by differentiation in the direction of the dipole axis. From this, we see that if the potential at P due to an element dS of a surface having a charge density a is
dV =
dS 4rer where r is the distance from dS to P, then
dS
4re is the potential at P due to a dipole of strength a- dS which has the direction of n. Therefore, if we have an electric double layer of moment (I3 per unit area, the potential at P due to it is
v=
14,
I 4,n • r dS = dS (1) 4rej s r3 an\r But we saw in 1.10 that n • r r-3dS = 6/0 where dfl is the solid angle subtended at P by the element dS. Thus we have (
V = f c3 d (2) 4re If the double layer is of uniform strength NI!, this becomes qf V= (3) 4ire where 12 is the total solid angle subtended by it on the positive side at P. 1.13. Electric Displacement and Tubes of Force. The product of the capacitivity by the electric intensity occurs frequently and, in an isotropic dielectric, this product is designated as the electric displacement D; thus (1) D = eE In the mks system of units, D is measured in coulombs per square meter and E in volts per meter. Lines of electric displacement, which are analogous to lines of electric —
§1.14
STRESSES IN AN ELECTRIC FIELD
15
intensity, will have the same direction as the latter in an isotropic dielectric, but if e is greater than E,, they will be more closely spaced. Taking a small element of area normal to the displacement and drawing lines of displacement through its boundary, we cut out a tubular region in space known as a tube of force. Applying Gauss's electric flux theorem to the charge free space bounded by two normal cross sections of such a tube, we see that, since there is no contribution Si to the surface integral by the side walls of the tube, the flux entering one end must equal that leaving the other, so that if the areas of the sections are Si and 82 the flux through the tube is
N = SID]. = 82D2 Thus we define a unit tube of force to be one in which the flux through any section is unity. Many diagrams showing tubes of force will be found in subsequent chapters. Since there are 41- square meters surface on the sphere of unit radius surrounding a point charge q and since the displacement on this surface is q/(47r), it follows that there are q unit tubes of force emerging from a charge q. Thus the charge on the end of a unit tube is 1 coulomb. 1.14. Stresses in an Electric Field.—We have introduced the conception of lines and tubes of force merely as an aid toward visualizing an electric field. It is possible to carry this idea considerably further, as Faraday did, and actually look upon these tubes as transmitting electrical forces. Since this is sometimes a -±q most useful method of attacking a problem, let us see if a system of FIG 1.14. stresses can be postulated to account for observed electrical forces. Let us determine what function of the electric intensity the tension along the tubes of force must be in order to account for the observed Coulomb force between two equal charges of opposite sign at a distance 2a apart. Call this function 43(E). From 1.05 (1), the intensity in the plane of symmetry is q cos 3 0 2aq E= 27rea2 4re(a2+ y 2)1 The geometry is shown in Fig. 1.14. The ring element of area is 2ira2 sin 0 dS — dO cos' 0
BASIC IDEAS OF ELECTROSTATICS
16
§1.14
Writing the Coulomb force on the left and the tension across the yz-plane on the right and dividing both sides by 27T-a2 give n2
3272ea4
—
f2 ( q cos' Oyin 0 de 1 f CE) dS = 2n-Ea2 cos3 27r C2
(1)
Let x = q/(27rea2), and expand cp in powers of E; then
e 2 8
1 2 7ra2
2 f cos3(n-1) 0sin 0 dB
C„E' dS = n =0
n =0
This equation must hold for all values of q and a and hence for all values of x. Thus C. = 0 except when n = 2. For this we have, canceling out x, r
2
e g = C2
- -
cos3 0 sin 0 d0 =
C2
4
Thus we find that
4,(E) = e
2
(2)
This is the tension along the lines of force required to account for the Coulomb law of attraction between charges of opposite sign. It is evident in the preceding case that if there were only a tension along the tubes of force they would shorten as much as possible and all would lie in the line joining the charges. Since we know that in equilibrium they fill the entire space surrounding the charges, there must be some repulsion between them to prevent this. To determine this pressure NI,(E), let us consider the force between two equal charges of the same sign. This differs from the case just considered because lines of force terminate at infinity. The tension per unit area across a spherical surface of large radius falls off as the inverse fourth power of its radius because of (2) and the inverse-square law. The area of the sphere increases only with the square of its radius so that no force is transmitted this way. Therefore the entire force may be considered as due to the repulsion of the lines of force across the plane of symmetry. From 1.05 (1), the intensity in this plane is
E—
= q cos2 0sin 0
2qy
2irea2 Proceeding as before, we obtain instead of (1) q2
471-6(a2
1 f
y2)i
2
(3)
cos2 0 sin 0)sin d0 24) 271-ta COS3 cos For the same reasons as before, 4,(E) must be of the form C2E2, and we evaluate C2 in the same way which gives 32,2,a4
271-a2
NP(E) dS
=
Jo
§1.15
GAUSS'S ELECTRIC FLUX THEOREM
17
r
2
-- = 8
C2f
Sins 0 cos 0 dO =
C2
Thus we find that
(E) = — eE
2
(5
)
This is the repulsive force per unit area between adjacent lines of force necessary to account for the Coulomb law of repulsion between charges of the same sign. These results may be put in the equivalent forms
eE2 E •D=D2 2 2 2€
(6)
Since (1) and NI/ are functions of E and E only, the origin or shape of the field is immaterial and they must have the same form in all fields.
1.15. Gauss's Electric Flux Theorem for Nonhomogeneous Mediums. We are now prepared to extend Gauss's electric flux theorem to isotropic mediums within which the capacidS tivity varies continuously from place to place. Let us suppose that a point charge q is situated at a point P somewhere inside a closed surface S in such a medium. We shall draw about P a sphere S' so small that e' is constant throughout it. Now take an element dS of the surface S, so small that E is constant over dS, and consider the tube of force, of which dS is a section, that cuts out an element dS' on S' and ends on q. Now apply Gauss's theorem to the uncharged dielectric inside the tube between dS and dS'. Since the normal component of D over the walls is zero, the only contribution to the surface integral is from dS and dS', so that
e'E' • n' dS' = eE • n dS Integrating over the two surfaces, we have
e' fs,E' • n' dS' = fseE • n dS since e' is the same for all elements dS'. But we have already proved in 1.10 that the left-hand integral is q, so that
fseE • n dS = q
(1)
where e and E are both functions of position. This may be extended as before, so that q includes all the charges inside S. A complicated field may result from simple sources. The application
18
BASIC IDEAS OF ELECTROSTATICS
§1.17
of (1) in such a case is often greatly simplified by calculating the flux from each source separately and adding the result. Thus j(Ei E2
+ • • • + En) • n
dS = jEl •n dS 5E2 • n dS + • • • + 5E. • n dS
(2)
This permits many problem answers to be written down by inspection.
1.16. Boundary Conditions and Stresses on the Surface of Conductors. When the electric charge on a conductor is in static equilibrium, there can be no fields in the interior or along the surface. Otherwise, since by definition charges are free to move on a conductor, there would be a movement of the charges, contradicting the postulated equilibrium. It follows that the whole conductor is at one potential and that lines of force meet its surface normally and end there. The charge density on the surface is a- coulombs per square meter, and there is one unit tube of force, pointing outward if 0- is positive, from each charge; so we have (1) D = eE = Since lines of force leave the conducting surface normally, they can intersect each other only at infinitely sharp points or edges. We have seen that this occurs at a mathematical point or edge. The converse also holds. At the bottom of a sharp V-shaped groove or conical depression, D and a are zero. We have seen in 1.14 that there is a tension along these lines of force of the amount 2 F = D2 a (2) 2e 2€ and so this is the outward force per square meter on a charged conducting surface. This force is independent of the sign of the charge. It should be noted that we have neglected the hydrostatic forces that may be present in the dielectric owing to its tendency to expand or contract in an electric field. An expression including such forces will be derived later in 2.09.
1.17. Boundary Conditions and Stresses on the Surface of a Dielectric.—Let us apply Gauss's electric flux theorem to the small disk-shaped volume whose flat surfaces of area dS lie on opposite sides of the plane boundary between two dielectrics e' and e" (Fig. 1.17a). The disk is so thin that the area of the edges is negligible compared with that of the faces. If there is no free charge on the surface and the normal components of displacement are D„ ", we find from 1.15 that ' and Da D'„ dS = D',,' dS
or
D' = D„" (1)
The stress on this surface due to the normal components of the displacement must be the difference between the stresses on the two sides, so
BOUNDARY CONDITIONS AND STRESSES
§1.17
19
that, from 1.14 (6), we have — e") K' — K" T =D'n2 D'.' 2 (2) 2e'e" 2E,, 2e' 2e" = KW" Let us consider the work required to take a unit charge around the path shown in Fig. 1.17b in which the length of path normal to the boundary is vanishingly small. Starting at any point, the work done in taking
7
/ / / 7 dS D
.,s' , ,\\
1,.,,
Et
a's / = , ====
ex if
X ds
\ ( \ \ , ,
Et
FIG. 1.17b.
FIG. 1.17a.
a unit charge around the path is, if energy is to be conserved, zero, so that E; ds = g' ds or E; = E;' (3) The pressure on the interface will be the difference in the pressure on the two sides; thus, from 1.14 (5), we have
— -€"g'2 = ig2(e — e") = riE;2e,,(K' — K")
P,. =
(4)
We may state then that at the uncharged interface between two dielectrics, the normal component of the displacement and the tangential component of the intensity are continuous. In terms of the potential these boundary conditions may be written
,av' an
E
„av an
"
K' a' v
or
n
Knav" an
(5)
V' = V"
(6)
where V' and V" are the potentials in e' and e". Equation (6) implies that the zero of potential is chosen in both mediums, so that at some point on the boundary V' = V". It then follows, by integration of (3), that (6) holds for all points on the boundary. The normal stresses directed from e" to e' on the interface between two dielectrics may also be written, from (2) and (4),
F = T. — P. =
K' — K"{D;2 2E ,K'
K'
e' — e"(D'2 t D'2) n
K" =
2e'
E'
e"
(7)
The foregoing formulas do not take account of the fact that some dielectrics tend to contract or expand in the presence of an electric field. In such a medium, there will be an additional force of a hydrostatic nature acting on the surface. An expression including this force will be derived later, in 2.09. At the boundary between two isotropic dielectrics, the lines of force and the lines of displacement will be refracted in the same way. Let
20
BASIC IDEAS OF ELECTROSTATICS
§1.18
the angle between E1 or D1, in the dielectric E1 and the normal to the boundary be al, and let the corresponding angle in €2 be a2 (Fig. 1.17c). Then from (1) and (3), we have D1cos al= 61E1 cos al = D2 cos a2 = €2E2 cos a2 DI ET' sin al = E1 sin al = D2€21 sin a2 = E2 sin a2 Dividing the first equation by the second gives E l cot al = €2 cot a2 (8) ,
This gives the law of refraction for both D and E at the boundary between two isotropic mediums with different FIG. 1.17c. capacitivities. 1.18. Displacement and Intensity in Solid Dielectric.—The capacitivity was first introduced in the statement of Coulomb's law [(1.02 (1)] as a factor that depends on the medium in which the electric force is measured. This hypothetical measurement is difficult to conceive in a solid dielectric, but by using the boundary conditions just derived we may devise a method of determining the displacement and intensity and hence the capacitivity in such a medium. To determine the displacement and intensity in a solid dielectric, let us excavate a small evacuated disk-shaped cavity whose thickness is very small compared with its radius. The intensity inside the cavity E,
FIG. 1.18.
far from the edges will be determined entirely by the boundary conditions over the flat surfaces, as shown in Fig. 1.18a. To determine the displacement, we orient the cavity so that the intensity inside is normal to the flat faces in Fig. 1.18a. From 1.17 (1), we know that the displacement in the dielectric equals that in the cavity. Thus by measuring the intensity in the cavity and multiplying by ev, we determine the displacement in the dielectric. To determine the intensity in the solid dielectric, we orient a long thin cylindrical cavity so that the intensity inside is parallel to the axis as in Fig. 1.18b. Then from 1.17 (3), we know that the intensity inside is the same as that in the solid dielectric. The ratio of the displacement to the intensity, the cavities being small compared with all dimensions of the dielectric mass and the external field remaining constant, gives the capacitivity of the medium.
CRYSTALLINE DIELECTRICS
§1.19
21
The values of displacement and intensity found in this way certainly do not represent the actual fields on a molecular scale found inside a dielectric but are some type of average. Other types of average might give results contradicting macroscopic observations. 1.19. Crystalline Dielectrics.—We shall now use the experimental tests of the last article to find the relation between displacement D and intensity E in a homogeneous crystalline dielectric. Let us cut three plane parallel plates of thickness d from the positive x, y, and z faces of a large cube of this material. We now deposit a conducting layer on each face of each plate and apply a difference of potential V. If we consider areas of the plate sufficiently far from its edges we see that the potential boundary conditions on all such areas of all plates are identical so that the potential distribution in the central areas of all plates is the same. Thus the equipotentials near the center of each plate are parallel to its faces and from 1.06 (4) the intensity E equals V/d. By experimenting with our disk-shaped cavity which is taken small compared with d so that it does not disturb the charge distribution on the conducting faces we find that D is proportional to E but not in the same direction. Thus we have in the x, y and z plates, respectively, (D.). = eilE., (D.). = e13E. (Dy). = 612E., (1) (D0y = E22Ey, (DA, = €21Ey, (D.)y = 623E/, (D.). = € 31E.,
(Dr). = E324
(D.). = € 33E.
Even when E is the same in all plates the normal components of D will in general differ. However in all cases it is found experimentally that (D.),E. = (Dy).Ey,
(D.).E. = (D.).E.,
(Dy).Ey = (D.)„E.
(2)
Superimposing the results in (1) gives
Dx = €11E. + €21Ey -I- €31E. €32E. D. = €13E. + €23Ey + €33E.
Dy = e12Ex + €22Ey +
(3)
Comparing (2) and (1), we see that €12 = €21,
€13 =
€31,
€23 = e32
(4)
Thus the single simple factor e connecting the quantities D and E in the case of an isotropic medium becomes, for a crystal, a quantity known as a symmetrical tensor, having nine components of which six are different. Let us see if it is possible to orient our axes so as to simplify (3). The product E • D, being a scalar quantity, must be independent of the choice of axes. Writing out in terms of the components of E, we have, from (3), using (4),
E • D = €11E1 + €22E,2, + €33E1 + 2€12E.Ei, + 2e13E.E. + 2€23EyEz (5) This is the equation of a quadric surface in E, Ey, and E. We can vary
22
BASIC IDEAS OF ELECTROSTATICS
§1.19
these coordinates in any way we please, keeping E! + E2y + Ei constant, by rotating our axes. We may therefore orient our axes so that the cross products EzE,,, EzEz, and E„Ez disappear. The equation of the quadric referred to the new axes is E3E! (6) E•D= E2E: The components of the displacement referred to these axes are Dz =€1Ez Dy = €2E,, Dz = €3E2 ( 7) The directions of the coordinate axes in (7) are called the electrical axes of the crystal. If €1, €2, €3are all the same, we have an isotropic medium; if only two are the same, we have a uniaxial crystal; and if all three are different, we have a biaxial crystal. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. The work required to bring a point charge q up to the centers, which are a distance b apart, of two thin parallel conducting coaxial rings, each of radius a, is W1 and W2, respectively. Show that the charges on the rings are
Crea
(a2 + b2)1[(a2 + b2)1W1.2 — aW2.1]
Q1.2 = b2q
2. Four equal parallel line charges lie at the corners of a square prism, those at the ends of one diagonal being positive and the others negative. Find the fraction of the total flux that enters the prism. 3. There is a charge q at x = +a, y = 0, z = 0. Find how large a charge must be placed at x = —a, y = 0, z = 0 so that a flux N passes in the positive direction through the circle x = 0, y2 + z2 = a2. 4. Two thin concentric coplanar rings of radii 1 and 2 carry charges —Q and +(27)1(2, respectively. Show that the only neutral points in the field are at x = 0 and ± 2-1. 5. Show that the equation of the lines of force between two parallel line charges, of strength q and —q per unit length, at x = a and x = —a, in terms of the flux per unit length N, between the line of force and the positive x-axis is 277-N)] 2
[y — a cot (— + x2 = a2 csc2 (— 4
6C. Charges +4q, —q are placcd at the points A, B, and C is the point of equilibrium. Prove that the line of force that passes through C meets AB at an angle of 60° at A and at right angles at C. Find the angle at A between AB and the line of force that leaves B at right angles to AB. (Write potential for small region about C in polar coordinates with origin at C.) 7G. Two positive charges qt and q2 are placed at the points A and B, respectively. Show that the tangent at infinity to the line of force, which starts from q1making an angle a with BA produced, makes an angle 2 thn-' (Ng' q2)-1sin with BA and passes through the point C in AB such that AC: CB = q2: 41.
PROBLEMS
23
8C. Point charges +q, -q are placed at the points A, B. The line of force that leaves A making an angle a with AB meets the plane that bisects AB at right angles in P. Show that a PAB sin - = 21sin 2 2 9C. If any closed surface is drawn not enclosing a charged body or any part of one, show that at every point of a certain closed line on the surface it intersects the equipotential surface through the point at right angles. 10C. Charges 3q, -q, -q are placed at A, B, C, respectively, where B is the middle point of AC. Draw a rough diagram of the lines of force, show that a line of force that starts from A making an angle a with AB > cos-' ( i) will not reach B or C, and show that the asymptote of the line of force for which a = cos-1 ( -I) is at right angles to AC. 11C. If there are three electrified points A, B, C in a straight line, such that AC = f, BC = a2/f, and the charges are q, -qa/f, and 471-eVa, respectively, show that there is always a spherical equipotential surface, and discuss the position of the equilibrium points on line ABC if 47r€V = q(f + a)/(f - a) 2and if 4ireV = q(f - a)/(f + a)2. 12C. A and C are spherical conductors with charges q + q' and -q, respectively. Show that there is either a point or a line of equilibrium depending on the relative size and positions of the spheres and on q' /q. Draw a diagram for each case, giving the lines of force and the sections of the equipotentials by a plane through the centers. 13C. An electrified body is placed in the vicinity of a conductor in the form of a surface of anticlastic curvature. Show that at that point of any line of force passing from the body to the conductor, at which the force is a minimum, the principal curvatures of the equipotential surface are equal and opposite. 14C. If two charged concentric shells are connected by a wire, the inner one is wholly discharged. If the force law were r-(2+P), prove that there would be a charge B on the inner shell such that if A were the charge on the outer shell and f, g the sum and difference of the radii 2gB = -Ap[(f - g) In (f + g) - f In f + g In g], approximately. 15C. Three infinite parallel wires cut a plane perpendicular to them in the angular points A, B, C of an equilateral triangle and have charges q, q, -q' per unit length, respectively. Prove that the extreme lines of force which pass from A to C make at starting angles (2q - 5q')7(6q)-' and - (2q q')7(6q)-1with AC, provided that q' > 2q. 16C. A negative point charge -q2 lies between two positive point charges qi and qa on the line joining them and at distances a, $ from them, respectively. Show that if the magnitudes of the charges are given by quErt = q3a-1 = q2X3(a + 15)-1and if 1 < X2 < (a + 0)2(a - ft)-2, there is a circle at every point of which the force vanishes. Determine the general form of the equipotential surface on which this circle lies. 17C. Charges of electricity qi, -q2, q3, (q3 > a.-e placed in a straight line, the negative charge being midway between the other two. Show that, if 4q2 lie between (q31 q11)3 and (q31 OA the number of unit tubes of force that pass from qi to q2 is q2 - q3) 3(214)-1(q31 - q13)(q11 - 210 + 18C. An infinite plane is charged to surface density f, and P is a point distant in. from the plane. Show that of the total intensity icr/e at P half is due to the charges at points that are within 1 in. of P and half to the charges beyond. 19C. A disk of vulcanite (nonconducting), of radius 5 in., is charged to a uniform surface density by friction. Find the electric intensities at points on the axis of the disk distant, respectively, 1, 3, 5, 7 in. from the surface.
24
BASIC IDEAS OF ELECTROSTATICS
20. Two parallel coaxial circular rings of radii a and b carry uniformly distributed charges Q1and Qz. The distance between their planes is c. Show that the force between them is
F—
ck 3(2,(22
E
1672e(ab)11 — k2)
where k2 —
4ab c2 -I- (a
b)2
and E is a complete elliptic integral of modulus k. 21. Show that at a great distance the field of a ring charge —Q of radius b concentric and coplanar with a ring charge Q of radius c is identical with that of a linear quadrupole in which the end charges —Q are at a distance a from the center charge 2Q, provided that b2 — c2 = 4a2. References Useful treatments of the material of this chapter occur in the following books: and R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Uses vector notation and treats simple cases. CORSON, D. R., and P. LORRAIN: "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Well-illustrated dielectric discussion. DuitAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Complete and well-illustrated treatment. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Berlin, 1927. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapters I and IV are especially pertinent. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Uses long notation but gives comprehensive treatment. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives an excellent treatment using vectors. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation with many good field drawings. OWEN, G. E.: "Electromagnetic Theory," Allyn and Bacon, 1963. Has clear and well-illustrated discussion of multipoles. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Chapter I discusses clearly the material in this chapter. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives extensive mathematical treatment based on Maxwell's equations. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Gives clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. ABRAHAM, M.,
CHAPTER II CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS 2.00. Uniqueness Theorem.—Before trying to find the potentials in a system of conductors when the charges are given or vice versa, it is well to be sure that there is only one correct solution. First let us assume that a surface density a on the conductors produces the same potentials as a different surface density a'. The potential at a point P on the surface of one of the conductors due to the surface density a — a' will be, from 1.06 (6),
VP
cr — =
s 4rer
ak)
1
1 f
„
= — — — — — aO ilire S sr 47re ,s r
where r is the distance from P to the surface element dS and the integration covers the surface of all conductors. The two integrals on the right were initially assumed equal, so that VP is zero. The surface density — a therefore makes all conductors at zero potential, so that there is no electric field and a — a is zero. Therefore a = a, and the distribution is unique. We have proved then that only one distribution of electric charge will give a specified potential to every conductor in an electric field. Now let us assume that a surface density a on the conductors gives the same total charge Q in each conductor as a different surface density a. The surface density a — a' will then give a total charge zero on each conductor. The density a — of may be zero over the whole conductor or negative on some areas and positive on others. If the latter is the case, the tubes of force ending on negative areas must originate at points of higher potential and those ending on positive areas at points of lower potential. This reasoning applies to every conductor in the field so that, with the density a — a', no conductor can be at the maximum or the minimum of potential. Therefore, all of them must be at the same potential, and the field vanishes, which gives a = a'. We have now proved that there is only one surface charge distribution which will give a specified total charge to each conductor in an electric field. 2.01. Capacitance.—One consequence of the fact (1.14) that all electrical stresses in a medium in an electric field depend on the intensity in the same way is that the equilibrium is undisturbed if the intensity everywhere in the field is changed by the same factor. Thus if we double the surface density at every point in a system of charged conductors, the 25
26
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.02
configuration of the field is unchanged but the intensity is doubled, and hence the work required to take a small charge from one conductor to another is also doubled. This constant charge to potential ratio of an isolated conductor is its capacitance, and the reciprocal ratio is its -21astance. These terms are not precise when other conductors are in the field unless all are both earthed and uncharged. If Q is the charge in coulombs, C the capacitance in farads, S the elastance in darafs, and V the potential in volts, then the definitive equations are Q = CV V = SQ
(1)
Two insulated conductors near together constitute a simple capacitor. If these two conductors are given equal and opposite charges, the capacitance of the capacitor is the ratio of the charge on one to the difference of potential between them. The ratio is always taken so as to make the capacitance a positive quantity. Thus, for a capacitor we have V1— V2 = SQ (2) Q = C(Vi — V2) 2.02. Capacitors in Series and Parallel.—If we take n simple uncharged capacitors, connect one member of each capacitor to a terminal A and the other member to a terminal B, as shown in Fig. 2.02a, and then apply a potential V between A and B, we have
Q = C1V,
Q2
Q. = C.V
= C2V,
where Q1, Qz, . . . , Q. are the charges on the capacitors whose capacitances are C1, Cz, . . . , C,,, respectively. The total charge is then
Q = Q1 + Q2 + • • + Q. = V(Ci + C2
+ • • • + Cn)
Thus these capacitors, which are said to be connected in parallel or "multiple arc," behave like a single capacitor whose capacitance is C
CI ± C2 + • • • + C.
(1)
Let us take n simple uncharged capacitors, connecting the first member of capacitor 1 to A and the second member to the first member of capacitor 2, then connecting the second member of 2 to the first member of 3, and so forth, finally connecting the second member of the nth capacitor to B as shown in Fig. 2.02b. These capacitors are now said to be connected in series or "cascade." The application of a potential between A and B gives V = V1 + V2 + • • + V. = S1Q1 + S2Q2 + • • • + S.Q. where Vi is the potential across the ith capacitor. Since each pair of connected conductors has remained insulated, its net charge is zero. Hence, if all the tubes of force leaving one member of a capacitor end on the other member of the same capacitor, we have Qi =
Q2 = • • • =
Q.
§2.03
SPHERICAL CAPACITORS
27
so that the charge factors out of the right side of the equation. Thus these capacitors, connected in series, behave like a single capacitor of capacitance C or elastance S where S
= SI + 82
1
+ • • • +5,,,
=1+
± • • • + c.
(2)
We observe that, in general, it will not be possible to confine the fields in the manner assumed so that this formula is an approximation. The additional "distributed" capacitance due to stray fields is usually negligible in cases where the members of a capacitor are close together, and when the capacitivity of the region between is high compared with that outside the capacitor. In the case of air capacitors with widely separated members, this formula may be worthless A
°11 141 -0240
3
A. 4
Ca
C2
C'
B FIG. 2.02a.
C4
2.02b.
2.03. Spherical Capacitors.—Consider a pair of concentric spherical conducting shells, the inner, of radius a, carrying a charge +Q and the outer, of radius b, carrying a charge —Q, the space between being filled by a homogeneous isotropic dielectric of capacitivity E. From symmetry, the electric displacement must be directed radially and its magnitude can depend only on r. Hence, applying Gauss's electric flux theorem to a concentric spherical surface of radius r where b > r > a, we have
fseE • n dS = 47rr2eE = Q
so that
aV car
Q
47rer2
The potential difference between the spheres is therefore Va
Q radr — — Lirejb r2
Q (1 47re \a
b
_ b — aQ 47reab
This gives the capacitance of a spherical capacitor to be
C= 4ireab b—a
(1)
The capacitance of a single sphere of radius a immersed in a dielectric of capacitivity e can be obtained from (1) by letting b —) Go giving (2) C = 47rea
28
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.05
It should be noted that in deriving (1) it is assumed that there is no charge on the outside of b, which requires that b be at zero potential. If this is not the case, the additional capacitance between the outside of b and infinity, computed from (2), must be considered. 2.04. Cylindrical Capacitors.—Consider a pair of concentric circular conducting cylinders of infinite length, the inner, of radius a, carrying a charge Q per unit length and the outer, of radius b, carrying a charge —Q per unit length, the space between being filled with a homogeneous isotropic medium of capacitivity e. From symmetry, the electric displacement must be directed radially outward from the axis and lie in a plane normal to the axis, and its magnitude must depend only on r. Apply Gauss's electric flux theorem to the volume enclosed by two planes, set normal to the axis and one meter apart, and the concentric circular cylinder of radius r when b > r > a. The plane walls contribute nothing to the surface integral; we have therefore
so that
fseE • n dS = 2rreE = Q Q
E_ aV ar
(1)
2irer
The potential difference between the cylinders is therefore
Q fadr 27rejb r
Q a 2ire b
Thus the capacitance per unit length of a long cylindrical capacitor is C—
2re In
(b/a)
(2)
If we let b—> co, we see that C —› 0. Therefore, if there is a finite charge per unit length on a circular cylinder of finite radius and infinite length, the potential difference between its surface and infinity is infinite. Since physically we deal only with cylinders of finite length, this difficulty does not arise, but it indicates that the results of this article apply only where the distance to the surface of the cylinder is small compared with the distance to the ends. 2.05. Parallel-plate Capacitors.—When two infinite parallel conducting planes, carrying charges ±Q and — Q, are a distance a apart, the space between being filled with a homogeneous isotropic dielectric, we see from symmetry that the field between them must be uniform and normal to the plates. If cr is the charge per square meter, then there must be, from 1.13, a unit tubes leaving every square meter of the plates. Thus we have for the displacement and intensity between the plates the relation
GUARD RINGS
*2.06
29
av D = eE = —e— =o
ax
The difference of potential between the plates V2
—
V1
is
= —Crefoa dx = va
Therefore the capacitance per unit area is e/a, and the capacitance of an area A is ry
eA
=— a
(1)
In practice, the field will be uniform only far from the edges of the plate, so that this formula is an approximation which is good if a is small compared with all surface dimensions of the plate and still better if, in addition, the capacitivity of the region between the plates is higher than that of the region beyond the edges. 2.06. Guard Rings.—The derivation of formula 2.04 (2) for the capacitance per unit length of a cylindrical capacitor and that of 2.05 (1) for the capacitance of a parallelplate capacitor both involve the / ( ( y ) hypothesis of conductors of infi11, (172 nite dimensions. To permit
FIG.
2.06a.
FIG.
2.06b.
the application of these formulas to actual capacitors, a device known as a guard ring is used. For the cylindrical capacitors, shown in Fig. 2.06a, the end sections of one of the members are separated from the center section by narrow cracks but are maintained at the same potential. Thus the distorted field near the edges does not affect the center section, except for a very small effect produced by the cracks, so that the charge on this section is related to the potential difference by 2.04 (1.1). A similar arrangement is used for the parallel-plate capacitor by leaving a narrow gap between the central section of one plate and the area surrounding it, maintaining both at the same potential as shown in Fig. 2.06b. The field between the central areas is now uniform except for the negligible effect of the small gap, and the charge on this section is related to the potential difference by 2.05 (1.1).
30
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.08
2.07. Energy of a Charged Capacitor.—We can compute the mutual energy of any system of charges directly from the definition of potential. The work in joules to put the jth charge in place will be, from 1.06 (3), n
IV; =
=
4re
where i
rii
j
The total work to put all charges in place is n
n
W= 1 ~~ q 1 871-E
where i
j
(1)
1=1 j=1
The factor 1 is necessary because the summation includes not only the work done in bringing the ith charge into its position in the field of the jth charge but also that done in bringing the jth charge into the field of the ith charge, which is the same. If Viis the potential at the spot where the ith charge is situated, this may be written, from 1.06 (3), (2) i=i When all charges lie on the same conductor a, they are at the same potential and if their sum is Qa, we may write W =
1
= Wa 2 q:Va = over a
Va •
q1 =
Va
(3)
over a
If the capacitance of a conductor is C, we have, from 2.01 (1), the following equivalent expressions for the energy of a charged conductor: W = IQ V = IQ2/C = iCV2
(4)
For a capacitor, the two members of which carry charges Q and —Q at potentials V1 and V2, respectively, the energy becomes TV = 1071 — rlQV2 = 1Q(T71 — T72) (5) From 2.01 (2), (5) has the same equivalent forms as (4). 2.08. Energy in an Electric Field.—We have seen that visualizing electric forces as transmitted by stresses in the region occupied by an electric field gives results consistent with the observed laws of electrostatics. Where stresses exist, potential energy must be stored. We shall now compute this energy density. Consider an infinitesimal diskshaped element of volume oriented in such a way that the two faces are equipotential surfaces. If this element is taken sufficiently small, the faces of the disk will be flat and parallel and the field inside uniform and identical with that in an infinitesimal parallel-plate capacitor. Let the thickness of the disk be ds; then the difference of potential between the
§2.081
PARALLEL-PLATE CAPACITOR
31
faces is (av /as) ds = —E ds. Let n be the unit vector normal to the faces so that E = En. The charge on an area dS of the face is
D • n dS — D • E dS and the capacitor volume is dv = ds dS, so that 2.07 (4) gives
dW — D • E dv This gives, for the energy density in the field,
dW _ D • E dv 2
(1)
In an isotropic dielectric, D • E = DE, giving
dW _ €E2 _ DE _ D2 dv — 2 — 2 — 2€
(2)
In a crystalline dielectric, we have, from 1.19 (5), dW
= 1(€11E1 + €22E2 + €33Ei 2612E1E2 + 2E13E1E3 + 2E23E2E3) (3) dv or if the coordinate axes are chosen to coincide with the electric axes of the crystal, we have, from 1.19 (6), dW = dv
2 + €2E1; + €3ED
(4)
2.081. Parallel-plate Capacitor with Crystalline Dielectric.—Let us now calculate the capacitance per square meter of a parallel-plate capacitor where the dielectric consists of a slab of crystal of thickness d. Let the capacitivities along the crystal axes x, y, and z be El, €2, and E3, respectively, and let the direction cosines of the angle made by the normal to the capacitor plates with these axes be /,., and n. Since, electrically, one square meter section is like any other, the equipotentials must be parallel to the plates and equally spaced and the electric intensity must lie along the normal. Thus we have
V E = (X2 ± Y2 4- Z2)-1 = 71where V is the potential across the capacitor. Thus
nv nt V Z= X = IV — Y= (1) d Substituting in 2.08 (4), multiplying by the volume d of a square meter section, and using 2.07 (4), we have d)21 2C1V2 2 1[6112G)2 €2m2G)2 e 2(V
32
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
so that the capacitance per square meter is 0€3 „,2,2 12 61
CI =
§2.09
(2)
2.09. Stresses When the Capacitivity Is a Function of Density.—In considering stresses in a dielectric heretofore in 1.16 and 1.17, we have ignored the possibility that the capacitivity may actually change with density T so that there may be a hydrostatic stress tending to expand or contract the dielectric. By working with a volume element of the shape and orientation used in 2.08, we can simplify the investigation to that of a small parallel-plate capacitor of area AS and spacing 3 in which the charge on the plates is considered as embedded in the dielectric at the boundary. Combining 2.07 (4) with 2.05 (1) and assuming an isotropic dielectric of capacitivity E, we have for the energy of our capacitor
mD2
Q2 _ Q2 _ AS 2e AS — 2er AS 2er where we have let m be the mass of the dielectric per unit area between the plates so that m = ra. If m is assumed constant and E is taken as a function of T, the force on an area AS of the plate is a(AW) a (A W) a, = D2 a(er) As AF = — AW=
as
as
Or
— 2E2 a,
Thus the stress or force per unit area pulling on the surface of the conductor is _AF _ D2a(ET) (1) AS — 2E2 a, Carrying out the differentiation and comparing with 1.16 (2), we see that the additional hydrostatic stress is €,E2 arc D2a€ E2 (2) — 2E2Tor = — 2 Tar = — 2 r a, At the surface between two dielectrics, we shall now have to add to those stresses already considered the difference in this hydrostatic pressure giving, in place of 1.17 (7), for the total stress directed from K' to K" the value 1 1-1(' — K"(g2 13'2) D'2r' aK' D"2," 0K" ] F = (3) K"2 n K' \K' K" K' 2 a,' a," At the surface between a dielectric and a vacuum set K" = 1 and alc"/ar" = 0, we have
F
lc - 1 N2 D ,2)
[
24 e„E' 2[
2
(K'
1)
D/2T'OKI
K' 2 a,' j
K' \K' ,aK'j T
a
eE;,2
1 (K' (K — 1))2
2
(4)
§2.11
FORCE ON CONDUCTOR IN DIELECTRIC
where En = g2
en2
=
33
ID;A2
The sign of aK'/ar' may be either positive or negative, and if this term is large enough to predominate in (4) the dielectric may either expand or contract in the field. The phenomenon is known as electrostriction and has been observed by Quincke and others. 2.10. Electrostriction in Liquid Dielectrics.—There is a relation known as the Clausius-Mossotti formula which connects the relative capacitivity with the density for a given liquid.
K — 1 = CT K-{- 2
(1)
C is a constant characteristic of the liquid. This formula, although theoretically derived on incomplete assumptions, has been extremely well confirmed in many cases experimentally (see Geiger-Scheel, "Handbuch der Physik," Band XII, p. 518). Differentiating, it gives 2)(K — 1) a K _ (K 2) 2 (K C — tar 3r 3 Substituting in 2.09 (2), we find that the hydrostatic pressure tending to contract the liquid is e E2(K -1- 2) (K — 1) P —' (2) 2 3 We have assumed throughout that the liquid is nearly incompressible so that r is nearly constant. Thus we see that if we had a charged sphere immersed in a large body of liquid dielectric the pressure given by (2) would vary inversely as the fourth power of the distance from the center of the sphere, and if the liquid were slightly compressible it would be most dense at the sphere's surface, when other effects such as gravitation are neglected. 2.11. Force on Conductor in Dielectric.—So far we have considered the force at the charged boundary between a dielectric and a conductor as if the charge were on the dielectric side of the boundary so that the entire field lies in the dielectric. This gives, from 1.16 (2), the force per unit area to be D2/(2€), neglecting electrostriction. We know from considering the energy of a charged capacitor that this result is correct. Now let us investigate the force if the charge is taken to be on the conducting side of the boundary, the capacitivity of the conductor being e'. The force then becomes D2/(269. The surface of the dielectric is now in the field so that there will be a tension on it pressing against the conductor whose amount is given by 2.09 (3). Neglecting electrostriction and remembering that the field is normal to the surface we obtain for the sum.
34
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.12
of these forces F2 —
K — Kr D2— 2€,,KK'
— e' D2 2c€'
The total force acting on the conductor will now be F = F1—
D2 K — K K' D2 = = 2E„KK' 2€
F2
(1)
Thus either point of view as to the location of the charge gives the same resultant force. If, instead of being discontinuous, the relative capacitivity is considered to change rapidly but continuously in crossing the boundary, the stresses must be determined by integrating. The total resultant stresses come out the same as before although their distribution near the boundary is different. 2.12. Green's Reciprocation Theorem.—We shall now prove that if charges Q 1, Q2, . . . Qn on the conductors of a system give rise to potentials V1, V2, . . . , V. and if charges Qi, Q3 . . . , Q, give rise to potentials VI, V2i . . . , V,„ then ,
ZQJT'i =
(1)
Let us first consider a system of point charges and write down in the form of a matrix n2 terms, each of which comprises the product of one point charge by the potential of a second point charge. Using 1.06 (3), we shall write the sum of the columns in the bottom row and the sum of the horizontal rows in the right-hand column. Thus we obtain
0 d- qq1
47rEr2i
gli g2 _F
-r
0
47r r12 qt q;q3 A ±
airEri 3
Viqn
4rer23
43rer3i
-F
-r
T Aqq2 A-ire 32
m
-F 0 + •
_F
+
q;q.
4rEr1n
47rEr2n
4rer3n
VIVI
V2 V2
g3 V3
— qi
'IV ggri En1
V42 — 47rErn 2
• . •
d_
q2-17
V43 . q3r3
ilrern
0 = qnn qfn Vn
The order of summation being immaterial, the sum may be obtained by adding the terms of the last row or last column. Equating these gives
1q;Va = Zgara e =1
13.°1
(2)
§2.14
INDUCED CHARGES ON EARTHED CONDUCTORS
35
It should be noted that for these point charges V, is the potential at qa due to all unprimed charges except q, itself. All charges located on the same conductor are multiplied by the same potential and so may be collected; thus IVY = = Q'V giving (1). One important application of this theorem is obtained by putting Q1, V3, • • • , Qn = 0, Q2, Q3, • . , Q. = 0, and Q1 = Qs in (1). This gives V', = V2. Thus the potential to which an uncharged conductor A is raised by putting a charge Q on B is the same as that to which B, when uncharged, will be raised by putting the charge Q on A. This theorem is still valid if dielectric boundaries are present as is proved in 3.07. 2.13. Superposition of Fields.—By adding MQ,,Vi or EQ',W, to both sides of 2.12 (1) and comparing with 2.12 (1), we see that if charges Qi, Q2, . . . , Q. produce potentials VI, V2, . . . , V. etc., then charges Vn Qi + VI, Q2 + (4 • • • , Q. + Q'n give V1 + VII, V2 ± A great many new problems can be solved by superimposing the solutions of known problems. As an example of this, suppose we are given the charges Qi, Q2, . . . , Qn on n concentric conducting spheres of radii r1, r2, . . . , r. and wish to find the potential of any sphere, say the sth. We may superimpose the potential produced by each of these charged shells separately, giving 42r€V. = (Q1 +
Q2 + • •
+ Qs)rTi
(23-1-177.4!.14- • • • + Qnr-n-1 (1)
since the potential outside a charged shell at a distance r from its center is independent of its radius and equals Q(471-Er)-1, and the potential inside a conducting spherical shell of radius a is equal to Q(4rEa)-1. 2.14. Induced Charges on Earthed Conductors.—If we place a point charge q at P in the neighborhood of a group of earthed conductors, induced charges will appear on them. We can find the induced charge Q on any one of these conductors if we know the potential V; to which the point P would be raised, the charge q being absent, by raising that conductor to a potential V'. For, from 2.12 (1), QV' ±
qc 0 ± q2 • 0 + • • • = Q' • 0 ± 0 • V ,
0 ± 0-F. • -
so that Q=—
V' 1
(1)
If a conducting sphere is the only conductor and P is at a distance r from its center, we have, from 2.03 (2), V' = qa4irEa)-1and V; = qa42rer)-', so that the charge induced on it by a charge q at P is Q=
aq r
(2)
36
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.15
If a point P lies between two conductors, one of which encloses the other, and if the potential at P is known when the conductors are at potentials V'1 and -172, respectively, then the charge induced on either, when both are earthed, by a charge q at P can be found by 2.12 (1). Let the induced charges be Q1 and Q2; then qV; = 0 Also since all tubes of force from q end on these conductors, we have the relation Q1 ± Q2 = —q. Solving gives
Q1 =
2
—
V1 -
V' -17;q
and
Q2 =
V' — V' 1 V i4 TP2 —
(3)
Thus from 2.03 if the point P lies between two earthed spheres, the charge induced on the inner and outer spheres are, respectively, Q1 = —
ri(r2 — r) r(r2 — r1)'2'
and
Q2
—
r2(r — r1) r(r2 — ri)q
(4)
where r is the distance of P from the center and r1 < r < r2. From 2.04 if P lies between two earthed cylinders, the charge induced on the inner and outer are, respectively, -
In (r2/r) In (r2/r1)
and
Q2 =
In (ri/r) In (ri/r2)q
(5)
From 2.05 if P lies between two earthed plates at a distance a from the first plate and b from the second, we have aq bq (2, = and (6) Q2 a ±b a±b 2.15. Self- and Mutual Elastance.—Consider n initially uncharged conductors, fixed in position and shape. We have seen that putting a charge on any conductor of the group will affect the potential of all other conductors in a definite way which depends only on the geometrical configuration of the system and the capacitivity. The ratio of the rise in potential V,. of conductor r to the charge Q, placed on conductor s to produce this rise is called the coefficient of potential or mutual elastance s.„.. The first application of Green's reciprocation theorem (2.12) shows that 8, = Sr,. A superposition of solutions for charges Q„ Q„ Qt, etc., on conductors r, s, t gives V1 = siiQi
821Q2
V2 = S12Q1
S22Q2
••+ • • •
Sn1Qn Sn2Qn
(1)
V = si.Qi T 82.Q2 - • + snnQ,, Thus ssr is the potential to which the rth conductor is raised when a unit charge is placed on the sth conductor, all the other conductors being „
ELECTRIC SCREENING
§2.17
37
present but uncharged. Putting a positive charge on a conductor always raises the potential of neighboring insulated conductors so s„ is always positive. The coefficient s„ is called self-elastance. 2.16. Self- and Mutual Capacitance.—We may solve the set of equations 2.15 (1) and obtain the charge on a conductor in terms of the potentials of neighboring conductors. The solutions are Q1 = C11V 1 + C21V2 + • • • + Cn1V n Q2 = Cl2V1
Qn
= cinVi
• • • 4" Cn2Vn
C22V 2
(1)
c2nV2 + • • • + c..Vn
where S22832 •
S21831 •
• • Sn2
1 S23S33 • • • Sn3 cll = — A
C12
=
C21
=
• • Snl
1 S23S33 • • • Sn3
S2nS3n • • • Snn
82nS3n • • • Snn
and S11S21 • • • Snl
A =
812S22 • • • Sn2
(2)
S1nS2n • • • Snn
so that c, is the cofactor of s„ in A divided by A. The quantity Cr, is called the coefficient of capacitance or the selfcapacitance and is defined as the charge to potential ratio on the rth conductor when the other conductors are present but earthed. Since the potential has the same sign as the charge, c, is always positive. The quantity c„ is called the coefficient of induction or the mutual capacitance and may be defined as the ratio of the induced charge on the rth conductor to the potential of the sth conductor when all conductors, except the sth, are grounded. The induced charge is always opposite in sign to the inducing charge so c„is always negative or zero. 2.17. Electric Screening.—Suppose Fm. 2.17. conductor 2 surrounds conductor 1 as in Fig. 2.17. All tubes of force from 1 end on 2 so that, if V2 = 0, the charge on 1 depends only on its own potential, which means that, in 2.16 (1), Cal = C41 = • • • = Cnl = 0
(1)
Thus there is no inductive effect between 1 and any of the conductors outside of 2. In this case, we say that they are electrically screened from
38
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.19
1. We notice also that, since Q2 = —Qi when all but 1 are earthed, = —C12. Because of the complete similarity of 2.15 (1) and 2.16(1), we see that we can express the elastances in terms of the capacitances by an interchange of s and c in 2.16 (2). When 1 is screened by 2, we put in the values given by (1) and obtain (2)
1
5 1r = 52r
Substitution of s2, for sir in the first equation of 2.15 (1) and subtraction of the second from it give V i—
V2 = (S11 — S12)Q1
If V1 > V2, tubes of force pass from V1 to V2 and Qi is positive so that sii— s12 is positive, and since, from (2), s12 = 522, we have 511 >
Si2
and
sii >
(3)
522
2.18. Elastances and Capacitances for Two Distant Conductors.—Let two conductors 1 and 2 have capacitances C1 and C2 when alone. When 1 is uncharged, let us bring up 2, which has a charge Q2, to a mean distance r from 1, r being large compared with the linear dimensions, of magnitude a, of either conductor. The potential to which 1 is raised is Q2(47-er)-', if we neglect the variation of this potential over the region occupied by 1 which is of the order a(470-2)-1. Comparison of this with the first equation of 2.15 (1) shows that 821 = (471-er)-1. A charge of opposite sign to Q2 and of order of magnitude Q2a(4rEr)-' will be induced on the nearer parts of 1 and an equal charge of the same sign on the more remote parts. Here we have equal and opposite charges, separated by a distance smaller than a which is small compared with r, so that the field at the distance r is essentially a dipole field, and the potential [1.07 (1)] at 2 due to the presence of 1 uncharged is at most of order Q2a2(4rer3)-1. To this order of magnitude, therefore, 1 does not affect the potential at 2 so that, from the second equation in 2.15 (1), we have S22 = V2W1 = CV- and similarly sii = Cr'. Thus to a first approximation, 511 =
1 Ci
812 = S21 = 4E 1 r'
8 22 =
1
(1)
C2
Solving for the self- and mutual capacitances by 2.16 (2), we have, neglecting r--3terms, 167,2,2r2c1 161.2,2r2c2 C,C2
Cu —
1e7r 2E 2r 2
-
ir; ; v2
C12 = C21 =
4rer
C22 = 162.2,2r2
C1C v2
(2)
2.19. Energy of a Charged System.—If the electrical intensity and the displacement were known at all points outside a set of charged conductors, it would be possible to get the energy of the system by integration from
§2.20
FORCES AND TORQUES ON CHARGED CONDUCTORS
39
2.08 (1); thus W = ifvD • E dv
(1)
where the integration extends over the whole volume outside the conductors. Frequently we do not know the field at all points but do know the potentials and charges on conductors and the self- or mutual capacitances. To determine the energy, suppose we add to the charges Qice, Q2a, • • • , Qnce on the conductors, by infinitesimal steps, (21 da, Q2 da, . . . , Qn da, starting with the initial uncharged state where a = 0 and ending with the final state where a = 1. Since the potentials are Via, V2a, . . . , V ,,a when the charges are Qia, Q2a, . . . , Q.a, the work done in a single step is da
dW = V aia da V2Q2a da -I- • • • +
The energy in the final state is then
W=
ViQifola da =
(2)
V iQi
If we substitute for from 2.16 (1), we have Wv = i(ciiVi
2c12V1V2
c22-V2
••)
(3)
• )
(4)
If we substitute for Vifrom 2.15 (1), we have WQ = Esial
40 -4-. 281219,11,2
1,1 4-.
s2210
•
2.20. Forces and Torques on Charged Conductors.—If the electrical intensity and displacement are known at all points on the surface of a conductor, the total resultant force on the conductor in the direction of the unit vector p can be obtained from 2.11 (1) by integration over the surface of the conductor, giving FP
—
fs
D• E p • n cl3 2
(1)
where n is the unit vector normal to the surface element dS. When the charges and the elastances of a set of conductors are known, the potential energy of the system is given by 2.19 (4) in which si b 812, etc., are functions of the configuration of the system. As in mechanics, we determine the force or torque tending to produce a change in any element of this configuration by taking the negative derivative of the potential energy with respect to this element.
aw an
=
1(asn.Q1 — 2 an
+ 2— anQ1Q2 +
•••)
(2)
This gives either a force or a torque tending to increase n depending on whether n is a length or an angle. In this case where the charge is kept
40
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.20
constant, the change in electrical energy equals the mechanical work done. If, on the other hand, we work with 2.19 (3), keeping the potentials constant by a battery or equivalent device, we may supply additional energy to the system. The force in both cases must be the same, since it depends only on the initial state of the system which may be expressed in terms of either charge or potential. To determine the force in terms of the potential, let us combine 2.19 (4), (2), and (3) in the equation TiTv— IV iQi = 0
=
(3)
Taking the total differential, we have
a* dQi aQi
= i=i
dn. = 0
dVi
(4)
i=i
But from (3), 2.19 (4), and 2.15 (1)
a* aQi
a WQ aQi
.177
Sii I —
Vi= 0
CiiVi
Qi
5=1
and from (3), 2.19 (3), and 2.16 (1) n
aq, a wv a vi avi
Vz
=
Substituting in (4) gives dn. -= 0 This sum is zero for any values of dn. we choose; so each term must be separately equal to zero. Substituting for NI,from (3) gives
axp , 7 aW V an. = 0 aW an. an. = (5) on. on. on. But we know that —a WQ/an. is the force or torque tending to increase n., so that in terms of the potentials this is
_LaWv = — 1(acii —Vi2 an.
2 an.
2aci2 viv2
an.
(6)
When the charges are kept constant, the work done making a small displacement is given by (2), and when the potentials are kept constant it is given by (6). The difference must represent the work done by the device maintaining the potential giving
Cawv aw,) an.
=2
aw, dn, an.
(7)
41
PROBLEMS Problems
Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Three identical spheres of radius a are placed with their centers in line, the intervals being r1 and r2, and are numbered in order. At first, only the central one 2 carries a charge Q. It is now connected with 1. This connection is then broken, and it is connected with 3. Show that, if the intervals are large compared with a, the final charge on 3 is are Q + 1] Q3 = 4 [ri (ri ±r2)(r2 — a) 2. Four identical uncharged conducting spheres are p aced at the corners of a square and numbered in rotation. A charge is given to 1, which is then connected for an instant by a thin wire to 2, 3, and 4 in turn. Show that finally Q Q4= Q
511 - 524
Q
Sji - 25,4 + S24
8 0
511 - 514
Q1 -
0 Sll - :314
3. If, initially, spheres 1 and 2 in the last problem had charges +Q and —Q, respectively, and if, as described, 1 is connected for an instant to 3 and 4 in turn, and if the side of the square r is large compared with the radius a of the spheres, show that, finally, Q4 = 2-1Q[2ir — (21— 3)a] / (r — a) approximately, and find Q3 and Qi. 4. Each of three similar spheres, situated a considerable distance apart at the corners of an equilateral triangle, has initially a charge Q. Each in turn is now earthed for an instant and then insulated. Show that the final charge on 3 is a2r-2Q(3 — 2a/r) and find the final charges on 1 and 2. 5. Four similar conductors are arranged at the corners of a regular tetrahedron in such a way that each is perfectly symmetrical with respect to the other three. All are initially uncharged. One is now given a charge Q by means of a battery of voltage V and is then insulated. It is now connected for an instant to each of the other three in turn and then to earth. The charge is now —Q1. Show that all the mutual capacitances are 56Q2Q1/[V(24Q1 + 7Q)(8Q1 7Q)] and find the elastances. 6. Two conductors are mirror images of each other and are initially uncharged. A sphere with a charge Q is now brought in contact with a certain point on one conductor and then with the image point on the other. If, in each case, it shares its charge equally with the conductor, show that after a large number of alternate contacts the charge will be equally distributed between the three conductors. 7. Three similar insulated conductors are so arranged that each is perfectly symmetrical with respect to the other two. A wire from a battery of unknown voltage is touched to each in turn. The charges on the first two are found to be Q1 and Q2. What is the charge on the third? 8. The inner and outer of three concentric spherical conducting shells of radii a, b, and c are grounded. The intervening sphere is split into halves and charged. Find how large a must be in order just to prevent the halves from separating if a
511 22.
822 $ 822,
Sil = S29
Sli,
911 + 912 ° Si, + 822
42
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
10. A pair of concentric spherical conductors of radii a and b are connected by a wire. A point charge q is detached from the inner one and moved radially with uniform velocity v to the outer one. Show that the rate of transfer of the induced charge from the inner to the outer sphere is dQ dt = —qab(b — a)-1v(a vt)--s. 11. A ring carrying a total charge Q and of radius a lies inside of an earthed sphere of radius b, its axis coinciding with a diameter and its plane at a distance c from the center. Find the potential at the center. 12. A charge q when placed on the axis of a thin earthed conducting ring of radius a at a distance b from the center is found to induce a charge —Q on it. Show that the capacitance of the ring is 47-EQq-1(a2 b2)1. 13. A conducting sphere of radius a is embedded in the center of a sphere of dielectric of radius b and relative capacitivity K. Show that the capacitance of the conducting sphere is 4re,,Kab(Ka b — 14. The conductor in the last problem is earthed and a point charge q is placed at a distance r from its center, where r > b > a. Show that the charge induced on the sphere is — Kabq I tr[b (K — 1)a]}. 15. A spherical capacitor with inner radius a and outer radius b is filled with two spherical layers of capacitivities el and ez, the boundary between being given by r = i(a b). If, when both shells are earthed, a point charge on the dielectric boundary induces equal charges on the inner and outer shells, find the ratio e1/c2. 16. An uncharged conducting spherical shell of mass M floats with one quarter of its volume submerged in a liquid of capacitivity e. To what potential must it be charged to float half submerged? 17C. If the algebraic sum of the charges on a system of conductors is positive, then on one at least the surface density is everywhere positive. 18C. There are a number of insulated conductors in given fixed positions. The capacities of any two of them in their given positions are C, and Cz, and their mutual coefficient of induction is B. Prove that if these conductors are joined by a thin wire, the capacity of the combined conductor is CI +C2 + 2B. 19C. In a system of insulated conductors, having been charged in any manner, charges are transferred from one conductor to another, till they are all brought to the same potential V. Show that V = E As' + 28z), where Si, sz are the algebraic sums of the coefficients of capacity and induction, respectively, and E is the sum of the charges. 20C. Prove that the effect of the operation described in the last question is a decrease of the electrostatic energy equal to what would be the energy of the system if each of the original potentials were diminished by V. 21C. Two equal similar condensers, each consisting of two spherical shells, radii a, b, are insulated and placed at a great distance r apart. Charges e, e' are given to the inner shells. If the outer surfaces are now joined by a wire, show that the loss of energy is, approximately (1670-1(e — e')2(b-1— r-1). 22C. A condenser is formed of two thin concentric shells, radii a, b. A small hole exists in the outer sheet through which an insulated wire passes connecting the inner sheet with a third conductor of capacity c at a great distance r from the condenser. The outer sheet of the condenser is put to earth, and the charge on the two connected conductors is E. Prove that the force on the third conductor is, approximately, ac2E2(47rer3)-,[47reab(a — b)-1
c]-2
23C. Two closed equipotentials VI, Vc) are such that V, contains Vo, and Vp is the potential at any point P between them. If now a charge E is put at P and both equipotentials are replaced by conducting shells and earth connected, then the charges
PROBLEMS
43
E1, E., induced on the two surfaces are given by E,(Vo — V8)-1 = Eo(Vp — V1)-1 = E(V1— Vo)-' 24C. A conductor is charged from an electrophorus by repeated contacts with a plate, which after each contact is recharged with a quantity E of electricity from the electrophorus. Prove that if e is the charge of the conductor after the first operation, the ultimate charge is Ee(E — e)-1. 25C. Four equal uncharged insulated conductors are placed symmetrically at the corners of a regular tetrahedron and are touched in turn by a moving spherical conductor at the points nearest to the center of the tetrahedron, receiving charges el, e2, es, e4. Show that the charges are in geometrical progression. 26C. In 25C, replace "tetrahedron" by "square" and prove that (ei— e2)(eie3— 4) = ei(e2e3 — e1e4) 27C. Two insulated fixed conductors are at given potentials when alone in the electric field and charged with quantities E1, E2 of electricity. Their coefficients of potential are s11, 812, 822. But if they are surrounded by a spherical conductor of very large radius R at zero potential with its center near them, the two conductors require charges E'„ E2fto produce the given potentials. Prove, neglecting R-2, that (.0; — E1)(E2' — E2)-1= (822 — 812) (811 — .312)-1 28C. Show that the locus of the positions, in which a unit charge will induce a given charge on a given uninsulated conductor, is an equipotential surface of that conductor supposed freely electrified. 29C. Prove (1) that if a conductor, insulated in free space and raised to unit potential, produces at any external point P a potential denoted by (P), then a unit charge placed at P in the presence of this conductor uninsulated will induce on it a charge — (P); (2) that if the potential at a point Q due to the induced charge is denoted by (PQ), then (PQ) is a symmetrical function of the positions of P and Q. 30C. Two small uninsulated spheres are placed near together between two large parallel planes, one of which is charged, and the other connected to earth. Show by figures the nature of the disturbance so produced in the uniform field, when the line of centers is (1) perpendicular, (2) parallel to the planes. 31C. A hollow conductor A is at zero potential and contains in its cavity two other insulated conductors B and C, which are mutually external. B has a positive charge, and C is uncharged. Analyze the different types of lines of force that are possible within the cavity, classifying with respect to the conductor from which the line starts and the conductor at which it ends, and proving the impossibility of the geometrically possible types that are rejected. Hence, prove that B and C are at positive potentials, the potential of C being less than that of B. 32C. A portion P of a conductor, the capacity of which is C, can be separated from the conductor. The capacity of this portion, when at a long distance from other bodies, is c. The conductor is insulated, and the part P when at a considerable distance from the remainder is charged with a quantity e and allowed to move under the mutual attraction up to it. Describe and explain the changes that take place in the electrical energy of the system. 33C. A conductor having a charge (21is surrounded by a second conductor with charge Q2. The inner is connected by a wire to a very distant uncharged conductor. It is then disconnected, and the outer conductor connected. Show that the charges V2 are now (m n)Q2-1- mnQ: (2; — nQ2 , (2; = n m m n mn
44
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
where C, C(1 m) are the coefficients of capacity of the near conductors, and Cn is the capacity of the distant one. 34C. If one conductor contains all the others, and there are n 1 in all, show that there are n relations between either the coefficients of potential or the coefficients of induction, and if the potential of the largest is Vo and that of the others V1, V2, . . . , V,,, then the most general expression for the energy is icv,1 increased by a quadratic function of V1— Vo, Vz — Vo, . . . , V. — Vo, where C is a definite constant for all positions of the inner conductors. 35C. The inner sphere of a spherical condenser (radii a, b) has a constant charge E, and the outer conductor is at potential zero. Under the internal forces, the outer conductor contracts from radius b to radius b1. Prove that the work done by the electric forces is E 2(b — bi) I (87 cobb 1) • 36C. If, in the last question, the inner conductor has a constant potential V, its charge being variable, show that the work done is 27r€V2a2(b — b1)/[(51 — a)(b — a)] and investigate the quantity of energy supplied by the battery. 37C. With the usual notation, prove that (sii 12 +SIRS + 8 13 ) , -11-23 > -12 S13. 823) > (, S 38C. Show that if srr, s”, s,, are three coefficients before the introduction of a new conductor and slc, 481 s,', the same coefficients afterward, then (srrs.. — srts:,) 4 (Sr, — s'.)2 39C. A system consists of p q + 2 conductors A1, A 2, . . . , Ap, B1, B2, • • • B5, C, D. Prove that when the charges on the A's and on C and the potentials of the B's and of C are known, there cannot be more than one possible distribution in equilibrium, unless C is electrically screened from D. 40C. A, B, C, D are four conductors, of which B surrounds A and D surrounds C. Given the coefficients of capacity and induction (1) of A and B when C and D are removed, (2) of C and D when A and B are removed, (3) of B and D when A and C are removed, determine those for the complete system of four conductors. 41C. Two equal and similar conductors A and B are charged and placed symmetrically with regard to each other; a third movable conductor C is carried so as to occupy successively two positions, one practically wholly within A, the other within B, the positions being similar and such that the coefficients of potential of C in either position are p, q, r in ascending order of magnitude. In each position, C is in turn connected with the conductor surrounding it, put to earth, and then insulated. Determine the charges on the conductors after any number of cycles of such operations, and show that they ultimately lead to the ratios 1: )3:#2 1, where /3 is the positive root of rx 2— qx p — r = 0. 42C. Two conductors are of capacities C1 and C2, when each is alone in the field. They are both in the field at potentials V1 and V2, respectively, at a great distance r apart. Prove that the repulsion between the conductors is —
—
C1C2(471-Er V l — C2 V2) (47er V2 — VI) (476)/[(1672E2r2 C1C2)9 As far as what power of r-1is this result accurate? 43C. Two equal and similar insulated conductors are symmetrically placed with regard to each other, one of them being uncharged. Another insulated conductor is made to touch them alternately in a symmetrical manner, beginning with the one that has a charge. If e1, e2 are their charges when it has touched each once, show that their charges, when it has touched each r times, are, respectively, —
e?
[1 ( 1 e1 e)1 [1 + (el e1 e2 and 2e1— ez ) 2e1— ez (For 43C and 44C see difference equations, 5.081.)
PROBLEMS
45
44C. Three conductors A1, A 2, and As are such that As is practically inside A2. A1 is alternately connected with Ay and A3 by means of a fine wire, the first contact being with As. A1 has a charge E, initially, A2 and As being uncharged. Prove that the charge on A lafter it has been connected n times with As is
ES [
a+
1+
a —
flyi]
fi(a ± 7) a ± 7
where a, 13, -y stand for sil — 81 , 822 — Si x,and 833— 812. 45C. A spherical condenser, radii a, b, has air in the space between the spheres. The inner sphere receives a coat of paint of uniform thickness t and of a material of which the inductive capacity is K. Find the change produced in the capacity of the condenser. 46C. A conductor has a charge e, and V1, V2 are the potentials of two equipotential surfaces completely surrounding it (V1 > V2). The space between these two surfaces is now filled with a dielectric of inductive capacity K. Show that the change in the energy of the system is 1e(Vi — V2)(K — 1)/K. 47C. The surfaces of an air condenser are concentric spheres. If half the space between the spheres is filled with solid dielectric of specific inductive capacity K, the dividing surface between the solid and the air being a plane through the center of the spheres, show that the capacity will be the same as though the whole dielectric were of K). uniform specific inductive capacity 1(1 48C. The radii of the inner and outer shells of two equal spherical condensers, remote from each other and immersed in an infinite dielectric of inductive capacity K, are, respectively, a and b, and the inductive capacities of the dielectric inside the condensers are K1, K2. Both surfaces of the first condenser are insulated and charged, the second being uncharged. The inner surface of the second condenser is now connected to earth, and the outer surface is connected to the outer surface of the first condenser by a wire of negligible capacity. Show that the loss of energy is Q2[2(b — a)K aK2] 8rb4[(b — a)K aK z]
where Q is the quantity of electricity that flows along the wire. 49C. The outer coating of a long cylindrical condenser is a thin shell of radius a, and the dielectric between the cylinders has inductive capacity K on one side of a plane through the axis and K' on the other side. Show that when the inner cylinder is connected to earth and the outer has charge q per unit length, the resultant force on the outer cylinder is q2(K — K') per unit length 72e„a(K K')2
50C. A heterogeneous dielectric is formed of n concentric spherical layers of specific inductive capacities K1, K2, . . . , K,,, starting from the innermost dielectric, which forms a solid sphere; also, the outermost dielectric extends to infinity. The radii of the spherical boundary surfaces are al, az, . . . , a,,_1, respectively. Prove that the potential due to a quantity Q of electricity at the center of the spheres at a point distant r from the center in the dielectric K. is 1 [Q (1 47re„ K„ r
1)
Q (1 + K,÷1 a ,,
Q 1 I 1 ) a+1 + • • • + K„ a,_1
MC. A condenser is formed by two rectangular parallel conduct ng plates of breadth b and area A at distance d from each other. Also, a parallel slab of a dieleo.
46
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
trio of thickness t and of the same area is between the plates. This slab is pulled along its length from between the plates so that only a length x is between the plates. Prove that the electric force sucking the slab back to its original position is E2 dbt'(d — t') xbt']2 24[A (d — t')
where t' = t(K — 1)/K. K is the specific inductive capacity of the slab, E is the charge, and the disturbances produced.bythe edges are neglected. 52C. Three closed surfaces 1, 2, 3 are equipotentials in an electric field. If the space between 1 and 2 is filled with a dielectric e, and that between 2 and 3 is filled with a dielectric e', show that the capacity of a condenser having 1 and 3 for faces is C, given by 1/C = (GI) e„/(e'B) where A, B are the capacities of air condensers having as faces the surfaces 1, 2 and 2, 3, respectively. 53C. The surface separating two dielectrics (K,, Kz) has an actual charge a per unit area. The electric intensities on the two sides of the boundary are F,, F2 at angles al, cz with the common normal. Show how to determine F,, and prove that K2 cot C2 = K1 cot — a/(e,K,F, cos ci)]. 54C. The space between two concentric spheres, radii a, b, which are kept at potentials A, B is filled with a heterogeneous dielectric of which the inductive capacity varies as the nth power of the distance from their common center. Show that the potential at any point between the surfaces is [(Aan+1— Bbn+1)/(a.+1— bn+1)] — (ab/r)n-"(A — B)/(0+1— bn+1) 55C. A condenser is formed of two parallel plates, distant h apart, one of which is at zero potential. The space between the plates is filled with a dielectric whose inductive capacity increases uniformly from one plate to the other. Show that the capacity per unit area is e„(K, — K,)/[h In (K2/K1)] where K1and K2 are the values of the inductive capacity at the surfaces of the plate. The inequalities of distribution at the edges of the plates are neglected. 56C. A spherical conductor of radius a is surrounded by a concentric spherical conducting shell whose internal radius is b, and the intervening space is occupied by a r)/r. dielectric whose specific inductive capacity at a distance r from the center is (c If the inner sphere is insulated and has a charge E, the shell being connected with the earth, prove that the potential in the dielectric at a distance r from the center is [E/ (47rEc)] In {[b/r][(c r) / (c b)]]. 57C. A spherical conductor of radius a is surrounded by a concentric spherical shell of radius b, and the space between them is filled with a dielectric of which the inductive capacity at distance r from the center is jle—P'71-3where p = ra--1. Prove that the capacity of the condenser so formed is Rare„lia/(eb2/"' — e). 58C. Show that the capacity of a condenser consisting of the conducting spheres r = a, r = b, and a heterogeneous dielectric of inductive capacity K = AO, cb) is e,,ab(b — a)-if ff(0, 0) sin fl do dqS. 59C. In an imaginary crystalline medium, the molecules are disks placed so as to be all parallel to the plane of zy. Show that the components of intensity and polarization (displacement in our notation) are connected by equations of the form f = eiiX + exiY,
g = tia
e22Y,
h = e33Z
60C. A slab of dielectric of inductive capacity K and of thickness z is placed inside a parallel-plate condenser so as to be parallel to the plates. Show that the surface of the slab experiences a tension (1a2 /e,)[1 — K 1— x d(K-1)/dx].
47
PROBLEMS
61C. For a gas K = 1 + Op, where p is the density and B is small. A conductor is immersed in the gas. Show that if 82 is neglected the mechanical force on the conductor is laz/e, per unit area. Give a physical interpretation of this result. 62C. The curve a— x a+x ± 1 y213} — a y9 / a)2 [(x — a)2 when rotated round the axis of x generates a single closed surface, which is made the bounding surface of a conductor. Show that its capacity will be 471-€,,a and that the surface density at the end of the axis will be e/(3ira2), where e is the total charge. 63. The potential ratios, of a system of n conductors may be defined by 9a 16{[(x
1
(x2 +y2)4
V1 = cril(21 Pi2V2 + • • • + V2 = P21 V, c21Q2 + • •
P2.V.
Pn2I72 + • • • + c,T,;(2. Show that Pr, is given in terms of the capacitances, if s # r, by Pr.
— CreC,7,1
References The author has found helpful treatments of the subject matter of this chapter in the following books: ABRAHAM,
M., and R.
BECKER:
"Classical Electricity and Magnetism," Blackie,
D., and P. LORRAIN: "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Second chapter includes much of this material. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Includes a well-illustrated treatment of this material.
CORSON,
J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives comprehensive treatment using long notation.
JEANS,
J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation and contains many good scale drawings of fields.
MAXWELL,
M. K. E. L.: "Theory of Electricity and Magnetism," Macmillan, 1932. Uses vector notation. RAMSEY, A. S.: "Electricity and Magnetism," Cambridge, 1937. Gives clear elementary treatment with problems. Uses vectors. RUSSELL, A.: "Alternating Currents," Cambridge, 1914. Gives applications to multiple conductors. THOMSON, J. J.: "Mathematical Theory of Electricity and Magnetism," Cambridge, 1921. THOMSON, W.: "Papers on Electrostatics and Magnetism," Macmillan, 1884. Solves many problems and gives numerical results on spheres and spherical segments. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Gives a clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. PLANCK,
CHAPTER III GENERAL THEOREMS 3.00. Gauss's Theorem.—We now proceed to find a relation between the integral over a closed surface S and over m — 1 other closed surfaces, inside it, of the normal component of a vector A, which is a continuous function of position in space, and the integral of its divergence defined as V • A, throughout the volume v between these surfaces. Let the components of A in rectangular coordinates be A, Au, Az so that
v•A
= aAz + aA, +aA. ay
ax
(1)
az
Let us suppose v divided up into slender prisms of rectangular cross section dy dz. One of these (Fig. 3.00) cuts elements dgi, dS'l from Si,
the coordinates of which are x; and x7 and the outward drawn normal unit vectors to which are n;'. The contributions of the elements d8; and d8;' to the surface integral are Az,,i •
dS;
But from Fig. 3.00, i • n; dS; = d8;' = dy dz so that the total contribution to all sections of that prism, cut from v, which penetrates surfaces p to q is
dy dz(A„,— Azi,) = 7 °P
dy dz f
z,,
aA ox
dx
7 =P
We now sum over all the prisms into which v is divided, and we have
1
si
Azi • nid51
= f aA z dx dy dz jv ax
48
§3.02
EQUATIONS OF POISSON AND LAPLACE
49
Working out similar expressions for A„ and A, and adding, we obtain, after substituting V • A from (1) and writing dv = dx dy dz, the expression
ZfsiA • ni dS; = fvV • A dv
(2) =1. This formula, apparently first given by Green, states a law widely known as Gauss's theorem. 3.01. Stokes's Theorem.—We can derive another important theorem directly from Gauss's theorem. Let us apply 3.00 (2) to an infinitesimal right cylinder of height h, base area S, curved area S', and peripheral length s. Let n be the unit vector normal to the base and n1 that normal to the side. Let F be some vector function and replace A by n x F. Since n is a constant in the following expression F•Vxn vanishes. V •nxF =F•V:n—n•V:F But [n x F] • n is zero on the flat faces so that the surface integral over S vanishes and 3.00 (2) becomes
fen:F • nidS' = f F•nl xndS' = — f n-V:F dv But dv = h dS and dS' = h ds, so that the surface and volume integral become line and surface integrals, respectively, giving
f F • ni x n ds =
—fsn • Vx F dS
Now n: ni= — n1x n is a unit vector directed along the boundary, so that if we choose it for the positive direction of s we have n1: n ds = —ds, and summing up for all the infinitesimal cylinders composing a surface we have
ZfF • ds = Zisn • V xF dS In summing up the line integrals, every interior contour is covered twice in opposite directions, so that they cancel each other leaving only the integral around the outer edge of the whole area. The sum of the surface integrals is of course the integral over the whole surface, so that
f F • ds = f n•V xF dS
(1)
This is Stokes's theorem and may be stated as follows: The line integral of a vector function F around any closed contour equals the surface integral of its curl over any surface bounded by this contour. 3.02. Equations of Poisson and Laplace.—In Gauss's theorem, let the vector A be the electric displacement D = EE and apply Gauss's electric
flux theorem [1.15 (1)] to the surface integral, and we obtain fvV • D dv = Q = fvp dv
GENERAL THEOREMS
50
§3.03
where p is the electric charge density. Thus we have, if dv becomes vanishingly small, dQ (1) div D = V • D = — = p
dv
If we write D = EE = —e grad V = — eVV, this becomes div (e grad V) = V • (eVV) = — p
(2)
This is Poisson's equation for a nonhomogeneous dielectric. If the dielectric is homogeneous, e is a constant and may be factored out, giving div grad V = V VV = V2 V =
(3)
If p = 0, we obtain Laplace's equation which may then be written, in rectangular coordinates, for a nonhomogeneous but isotropic dielectric
a aX
E
ax
,a( -r 6 ay ay
,— a(6-- _— az az
(4)
If the dielectric is homogeneous but not isotropic and if the coordinate axes lie on the electric axes of the crystal as in 1.19 (7), this becomes
a2 v E1
ax2
a 2v
+
a2v ( + E 3---T
= 0
(5)
If the dielectric is both homogeneous and isotropic, the equation becomes
a 2v ax2
a2va2v =o az2 ay2
(6)
3.03. Orthogonal Curvilinear Coordinates.—In most electrostatic problems, the data given are the charges on, or the potentials of, all conductors involved, the size and location of all other charges, and the capacitivity at all field points. The problem is considered solved when the potential at all points has been determined. To do this, it is necessary to find a solution of Laplace's equation that can be made to fit the given boundary conditions. There is usually one system of coordinates in terms of which these conditions can be most simply expressed, and it is therefore desirable to solve Laplace's equation in this coordinate system. All commonly used coordinate systems can be classed as orthogonal curvilinear coordinates. Consider three families of mutually orthogonal surfaces, one member of each family passing through every point of the region under consideration. Any member of the first family can be specified by giving the proper numerical value to u1and similarly for the other families by u2 or u3. An infinitesimal rectangular parallelepiped will be cut out of space by the six surfaces 21,1, u14- dui, u2, u2 duz, u3, u3 dui. Since the quantities u1, u2, and u3will represent distances in a few cases only,
CURL IN ORTHOGONAL CURVILINEAR COORDINATES
§3.04
51
it will, in general, be necessary to multiply dui, du2, and du3by certain factors* h1, h2, and h3to get the actual length of its edges. These factors may be variable from point to point in the field, and so they are functions of u1, U2p and u3. The lengths of the edges of our parallelepiped are therefore F ds3 = h3 du3 (1) ds2 = h2 du2, dsi = hi dui, These are shown in Fig. 3.03. If V is a scalar, then, by definition, components of the gradient are
av 0.32
av = av as, h, au,
—
av
C
ay
.33
h 2 0142'
(2) — hL3 ., OU3
0
Fro. 3.03.
To calculate the divergence in this coordinate system apply Gauss's theorem [3.00 (2)] to the infinitesimal volume shown in Fig. 3.03. The outward normal component of the flux of the vector A through the faces OCGB and ADFE is
a a a —(A, ds2ds3) ds, = as, ou,(h2h3A dui du2 du3— h,h2h3 au,(h2h3A,) dv Adding the two corresponding expressions for the other four faces and comparing with 3.00 (2), we see that div A = v • A — Now letting A = we have 1
1 r a (h2h324 1) hih2h31_ aul
— (h3h1A 2)
au2
a — (h,h2A3)] au3
(3)
VV and substituting in Poisson's equation [3.02 (2)],
r a (h2h3e
av)
a 03hie av)
hih2h3L au,‘ hi au, + au2
au2
a eih2€ aV au3 h 3 au3)] = — P
v • (e VV) = —p
(4)
Putting p = 0 gives us Laplace's equation for a nonhomogeneous isotropic dielectric. If the dielectric is homogeneous and isotropic, e as well as (h1h2h3)-1may be factored out of the equation. 3.04. Curl in Orthogonal Curvilinear Coordinates.—Let us apply Stokes's theorem to the face 0 ADC of the elementary curvilinear cube of Fig. 3.03. Let F1, F2, and F3 be the components of a vector along u1, u2, and u3. Then, from 3.03 (2), the line integral along OA and DC is
u2)F
u2) —
u2
du2)Fi(ui, u2 du2)] dui (h ,F7 au2
duidu2
* Our notation is that used by Houston, "Principles of Mathematical Physics,"
and Abraham-Becker, "Classical Electricity and Magnetism." The Peirce and Smithsonian Tables and Jeans, "Electricity and Magnetism," use h1, h2, and ha for the reciprocal quantities.
§3.05
GENERAL THEOREMS
52
and along AD and CO it is [h2(ui -1-- dui, u2)F2(ui + dui,
u2) — h2(UI, U2)F2 (U1, U2)] dU2 —
(h2F2)dUl dU2
aui
Adding these gives the line integral around this face, and, by Stokes's theorem, this equals the integral of the normal component of the curl of F over the area hih2 dui du2 of the face. Canceling out the factor dui du2, this gives ra(h2F2) a(h1F1)] (1) (V x F)3 = 1 au2 hih21_ aul and similarly for the other faces, we have 1 [a (haF3)
x F = h2h3 (V x F)2 =
a (h2F2)1
(2)
au3
au2
1 [a (hiF i) h3h1 au3
a(haF3)]
3)
(
au,.
3.05. V • (e VV) in Other Coordinate Systems.—In spherical polar coordinates where r = u1is the distance from the origin, 0 = u2 is the
(a) FIG. 3.05.
colatitude angle, and q5 = u3 is the longitude angle, we have h1 = 1 dsi = hi dui = dr ds2 = h2 du2= r d0 h2 = r h3 = r sin 0 ds3 = h3 du3 =r sin 0 det. From 3.03 (4), Laplace's equation becomes
1 a 2av T *2 ar \Er ar
1 a
.
r2 sin ao tsin
0a_ u
Ae
1 + r2 sin2 e
ad,
=0
(1)
In cylindrical coordinates where p = u1 is the distance from the z-axis, 4) = u2 is the longitude angle, and z = u3is the distance from the xy-plane, we have h1 = 1 dsi = h1dui = dp ds2 = h2 du2= p d4) h2 = p h3 = 1 ds3 = h3 du3 = dz
GREEN'S THEOREMS
§3.06
53
This arrangement is shown in Fig. 3.05b. From 3.03 (4), Laplace's equation becomes V • (e V V)
_1 a cay) + 1. a ( av) + ean = 0 pap\ apl p2 ao\ ail az\ az/
(2)
We shall have occasion to use still other coordinate systems, such as the confocal system. These will be taken up later in connection with special problems. 3.06. Green's Theorems.—In 3.00 (2), let A = (e grad 43)* = *e V43 where if and (13 are scalar quantities which are finite and continuous in the region of integration and can be differentiated twice, and e is a scalar quantity which may be differentiated once and which may be discontinuous at certain boundaries in the region. We shall exclude these discontinuities by drawing surfaces around them which fit them closely on both sides. Let n; and np" be the unit normal vectors drawn into the pth boundary from the two sides, and let the values of A on the two ' and Then if there are q such surfaces enclosing q dissides be Ap continuities, the integral over them is
fs,(A; • n; Ap • n';) dS, p=i where dS, is an element of the original surface of discontinuity. Adding these terms to 3.00 (2) and substituting A = \Fe In, we have
e* &to Is, an' dS•
f (epq/pZ epV s ,
P=1
=f
1r
dS, )
V. (e Vt.) dv e V* • V43 dv ± f ‘1,17 v
(1)
If we write a similar equation with' and ' interchanged and subtract the two, we obtain q
p=1
ac; a& — (1)"°*;:)1 dS„ an II eil`P' Panp ' 1 + e;:(41 \ an" \ van;— V P an", P P i m
1••1
fsi
€(* `L — an;
)d = f [4/V • (e V') — (1)17 • (e VNP)] dv (2)
an;
When e is a constant, without discontinuities, (1) becomes C13 d S = f (vim • 7.13 + s, on;
j.•1
'724,) dv
(3)
GENERAL THEOREMS
54
§3.08
and (2) becomes m
fs,(1/wi aci) — ca ll dSi =fv NI V 24) — cl) V2 T) dv
(4)
i=1 A useful vector analogue of (3) and (4) has been proved by Stratton as follows. In 3.00 (2) let A = x (V x 41) where Ar and 43 are vector quantities that are finite and continuous in the region of integration and can be differentiated twice. Then E [117 x (V x OA • ni d$5 = fvv • ['ir x
(v x 4,)]
dv
si i
lr • [(V 41)x n] dS; = fv [(V x w) • (V x 4)) — w • v x (V x 0)] dv (5) 5-1 Subtraction of a similar equation with AP and interchanged gives m
fsi
[Ar x (V x 4)) — x (V x I•)] • n dS;
5-1
=
rt
Jv
cp • [V x (V x 11?)] — AP • IV x (V x
dv (6)
3.07. Green's Reciprocation Theorem for Dielectrics.—In 3.06 (2), let = V be the potential of one distribution of electricity 1' = V' that of another and let e be the capacitivity. If the contact surfaces of the different dielectrics are uncharged, we have, from 1.17 (5), , air €7) an" € Pan' = ev an; = and, from 1.17 (6), = 4,,," and cP',, = c13„'' and a similar relation for Alf, so that the integrals over the surfaces of discontinuity disappear; and if there are no charges throughout the volume, V • (e VT) = 0 and V • (e V (I)) = 0, so that the volume integral vanishes, leaving
Or
Or
i=i
-1
fv
av
si\" i6 an;
Vifsi cr'
*a7j dSi = 0
dS; — V; fe d8i) = 0
171
EQ; V; = EQ;Vi i -1
(1)
Thus we have proved Green's reciprocation theorem 2.12 (1) to hold when dielectrics are present. 3.08. Green's Function.—Let NI,be the potential due to a unit charge at the point P, and let c be the potential due to the induced surface
SOLUTION OF POISSON'S EQUATION
43.09
55
density o on some closed surface S at zero potential, which surrounds P. We may consider the unit charge at P as consisting of charge density p which is everywhere zero except in an infinitesimal volume dv enclosing P. This element is so small that has a constant value c13, throughout it. On the surface 43 = -if, at a distance r from P, = (471-er)-1; also, V2c13 is zero throughout v, whereas V24, = -p/e in dv and is zero elsewhere. Substituting in 3.06 (4) gives
(47er)-11 dS = 4,„ f p dv = O n where, as in 3.00 and 3.06, the positive direction of n is from v into S. (47rer)-1, this becomes Writing G for {ON)
f
aG = 1 f 1 T dS = — 1 f 47 sr S an n. sr dS 4re
(1)
We shall refer to G as Green's function although some authors designate e by this name. It is evident that G is a solution of Laplace's equation which is zero over a given boundary and has a simple pole at a point P inside. Electrically, it represents the potential inside an earthed conducting surface under the influence of a unit charge at a point P within it. Equation (1), which is identical with 1.06 (6), is usually worthless for determining G as we rarely know a on the conductor. Many methods of determining G for various types of boundaries will appear in subsequent chapters. The force F1 acting along u1on a charge q at ui, 4, u3 in the coordinate system of 3.03 can be found from G(ui, u2, u3) by the formulas
q iim frOG(ui,4 ILO
L
1,,±3 [
OG(ui,
u01
q OG(u'i, u'2,
Jue_a} = 2
-hiaz4
(2)
The first is valid because, when 8 is small, the fields of the induced charges at and uc - 8 are the same, but those of the charge itself are equal and opposite and cancel out. The second form follows from 2.07 (2) because the field energy of the charge itself is unaltered by a change in u'1 so only the induced charge energy is affected. 3.09. Solution of Poisson's Equation.—In a vacuum, the potential at the point P due to a charge density p in the element of volume dv is, from 1.06 (2), dV = p(47rer)-1dv, where r is the distance from the element dv to P. Thus the potential at P due to all charges in space is
v P
1 p dv
47ejlr r
(1)
But we know from 3.02 (3) that this potential satisfies Poisson's equation
= -12 Thus (1) is a solution of (2).
(2)
56
GENERAL THEOREMS
§3.10
We may also solve (2) directly by means of Green's theorem. Let us apply this theorem to the region v between a very small sphere of area a and a larger sphere of area and radius R, both centers being at P. Write x1,= r--1and (I) = V in 3.06 (4), and we have I pi:: an
dS .111. az —
7. 1
T l(r)]
dS
= f [1V 2V — VV 2(1)1 dv (3) vr Consider the first of these integrals. On the small sphere, a/an = — a/ar and the solid angle subtended by any area dS is dS2 = 7-2dS, so that, since a V/ar is finite, the integral is
av
fV —rf — dS2 —dC2
ar
— 47V,
(4)
If the total charge Q producing the potential V lies in the finite region, then V —* Q(471-cR)-1as R —> .0 so on the surface Z, both terms in the integrand approach Q(477-ER3)-1whereas the surface area is 471-R2. Thus the second integral becomes zero as Q(ER)-1. Since r-1is a solution of Laplace's equation, v2 (r— 1) = 0 and (3) becomes, if (2) is used,
- =
v2v
1 p
dv = — dv v r e vr
(5)
f
which gives us (1) again. 3.10. Uniqueness Theorem with Dielectrics Present.—If there were several solutions to Laplace's equation satisfying the same boundary conditions so that we had to resort to experiment to choose the correct one, there would be little use for potential theory. We shall now prove that if the location and size of all fixed charges inside a certain region are specified, as well as the value of the potential over all its boundaries, with the possible exception of certain closed conducting surfaces on which only the total charge is given; then the value of V is uniquely determined in this region. Suppose there are two values V and V', both of which satisfy the boundary conditions. Since both satisfy Poisson's equation, their difference is, from 3.02 (2), V • [eV(V — V')] = 0 everywhere, since fixed charges cancel. Put = NI/ = V — V' in 3.06 (1), and remember that, as in 3.07, the integrals over the boundaries between dielectrics vanish if these are uncharged. The substitution gives m
c(V —
— V') dS; =
f[V(V — Vi)]2 dv (1) j=1 The surface integrals vanish in those terms on the left representing boundaries on which the potentials are specified, since V = V' at every s,
an,
GREEN'S EQUIVALENT STRATUM
§3.12
57
point. All surfaces on which the total charge is specified, but not the actual potential, are conducting surfaces so that V and V' are constant and may be taken out of the integrals, giving, for the sth such surface, by 1.15 (1),
- V;) dS, = (V, - V;) (Q, - Q;)
(V, - V's ) f s.ur4
By hypothesis the charge on this surface is fixed so that Q, - Q; is zero. The whole left side of (1) is, therefore, zero and, since the integrand on the right is always positive, V(V - V') must vanish everywhere. Thus V - V' must be a constant and since it is zero on the boundaries it must be zero everywhere and the theorem is proved. 3.11. Introduction of New Conductor.—We shall now prove that if a new uncharged or earthed conductor is introduced into the electric field produced by a system of charged conductors, the charge on all conductors remaining unchanged, the energy of the system is decreased. Let the energy, electric field, and the space occupied by this field be designated by W, E, and v, respectively, in the original system and by W', E', and v' in the final system; then we have W - W' =2 vE2 dv - !-f E'2 dv 2v
(1)
= 2fv-v•E2dv + --2efv,[(E - E') 2- 2E' • (E' - E)] dv Now putting vIf = V' and I. = V - V' in 3.06 (3) and remembering that V243 = 0 throughout the volume give
fv•
•v(v —
dv = f E' • (E - E') dv
f
v•
i -1
= 111V; ej
s,
v,(aV s/
- a') dS, =
an;
av) '3' an; (1
V;(Q, - Q;) = 0 —1
since Q; = Q,, so that the last term in the last integral (1) vanishes, and W - W' is a positive quantity. Since we do no work in making an earth connection, any motion of electric charge caused thereby must be at the expense of the electric field and so cause a further decrease in energy. 3.12. Green's Equivalent Stratum.—Let Vp be the potential at some point P, outside a closed surface S, due to an electric charge density p(x, y, z), throughout the volume v, enclosed by S. In 3.06 (4), let = r-1and 4 = V, so that V24. = -p/6 by 3.02 (3). This gives
lay
- — dS f V
an
a(
s an r dS =
f
p dv = — -4TV,
e v r
(1)
58
GENERAL THEOREMS
§3.13
by 1.06 (6), where r is the distance from P to dv. Thus we see that the potential Vp at P can be expressed either as a volume integral of the charge density in v or by two surface integrals over the surface, S, enclosing v. Referring to 1.06 (6) and 1.12 (1), we see that this means that Vp will be unchanged if we remove the volume distribution p and replace it by a surface stratum which coincides with S and carries a charge — €(8V/an) per unit area and an outward pointing dipole distribution of moment eV per unit area. If S is an equipotential surface, the second integral becomes, by 3.00 (2), when V is taken out, 1) dS = V f V 2(-Vi n • V( 1)dv v r
(2)
This is zero since 1/r is the potential of a point charge 4ire at P, and therefore V2(1/r) = 0 inside v. In this case, no dipole layer is required. This shows that an electric field is unchanged when any area of an equipotential surface is replaced by a very thin uncharged conducting sheet, because such a sheet may be regarded as two equipotential surfaces infinitely close together enclosing this area. By (1), they must have equal and opposite densities on the exterior faces to leave Vp unchanged. 3.13. Energy of a Dielectric Body in an Electric Field.—From the last article the forces exerted on external fixed charges by a given volume distribution and by its equivalent stratum are the same. It follows from Newton's laws that the reverse is also true. This fact facilitates the calculation of the work done in placing an uncharged dielectric body in the field produced by fixed sources in a region of capacitivity 6.. If the stratum is brought into the field fully formed, 1.06 and 1.071 (2) give the energy of an area dS to be dW = [aV (m • V)V] dS
(1)
where o is the charge density, M the dipole moment density, and V the potential of the external fixed charges. The dielectric polarization and hence the strength of the equivalent stratum are proportional in this case to the external field and are built up by it from zero to its final value so that the total work done is one-half that given by (1). Substitution for a and M from the last article and integration over the surface of the
stratum give W
= -14( — Vn • VV. ± Von • VV) dS
(2)
where V. is the final potential just outside the equivalent stratum or the dielectric surface. By 1.17 (5) and (6) the potential just inside the dielectric surface is related to V„ by the equations V„ = V,
and
eon • VV. = en • VV;
(3)
§3.14
EFFECT OF AN INCREASE OF CAPACITIVITY
59
where e is the capacitivity of the body. Substitution of (3) in (2) and application of Gauss's theorem 3.00 (2) to the result give W = ifs (e.,Vin • VV — eVn • VVi) dS = ifv(e0 — e)VV • VVi dv (4) since V 2V = V • eV V, = 0. Thus if the electric intensity produced in the volume v of a uniform isotropic medium of capacitivity eo by a fixed charge distribution is E, and when v is occupied by a uniform isotropic body e it is Ei, then the energy difference is Es dv
W = ifv (E0 —
(5)
The torque or force acting on the body in the direction 0 is F=—
OW
(6)
ao
3.14. Effect of an Increase of Capacitivity.—If the capacitivity at any point in the electric field produced by a system of charged conductors is increased, the energy of the system is decreased provided the charges on the conductors are kept constant. To prove this, let W be the energy of the system, Q;the charge on the jth conductor, p the volume density of charge, V the potential at any point, and e the capacitivity. We assume that Q, and p remain constant when E is varied but that V and W vary. Thus, W = ifv€E2 dv = krvE(VV)2 dv SW = ifvSe(VV) 2 dv fveVV • V(W) dv
(1)
Substitute SV for and V for 4 in 3.06 (1), set V • (E VV) = —p, and assume uncharged dielectric surfaces over which the integral vanishes. LEV(SV) • VV dv =
f 3Ve— av dSan'I 7 si
SV p dv V
i=i 6-17,5 dS ;
SV p dv
j-i Q; SV; f SV p dv = 2(5W
= -1 since we also have
1 W = QiVi i -1
1
.fvpV dv
GENERAL THEOREMS
60
§3.15
Substituting in (1) and transferring 23W to the left side, we have
SW = --f Oe(VV) 2 dv (2) 2 v Thus STV is negative if Se is positive. 3.15. Potential of Axially Symmetrical Field.—It can be verified by substitution in 3.05 (2) and integration by parts that a solution of Laplace's equation, if e is constant and V does not depend on 4), is
f
=
V(z,
jp sin 0) dB
0 I(Z
(1)
where cl)(z) is a real function of z whose Taylor expansion is, by Dw 39, 4)(z, p) = 1'(z)
(1)'(z)jp sin
0 + (2!)-4"(z)(jp sin 0)2
• • • (2)
Substitution of (2) in (1) and integration from 0 to 271- give C0
p 2n
V(z, p) = n
=o
(-1)' a")(z) 2 (n !) 2 az2n
(3)
It is evident that V(z, 0) is identical with 4'(z). If no axial charges exist, (3) gives V uniquely at all points attainable from the axis without crossing a charged surface. The proof resembles that used for 7.07 (5). Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Show that the components of the curl in cylindrical coordinates are ,
Mid pA =
aA. 1 OA — — p acp az
aA, aA,
=az -
aP
aA p] acs
i[a(pAgs)
curl.A
aP
P
2. Show that the components of the curl in spherical polar coordinates are [ (sin OA.)
1
curl,A
r sin =
o
1 [OA, r sin 0
l[a(rA8) curloA = r ar
ao sin
i
aA0 act,
]
(I a( rA,
or
aAr ao
3. Show that for ellipsoidal coordinates, as defined in 5.01, we write in 3.03 (1) 47,1 = (.2 - .0(.2 - .3)D2,
= (u2 - .2)(.2 - .2)D2,
= (u3 - .2)(., - .2)D3,
PROBLEMS
61
where D1.2,3 = [(a' + u1.2.3)(b2+ u1,2.3)(c2+ u1.2.3)]-13 c > b > a, -b2 < u2 < -a2, and -a2 < ul< co. For special cases of oblate and prolate spheroids, see 5.27 and 5.28. 4. If three sets of orthogonal surfaces are defined by the concentric spheres ui = x2 +y2 + z2 and by the two cones eu-i2 + y2(4 - b2)-1 z 2(4 - 0)-1 = 0 and x214, 2 + y2(u: - b2)-1+ z2(4 - c2)-1= 0, show that we must choose h1 =1, hs = u?(L4 — 4)[(4 — b2)(0 — 41-1, and h: = ul(u: - u:)[(b 2 - ul)(c2 - 4]-1. 5. If the three sets of orthogonal surfaces are x2c-2ur2 y2c-2(ul - 1)-1= 1, x2c-2uT' - y2c-2(1 - 4)-1= 1, and z = u3, show that hl = c2 (ul - 4)(ul h: = c2(u: - 4)(1 - 4)-1, and h3 = 1. 6. If the orthogonal surfaces are y2 = 4cuix + 4c24 y2 = -4cu2x + 4c24, and 1L z = u3, show that hl = c2'(ul + u2), = (33 + u2), h3 = 1. 7. If the sets of orthogonal surfaces are z = c(u1 U2), x2 +y2 = 4c2u1u2, and y = x tan u3, show that h1 = cIaTi(u1 + h2 = c[ici(u1 + u2)]1, h3 = 2c(a1a2)*. 8. If the orthogonal surfaces are (x2 + y2)1 = c sinh u1(cosh al - cos u2)-1, y = x tan u3, and z = c sin u2(cosh ui - cos u2)-1,prove h1 =c(cosh ai - cos u2)-1, h2 = h1, and h3 = c sinh ui(cosh ul - cos u2)-1. These are known as toroidal coordinates and give orthogonal anchor rings, spheres, and planes. 9. If the orthogonal surfaces are (x2 + y2)1 = c sin u2(cosh ul - cos u2)-1, y = x tan u3, and z = c sinh u2 (cosh 14 2 - cos u2)-1, prove 112 = c(cosh u1- cos u2)-1, h2 = h1, and h3 = c sin u2(cosh u1- cos u2)-1. These are called bipolar coordinates, because if 7.2 and 7.2 are the vectors from a point P to the points z = +c and z = -c, respectively, we have ul = In (r2ri-1) and cos u2 = (7'1- r2)771rii. 10C. If the specific inductive capacity varies as e-il', where r is the distance from a fixed point in the medium, verify that the differential equation solution satisfied by the potential is a2r-2[eria - 1 - ra-1 - r2(2a2)-9 cos 0, and hence determine the potential at any point of a sphere, whose inductive capacity is the above function of the distance from the center, when placed in a uniform field of force. 11C. If the electricity in the field is confined to a given system of conductors at given potentials and the inductive capacity of the dielectric is slightly altered according to any law such that at no point is it diminished and such that the differential coefficients of the increment are also small at all points, prove that the energy of the field is increased. 12. Find the condition where a set of two dimensional equipotentials V2 = f(z, y) can generate a set of equipotentials when rotated about the z axis. Show that if this is possible the potential is V = A f e-IF(V2)dV2 dV 2 + B
where F(V2) -
1
3V2
y(vV2) , ay
References The subject matter of this chapter is treated in the following books: ABRAHAM, M., and R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Gives lucid derivations of general theorems using vector notation. BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Gives a very complete treatment. Cowx, R. E.: "Field Theory of Guided Waves," McGraw-Hill, 1960. Chapter 2 includes a fine detailed treatment of Green's functions. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Extensive treatment in first four chapters. GEIGER-SCHEEL: "Handbuch der Physik," Vols. III and XII, Berlin, 1928 and 1927.
62
GENERAL THEOREMS
JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapter 1 treats this material. Static Green's functions are applied neatly in Chap. 3. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives comprehensive treatment using long notation. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives an excellent treatment using vector notation. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. First two chapters cover this material. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives a very extensive and rigorous treatment. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Chapter 1 contains clear and detailed treatment. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930.
CHAPTER IV TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS 4.00. Field and Potential in Two Dimensions.—A potential problem is said to be two-dimensional when all equipotential surfaces in the field are cylindrical, which means that each can be generated by moving an infinite straight line parallel to some fixed straight line. The unit of charge is now a uniform line charge parallel to this axis and having a strength of 1 coulomb per meter length. We have already seen in 2.04 (1) that in a homogeneous isotropic dielectric the field intensity at a distance r from such a charge is radial and its magnitude is E
q 2rer
(1)
The potential obtained by integrating this is
V =
2re
In r
C
(2)
Clearly the zero of potential cannot be chosen, conveniently, at infinity as this would make C infinite. We usually give C a value that makes the computation as simple as possible. Theoretically, a two-dimensional electrostatic problem can never occur as all conductors are finite. However, there are a vast number of important cases in which the lengths of the parallel cylindrical conductors are so great compared with the intervening spaces that the end effects are negligible, and the problem then becomes two-dimensional. 4.01. Circular Harmonics.—In its most general sense, the term "harmonic" applies to any solution of Laplace's equation. In the more usual but restricted sense, it applies to a solution of Laplace's equation in a specified coordinate system, which has the form of a product of three terms, each of which is a function of one coordinate only. The solution required to fit a given set of boundary conditions is then constructed by adding up a number of such harmonics which have been multiplied by suitable coefficients. In the ordinary cylindrical coordinates described in 3.05, we should therefore have cylindrical harmonics of the form V = R(p).13(0)Z(z) (1) In the special case where Z(z) is a constant, this reduces to a two-dimensional problem and the harmonics are called circular harmonics. For 63
64
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.01
this case, we usually write r for p and 0 for so that for a uniform isotropic dielectric Laplace's equation, 3.05 (2), becomes, when we multiply through by r2,
a (al , ao2 avv —
rOr r ar
(2)
0
We now wish to find solutions of this equation of the form
V = R(r)O(0)
(3)
Substituting in (2) and dividing through by (3) give 1(
aR
R r
2a2R) _1_329 e or. + r07.2
0 —
This equation will evidently be satisfied by solutions of the equations c/20
,7 = — n20 , d 2R _L 1 dR 2R
dr2
2- dr =
(4) (5)
A solution of (4), the simple harmonic motion equation, is e = A cos nO + B sin nO
(6)
and it is easily verified that the solution of (5) is R = Crn where n
Dr-n
(7)
0. If n = 0, we have the solutions
0 = AO + B R = C ln r + D
(8) (9) The number n is called the degree of the harmonic. The circular harmonics then are
V = (AO + B)(C In r D) V = (A cos nO + B sin n0) (Cr" + Dr-n)
Degree zero Degree not zero
(10) (11)
A sum of such terms, with different constants for each n, or an integral with respect to n is also a solution of (2). Thus
V = Zenitm
or
V = f f(n)O.Rm do
(12)
n
It should be noticed that we have placed no restrictions on n. If the equation 1(An cos nO + Bm sin nO) = l(Cm cos nO + Dm sin nO) (13) holds for all values of 0 where n is an integer, then we have the relations
Am = Cm
and
B. = D.
(14)
65
CONDUCTING CYLINDER
§4.03
To prove this, it is necessary only to multiply both sides of (13) by cos m0 and integrate from 0 to 2r, giving
Z(A,1027 cos nO cos m0 dO + B„f021.sin nO cos m0 do) = VC4027 cos nO cos me do + D„f021- sin nO cos m0 dO) If m n, using Pc 361 and 360 or Dw 445 and 465, we see that all terms on both sides drop out. If m = n, we use Pc 489 and 364 or Dw 450.11 and 858.4 and obtain frA„ = irC„ A similar procedure using the sine proves the second part of (14). 4.02. Harmonic Expansion of Line Charge Potential.—In using circular harmonics, it is frequently necessary to have the expansion due to a line charge whose coordinates are ro, 00. If R is the distance from the line charge q to the field point P (Fig. 4.02), then, from 4.00 (2) when C is zero, the potential V at P is given by treV = - 2q In R =
-
q ln [r2 -I- ro — 2rro cos (0
-
00]
= -2q In r - q In [1 - /le l i(4-6)][1 - ne-i(" 01 Using Pc 768 or Dw 601 where r 0 is less than r, we obtain
ro 1 ro 2 -2q In r + 4-[ei(e--9) + e-i0-60] + (-) [020-80-I- e-120-00)] -I- • - - 1 r r 1 7:2 = -2q[In r - 7)cos (0 - 00) - 2 cos 2(0 — 00 ••• r 2r -
Writing as a summation and expanding cos n(0 Dw 401.04 give
-
Bo) by Pc 592 or
40
n
V = j–-
-I- -
nr
27re
(cos nOocos nO + sin nO0 sin nO) - In r
I
(1)
nml.
This holds when r > ro. When r < ro, the same procedure gives, factoring out In r0 instead of In r, [.
V = -C 1-1(I. . (cos nO0 cos nO } sin nO0 sin nO) - In ro 24r E n\r 0)
(2)
n ∎1
These are the required expansions. 4.03. Conducting or Dielectric Cylinder in Uniform Field.—As an example involving boundary conditions at both dielectric and conducting
66
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.03
surfaces, let us find the field at all points when an infinite conducting cylinder of radius a surrounded by a layer, of relative capacitivity K and radius b, is set with its axis perpendicular to a uniform field of strength E. The original potential outside the cylinder is of the form V=Ex=Er cos U At infinity, the potential superimposed on this, due to the induced charges on the cylinder, must vanish so that no terms of the form r" can occur in it. Since this potential must be symmetrical about the x-axis, no terms involving sin nO can occur. The final potential outside must therefore be of the form Anr-n cos nO
V. -= Er cos 0 ±
(1)
n =1
Since the values r = 0 and r = 00 are excluded from the dielectric, both r" and r-n can occur there but sin nO terms are thrown out as before. The potential in the dielectric must therefore be of the form (B"r" Car-n) cos nO
=
(2)
n =1
If we take the center of the cylinder at the origin, then V = 0 throughout the conductor. We have therefor• found solutions of Laplace's equation that satisfy the boundary conditions at infinity and the symmetry conditions. It is now necessary only to determine A, B, and Ch so that the boundary conditions at the dielectric and conducting surfaces are satisfied. From 1.17 (5) and (6), at r = b the boundary conditions are
ay. €v
ar =
avi ar
or
avo Kavi ar = ar
and
V. = Vi
(3)
Substituting (1) and (2) in (3) gives E cos 0 — XnAhb-n-1cos nO = Mn(Bhbn-1— C,,b-n-1) cos nO Eb cos 0 ± ZA.b-n cos nO = Z(B.bn C.b-n) cos nO From 4.01 (14), we have, if n
1,
— —Anb-n--' = Anb-n = B.bn +
(4) (5)
At the conducting surface, r = a, Vi = 0, so that 0 = &an T C.a-n (6) Adding (4) X b to (5) gives (K 1)Bn = (K — 1)C.b-2n (7 ) The only way (6) and (7) can be satisfied is to have B. = C. = 0 or (K 1)/(K — 1) = — (a/b)2n. The latter condition is impossible since
§4.03
CONDUCTING CYLINDER
67
the left side is positive and the right side negative so that the first holds, and substituting in (5) we have An = B„ = C. = 0 (8) When n = 1 we have, instead of (4), (5), and (6),
E — Alb-2= KB,. — KC1b-2 E + Alb-2= B1+ C ib-2 0 = Bla + Cia-' Solving these equations for A1, B1, and C1gives
(K +1)a2 + (K — 1)b2 (K + 1)b2± (K — 1)a2 2Eb2 B1— (K + 1)b2+ (K — 1)a2
Al = — Eb2
Cl -
(9)
—2Ea2b2
(K + 1)b2+ (K — 1)a2
The potentials (1) and (2) then become
Ai V. = (Er ± 7.) cos 6,
C J. Vi = (Bir + --i . ) cos 0
(10)
The lines of electric displacement are shown in Fig. 4.03a.
a
b
FIG. 4.03.
We may get the case of a conducting cylinder of radius a by letting
K = 1 in (9), so that 1 =E A1= C1= —Ea2 and B and the potential is
V . = E(r — 1 cos e r
(11)
The field is shown in Fig. 4.03b. We get the case of a dielectric cylinder of radius b by letting a = 0 in (9), so that
68
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Al =
b2E K+1 '
El = K + l'
§4.04
C1 = 0
and the potentials are vo
1.(7. _ K .
1b2) cos 0, K±1 r
2E V. — K ±ir cos 0
(12)
We notice that the field inside the dielectric is uniform. The displacement is shown, for K = 5, in Fig. 4.03c.
FIG.
4.03c.
4.04. Dielectric Cylinder. Method of Images.—Let us consider a dielectric cylinder of radius a under the influence of a line charge at r = b, B = 0. The potential due to the line charge alone may be obtained from 4.02 (2) by putting ro = b, 00 = 0. We must superimpose on this a potential, due to the polarization of the dielectric, which vanishes at infinity and is symmetrical about the x-axis. The final potential when a < r < b is therefore re,{ V° = :
nOn + Ann] [
cos nO — In b ± C1
(1)
n =1
Since the potential inside must be finite when r = 0 and symmetrical about the x-axis, it is of the form „rn cos nO + C2 V =./3 q ' 276,( n
(2)
=1
Setting V. = Vi when r = a, we have ()"
nB — a n
Setting Eva Vo/ar = Ea V,/ar or an--1 bn
and
C2 = —In b
C1
(3)
avo/ar = Kavi/ar when r = a, gives
n — an+1An = nKan-1.13.
(4)
69
IMAGE IN CONDUCTING CYLINDER
§4.05
Solving (3) and (4) for A. and B. gives
1 — K a2n
2 B. — (1 + K)nbn
A n = 1 + K nbn'
The potential outside then becomes
= 27qr€.1 n=1 1[(rb y ±
l lqa r
cos nO — In b
CI
(5)
and that inside becomes =
q
1r€v(1 -F
K)
l(ry n b cos nO —
b — Ci)
(6)
n =1
If we let
C=0=—
1 — K 1 +K
ln r +
K 1 ln 1+K
then (5) becomes exactly the expansion given by 4.02 (1) and (2) for the potential due to three line charges on the x-axis, one of strength q' at x = a2/b, one of strength —q' at x = 0, and one of strength q at x = b. Also, (6) gives the expansion due to a charge q" situated at x = b, where = -F — Ifq
and
2
1 ± Kg
(7)
Thus when an uncharged dielectric cylinder of radius a is brought into the neighborhood of a line charge q with its axis parallel to the charge and at a distance b from it, the additional potential in the region outside the cylinder due to its presence is the same as if it were replaced by a parallel "image" line charge q' located between q and the axis at a distance a2/b from the latter, plus a line charge —q' at the origin. The potential inside the cylinder is the same, except for the ac ,litive constant, as if it were absent and q were replaced by a charge q". Some authors write 2K instead of 2 in the numerator of q", in which case the potential inside the cylinder must be calculated as if all space were filled with the dielectric K. When a dielectric cylinder is introduced into any electric field produced by a two-dimensional charge distribution parallel to its axis, it follows from the last paragraph, since such a distribution can always be built up from line charges, that the form of the field inside the cylinder is unchanged but its intensity is reduced by the factor 2/(K + 1). 4.05. Image in Conducting Cylinder.—It follows from 1.17 (8) that as K co the lines of force become incident normally at the dielectric surface. This is the condition at the boundary of a conductor, so that we may get the law of images in an uncharged conducting cylinder by
70
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.07
letting K 00 in 4.04 (7), giving q' = —q. Thus when a line charge of strength q is placed parallel to the axis of an uncharged conducting cylinder of radius a and at a distance b from it, the additional field outside the cylinder due to its presence is the same as if it were replaced by a parallel line charge —q located between q and the axis at a distance a2/b from the latter plus a charge q on the axis. As will be noted from Fig. 4.05, the image position can be located by drawing a tangent to the dielectric or conducting cylinder from the point q. The line through this point, normal to —qq', intersects —qq' at q'.
4.06. Image in Plane Face of Dielectric or Conductor. Intersecting Conducting Planes.—If we let the radius of the cylinder become infinite, keeping the distance d = b — a of the line charge from the face constant, then the distance of the image q' from the face is a — a2b—' = ab-1(b — a) d Thus the image laws derived in the last two sections apply to a line charge lying parallel to the faces of a semi-infinite dielectric or conducting block. The image charge is the same distance back of the face as the actual charge is in front. For a conductor, q' = —q, and for a dielectric, q' and q" are given by 4.04 (7). It is evident from Fig. 4.06 that if two planes intersect at the origin at an angle 7r/m, where m is an integer, both planes will be equipotentials under the influence of line charges parallel to the intersection lying in the cylinder r = r0 and arranged with +q at 00, 27rm—' ± 00, 4rm1 00, . . . , 2(m — 1)7rm-1 0 0 and —q at 27rm-1— 00, 47rm—' — 00, . . . , 2,r — 00. 4.07. Dielectric Wedge.—Another solution of 4.01 (2) is obtained if n 0 by writing jn for n in 4.01 (4), (5), (6), and (7), so that 0. = A cosh no + B sinh nO R. = Ci er n = C1 2e±in r = C cos (n In r) D sin (n In r)
(1) (2)
This solution is periodic in In r instead of 0 so that in 4.01 (12) Rn not On is now the orthogonal function. These harmonics can be used to solve
§4.07
DIELECTRIC WEDGE
71
the problem of a dielectric wedge of capacitivity e2 bounded by the planes 0 = —a and 0 = a under the influence of a line charge q at 0 = 7, r = a 11♦ in a medium of capacitivity el as shown in Fig. 4.07. There are no cylindrical boundaries where the sine and cosine terms in (2) must vanish so n is not restricted to discrete values but may vary continuously which indicates the integral form of 4.01 (12). We write for (el — e2)/(€1 -I- €2) and choose potentials of the forms V1= (1 +0)fo [A(k)ek°
B (k)e-kej cos [k In (r / a)] dk-FC 0
V2= fo'[C(k)ek°+ D(k)eke] cos [k In (r/a)] dk + Co V3
=
JO
'[E(k)eke + F(k)e-h°1 cos [k In (r /a)] dk
Co
(3) (4) (5)
according to whether —a < 0 < a (3), a < 0 < y (4), or 7 < 0 < 27r — a (5). The constant Co can be chosen to make V zero at any specified point. For this point at r = a, 0 = 0, it has the value —(1 + 13)fo[A (k) + B(k)] dk The circle r = a passing through q is a line of force because aV /ar is zero there. Thus half the flux from q goes to r = 00 and half to r = 0 where there is a charge By Fourier's integral theorem, if two of the integrals in (3), (4), and (5) are equal to each other over the whole range of In (r / a), then their integrands are equal. Application of 1.17 (5) and (6) to the integrands of V1and V2 at 0 = a gives, after rearrangement,
D = 8e2kaA + B C = A + /3e-2kaB, Similarly joining V1 at 0 = —a to V3 at 0 = 2ir — a gives F = (gAe-2ka + B)e2k. E = (A + 13Be2ka)e--2kr,
(6) (7)
To meet the remaining boundary conditions at 0 = y, we write down an expression for the flux density originating anywhere in the 0 = 7 plane and evaluate it by Gauss's theorem, thus
4rav2 av3 = L ao ae el
)
ro
.' k[Cek'Y — De-h7 — Eek7 + Fe-k1 cos k In --r dk a
When both sides are multiplied by cos [t in (r /a)] dr and integrated from r = 0 to r = 00, the left side is simply q by Gauss's theorem because the integrand vanishes except near r = a and the right side is found by Four-
72
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.08
ier's integral theorem so that q = iireik[(C - E)eki - (D - F)e-k7] Also V2 is equal to V 3 at 0 = y so that (C - E)ek7 = -(D - F)e-k7 Elimination of D and F or of C and E from these equations gives C = E
ireik
'
D=F-
ire
(8)
The solution of (6), (7), and (8) for A and B gives A B - q[e'k(7-r) sinh kr - elk (7-7) Qsinh k(ir - 2a)] , (9) - 2a)] 2ifeik[sinh2 kr - 02 sinh2 where the upper signs go with A and the lower with B. In the special symmetrical case where 7 = 7r the potential in the dielectric wedge is ,
[k In (r /a)] - 1 dk (10) sinh k(ir - 2a)] The integrand, which is finite at k = 0 and falls off exponentially as k increases, can be plotted as a function of k and the integral then evaluated with a planimeter. If the only charge is to be at r = a, 0 = -y, then the charge at the origin may be canceled by addition to V1, V2, and V3 of the term cosh Ice cos rei jo k[sinh kir
1 + i3)
Vl = q(
iq In a [a(Ei — Es) —
7rEi]-1
(11)
If the cylinder r = b is at zero potential, then we must superimpose two solutions of the form just derived, one for q at r = a, 0 =- 7 and the other for -q at r = b 2/a, 0 = y. If the cylinders r = b and r = c are at zero potential, then we shall need discrete values of n in (2) and the potential function will be a series instead of an integral. 4.08. Complex Quantities.—Before taking up the general subjects of conjugate functions and conformal transformations, it will be well to review some of the more important properties of complex quantities. If z = x jy, it is clear that to every point in the xy-plane, usually called, in this connection, the z-plane, there corresponds one value of z. In polar coordinates, we have, using Pc 609 or Dw 408.04, z = x jy = r cos 0 jr sin 0 = (1) The magnitude of the vector r is known as the modulus of z and written izi. The angle 0 is known as the argument, amplitude, * phase or angle * This term, universally understood by mathematicians to represent the angle in this case, is also, unfortunately, in general use to specify the maximum fluctuation of any alternating quantity in connection with alternating currents. Thus, when the complex notation is used for such quantities, it becomes practically identical with the term "modulus" as used here.
§4.09
CONJUGATE FUNCTIONS
73
of z. When a complex number z is raised to a power n, we have zn = rnein°
(2)
so we may say that the modulus of zn is the nth power of the modulus of z and the argument of z. is n times the argument of z. When we take the product of two complex numbers, we get zzl = me; ("0
(3) so that the modulus of the product of two complex numbers equals the product of their moduli and the argument is the sum of their arguments. By putting zi1for ziin (3), we see that the modulus of the quotient of two complex numbers equals the quotient of their moduli and the argument equals the difference of their arguments. If we have zi = xi + jyi = f(z) = f(x jy), then the quantity zi = xl — jyi = f(x — jy) is called the complex y conjugate of z. That the latter relation holds if f(z) is an analytic function can be shown by expanding f(x ± jy) in a power series with real coefficients, for wherever ±j occurs to an even power the term is real and the choice of sign FIG. 4.08. disappears, whereas whenever +j occurs to an odd power the term is imaginary and the choice of sign remains as before. Thus we have Iz11 2 = xi + yi = zizf = f(x jy) • f(x — jy) This gives us the rule that to obtain the modulus of a function of a complex variable we multiply it by its complex conjugate and take the square root of the product. 4.09. Conjugate Functions.—Laplace's equation in two dimensions and rectangular coordinates is written
a2U a2U axe + ay2 =
(1)
Since this is a partial differential equation of the second order, the general solution must contain two arbitrary functions and it is easily verified by differentiation that such a solution is U = c1)(x jy) ''(x — jy) It should be noted that to be a solution of (1), and '' must possess definite derivatives where (1) holds, and hence are analytic functions in this region and capable of expansion in a power series (see Whittaker and Watson, "Modern Analysis," Chap. V). Since U is to be the electrostatic potential, it must be a real quantity and so the imaginary part of (13 must be equal and opposite to that of NP, so that, if c1)(x jy) = u jv, then '1,(x — jy) = w — jv, where u, v, and w are real quantities. Since
74
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.11
both 7 and are analytic functions, we may expand them in series; thus we have 43(x
jZ A„rn sin nO n=0
n=0
n=0
P (x — jy) = xli(re-2°)
A nr' cos nO
Ad-nein° =
jy) = 43(reig) =
ZB.rne—intl = ZB.r" cos nO — jZBnrn sin nO n=0
n=0
n=0
Since the imaginary parts of 43 and •If are equal and opposite over the entire range of 0, we see from 4.01 (14) that An = Bn so that the real parts of cf. and if are equal or u = w. Thus we have U = 2u. Let V be another real quantity such that V = 2v, and we have
U jV = 2(u -I- jv) = 240(x
jy) = f(x
jy)
(2)
The function V also satisfies Laplace's equation as may be shown by noting that the above expansion is in circular harmonics or by multiplying through by —j, giving jy)
V — jU = —jf(x jy) = F(x
Thus V is the real part of F(x jy), just as U is the real part of f(x We shall write W for U jV and z for x jy, giving
jy).
W = f(z)
(3) The functions U(x, y) and V(x, y) are called conjugate functions. 4.10. The Stream Function. Differentiating (3) with respect to x and y gives OW au .av , az —
ax = ax + 3ax = f (z)79X. = f/(z) ., aw _ au .av , az 5 = (z) ay ay + ay = f (z),
Multiply the second equation by j, add to the first, equate real and imaginary parts, and we have
av = au ax ay
au — = anddav ay ax
(1)
This is the condition that the two families of curves U(x, y) = constant and V(x, y) = constant intersect each other orthogonally. As we have seen, we may choose either set to represent equipotentials, in which case we call this the potential function. The other set which intersects this set everywhere orthogonally then represents the lines of force and is known as the stream function. 4.11. Electric Field Intensity. Electric Flux. Let us consider the derivative —
dW dz
dU-Fj dV dx-Fj dy
(a U / ax)dx+ (0 U/ ay)dy+jRav/ax)dx-F (a V / ay)dy] dx j dy
§4.12
75
FUNCTIONS FOR A LINE CHARGE
Substitute for a U/ax and aU/ay from 4.10 (1). dW = dz
=
(av/ay)(dx
j dy) j(aV /ax)(dx j dy) dx j dy
ay AV ay .ay = 73x —
(1)
ay
Thus if V is the potential function, the imaginary part of —dW/dz gives the x-component of the electric field intensity and the real part gives the y-component. Regardless of whether U or V is the potential function, the absolute value of dW/dz at any point gives the magnitude of the electric field intensity at that point. If dn is an element of length in the direction of maximum increase of potential and ds is the element of length obtained by rotating dn counterclockwise 4 7r radians, then we get from (1) -
dW dz
By =_- ay = an as
or
dW dz
= By = an
au as
(2)
according as U or V is the potential function. In the latter case, if we wish to know the flux through any section of an equipotential surface between the curves U1and U2, we integrate, giving by 1.10 (1), m
Flux =
f
av ds = el ul — ds ui as as = e(U2 —
ui an
U1)
(3)
Thus just as the difference of potential between any two points in the field is given by the difference in the values of the potential function at the two points, so the total flux passing through a line joining any two points in a field equals the product of the capacitivity by the difference in the value of the stream function at the two points.. If the surfaces V1and V2 are closed and all lines of force in the region they bound pass from one to the other, then, from 2.01, they form a capacitor. The charge Q on either is the total flux per unit length. From (3), this flux is the product of E by the increment [U] in U going once around a V-curve. Since the potential difference is V2 the capacitance and field energy per unit length are C —
[Q]
4U]
(4)
1 -172 — I Va — Vl Field energy = ICI V2 — V1 2= 4-€1 U2 - U1~ V2 (5) 4.12. Functions for a Line Charge.—Before taking up the methods available for finding the required f(x jy) to fit a given problem, let us consider a simple case where the form of this function is evident by inspection. In polar coordinates By Pc 609 or Dw 408.04, we have (1) z= x jy = r cos 0 jr sin 0 = U = --k(re)-1 In r from The potential of a line charge at the origin is 4.00 (2). Clearly from (1) this is the real part of ---k(re)-- i In z, so —
76
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.13
e ,q0 2 z = _ qnr W = U ± jV = qln q In (x + jy) _
=
2re
q In (x2 + y2)1 jqtan-1(y/x) 2re 276
(2)
is the desired function. To check this we observe that, as 0 goes from 0 to -21r, the function eV goes from 0 to q and so is the flux emerging from a charge q as required by 4.11. We may now write down f(z) when there is a line charge at ro, 00, or zo. The potential is
q FIG. 4.12.
U = -1q(7rE)--' In R = —.1q(7re)-1In [r2 + 4 - 2rro cos (0 — 00)] = ---1 1q(7re)-1In [(r cos 0 — ro cos 002 + (r sin 0 — r0 sin NM = —1q(71-€)-1In (A2 + B2)1 But from (2), substituting A for x and B for y, we see that In (A' is the real part of In (A + jB) so that the required function is
+ BY
W = y77 In (A ± jB) = 27eq In (rei° - roei 09 =-7 ---re In (z - zo) (3) The function for n line charges situated at z,, z2, . . . , z. is therefore n
W=
In (z - z.)
2--re 8
(4)
=1
4.13. Capacitance between Two Circular Cylinders.—We have already seen in 4.05 that the equipotentials about two equal and opposite line charges are circular cylinders. Let us therefore superimpose the fields due to charges 2re at y = a and -2re at y = -a. This choice of the charge q simplifies the coefficients. From 4.12 (3), the expression for W becomes, using Pc 645 or Dw 601.2 and 505.1, W = In z + .7 a = 2j tan-1 a = 2j cot-1
z - 3a
z
a
(1)
Solving for z and using Pc 601 or Dw 408.19, we have z = a cot
U + jV = -a sin(U/j) ± a sin V cos (U/j) - cos V 2j
Using Pc 606 and 607, we can now separate real and imaginary parts, giving
a sink U Y _ cosh U - cos V Eliminating V from these equations gives x-
a sin V
cosh U - cos V
x2 + y2 - 2ay coth U + a' = 0
(2)
(2.1)
§4.13
CAPACITANCE BETWEEN TWO CIRCULAR CYLINDERS
77
This may be written
x2 (y — a coth U) 2 = a2 csch2 U
(3) Thus, as anticipated, the equipotentials are a set of circles with centers on the y-axis, positive potentials being above the x-axis and negative
1
1 i
1
i
% %
)
I \ / • ` `. \
/
x
1
// a
/ ,
1
(b)
(a)
FIG. 4.13.
below. Eliminating U from (2) gives x2 — 2ax cot V + y2
a2 = 0
This may be written (x — a cot V)2 + y2 = a2 csc2 V
(4)
Thus the lines of force are also a set of circles, all of which pass through the points y = +a and y = —a on the y-axis. To determine the capacitance per unit length between the cylinders U = U1 and U = U2, it is necessary, from 4.11 (4), only to divide the charge 2re by the difference of potential U2 — U1. We are given the radii R1 and R2 of the two cylinders and the distance D between their axes, from which we must determine a, U1, and U2. From (3), R1= alcsch U11, R2 = *soh U21 and D = a(Icoth U11 ± tooth U21), taking the lower sign if U1 and U2 are both positive, giving one cylinder inside the other, and taking the upper sign if Uiis negative, giving one cylinder outside the other. We now write by Pc 661 or Dw 651.02, using the same sign
rule,
78
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.15
cosh (U2 — U1) = cosh U2 cosh U1± Isinh U2 sinh U11 = (bcoth U2 coth Uli ± 1)Isinh U2 sinh Ulf Now apply Pc 659 or Dw 650.08 to the ± 1 = ± ± term, giving cosh (U2 — U1) = [Icoth U1coth U2! ± i(coth2 U2 — csch2 U2) (coth2 csch2U1)]Isinh U1 sinh U21 (1coth U21 ± 1coth U11) 2 1csch2 Ui csch2 U2 2ICSCh U1csch U2I Putting in values of D, R1, and
R2
gives
cosh (U2 — U1) — ±
D2 — R? — RI 2R1R2
Thus the capacitance per unit length between the two cylinders is
D2 — R7 — C = 24cosh-1( -12R1R2
(5)
where the lower sign is taken when one cylinder is inside the other and the upper sign when they are external to each other. The two cases are shown in Figs. 4.13a and 4.13b.
4.14. Capacitance between Cylinder and Plane and between Two Similar Cylinders.—In Fig. 4.13a, let RI = D h co so that the outer cylinder circles through infinity and coincides, in the finite region, with the x-axis. Then we have, neglecting R2 compared with fel and D2, and h compared with 2R1, R1 D = 2R1 — h 2R1,
Ri + RI — D2 (Ri— D)(Ri + D) h 2R1R2
2R1R2
Thus the capacitance per unit length of a conducting cylinder of radius R with its axis parallel to and at a distance h from an infinite conducting plane is
C = 274cosh-1
R
(1)
Two similar cylinders with their centers at a distance D = 2h apart have half the capacitance per unit length given by (1) since they consist of two such capacitors in series. The resultant expression is in somewhat simpler form than that obtained by putting R1 = R2 and D = 2h in 4.13 (5)
C = VE(COSh.-1
D 2R
(2)
4.15. Conformal Transformations.—Evidently conjugate functions furnish a powerful means for solving two-dimensional potential problems, provided we can find the proper function. Before taking up methods of doing this, we shall investigate certain special properties of functions of a
§4.16 GIVEN EQUATIONS OF BOUNDARY IN PARAMETRIC FORM
79
complex variable. Suppose we plot values of z = x jy on one plane and values of z1= x1 + jyi on a second plane and let z be an analytic function of zi so that at least one point in the z-plane corresponds to each point in the zi-plane. Then as we move the latter along some curve, the corresponding point in the z-plane will also describe a curve, provided z = f(zi) is continuous, otherwise the second point may jump from place to place. If the z curve returns to the starting point when the z1curve does, then f(zi) is said to be single valued in this region of the z-plane. From the rule for quotients of complex numbers 4.08, we have
dz ds = = =h (1) dzl jdzil dsi where ds is the length of an element dz of a curve in the z-plane and ds1 is the length of the corresponding element dz1of the corresponding curve in the z1-plane.Thus the modulus of dz/dzimeasures the magnification of an element from a certain point in the z-plane when transformed to the corresponding point in the z1-plane. Let us draw the infinitesimal triangle made by the intersection of three curves in the zrplane, and let the lengths of the sides be dsi, Then the lengths of the sides of the transformed triangle will be ds = h ds1, Thus dsi:ds'I :ds'i' = ds:ds':ds", so that the ds' = h ds;., ds" = h two triangles are similar and the angles of the intersections of corresponding curves in the two planes are the same. Such a transformation is said to be conformal. Since the argument of the quotient of two complex numbers is the difference of their arguments, we see that the argument of (dz/dz1) is the angle through which the element has been rotated in transforming from one plane to the other. 4.16. Given Equations of Boundary in Parametric Form.—If f(x, y) = 0 is the equation of one of the desired equipotential boundaries and x and y can be expressed as real analytic functions of a real parameter t whose total range just covers the conductor, then there is a simple method of obtaining a solution of Laplace's equation fitting this boundary. Let x = fl(t) Then the desired solution is
y = f2(t) (1)
x jy = i(bW) jf 2(bT47)
(2)
This gives V = 0 as the conductor, for substituting this value gives exactly the parametric equation of the conductor with b U in place of t. Unfortunately, the number of cases where this method is useful is very limited. Among them may be mentioned the confocal conics and the various cycloidal curves. As an example, let us find the field on one side of a corrugated sheet of metal whose equation is that of a cycloid
80
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.17
and which forms one boundary of a uniform field. The equation of the surface is y = a(1 — cos 0) x = a(8 — sin 8) (3) j — so that z = a(bW — sin bW) aj(1 — cos bW) = a(bW giving x = a(bU — ebv sin bU), y = a(bV 1 — ebv cos bU). When V is large and negative x = ab U and y = d-abV, so that we have a uniform Substifield in the —y-direction of strength E = +9V /ay = tuting for b gives / TR V . -aE '— z = AcT + 3— 36 To find the field at any point, we differentiate, and
1=
,- cLE dW — —e_:kldw E dz dz
I
dW IE 1 — e-lawE = E 1 — 24 cos U e 2c 8. aE dz = To get the surface density on the sheet, we note that on the conductor V = 0 and y = a[1 — cos (U/aE)], and =
1 a\1 dW dz v-o— EE
This result gives the field on the side of the sheet with sharp ridges. The solution on the other side gives lines of discontinuity so that it is of no value there. 4.17. Determination of Required Conjugate Functions.—In most cases, the search for a function W which fits the given boundary conditions in the z-plane begins by looking for a transformation which reduces the boundaries to simpler shapes. If the new boundary conditions are unfamiliar, we endeavor to find a second transformation which will still further simplify the boundary conditions. Eventually, we should arrive at a situation where the solution can be written down by inspection. We then proceed backward along the steps we have come to the solution of the original problem. It is frequently possible to jump these steps and write f(W, z) = 0 by eliminating the intermediate complex variables. If this is impossible, those variables serve as parameters connecting TV and z. In manipulating these transformations, it is often helpful to visualize that portion of the z,-plane between the boundaries as an elastic membrane which possesses the property that no matter how we distort the boundaries the angle of intersection of any lines drawn on the membrane remains unchanged. The membrane may not separate from a boundary
§4.17
DETERMINATION OF REQUIRED CONJUGATE FUNCTIONS
81
but may slide along it and be expanded or contracted indefinitely. Vor example, suppose the boundaries in our problem are two nonconcentric nonintersecting circles or two intersecting circles or two of one type and one or two of the other intersecting orthogonally. The region with such boundaries may be transformed into a rectangle by 4.13 (1), the relation being
, z zi = in (1) z — ja 1 for W = U jV, since we are Here we have written zi = x1 jy attaching no electrical significance to x1and yiat present. From 4.13 (3) and (4), we see that, when — 00 < x < cc and — 00 < y < 00, then —0 < yi < 27r and — 00 < x1< 00. Thus (1) transforms a horizontal from y'-oo to y =a from from C X'. -IT x'=rt toroo tura)
Ait■
.4111. fox=-co
■ 41111/■ \A,
•11 from C' x'vt toy=-co T toy.-co from y'=-co to y=-a from I B
PIG. 4.17.
strip of width 2r in the z1-plane into the whole of the z-plane. Vertical lines in this strip, from 4.13 (3), go into circles given by
x2
(y — a coth x1) 2 = a2 csch2 x1
(2)
and horizontal lines go into circles passing through y = ±a, x = 0 given by 4.13 (4) to be (x — a cot yi)2 -F y2 = a2 csc2 yi (3) This transformation can be visualized by imagining an infinite horizontal strip of elastic membrane of width 2r being rotated counterclockwise to a vertical position in a z'-plane, so that points xi = 0, yi = 0, and xi = 0, yi = 27 go to AA' and BB' and then pinched together at y' = 00 and y' = — co . These points C and C' are then brought toward each other along the y-axis while the center of the strip is expanded horizontally. The lines CA, CB and C' A', C'B' are opened out, as a fan is
82
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.18
opened about the point C and C', respectively, until CA coincides with CB and C'A' with C'B'. The membrane thus expands to fill the whole z-plane, the infinitesimal arcs AA' and BB' being stretched into infinite arcs bisected by the x-axis. The operation, after the first rotation, is pictured in Fig. 4.17. If there are to be no discontinuities where CA folds against CB and C'A' against C'B', then the potential used in the horizontal strip in the 4-plane must be periodic in y,with a period 27. Problems involving line charges and rectangular boundaries or ones where different sections of such boundaries are at different potentials can now be solved by images. In other cases, it may he necessary to unfold the rectangle into a half plane by a Schwarz transformation, giving, in general, elliptic functions. 4.18. The Schwarz Transformation.—One of the most useful transformations is that which transforms the upper half of the zrplane, bounded by the real axis and an infinite arc, into the interior of a polygon in the z-plane, or vice versa. If the latter is finite, it will be bounded entirely by the deformed real axis yl = 0 of the zrplane. If not, portions of its boundary at infinity may be formed by expanding or contracting the original infinite arc of the zr-plane. To find the transformation that will bend the real axis of the zrplane into the specified polygon in the z-plane, consider the complex derivative dz
dzr
= Ci(zi — u1) 44(21 — u2)13' • •
(z, — un)P^
(1)
where /LI, u2, . . . , u. and 3 27 • • • , are real numbers, Cr is a complex constant and u. > u._, > • • > u2 > ur. As we have seen, the argument of the product of several factors raised to certain powers equals the sum of the products of the argument of each factor by its exponent, so that dz arg — = arg Ci arg (z — ur) ± • )3. arg (zi — u.) (2) d zi When dz, is an element of the real axis in the zrplane, we may write it dx1and we have dz dx j dy —, dy arg -d-4 = arg tan — (3) dx dxr This is the angle that the element dz, into which dzi is transformed, makes with the real axis y = 0 in the z-plane. When z1is real and lies between u, and ur+i, then (z1— (z, — u2), • • • , (zi — ur) are real positive numbers whose arguments are zero, and (z, — ur+i), (zi — 24+2), . . . , (zr — u„) are real negative numbers whose arguments are T. This gives, from (2) and (3)
83
THE SCHWARZ TRANSFORMATION
§4.18
Br = tan-'
= arg C1 + ((I,-÷2
i3r+2
± • • • +
gn)ir
(4)
Thus all elements of the xi-axis which lie between u, and u,..+.1 (Fig. 4.18) have the same direction after transformation and form a straight line, whose slope is given by (4), as shown. Similarly, the elements lying between ur4.1 and ur+2 form a straight line having the slope dy 0,.+1 = tan-' — = arg C1 + (13r+2 dx
Or+3
• • • ± i3n)r
(5)
The angle between these two lines is given by Or+1 — Or = —163r+1
We have now formed two sides of a polygon, and in a similar fashion we can evaluate the remaining (3's and u's to give us the desired vertex angles and lengths of sides of our polygon.
Ur.t
FIG 4.18.
Suppose that the field is to be on the upper side of the broken line in the z-plane shown in Fig. 4.18. Then any angle, such as cer+i, measured between two adjacent sides of the polygon on the side of the field is called an interior angle of the polygon, and since r - ar+7 = -Or+7 we have, to get this angle, to choose =
ar+ 7—1 7r
(6)
Substituting in (1) gives dz = Cl(ZI dzi
-ai -1
u1) 7 (Z1 — u2) 7
• • • (Z1 —
un ) r
(7)
Integrating this expression gives al Z = Cif [(Z
—
02
u1) 7 (Z1 — u2)7
• • •
dz1 + C2 = Clf(21) + C2 (8)
This is the desired transformation. From (4), we see that we may orient our polygon in the z-plane any way we please by giving C1the proper argument. The size of the polygon
84
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.19
is determined by the modulus of C1. The polygon may be displaced in any direction without rotation by choosing the proper value of C2. To show that we have measured a, on the correct side of the boundary, let us set z1= W so that the real axis yi = V = 0 represents an equipotential line. If ar < or, the field dW/dz = dzi/dz must be zero in the angle; and if a, > r, it must be infinite. Setting x1 = U = u, in (7), we see that this is true which verifies the result. 4.19. Polygons with One Positive Angle.—When the real axis is bent at only one point, no generality is lost by taking this point at the origin, so that ui = 0 in 4.18 (8) giving, if a > 0, z
+ C2
(1)
If we choose C2 = 0 and C1real, the polygon formed has its apex at the origin and is bounded by 0 = 0, 0 = a and by an arc at infinity. From 4.08, the modulus of z" is the nth power of the modulus of z, so that circles given by r1 = al in the z1-plane transform into circles r = a in the z-plane. Thus a problem involving the region bounded by two radii of a circle and the included arc can be reduced by (1) to one involving a diameter and a semicircular arc. When a = 2r, the real axis is folded back on itself and the upper half of the z1-plane opened into the full z-plane. Separating real and imaginary parts gives y = 2Cix1yi and x = Ci(xl — yl). Eliminating yi and xiin turn gives y2= —4C1x1(x — Cixi) and y2 = 4CIYI(x + CIA) which are the equations of two orthogonal families of confocal parabolas. Thus, if we plot the uniform field W = zion the z1-plane, we find it transformed in the z-plane into the field about a charged semi-infinite conducting plane. If we plot the field between a line charge and an earthed horizontal conducting plane passing through the origin in the z1-plane, then we find it transformed in the z-plane into the field between a semi-infinite conducting plane and a line charge parallel to its edge. When a = 3r/2, if we set W = z1, we have, in the z-plane, the field near the edge of a charged conducting 90° wedge whose sides coincide with the positive x-axis and the negative y-axis. The field between such a wedge and a line charge can be found in the same way as when a = 2r.
When a = r, there is, of course, no change in the coordinate system except that it is magnified by the factor C1. When a = fr, if we set W = z1, we find, in the z-plane that W represents the field in a 90° notch in a conductor, the positive x-axis and the positive y-axis forming the sides of the notch. In this case, cl U = x2 — y2 and C1V = 2xy, so that the equipotentials and lines of force are orthogonal families of equilateral hyperbolas. The field due to a charged filament parallel to the edge of the notch is obtained as when a = 27r.
-
§4.20
POLYGON WITH ANGLE ZERO
85
4.20. Polygon with Angle Zero. In this important case, the xl-axis is folded back parallel to itself and the upper half plane is compressed into the space between. In order that two parallel intersecting lines have a finite distance between them, it is necessary that the point of intersection be at infinity. Instead of 4.19 (1), we now have = Ci In zi ± C2 (1) If, as before, we take C2 = 0 and C1real, the origin zi = 0 is transformed to z = — 00 and the new origin is at 21= 1. Writing 21 = rieje., we see that when 01= 0, z = C1 ln r1, which is the real axis in the 2-plane, and when 01 = r, x = C1 ln r1and y = Cir, which is a line parallel to the real axis and at a distance Cur above it. Thus the upper half of the 21-plane is transformed into a horizontal strip in the 2-plane. Radial lines 01= constant are transformed into horizontal lines y = constant; the semicircles r1= constant go into vertical lines, of length Cur. Frequently, we have problems in which a configuration is periodic ; i.e.,the field may be split in identical strips. In such a case, the transformation (1) is useful. Suppose, for example, we have a charged filament between two parallel earthed conducting planes. Going back to the 21-plane, we see that this is reduced to a filament parallel to a single earthed plane. By images 4.06 and 4.12 (4), the required function is —
(ln zi ele° A 2re z i— e—i8° To make W = 0 when zi = +1, take C = — In (—eie.) so that we have = q in 21— eie. (2) 2re 1 — z1eiB0 Transforming into the 2-plane taking C2 = 0, Cir = a, and C100 = b, we have, substituting for 21from (1), W=—
-07
eiwa
eprbia
,y
eircz—Ava
In e_ ir(z+2b)/a eir(z+10/a 1 — er(z+3')/a Now using Pc 652, 660, and 645 or Dw 654.1, 655.1, and 702, we have sinh[1--n-(2 — jb)/a] W = — q In 2r€ sinh [1-.7r(z j5)/a] j tanh (4-71-2/a) cot (1-n-b/a) = — q ln 1 2re 1 — j tanh (1-71-2/a) cot (1--rb/a) = —iqtan-1[tanh (jrz/a) cot (irb/a)] (3) 'WE We can separate real and imaginary parts and get 2re V — sin (rb/a) sinh (rx/ a) tan (4) — cos (rb/a) cosh (rx/ a) + cos (ry/a) q 2reU sin (irb/a) sin (ry/a) tanh (5) cos (irb/a) cos (ry/a) + cosh (rx/ a) q W=—
2re
In
-
—
86
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.21
The z1- and z-planes are shown in Fig. 4.20. If the plates are to be at a difference of potential Uo, we need superimpose on this only a vertical field given by W' = —jUoz/a. The final solution will then be W" = W+ W' = — [jq/(7re)] tan-1[tanh (firz/a) cot (-7rb/a)[—jUoz/a (6) where U is the potential function.
I& WV,
z plane
z, plane
Fro. 4.20.—Line charge between earthed plates.
When a plane grating of parallel wires, spaced a distance a apart, forms one boundary of a uniform field, we can find the conjugate functions in the same way. It is necessary only to put +q at both z1 = j and zi = —j and proceed as before. We obtain, in the z-plane, a section of the field including one wire and bounded by lines of force proceeding from the wire to x = + co. It may be noted that if we plot a uniform field in the z-plane and transform back into the z1-plane, we get the field about a line charge already worked out in 4.12. z plane
z plane
z plane
y
y Z1=0
,=-00
z
Zi=0
z.z .#00
x
a -0 Fro. 4.21.
4.21. Polygons with One Negative Angle. Doublet. Inversion. It is now natural to investigate the meaning of Schwarz transformations with negative angles. The transition from positive to negative angles is clearly illustrated in Fig. 4.21. The ends of the real axis of the z1-plane are now brought together at an angle a at the origin in the z-plane, if C2 = 0 in 4.19 (1). The approximate form of the lines yi = constant in the z-plane is shown in the figure. The most important cases, by far, in this category are those in which a = --7r. Suppose, for example, we start with a uniform field W = z1. If U is the potential function, we may think of this field as produced by
§4.21
POLYGONS WITH ONE NEGATIVE ANGLE
87
an infinite positive charge at x1= 00 and an infinite negative charge at x1 = — . Inspection of Fig. 4.21 shows that the transformation in question brings these two charges infinitely close together on opposite sides of the y-axis in the z-plane. By definition, a two-dimensional dipole consists of two equal and opposite parallel infinite line charges, infinitely close together, the product of the charges per unit length by the distance between them being finite and equal to the dipole strength per unit length M. We can get the transformation directly from 4.13 (1) by writing a for ja and using Pc 769 or Dw 601.2; thus 1 — —a
z
W= — A- lIn tae
1 ± (1z
—>
aq= m 2rez
—
, 0 rez
(1)
—
whence
U—
m cos 0 2rer
V_ —m sin 0 2n-Er
also X2 +y2
MX
271-6U
—0 '
x2 -I- y2 +
My
271-€T7
= 0
(2)
Thus the equipotentials are circles tangent to the y-axis at the origin, and the lines of force are also circles but tangent to the x-axis at the origin. Another important case where a = —7 is obtained by letting C1 = a2 in 4.19 (1) giving a2 Z — (3) Zi Write this in polar coordinates, and separate real and imaginary parts, and we have rri = a2 and 0 = —01 Thus every point outside the circle r1= a in the z1-plane transforms into a point inside the circle r = a in the z-plane. If W = f(zi) is a solution of Laplace's equation, then W* = f(4), where zr is the complex conjugate of z1, is also a solution and this field is the mirror image of the other, giving r1= rf and 01 = —Of. Comparing the z-plane and the zr-plane, we have and 0 = 0i (4) rrr = a2 When such a relation holds, r, 0 and r r, Or are said to be inverse points and a is the radius of inversion. If the sum of the charges in the z-plane is not zero, then there must be an equal and opposite charge at infinity, on which the excess lines of force end, and which will appear at the origin on the z1- or zr-plane. Hence we have the rule for inversion in two dimensions.
88
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.211
If a surface S is an equipotential under the influence of line charges q', q", etc., at z', z", etc., then the inverse surface will be an equipotential under the influence of charges q', q", etc., at 4*, z','*, etc., plus a charge — Mq at the origin. This method is useful for obtaining, from the solution of a problem involving intersecting plane boundaries, a solution of a problem with intersecting circular cylindrical boundaries. In polar coordinates, the equation of the circle in the z-plane is u2 r2 — 2ur cos (0 — a) = R2 (5) where the center of the circle is at u, a and R is its radius. Multiplying through by rt2 and writing or for 0 and ar for a from (4), we have u2)7.512 (rrf )2 — 2rrfurr cos (Of — a t) = (R2 Writing a2 for rrf from (4) and rearranging give
a2u
rr2 + 2R2
Or
where
(Cr — ar) — R2 _ u2rcos 1*
212 — 2u1r1 cos (OP —
a4 u2
an = Ri — u?
azu — —1 2 and R1— — u2
(6)
a2R
Iry - u21
(7)
Thus the circle inverts into a circle. If the original circle passed through the origin, then = 1? so that lull Ri = co . Thus the inverted circle is of infinite radius with its center at infinity, which means a straight line. The nearest approach of this line to the origin is a2 a2 —2 1 2/11 — IR11 u R R and the equation of the radius vector normal to it is Or = «T. Conversely, a straight line inverts into a circle passing through the origin. 4.211. Images by Two-dimensional Inversion.—At this point, we shall pause in the discussion of Schwarz transformations long enough to give an example of two-dimensional inversion. We shall use the rules just derived to find an expression, in terms of images, for the field between an infinite cylindrical conductor, carrying a charge —q per unit length and bounded by the external surfaces of two circular cylinders intersecting orthogonally and a parallel infinite line charge of strength +q per unit length. In Fig. 4.211a, representing the z-plane, the charge is at P and the conducting surface is indicated by the solid line. From 4.21, we see that, if we invert about the point 0, the circles will become straight lines, and since this process is conformal, these lines will intersect orthogonally. For simplicity, we take the circle of inversion, shown dotted, tangent to the larger cylinder. Figure 4.211b shows the inverse system in the zr-plane. The latter system presents the familiar problem,
89
POLYGON WITH TWO ANGLES
§4.22
treated in 4.06, of a line charge parallel to the line of intersection of two orthogonal conducting planes. From 4.06, we know the field in the angle between the planes to be identical with that produced in this region when all conductors are removed, but we have, in addition to q at P', line charges -q, +q, and -q at P'1, P'2, and P3 respectively. The rules of inversion tell us that the surface inverse to these planes, which is the surface of our conductor, coincides with a natural equipotential in the field of charges +q, -q, +q, and -q at P, P1, P2, and P3, respectively, where P, P1, P2, and P3 are inverse points to P', P'„ and P3. We see that if CP = rc and BP = rb, then P1is on BP at a distance b2/rb from B, P3 is on CP at a distance c2/r, from C, and P2 ,
.p' )31• \ • 2 \ z plane
••
/
■
(b)
(a)
Fm. 4.211.—Two-dimensional inversion.
is at the intersection of BP, and CPI. Since P', Pi P2 and P', lie on a circle orthogonal to the lines Q'QI and Q'Q'2in the zf plane, then P, P1, P2, and P3 lie on a circle orthogonal to the circles in the z-plane. 4.22. Polygon with Two Angles.—From the large number of cases that come under this class, we shall select as an important example that in which the real axis is bent into a rectangular trough of width 2a. Taking a1 = a2 = it in 4.18 (7), ui = u2 = -a1, we obtain ,
,
dz A dzi (zf Integrating by Pc 126b or 127 or Dw 260.01 or 320.01 gives
(1)
z = A cosh-' (z CI = jA sin-1 ( z al) + C2 ai Taking C2 = 0 gives, if the second form is used, z = 0 when zi = 0. In order that z = fa when z1= + al, we must have a = ijkir or
jA = 2a/ir, giving z=
2az1 sin-1 -
—
7
C. i
or
z1 = a1sin —z 2a
(2)
90
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
0.22
The most useful application of this is obtained by applying (2) to a uniform field in the z-plane. Taking a = lir and writing W = z, we have zi = ai sin W ( 3) The vertical strip of the uniform field is unfolded as shown in Fig. 4.22a. Thus, if V is the potential function, the field in the z,-planeis the upper half of that due to a charged strip of width 2al, and if U is the potential, it is the upper half of that due to two semi-infinite coplanar sheets at a difference of potential r with a gap of width 2a, between them. Separating real and imaginary parts of (3), if Pc 615 or Dw 408.16 is used, gives xi = al sin U cosh V yi = al cos U sinh V Divide the first equation by alcosh V and the second by alsinh V, square, and add, and we have
x?
ai cosh' V
a? sinh' V
=1
(4)
This shows the curves on which V is constant to be confocal ellipses with major and minor axes 2a, cosh V and 2a1sinh V, respectively. In the same way, dividing by al sin U and al cos U, squaring, and subtracting, we have X?
ai sin' U
2
Y1 -
ai cos' U
1
(5)
The curves on which U is constant are confocal hyperbolas. We note that by giving V values from 0 to 00 and U values from —7r to —+.71- and r to lor in the lower half plane, the field due to the strip is represented everywhere in the plane. The hyperbolic lines of force are then discontinuous in passing through the conducting strip. On the other hand, by giving U values between --br and -A-71- and V values from 0 to — 00 in the lower half plane, we represent completely the field due to the two planes. The elliptic lines of force now are discontinuous in passing through the conducting planes. Suppose that in place of a single charged strip we have a great number of similar strips lying uniformly spaced and parallel to each other in the same plane. It is clear that we may obtain a typical cell of such a field by folding up the xi-axis in Fig. 4.22a at ul = -bbl and u2 = — b where bi > aland by applying (2), with bisubstituted for aland b for a, to (3), giving zi = al sin W = b1sin (brz/b) where 2b is the width of the cell in the grating. This is shown in Fig. 4.22b. Since x = ±a, y = 0 when V = 0, U = +fir we must have al = bi sin (jra/b), so that the transformation is W = sin-1
sin (rz/b)1 sin (fra/b) j
(6)
91
SLOTTED PLANE
X4.23
Although derived for the region above the x-axis, we notice that this formula represents the field equally well when 0 > y > - co, where choosing positive values of V requires that, when 0 < x < b, 37r < U < r and, when 0 > x > -b, < U < The strength of the field at y = co is --br /b, so that to get the transformation for a field of strength E' we must multiply by the factor 2bE'/ir. If we used this grating as the boundary of a uniform field of strength E extending to y = 00 , then we need only superimpose a vertical field of strength E' when E' = iE, U= - rr
U= 0
z plane
U= + f
;
I -I
1 pr I
I
u
I 1
I
I
LL
x
U=--
U= +
2
Fxo. 4.22a.—Transformation of semi-infinite vertical strip into upper half plane.
=+—
2cei
2a —p-1
2b U2
ur
4.22b.—Transformation of field of single charged flat strip into one section of field of charged grating composed of coplanar parallel flat strips.
which will cancel the original field at y = - 00, and add to that at y = + co giving , sin (4-rz/b)R 2b { z — sin= (7) sin (fra/b) jf The field intensity at any point is then given by I Iddz I
cos (irz/b) = .E -1 + 2 - [cos2 (frz/b) - cos2 (fra/b)]il
(8)
4.23. Slotted Plane.—To obtain the field in the neighborhood of a plane conducting sheet with a slit of width 2a in it, we may bend down that portion of the real axis lying between x = +a and x = -a and
92
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.23
expand it to enclose the whole lower half of the z-plane, at the same time drawing that area of the zrplane near the origin through the gap. By examining Fig. 4.23, we see that al = 2r, a2 = —r, as = 21r, ul = d-a1, u2 = 0, us = —as, giving, from 4.18 (7),
dz — dzi Integrating gives 4.18 (8) to be z
2— Zi
z,2
= CI(Zi ±
a,2
± C2
z l/
When z1 = al, we wish z = ±a. This gives C2 = 0 and 2a1C1 = a, so that the result is zi , al z= — (1) 2 al zi If we start with a uniform vertical field W = —Ezi in the zrplane and take 2a1= a, the transformation leaves the field at co in the z-plane
x
-4 Flo. 4.23.—Plane with slot forming one boundary of uniform field.
unchanged but it now terminates on a sheet with a slit in it and part of the field penetrates the slit. From (1), we have W = —Ez1 =
(2) ± (z2 — a2)1] where V is the potential function. If the imaginary part of the square root is always taken positive, then, to make the fields add above the x-axis and subtract below, we must use the positive sign in (2). The surface density on the sheet is given by eE ( x IdW + 1) — 2 \(x 2— a2)1— dz ,-0
= ei
(3)
where the sign is chosen so that the two terms add on the upper surface and subtract on the lower. The field between a charged filament and an earthed slotted sheet in the z-plane is easily obtained by (1) from the field between a charged filament and an earthed plane in the z1-plane.
RIEMANN SURFACES
§4.24
93
Writing the right side of (1) in polar coordinates and solving for x and y, we have a cos 0 () 4 ri + 2r1a1 a sin 01 (5) rl — al 2r1a1 Squaring and adding give x2 y2 a2 (6) (r/ + 2)2 + (7,2 1 a?)2 a 4rla1 Thus, if ri > al, the semicircle of radius r1and that of radius al/r1transform into the upper and lower halves, respectively, of the ellipse in the z-plane given by (6). The semicircle r1 = al flattens into the real axis between x = +a and x = —a. B
4.24.—Riemann surface.
4.24. Riemann Surfaces.—To visualize completely the possibilities of a given transformation, it is often valuable to use the concept of a Riemann surface. This is well illustrated by the example in the last article. Although we have already used up all points in the z-plane to represent positive values of yl equation 4.23 (1) gives us a value of z for every negative value of yi, i.e., when —r < 0 < 0 as well. From 4.23 (5), we see that if 0 < r1 < aland —7r < el < 0 we arrive in the upper half of the z-plane, and if al < r1 < 00 and —7 < 0 < 0 we arrive in the lower half. Thus there are two points in the z1-plane corresponding to each point in the z-plane. We can eliminate this double value by making the z-plane a double sheet. We must be careful, however, to connect these two sheets in such a way that, in tracing a continuous circuit in one plane, we trace a continuous circuit through corresponding points in the other. The central portion of such a surface, known as a Riemann surface, is shown in Fig. 4.24 at the right. Lines in the lower surface are dotted, and the sheets are spread apart to clarify the diagram When x2< ce, it is possible to pass only between A and B or C and D, and when x2 > a2it is possible to pass only between A and C or between ,
94
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.26
B and D. The circle r1 = ei where ei—+ 0 becomes a circle at infinity lying in and surrounding the AC sheet. It is possible to construct Riemann surfaces for a great many transformations. 4.25. Circular Cylinder into Elliptic Cylinder.—The last two articles show that the region outside the circle r1 = al in the z1-plane can be transformed into the region outside that part of the real axis lying between x = +a and x = —a, in the BD sheet of the z-plane, by means of the relation z= 2 a(zi ± zi a) (1) al Since, in the zi-plane we are restricted to a region outside the circle ri = al, we cannot cross the xi-axis where —al < xi < +a1. Consequently in the z-plane, we cannot cross the x-axis where —a < x < +a and the line joining +a and —a is said to be a cut in the z-plane. Electrically, this enables us to transform any field which, plotted in the zi-plane, makes r1= al an equipotential or a line of force into a field, plotted in the z-plane, in which the line joining x = +a and x = —a is an equipotential or a line of force. More generally, since from 4.23 (6) any circle r1 = b1where b1 > al transforms into an ellipse, we may obtain from the solution of any problem involving concentric circular cylindrical boundaries the solution of a problem involving confocal elliptical boundaries. If the circle in the z1-plane is taken eccentric to ri = a1in the proper way, it transforms into an airfoil. This is therefore the famous "airfoil transformation" used in aeronautics. 4.26. Dielectric Boundary Conditions.—With the aid of 1.17 (5) and (6) and the relations 4.10 (1), we shall now determine the boundary conditions that must be satisfied by the potential and stream functions when WI = Ui +jV1 = f1(z) and W2 = U2 + jV2 = f2(z) are functions representing the electrostatic fields on the two sides of a boundary where the capacitivities are ei and €9, respectively. Let Van and a/as represent differentiation normal to and parallel to the boundary, respectively. Then, from 1.17 (5) and 4.10 (1), if U is the potential function,
aui an
a U2 an
= €2-
or
3V1
el
as
= €2
a V2
as
(1)
If we take the zero stream line of W1 and W2 to join at the boundary, then we may integrate (1) along the boundary from this line to any other point V1, V2 and obtain in terms of the capacitivity (2) or K1V1 = K2V2 e1V1 = 62V2 in terms of the relative capacitivity. From 1.17 (6), we have = U2 (3) These are the boundary conditions when U is the potential function.
95
ELLIPTIC DIELECTRIC CYLINDER
§4.261
If V is the potential function, the boundary conditions are or K1U1 = K2U2 (4) €1U1 = e2U2 4.261. Elliptic Dielectric Cylinder.—Conformal transformations can be applied not only to cases involving equipotential or line of force boundaries, but also to many cases involving dielectric boundaries. The fact that such an operation preserves angles means that the law of refraction of the lines of force [1.17 (8)] will still be satisfied. Suppose, for example, our problem is to find the conjugate functions giving the field at all points when an elliptic cylinder is placed in a uniform field of force making an angle a with its major axis. Let the equation of the elliptic dielectric boundary be
V1 = V2
and
x2 971, 2
y2 ±— n2 = 1
(1)
The transformation of 4.25 immediately suggests itself as a means for deriving this elliptic boundary, represented in the z-plane, from a circular boundary r1= b, represented in the z1-plane. For simplicity, let us take al = 1 in 4.23 and 4.25, so that the z-plane represents the area outside the unit circle in the z rplane. Writing 4.23 (6) in the form of (1) and equating coefficients, we see that m n b2 a2 = - m m2 - n2 and (2) m-n It is often convenient, when elliptical boundaries are involved, to use confocal coordinates u and v instead of rectangular coordinates x and y. Such a system is shown in Figs. 4.22a and b. From 4.22 (3), we see that the relations are z = a sin w,
x = a sin u cosh v,
The transformation of 4.25 (1) then becomes z zi
=
1 je-iw = je-iu+v = -[z a
y= a cos u sinh v
(3)
zr1) or
c 2)i]-1 (4) (z2 a2)i ] = a[z - (z2 -
This equation shows us that at a great distance from the origin, where ri co, z = jaz1, so that a uniform field W = laEzitransforms into the uniform field W = Ez in this region. It is clear that the desired field may be formed by superimposing a vertical field of strength E sin a on a horizontal field of strength E cos a. These component fields are shown in Figs. 4.261a and b. Since the axis between the foci is an equipotential in a and a line of force in b, we see that in the zi-plane the situation is as shown in Figs. 4.261c and d, respectively. The field of Fig. 4.261c has already been solved in 4.03, from which we have, taking U' for the potential function and U.: = 0 on the unit circle,
96
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS z pla ne
Z
§4.261
pion
n
(a)
z, plane
(c)
(b)
z, plane
(d)
FIG. 4.261.—(a) and (b) show elliptic
cylinder, K = 9, m/n = 3, in uniform fields 2-4E. Equation (4), Art. 4.261, transforms (a) and (b) into (c) and (d), which involve only circular boundaries. Superposition of (a) and (b) gives (e) where cylinder is in field E at 45° to the axis.
(e)
§4.262
TORQUE ON DIELECTRIC CYLINDER
97
"r, = ijaE sin a(—zi = —ijaB'E sin a(zi zi -1) where, using (2), we have 2[(K + 1) (K — 1)b2] — (m n)(Km — n) A' — —b (K — 1) (K + 1)b2 (m — n)(Km + n) 2b2 m n B' (K — 1) (K 1)b2 Km — +n
(5) (6)
(7)
(8)
For Fig. 4.261d, we see similar harmonics must occur. Since T7:' = 0 when zi= eo, the solutions must be of the form = --aE cos a(zi A"zr1) = iaB"B cos a(zi z-fi) Since, from 4.26, we must have we have b2 + A" = B"(b2 + 1)
= U' and
and
(9) (10) = KV.' when r = b,
b2 — A" = KB"(b2— 1)
Solving for A" and B" gives A" _ b2[(K + 1) — (K — 1)b2] (m
n)(m — Kn) b2(K + 1) — (K — 1) (m n)(m Kn) 2b2 m n B" b2(K + 1) — (K — 1) — m Kn
(11) (12)
Superimposing and using the transformation (4), we have, outside the cylinder, We = + (A" cos a jA' sin a)zr9 = lajE[e-gc"-.) — (A" cos a + jA' sin «)eil = iEle-ja[z±(z2 —a2)4]+ (A" cos a+jA' sin a)[z— (e—a2)i]1 (13) Similarly, inside the cylinder, we have 1) = -aE(B" cos a — jB' sin a)(zi zr ( cos a . sin a \
= E(m n)
m Kn
3Km -1- nj
z
(14)
We notice that the field inside the cylinder is still uniform. The lines of electric displacement when the external field makes an angle of 45° with either axis of the ellipse are shown in Fig. 4.261e. 4.262. Torque on Dielectric Cylinder.—The torque T per unit length on an infinite dielectric cylinder perpendicular to a uniform electric field with which its major axis makes an angle a may be found from the last article and from 3.13 (5) and (6). Thus
T=
a [1.2e„(1 — K) f EE;cos 4, dv]
(1)
where 4, is the angle between the original field E and the final inside
98
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.28
field E, and cos cp are constant in v and may be taken outside the integral leaving f dv = irmn per meter. The relative directions and the magnitudes of E and Eiare given by 4.261 (14) with 1 or K replacing K.
(9W, = = = az K az 1 Simplification of the 4.261 (14) product and substitution in (1) give I E I IE;I cos (0 - = EEP (real part) = 1E(T)EN 1E(i)EN cos2 + i 2 ) T = ievE2(K - 1)7rmn(m n) ( aa m + Kn Km n _ re„E2(K - 1)2mn(n2 - m2) sin 2a n)(m + Kn) 2(Km E=
(2)
(3)
(4)
This torque acts to align the major axis with the field. 4.27. Polygon with Rounded Corner.—There are several methods of replacing the sharp corner in a Schwarz transformation with a rounded one. By one method, we replace the factor zr°-4 in
--1 a-1 dz ir (zi - u2)w •• dzi= z by [(z1 + X(zi - 1)/)](1r)-', where Ili.' > lun-1I > • • juzi j 1 and X < 1. By another method, we replace the factor zro-1by (z1
1)ir
+ X(zi -
In both cases, the argument of the new factor is zero when zi j 1 and is a - r when z1< -1, so that the faces of the polygon on either side of the region -1 < z1< +1 make the same angle with each other as if the factor zr°-1were used. Between z1= +1 and z1= -1, we now have a curve whose shape can be adjusted somewhat by means of the factor X. 4.28. Plane Grating of Large Cylindrical Wires.—We have already seen in 4.20 how the problem of a plane grating formed by cylindrical wires of small radius may be solved by taking for the cylindrical surfaces the natural equipotentials surrounding a grating of line charges. When the diameters of the wires become comparable with the spacing, this approximation breaks down completely. We may solve this case however by the method of the last article. Let us take the "cell" of the grating, outlined in Fig. 4.28a. From the figure, we see that the differential expression connecting z1and z is
dz (z1 + 1) 3 +X (z1 - 1)1 C dz1 - 1[(zi - 1)(zi 1)(zi al)] Cl AC1 ai)11 [(z1 1)(z1 Rzi - 1)(zi
*
(LOP
(1)
PLANE GRATING OF LARGE CYLINDRICAL WIRES
§4.28
99
We illustrate here a method we have not hitherto used for evaluating the constant Cl. Since z1 = rleith if we keep r1constant, we have
dz1 = jrieigi clOi= jz1 dei We notice that when r1 00 and Oi goes from 0 to r then z goes from y = 0 to y = b. Substituting jzidOi for dz1in (1) and letting zi co we have foal' dz = jC1(1 + X) f: dOi C1
,fib = jCir(1 + X)
or b
+A)
—
(2)
Frequently, this method can be used to evaluate a constant by an elementary integration of a special case, usually r1 —> 0 or r1 —> 00, before -- r
t
2b
(c)
(b) FIG. 4.28a, b, c.
(a)
performing the general integration. Integrating (1) using Pc 113 or letting z1 = (a1u2± 1)(1 — u2)--1and using Dw 140.02 gives
2b
1 1)i + tanh-1 z1 )+ + al zi + al When z = c, zi = +1, so that C2 is not real and Z —
c
—
+ A)
[t a nh-1 (
2bX tanh-1 ( 2 )1 or 7(1 + A) 1 + al
When z = jc, zi =
c
Zi
—
+2b 7(1 + X)
—
1, so that
tan-1 ( al 2— 1
C2
al + 1 2
— COth2
C2
(3)
[7c(1 (4) +A) 2bX ]
is not imaginary and or
—
2
1— cot2 r7c(1
L
X)]
2b
(5)
Solving (4) and (5) for (al — 1)/(al + 1) or eliminating al gives al — 1 rc(1 + A) rc(1 + A) irc(1 + A) — cot2 tanh 2 — cos2 al + 1 2b 2bX 2b cscrc(1 + rc(1 + A) A)— coth 2b 2bX
(6) (7)
100
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.29
To determine X in terms of b and c, we can plot the ratio of the left to the right side of this equation as a function of X and choose that value of X for which the ratio is unity. Having determined X, we now add (4) and (5) and obtain a1 = coth2[7c(1 2bX X) ]
cot2
[ 7C(1
] (8)
x)
2b
which determines al. The question of how close the 0 curve approximates to r = c has been investigated by Richmond,* \\ 1, and its distance from the origin has —— -4-4--/ been found to vary from c by less than 2 per cent if 2c < b. (d) To obtain the solution when the grating forms one boundary of a -a, uniform field (Fig. 4.28f), we can 0 -3•/ -------superimpose Fig. 4.28d and Fig. 2 4.28e where lines of force are dotted and equipotentials solid. For case — ------ 4.4 ).5 d, the desired function in the zr 4-1r plane is evidently, from 4.22 (3), re) W= Asin' zl where V is the potential function. -a, Since we wish iEb lines of force in _---- --> 2 the strip of Fig. 4.22a instead of / we must take A equal to *Eb/ir, giving ---Eb . W=— 2-ir in -1 z1 (f) (9) z planes (approximate) FIG. 4.28d, e, f.
s
Similarly for case e, we need the xraxis between —al and +1 at potential zero; so, shifting the origin,
we have Ebz 2 1+al — 1 . W = Z-- sin 27r al + 1 When these fields are superimposed, we get case f with a uniform field of strength E at x = + co and strength 0 at x = — co . The potential function, as in 4.22 (3), is V. Accurate drawings of such fields may be seen in the article mentioned. 4.29. Elliptic Function Transformations. Two Coplanar Strips.—The integral in 4.18 (8) of the Schwarz transformation often leads to Jacobi * Proc. London Math. Soc., Ser. 2, Vol. 22, p. 389, 1924.
101
ELLIPTIC FUNCTION TRANSFORMATIONS
§4.29
elliptic functions when the polygon angles are not integral multiples of r or when more than two fir bends are involved. Examples of these appear in Kober and other references. The functions for two similar coplanar parallel conducting strips with equal and opposite charges is a starting point for the solution of several important practical problems. The data given are the strip boundaries + a, + b and the potential difference 2 Uo. Figure 4.29 shows that four ir real axis bends in (a) form a rectangular box in (b) which encloses an area of the uniform field bounded by V = 0, V = V 1, U = Uo, and U = — Uo. Thus 4.18 (8) takes the form Dw 781.01.
f
w
Cl
[(z2
dz
a 2‘1 LI
b2) (z2
—
b\
1 _1
(1)
a)
C2 Sil
There is no additive constant if z = 0 when W = 0, so
z = b sn (Cow, b)
(2)
Special values of the elliptic sines are (HTF II, page 346) sn (K + jK', k) =
sn (K +
k) = k--+
(3)
where k = b/ a is the modulus, k' = (1 — k2)1is the complementary modulus, and K and K' are the corresponding complete elliptic integrals y
V
F U0
,1V1
/
17 —
b
I
0
17
—U0 —a
r
• VT
v=0 a (a)
in
1 \
r1
+U0
VI
FIG.
U
V=0
b 4.29a, b.
(b)
(Dw 773.1). Thus when z = a in Fig. 4.29a
C2(Uo + jV = K jK',
C2 = -u-
=
KU ' o
(4)
The final transformation may then be written
= b sn (KW, b) Uo a
(5)
When y = 0, on the strips Uo and — Uo or the strip V = 0 and the plane Vi,
aw az
Uo a z b) — sn-1 (b' - K az a
vo
TC-I{(x2
a2)(x2
V)]-11
(6)
102
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.30
For the two strips at potential ± Uo, b < I xl < a so that, as in 4.23 (3), b2)(a2 x2)1—i = ± U0(e1C)-1[(x2 (7) For a strip at potential 0 coaxial with a slit in a coplanar plane at Vi, x2)(a2 x2)1—} 0-8 = — Vi(eK')-1[(b2 (8) cr7, = V i(eK')-1[(x2 — b2) (x2— a2)]-1 It is evident that a and b are inverse points for the circle of radius r = (ab)i so that inversion of Fig. 4.29a about this circle leaves the equipotential strips Uoand — Uo, and hence their fields, unaltered. This is impossible if a field line crosses the circle diagonally, so the circle itself must be a field line of value iVi. iop
O
Z Vi
Vi
+ Uo
—ico
( c)
(d) FIG.
4.29c,
d.
The transformation (5) also solves the problems of two coplanar strips or a plane and strip (c) and a strip in a coaxial cylinder or a cylinder or strip coaxial with a slit in a plane (d) . For coplanar strips the capacitance per unit length is, from 4.11 (4),
C =
e[V] eK' = K 2 Uo
6K[(1— b2a-2)1] K (ba-1)
(9)
For strip and plane it is twice this, and for strip in slit it is C =
E[U]
V1
=
4e U 0 V1
4eK (ba--1) K[(1 — b2a-2)i]
(10)
For a strip (2b) in a cylinder [ro = (ab)i] or a cylinder [ro = (ab)i] in a slit (2a), the capacitance per unit length is twice (10). A logarithmic transformation, zi = In z, applied to (5) will give the fields of an infinite stack of strips charged alternately to potentials — Uo and + Uo, and includes a "strip line," which is a strip at potential U0 halfway between and parallel to two infinite earthed planes. Useful formulas for Jacobi elliptic functions of a complex variable appear in HMF, Dwight, Whittaker and Watson, Erdelyi et al., and elsewhere. Jahnke and Emde give some excellent graphs. 4.30. Unequal Coplanar Strips by Inversion.—The inversion method of Art. 4.21, applied to the results of the last article, can solve directly
§4.30
UNEQUAL COPLANAR STRIPS BY INVERSION
103
the unsymmetrical case of two unequal and oppositely charged strips, but the solution takes a much simpler form if we first solve the problem of a strip of width B at potential zero separated by a gap of A — B from a semi-infinite plane at potential V1 as shown in Fig. 4.30a. This is done by folding the real axis in Fig. 4.29a back on itself at 0 so that in 4.18 (7) al = 21r, ui = 0, and a2 = a 3 = • = r. Then if the
U=0
(a) Yi
U=0
h (b) FIG.
4.30a, b.
origin is kept at W = 0, 4.18 (8) becomes z = Czi. If z = B when zi = b, then C is B/b2. Substitution of zifor z in 4.29 (5) gives
z = B sn2 [KW (By]
(1)
'
The modulus is (B/ AP from 4.29 (3), for when W = Uo jVi, then KW/ Uo is K jK' and z = A. From 4.11 (4) and 4.29 (4) the capacitance per unit length between strip and plane is
C =
2e Uo V1
2eK[(B/A — (BMW} )]
(2)
Since the cylinder of radius R = (AB)i is at potential 4-V1, the capacitances between this cylinder and an enclosed strip of width B with one edge on its axis and between it and a semi-infinite plane whose edge is a distance A from its axis both equal 2C or twice that given by (2). Inversion of Fig. 4.30a about point z = w with an inversion radius w(2w < B) will solve the problem of two coplanar strips at potentials U = 0 and U = Uo when one strip is of width w, the inside gap is g, and
104
TWO-DIMENSIONAL POTENTIAL' DISTRIBUTIONS
§4.31
the distance between outside edges is h. That part of Fig. 4.30a lying outside the circle of inversion is shown in Fig. 4.30b, along with that part of the inverse system that lies inside it. From the inversion law w2
zi =
A—
z—w
W(W
g) g
B—
wh h—w
[ (w
k = (By A
(3)
g) (h
So that the transformation is, from (1) and Dw 755.1,
w(h — w) zi — w — h on' (KW/Uo,k)
(4)
From 4.29 (4) the capacitance between the strips is
C —
2E[V1]2€K[(1 — k2)1]
(5)
K(k)
Uo
Let the distance from the center of the iV1 cylinder of radius R to the edges of V = V 1be ci and c2 and to the near edges of V = 0 be d1 and d2 . y
U=0
i
V:=0 --1
Uo
V = Vo
Lx
FIG. 4.31
Then cidi = R2, c2d2 = R2, d1 = ci w, d2 = c2 h — g — w, and ci c 2 = g. With these relations h, w, and g may be eliminated in (3) so that R(ci R(d1 d2) c) k= (6)
R2+ cic2
R2+ cl1c12
The capacitance between the cylinder iV1 and the strip V1equals that between 4-V1 and the two semi-infinite planes V = 0 and is given by
C=
2E U0
iV1
LIEK(k) K[(1 — k2) 1]
(7)
4.31. Charged Circular Cylinder between Plates. The problem in which a conducting cylinder of radius c at potential Uo lies midway between two infinite parallel earthed conducting planes at a distance 2b apart can be solved by using 4.30 (1) and 4.28 (3). The solution is b. First Fig. 4.30a must be almost exact for c < ib but fails as adjusted so that Uo lies between +1 and —1 in Fig. 4.28b and W = 0 is —
§4.31
CHARGED CIRCULAR CYLINDER BETWEEN PLATES
105
at - a. Evidently the origin in Fig. 4.30a must be shifted a distance ai = (A + B) to the right where B = al - 1 and A is ai + 1. Thus 4.30 (1) becomes zi + al = (ai - 1) sn2 ( KW , k) Uo
(1)
The modulus, from 4.28 (6), is then k = (13\1 (al 1)1 - cos +1
firc(1 ± X) ]
2b
(2)
Substitute (1) in 4.28 (3), use the relations between en' u, dn2 u, and sn2u in Dw 755.1, 755.2, and 755.3, and note that tanh-1(jy) = j tan-1 y from Dw 722.3 and HMF 5.07.32 and that C2 = 0. Then 4.28 (3) becomes 2jb dn u en ) ± X tan7(1 + X) an k sn u 1 sn u = 2jb[cos-1(k sn u) + X sin-1(0n u)][7(1 + X)]-1
z -
(3) where u = (KW / U 0,k) . The parameter X must be chosen to satisfy 4.28 (6). From Fig. 4.31 and 4.11 (4) the capacitance per unit length is C -
e[V] 4eK[(1 - k2)l] K(k) Uo
(4)
This is a very close upper limit if c < 2 b, and the error is less than 1 per cent when c = 2 b. Numerical tables and formulas for the real and imaginary parts of en u, do u, and sn u are given in HMF, pages 574, 575. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans, by permission of the Cambridge University Press. 1. That half of the xz-plane for which x is positive is at zero potential except for the strip between x = 0 and x = a which is at potential Vo. The whole of the yz-plane is at zero potential. Show that the potential at any point for which x and y are positive is rV = Vo(tan-1
x — Y a
2 tan Y
x
Y
a
2. The positive halves of the x and y planes are conducting sheets. All the surface is earthed except the region near the intersection, bounded by the lines x = a and y = b, which is insulated and raised to a potential V. Find the surface density at any point on the sheet. 3. By inversion, establish the image law for a line charge in a parallel circular conducting cylinder. 4. A conductor is formed by the outer surface of two similar circular cylinders, of radius a, intersecting orthogonally. It carries a charge q per unit length. Show that q(2r 2— a') (47rar2)-1is the surface charge density, where r is the distance from the axis.
106
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
5. Show that when an elliptic conducting cylinder is charged, the ratio of the maximum to the minimum surface densities equals the ratio of the major to the minor axis. 6. A filament carrying a charge q per unit length passes vertically through a vertical hole of radius a in a block of relative capacitivity K. The filament is a distance c from the center of the hole. Show that the force per unit length pulling it toward the wall is [(K — 1) / (K + 1)]cq2/[27re„(a 2— 0)]. 7. Two fine fiber leads are to be run vertically through an earthed conducting circular cylindrical tube of radius a. Show that the spacing of the fibers must be 2(51— 2)la in order that, when charged to equal and opposite potentials, there will be no force on them. 8. An infinite circular conducting cylindrical shell of radius a is divided longitudinally into quarters. One quarter is charged to a potential V, and the one diagonally opposite to — VI. The other two are earthed. Show that the potential at any point inside is 2ax 2ay + tan ' (tan-1 a' — r' a' — 9. Consider the region of space between the cylinder x' + y2 = b' and the xzplane. All the curved boundary and that portion of the plane boundary for which a < Ix' < b is at zero potential. The part of the latter for which —a < x < +a is at the potential VO. Show that the equation of the lines of force inside is, when a < r b U = 2V ova' [ — ,.— 1 + 7. b' 11- .••1 n f"
cos nO
where only odd values of n are taken. 10C. Three long thin wires, equally electrified, are placed parallel to each other so that they are cut by a plane perpendicular to them in the angular points of an equilateral triangle of side (3)1c. Show that the polar equation of an equipotential curve drawn on the plane is r6
c6 — 2r3c8 cos 30 = constant
the pole being at the center of the triangle and the initial line passing through one of the wires. 11C. A long hollow cylindrical conductor is divided into two parts by a plane through the axis, and the parts are separated by a small interval. If the two parts are kept at potentials Vi and V2, the potential at any point within the cylinder is 1 2ar cos 0 V — V, tan-1 -(V1 +V2) + 2 a2 7.2 where r is the distance from the axis, and 0 is the angle between the plane joining the point to the axis and the plane through the axis normal to the plane of separation. 12C. An electrified line with charge e per unit length is parallel to a circular cylinder of radius a and inductive capacity K; the distance of the wire from the center of the cylinder being c. Show that the force on the wire per unit length is a2e2 K—1 K + 1 2re,c(c2— a2) 13C. A cylindrical conductor of infinite length whose cross section is the outer boundary of three equal orthogonal circles of radius a has a charge e per unit length.
107
PROBLEMS Prove that the electric density at distance r from the axis is e (3r' 67ra
+0)(3,2 — a2
filar) (374— a2 r2(9r4 — 307.2 + a4)
61ar)
14. A horizontal plane at potential zero has its edge parallel to and at a distance c from an infinite vertical plane at potential 17. Show that the charge density on the vertical plane is E(y 2 c 2)-1and on the horizontal plane — e(x 2— c2)-1, where x and y are measured from the line in the vertical plane nearest the edge of the horizontal plane. 15. Show that the capacitance per unit length between a flat strip of width 2c and an elliptic cylinder whose foci coincide with the edges of the strip and whose semimajor axis is a is 27re[cosh-1(a/c)]-1. 16. Show that the force of attraction per unit length between two similar parallel wires of radius a and carrying a charge +q and —q per unit length, respectively, is q2/[27ro(c2 — 40)1] where c is the distance between centers. 17. A plane grating is formed of parallel flat coplanar strips of width 2a with their center lines at a distance 2b apart. If the grating is charged, show that the equipotential surface which is at a mean distance b from the plane of the grating deviates from a true plane by approximately .028b cos' (.1-7ra/b). 18. Find, approximately, the field about a grating composed of parallel wires of radius a at a distance 27r apart, charged to a potential Uo and a parallel earthed plate at a distance b from the center of the wire, if a << b and a << 2r. U—
2' — 2cex cos y —U 0 In e c2e2= — 2ce' cos y U
c2 1
where U, =
In e
2b — 2ceb cos a + c' 2ceb cos a + 1
C 2e2b —
and c = cosh b cos a + (cosh' b cos' a — 1)1. Here we have taken 2a for the maximum thickness, measured parallel to the earthed plate, of a closed equipotential about a line charge, the distance from 2a to the plate being b. 19. Using the results of Arts. 4.03 and 4.25, find the transformation which gives the field about a conducting earthed strip of width 2c whose plane lies in the direction of a uniform electric field. Let U be the potential function. W = + E(z 2— c2) 4 20. Superimposing a uniform field on the result of the last problem, find the transformation giving the field about a flat conducting strip of width 2c, lying with its axis perpendicular to a uniform electric field and its plane inclined at an angle a to it. Find the torque per unit length acting on it. W = E[ ± (z' — c2)1cos a — jz sin a],
T = -rec'E' sin 2a
21. By an inverse Schwarz transformation, find the field about two semi-infinite coplanar planes charged to potentials ± Uoand — U,, whose edges are parallel and at a distance 2c apart. Let U be the potential function. Show that z = c sin (1rW/Uo)• 22. By rotating the transformation of problem 21 through 90° and applying a Schwarz transformation, find the field about a freely charged horizontal grating composed of similar vertical strips of width 2a parallel to each other and at a distance 2b
108
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
apart. Let V be the potential function. The charge on the strip is eUo.
TV =
0 27r
sin1
sinh [rz / (2ib)] sinh (17ra/b)
23. By shifting the origin of problem 21 to the left and applying a logarithmic transformation, find the transformation which gives the field about the grating of problem 22 when it forms one boundary of a uniform field. Let the potential function be V. The charge on the strip is eUo. W=
Uo 27r
sin-,
e'r" — cosh (7ra/b) sinh (7ra/b)
24. Taking U for the potential function and taking c = 1 in problem 21, apply a logarithmic transformation and obtain the field about a set of semi-infinite conducting planes spaced a distance b apart whose edges lie on the y-axis and which are charged alternately to potentials -1-Uo and —
b irW z = - ln sin 7r
2 Uo
25. Show that it is possible to shape the edges of the plates in a many-plate parallelplate capacitor so that the field over the whole plate surface including the edges is constant. Show that this surface is the surface U = Uo/2 in problem 24 and that its equation is
b
X
= -- In 1 2 cos
ary)
26. A variable air capacitor consists of thin plates which slide between other fixed plates. Show, using problem 24, that the additional capacitance due to the bulging of the lines of force at the edges is equivalent to adding a strip of width (b/r) In 2 to the edges of the plate, where the spacing between fixed plates is b. 27. By rotating the transformation of 4.22 through 90°, letting b = 7r, and applying a logarithmic transformation, find the field about a cylindrical trough of radius 1 which subtends an angle 2a at the center of curvature and carries a charge —Q, in the presence of a charge +IQ at the origin.
Q
W = — sin -1 27re
[j(r — 1) cos (i0) — (r
+ 1) sin (40)
274 sin (la)
28. By superimposing a potential due to — W at the origin in the last problem, get the potential due to a freely charged circular cylindrical trough of unit radius, subtending an angle 2a at the origin. W=
47rE
1) sin (20) j(r — 1) cos (10) — (r [ 2 sin-, 2r1sin (la)
j ln r — 0]
29. Show that the ratio of the charge on the convex surface to that on the concave surface in the last problem is (77a)/(7r — a). 30. Show that the charge density on the surface in the last problem is -T-1(Q/70[1 ± cos 10(cos2 ZB — cos' 2ee)-1] where the positive sign applies to the convex surface.
109
PROBLEMS
31. Two semi-infinite straight-edged thin conducting sheets are arranged with edges parallel, so that one is the image of the other in the y-plane. They are at zero potential and form one boundary of the field. Find the transformations for the following cases: a. Sheets parallel, field outside,
b0)( jw2a_2
2j ln W —
z _(7r
- j ± 2j ln a)
b. Sheet at 90°, field outside, z = (1
j)
3boa-I 8
2" +2a2W-*
3
c. Sheets at 90°, field inside, z = (1 +j)lboa-1W-4(W2 a 2 ) — jbo 32. Find the transformation that gives field at all points when a filament carrying a charge q per unit length is placed at a distance d in front of the center of a slit of width 2b in an earthed conducting plane.
q z ± (z 2— b2)1— j[d ± (d2 + b2) 1 W= — — 2re z ± (z2 — b2)1 j[d ± (d 2 b2) 1] ]
In
33. A Wolff electroscope consists of two similarly charged fibers of radius c at a distance 2d apart placed symmetrically inside and parallel to the axis of an earthed cylinder of radius b. Show that the capacitance per unit length between the two fibers and the case lies between
C = 87refln[4cb4 1} . 2d2
and
C = 8re ln
b4 4c2(d2
c2)
neglecting c4 and d4compared with b4. 34. Two similar parallel conducting cylinders are in contact and carry a total charge q per unit length. Show that the capacitance per unit length between these cylinders and a third large cylinder whose axis is their line of contact is approximately 2rE/ln (2b/z-a), where b > a. 35. One plate, of thickness 2A, of a variable air capacitor moves midway between two other plates at a distance 2B apart. Show that the transformation for the field in the neighborhood of the edge is (B — A) sinh (47T-W) 2jB B tanh (17W) 2(B — A) z = A — — cosh-, + cosh-, [A (2B — A)]1 [A(2B — A)]* 7 and show that the apparent increase in width of the plate due to the bulging of the field at the edge is approximately Dy =
{ 71-
B ln
[A (2B — 4)]l} 2B — A A ln B—A B—A
36. A thin infinite plate at potential V = 1 has in it an infinitely long slit of width 2K. It lies parallel to and at a distance h from another plate at potential V = 0.
110
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Show that the field at any point is given by [
x =h
2a2
sinh 7U
r(1 - a2) cosh 7I-U + cos irV
[ y=h
2a2
r(1 -
sin7V
a') cosh 7U + cos 7 V
+ Ul + V]
where U is the stream function and a is determined from the relation aK = 4-2h1tanh-1a + a/(1 - a2)] 37. It is desired to find the field near a charged plate of thickness 2h having a rounded edge of radius h. Show that a suitable transformation is
z = h[W2
27-1147(W2 - 1)1-I- 2j,7-1sin-1W]
where V = 0 on the plate whose surface is given by y = (2h/7)E(x/h)1(1 - x/h)1+ sin-1(x/h)1]
when 0 < x < h and by y = ±h when x > h. Show that this curve deviates from the circle of radius h by less than .14h. 38. Show that when a conducting cylinder whose cross section is a hypocycloid of n points is at the potential V = 0, and carries a charge Q per unit length, the potential at any point outside is given by the transfor.nation Z
2hreW a = -e Q [(n - 1)
n
2npre w]
e
where a is the distance of an edge from the axis. 39. Show that a regular prism of n sides each of width 2a sin (17/n)[cot (i7/n)]1 //(1 - 2/n) sec (7/n) - 1] may be circumscribed on the equipotential V = Q/(27€n) In tan (17/n) in the preceding problem in such a way that each side touches it along two lines at a distance 2a sin (17/n)[cot (17/n)]1/n apart. Show that the two surfaces come nearer coincidence as n increases and that when n = 5 their maximum axial distances differ by less than 6 per cent and their minimum by less than 1 per cent. 40C. A cylinder whose cross section is one branch of a rectangular hyperbola is maintained at zero potential under the influence of a line charge parallel to its axis and on the concave side. Prove that the image consists of three such line charges, and hence find the density of the induced distribution. 41C. A cylindrical space is bounded by two coaxial and confocal parabolic cylinders, whose latera recta are 4a and 4b, and a uniformly electrified line which is parallel to the generators of the cylinder intersects the axes which pass through the foci in points distant c from them (a > c > b). Show that the potential throughout the space is 71-0 sin 10 - ci) 77.1 cos 10 cos cosh a1 - b1 al-b1 A In 7(r1sin 10 + c1 a1 - 51) 7r1cos 10 ± cos cosh a1 - b1 at - b1
PROBLEMS
111
where r, CI are polar coordinates of a section, the focus being a pole. Determine A in terms of the electrification per unit length of the line. 42C. An infinitely long elliptic cylinder of inductive capacity K, given by E = a where x + jy = c cosh (E jn), is in a uniform field P parallel to the major axis of any section. Show that the potential at any point inside the cylinder is -Px(1 + coth a)/ (K + coth a) 43C. Two insulated uncharged circular cylinders outside each other, given by in), are placed in a uniform field of n = a and n = -13 where x jy = c tan a(E force of potential Fx. Show that the potential due to the distribution on the cylinders is en(n-a) sinh nO + e-n(n+is) sinh na sin n -2FcZ( 1)n sinh n(a + 0) 44C. Two circular cylinders outside each other, given by n = a and n = -# where x jy = c tan 1(E jn), are put to earth under the influence of a line charge q on the line x = 0, y = 0. Show that the potential of the induced charge outside the cylinders is q.k1 e-na sinh n + 3) + e-'13 sinh n (a re ,A'n sinh n(a (3)
-
n)
cos n
C
the summation being taken for all odd positive integral values of n. 45C. The cross sections of two infinitely long metallic cylinders are the curves (x 2
y2
2x2= a4 + c2)2 4c
and
y2 +c2)2 4c 2x2
(x2
= b4
where b > a > c. If they are kept at potentials VI and V2, respectively, the intervening space being filled with air, prove that the surface densities per unit length of the electricity on the opposed surfaces are e(Vi -V2)(x 2 y2)1 a2 ln (b/a)
and
e ( V2
—
Vi) (x2 +y2)1
b2 In (b/a)
respectively. 46C. What problems are solved by the transformation d(x + jy) dt
c(t2 1)i a' - t 2
(4,± j4.) - ln
(a a - t
where a > 1? 47C. What problem in electrostatics is solved by the transformation jy = en (4, ± j4,) where 4,is taken as the potential function, 4,being the function conjugate to it? 48. A uniform field Eo is bounded by the upper face (y = 0) of an infinite plate conductor of thickness b. The portion between x = a and x = -a is removed leaving an infinite slit. If U is the potential function, show that the transformation for the
112
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
field may be written in the alternative forms: Eoz = -(1 - W2)1 m22 - 1 + (1 - m2)F
712 —
2E sin-1— W, m
CiEo
1 W2Y, (1 - m2)1] jEoz = (1 - W2)1(1 - 921+ (1 + m2)F[sin-1 ( w2 1 - m2 - 2E[sin-' (1 W2)+, (1 - m2)1] ± jaEo 1 - m2 Eoz
= (_ 1
m2 )1 _ , _ . 1 m) .147i (W2 — 1)1 - (1 - m2)F sin m
2E sin
1
)
m + c2E.
where F(p, q) and E(p, q) are elliptic integrals of modulus q of the first and second kinds, respectively. The constants Cl, C2, and m satisfy the equations C2 - a = a + jb - Cl = [(1 - m2)K(m) - 2E(m)]/Eo bE0 = - (1 + m2)K[(1 - m2)11 2E[(1 m2)1] where K and E are complete elliptic integrals of modulus m. 49. An infinite conducting sheet at potential zero occupies the x = 0 plane except for a groove or ridge whose section is a circular arc of radius b with its center at x = c. This surface forms one boundary of a field E that extends to x = co and is uniform except near the groove or ridge. Show that the potential is the imaginary part of W=
(b2 - c2)17rE [z a
[z
j(b2 - c2)1]7/' [z - j(b2 - c2)1pri. j(b2 - c2)1]nia - [z - j(b 2 - c2)+W"
where cos a = c/b and for a ridge 0 < a < r and for a groove 7r < a < 27r. 50. Show that in two-dimensional rectangular harmonics a solution of Laplace's equation is V = (A sin mx B cos mx) (C sinh my + D cosh my) when m is not equal to zero and V = (A + Bx)(C Dy) when m is zero. 51. The walls of an infinitely long conducting prism are given by x = 0, x = a and y = 0, y = b. A line charge of strength q per unit length lies at x = c, y = d, where 0 < c < a and 0 < d < b. Show that the potential inside the prism is .
2q
V' = — 'WE
miry . mrc . m71-x 1 mrb . mr sin sin - csch sinh (b - d) sinh , where 0 < y < d M
m=1
a
a
2q z 1 mirb mind . mr (b V" = — - csch sinh sinh m 71-E a a a m=1
a
y) sin
a
a
mirc minx sin where d < y
Show by 3.08 (2) that the force per unit length is
-i
.. az mrb 71171-C mra mr(2c - a) m7rd j mr(2d - b) 1. sine sinh sin' + - csch ' Z E bcsch b sinh b a a a a b
e m=1
Note that, unless c and d are very small, this series converges rapidly. 62. Show that the set of three line charges q at zio, q at zoo, and -q at the origin in the zi-plane of Art. 4.261 transform into the undistorted field of a single line charge q, at zoin the z-plane where zo = zio 53. Using the image law of 4.04 for a circular dielectric cylinder and the image law established in the last problem for the unit circle, find the conjugate functions
113
PROBLEMS
giving the field when a line charge q is placed parallel to the axis of an elliptic dielectric cylinder K. In the confocal coordinates of 4.261, the charge is at uo, vo outside the cylinder whose surface is given by v1. Obtain the result, inside the dielectric. W, =
x.k K 1), —q In (2 (sin (u 7,€,,(K + 1) l (K +1
jv) — sin [uo ± j(2nv1
vo)]})
n=0
and outside, due to polarization alone, K — 1 In [1 — egu-u0)+2v1uo-v] 27re. K + 1
= q
W°
{
xN — 4K (K + 1) 2 %1 K + 1 n
In [1 ± eicu±o0)-(o+o0+2ool)]
where the upper sign is taken when n is even and the lower when n is odd. 64. A line charge q is placed parallel to the generators of a parabolic dielectric cylinder. In terms of the parabolic coordinates y = 21;n and x = Z 2 — n2, where — < t < co and 0 < n < co, the coordinates of the charge are to and n o and of the surface of the cylinder n i. Show that inside the dielectric the transformation giving the field is W.
K— —q In 1(t 7,€,(K + 1) t'(K + 1
in) 2 —
± j(no
2non)]2 )
n=0
and that outside, due to the polarization alone, it is W°
__. q (K — 1 In it — to — j(2(ri — no — n)1 2orc,, K ± 1
4K (K + 1)2
n= 1
— K+1
ln 1E T- to +./[77 + (no + 2nna
where the upper sign is taken when n is even and the lower when n is odd. (y/b)2n = 1 carries a charge —Q per unit length. 56. The cylinder (x/a) 2n Show by applying Gauss's theorem to the surface V = 0 that 1
1
z = a(os
2oreW)n
. 21-€W)n jb(sm -
Q
Then show by applying it to the cylinder r = oo that the enclosed charge per unit length is nQ. Hence the space outside the cylinder is charge free only when n is 1. 56. A conductor is bounded by 0 = a, 0 = —a, and r = 1, and carries a charge 71-6 per unit length. Show that V is the potential function in the transformation In z = — 2j {sin-1
or sin W La(2v — a)I1
+n— a or
tan
(ir — a) sin W
}
[a(2or — a) — v2 sine MI
57. An infinite plane conducting plate with a thin ridge of uniform height a at x = 0 forms one boundary of a field which is uniform and of strength E far from the ridge. By a simple Schwarz transformation show that W = E(c1,2+ z2)1.
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
114
58. A conductor of star-shaped section is formed by 2m perfectly conducting strips of width a with one edge on the axis and with uniform angular spacing. The total charge per unit length is q. Show that the complex potential function is W = [q/(2mre)] sin-,(z/a)m. 59. One plate at potential Vo of a capacitor lies at y = a, - co < x < 0, and the whole x-axis is at zero potential. Show that the complex potential function is given by (W z=a - — + e-7rwlvo ± 1 ir Vo
60. The surface V = 71" is the negative x- and y-axis, and the surface V = 0 is x = h when y < k and y = k when x < h. Show that the transformation giving the field is =h
1)1(h2e 5' - k2)-1] - k tanh-,[h(e5'
[k(ew
1)1(h2e5' - 10)-1]
Show that the additional capacitance per unit length over what would exist if the only fields between V = 0 and V = 7r were uniform and in the regions x < 0 and y < 0 is 2( e h h 2 k2 k In + tan-, - + -tan-,LC h4hk h k k
)
61. A conducting flat strip of width 2a lies on the real axis midway between and parallel to two infinite earthed conducting planes at y = b and y = -b. Apply a logarithmic transformation to 4.29 (5) and thus show that the complex potential function and its derivative are W
dW
=
U0 sn-i K(e-'.16) i7rU0
bK(e-'am )
(eilr(z+a)111, e—ra/b)
( 2 cosh
- 2 cosh Ira -1
where K(k) is a complete elliptic integral. Show from 4.29 (9) that the capacitance per unit length of this strip line is C
= 2eK[(1 - e-27./b)1] K(e-ram)
The origin is at the center of the strip. 62. An infinite plane at potential V1occupies the y = 0 plane except for a gap bounded by x = d and x = -d. A flat strip at potential 0 occupies the y-axis between y = -h and y = h. By application of a Schwarz transformation to Fig. 4.29a, show that z = jh en
[KW U0
h (d'
h2)1]
and that the capacitance per unit length is C-
elf[h/(d 2
h2)4]
K[d/(d 2
h2)1]
115
PROBLEMS
63. Two right-angle bends in the real axis of Fig. 4.29a solve the problem of a horizontal conducting strip at potential V = 0 with its edges at x = c and x = -c between two infinite vertical conducting planes at potential V2 located at x = d and x = -d. Show that the transformation is 2d (KW cr)] z = — cos s- do , sin 712d UO and that the capacitance per unit length is, if K(k) is a complete elliptic integral,
C =
4eK[sin (ic7r/d)] K[cos (ior/d)]
64. Apply a Schwarz transformation to Fig. 4.30a with 17r bends at B and A and a 27r bend at 1(A B). Thus show that the complex potential function giving the field when the negative x-axis is at potential zero and a flat strip with one edge at x = b, y h and the other at x a b, y = h is at potential U0 is -
z = b[l dn (KW, k) cn (K , k)] Uo Uo where k is [(h2
b2)1 - h]/b. Show that the capacitance per unit length is
C
= 2EKR1 - k2)1l K (k)
65. Write z1for z in 4.30 (1) and invert the result about z1= 2R (cot y — j) with a radius of inversion 2R. Thus solve the problem in which that part of the surface of a cylinder of radius R which lies between B = 0 and B = a is at potential V1 and that part between B = 13 and B = y at potential zero. Let a < p < 17r and 0 < y < Take z = 0 at the center of inversion and let the x (8 = 0)-axis parallel the x1-axis. Show that the complex potential function is -
z = 2R {M
U0 (MY] - cot 0 + j sn2 [KW, o N '
where M is cot y + cot 0 and N is cot -y + cot a. Show that the capacitance per unit length is C = (e E,)K(MIN-1) {KR]. - MN-011) -1 if the cylinder is of dielectric constant e. 66. The centers of two parallel flat conducting strips lie at x = d and x = -d and the edges at x = d, y = ±h and x = -d, y = ±h. Show, by use of a Schwarz transformation on Fig. 4.29a, that the capacitance is given by 4.29 (9) where b /a is found from k in E(k)F (0,k) - K(k)E(q5,k) = —; sin q5 [
and a is then given by a = 2K(k)d/r. 320, to simplify relations.
by k2 = 1 - a
K(k) - E(k)]i k 2K(k) Use Legendre's relation HTF II (15), page
116
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
67. A rectangular bar with sides 2a and 2b carries a charge -27re per unit length. By the application of a Schwarz transformation in which the real axis bends are 1r at x1 = ±k and fir at x1= ±1 to 4.22 (3) with a = 1, derive the transformation E[sin-, (k-,sin W),k] - (1 - k2)F[sin-' (k-' sin W),k + 3b E(k) - (1 - k 2)K(k) ]
z -
where the origin is at the center of the rectangle and k is found from a b
E(k) - (1 - k 2)K(k) k 2K[(1 - k 2)i] - E[(1 - k 2)4]
On the top surface 0 < sin U < k and on the side surface k < sin U <1 so that the charge densities are =
E(k) - (1 - k 2)K(k) a(k 2 sin2 U)
e
ab
-
E(k) - (1 - k 2)K(k) a(sin2 U - k2)1 6
68. A rectangular conducting tube bounded by x = - b, x = b, y = 0, and y = b contains a line charge of strength q per unit length at xo, yo. Show that the complex potential is potential is
q
In
W = 2re
sn (z/C,k) - c - jd sn (z/C,k) - c jd
where C and k are determined by b = CK(k) and a = Clf[(1 - k2)1], and c and d are given by c
jd =
sn (u,k) dn (v,k') + j cn (u,k) dn (u,k) sn (v,k') en (v,k') cn 2 (v,k') - k2 sn2 (u,k) sn2 (v,k')
where u = xo/C, v = yo/C, and k' = (1 - k 2)4. 69. A slit electron lens consists of two earthed planes at y = h and y = -h, each of which has a slit with edges at x = k and x = -k and a plane at potential Vo occupying the x-axis except for a slit with edges at x = 1 and x = -1. Show that the complex potential function is given implicitly by z = [1,2
b2
a2
-
11
c2 -b 2 b
sin' (lirW/jVo)
cosh-'
[b sin (ir/VtiVo)1
(b2
a2)1
-
where b, a, and c are given by k =[b(b
11
2h ,r
, 2h sink-1
iorb
ir / J1
2h a 1 = a — tanh
1) c =[b(b + —
7r
70. An ion source consists of a thin flat earthed strip oppos.te a thin infinite plate at potential V1 with a slit in it. The slit lies on the real axis between x = a and x = -a and the plane is at p = - b with the slit between x = g and x = -g. Apply a Schwarz transformation to Fig. 4.29a with 27 bends at x = c, x = -c, x = d, x = -d; ilr bends at x = b, x = -b; and -17r bends at x = a and x = -a where 0
2h zl = —[K(k)Z(u)
E(k)k 2ABsn u cn u dn u (A - B)(1 -
117
PROBLEMS
where u = KW/U o, Z(u) is a Jacobi zeta function, and A, B, and the modulus k are found from [1 — (1 — k 2)-1AB]E(k) = (A — B)K(k) E(k)AiB(1— A)1(A — 1 + k2)il 2h f = — [K(k)Z(u1) (A — B)(1 — k 2) 7r
2
g
E(k)ABI(1
K (k)Z (U2)
7r
B)i(B +1 — k2)ii
=
(A — B)(1 — k 2)
(1 B)-i. Show that the capacitance where ut = sn-1[(1 — A)ific] and u2 = per unit length is K(k) /K[(1 — k 2) 1]. Milne-Thompson gives tables of all functions used and all formulas are in Whittaker and Watson. 71. Show that the Green's function for a line charge q per unit length at zo in an equilateral triangular conducting tube whose base lies between x = 1 and x = —1 is — f (cz) —f (czo) W = —q In f(cz) — f(cz'o`)
where f(cz) —
sn (cz,k) do (cz,k) [1 + en (cz,k)]2
The modulus k is 2(1 + 31)2-1or 0.96593, and c must be chosen so that sn (c,k) is 2(1 + 31)-131, which gives c = 1.9138. 72. A cylindrical conductor has a lens-shaped section bounded by two circular arcs and carries a charge q per unit length. The exterior angle of intersection of the arcs is a and the exterior angle made by the upper convex surface A with the plane passing through the edges is O. Let the distance between the edges be 2c, the distances between a point on the A surface and the two edges a and al and between a point on the B surface and the edges b and bi. Show that the charge densities on the two surf aces are 2qc(aai)orico-1sin (ir(3/a) ga a[a2ria — 2(aai)'/' cos (0/a) + 2qc(bbi )orw-1sin (irp/a) B
=
a[b2ria
2(bbi)'"c' cos (irf3/a)
This result can be found by inverting about the charge the Green's function for a line charge parallel to the edge of a conducting wedge. The latter is found either by circular harmonics or by a conformal transformation. 73. Solve the preceding problem by a conformal transformation taking the edges of the lens-shaped section at x = c and x = —c. Thus show that the complex potential function outside the cylinder is W=
(z c)ria — (z — c)ria — q In
2,re
(z
c) ria
Oricee220/a
(z
Show that the potential U may be written U=
47re
In
7Tr/' — 2(r ir2 )ria cos
r ,2 '/'
(r /a)
a] 71" / a — 2(rir2)"la cos [r(2,3 — 4,)/
where r1 and r2 are the distances from the field point to x = c and x = —c and 1,t. is the angle between them. The surfaces of the cylinder U = 0 are given by = and yt, = — a. 74. The cylindrical conductors of lens-shaped sections of the last two problems are earthed in the presence of a charge q per unit length at zo outside the cylinder. Using the symbols of the previous problems, show that the complex potential is W=
q 277-E
[(z
ln [(z
c) (zo — 01"./'— C)(2: — C)]T/a
[(z
— c)(zo
[(z — C)(2:
OP"' C)Prk‘e22Pria
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
118
Show that the electrostatic potential is
U =
-q Lire
ln
(rir20)21ria - 2(rir2riono)'/' cos (rir20) 27r'a
-
[7(0
-
2(rir2riorzo)"i' cos [r(2)3
(r2r2o)zria
s¢o) /a] -
ito)/a]
(r2r2o) 2'""
where no and no are the distances from q to x = c and x = -c and %Go is the angle between them. When t/, = p or 1,t = p - a, U = 0. 75. A conductor at potential zero lies at y = 0 from x = h to x = ot, and at x = h from y = 0 to y = - .0. A second at potential Vo lies at y = k from x = -h to x = - 00 and at x = -h from y = k to y = - co. Show that the complex potential function is ai ) In z = (z1 - ai)/(zi + b1)1+ (b1 -
(zi
- ai)/ + (z1+ bi)/ (ai + b1)1 (bi - ai)zi + 2a1b1 (al + bi)zi
- (aibi)1
where zi = erwivo, a1= 2[(h2 +k')1 k]/,r, b1 = 2[(h' - k')1 +k]/,r. 76. With a Schwarz transformation, bend the real axis in the zi-plane through angles 3r/2 at the origin, 2,r at xi = 2 1, 71-/4 at xi = 312 and xi = -312 and require that•zi = 0 and zi = +312 all fall at x = y = 0. Thus show that the transformation z = bz11(34 - 4)1forms the xi-axis into an arrow with the point at x = y = 0, the shaft on the positive x-axis, and the barbs of length b at 45° with the shaft. 77. Solve the problem of a thick cylindrical plate with a rounded edge by the first method of Art. 4.27 with a = 2r so that z-
- Xzi (4 - 1)1+ X cosh
When x1 > 1, - c0 < x < -1 and y = 0; when x1 < -1, - c0 < x < -1 and y = 2; when xi = 0, x = 0 and y = 1. Show that if y = 3.735, points x = -1, -1, 0, I Z, +1 all lie on a circle of radius 1 centered at x = -1, y = 1 in the z-plane. 78. A semi-infinite conducting sheet at potential Uo occupies the plane x = 0 from y = 0 to y = .0. The plane y = 0 is a conducting sheet at potential V = 0 except for a gap between x = -a and x = +a. Show that the force per unit length sucking the semi-infinite sheet into the gap is 2eUVra. 79. A circular cylinder of radius a centered at the origin is at potential U = U,. A set of four rectangular hyperbolic cylinders given by xy = +1,2 is at potential U = 0. Show from 4.31 that the transformation giving the field is --
KW z = 2b {j cos-1En k(—,k)]-1- jX Uo
[cn
KW 4)1} Uo
+ X)]-1
where the modululus k is cos [-kb-27rc2(1 + X)] and the parameter X is found from 4.28 (6) with a2 in place of 2c and b2 in place of 2b. Show that the capacitance per unit length is C = &K[(1 - k2)1][K(k)]-1
For good accuracy c/b 2-1. 80. A circular cylinder of radius a centered at the origin is at potential U = Uo. A parabolic cylinder given by y2 = 4p(x + p) is at potential U = 0. By folding the x-axis back on itself in 4.31 show that the transformation giving the field is z - - 8p[cos-1
sn
KW Uo
+ X sin-1(cn
KW)T
[r(1 X)]-'
PROBLEMS
119
where the modulus k is cos [71-2)-12a2 (1 + X)] and the parameter X is found from 4.28 (6) with 2a2 in place of c and 2b2 in place of b. Show that the capacitance per unit length is C = 2€K[(1 — k 2)1J[K(k)]-1. For good accuracy a/p 1. 81. The equation of a parabolic cylinder at potential U = 0 is y2 = 4p(x p). An infinite thin strip at potential Uo lies between x = 0 and x = a. Show by folding the x-axis in problem 61 back on itself that the complex potential function for the field between parabola and strip is 1 W = U o[K(k)]-2 sn-1{exp [iirpt(zi — al)],k), k = exp (-7rabp1) Show that the capacitance per unit length is eff[(1 — k2)1][K(k)]-2. 82. A rectangular bar at potential Uo is bounded by x' = ±b, y' = ±c and lies between earthed conducting planes at y = ±a. Form one-quarter of this system by applying a Schwarz transformation to Fig. 4.30a with a bends at x = 0, x = B, x = A and a fr bend at x = 1, where B <1 .< A. Show that CI in 4.18 (7) is ahr by the method used for 4.28 (2), and for the main integration substitute a new variable u where z is (B — u)/(1 — u). The resultant integrals are in Milne-Thompson, page 29. Thus show that the relations needed to evaluate B and A in terms of a, b, and c are
A = 2k{ (1 p2)k [(1 (a — c)K(k) = bK[(1 — k 2) 1], irbp = 2akK(k), A = (1 — B)p 2
p2)2k2 — 4701
The first equation is solved for k by noting the intersection of the plotted curves for left and right sides as a function of k. The full accuracy of the K(k) table may be attained by successive plots with increasing scales. Now p, A, and B are found in terms of a,b,c. The capacitance per unit length is, from 4.29 and 4.30,
C =
4E171
4€K [1 — (B/A)]1
Uo
KRB /AA
83. One edge of a conducting strip at potential Uo lies at a distance b from the edge of a conducting wedge at potential zero and the other at a distance a, where a > b. The plane of the strip bisects the wedge angle making an angle a with each face. From Art. 4.30 show that the transformation for the fields is b ir/a 2a/r z = b{sn [KW, (
Uo a
Show that the capacitance per unit length between wedge and strip is C
=
20V1 Uo
(b/a) 7/a ]1} 2€K1[1 K[(b/a) 1""] —
84. The sharp edges of two identical infinite conducting wedges at potential zero, split by the x1-axis, face each other at a distance 2a apart. A strip at potential Uo occupies the x1-axis midway between them leaving a gap of width b on either side. Apply a Schwarz transformation to Fig. 5.30a with a bend a, between a and 2r, at the origin and 2a at z = A to form one-quarter of the field, so that if V = B / A
120
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and B (M,N) is an incomplete beta function, 1 zcatir)-1 }-1 zcalr)-1 b dz dz a o (1 — z)1 {fo (1 — z)1 cf.
1+
BRa/r), ()J
(ahr),n + 1] k2n+2}
rk2a/7(1 — k2)1{
an(4/r),( )]
Bk2 Ra/r)i (4)]
n=0/3q) + (a/lr),n + 1]
by HMF 6.6.1, 26.5.1, and 26.5.4. This gives k2 implicitly in terms of b/a. Show that the capacitance per unit length between strip and wedges is C=
4eVi Uo
4elf[(1 — k 2) 1]
K(k)
References The following books contain material pertinent to this chapter. H.: "Partial Differential Equations," Cambridge, 1932. Gives some interesting bipolar transformations. BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Has very complete and detailed treatment of all methods. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Gives extensive, well-illustrated treatment in Chap. X. ERDELYI, A., W. MAGNUS, F. OBERHETTINGER, and F. G. TRICOMI: "Higher Transcendental Functions," 3 vols., McGraw-Hill, 1953. Vol. II has extremely useful elliptic function formulas. (HTF) FLUGGE, S., "Handbuch der Physik," Vol. XVI, Springer, 1958. Pages 40 to 82 give two-dimensional fields including many conformal transformations. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Berlin, 1927. JAHNKE, E., F. EMDE, and F. Loscif.: "Tables of Higher Functions," McGraw-Hill, 1960. Numerical tables and graphs of many functions. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Follows Maxwell and gives examples. KOBER, H.: "Dictionary of Conformal Transformations," Dover, 1952. The most extensive collection. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Considers images inversion and conjugate functions and gives good field drawings. MILNE-THOMPSON, L. M.: "Jacobian Elliptic Function Tables," Gives all needed manipulations formulas and numerical tables of sn u, en u, do u, and Z(u). MORSE, P. M., and H. FESHBACH: "Methods of Theoretical Physics," 2 vols., McGrawHill, 1953. Many conformal transformations scattered through both volumes. NATIONAL BUREAU OF STANDARDS: "Handbook of Mathematical Functions," U.S. Bureau of Commerce, 1964. Good for all functions. Has both formulas and numerical tables. (HMF) ROTHE, R., F. OLLENDORF, and K. POHLHAUSEN: "Theory of Functions," Technology Press, 1933. Gives theory with important practical applications. RYSHIK, I. M., and I. S. GRADSTEIN: "Tafelin," Deutscher Verlag der Wissenschaften, 1957. Formulas for most functions. Very good on elliptic functions. WHITTAKER, E. T., and G. N. WATSON, "Modern Analysis," Cambridge, 1920. Gives very lucid treatment of elliptic functions. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and references. BATEMAN,
CHAPTER V THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS 5.00. When Can a Set of Surfaces Be Equipotentials?—At first glance, one might think that the class of three-dimensional potential distributions in which there is symmetry about an axis could be obtained by rotation of a section of a two-dimensional distribution provided that, in so doing, the boundaries in the latter case generated the boundaries in the former. This is not true in general. We shall now find the condition that a set of nonintersecting surfaces in space must satisfy in order to be a possible set of equipotential surfaces. Let the equation of the surfaces be
F(x, y, z) = C (1) Since one member of the family corresponds to each value of C, if it is to be an equipotential, we must have one value of V for each value of C, so that V = f(C) must satisfy Laplace's equation. Differentiating results in
a2v = f"(C)(-ac)2 -I- (C)a2C
ac av = f ,(C)a--i, etc.,
axe
ax
OX
O X2
etc
Substituting in Laplace's equation gives
a2v v2v = a2v — aye axe
a2 v az'
= f"(c)(vc)2 + r(c)v2c = 0
giving V 2C (VC)2
f " (C)
(2) f'(C) The condition then that the surface F(x, y, z) = C can be an equipotential is that V2C/(VC)2 can be a function of C only. By integration of (2), we can now obtain the actual potential. Since f " (C)/ f'(C) = d[ln f'(C)]/ dC, we have
4.(C) dC = — in [f'(C)] + A' or
f'(C) = Ae-fcc) dc Integrating again gives
V = f(C) = A f e-f cc) dC dC B 121
(3)
• 122
• c
-
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.01
The constants A and B can be determined by specifying the values of the potential on any two of the surfaces given by (1). 5.01. Potentials for Confocal Conicoids.—As an application of the formula just derived, we shall now show that any one of the three sets of nonintersecting confocal conicoids, given by the equation 112 22 x2 ± —1 (1) c2 + 0 a2 b2 where c > b > a and — c2 < 8 < 00, is a possible set of equipotential surfaces. To get a picture of these surfaces, let us vary 8 over the given range. For the range —a2< 8 < 00, every term in (1) is positive so that it represents an ellipsoid. When 0 = 00, we have a sphere of infinite radius, and when 0 = — a2, the ellipsoid is flattened to an elliptical disk lying in the yz-plane. When 0 passes from —a2 3 to —a2 — 3, we pass from the region of the yz-plane inside the disk to that outside. The latter is one limiting case of the hyperboloid of one sheet which (1) represents when —b2 < 0 < —a2. When 0 = —b2, the hyperboloid of one sheet is flattened into that region of the xz-plane which includes the x-axis and lies between the hyperbolas cutting the z-axis at z = + (c2 — b2)1. (3 to —b2 — 3, we pass to the region of the When 0 passes from —b2 xz-plane on the other side of these hyperbolas which is the limiting case in which the hyperboloid of two sheets, represented by (1) when —c2 < 0 < — b2, is flattened into the xz-plane. When 8 = —c2, we have the other limiting case in which the two sheets of this hyperboloid coalesce in the xy-plane. Thus one curve of each set passes through each point in space; and, since it can be shown that the three sets are orthogonal, we can apply to them the theory developed in 3.03 for orthogonal curvilinear coordinates which leads to ellipsoidal harmonics. The latter are too complicated to be treated here, although later we shall treat the special cases of spheroidal harmonics. To return to our problem: Let x2
Mn — (a2 0)n
Y
2
(b2 + 0)n
Z2
( 2 + 0)n
and
1 N — a2 +0
1
1
b2 +0 c2 ±0 With this notation, (1) becomes M1 = 1, and differentiating this we have 2x ao a0 2x M2 = 0 or a2 0x M2(a2 0)' etc. aX so that 2 (1 2 4M2 — 4 (v0)2 =
a e)2 (1+ +
=
M2—M z
§5.02
CHARGED CONDUCTING ELLIPSOID
123
Differentiating again gives
a20
2
a x2
M2(a2 + 0)
2x
2 M2(a2 +
0)
00
a2
M2(a2 + 0)2 ax 4x2
2x
12
2x
0 m3 (a2 + 0)2
4x2
ax
8x2M3
m2(a2 + 0)3 m2(a2 + 0)3 + (a 2 0)2mg, etc.
Adding similar expressions for y and z gives
8M3 8M2M3 - 2N
v20 = 2N
M2 Mi
MI M2
Substituting in 5.00 (2), we have
N V202N .11/2 (V0)2 _ M2 4 - -2-
1( 1 so 43(0) - 2 a2
1
1
) (2)
b2 ± 0 c2 + 0
This proves that such a set of equipotentials is possible. We now find the potential by 5.00 (3) to be
V = A f e[(a2
0)(b2
0)(c2
d0 B
(3)
This is an elliptic integral given by Peirce 542 to 549 with x = - 0. The constants A and B may be taken real or imaginary, whichever makes V real. 5.02. Charged Conducting Ellipsoid.—If we choose V = 0 when 0 = CO, 5.01 (3) takes the form V = -A f 'e[(a2
0)(b2
0)1-1 de
0)(c2
(1)
If we choose V = Vo when 0 = 0 then, substituting in (1) gives
-A = V 0 ) fo 'eRa2
0)(b2 +0)(02 + 0)]-1 de
(2)
The field at infinity due to this ellipsoid, if its total charge is Q, will be Q/(47rEr2). We see from 5.01 (1) that as 0—* co, x2 + y2 +z2 = r2 -> 0, and so a0/ar -> 2r giving
ay ar
2A ay ae A ae ar = r3 2r = r2 =
4rEr2
(3)
Hence The capacitance of the ellipsoid is, from (2),
Q Vo
8/rE A
, = 8re{f [(a2 vo
0)(b2
0)(c2
4,e(a2 - b2)IFRa2 -b2)1(a2 - c2), sin I (1
0)]-i dB c2a 2)11
The surface density is given by OVI T7 to
o = -e(VV)0-0 = -e(— 490
v vi
)o-o
(4)
124
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.04
From (1), (OV/00)0-0 = A(abc)-1and, from 5.01 (1.1), IVOI = 2M2-1 so that Q (x2 y2 ,_ (5) = 4irabc
+
5.03. Elliptic and Circular Disks.—The capacitance of an elliptic disk, obtained by putting a = 0 in 5.02 (4), is still an elliptic integral. To get the surface density, we write 5.02 (5) in the form a 2z2y a 2y2 Q (x2 c4= 471-bC a2 b4 Now let a -> 0, and the terms involving y and z can be neglected. Since both x and a are zero, the first term must be evaluated from 5.01 (1) where 0 is put equal to zero, giving Q = 4.71-bc\
y2
z2 --3
b2
c2
(1)
The capacitance of a circular disk is obtained by putting a = 0 and b = c in 5.02 (4), giving by Pc 114 or Dw 186.11 1-1 = 8 tan-lei (2) -1c/0 7e0 = C = 8ire[f 0-1(b2 0) 0 b Letting p2 = y2 -I- z2 and b = c in (1), the surface density on each side is Q
(3) p2)i The potential due to such a disk given by 5.02 (1) with a = 0 and b = c is 4irb(b2 -
)= 2V0 1b 2V 1 oi tan_ V =( 2 Oi Putting in the value of 0 obtained by letting r2 = x2 4. y2 + z2, and b = c in 5.01 (1) gives 2 Vo 52)2 4b2x2P1-1) V = — tan-1(2ibIr2 - 52 + [(r2 -
a=
(4)
This problem can also be solved by oblate spheroidal harmonics (5.271). 5.04. Method of Images. Conducting Planes.—An application of the test of 5.00 shows that in no case involving more than one point charge can we obtain the potential from the analogous two-dimensional case. Nevertheless, two of the methods used in such cases can also be applied to three-dimensional problems. One of these is the method of images. Any case in which the equation of a closed conducting surface under the influence of a point charge can be expressed in the form n
a .■ 1
§5.05
PLANE BOUNDARY BETWEEN DIELECTRICS
125
where r is the distance from q to any point P on the surface and r, is the distance from some point s on the other side of the surface to P can be solved by the method of images. We shall consider only simple spherical and plane surfaces. It is evident from symmetry that a single infinite conducting plane or two such planes intersecting in the z-axis at an angle ir/m under the influence of a single point charge q in the xyplane can be solved by locating point images in the xy-plane at the same places as in the two-dimensional case shown in Fig. 4.06. Adding up the potentials due to the point charge q and its point images gives exactly the same potential V in the region between the intersecting planes as would be obtained by adding up the potentials due to the charge q and the equal and opposite induced charge distributed over the planes. Hence we can determine this induced surface density o on the conductor by finding —69 V/On on the plane to be replaced by the conductor. In the case of a point charge q at a distance a from an earthed conducting plane from 1.14, and 1.16 (1), the induced density at P is
0- —
—aq 27rr3
(1)
where r is the distance from q to P. From 2.17, we know that whatever the actual distribution of charges in the space separated from q by the earthed plane this cannot affect c on the side facing q. 5.05. Plane Boundary between Dielectrics.—Since the case of a uniform line charge parallel to the plane face bounding two dielectrics may be considered as built up of equal point charges uniformly spaced along the line and since the images can be built up in the same way, it is plausible to suppose that the same image law holds in both cases. Let the relative capacitivities of regions of positive and negative z be K1 and K2, respectively. Consider any configuration of charges in dielectric K1whose potential, with no dielectric present, would be given by
Vva, = f(x, y, z) so that, if only K1were present, it would be, from 1.06 (3), 1 y, z) V1 = — K1f(x, The potential of an image in the z = 0 plane would be = Ci f(x, y, —z) The law of images of 4.04 suggests that when K2 is present the fields in K1and K2 should have the forms V1 = rif(x, y, z)
Ci f(x, y, —z)
(1)
126
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.06
and (2)
V2 = C2f(x, y, z)
When z = 0, V1 = V2 and KiaVdaz = K2OV2/az from 1.17 (5) so that 1 + K1C1 = K1C2
and
1 - K1C1 = K2C2
solving and substituting in (1) and (2), we have 2 f(x y, -z)] VI = —[f(x y Ki "Ki ± K2 "
(3)
2 V2 - Ki K2f(x, y, z)
(4)
and
Referring again to 1.06 (3), we see that if we have an actual charge q at P1in K1, then the field in K1is the same as if the whole region were occupied by K1and we had an additional charge q' at the image point P2, and the field in K2 is that which would exist if the whole region were occupied by dielectric K. and there were only a charge q" at P1, where q,
K _ K1 K1+ K2q 2
r K 2K„ 1 K2q
and
(5)
For calculating the field the choice of K. is immaterial but all authors take K1, 1, or K2. We use K1. 5.06. Image in Spherical Conductor. We saw in 4.05 that the cylinder p = a is an equipotential under the influence of a line charge q at p = b and a second line charge -q at p = -I-a2/b. Likewise we shall now see that it is possible to have the sphere r = a at zero potential with a point charge q at r = b and a point FIG. 5.06. charge q' at r = a2/b, and we shall determine q', which does not, as in the two-dimensional case, equal -q. The potential V' at r = a due to q' is, from Fig. 5.06, a2 471-E V' = q,(a4b-2 2a3b--1cos 0)-1 = ba-lq'(a2 +52 - 2ab cos 0)-i = ha-T-1g' —
The potential at r = a due to q is q/(471-€R). In order that the sphere be at zero potential, these must be equal and opposite, giving q,
= _aq
The potential at any point is then b2 - 2br cos 60-i - a(b2r 2 V = (41-€)-icr2
(1) a4
2a2br cos 0)--i] (2)
The surface density on the sphere will be
av\ CT
=
ar
2 b2 qra-b cos 0 b(b - a cos 0)1 aR3 a 47raR3 q (3) R3 -
— 44
;5.07
EXAMPLE OF IMAGES OF POINT CHARGE
127
We can evidently raise the potential of the sphere to any desired value V by adding the potential due to a point charge q at the center where q = 47reaV
(4)
If we wish the field between a point charge q and a conducting sphere having a total charge Q, we need only add to (2) the potential due to a charge Q aq/b at the center. We find that there is not, as in the analogous two-dimensional case, a simple image solution for the dielectric sphere and the point charge. We shall have to use spherical harmonics to solve this problem.
5.07. Example of Images of Point Charge.—As an example illustrating ev, the last three articles, let us locate the -q. aq -ce images in the case where the earthed b conductor consists of a plane sheet, / // /:/ /:,,AV:// o /f /:/•-erbq lying in the yz-plane with a spherical / / / boss of radius a centered at the origin, q r•//, // // and the region below the xz-plane is filled _ // // with a material of relative capacitivity ',/ ' //,/ K. The point charge q lies at xo, yo, zo / / , = V. To locate the and 4 + FIG 5.07. required images, we complete the boundaries with dotted lines as shown in Fig. 5.07. For the field above the xz-plane, the potential can be found by superimposing that due to the original charge plus seven images located as follows: —
q at xo, yo, zo q at —xo, yo, zo
aq —
a)2 (
a)2 at ( xo,
—
(a)2 yo, 7) zo
b
aq
75-; at —
(a)2 (ay (ay xo, yo, zo
q' at xo, —yo, zo q' at — xo, — yo, zo
— aqi at (Ix 412 ( a)2 b b °' b Y°' -6 z° aq' a)2 (a)2 --b— at 4-6 xo, — yo, zo
where, from 5.05 (5),
, 1 —K q — 1 + Kg Below the xz-plane, in the dielectric, the potential can be obtained using only the images above the xz-plane and substituting q" for q where, from 5.05 (5), _ 2q q1 + K
A section of the conductor by the xy-plane together with a projection of the images upon this plane is shown in Fig. 5.07.
128
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.08
5.08. Infinite Set of Images. Two Spheres.—Frequently, it is impossible to get equipotentials of the desired shape by any arrangement of a finite number of point charges. In some cases, however, one surface can be made an equipotential by some interior configuration of point charges. Then, by a suitable arrangement of image charges, the other surfaces may be made equipotentials, at the same time, however, distorting the first surface. By a third set of images, the latter is restored to its original shape at the expense of the other surfaces, and so forth. If, because they become more remote or grow smaller or tend to cancel each other, the effect of each successive group of images diminishes, we may get as close an approximation to the exact solution as we wish by taking enough of them.
V=0 (a)
(6)
FIG. 5.08.
This method can be used to find the self- and mutual capacitances for two spheres. Let the radii of spheres 1 and 2 be a and b, and let the distances between their centers be c. From 2.16, the coefficient cii is the charge on 1 and c12 that on 2, when 2 is grounded and 1 is raised to unit potential. By interchanging a and b in c11, we can obtain c22. The two cases, when c < (b — a) and when c > (b + a), are shown in Figs. 5.08a and 5.08b, respectively. In these figures, m = a/c and n = b/c. First, we put sphere 1 at unit potential by placing a charge q = 4rea at its center 0'. We then put 2 at zero potential by placing an image q' = —217rEab/c = —4irEna [see 5.06 (1)] at a distance b 2/c = nb to the left of 0. Next, we restore 1 to unit potential with the image -T aq' +471-Emna = (c — nb) — (1 — n2 ) at a distance a2/(c — nb) = ma/(1 — n2) to the right of 0' and then q
restore 2 to zero potential by placing an image
r, =
— bq" c — ma/(1 — n2
— )
-T 4remn2a 1 — m2 — n2)
(
▪•▪ §5.081
DIFFERENCE EQUATIONS. TWO SPHERES
129
at the proper distance from 0, and so forth. We note that the size of each successive image decreases so that we are approaching the exact solution. Adding up the charges on 1 gives m 2n2 n cii = 4rea 1+ — m2± •I • • (1) —1m — n2 (1 — n2)2 where the upper sign refers to case b. in case b, c12 = 4re[—na —
Adding up the charge on 2 gives, mn2a
(2)
1 — n2 — m2
In case a, from 2.17, c12 = —cii. From symmetry, we can write down, in case b, mn m 2n 2 c22 = 47rEb[l (3) 1 — m2 (1 — m2)2 — n2 In case a, c22 is of no importance. From 2.20 (6), the force of attraction between the two spheres will be, since V2 = 0, acli F V2 == 2 ac
47rEa V2 mn {(1_n2)2
m2n2[2(1 —n2) — m2] [(1 — n2)2 — m92
••
(4)
If we wish the potential at any point P, we must add up the potentials at P due to each of the image charges. If the spheres are at potentials V1 and V2 where V2 0, we shall have, in case b, a second set of images, for we must now start with a charge qi = 4reaV1at the center of 1 and q2 = 4n-Eb V2 at the center of 2. The sizes and positions of the second set of images can be obtained by interchanging 0 and 0' and a and b in the values for the first set. 5.081. Difference Equations. Two Spheres.—The formulas of the last article are not very convenient for precise calculation unless m and n are small. Better formulas, involving hyperbolic functions, can be obtained by getting the general relation between successive images and then solving the resulting difference equation. Let us designate the nth image in 1 by qn so that the initial charge at its center is q1= 471-Ea and the image of qn in 2 by pn. In Fig. 5.08, the distance from 0' to qn is sn and that from 0 to pn is rm. Then, using lower signs for case a and upper for case b, we have b2 — bqm — _ (1) pn = —
c + Sn + sn + abqn apn _ c(c sn) — b2 c — rm +a2(c T s,) a qm+i, _ sn) se.) b qn sn) — b2 c(c
c
q.+1 Sn+1 =
c
+ a2 rm —
(2)
-
(3)
130
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.081
Eliminating c snfrom (2) and (3), we have qn s7L qn = -d2 q.+1 qn+i
so that
qn-i qn
= C qn-1 a22 qn
_ b
a n
(4)
-
Eliminating s. from (2) and (4) rearranging give b2 1 1 = ,2 _ a2 1 (5) ab qn qn+ qn-i This is known as a difference equation of the second order with constant coefficients. The general method of solution is to substitute 1/qn = divide through by u"--1, and solve the resultant quadratic equation for u algebraically. If the two solutions are u1 and u2, the solution of the difference equation is
1
q.
(6)
= At? ± Bu4
where A and B are to be determined by the terminal conditions. The present equation is specially simple because the coefficients of 1/q.+1 and 1/qn_i are the same. Comparing with Dw 651.03 and 651.04 or Pc 669 and 671, which give sinh (n cosh (n
1)0 sinh (n - 1)0 = 2 cosh 0 sinh nO 1)0 + cosh (n - 1)0 = 2 cosh 0 cosh nO
we see that a solution of (5) is 1 - = A cosh na qn
if we choose cosh « = ±
c2
B sinh na a2
(7)
b2
(8) 2ab To evaluate A and B, let us write down for the first two images 1 1 = _= A cosh a± B sinh a 421-Ea qi c2 _ b2 1 ± + - (2 cosh a + A cosh 2a + B sinh 2a 3 421-Ea _ = 1 = ab 4rea q2 Multiply the first of these equations by 2 cosh a so that it can be written in terms of cosh 2« and sinh 2«, and solve for A and B, and we have _ 1 b + a cosh a B - . (9) 4reab sinh a A = -1-zlreb Substituting in (7) and using Dw 651.01 or Pc 660, we have 1 b sinh na ± a sinh (n - 1)a = (10) 42reab sinh a qn
§5.082
SPHERE AND PLANE AND TWO EQUAL SPHERES
131
Adding the charges on 1 then gives 02
c11 = 4reab Binh a E [(b sinh na ± a sinh (n — 1)a]-1
(11)
n=1
where the lower sign refers to case a (Fig. 5.08a) and the upper to case b (Fig. 5.08b). In case a, from 2.17, c12 = —ca. In case b, to get c12, we must determine p„. Taking the upper signs and eliminating c — from (1) and (2), we have 1 a 1 b 1 — c c Substituting for 1/q.+1and 1/q„ from (10), using (8) and Dw 651.06, we have 1 = c sinh na (12) 47reab sinh a p. so that adding up the charges on 2 gives 4reab sinh
csch na
C12 =
(13)
n=1
We can write down c22 from (11) for case b, by symmetry, thus: c22 = 4n-eab sinh a E [a sinh na b sinh (n — 1)a]-1
(14)
n=1
5.082. Sphere and Plane and Two Equal Spheres.—A special case of some interest is that of a plane and a sphere. We can get this case by letting d c = b —> co in case a or c — d = b —> co in case b, where d is the distance from the plane to the center of sphere 1. In either case, n —> 1, m —> 0, and m/11 — n1 = a/d so that we have OD
c11 = 41-ea( ±
a2 2d
4C12
—
a2
•
csch na (1)
= 4irea sinh n=1
where d = a cosh a and a is negligible compared with b. In either case, In the case of the sphere and the plane, the force is C12 = C11. 8ad ? acii = —27rea2 1 a 2)2 v.{2d2 + (4d2 F = 2 Od
= 2n-eVfE[csch na(coth a — n coth na)]
}
(2)
n-1
In the case of two equal spheres of radius a at a distance c apart, the formulas of 5.081 are somewhat simplified by writing in terms of
132
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.10
= ia where, setting a = b, we have, by Dw 652.6 or Pc 667, cosh g =
c
2
+ cosh a)] = 2—a
(3)
In terms of 0, 5.081 (11) becomes, setting a = b,
c11 = c22 = 41-ea sinh
csch (2n — 1)I3
(4)
n= 1
and 5.081 (13) becomes c12 = — 4rea sinh OE csch 2n0
(5)
n =1
5.09. Inversion in Three Dimensions. Geometrical Properties.—We have already seen in 4.21 that if, in a plane, we draw a circle of radius K about the origin and then draw a radial line outward from its center, any two points on this line, distant r and r' from the origin, are said to be inverse points if rr' = K 2 (1) In 4.21, the circle represented a section of a cylinder, but it can just as well represent a principal section of a sphere. Thus for every point of any surface in space there corresponds an inverse point, and the surface formed by these inverse points is said to be the inverse surface. If the equation, in spherical polar coordinates, of the original surface was f(r, 0, 0), then the equation of the inverse surface is f(K 2/r, 0, 0). We proved in 4.21 that straight lines invert into circles lying in the same plane and passing through the center of inversion, and vice versa, whereas circles not passing through the origin invert into circles. Hence, in the present case, planes invert into spheres passing through the center of inversion and spheres not passing through the center of inversion invert into spheres. Since, in 4.21, we arrived at the laws of inversion by a conformal transformation, we know that angles of intersection are not altered by inversion. It is evident from this that when a small cone of solid angle dSt with its vertex at the origin cuts out area elements dS and dS' from the surface S and its inverse surface S', the angle 0 between the axis of the cone and the area elements is the same, so that
dS _ r2dO cos 0 _ r2 (2) r" dSt cos 0 — r' 2 dS' 5.10. Inverse of Potential and Image Systems.—We shall now show that it is possible to formulate laws for the inversion of electrical quantities such that we can obtain from the solution of a problem with a certain boundary surface the solution of a second problem with the inverse boundary surface.
§5.101
EXAMPLE OF INVERSION OF IMAGES
133
Consider Fig. 5.10 in which P', R', and Q' are the inverse points of P, R, and Q, respectively, and 0 is the center of inversion. Then a charge q at P gives a potential V= q/(4rePQ) at Q, and a charge q' at P' will produce a potential V' = q' /(47EP'Q') at Q'. Triangles OQ'P' and OPQ are similar since K 2= OP • OP' = OQ • OQ', and 0 the included angle of both is a. The necessary relation between the potential V at Q before inversion to the potential V' at Q' after inversion is then
V' q' PQ q' OP V q P'Q' q OQ' In order to make this relation useful, we must find a suitable law for the inversion of charges. It was shown in 5.06 that a sphere of radius K is at zero potential under the influence of a charge q at r = b and a charge jq'I = --FKIql/b at r' = K 2/b which is the inverse point. If the sphere is to remain at zero potential after inversion about itself, which will interchange the charges, the law of inversion of charges must be
q' _ K K _ _ OP' q b OP K
(1)
We give the ratio the positive sign because we wish the inversion of charge to leave its sign unchanged. Substituting this value in the equation for the inversion of potential, we have the relation
V' K OP KOQ V OP .OQ' OQ' = K
(2)
This shows that V' = 0 if V = 0 provided OQ is finite, which means that if any surface is at zero potential under the influence of point charges a, a9 • • • , qn at Pi, P2 • • • , P,, which are at finite distances, not zero, from the center of inversion, then the inverse surface will be at zero potential under the influence of the inverse charges qc, q;, . . . , at P'„ P2, . . . , P. The qualification is necessary since (1) does not provide for inverting charges when OP is zero or infinite. It also follows from this rule that if any problem can be solved by images, the inverse problem can also be solved by images. 5.101. Example of Inversion of Images.—Let us now compute the force acting on a point charge q placed in the plane of symmetry at P at a distance b from the point of contact of two earthed spheres of radius a. Clearly, the two spheres can be obtained by inverting the system of planes shown in Fig. 5.101a. Taking the circle of inversion, shown dotted, tangent to the planes simplifies the computation. Before inver-
134
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.102
sion, the potential at P' due to the images alone is co 1 V'= = — 27re
(
q' 4na
— 1. )n—=—
ql ln 2
8rea
n = 1
' of The field of the images has a neutral point at P' so that the value V, V' inside a circle of radius S' around P' is constant if terms in S'n are neglected when n > 1. Thus inversion by 5.10 (2) gives the potentials at
+4a > .< 4a - 4 4a-►
#q .1
-
Q.
f#.7'
0 ..cil
6'
/ -1-'• •
9
Inverse System (a) Fia. 5.101.
P and at a distance 8 above P of the images that lie in the spheres on a circle through P and the point of contact to be, respectively,
2a
Vb = -6- V p,
V6+5
+6 2a
-
b
, 2ao , 172, ',z--,Vb — -p--V p
From 5.10 (1), q' inverts into 2aq/b so that the force on q is
F = qE =
lim 5—. 0
a) _ _ aq2
q(Vb
+
a
b
aireb 3
ln 2
(1)
5.102. Inversion of Charged Conducting Surface.—Let us now consider a conducting surface S charged to potential 1/K, which produces a surface density a, and let Q be any point on this surface. From 5.10 (2), the potential at the corresponding point Q' on the inverse surface AS' is V'Q= 1/0Q' . Since the potential at Q' due to a negative charge 4re at 0 is —1/0Q', it is clear that we can make the potential over the inverse surface everywhere zero by superimposing the potential due to such a charge. Reversing the procedure, we have the useful rule that, if we can solve the problem of a conductor at potential zero under the influence of a positive charge 4re, then we obtain by inversion, with this charge as a center, the solution of the problem of the inverse conducting surface raised to a potential —K—'.
-
THREE-DIMENSIONAL HARMONICS
§5.11
135
For the inversion of the surface density at P, we have, from 5.09 (1) and (2) and 5.10 (1), the relation
• = cr
dS K . (OP) 2 _ (OP) 3 q OP (OP')2 K3
q'
K3
—
dS'
(1)
(OP') 3
5.103. Capacitance by Inversion.—As an application of the rule of 5.102, we shall compute the capacitance of a conductor formed by two spheres of radii a and b which intersect orthogonally. Clearly, the inverse system is that shown in Fig. 5.103a, in which two earthed planes,
/1"qf I I a - /
/
%\
\ \
1/ I
-q'.
11 1
/
/
/ \
\
/
-q .
\
/
//
/
I
I
I
I
\ \
/
/
1
// /
/
.471
\
/
\
\
\
\
ab .
/
/
/ /
Inverse System la)
(61 FIG. 5.103.
intersecting at right angles, are under the influence of the point charge q' = 47re. This field can be replaced by that due to the image charges shown. The desired conductor, obtained by inversion, will be at potential V = —1/K = —1/2a. By 1.10 (1), the charge on this surface equals the sum of the image charges, qi, q2, and q3, giving, by 5.10 (1), = qi q2 q3= 49re i —a + ab L 2a 2a(a2 b2 )1 and the capacitance is Q
C
r
= Q =, (a ± b)(a2 b 2)i — ab] ± V "L e
b
(1)
The potential around the conductor can be computed directly from the image charges ql q2, q3or by inversion of the potential due to image charges alone in the region in the angle of the planes. 5.11. Three-dimensional Harmonics.—We saw in 5.00 that it is not, in general, possible, by rotating the orthogonal set of curves representing a normal section of a two-dimensional field, to obtain a set of three-dimensional equipotential surfaces. It is, however, possible to generate ,
136
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.11
in this way a set of surfaces which, together with the planes intersecting on the axis and defined by the longitude angle, form a complete set of orthogonal curvilinear coordinates which can be treated by the method of 3.03. When the axial section of our three-dimensional boundary problem gives a two-dimensional boundary solvable by a conformal transformation, this provides a method of finding a coordinate system in which the original boundary conditions are very simple. The problem then becomes that of obtaining a general solution of Laplace's equation in this coordinate system. Let us suppose that u1 = fi(x, y) and u2 = f2(x, y) are conjugate functions in the z-plane, given by z = x ±k
= f(ul + ju2) = f(u)
(1)
Then, by 4.11 (1), we have
, dz du= f (u)
or
dx ± j dy = I (u)(dui± j due)
Multiply this by the conjugate complex, and we get
dz 12 (dui + dul) du
ds1 = dx 2 + dy2 = 1
If this system is rotated about the y-axis, we have, for the element of length,
ds2= ds? + x2(4)2 =
dz 2 du (dU!
+ dui) + x2(dcb) 2
(2)
Comparing with 3.03 (1), where dcl, = dui, we see that h1 = h2 =
dzi
dui
_ 3 = .._
and
h
It
so that, from 3.03 (4), Laplace's equation is dz1-21- a i an + ±a( all + a2 V 0 (3) x Idu Laui\ xauif ' au2 xau2 j ' aq52 = We can immediately split off the last term by the method of 4.01 by assuming a solution of the form (4) V = U(uI, u2)c13(95) Dividing through by V and setting the last term of (3) equal to -m2, we have for cf the form, as in 4.01 (6), cI,= Aiefro + Bie-fro = A cos mcp + B sin mcp
(5) The partial differential equation to be solved for U(ui, u2) then becomes
1— a (au) 1 a (al — -I- — x— x aui x814 x au2 au2
- m2 dz 2
x du U = 0
(6)
137
THREE-DIMENSIONAL HARMONICS
§5.11
The difficulty of solving this equation depends on the form of x(u i, u2) and of Jdz/duJ. In many important coordinate systems, including all those treated in this chapter, x has the form x
= g1(u1)g2(u2)
(7)
and, when 4.11 (1) is used,
dz 2 du
ax _ .ax 2
aui
"au2
( ax
)2
au'
aX )2
au2
= 912g2
gIg;2
(8)
Assuming a solution of the harmonic form U(ui, u2) = Ui(ui) U2(u2)
(9) and substituting (7), (8), and (9) in (6), we find that, when divided through by U1(u1) U2(u2), each term contains only one variable so that, putting terms containing u1equal to a constant +(s m2c) and those containing u2 equal to — (s m2c), we obtain the total differential equations 1 d drli) M2[ (12 s=0 (10) gl dui g1 + c ,1 2 1 d (dU2) C] s=0 g2 U2 dug g2 dU2 m2[ g2) ]
(
The remainder of this chapter will be concerned with the solving of these equations and with applying these solutions and (5) to potential problems. We obtain spherical polar coordinates from 4.12 (2) by interchanging x and y so that
gi = eui = r and g2 = sin u2 = sin B (12) If we put c = —1 and s = n(n 1) in (10) and (11) and write in terms of r and 0, these equations become 5.12 (2.1) and 5.14 (2). We get oblate spheroidal harmonics from 4.22 (3) giving and g2 = cosh u2 = (1 + '2)1 (13) gi = a1sin u1= ci(1 — 2)1 If we put c = +1 and s = —n(n + 1) in (10) and (11) and write in terms of t and 1 these equations become exactly 5.271 (2). We obtain prolate spheroidal harmonics from 4.22 (3) with x and y interchanged, giving ,
and g2 = sinh u2 = (n2 — 1)i (14) 91 = a1cos ui = c2(1 — t2)i If we put c = +1 and s = —n(n + 1) in (10) and (11) and write in terms of t and n, we get the standard form of Laplace's equation in prolate spheroidal coordinates referred to in 5.28. We obtain cylindrical coordinates by rotating the transformation z = u so that we have gi = ui = p = k-1v and g2 = 1. If, in (10), we put c = 0, s = —k2, m = n and divide through by k 2, we obtain
138
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.111
Bessel's equation [5.291 (3)]. The same substitution in (11) gives 5.291 (2). 5.111. Surface of Revolution and Orthogonal Wedge.—In the particular case where m 0 and Eq. 5.11 (7) is satisfied, we can obtain the solutions of 5.11 (10) and 5.11 (11) in a simple form. Let the conjugate functions to be rotated be w1 = fi(x, y) and w2 = f2(x, y). Then let us choose new orthogonal coordinates u1and u2, such that (1) = F1(wi) gi(ui) = (A le", (2) g2(u2) = (A2 cos U2 + B2 sin u2) = F2(w2) We have from 5.11 (7), by hypothesis, using p for the radius of rotation of any point, p = gi(ui)g2(u2) = Fl(w1)F2(w2) (3) If we set me = 1 and s = 0 in 5.11 (10) and 5.11 (11), we can easily verify by substitution that solutions of these equations are U1= C win
(4)
and U2 = C2g7
(5) The differential equations are of the second order so that there must be a second solution of each. Since the method of finding this from the known solution will be useful in other cases besides the present one, we shall work out the case with general coefficients here. Suppose that y = v is a particular solution of 2y d — Maxy = 0 (6) dx dx 2 where M and N are function of x. Substitute y = vz in this equation, and write z' for dz/dx, and we have, after eliminating v by (6), the result
dz' v— dx
dx,
ivi
, v)z = 0
Multiply by dx, divide by vz', and integrate ln z' ln v2 f M dx = C or
dz = By-2e–f mdx dx Integrating gives
z= A+ Bfv-26–f M dx dx so that
y = v(A B fv-2e– f " " dx) (7) In the present case, putting 5.11 (10) and 5.11 (11) in the form of (6) by carrying out the first differentiation and multiplying through by
SPHERICAL HARMONICS
§5.12
139
U1,2, we see that M1,2, has the form 2
M1,2 — ,
_ d In
gi,2
g1,2
dui,2
Using for v the solutions of (4) and (5), we obtain the second solutions from (7), setting A = 0, to be dui U1 = D19i 2.+1
(8)
U2 = D2V3 f 92du_!, L 2m,
(9)
and
Thus a solution of Laplace's equation is given by
/ du1 du2 V = gi l + Di f gi u1 g'2" C2 ± D2 grn+i cos (mch -I- 1) (10) This form of solution is particularly useful when the surface of a charged conductor is formed by a figure of rotation and a wedge with its edge on the axis of rotational symmetry. A simple example will illustrate an application of this solution. A charged infinite conducting wedge of exterior angle a has a spherical conducting boss of radius a intersecting both faces orthogonally. Here, taking the center of the sphere as the origin and the faces of the wedge to be 95 la, = we + have p = r sin 0, gi = = r, g2 = sin u2 = sin O. All the boundary conditions are satisfied by taking m=
, DI = (1 —
1+ la a ,
C2 = 1, D2 = 0, and = 0
Thus we obtain
V = Cirfl —
ay +1 74, (sin 0)a cos a
(11)
This gives V = 0 on the surface of the wedge and sphere and agrees with 4.19 when we are far from the origin. 5.12. Spherical Harmonics.—When the boundary conditions of a potential problem are simply expressed in spherical polar coordinates, it is useful to have a general solution of Laplace's equation in this system. To get this solution, we proceed in exactly the same way as in 4.01. In terms of the distance from the origin r, the colatitude angle 0 measured from the z-axis, and the longitude angle measured about the z-axis from the zx-plane, Laplace's equation, from 3.05 (1), takes the form
a av ar ar
1 av 12 a av sin 0 y.) =0 sin 0 ae (. u sin2e a o2
We wish to find solutions of the form V = ROI) = RS
(1) (2)
140
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.13
where R is a function of r only, 0 a function of 0 only, and 4,a function of 4) only. The function S = 04,is called a surface harmonic, and the function 0, when 4 is a constant, is called a zonal surface harmonic. Substituting RS for V in (1) and dividing through by RS, we have 1 d ( 2dR\ _L 1 a ( 1 agS sin 0°S + ae S sin e ae s sing 0 a¢2 - 0 dr\ r dr The first term is a function of r only, and the other ones involve only the angles. For all values of the coordinates, therefore, the equation can be satisfied only if 1 as 2 — -K 1 sin °as) ± a ( 00 S sing 0 act.2 S sin o 00
1 d 2dR) K (2.1) R dr r dr ) The solution of the second equation is easily seen to be R = Arn Br-n-1 + (3) where K = n(n + 1). Substituting this value of K in the first equation and multiplying through by S give 1 0 sin1
a (sin as)
1 as 2 sing 0 002
n(n + 1)S = 0
(4)
Equation (2) therefore takes the form
Br-"--1)S„
V=
(5) It is evident that this is a solution of Laplace's equation only if the same value of n is used in both terms, hence the subscript on Sn. Any sum or integral with respect to n of terms like (5) is also a solution. In the special case where n = 0, (4) becomes
(. aso ae
sm n — sin 0— si
e ae
)
a,s0
2 =0
(6)
In 6.20 it is shown that either U or V satisfies this equation, provided j sin 43) tan 10] U jV = F[(cos (7) Thus every conjugate function of the last chapter yields two solutions of Laplace's equation in three dimensions by substitution of cos yt, tan 1-0 for x, and sin ct, tan 10 for y, and multiplication by A + Br-1. Especially useful solutions obtained by choosing W = zI'n and W = In z are, respectively, V = (A + Br-1)(C cot'n 1-0 D tan'n 1-0) cos (mci, + S.) (8) V = (A + Br-1)(C In tan i0 + DO) (9) 5.13. General Property of Surface Harmonics.—Before obtaining solutions of (4), we may derive, by Green's theorem, a useful property of
15.131
POTENTIAL OF HARMONIC CHARGE DISTRIBUTION
141
S,,. Let us again write down 3.06 (4), giving 9*) dS (PV2c13 — (1)V2T) dv = fs(1a(13 );-z— 11-
(1)
Let us take NI, = r'n8. and = rnS„ so that V21. = V2NY = 0 and the volume integral vanishes. Taking the surface to be a unit sphere and writing dig for the element of solid angle, we have
a* an
a (rmS,n) = mrm—iS„, = mS„, ar
and similarly a1/an = n8n. Substituting in (1) gives
fo (n8„8. — mS„S„,) d 1 = (n — m) feS„S„, dig = 0 so that if n m we have the result foS„8„, d it = 0
(2)
5.131. Potential of Harmonic Charge Distribution.—Let us suppose that, on the surface of a sphere, we have an electric charge density, cr,,, which is finite, continuous, and of such a nature that we can choose a small area, AS, anywhere on it so that, over AS, an — an is negligible compared with the mean value, an, of an over AS. This charge gives rise to a potential, V0, outside the sphere and the potential, inside it. Applying Gauss's electric flux theorem [1.10 (1)] to a small box fitting closely this element of the shell, we have =
avi av)
(
ar
aT
aft
(1)
Let us consider that an is such that 1 a0+2 2n + 1 rn+' "
E Vo =
(2)
Then, since V° = V when r = a, we must have EVi —
1 2n + 1 an--1811
(3)
Substituting (2) and (3) in (1) gives an
(4)
= Sn
In studying 5„ we shall see that when 0 enters as FT (cos 0), it meets the conditions imposed on anat the beginning of this article. By using these equations and 1.06 (6) we can evaluate two useful integrals.
' dS
SAS
_ 4/r rnS„ 2n + 1 an-1'
sR: dS =
4ir an+2S. 2n + 1 rn+1
(5)
142
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.151
The angles in S, are the coordinates of the terminal point of R, or R0. Thus by superposition, the potential due to any surface distribution which can be expanded in the form (6)
cr = So ± Si + S2 + • • •
is given by
0113-1 ± a
= 120
/20
-a 5
±•••I
if r < a
Vo = Lig-So ±
if r > a (7) r 5 +•'• r r 3 5.14. Differential Equations for Surface Harmonics.—The variables 0 and 40, in the differential equation, 5.12 (4), for the surface harmonics S = 043, may be separated by the process already used. Substitute 043 for S in 5.12 (4), and divide through by 0,10/sin2 0 giving sin e 0d (. de) sin 0 —
dO
dO
1 d2c1) — 4)(1 2
n(n + 1) sin2 =0
For all values of 0 and 95, this equation can be satisfied only if sin 0 d ( 0 sin sin 0a -cr o
n(n + 1) sin2 0 = K1
-)
and
1 d2c13 dq52
= —K1
If we put K1 = m2, the solution of the second equation is easily seen to be
(13 = C cos m¢,
D sin ing5
(1)
except when m = 0, when it becomes c13 =
N
Putting K1 = m2 in the first equation and multiplying through by 0/sin2 0 give 1 d ( dO m2 [n(n + 1.) sin 0 ) 0=0 (2) Sin.2 0 sin 0 dO dO This is the differential equation for 0. 5.15. Surface Zonal Harmonics. Legendre's Equation.—Before considering a more general solution of 5.14 (2), let us solve the most important special case in which V is independent of 4, so that t is constant and, from 5.14 (1), m is zero. Equation 5.14 (2) then becomes, when A is written for cos 0, d
[
d0„ (1 — I.42) -1+ n(n 1)0„ = 0
(1)
This is Legendre's equation, and its solutions are surface zonal harmonics. 5.151. Series Solution of Legendre's Equation.—To obtain a solution of 5.15 (1) in series, let us assume the solution 0„ = /ariir (1)
SERIES SOLUTION OF LEGENDRE'S EQUATION
§5.151
143
Substituting this in 5.15 (1) gives 1{r(r — 1)argr-2+ [n(n + 1) — r(r + Martel = 0 To be satisfied for all values of A, the coefficient of each power of A must equal zero separately so that (r + 1)(r + 2)ar4.2 + [n(n + 1) — r(r + 1)1a,• = 0 (r + 1)(r + 2) (r + 1)(r + 2) ar = n(n + 1) — r(r + 1)a1.42 (n — r)(n + r + 1)(4+2 (2) To expand in increasing powers of we notice that if a, = 0 then ar-2 = ar_4 = • • • = 0 and from (2) a_1and a_2 are zero if ao and al are finite, so that all negative powers of A disappear. Hence, if we choose ao = 1 and use the even powers, we have the solution pn = 1
n(n + 1) 2n(n —
1)(n + 3)
µ4• 2)(n4 If we choose al= 1 and use the odd powers, we have gre
—
12
(n — 1)(n + 2) g 3!
(3)
(n — 1)(n — 3)(n + 2)(n + 4) 5 A — 5! (4)
A complete solution of 5.15 (1), if —1 < µ < + 1, is
On = A nP n Bnqn regardless of whether n is an integer or a fraction, real or complex, provided the series converges. Recurrence formulas for pn and qn may be obtained by subtracting pn±i from pa_i, giving pn-l
[(n + 1)(n + 2) — n(n — 1)µ2 2! (n + 1)(n + 4 — n(n — 3) (n — 1)(n + 2)µ4 3! (n — 1)(n + 2) 3 = (2n + 1)/2(./ 1.1 + • • • ) 3! = (2n + 1)Aqn
— p7H-i =
)
(5)
By a similar procedure, we obtain (n + 1)2qn+1 —
(6) = (2n + 1)APn Differentiating (4) and adding n dqn_i/c112 and (n + 1) dq.÷i/c111 give
(n — 2 + n + 3)n(n + 1) 2 p 2! (n — 4 + n + 5)n(n — 2)(n + 1)(n + 3)ih4 4! • • • 1 n(n + + n(n— 2)(n + 1)(n + 3 ---- (2n + 1)(1 2! 1)/42 4! (2n + 1)p.
nq',1+ (n + 1)q;,±1 =[2n + 1
) µ4
144
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.152
By a similar procedure, we obtain np4, = —n(n + 1)(2n + 1)qm
(n
(8)
5.152. Legendre Polynomials. Rodrigues's Formula.—If n is a positive even integer, the series in 5.151 (3) evidently terminates and has -En + 2) terms and may be written in
[( 71)!P pm = (— 1)22n n!
1)r—in
2n[-En
2r)! µ 2* (n — 2r)]! [En + 2r)] (2r)
r=0
In this case, we define the polynomials P,,(µ) to be (-1)1"n! 2'1(i) q2Pn
(1)
If n is a positive odd integer, the series in 5.151 (4) evidently terminates and has -(n + 1) terms and may be written —1)
qn =
1)1(n-1)2n-1{[En—
1)1!) 2
n! r =0
2r ± 1)!1,2,4-1 (n 2"[En — 2r — 1)] ! [En ± 2r ± 1)] !(2r In this case, we define the polynomial P,,,(µ) to be (_1)1(n—i)n t Pn(P) 2„-1{{En — 1)1! }2qn
1) !
(2)
Legendre's polynomial, given by P n(A) where n is a positive integer, in ascending powers ofµ by (1) and (2) may be written in reverse order by substituting s = in — r in (1) and s = En — 1) — r in (2), both giving the result m
(2n — 2s)! un-28 1)8 2n (s!) (n — s)!(n — 2s) l
P.(µ) =
(3) s=o where m = in or i-(n — 1), whichever is an integer. An expression for Pn(A), known as Rodrigues's formula, may be obtained by writing (3) in the form m
Pn(P)
n! (2n — 2s)! „2a 1 2n1(1)8s!(n — s)! (n — 2s)! '21 8=0
n
1 dn 2nn ! d pn
( 1)8
nI „ 2n-2e (n — s)!'”
§5.154
RECURRENCE FORMULAS FOR LEGENDRE POLYNOMIALS
145
The last summation is the binomial expansion of (p2— 1)n so that
1 cln (4) 2r —Tzrdr tn(A2 — 1)n Equations (3) and (4) are valid solutions of Legendre's equation [5.15 (1)] whatever the range of the variable A. In prolate spheroidal harmonics, we have 0 < µ < . For very large values of A, the highest power outweighs all others; so we have (2n)! Pn(11) A—. —÷ (5) co 2n(n!)21A Pn(P) =
5.153. Legendre Coefficients. Inverse Distance. The polynomials of 5.152 are also known as Legendre's coefficients, the reason being evident from the following considerations. The reciprocal of the distance between two points at distances a and b from the origin, where b > a, when the angle between a and b is 0 and A = cos 0, can be written: —
1 _ (a2 + b2 —
=
a' —b22abAy
R
1 l 1 a' — 2abiA 1 • 3(a2 — 2abily b[ 2 b2 -1- 2 4 b2 ••• a 3122 1(4 _ 1 i_ 5/13 — 3p(ay = 4 -42 \b/ 1;[ • • b 2 \b —
-
u
We see that the coefficient of (a /b)Th is exactly the expression for P, (µ) in 5.152 (3) so that we may write 2 (1) =— bl[Po(A) (bPI(P) (0 P20.0 + • • • — R 1
We shall refer to this expansion many times in solving problems. 5.154. Recurrence Formulas for Legendre Polynomials. If n is an odd integer, we may substitute for pn_i, pn_Fi, and qn in 5.151 (5) from 5.152 (1) and (2) and obtain, after dividing out the factor —
{[-En — 1)]!12 n!(-1)i(n-") the result nPn_i (n 1)P.+1 = (2n 1)AP.
(1)
An identical expression is obtained, when n is even, from 5.151 (6). We omit the argument of these polynomials when there is no ambiguity in doing so. If n is an even integer, we may substitute in 5.151 (7) for pn, en,_„ and q'n4.1from 5.152 (1) and (2), divide out the factor 2n(2n 1)[(n)!]2 (-1)inn!
146
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.155
and obtain = (2n + 1)Pn P41 An identical expression is obtained, when n is odd, from 5.151 (8). The integral of P.(.4) is given by integrating (2), the result being Pn(P) diA =
Pn+1 Pn-1 2n + 1
(2)
(3)
The derivative of Pn(u) is given, by adding successive equations of the type of (2), to be P'n (g) = (2n - 1)P,1 +(2n - 5)P,3
•••
(4)
Another useful expression for the derivative may be obtained by differ1by (2), giving entiating (1) and eliminating = /AP:,
(n
1)Pi,
=
or
nPn_i (4.1)
Eliminating Pn_1 and P,',±1between these equations and (2) and combining with (1), (2), or (3) give the following equivalent forms:
P'.
1)(I.IP„ P.±1) _ -n(tiPn - P.-1)
_ (n
1 - 112
1-µ2
i,n„ ,
n(n + 1) P„_1 n-1 — Pn+1 -n(n 1) 1 - /22 2n 1 1 —µ 2
UI)
(5)
5.155. Integral of Product of Legendre Polynomials.—In using Legendre polynomials, the integral of their product over the range 0 = 0 to r, orµ = -1 to +1, is important. We saw in 5.13 (2) that
j_ i r-n(-L)P.(A) dµ = 0 If m = n, we substitute for one
11 [Pn(a)]2
(1)
if m 0 n
P. from 5.152 (4)
1 f+t =— Pn(o —I cit-( 12 2fin!
dun
1)n dpi
Integrate the right side n times by parts, letting u be the first term and dv dr(;2 2 — 1)n the second each time. Since always contains the factor dAr ().,2 1)n—r, v will always be zero when the limits are inserted and so the product uv drops out and we have finally +1
f-1 [p..}2 dµ _
(-1- rid,.Go — 1)n
dA
)
2nn ! J
eV
But from 5.152 (3), dnP,,(p) _ (2n) !n !— (2n,) !
di‘n
2nn !n !
2nn !
(2)
TABLE OF LEGENDRE POLYNOMIALS
§5.157
so that [p.(/1)]2
_
22(n
(nn)1)2i
147
- p2)
(2n — 1)!!.1 (2n) !!
0 sin2n+i 8 dB
(2.1)
Integration by Pc 483 or Dw 854.1 gives
r+,
2
[Pn(A)]2 (3) 2n + 1 J --1 5.156. Expansion of Function in Legendre Polynomials.—Any function that can be expanded in Fourier's series in the interval —1 < < +1 can also be expanded in a series of Legendre polynomials in the same interval and by a similar method. Assume the expansion
= aoPo(A) + aiPi(u) + • • • + a,,P.(A)1)+ • • • Multiply by Pm(u), and integrate from µ = —1 to +1. By 5.13 (2), all terms vanish except the mth term; so we have, from 5.155 (3). am = 1(2m + 1)f+11f(p)Pm(A) du
(2)
Note that if AA) = 0 when —1 < µ < +1 then am = 0. This means that if we have an expansion in Legendre's polynomials equal to zero, the coefficient of each term must be separately equal to zero. As in the case of a Fourier's series, at a discontinuity this expansion gives half the sum of the values of Au) on each side. A formula for am which is frequently more convenient than (2) can be obtained by substituting Rodrigues's formula [5.152 (4)] in (2). Thus am
2m + l +1.. — 2) m = ( -1)' 2„,÷ Im ! f_1 (A)dinadin
Integrating this by parts repeatedly, always letting the first member be u and the second dv, we find that luv1+1 is zero and f +1u dv alternates in —1 sign until finally we are left with am
2m + I-SI-lc/1(1)(1 = 2"1+1M _
difn
A2)17' cipt
(3)
If the derivatives of f(g) are simple, this gives usually an easy integration. 5.157. Table of Legendre Polynomials. A table of values of P.(12) can be computed from 5.152 (3) or (4). The values of n < 9 are —
Po(A) = 1,
Pi(A) = it,
1 P3(m) = — 2(5/13— 3/2),
P2(ti) = EU' — 1) (35m4 _ 301.0 3) P4(A) 8
148
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Pb(A) =
(63/46— 70/.4 3
15/4
8
,
P6(Y) =
§5.158
(231/.46 — 315124 + 105/22 — 16
5)
(429/47 — 693125+ 3150 — 35/4) 16 6930/.44 — 1260442 + 35) (6435/.48 — 12,012/46 P801) — 128 P7(A)
Other useful values of P„(i.L) are (n odd)
Pn(0) = 0
(n even)
Pn(0) = (-1)inl
(any n) (any n) (any n) (any n)
Pn(1) = 1
3 5 • • (n — 1) 2.4.6 • • n
Pn( — = (-1)P.(14)
P'n (0) = —(n P'„(1) = in(n
1)Pn+1(0) 1)
[from 5.154 (5)] [from 5.15 (1)]
The values of Pn(A) for 0 n 7 are shown in Fig. 5.157. 5.158. Legendre Polynomial with Imaginary Variable.—We shall have occasion, in oblate spheroidal harmonics, to deal with Pn(j0 where j = (-1)1 and 0 < 00. Substituting for A in 5.152 (3), we have 13, (
=
m )in: A‘J s= 0
(2n — 2s)! 2" (s) !(n — 8)!(n — 2s)!'
2s
(1)
where m = ?tn or -En — 1), whichever is an integer. A similar substitution in 5.152 (5) gives P„(j0
(2n)!
2.(n!)2
(-1)in?-.
(2)
§5.17
CHARGED RING IN CONDUCTING SPHERE
149
5.16. Potential of Charged Ring.—If V is symmetrical about the x-axis and its value is known at all points A on this axis and if this value can be expressed by a finite or convergent infinite series involving only integral powers of x, then the potential at any point can be obtained by multiplying the nth term by P„(cos 0) and writing r for x. The result holds for the same range of values of r as the range of x in the original expansion. Let us apply this to a ring carrying a total charge Q (Fig. 5.16) giving 471-EV4 = Q(c2 +
x2
0
2cx cos a)-I
Fm. 5.16.
Expanding this by 5.153 (1) gives X C
VA
a)
= Lircc n =0
x< c
VA =
(C os
4irEc
a)
n =0
The potential at any point P at r, 0 is given by r > c, or e
a, r = c
VP
r
= 471-€c
n+1 P„(cos a)P„(cos 0)
n =0
r < c, or 0
a, r = c
Vp
P„(cos a)P„(cos 0)
4/rEc
(1)
n=0
Other examples of this method will appear at the end of this and subsequent chapters. 5.17. Charged Ring in Conducting Sphere.—If the value of the potential due to a given fixed charge distribution is known in a certain region, then the values of the potential when an earthed conducting spherical shell is placed there can be found. Expand the original potential in spherical harmonics, and superimpose a second potential, similarly expanded, due to the induced charge such that the sum is zero over the sphere. The latter should vanish at infinity if the original distribution is outside the sphere and be finite at the center if it is inside. As an example, let us find the potential at any point inside a spherical ionization chamber of radius b, if the collecting electrode is a thin concentric circular ring of radius a. Let us set a = fr and take r > a in the last problem, inserting the value for P.(0) from 5.157 and writing 2n
150
§5.18
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
for n, since only even powers remain, we have for the potential due to the ring alone 4/1-ea
'In
1 3 • 5 • • (2n. — 1)(aYn+i P2n(cos 0) \ r) 2 • 4 • 6 • • • 2n
n=o The potential of the induced charge must be finite at the origin; so it has the form = 4rea l
A r2nP (cos 0)
2n 2n
n= o
But Vi V,.= 0 when r = b, and from 5.156 (2) we may equate the coefficient of each Pn(cos 0) separately to zero, so that 1 3 5 • • • (2n — 1) 1 (aYn+1 2.4.6 • • 2n b 2n\bf
A
= — ( —1)n
and if a < r < b or r = a, 0 'yr, the potential becomes
v= Q 47rea
n= o
( _ 1),,(2n — 1)!! / a)2n+1 (2n) !!
a 2.+1 r 2n
-6
]P2.(cos 0) (1)
If r < a or r = a, 0 0 Fr, we have
= ___1( _ 1)„(2n
— (2n) !!
4rea
ryn
(ayn+102n —
\bj
]P 2n (cos
0)
(2)
n=o
Care should be taken in the application of this method to fields produced by a distribution of charge on conductors. In general, in an actual case, the field of the induced charges will cause the inducing charges to redistribute themselves and invalidate the result. 5.18. Dielectric Shell in Uniform Field.—We now compute the field inside a dielectric shell of internal and external radii a and b, placed in a uniform electrostatic field of strength E. As in the last problem, we may consider the potential outside as that due to the original field Er cos 0 plus that due to the polarization of the dielectric. The latter must vanish at infinity so that it can contain only reciprocal powers of r. Furthermore, the only surface harmonic appearing in our boundary condition at infinity is Pi(A) = cos 0 so that the potential outside must be of the form A = (Er + —2) cos 0 In the dielectric of relative capacitivity K, since r is neither zero nor infinite, we include both terms, giving
OFF-CENTER SPHERICAL CAPACITOR
§5.19
V2 =(Br
151
+ —1 cos 0 r2
In the cavity, the potential must be finite so our only choice is V3 = Dr cos 0 The four boundary conditions necessary to determine A, B, C, and D are 2' = b,
=K aviav2 r = a,
2A
Or
Or
Or
V2=
A C Eb -I- 172 = Bb ± p
or
VI = V2
V3
,av2 av3
.n.— =
UT
Or or Solving these equations, we obtain
E — — = KB — b3 C
or
2KC
(2)
0
Ba + — = Da a2
KB —
2KC a3
(1)
(3)
=D
(4)
9KE
D=
(5) 9K — 2(1 — K) 2[(a/b) 3— 1] Looking at the expression for V3, we see that this is the strength of the electric field inside the shell. 5.19. Off-center Spherical Capacitor.—As an example of boundary conditions involving surface harmonics, let us compute, approximately, the charge distribution on the inner shell of a slightly off-center spherical capacitor. The formulas derived by images in 5.08 require many terms to give accuracy in this case if the inner radius a is nearly equal to the outer radius b. Let us choose the origin at the center of the inner surface; then the approximate equation of the outer surface is
r=b
cP1(.1) (1) where c is the distance between centers and A = cos 0. This is obtained from 5.153 (1), the reciprocal of which may be written b = r[1 cr-1P1(1)]-1= r — cPi(A) when terms in c" are neglected if n > 1. Since here the boundary conditions involve both Po (A) and Pi(A) and since both r = 0 and r = 00 are excluded from the field, the potential must be of the form
B
V =A+—
(Cr
—NA) r2
where C and D are small correction terms of the order of c. ary conditions are r=a r=b
cP1(.1)
(2) The bound-
V=A+
(C a ± I a 3 B c P (Cb V2 = A +— T4)]
Li)P101)
152
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.21
where the products Cc and Dc have been neglected. From 5.156 (2), we may equate the coefficients of P0(u) and Pi(A) separately to zero, giving
A + — — Vi = 0 a A + — V2 = 0
Ca + - = 0 a2 Bc D - + Cb + — = 0 b2
Solving, we have B —
ab(V1— V2)
b—a
C=
'
abc(V1. — V2) (b — a)(b3— a3)'
a4bc(V i— V2) (b — a) (b3— a')
We may now obtain the surface density
aV
0. = —E(—) = ar
eab(Vi — V2) (1
b—a
3c cos 0) a2 b 3— a3 \
where terms in cn are neglected. We notice that if we integrate this over the surface of the sphere, the correction term drops out so that the capacitance is the same as if the spheres were concentric, if only terms in c are included. 5.20. Simple Conical Boundary.—We have already seen in 5.131 how the potential due to any charge distribution on the surface of a sphere can be expressed in spherical harmonics. It might also be pointed out that if the value of the potential V on the surface of any cone of revolution, whose equation is 0 = a, can be expressed in the form /3nr--n-1) V = /(Anrn where n is an integer, then, at any point inside the cone (Anrn + Bar—n—i)P„,(cos 0) V P,(cos a) —
(1)
for it can be seen by inspection that this solution satisfies Laplace's equation and the boundary conditions. 5.21. Zonal Harmonics of the Second Kind.—The second solution of Legendre's equation given by the infinite series in 5.151 (3) or (4) is called a zonal harmonic of the second kind and is denoted by Qn(A). We Alan define it by expressions similar to 5.152 (1) and (2) to be 22.-1. Qn(A) = (-1)1(..+1)1[ En — 1)M (n odd) (1) P. n! run)!122n Qn(A) = (-1)int " ' (2) (n even) n! qn These hold for —1 < µ < +1. Although at present we are dealing with solutions of Legendre's equation in which p = cos 0, we shall find later, in using spheroidal harmonics, that we need solutions in which p2 > 1. This requires the
LEGENDRE FUNCTIONS OF THE SECOND KIND
§5.212
153
expansion 5.151 (1) with negative values of r. We may write 5.151 (2) in the form (n — r)(n + r + 1) (r + 1)(r +2) ar ar+2 (3) We observe that if ar.+.2 = 0, ar+4 = a7+6 = • • = 0 and if a_,.+1 = 0, = ct_r+5 = • • --- 0. But (6+2 is zero if an is finite, and a_n+i is zero if a,1is finite. If we let an =2n!/[2n(n!) 2], we get 5.152 (3), a polynomial defining Pn (A), but if we take a_n_1 = 2n(n!) 2/(2n + 1)! we obtain, by writing r = —n — 3, r = — 5, etc., in (3), the coefficients in the series 2n(n!) 2[ 1 (n + 1)(n + 2) 1 (2 .(i) = (2n + 1) ! An-FI 2(2n + 3) An+3 03
2n(n!)2 (n + r)!(n + 2r)!(2n + 1)! (2n + 1)!LJ r!(n!)2(2n + 2r + 1)!
—n-2r--1
r=0
Writing s for r to agree with the notation of 5.152 (3), we have (in(A)
*1(n + s)!(n + 2s)! — 2n .ZJ s!(2n + 2s + 1)!A
(4)
a =0
This series evidently converges when A2 >1 and defines Qn(u) in this range. When fl is very large, the smallest negative power in (4) outweighs the remainder so that we have
(2.01)
2n(n!)2
(5)
(2n ± 1)
5.211. Recurrence Formulas for Legendre Functions of the Second Kind. The following formulas connecting different orders of Q, (µ) can be obtained from 5.151 (5) and (6) by substituting for pn and q>, the values of 5.21 (1) and (2) in exactly the same way as 5.154 (1), (2), (3), (4), and (5) were derived. —
nQn_i + (it + 1)Qn+1 = (2n + 1)4. Q'n+1— = (2n + 1)Qn
= Qn +i Qn(A) 2n + 1 J + • • • + p(1 — Q'„ = (2n — 1)Q,„.1+ (2n — 5)(20_8 (n + 1)(AQ. — Qn4-1) 1— u —
(1) (2) (3) (4) (5)
5.212. Legendre Functions of the Second Kind in Terms of Legendre Polynomials. A useful expression for Qn (A) may be obtained from Legendre's equation if we know that Pn(A) is a solution. In 5.111 (6) —
154
§5.212
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and (7), we saw that if v is one solution of the differential equation 2y dx 2
dy M— dx
(1)
Ny = 0
where M and N are functions of x, then a second solution is y = v(A
Bfv-2e— fm dx dx)
(2)
In Legendre's equation [5.15 (1)] v = P„(A) and M = —2A(1 — 1.12)-1 so that 5M diA = In (1 — 2) and e—I m dµ = (1 — /22)-1giving (2) the form d11 (1 — ) [P.(A)] 21 To determine the constants A and B, let n = 0 and n = 1, and integrate by Dw 140 or Pc 48 and by Dw 152.1 or Pc 57, and expand the logarithm by Dw 601.2 or Pc 769. This gives Q.(ii) =
P.(A){A
Bf
1 + — A + B( + 3 1 Qo(A) = A ± —B ln 1 -44 5 + 2 3 +-115 A — Qi(A) —
=
2
ha l + 1—
B = B(-1
142 ±
3
+ 1A2
• • )
5
From 5.21 (2) and (1), we see that Wiz) = qo and Qi(A) = — pi, and comparing the values of qo and pl. in 5.151 (4) and 5.151 (3) we see that B=1
A = 0 and so that the general expression for Q„(A) is
and in particular
dA Qn(A) =Pa(A).) (1— 42)[Pn(14)]2 Qo(A) = 1 In
(3)
1+A
(4)
and 1 1±A WA) = ? 1n 1 _ A
1
Applying 5.211 (1), we obtain 1 1 ± A 3A 1 —(3/42 — 1) In 1 A -2 (22(11) = 2 —AQi(A) — 2 —Qo(A) = 4 1 ± I.L 1 = -P2(.1) In A 2 1— By repeated application of 5.211 (1), we get 1 +A 2n — 1 in l_ /I 1 . n P.--101) Qn(A) = 1 2Pn(A)
(5)
3 2Pi.(
)
(6)
2n — 5 3(n _ i)Pn-3(ii) — • • • (7)
This holds for 42 < 1. The general solution of Legendre's equation is 0 = AP.(A) + BQn(A)
15.214
LEGENDRE FUNCTION OF THE SECOND KIND
155
If we write A' = A - iB In (-1) and substitute in (7), we have 0 = A'Pn(p) + BrP n(A) in 2
+ 2n -1 P„_1(/.1) - • • 1 1•
So that when j.c2 > 1, we may define Qn(i.t) by the equation 1 2n - 1p 2n - 5 Qn(A) = -12Pn(A) In A + — 1 1 • n n-1 \Aj — 3(n - 1)A
( — • •
'
(8)
Putting n = 0 and n = 1, expanding the logarithm by Dw 601.3 or Pc 770, and comparing the values of Qo(A) and (21(12) with 5.21 (4), we see that the formulas agree. 5.213. Special Values of Legendre Functions of the Second Kind. The actual values of Qn(A) may now be found easily from 5.212 (7) or (8), but it is frequently convenient to know certain important special values for real arguments as follows: (n even)
Qa(0) = 0
(n odd)
5' Qn(0) = (-1)4(n+1)2 41 63 :
(any n)
(1) n 1)
Qn( --A) = ( -1)n+I(24) Q„(1) = co Q„( co ) = 0
(2) (3) (4)
5.214. Legendre Function of the Second Kind with Imaginary Variable.—In connection with oblate spheroidal harmonics, we shall have to deal with Qn(ji-) where j = (-1)1and 0 < < 03. No difficulty is experienced when ?- > 1, for substituting jl forµ in 5.21 (4) we have Qn(i0 = (—P1+12'.2 s =o
(-1).(n s!(2n
2sn s)! 2s '+ 1)! '
(1)
If we make the same substitution in 5.212 (8), the result is ambiguous owing to the fact that the logarithm term is multiple valued. Using Pc 770 and 780 or Dw 601.2 and 506.2, we may write In
±1 1
1 1 .( 1 - 23 ± - — 50 + •
= -2j cot-1I-
If we expand the coefficient in 5.212 (8) by 5.158 (1) and substitute the above series for the logarithm, we find the result is identical with (1) so that a consistent value has been chosen and we may write Qn(j1') = —j cot-1-Pn ii-) (
. 2n 5 . 2n-1 n Pn-1(3° -3(n -1)Pn-3(3)
• • • (2)
where the arc cotangent ranges from 0 to a. We may use this expression to define Qn(k) over the whole range - < < co since it has no peculi-
156
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.22
arities at /' = 1. We note that Qn(j0 is finite when ;- is finite and that (n even) Qn(,/ 0) = — 1'1-P.( •
P in
—
2
1 • 3•5• • •(n—1) (3)
24• 6• • n )(n+1)2 4 • 6 • (n — 1) (n odd) (2„(j • 0) = (-1 1•3 5 • n )
(4)
When is very large, the smallest negative power in (1) outweighs the remainder so that we have 2n(n!)
(2n
1 ( j)"+1 1)!
(4.1)
An expression that is sometimes useful is obtained by differentiating 5.212 (3) and eliminating the integral from the resultant equation by 5.212 (3). Substituting for A gives
P.(ft)V,(.10 — PW0Q.Ci0 = (1+ ?'2)-1
(5)
5.215. Use of Legendre Function of the Second Kind in Potential Problems.—The most important applications of zonal harmonics of the second kind occur in connection with spheroidal harmonics. Owing to the fact that Qn(A) is infinite when 1.1 = 1, they are used as spherical harmonics chiefly in problems where conical boundaries exclude the axis from the region of the electrostatic field. Let us consider the case of two coaxial cones. Let the potential be zero over the cone B = )3, and let V„ = M(Anrn Bnr—n-1) over the cone 0 = a. Then, at any point in between,
v=
(A,,rn Bnr—' 1)[(2.(cos i3)Pn(A) — Pn(cos S)Qn(A)] Pn(cos a)Qn(cos 0) — Pn(cos g)Qn(cos a)
(1)
where /A = cos 0. It is evident by inspection that this solution satisfies the boundary conditions. A particular case of some interest is that in which one cone is at potential zero and the other at potential V1. We then have, since Po(A) = 1,
V1
A Qo(cos 13) — Qo(cos a)
ri()
Qa(cos t3) — Qo (cos a)
—
A0
Also Qo(cos 0) = 4 In [(1 + cos 0)/(1 — cos 0)1 = — In tan 40 from Dw 406.2 or Pc 578 so that the potential has the form tan 13)
V = V1(ln tan 13)/(ln tan tan 40
(2)
5.22. Nonintegral Zonal Harmonics.—For many problems involving conical boundaries, we find harmonics with integral values of n inadequate. In such cases, we must expand in terms of harmonics with values
§5.23
157
ASSOCIATED LEGENDRE FUNCTIONS
of n so chosen that the roots of P n(kt) or Qm (A) are zero on the cones in question. Many of the equations already used, such as the recurrence formulas for P„(/.4) and (2,,(A) are equally valid for nonintegral values of n but the definitions and derivations need modification to fit this case. A convenient expression for Pn(A) good for all values of n is the series — n) • • (r — 1— n)(1-
(n + 1)(n + 2) • • • (n + ( — (r!) 2
2
1.4Y (1)
r=0
This series converges for A = cos 0 except at 0 = a. An example of this type of harmonic is given in 5.26, 5.261, and 5.262, if we take the charge on the axis of the box so that m = 0. When v is not an integer, Pp(A) and 13,(— A) are independent solutions of Legendre's equation and are connected with Q,(µ) by the relation
(2,(A) =
ir[cos
virP,(A) 2 sin
— P,(-12)]
(2)
V7
When n is integral, the (2.(u) of 5.21 is the limit of Q,(1) as v n. 5.23. Associated Legendre Functions.—We have seen in 5.12 and 5.14 that a solution of Laplace's equation, in spherical coordinates, of the form Rect. is possible when 5.12 (3) 5.14 (1)
R = Arn + Br —n-1
=
C cos m(j) +
(1) (2)
D sin mq5
and 43 is a solution of 5.14 (2) which becomes, when /.1. is written for cos 0, ni2
di.4[
(1 — µ2) dµ
dA ]
+ [n(n + 1)
1—
12 2
0 = 0
(3)
To obtain a solution of this equation, we start with Legendre's equation, obtained by putting m = 0. Differentiating the product composing the first term, we may write Legendre's equation 2y dy (1 —µ2)d µ-2µ + n(n + 1)y = 0
d
We know solutions of this to be y = Pn(A) and y = Qn(A) this m times, and write v for dmy/di.en, and we obtain
Differentiate
dv (1 — A2)— d 2v — 2A(m + 1)- + (n — m)(n m 1)v = 0 41
(4)
Write w = (1 — /22)Pnv or v = (1 — 112)—onw, and this is (1 —
d2w
dw
ZAT., En(n + 1)
2
I"1.,2 1v ( =0
1
4.1)
This equation is identical with (3), when the first differentiation in (3) is
158
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.23
carried out, so that a solution of (3) is 'fly
e = w = (1 — A2)1mv = (1 —
Since y is a solution of Legendre's equation, a complete solution of (3) is
GO
0 = ATT(A) +
(5)
where if —1 < /./ < +1, PT (A) and QT (µ) are defined by
PZ(1.1) = (1— P2)1m
d'nP„(A)
Q;n
m
i(A) = (1—µ2)1m
(6)
.(12)
(7)
c/Aen
For —1 < µ < +1 Hobson introduces the factor (-1)m on the right. Whenµ is imaginary or real, but greater than unity, we define these functions by the equations PT (µ) = (.12 —
(27:(1 ) = (µ,2 —
(8)
d imµ
1)imdm(0z)
(9)
These functions are known as associated Legendre functions of the first and second kind. We can compute them from the expressions for P„ (µ) and Q.(A) already obtained by using (6) and (7). Forµ real but less than unity, (6) and (7) give /1(A) = 15(1 — /42)1 Pi(A) = (1— /42)i 1:10.0 = -(1— 1.42)1 (71.13— 31i) = 3(1— Pi(A) = -V(1— ih2)(7A2 — 1) ITO = 3 (1— 1) /1(A) = 105(1 — A TAL Pa(m) = 4(1— ii2)1(5A2 MA) = 15(1 — ;12) P,10.0 = 105(1 — p2) 2 Q1( 1) = (1 —
2)1(1In 1 +
2
1
in 1+ Q1 1) = (1 — 112)i01, 2 1— (
(A (A) = (1 —
A
+ 1
/.1 2 )
— + 3;12 1—
(10)
512 3A '3) 2)(3 2 in 1 + + (1 112)2
For higher values of m and n, use the recurrence formulas of 5.233. For Az real but greater than unity, the values given by (8) and (9) can be obtained by substituting in the above formulas for the first factor (I.12— 1)im instead of (1 — /.12)1m and by writing, in addition, in the logarithm term of Q',,n(A) the factor /./ — 1 instead of 1 — A. For higher values of m and n, use the recurrence formulas of 5.233. For the imaginary values of the argument, we have, from 5.157, 5.214, and (8) and (9),
§5.231
P1(j0
INTEGRALS OF PRODUCTS OF ASSOCIATED FUNCTIONS
= ji
cot'
Qi(k) (21(j0
Pi(k) = — 3(1 +
Q2(k) =
Pi(k)
= —3(1+
—1
= (1 ± .* 2)1(cot-1 Q2(if) = i.7[(31.2 + 1) cot'
= i(1 .2)4 P2(k) = — 01' 2 + P1(..g)
3.2)
Qi(j•
1-2) (11) — 3fl n2 + 2)
+ 0)1(33" cot-1g.
= j(1
159
i-2) 3 cot-1
1 + 1-2
(1± i-2)2
In the formulas for Q:(k), the arc cotangent ranges from 0 to a as ranges from ± ,c) to — . co, we have, using 5.152 (5) and 5.21 (5) in (8) and (9), When a
r. (
(2n)! „, 2'in !(n — m)!/1 (_1).n!(n m) !2. Q:(A)--> (2n ± 1) t,.+1
(12) (13)
When m = n, insertion of 5.152 (3) in (6) leaves only the s = 0 term and yields the solution used in 5.111 (11). P'„n,(cos 0) = (2m — 1)!! sin" 0 An integral, useful for all values of n. when x
1-Pz(x) = (n + 1)(n
(14)
1, is
2) • • (n m)E[x + (x2— 1)i cos ckin cos m4 dqh (15)
Substitution in (3) and two integrations by parts prove that it satisfies Legendre's equation. The constant factor is verified by letting x —> 00 so that the integral can be evaluated and the result compared with (12). 5.231. Integrals of Products of Associated Functions.—Equation 5.13 (2) shows us that, if n n', f +1t.f 0*2VT(A)P:;(A)(A cos m4
B sin mck)(A' cos m'4) B' sin m' cliz c14) = 0
(1)
Because of the trigonometric products occurring, this integral is also zero if m is an integer, and m m', by Pc 359, 360, and 361 or Dw 435, 445, and 465, regardless of the value of n and n'. To get the result when n = n' and m = m' we must determine the integral of the square of P„'n(A) over a unit sphere. Substitute from 5.23 (6) and 5.152 (4) for P',.n,(A) and rearrange the factors and we have, integrating by parts,
160
§5.231
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS 1
1
+1
[P7,',(2)]2 (n1) 2
22n
=
J-1
u dv dnmdn+m 1)In di.o+.(112 — 1)nid[ d
f +1 [(112
(-1)ni
2
r+1 d
„
Cln+m
i'mdihn+.(112
2".(n!)2 _1 Idtd_
(A2 —
1)n]
Cln+m-2
1)"]}4d 1.01- m-2 (u2— 1)n]
We now integrate repeatedly by parts, the general form of u and v being U=
d8-1 [ ( 2
(
Mdte+M
48-1 °.4
—
1)n ],
V=
d m+n— a („2 dttm+n—s'
1)n
The product uv always vanishes at both limits since u contains the factor 112— 1 when m j s and v when m < s. Thus after m + n integrations by parts we obtain +1 da+m 5+1( 1 _ ti2)n {dn+m 2 (2) (12 1)d} [PT(A)P = 22n(n!) 2 I d µn+,, J-1 The second factor in the integrand is evidently a constant, A being eliminated by the differentiations. Keeping only the highest power of A, by substituting /22for 142 — 1, we see by inspection that its value is [2n(2n — 1)(2n — 2) • • • (n — m + 1)](n m)! = •
(2n) !(n + m)! (n — m)!
The equation now becomes, using 5.155 (2.1) and 5.155 (3),
[13n(µ)]2 1
J-1
m)! (2n) ! f +1 1— A2)n dµ (n — m)! 22"(70)2 1 (
= (n
= (n (n MI)) 1:11[13n(g)P
— 2n 2 + 1(n (n — ±m) ! (3)
From (1) when n n', +1
P'n'.( A)P;',%(11) dµ = 0
(4)
When m m', another integral relation that is sometimes useful is
f +1 1 P7(A)PV(A) = 0 1 — ;12
(5)
To obtain this formula, write 5.23 (3) with 8 = y, m = m and with 8 = y', m = m', multiply the first by y', the second by y, subtract and integrate from —1 to +1. To integrate when m = m', transfer the middle term of 5.233 (4) to the right side, square, write n — 1 for II throughout, multiply by n m, and eliminate 437+,113"„t_72. from the result by using the square of 5.233 (7) multiplied by n — m. When integrated from —1 to +1, all terms not involving 1 — /12 are integrated by (3) and cancel
§5.232 ASSOCIATED FUNCTIONS WITH IMAGINARY ARGUMENT
161
out, leaving +1(07,
i
+1(07_0 2 d/I
n
)2
_1 1 - p2
n - m _1 1 - /12
(n
m)! f +1(0,:02dau
(2m)!(n - m)! _1 1 - /.42
Substitution of 5.23 (14) for e: and integration by Dw 854.1 give 1. + V: (A)12 , (n + m) ! (2m - 1) !! 1 (fl ± m)! 'r sin 0 dO = (6) j_ 1 1 -i.L 2 u'm = (n - m)! (2m,)!! o m (n - m)! In dealing with vector potentials, we shall need to use the orthogonal properties of the surface vector function of cos 0 defined by PZ(A) = (1 - µ2)1P7'(µ) ± (1)m(1 - iz2)-IP'."(A) 0 40 Integration of the scalar product of two such functions gives r-Fi ri .1 1PZ(A) • 137:(A) dil = j 1 [(1 1.42)13 7' (14)PP
(7)
-
+ M 2(1 - 112)-1PT(A)P;(1-)1 dil
Substitution for the second term from 5.23 (3) and rearrangement give ±i{ H. R i - 2>p,,,,p,,, + 4[0 - ,u2)P71P4 dp + n(n + 1) 5 1 PT137; dp
r
J-1
„
11PT.P7,1 dii = n(n ± 1)f 11 PT,Pi; diA 1)f By (1) this is zero if n / p and by (3) if n = p it is f +1 n. 2n(n ± 1) (n + m)! (8) P n(-4) • PIZ (A) dA - 2n + 1 (n nt)!
--- [(1 -
p2)1';,7=7]+: + n(n +
-
6.232. Associated Functions with Imaginary Argument—In connection with oblate spheroidal harmonics, the functions P':(j?-) and (27,'(k) occur, where ranges from 0 to 00. Series for these functions can be found by applying 5.23 (8) and (9) to 5.158 (1) and 5.214 (1). It will be noted that the resultant series for P',n,(j0 has only zero or positive powers of ;- so that P7(j • 0) is finite and PZU • 00) is infinite. The series for QZ(k) contains only negative powers of so that (27,(j • 00) is zero. We shall obtain an expression analogous to 5.214 (5) which is sometimes useful. From 5.212 (2) we have (27(14)
=1:1(11){A +Bf (1 - il2d)11[PZ(11)]2}
Eliminating the integral from this equation and the equation obtained by differentiating it and substituting ..)* for ,u, we obtain d
jB
PZ(ii")— di.Q.7:(:70 - (27(i0d—.P7,(in - 1 + r
(1)
162
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.233
Letting ;.
co and using 5.23 (12) and (13) we find that (n + m) B = ( 1)m (n m)!
(2)
5.233. Recurrence Formulas for Legendre Associated Functions. We have seen in 5.154 and 5.211 that Pn(,u) and Qn(A) have identical recurrence formulas. The signs in the corresponding formulas for 0: = AP: (A) + depend on whether —1 < µ < +1 or whether > 1 or is imaginary. We shall now obtain these formulas, the upper sign referring to the case At = cos 0. Differentiate 5.154 (4.1) m times, multiply through by sinn'+' 0 or (i.t2 — 1)i(m+1), and then by 5.23 (6) and (7) or 5.23 (8) and (9) we have = (m + n + 1)[ ± (1 — /22)]10',,n + (1) 1.4 2)r ( m+ 1), Now differentiate 5.154 (2) m times, multiply by [+ (1 — and we have, by 5.23 (6) and (7) or 5.23 (8) and (9), — Oa±i = (2n + 1)[ ± (1 — ,u2)]10:
(2)
Subtract this from (1), and we obtain = (In — n)[ ± (1— 1/2)]18,11 + 1107+1 (3) Write n — 1 for n in (1), and eliminate 0:,±1.and 0,7_1by using (3) and (3) with m — 1 substituted for m, respectively. Dividing the result through by — (1 — ,u2), we get 07.±2
0:+1-T 2m4.4± (1 — /22)]-10: ± (m + n)(n — m +
=0
(4)
This is the recurrence formula in m. Now multiply (1) by m — n and (3) by m + n + 1, and subtract. Writing m for m + 1 in the result, we have (m — n — 1)0:+1+ (2n + 1)A0: — (m + n)0:_1 = 0
(5) This is the recurrence formula in n. Differentiating 5.23 (6) and (7) or 5.23 (8) and (9) and using the above formulas, we have [± (1— 112)POZ = d mg± (1— 112)]-107: = _TEm n)(n — m + 1)01:-1+10:+1 (6) =± ± (1 — /22) ]-4 0: 4-. (m n)(n — m + 1)0;i-1 We sometimes wish to express [-I- (1 — ,u2)]-40: in terms of Legendre functions. To do this, write m — 1 for m in (1), divide through by [ ± (1 — /12)]1, substitute for iz[± (1 — A2)]-107,1from (4), and write n — 1 for n in the result, giving 2m[+ (1 — A2)]-10: =
+ (m + n — 1)(m + n)07,1:.-i (7) Writing m — 1 for m in (2) and substituting in the last form of (6) give (1 — 1.42)9:' = ±
T- (m n)(n — m + 1)(2n + 1)-1(0,7+1 OZ-1)
NEUTRAL POINTS AND LINES
§5.235
163
Substitution of (5) in the last, first, or middle term gives, respectively, (1 — A2)0':' = ± (n 1)1.4017: T- (n — m 1)0;4, = ± (2n + 1)-1[(m — n — 1)neT4-1 (n = 1 nµ8n ± (m n)(31,7_.1
1)(m
+ n)07-1] (8)
5.234. Special Values of Associated Legendre Functions.—With the aid of the recurrence formulas of the last article, especially 5.233 (4) and 5.233 (6), combined with 5.157 and 5.213, we obtain the following relations m — 1) F":(0) = (-1)4(n—r )1 • 3 5• • • (n (n m even) 2 • 4 6 • • (n — m) (n m odd) P':(0) = 0 • Q7(0) =(-1)t(n—m+1~ 2 4 • 6 " • (n m 1) (n m odd) 1. 3.5 • • • (n — m) Q:(0) = 0 (n m even) —
[— dr PZ(A)]= PZ-Fr(0) de.
[d— Cit:' A7(1)]
1:1
= (27±r(0)
P7( — µ) = ( -1)n±mPZ(A) (— = (-1)n+ni+1(27(A)
5.235. Neutral Points and Lines.—The nature of neutral points is clarified by taking one as the origin for a spherical harmonic potential expansion. Inside a sphere so small that it includes no charge, the potential must be everywhere finite and so of the form A„,„rnP":(cos 0) cos m(4) — (5,n)
V =
(1)
n--0 m=0
At a neutral point, VV is zero so that as r 0, aV/ar —> 0, and Amt is zero. Neutral points may be classified according to the highest p value for which all Amn's (n < p) vanish at r = 0. Thus for a neutral point of the pth order, p is the largest integer for which, for all values of 0 and 0,
a2V ar r-+O art NO
av
aPv
0
ary
(2)
Sufficiently close to a neutral line or point of the pth order , p+1
V
rP÷' Z A.,p-1-1P7F1(cos 0) cos (m0 + S.)
(3) m=o The potential at the neutral point is AN. The equipotential surfaces intersecting there are given when r is small, by equating the summation to zero. For an axially symmetrical field, m is zero so the surfaces are = A00 +
164
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.24
circular cones of half angles a given by P2 (cos a) = 0. From 5.157, for a first-order point, a is 54°44' and 125°16'; for a second-order point, a is 39°14', 90°, and 140°46', etc. The simple neutral line occurs when all but the m = p + 1 term are zero. In this neutral line p + 1 planes intersect, the angles between adjacent planes being 7r/(p + 1). 5.24. Biaxial Harmonics.—It is sometimes convenient to be able to express a surface zonal harmonic P.(cos 0') in terms of general surface harmonics S,,(0, 0) referred to another axis. Let the two axes intersect at the origin and let the coordinates of the 8'-axis, referred to the 0, j system, be 0 = 0 and ¢ = 0. We must then determine the coefficients in the expansion , 4. 1
m=n
A„,P',;(cos 0) cos m4
P„(cos 0') =
(1)
m= o Multiply both sides by P:,(cos 0) cos s4,, and integrate over the surface of a unit sphere. On the right side, by 5.231, all terms go out except the S. terms. By 5.231 (3), this becomes
A,,,f [PT (cos 0)12 cost me dS =
2r (n + m)! 2n + 1 (n — m)! A'n
(2)
on the left side, we must evaluate the integral fsP.(cos 0')S„(0, 4,) dS This identical integral appears as the coefficient of bn when we find the potential V,, at the point 0' = 0, r = b due to a distribution of charge density S.(0, q5) on the surface of a unit sphere. We have solved this problem in 5.131 so that we can evaluate the integral by equating it to the coefficient of bn in that solution which must hold for all values of b < 1. Thus we have
4.71-6Vp = f NdS = f S„(0, cp)(1 b2— 2b cos O') —'3 dS s PQ When expanded by 5.153 (1), the integrals of the products of the 'orm Pr(cos 0')S„(0, 4)) go out by 5.13 (2) leaving only the term 47r€Vp = bnfsPn(COS 0' )Sn(0, 4)) dS But we have already evaluated VP in 5.131 (5) in the form 47-€Vp —
4r b1S„(0, O)]P 2n + 1
In the present case, the coordinates of P, referred to the 0-axis, are 0 = 0, cfr = 0 and so S,,(0, 4)) = P7 (cos 0) and equating coefficients
§5.26
NONINTEGRAL ASSOCIATED LEGENDRE FUNCTIONS
165
of b" in the two expressions for Vp gives
is
P„(cos 0')/:';',1(cos 0) cos mct, dS =
F';,'(cos 0) 2n + 1
(3)
Equating (2) and (3) and solving for A„„ we get Am
— 2(n—ni) •Pz(cos 0) (n + m)!
(4)
when m = 0, (2) and (3) become by 5.155 (3) and 5.131 (5) 47rAo — f Pn(cos 0')P,,(cos 0) dS 2n+ 1 s Substituting in (1) gives A of [Pn(cos 0)]2 dS =
47/3„(cos 0) 2n + 1
m
Po(cos 0') =
(2 — S° )
(n — m) ! (cos e)Pr, (cos 0) cos mcb (5) (n + m)!Pmn
m=0 where Sni ° = 1 if m = 0 and gn =0 if m 0. This symbol is called the Kronecker delta and is more generally written ST where 3,7 = 1 if m = n and 3: = 0 if m n. 5.25. Conical Boundaries.—The potential inside a cone produced by an arbitrary potential distribution over its surface is found by choosing separation constants in 5.12 (1) so that R and (bare orthogonal functions. Thus, if n is jp — -I, then K is —p2— and R becomes, from 5.12 (3), R, =
riP—t
= r-424. cos (p In r + 82,)
(1)
The product RpRp, dr then becomes cos (74 + 3„) cos (p'1,G + 6„,) d,& where ¢ is In r and leads to Fourier's series or integrals in In r. There now appear in the 0-factor Legendre functions of order jp — known as cone functions and treated by Hobson, Heine, and others. Hobson gives the series 4292 + 12 (1T± = 1 ± 22 2 12)(42 ± 32) (1 _T_ (4p2 • • • (2) 22 . 42 2 The upper sign solution, infinite atµ = 1, is useful outside a cone, and the lower sign one is useful inside it. The function is not periodic. No assumptions regarding n were made in deriving 5.23 (6) so that
Prp-A(±1.0 = (1 —
a"P
ate,
(3)
5.26. Nonintegral Associated Legendre Functions.—As mentioned in the last article, when working with conical boundaries we may use Pn (12) or Qm„ (i.,) where n is not an integer. To use these functions, we
166
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5261
need expansion in series of Legendre functions that include all values of n for which OVA) = 0. For this purpose we must have formulas analogous to 5.13 (2) and 5.231 (3). Let OVA) = y and TAIL) = y' be solutions of 5.23 (3) such that (1) 07, (Ao) = 0,7(.10) = 0 Thus we have in ,,)d2 — 2 dY + [n(71 + 1) 1 2 .1Y = 0 (1— 42 Adg — ,dy' 2 dy' _L + 1) 1 m 2 1y, 0 (1
122) d/12
d/1
—
Multiply the first by y' and the second by y, and subtract, and we obtain
r (1
_dy' ] ydµ)] + (n — n')(n n' 1)yy' = 0
Integrate from go to 1, and we get
r
i
J
=
0":01)07,(A)
(2)
if) eZdcle 2)iddelf (en (1— (n — n')(n n' + 1)
From (1), we have, if n X n',
r
i
J 0:GI) 0:
(1.1)
A.
dµ= 0
If n = n', we must proceed as follows. Let n — y = y' (ay' /on')An' in (2), and we get dy'
,dy _ ,ay'
Since at A =, Ao,
+
Oy' Oy' A
,
an''
— y,aY' — —
(3) = An'. Substitute aY
apn,An
t
= 0, we have, as n n',
fo [43
7(4)]2 diA
(1 —p.g)(ae:ao;,4\ 2n + 1 \ ap. an „_„.
(4)
To calculate values of ae'n ,lan, one may use series such as 5.151 (1) or definite integral expressions which may be found in works cited at the end of this chapter. 5.261. Green's Function for a Cone.—We shall now solve the problem of an earthed cone B = a under the influence of a point charge q at r = a, 0 = 13, ¢ = 00. By a point charge, we mean one having dimensions too small to measure physically but mathematically not zero so that the field-intensity and potential functions are mathematically bounded. Our boundary condition V = 0 on the cone will be satisfied automatically if we use a series of Legendre functions in which we choose such values of n that Pm „ (go) is zero where go = cos a. Clearly, from
§5.262
167
GREEN'S FUNCTION FOR A CONICAL BOX
5.23 (1), (2), and (6), with such a choice of n, solutions of Laplace's equation which are finite at r = 0, r = co , agree at r = a, and have the proper symmetry about cko are r
An,„( a)nl''',',1 (it) cos m(g5 — 00)
Vi =
(1)
n m=o
r>a
aln+1 P7,;(1.) cos m(cb — 950) V =2=1.,,,,,(r — i
(2)
n =0
To determine the coefficients Amn, we use a new variable (45' = sb — 95o and proceed, as in 4.07, to write #VWar — 3V0/ar, multiply both sides = 2r and from by Mkt) cos pq5' dµ d4,', and integrate from e = 0 to = go to A = 1. Then, by Pc 361 or Dw 445, all terms on the right are out unless m = p and, by 5.26 (3), they are also out unless n = 8. Thus we have, multiplying through by a2, 2w+ f 1 ay, avo r_adµ d¢' Pni) cos p¢i( a2.1
or — Or
o
+1
2r
= aA„„(2s + 1) f [1:10.0i 2 f cos2p¢' d¢' 0
(3)
The integrals on the right are evaluated by 5.26 (4) and, except when p = 0, by Pc 363 or Dw 440.20. To evaluate the left side, we note that a2 = —a2 sin B d0 de = —dS. The field is continuous across the sphere r = a, except over the infinitesimal area AS at 0 = 0, 4, = 400 or 01 = 0 where the charge is located. Thus OVi/Or = aVo/ar except over AS and the integral vanishes elsewhere. We take AS so small that we may write Pis' (Ai) for Pf(i./), where AI = cos )3, and 1 for cos pe. We note that on the inner face of AS, #Vi/#r = --av/an and that on the outer face of AS, aVo/or = ay/an so that the left side of (3) becomes, by Gauss's electric flux theorem [1.10 (1)] —
ay dS = eq-Pf(i.q)
Pf(A1)
(4)
or p and n for s to fit the notation of Solving (3) for A„, and writing Asa mfn (1) and (2), we have
Am,, —
— (2 — 31)q 27rea(1 —
) n
.913;7 (L)aPT [aµ
1
(µ)
an
=1,0
(5)
where 3„, ° = 1 if m = 0 and S;)„ = 0 if m 0. 5.262. Green's Function for a Conical Box.—Suppose that the charge q lies between the earthed spheres r = d and r = c and inside the earthed cone 0 = a, so that c < a < d and 0 < # < a. We must then superimpose on the Green's function for the cone a potential that will be zero
168
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.27
on the cone and will give a resultant zero potential when r = d and r = c. Since both r = 0 and r = co are excluded, such a potential will have the form of 5.261 (1) or 5.261 (2) where the terms (r/a)n or (a/r)n±' are replaced by Cnrn + Dnr-n-1. When this is added to 5.261 (1) and we put r = c, the result must be zero so that + Cne
„
=
(1)
When we add it to 5.261 (2) and put r = d, the result must also be zero 60 that n+1
Ccdn
dl 7 1 = 0
(2)
4 ?
Solving (1) and (2) for C'n and Dfl, and adding the new potential to 5.261 (1) give, if r < a,
a 2n-1-1
TT =
Amnan
c 2n+1
d2n-F1 d2n+1)
rn
c2n+1) n+l P7:(A)
cos
— 00) (3)
n m=0
Similarly, if r > a, we have, from 5.261 (2), A
V° =
a2"-F1— c2n+1
-fnan(c2n1
d2n+1) rn
d2n-I-1)
rn+1 P7(-1) cos ni(4)
(4)
n m=0
If, in addition, the planes 4. = +7 and q = 0, where 7 > cbo > 0, are at zero potential, and if 7 = 7/8 where s is an integer, we can obtain the potential by the method of images, superimposing solutions of the type (3) and (4). If 7r/ s, we would need to use nonintegral harmonics of the form sin (mT-0/7). This would introduce in A.,„„ the factor 2r/-y, and in (3) and (4) the factor sin (mr4)0/-y) sin (mr0/7) would replace cos m(4, — cbo). 5.27. Oblate Spheroidal Coordinates. Common geometrical forms occurring in electrical apparatus are the thin circular disk and the thin sheet with a circular hole. None of the coordinate systems so far studied gives these forms as natural boundaries except the confocal system of 5.01 and 5.02. In this system, we can specify a single value of one coordinate that gives exactly the desired surface, and no more, for the whole range of the remaining coordinates. Our problem is greatly simplified by the axial symmetry which gives oblate spheroids instead of the general ellipsoids. We shall now investigate the solution of Laplace's equation in such a system that introduces the functions called oblate spheroidal harmonics. In 5.01 (1), let the longer semiaxes b and c be equal and let y = p cos 41 and z = p sin gb. We then have the equation —
§5.27
169
OBLATE SPHEROIDAL COORDINATES X2
P
a2
2
b2
0
—1
(1)
(b2 a 2)0? = c?81. Then we have In this equation, let a2 = confocal oblate spheroids when —a2 < 0 < 00 or 0
P
2
CW CIW + 1) and that of the hyperboloids is
=1
p2
x2 —c?E2
(2)
—1
c?(1 — 2)
($)
Eliminating p from these equations gives x =
(4)
and eliminating x gives P = Ci[( 1 2) (1— E2)]* (5) The coordinate p is always positive, but x goes from — .0 to + 00. There-. fore if we choose 0 < < 00, we must take —1 < < + 1 as shown in Fig. 5.273, and if we choose — 00 < < .0 we must take 0 < < 1 as shown in Fig. 5.272. To write Laplace's equation in this system, we must determine h1, h2, and h3 in 3.03 (4). From 3.03 (1),
— (9a—nE —_ae,
1_ h1
1
and
_
h2 —an
_ae, — an
With the aid of 5.01 (1.1), we have
a°, 1ae Ivo' an = 2cm1an 2cle1 Writing the coordinates in the order E, from (4) and (5) give
X2 (__ clE2)2
h1 =
ce1M12 and substituting for x and
p
p2
cla
(6)
E2)2
p2
X2
h2 = cit-cy4
1
4(1
(7)
i'2) 2
ck ha = pd = p = c1[(1
dcp
2)(1 — t2)]i
(8)
We may now write Laplace's equation by 3.03 (4).
a
av] ,_ a —
4-)w-
, ,_2\av
-r
a 2T7
E2 -I- (1
2) (1 — E2)
a02 =
0 (9)
170
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.271
5.271. Oblate Spheroidal Harmonics.—A solution of 5.27 (9) in the form V = ::7,Z31) where 4-7, Z, and 43 are functions of E, and only, respectively, is called an oblate spheroidal harmonic. Multiplying -2)]-1, we find that the 5.27 (9) through by (1 + /.2)(1 — last term involves ci5 only. We can get a solution, as in 5.12, by setting this term equal to —m2 and the other terms equal to d-m2, giving 1 d24,
dq12
= —m2
(1)
and
1 d [ (1 dt
1 d dZi M2a2 g-2) --[(1 + 0)—4 = (1 + -2)(1 — 2) —Z 4
2) d5,1
M2 1—t2
M2
1+
As before, this can be satisfied by taking v_ c/
dd
t2 dErl —
6 I dd
n(n
m22
1—
, 2
1)4, = 0
(2)
and ci [ (1 +
dZ1
m2Z 1 ± 1-2
n(n ± 1)Z = 0
Putting r = j3 in the last equation gives
4/L
'
1
1
m2z,
2
n(n ± 1)Z = 0
(3)
The solution of (1) is, from 5.23 (2), = C cos m.41
D sin mcb
(4)
Since (2) and (3) are identical with 5.23 (3), the solution is given by 5.23 (5) to be = APT () 13(21:() (5) (6) = A'137(i0 B'QZ (ii") = ATZ(r) B'QZ The general solution is then
V =
(7) m n
Since a sphere is a special case of a spheroid, we should expect spherical harmonics to be a special case of spheroidal harmonics. Let us see how the solution given by (4), (5), (6), and (7) passes over into 5.23 (2), 5.23 (5), and 5.23 (1) as the eccentricity of our ellipsoid becomes zero. By letting 0, the disk given by ?* = 0 shrinks to the point given 0, by r = 0. From 5.27 (4), we see that, since 1 00 as ci ---> 0, we may neglect 1 compared with 2in 5.27 (5) and have P ai
c13-(1 — t2)i =
—
§5.272
CONDUCTING SHEET WITH CIRCULAR HOLE
171
Solving for t gives ci-oo
p2)—i = cos 0 =
x(x 2
(8)
Thus (5) becomes 5.23 (5). From 5.27 (4), we now see that x
ci cos 0
ci
(9)
Referring to 5.23 (12) and 5.23 (13), we have A IP":(.i")
c,-> o
A 2Tn
and
-81.Q:(.i)
B2 ci--) 0
(10)
Thus (6) becomes 5.23 (1). By means of (8) and (10), we can frequently choose, at once, the proper oblate spheroidal harmonics to use in a given problem from a knowledge of the form of the solution for the corresponding problem in spherical harmonics. 5.272. Conducting Sheet with Circular Hole.—Let us now consider the case of an infinite thin plane conducting sheet, with a circular hole, which either is freely charged or forms one boundary of a uniform field.
--------
In order to have our coordinate continuous in the region of the field, we take 0 < t < 1 and — 00 < < + co , 3having the same sign as x. The equation of the sheet is t = 0, as shown in Fig. 5.272. Let us work out the case where such a sheet at potential zero forms one boundary of a uniform field. At x = co, the field will still be uniform and undisturbed by the hole, and at x = — cc the field will be zero. Equations 5.27 (4) and (5) simplify when 2—÷ co, since 1 can then be neglected compared
172
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
with j giving r 2 = x2 + p2
§5.273
,
cif
c?re = c2ir
2t2 + 43-2
Ixl/r = Icos 01. Now the equation so that ?' —> ± r/ci and t = x/(ci?-) of the uniform field at x = co is V = Er cos 0 so that t can enter as cos 0 only which permits only Pi(a), and hence only in = 0, n = 1. Then the potential from 5.271 (7), (5), and (6) is (1)
V = Pi(E)[A/Pi(i0
B'Qi(jOi Writing in the values of P, and Q, from 5.23 (11), we have
V=
g-— 1)]
B'(?'
To evaluate A' and B', consider = ± 00. When /' = ficient of B' is negligible and jA'r Er cos 0 = — cos 0 ci
(2) 00, the coef-
Eci = jA'
giving
(3)
when = — 00 the constant term is negligible and jA'r +.21-Hr 0 = cos 0( ci ci
giving
7r-B' = —jA'
(4)
Or
Eci
B' =
(5)
If the edge of the hole is given by p = a, we have, from 5.27 (5), since both ?' and t are zero there, ci = a, giving for the potential V = aEEk —
1 .7r
cot—'
— 1)]
(6)
To get the charge density on the plate, we have ay av = — tax x-0 +€ ag
(7)
From 5.27 (5), when t= 0, a?- = ±(p2— a2)i and also cot-1 [ ±
a2)1
(p2
a
]
1
=2
,
-T cos--
so that from (6) and (7) cr
—2
l[cos---1 p
+ (p2
a ay ]}
(8)
where we take the plus sign on the upper side of the plate and the minus sign on the lower side. 5.273. Torque on Disk in Uniform Field.—The case of an uncharged conducting disk of radius a whose normal makes an angle a with an electric field E which was uniform before the introduction of the disk
§5.273
TORQUE ON DISK IN UNIFORM FIELD
173
may be considered as the superposition of two cases, the first being a field E cos a normal to the plane of the disk and the second a field E' = E sin a parallel to the plane of the disk. To solve the latter case, we may use oblate spheroidal harmonics. In this case, in order that the coordinates will be continuous in the region of the field, we shall take 0 < oo and —1 < t < +1. The coordinates are then as shown in Fig. 5.273. As in the last article at r = co, --* r/a and t —> ± cos 0,
FIG. 5.273.
having the same sign as x. At infinity, where the field is still undisturbed, we have V., = E' p cos 4, = E'a[(1 — E2)(1 ± 3-2 )]1cos 4, (1) Since 4) enters only as cos 4., we must have m = 1 in 5.271 (4), (5), (6), and (7). Since n j m, the smallest value of n that can occur is n = 1. Furthermore, when r = oo , E and g- can enter only as in (1) which is given by the factor (1 — A2)1'n in 5.23 (6) and (7). We can therefore not have n > 1 since the differentiation in 5.23 (6) and (7) would then bring in t and i-as factors. From 5.23 (11), Q1(.i) i = (1± i-2)1(cot -1?" 1 ±r 2) and from 5.23 (10), PI() = (1— e)I Since the potential is finite when E = +1, the factor (21(t) is excluded and the potential must be
V' = Pi(t)[APi(90 + BQi(ini cos 4, ... (1 — 2)i[j A(1 + 0)i + B(1 + i-13 -1
1 ± r 2)] cos 4. (2)
174
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.273
When /- = 0 V' = 0, therefore jA = —}7rB. When ?" = 00, the last two terms in are negligible compared with the first term, and equating to (1) 2E'a 1 B=— E'a = Or 2 so that the final expression for the potential is ± 1 ±i_2) cos q5
V' = 2r—IE'a[(1 — 2)(1
(3)
Substituting p from 5.27 (5), this may be written ± 0)cos q5
V' = 27-1Eup(tan-1
(4)
Since the component E cos a of the field perpendicular to the disk is not disturbed by the presence of the uncharged disk, its potential is given by V" = Ex cos a, so that, superimposing on the solution (4) gives V = V' + V" the form
2p 7
) sin a cos
V = E[-(tan—'
r
1± is given in terms of p and x by 5.27 (4), (5).
x cos a]
(5)
whe e r We could get the torque from (5) by finding the surface density a whence the force per unit area would be lcr2le, etc. An easier method, however, is to observe that the surface density induced by E cos a produces no torque when acted on by E sin a because the lever arm is zero. Hence, the entire torque must be that due to cr', induced by E sin a, and acted upon by E cos a. From (4),
=
[acT,xT'1=0 _
Jr
2eE' p
1
1 -
r tLi + 1 + -2) 2 ax}r=0 cos cA so that c31'/ax = (at)-1. When t = 0, x/(a) But from 5.27 (4), ?" = a2e = a2 — p2 so that from 5.27 (5), 4€E sin a pcos (6) cr' = (a2 (
For the element of area p dp d4), the lever arm is p cos 4,, and the field is E cos a so that the total torque is, including charge on both sides,
T=
f
8€E2 sin a cos afa
p3cost 4 (a2 p2 ) dcA dp
Since 2 sin a cos a = sin 2a, we have upon integrating by Dw 858.3 or Pc 489 a 3d
T = —4€E2 sin 2af
(a2P
P
§5.274
POTENTIAL OF CHARGE DISTRIBUTION ON SPHEROID
175
Integrating again by Dw 323.01 or Pc 159 gives p2)1 - 861,3E2 sin 2« (7) 3 o 5.274. Potential of Charge Distribution on Spheroid.—Let us suppose that, on the surface of an oblate spheroid, = l.o, we have an electric charge density an of the nature described in the first paragraph of 5.131. This charge gives rise to a potential V. outside the spheroid and Vi inside it. Applying Gauss's electric flux theorem, 1.10 (1), to a small box fitting closely an element of the spheroid gives 1 p2)1 - a2(a2 T = -4€E2 sin 2«[3(a2
-
cr_L, E
1avi - ( an
avA
r 1(avi
an ) 8
L112\
n
avo
n
i-91-0
(1)
Let us consider that crn is such that eVo =j( 1)1n((:: 1 " m)) ;(1+ a)P:(il- o)Q:(j;-)87,
(2)
where
ST = C.,„Pinn() cos m(q5 - 0.) (3) This choice makes V. finite at /• = 00. In order that VVibe finite when t- = 0, = 0, and Vi equal V. when i- = -,a, Vi must be of the form (n - m) ; EVi = ,i(-1)1n (n ± m ) (1 + il)Q:(ko)P7(..081:
(4)
Substituting (2) and (4) in (1) and using 5.232 (1) we obtain an=
Setting
A" =
(5)
{172 1}ro877
j(-1)m (n E (n
- m)!(1 4_ a) in)!
(6)
we can, with the aid of (2), (4), (5) and 1.06 (6), evaluate two useful integrals, which are
S: dS = A „PmW-01Q-(j08'n n n ny froh2R.
I
AST
jr0h2R-i dS
= A.,„Q:(ji-0)13:(jOS: (7)
Now consider a surface density which is a superposition of densities of the form 00r so that the potentials are superpositions of potentials of the form of (7). Thus we have = IMItylS; =
ZhylC„PgE) cos s(cb - Os)
To determine the coefficients C.., we multiply by
P:(E) cos m(ct, - 0.)hih3d dcb
176
§5.275
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and integrate over the spheroid. Exactly as with (5), all terms go out except those for which m = s and n = r. For these terms, we use 5.24 (2) and so obtain, when m 0, c
2
r+1
= 2n + 1 (n — m)! 1 27rci (n + m)! 1
o
„
opm( ) cosm(0_,,m)hili3 d n
c/q5 (8)
The potentials due to the distribution a are then given by
<
mm,,P,7(.10P:(E) cos m(r/5 — 0.)
=
?.0
(9)
n=0 m=0 n
>
3o
v.
=
N 7,,,,QT(j0P:(t) cos m(cf, — (15.)
(10)
n=0 m=- 0
where, by combining (4) and (8),
Mmn
= j( _ 1).(2 — 31)(2n +
(n — m)!12 . L (n m)! J n g 0)
41rEc1
2T-
-1 0 f-F1f
aPZ () cos m(0 — q5m)hih3A dO
where SI = 1 if m = 0 and SI = 0 if m 0. Nmn
(11)
From (4) we have
Pm (ko) = m- ran •
(12)
Cg 0)
The 2 — al factor is required because if m = 0 the integration with respect to cb used to obtain (8) introduces a factor 27 instead of 7. 5.275. Potential of Point Charge in Oblate Spheroidal Harmonics. We can use the results of the last article to obtain the potential due to a point charge at Eo, (bo. By a point charge, we mean one whose dimensions, although too small to measure physically, are mathematically not zero so that the field-intensity function and the potential function are everywhere bounded. We take the charge density a everywhere zero except in an area S, at 4 = 00, E = Eo, which is so small that Pr() has the constant value Pm ,, (to) and cos m(4 — 0.) = cos m(4) — cpo) = 1. The integral in 5.274 (11) then becomes rhih3 d dck = Pn (to) f dS = P:(o)ffc
(to)
The coefficient of 5.274 (11) now becomes
M. = ./j(2
61)q(
2n + 1)m 4reci
By 5.274 (12) we have 2n +
Nmn
(n
m)
(2"1(./NP7(to)
ir(n — m)!12.nm„,.0,
= j(2 32q ( 1)m 4reci L ( z + m)!J i-n""1-
(to)
(1)
(2)
177
PROLATE SPHEROIDAL HARMONICS
P.28
The potentials due to q are then given by
<
Vi=EZ M.P:(i0P:(E) cos m(4 —
00)
(3)
N..(2:(i)PT(E) cos m(4) — 950
(4)
in=0 m n
p
>
.=zz
v
n=0 m=0
With the aid of these formulas, we can now get Green's function for regions bounded by surfaces of the oblate spheroidal coordinate system, using the same methods as with the spherical harmonics. 5.28. Prolate Spheroidal Harmonics.—A common form of spark gap consists of two coaxial rods with rounded ends which can be well approxi-
FIG. 5.28.
mated by the two sheets of an hyperboloid of revolution. We also find cases involving elongated conductors which resemble prolate spheroids. These surfaces are natural boundaries in the prolate spheroidal system of coordinates since a specified value of one coordinate will give one of these surfaces, and no more, for the entire range of the remaining coordinates. The solution of Laplace's equation in this system leads to functions called the prolate spheroidal harmonics. We proceed exactly as in 5.27 except that we now let the two shorter semiaxes a and b in 5.01 (1) be equal and let x = p sin 4) and y = p cos 0. We then have the equation p2
z2
b2 + + c2 + 0
=.-.1
178
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.281
= (c2 b2)02 = eV. We have now confocal prolate spheroids Let c2 when —b2 < 0 < 00 or 1 < .12 < co, where 4 = n2, and we have confocal hyperboloids of two sheets when —c2 < 0 < —b2 or 0 < E2 <1 where 4 = t2. Figure 5.28 shows an axial section of this system. The third coordinate is the longitude angle 0. The equation of the spheroids is P
2
Z2
i
4(712 —
(2)
1) CP7 — 1
and that of the hyperboloids is 2 Z2 P 1 — 4(1 — E2) CTE2 =
(3)
Eliminating p from these equations gives z = C211
and eliminating z gives
(4)
p = c2R1 — t2)(772 — 1)]i
(5) As in 5.27, if we take 1 < n < 00, we must take —1 < +1, but there is no reason to give negative values to n, since these coordinates are everywhere continuous. Taking the coordinates in the order t, n, cb, we determine h1, h2 and h3as in 5.27, giving 2 2 it clE4
p2
h1 =4(
4(1 - t2)2
h2 =
C2
y
e
1 — E2 c 2( 772 Z2 ]1 = 7.12 1 4n4
P
cl(n2
(n 2
=
1)2
(6) (7)
h3 = p = c2[(1 — t 2)(n2 — 1)Ii (8) Laplace's equation is obtained exactly as in 5.27 (9) except that we have n2 — 1 for 1 + r and 712 — E2 for r. Proceeding as in 5.271, we find that, assuming a solution of the form ZHF, we have the same differential equation 5.271 (2) for both 4 and H, and as before is given by 5.271 (4) so that Hmn = AP: (E) BQZ() (9)
H., = AT;7(77) clam = C cos m4
BV (n) D sin mcp
(10) (11)
The general solution is given by
V= n
(12)
5.281. Prolate Spheroid in Uniform Field.—As an application of the results of the last article, let us find the field where an earthed plane, having a conducting spheroidal boss of height c and base radius b, forms one boundary of a uniform field. This situation might arise, for example, in the field between a flat horizontal electrified storm cloud situated above a wet haystack or fir tree on the earth's surface. Let the field
P.29 LAPLACE'S EQUATION IN CYLINDRICAL COORDINATES
at z = c0 be Er cos O. n
As before, when n—* r
CO
179
C2
and—> - = cos 0 r
Thus, referring to 5.23 (12), we see that t can enter only as Pi(t); so the potential is, using 5.212 (8), tn[A B(1 n ± 1 ly1 V = Pi(t)RAPI(n) -I- 13(21(n)] = — 1 ni] 2
►
=c2 + 13(coth-, —
)1
at z = 00, n = 00, 21-1. = 0 and coth-1n = 0 so that
Az
Ez = —
Or
A = c2E
C2
The spheroid with semiaxes c and b is obtained by putting 0 = 0 in 5.28 (1) or cing = O. Thus
0 = c2E
1) + B coth-1no — — no (
Or
B = — c2E(coth-0no —
Thus the final expression for the potential is coth-1n — 177
V = Ez 1
coth-1 770 — 1/ 77o/
where 770 = e2 = (e2
by
(2)
and n, t, and z are connected by 5.28 (4) and (5). 5.29. Laplace's Equation in Cylindrical Coordinates.—For potential problems where the boundary conditions are most simply expressed in a cylindrical coordinate system, we need a solution of Laplace's equation in such a system, where it takes the form
aw ape
1_a2v a2v = 0 _ av ap p2 (902 az 2
p
(1)
Here x = p cos 4 and y = p sin cp. We have already in 4.01 investigated the special case in which V is independent of z and found it gives rise to circular harmonics. We now try a solution of the form V = where R, cD, and Z are, respectively, functions of p, 4, and z alone. This solution will be called a cylindrical harmonic. Substituting in (1) and dividing through by R4'Zp-2, we have
p d (dR) d2.13. f,__ 4_1d dp dp ' (13 q52
p2 d2Z = 0 Z dz2
(2)
180
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.293
5.291. Bessel's Equation and Bessel Functions.—In 5.29 (2), we set 1 d2c1c. _ —n2 71; d4)2 and
(1)
1 d 2Z
(2)
Z de = Then R satisfies the equation
(dR
crp pcTp ) (k 2p 2—
n2)R = 0
If we set v = kp, this equation takes the form
n2 dR ± 1dR . +1—— R=0 v2) v dv
(3)
Complete solutions of 1 and 2 are
B' sin nq Be 1' = A' cos n4) D' sinh kz Z = Gek' De—hz = C' cosh kz
(13 = Aelno
(4) (5) A solution of (3), which is known as Bessel's equation, is called a Bessel function of the nth order. The cylindrical harmonic may now be written
V = R.(kp)(1)(n.45)Z(kz)
(6)
except when k = 0 when, from 4.01 (5) and 4.07 (2), solutions are Vo = (Mpn N p—n)(Cz D)(A cos ncp B sin nq) (7) Vo = [M cos (n In p) N sin (n In P)i(Cz D)(Acosh nO B sinh n4)) (8) When k and n are both zero, we have Voo = (M In p N)(Cz D)(1141 B) (9) 5.292. Modified Bessel Equation and Functions. A somewhat different equation is obtained by substituting jki for k in 5.291 (2). We then get in place of 5.291 (3) —
d2R° j _ 1 dR° d 2 -1- v du
n2 7? v2°
(1)
which is called the modified Bessel equation and whose solutions are called modified Bessel functions. In place of 5.291 (5), we now have Z = Cello = C' cos kiz D' sin kiz (2) The cylindrical harmonic is now written
V = In(kiP)e±inoe+-A0 (3) Solutions of this type will be considered in Arts. 5.32 to 5.36. 5.293. Solution of Bessel's Equation. Let R. = Ects0+3 in 5.291 (3), and we obtain v'E[s(2n s)a„v*--2 a„v1 = 0 (1) —
SOLUTION OF BESSEL'S EQUATION
§5.293
181
Whence, a, = —[s(2n
s)r1a,_2
(2)
If ao is finite, then a_2, a_4, etc. are zero. Let ao be [2nr(n 1)]-1, which reduces to (2'n!)-1when n is an integer, then we have, by repeated use of (2), writing J„(v) for R,,(v), V2
Vn
1)["" — 22(n ± 1)
2'r (n
jn(1))
+
v4
242!(n + 1)(n + 2) (-}V) n+2r —
(n
,o
r
1) (3)
The function J„(v) is called a Bessel function of the nth order of the first kind. Evidently J„(0) = 0 when n 0, and it will be shown in 5.303 that J.(00) = 0 also. Since 5.291 (3) is a second-order differential equation, there must be a second solution. Where n is not an integer, this is J _„(v); but when n is an integer, J_„(v) and .19,(v) are not independent solutions. To show this, write —n for n in (3) then, since [r(-n)]-i is zero, the series for J_„(v) and ( —1)n./„(v) are identical. When n is not integral, the formula Y9 (v)
J p(v) cosor — J_,,(v) sin VT
(4)
defines a second solution. When v is integral, this becomes 0/0. To find its value, write — v for n in (3), split the 0 to cc r-summation into one s for r in the former, and from n to co and one from 0 to n — 1, write n 7r-Jr(v — r) sin VW for ( —1)r/r[l — (v — r)] by Dw 850.3 in the latter. When J _9 (v) in (4) is replaced by this result and cos vr by ( — 1)n, it becomes .
( — 1)n [
sin yr
J9
,„.1 (v)
(_1)8(1v) r+28 1 2 (n ± s)!r(n — v ± s ± 1) i
n-1 (iv, )2r— r r ( V —
r)
Tr ! r ■=o
8 =0
The bracket is zero when v = n from (3) so the first term still has the form 0/0. To evaluate this, differentiate each factor with respect to v and write n for v. By referring to page 19 of Jahnke and Emde we find z
d r 1 = —1 d In r (z) _ —1
r (z)
r(z)
dz
r(z)
J712,1
m ®1
where C = .5772157 is Euler's number. With this formula and Dw 563.3 or Pc 828, the 0/0 form is evaluated and writing In a for C — In 2 gives
182
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS n -1
2 Y„(v) = -J.(v) In (av)
(ivn
§5.294
1)!
)2r
-n( 77.1 r -
r-o
r
irr!(n
+
n+r
1 11,
r)! m ■=1
r=0
172)
(5)
m=.1
A complete solution of Bessel's equation when n is an integer is
R.(v) = A'.1.(v) B' Y,,(v) (6) We note that Y„(0) is infinite and shall see in 5.303 that Y„( co ) is zero. There are many notations for the function defined by (5). Watson and the British Association Tables use Yn(v), Jahnke and Emde, Schelkunoff, and Stratton use N.(v), and Gray, Mathews, and MacRobert use /7.(v). Setting M = 1/v and N = 1 - (n/v) 2in 5.111 (6) reduces it to 5.291 (3) so that, if J.(v) is the known solution, 5.111 (7) becomes
B f[v4(v)]-1dv)
Y.(v) = J.(v) {A
(7)
Differentiating this and omitting the argument give /
B
' Tn n)
(8)
=
From the recurrence formulas, B is independent of both n and v and so may be evaluated for integral n's by taking the simple case where n is zero and v is very small. Then only the In v term in (5) is important, so it and its derivative may be used for Y.(v) and Yav) in (8). The logarithm terms cancel and B comes out 2/ir so (8) may be rearranged in the form
Y.(v)J.(v) - Jfi (v)Y.(v) = 772:v
(9)
For cylindrical electromagnetic waves we need the "Hankel" functions which combine with ei't to give traveling waves and are defined by
HT(v) = J.(v)
jY.(v),
H;,?)(v) = J.(v) - jY.(v)
(10)
5.294. Recurrence Formulas for Bessel Functions.—If we multiply 5.293 (3) by vn and differentiate, we obtain d[vnJ.(v)]
dv
v 2n-1 [ i 2n 1r(n) -
1
V2
V4
22n
242!n(n + 1)
•••
Factoring out vn on the right side and comparing with 5.293 (3), we see that d[v"J.(v)] - vnja_i(v) dv
§5.295
VALUES OF BESSEL FUNCTIONS AT INFINITY
183
Differentiate left side, rearrange, divide out vn, and indicate differentiation with respect to v by a prime, and we have
-( 31.
4=
1)
If we use the same procedure multiplying by trn instead of by vn, we obtain (2)
— in+i)
+ 1171../. =
=
Subtracting (1) from (2), we have —JJ an
0
T 11--1 "1" d
n4-1
(3)
Writing (-1)nJ....n(v) for J„(v) in (1), (2), and (3) gives the recurrence formulas for J,(v). Differentiation of 5.293 (4) and substitution for 4(v) from (1) and for .11p(v) from a similar formula give, as v —> n, so that cos vir may be replaced by — cos (v — 1)r and sin vr by — sin (v — 1)r,
J'„ cos V7 „ p-1 cos (V — 1)71 — J -1 sin (v — 1)7r sin vr A similar process can be applied to (2) so that
Yn
v sin
J_„)
mr.
= 17.-1 —
(4)
= — Yn+.1
(5)
Subtraction of these equations gives 2n, V
v(J,, cos vir —
n = Yn-1
170-1-1
(6)
Two useful integral formulas are got by integrating the equation from which (1) was derived and the analogous equation associated with (4).
f vnJ n_i(v) dv = vnJ n(v) (7) fvnYn_i(v) dv = vnY n(v) (8) A similar integration of (2) and (5) gives dv = —v—nJ n(v) f v—nJ (9) v—nIc+I(v) dv = —v—n17.(v) (10) 5.295. Values of Bessel Functions at Infinity.—In potential problems involving p = co, one must know how J.(kp) and Yn(kp) behave there. To find this limiting value, we use a trick often employed by Sommerfeld. In 5.291 (3), as v —> co let us, as a first approximation, drop the terms containing v--' and V-2. This gives the approximate differential equation , CIV2
R= 0
(1)
184
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.296
The solution of this equation is (2)
R=
We now insert this trial solution in Bessel's equation 5.291 (3) and consider R' to vary so slowly with v that d'R' /dv 2, v1dir /dv, and can be neglected compared with dir /dv and v--'R', and we obtain
2 dR' c 27 +R'= 0
R' = Cv-i
Or
so that, from (2) our asymptotic solution becomes R = Cv-le±29
(3)
We now see that the largest term neglected was of the order v--1. J„(v) and Y „(v) must be real linear combinations of the two solutions given by taking the plus or minus sign so that they must be of the form Jfi(v) --> Av-i cos (v + a)
(4)
t)-> 00
By-4cos (v
Y fi(v)
fi)
(5)
To find how A and a depend on n, put (4) into 5.294 (1) and 5.294 (2) co, J,' = J, _1 and which give, as v = —J,+1, respectively. This which is satisfied by afi -= gives the relation an±1= an T+ and shows that A does not depend on n. Because n need not be an integer, we may write n = in (4) and compare it with 5.31 (2) which shows that -y = —fir and gives 2
J fi(v) —> () cos (v 1 — 1 )
(6)
where terms in v-im have been neglected if m j• 3 and n is real. To get Y „(v) substitute (6), with v and — v for n, in 5.293 (4). The result gives 0/0 when v is integral, but replacement of numerator and denominator by their derivatives with respect to v gives, when v = n, Y fi(v) —> co
2 y sin (7, — 1-n7r —1-7r) 2
7r1)
4
(7)
Thus both Bessel functions vanish at infinity. From (6), (7), and 5.293 (10) we find that for the Hankel functions 2)1e,(fi-Inr-ir), 4e-3(u—in.—I.-) — 2 ) HO (V) (8) irV v--) 5.296. Integrals of Bessel Functions. In 5.261, we made an expansion in spherical harmonics satisfy the condition V = 0 on the cone 0 = a by choosing only such orders n of the harmonics en'(cos 0) as made 0„m(cos a) = 0. To determine the coefficients in this expansion, it was first necessary, in 5.26, to evaluate the integral of the product of two such HT (v)
—
185
INTEGRALS OF BESSEL FUNCTIONS
§5.296
harmonics Over the range of 0 from 0 to a. In the same way, if we are to get an expansion in Bessel functions that meets the condition V = 0 or E = 0 on the cylinder p = a, we must evaluate the integral of the product of Ri,(k,p) and Ra(k,p) over this range, where 1c and k g are chosen to meet the boundary conditions. Let u = .1?7,(k,p) and v = Rn(k,p) be two solutions of Bessel's equation. Then from 5.291 n2 pkt) c
1cli ap, PA
P
+ (k; — -2 )u = 0
P
+ k 2— - - v=0 0 — dp\pirp) + ( g p2
Multiply the first by pv and the second by pu subtract, and integrate, and we have
u
a d (pT du) d ( dv\-1 d a[vir k ,2i)i puv dp = — — dp\PTpij P o aP Integrate each term on the right side by parts, and the integrals cancel, leaving
(4
-
5
-
k:)f p uv dp
—
dvy Puddo = —a[k„Rn(k ga)R'„(k,a) — k gRn(kpa)1=e;,(k ga)]
( du --\Pvdp
This is zero if
R„(kpa) = Rn(k,a) = 0
(1)
k(kpa) = Rak,a) = 0
(2)
kpak(kpa) BR„(k,a) = k gagi (k,a) BR.„(k,a) = 0
(3)
or if or if Thus if 1c
k g, we have the result
foapfen(k,p)Rn(k,p) dp = 0
(4)
If R„(kpa) = Rn(k,a) = R„(kpb) = Rn(k,b) = 0 we have, since
the result
fabf(x) dx = fobf(x) dx — foaf(x) dx fobpRa.(k,p)R.(k gp) dp = 0
(5)
To evaluate the integral when k, = k, multiply Bessel's equation [5.291 (3)] through by v2(dRn/dv) dv giving
odl:endORA dv \ dv
ORA2 d + v\ dv )
v
2„ dR„ dv v
_ n
—
186
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.296
Integrate from 0 to a using integration by parts on the first and third terms. We thus find the following expression to be zero; 2 2 av(dR„ v2 ,, n2R! a v2(dRI a ) dv+ f av(--) dR dv+ R2 2 " dv o Jodv 2 0 dv
r
Canceling the second and third terms and solving for the fifth give a a 1 v2[d civ(V)]2 (1) 2 — n 2 )[Rn(v)]2 v[R„(v)J 2 dv - 2
Jo
(6)
0
Substituting for the derivative from 5.294 (2) gives
fav[Rn(v),2 dv ==1-a2ia2{[Rn(a)]2 { [Rn(a)]2
[R,H-1(a)121 - naRn(a)R,,+i(a)
[R"-i(a)P1 - naRn(a)Rn-1(a) (7) In dealing with the vector potential we shall have occasion to make use of the orthogonal properties of the vector function defined by R„(k,p) = R' (k p) ± k R (k p) (8) P e pP P The integral of the scalar product of two such functions from 0 to a is
a J:
R„(k„p) • Ra(k gp)p dp a n2 = f [R'„(k„p)k(k,p) + kpicQp2Rn(k,,p)Rn(k gpd p dp
(9)
With the aid of (1), (2), (3), (4), (5), (6) of Art. 5.294 we may write this as the sum of two integrals of the form of (4). Thus tfoaRn+I(kpp)R.+1(kaP)P dP
foa Ra-i(kpp)Ra-i(k gP)p dp
(10)
Evaluate each integral by the formula for f puv dp already given and add the results, then eliminate the derivatives by 5.294 (1) and 5.294 (2), cancel terms not involving the nth order, and combine the resultant n + 1 and n - 1 orders by the same formulas. These operations give, if v = k„a and v' = k ga, foa Rn(k,p) • R„(k,p)p dp = (14-k:)-1[vRn(v')R:,(v)-v'R„(v)K,(V)]
(11)
Thus, if k, k g, the integral vanishes under conditions (1), (2), or (3). But, if k, = k g, evaluation of each integral in (10) by (7) gives for their sum (a/k„)R'„(k„a)R„(k„a) (12) -i[a2 -(n/k„)9[R,,(k,a)]2 + i[aR'n (kpa)]2 Thus a surface vector function of p and 0, one component of which vanishes at p = a, may be written as a sum of terms of the form Rn(ko) sin (nit, +
5n)
GREEN'S FUNCTION FOR CYLINDER
§5.298
187
5.297. Expansion in Series of Bessel Functions.—Consider a function f(v) which satisfies the conditions for expansion into a Fourier series in the range from v = 0 to v = a and which fulfills one of the following boundary conditions: (a) f(a) = 0. This case arises, if f(a) is a potential function, when the boundary is at zero potential. (b) f'(a) = 0. This case occurs when the boundary is a line of force. (c) af' (a) + Bf(a) = 0. This case reduces to (a) if B = co and to (b) if B = 0. An example of its use is given in 11.08. The function f(v) may be expanded in the form f(v) = ZA,,1„(1.4v)
(1)
r =1
where the values of 14 are chosen so that in case (a) J.(ura) = 0, in case (b) J (Ara) = 0, and in case (c) Ara4(Ara) = 0. To determine A,., multiply both sides of (1) by v./.(Asv), and integrate from v = 0 to v = a. By 5.296 (4), all terms on the right vanish except Asf v[J „(A„v)12dv so that a
vf(v)J,,,(Aav) dv (2)
fo v[J„(p4v)P dv We can evaluate the lower integral by 5.296 (7) giving Aso ov[J„(12,v)12 dv = pc-2 f x[J„(x)]2 dx JO
1 = cl21{4,(1A.coi2
{efn±j(Asa)12 —
na
J (ti.a)./.±i(Aaa) (3)
In case (a), substituting (3) in (2) gives
2 vf (v)./.(11,v) dv [aJ n±i(gsa)12/0 In case (b), substituting (3) in (2) gives 2 — (a2 _ n2i.c2)binoisa)12f0 vf(v)Ja(p.v) dv As =
(4)
(5)
In case (c), substituting (3) in (2) gives
a vf(v)Jn(A,v) dv (6) [a2 + (B2 — n2)/4 ./.(Asa)Pio 5.298. Green's Function for Cylinder. Inverse Distance.—We shall calculate by the principles of the last few articles the potential when a point charge q is placed at the point z = 0, p = b, = ¢o, inside an earthed conducting cylinder. By a point charge, we mean one whose As —
2
[
188
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.298
dimensions, although too small to measure physically, are different from zero, so that the field-intensity function and the potential function are everywhere bounded. From 5.291 (6), a solution that vanishes when z = 00, is symmetrical about the cP (Po plane, gives V = 0 when p = a, and is valid for positive values of z, is CO
Arae —A,'.1.(14P) cos s(g5—
V=
o)
(1)
r=1s=0
where Ar is chosen so that Js(pra) = 0. Y8(pr p) is excluded because it is infinite on the axis. From symmetry, the whole plane z = 0 is formed by lines of force except at the point charge itself. To complete the boundary condition in this plane, we shall consider (aV / az) 0to be zero except in a small area AS at cp = cpo, p = b. Differentiating (1) and setting z = 0 give
av\Jo
ii„A„..1.(Arp) cos s(4) — 4)o)
\az
r = 1 s =o
We determine A„ in this expansion as in 5.296 and 5.297 by multiplying through by p4(1. 0p) cos p(iP — q5o) and integrating from p = 0 to p = a and from 4) = 0 to 4) = 2r. By Dw 858.2 or Pc 488 and 5.296 (4), all terms on the right disappear unless p = s and q = r. In the latter case, we see by Dw 858.4 or Pc 489 that the 4) integration introduces a factor 7 on the right if s > 0 and 2r if s = 0, so that we have, by 5.297 (4),
,,
(2 f f az op. ,, Arp) cos s(4) — 4)o) dp d4) (2) [ctsia+ (Ara)] 2 where 62 = 1 if s = 0 and 62 = 0 if s O. In the z = 0 plane, the area AS in which (a V/az)0 0 is taken so small that in it Js(prp) has the constant value Ja(prb) and cos s(4) — cko) = 1. The integral then becomes
A„ =
J.Carb)
—
ff
—
av aZ 0
dqS
an
— dS — 1.1.(11,b) dp = .18 (Arb)Asf an 2E
from Gauss's electric flux theorem [1.10 (1)] it being remembered that only half the flux passes upward. Substituting for the integral in (2) gives g)J,,(prb) q(2 A, (3) 27E,Ura 2[Js+I(Ara)]2 —
—
Thus we obtain, for the potential,
V=
(2
271-Ea2 r=1 a=0
Vs)e—Adzi
Cub),1(AL P)
s(4) il,[4+1(mfa)]2 cos
rko)
(4)
GREEN'S FUNCTION FOR CYLINDRICAL BOX
§5.299
189
This is Green's function (see 3.08) for a circular cylinder. If the coordinates of q are p = b, z = z o, and 4, = 00, we should substitute Iz - zo l for iz! in the above formula. If the charge is on the axis, all but the first term of the s-summation drops out and Jo(Arb) = 1. When a -> 00 in (4), it gives the potential of a charge q in unbounded space. From 5.295 (6) J„(Ara) oscillates sinusoidally as Ara -> oo so that its zeros are uniformly spaced at intervals 7r = Ar+ia - Ara = a Ap. and so that, when J.(.1,,.a) is zero, Jn+I(Ara) is (lirpira)-1. Thus (4) becomes V-
27rect2
(2 - g,) cos s(4, - 950)e-rAmi'-'.1 J„(r Ailb).1,(r 8=0
If k = r OA then as Ai..(
47reR
47re
co this takes the integral form
0 and a
,
=
a2 Aps
r —1
(2 - (52) cos s(4, - 4)o)f e-kk-z01.1.(kb)J.(kp) dk
(5)
8=o
where R2 = (z - zo)2 + P2 ± b2— 2pb cos (4) - cbo). gives
If b = z o = 0, this
zz)-i =r y eki'lJo(kp) dk
( p2
(6
With z, zo, and 00 zero, write R for p in (6) and compare with (5). Thus OD
Jo[(p2 + b2 - 2pb cos OA = Z(2 - Vs)Js(P)Js(b) cos .34,
(7 )
s=o 5.299. Green's Function for Cylindrical Box.—Superimposing potentials of the form of 5.298 (4) according to the principle of images 5.07 and 5.101, we may make the planes z = 0 and z = L at zero potential as well as the cylinder p = a. If the coordinates of the positive charge q are z = c, p = b, and 4) = 4>o, there are positive images at z = 2nL c and negative images at z = 2nL - c where n has all integral values from - .0 to + 00. If z < c, the factor involving z in the resultant potential is eA,(2nL-c+z) n =0
Z e—p,(2nL—e—z)
n =1
n =1
= 2(0—Ar c
e-271)1,1
e—ur(2nL-1-e-1-2) n =0
eAreZ
sinh mrz
n =1
n=0
2[ (e-Pre - e+mre)Ze-2"i"
emrc] sinh
n=0
Summing the series, by Dw 9.04 or Pc 755, multiplying numerator and
190
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.301
denominator of the sum by eArl, and putting over a common denominator give 2 sinh ar(L — c) sinh kirz sinh Substituting this value in 5.298 (4) gives, when z < c, the potential
(2 _
V= q rea2 r=1 a=o
sinh Ar(L — c) sinhtiTzJ .(m,b)J .(mTP) cos s (4, — (bo) (1) sinh prL fi(Ara) P
When z > c, substitute L — z for z and L — c for c. If the charge is on the axis of the cylindrical box, drop the summation with respect to s in (1), retaining only the s = 0 term. If, in addition, the planes 4, = 0 and 4, = 4,1, where 0 < (Po < 4)1, were at zero potential, and if chi= 2r/n, where n is an integer, we could get the Green's function by superimposing, according to the principle of images (see 5.07), 2n solutions of the type of (1). 5.30. Bessel Functions of Zero Order.—In the important case where we are dealing with fields symmetrical about the z-axis, the potential is independent of 4) so that Bessel's equatio4 becomes, from 5.291 (3), d 2R 1 dR dv 2
vdv
R=0
(1)
and the solution [5.293 (3)] becomes v2 V 4 + 22 24(202
Jo(v) = 1
V6
(2)
26(3!)2
This series is evidently convergent for all values of v. As with Jn(v), Jo(co) = 0 but Jo(0) = 1. Equation 5.293 (5) becomes, when n = 0, Yo(v) = —2[Jo(v)
In av
v2
v4(1 + 1)
v6(1 ± + i)
22
24(2D2
26(3!)2
•••
(3)
where In a is —0.11593. 5.301. Roots and Numerical Values of Bessel Functions of Zero Order.—If we plot the values of Jo(v) and Yo(v) given by 5.30 (2) and (3), we get the curves shown in Figs. 5.301a and 5.301b. We see that they oscillate up and down across the v-axis. It can be shown that both Jo(v) and Yo(v) have an infinite number of real positive roots. The same is true of J.(v) and Y„(v). As we have seen in finding the Green's function for a cylinder, the existence of these roots is very useful as it makes it possible to choose an infinite number of values of lc which will make J.(kp) or 177.(kp) zero for any specified value of p. Many excellent tables exist which give numerical values, graphs, roots, etc., of Bessel functions. Care should be taken in observing the notation used as this
§5.302 DERIVATIVES AND INTEGRALS OF BESSEL FUNCTIONS
191
varies widely with different authors. Asymptotic expansions provide an easy method of evaluating Bessel functions of large argument.
.0.,
Jo (x) D.5
A PA■ Jow AI PAlb, . wir ./0,4x.
r
0
.10,(x) Og
5 FIG. 5.301a.
10
1..) Yoi(x)
lk() AlSW-Alk
0.5
0
0.5
_
5.301b.
5.302. Derivatives and Integrals of Bessel Functions of Zero Order. Putting n = 0 in 5.294 (2) and (5), we have Yo(v) = —Y1(v) (1) and JO(v) = —J j(v) From 5.294 (7) and (8), we have
fovvJo(v) dv = vJ i(v)
and
fovvYo(v) dv vY i(v) +27r-1 (2)
There are several definite integrals involving Jo(v) which will be useful. From 5.30 (2), using Dw 854.1 or Pc 483, we have e)
(2n)! (2n)! 22 n(n)2
(-1)nv2n
Jo(v) =
22n(n )2
n=o
= n=o
= 14( — 1.)n2V2n (2n)! 2
3 • • • (2n — 1) 2 4 • • • 2n
n
(2i vZ f: coent dt n∎O
192
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.303
Interchanging integration and summation symbols and using Dw 415.02 or Pc 773, we obtain
Jo(v) =
( 1)n(v cos t)2' dt = 1 - f cos (v cos t) dt (2n) ! 7 0
7r o n=0
so that 1 cos (kp sin t) dt cos (kp cos t) dt = (3) o o We can show easily, by using the trigonometric formulas for sums and differences of angles and integrating by parts, that an expression that satisfies the recurrence formulas 5.294 (1), (2), and (3), is 1 Jo(kp) = -
1o Jn(v) = - f cos (nt - v sin t) dt
(4)
This evidently reduces to (3) when n = 0. Direct substitution in 5.291 (3) and integration by parts show that
In(v) -
(1v)n
r(i)r(n +)
o
cos (v cos 0) sin2n 0 do
(5)
satisfies Bessel's equation if n > Comparing the value of (4v)-nJ„(v) given by (5) as v ---> 0 with that given by 5.293 (3) fixes the constant. It can be shown that 5.298 (6) still holds if we write jz for z, thus
c-ikzJo(kp) dk =
(1,2
z2)---1
Jo whence, equating real and imaginary parts, we have, when p2 > z2,
o. cos kzJo(kp) dk = (p2 f
-
(6)
sin kzJo(kp) dk = 0
(7)
cos kzJo(kp) dk = 0
(8)
sin kzJo(kp) dk = (z2 -
(9)
When p2
10
5.303. Point Charge and Dielectric Plate.—As a first application of the last two articles, let us find the potential on one side of an infinite plate of material of thickness c and relative capacitivity K, due to a point charge q on the opposite side. We shall take the point charge at the origin and the z-axis perpendicular to the plate. The equation of the surfaces of the plate will be z = a and z = b where b - a = c. From 5.298 (6), the potential due to the point charge alone is
V- q = q f J (kp)e-ki'l dk 4revr 47rea, ci °
(1)
POINT CHARGE AND DIELECTRIC PLATE
§5.303
193
Since this is a solution of Laplace's equation, which involves z and p only, it is evident that we shall still have a solution if we insert any function of k under the integral sign. Let V1 be the potential when write' — 00 < z < a, and write V1 = (47retri
'ci)(k)Jo(kp)e±k' dk]
(kp)e—kl'1 dk
(2)
The second term represents the potential in the region below the plate due to the polarization of the plate and is finite everywhere in this region. The potential in the plate may be written
V2 =
(k) J ° (kp)e—kz dk
(4re„)—'q[
e(k).10(kp)e+kz dk]
(3)
This evidently is finite if a < z < b. The potential above the plate, where b < z < 00, must vanish at z = 00 and so may be written
V3 =
(47-e„)—lqlo
(k), o(kp)e—k2 dk
(4)
We must now determine (13(k), •If(k), 0(k), and 2(k), so the boundary conditions are satisfied for all values of p between 0 and 00. This requires that the integrands alone satisfy these conditions. To prove this, we make use of the Fourier Bessel integral which states that, with suitable restrictions on n and the form of f(x), f(x) =
f 'tJ„(tx)[fo 'uf(u)J„(tu) du] dt
Thus, if we have f IRV" o(kP) dk =
JO
f2(k)Jo(kp) dk
we may multiply both sides by pJ o(mp) dp and integrate from 0 to 00
and obtain, after multiplying through by m, ..fi(m) = .f2(m) Applying this to the present case, at z = a we have, after canceling out Jo(kp), V1 = V2, giving e—"
cI)(k)eka — NP(k)e—ka — 0(k)eka = 0
(5)
and aVi/az = KaV2/az which gives, canceling J o (kp), —e—ka cl)(k)eka KT(k)e—ka — Ke(k)eka = 0
At z = b, we have in the same way V2 = V3, 4/(k)e—kb
(6)
giving
0(k)ekb — S2(k)e—kb =
0
(7)
and KaV2/az = aV3/az which gives —KT(k)e—kb KO(k)ekb
12(k)e—kb = 0
(8)
194
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.304
Solving (5), (6), (7), and (8) for 12(k) gives St(k) =
4K 1) 2 - (K - 1) 2e2k(a---0
(K
To simplify this we write b - a = c and 1 - 02 = 4K(K 1)-2which gives
g=
(9)
(K - 1)/(K 1) so that
1 -(32 St(k) - 1 - 02e-27. Substituting in (4), we have, for V3
V3,
the result
q(1 - 2) f*vo(kp)e—kzdk 3 4re,, jo 1
(10)
1c 0 2e-2e
If we wish to expand this as a series, we may expand the denominator by Dw 9.04 or Pc 755. 02fo 'Jo (k p)e-k(z-1-2e) dk
4wei,V3= q(1 - 132)[fo 'e Jo(kp)e-'4 dk
. . .
Substituting from 5.298 (6) for each integral gives V3 -q(1 °2) f 1 47E2, (z2 p2)*
+ $2
[(1 + 20 2 + P2P 04 [(z
4c)2
±•••
}
which may be written
n(1 _ V3
02).1
47ret, .24 [(z n=0
02n
2nc)2
p2P
Another, but more tedious, method of obtaining this result would have been to use images. The potential in the other regions can be obtained in the same way by solving (5), (6), (7), and (8) for c1)(k), (k), and (k) and substituting in (2) and (3). Clearly this method could be applied to any number of plates. 5.304. Potential inside Hollow Cylindrical Ring.—As another example, let us find the potential at any point in the region bounded by the two cylinders p = a and p = b, both of which are at zero potential, and the two planes z = 0 at potential zero and z = c at potential V = f(p). Since both p = 0 and p = co are excluded, we may have both J o (kp) and Yo(kp). From 5.291 (5) and 5.293 (6), it is evident that a solution of Laplace's equation which satisfies the boundary conditions at z = 0 and p = b is given by Vk = Ak
sinh (ukz)ko(AkP)
Jo(Akb) Yo(Akb) 17°(I4P)
(1)
NONINTEGRAL ORDER
§5.31
195
We can satisfy the conditions Vk = 0 at p = a by choosing values of IA so that (Akb)
yo(ukb)OULka) = 0
(.10)
Hence, all the boundary conditions except that at z = c will be satisfied by a sum of such solutions V = ZVk (2) k=1 The final boundary condition V = f(p) at z = c will be satisfied if we choose Ak so that f(p) =
J o(pkb)
Aksinh A1kc[J00A P)
Yo (4)170(mkP)]
k=1 Represent the term in brackets by Ro(Akp), since it satisfies Bessel's equation, multiply through by pRo(A„p) where Ro(A.a) = R0(.4b) = 0, and integrate from a to b. Since Ro(pka) = Ro(mkb) = 0, we see from 5.296 (5) that all terms on the right drop out except the one for which k = s. For this term, we have, by 5.296 (6),
fa
Pi(P)RoWP) dP =
[ 62( dRo \d(iikP)
p =b
(dc/ohRkpo))p2 _ aljAk sinh Akc — a2 2
Differentiating Ro(mkp) by 5.302 (1), we have RL(AkP) = — {Ji(iIkP)
Jo(pkb)
( )} Yo(b) Y1-"P
Solving for Ak gives 2 rbpf (P)Ro(mkp) dp .1 a
Ak =
(3) {b2[R'(.4kb)]2 — a2Pro(Aka)J 2 sinh Akc Substituting (3) in (1) and then (1) in (2) gives the required solution. If we wish the potential inside the earthed cylinder p = a with V = 0 when z = 0 and V = f(p) when z = c, we drop the Yo term in (1), and (3) becomes
Ak
2f aPf (P)J o(AkP) dP °
a2[J 1(gka)] 2 sinh ykc
(4)
5.31. Nonintegral Order and Spherical Bessel Functions.—When n is not an integer, a complete solution of Bessel's equation is 1?„(v) = AJ.(v)
BJ,(v)
(1)
We no longer need Yn(v) because it must be merely a linear combination
196
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.31
of Jn(y) and J,(v). The formulas of 5.294 are equally valid for these values. The function J±(n+i) (v) where n is an integer are especially simple, for if we substitute 72 = and n = in 5.293 (3) and compare with the sine and cosine expansion Pc 772 and 773 or Dw 415.01 and 415.02, we have 2 )1
Ji(v) =GT) sin v
and
2y J_i(v) = (rv cos v
(2)
If we then put n = - and n = --i in 5.294 (3), we get 2 i 1 —) G sin v — cos v) Ji (v) = (irv
(3)
2 i 1 J__,I (v) -- G--) ( — sin v— —vcos v)
(4)
Putting n = 1 and n = --i gives i 3 3 JI (v) = (.=) R—i— 1 ) sin v — — cos v] in)
(5)
V
V
3sin J_$ (v) = (--li[ir.) V
v ± (32.— 1) cos v] v
(6)
In this fashion we could eventually obtain the formula
c in J±(n _i)(V)
(11 7rV
= —.(sin
[v T 1(2n
+ 1 -T 1)1-]
( —1)8(n + 2s)! (2s) !(n — 2s) !(2v) 28
8
=0
(n-1)
(-1)8(n ± 2s ± 1)! ) (2s ± 1)!(n — 2s — 1) !(202.+1 (7)
+ cos [v T- -1(2n + 1 ± 1)7r] a= 0
where the upper limit of each summation is the nearest integer to the value given and the upper or the lower sign is to be used throughout. We observe that, because cos (n -Dr is zero, 5.293 (4) becomes 17n+i(v) = ( —1)"-"J_(n+i )(v) (8) v—C1
When n
is written for n in 5.293 (3) and v 0, there results
(211 vn vn-Ei ) (2n + 1) !! 2n±1r(n The spherical Bessel functions j,i(v) and nn(v) are defined as J,„+i(v)
ir 1 in(V) = (Toy Jn±i(v) = —vin(v),
nn (v) =
Y„+I(v) =
1
v19 (v)
(9)
(10)
where 3n(v) and gn(v) are the functions used by Schelkunoff. They arise in connection with spherical electromagnetic waves and have recently
197
MODIFIED BESSEL FUNCTIONS
§5.32
been tabulated. From (10), 5.293 (9), 5.294 (2), and 5.294 (3) it follows that n'n (v)j„(v) n„(v)f„(v) = 2)-2 (11) (2n + 1)v—ljn = jn+i, = ±jnTi -T ± 1-)v-1jn (12) —
Similar formulas hold for n„(v). Differentiation of (12) and elimination of jn_i(v) give the differential equation for jn(v) or n„(v).
d
d.
4v2cWin(v) ]
n(n
[u2
1)]j„(v) = 0
(13)
5.32. Modified Bessel Functions.—We can obtain a solution of 5.292 (1), the modified Bessel equation
a
dR° R° 1 v dv
v2
2 —
+
(1)
v 2 r°=0
by substituting jv for v in 5.293 (3). We see that the bracket is real but the coefficient j" occurs. Since we wish a real solution, we define it to be t+ 2r
I,(v ) = j-vJ,(P) =
!F(v
r
(2)
1)
r=0
When v is not an integer, a second solution is (p—p-1-2r
I,(v)
r
(3)
1)
r=-0
When v is an integer, we may write (n r)! for F(v r 1) and find that in(v) equals i_n(v) so that we need a solution which we may define by (4) IC.(v) = -iiri'+1[Jn(iv) This with the aid of 5.293 (3), 5.293 (5), and (2) becomes n —1
— 1)r (F))—n+2r (21 —
K,(v) = (- 1)n+11„(v) In av
r — 1) !
2r! r =0 n
gv) ,,-Fzr r)! i'.!( n r=0
(5) m=1
m
where In a = —0.11593. A complete solution of (1) when n is integral is
RZ(v) = AI„(v)
BK„(v)
(6)
From (2), (4), and 5.293 (9), we have
I:,(v)K„(v) — K'n (v).1„(v) = thr[V„(jv)J„(jv) — J,:(jr,)Y,‘(jv)] =
(7)
198
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.322
We observe from (4) and 5.293 (10) and from (2) that (iv) = ( —3)n-1-WM 4 (—jv) /n(0 = j—n,In(jv) (2/71-)K.(v) =
(8)
These are the functions tabulated in Jahnke and Emde.
5.321. Recurrence Formulas for Modified Bessel Functions.—Equation 5.294 (1) may be written d,In(jv) = 1140 — -!.-1'J„(jv) `" 3v j dv Substituting jn/n(v) for ,In(jv) and dividing out jn-1give — nvin
In =
(1)
and similarly from 5.294 (2) we get = /.4-1 -}- 71vin
(2)
Subtracting (1) from (2) gives 2n - 271„ =
—
In+1
(3)
Substitution of the recurrence formulas of 5.294 in 5.32 (4) gives
—
— K'n =
(4)
-
(5)
2n
= Kn.o. — 7/1Cn Kn_i —
Kn-Fi
(6)
We use the notation of Watson and of Gray, Mathews, and MacRobert. Some authors omit the (-1)" in 5.32 (5) in which case the recurrence formulas for In and K„ are identical. Two integral formulas that are sometimes useful can be obtained by integrating (1) and (4), as in 5.294. These integrals are fvnIn_i(v) dv = vnIn(v) fvnKn_i (v) dv = —vnIf n (v)
(7) (8)
A similar integration of (2) and (5) gives fv—nIn+i(v) dv = tnnIn(v) f v —nKn+i(v) dv =
(9) (10)
5.322. Values of Modified Bessel Functions at Infinity.—We can obtain the values of /„(v) and K„(v) as v —> co from those for J„(v) in 5.295 by substituting jv for v. The results are good to the same order, i.e., neglecting v-1 compared with v—i. Putting jv for v in 5.295 (6),
§5.324 GREEN'S FUNCTION FOR A HOLLOW CYLINDRICAL RING
199
writing e— },A for j-1, using the exponential form Dw 408.02 or Pc 613, and neglecting the e-v term, we have
I n(v) = j—nJ n(iV)
11evineijnr (-1 27111 j
Evaluating the last term, n being an integer, by Dw 409.04 and 409.05 or Pc 609 we have 1 1 /n(v) ev (1) From 5.32 (8) and 5.295 (8), we have
K,(v)
(7r ) e-v 2v
(2)
Although we have derived these equations assuming n to be an integer, we see, by substitution in 5.32 (1), that they hold for any value of n. 5.323. Integral of a Product of Complex Modified Bessel Functions. In Chap. X, when we calculate the power dissipated by eddy currents, we frequently find it necessary to evaluate the integral of the product of a modified Bessel function R„ °[(jp)ix] by its conjugate complex Rn°[(-jp)ix], where (j)i = 2-i(1 j) and ( -j)i = (1 - j). In cylindrical problems, n is usually an integer, and in spherical problems it is usually half an odd integer. We may evaluate this integral, for any value of n, by using the equations of 5.296. Let us take
k, = -j(jp)i = (-jp)i and k g = j(-jp)i = (jp)I u = R4-Ai0x1 = v = Rnt A - :70x1 = inn( Kf -A:70xl = (- D' int(i0x1 - i0xl = i0x] Substituting these values in the equation preceding 5.296 (1), we obtain, writing x for p, foax.14,[(jp)441??,[(- jp)ix] dx = f xl??,(k,x)M(k,x) dx = raj'73-i[jil??,(k,a)R?: (k ga) - (- j)iRg(k,a)R?: (k,a)] = (2p)-1[kpal??,(k,a)14,_1(k,a) k gaMi(k,a)RL,(kpa)] (1) = (2p)-1[kpar??,_2(k,a)M,_1(k ,a) k ,al??,_2(k ,a)RLi(kpa) - 4(n. - 1).1??,_1(k,a)Mn_1 (k ,a)] 5.324. Green's Function for a Hollow Cylindrical Ring.—As an example of the use of modified Bessel functions, we shall find the potential due to a small charge q at z = c, p = b, and cp = q50 inside the hollow conducting ring whose walls are given by z = 0, z = L, p = d, and p = a where a > d. The special case, when d = 0, gives us the cylindrical box which we have already solved in terms of Bessel functions in 5.299.
THREE—DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.324
200
Since neither /.(v) nor Km(v) has any real roots, it is evident that we shall need a combination of the two to get a function that is zero for a given value of p. Clearly, such a function is
(k, s, t) = lf,„(ks)I.(kt) — .I.(ks)K.(kt)
(1)
Since, in general, this function is zero for only one value t = s of t, we must use different functions for the region near the inner face and that near the outer, but they must have the same value at p = b, where the regions meet. We can now write down, by inspection, two functions that are zero over the conducting surfaces and identical when p = b. These are
d p
p = b,
or CD
-Trzcos m(4) — 40) C'vm,,,R1(917-, a, b)14,1,(124-7, d, p1 sin n
V' =
(2)
n=1 rn=0
b
a
00
p = b,
or
nr
V" =
nrz (nr d, b R„, ° -E, a, p sin TT:cos m(4, — 4)0) (3)
n=1 7n=0
These solutions have the proper symmetry about (1) = 4)0. To determine C,„„, we shall apply Gauss's electric flux theorem to the region surrounding the charge. Consider the charge q concentrated in a small area of the cylinder p = b at z = c, 4, = 4,0. This area is taken too small to measure physically but not a mathematical point, so that the potential and intensity functions are everywhere bounded. By 1.10 (1), the integral over the two faces 82 of this cylinder is —
f
av „ = s(ap
ay") dS = g
(4)
By (2) and (3), we have
ay' _av" = ap
Y..J C„,„1-7r[M„(77-
Op
d, b)fq,:(17- a, b)
n=1 rn=0
— R?„(1, a, b)1V,),:(71--, d, b)] sin nrzcos m(4) — 4'0) With the aid of 5.32 (7), this may be written
aviav" — cap
ap = —
C„, n=1 rn=0
1
nr
Ld
, a sin
nrz
L
cos m(4, — 4,0)
(5)
§5.324 GREEN'S FUNCTION FOR A HOLLOW CYLINDRICAL RING
201
Now let — (bo = 4,', multiply both sides by sin (prz/L) cos qq5' b do' dz, and integrate over the cylinder p = b. On the right, all terms go out except those for which p = n and q = m, by Pc 488 or Dw 858.1 and 858.2. For these terms, the integrals are evaluated by Pc 489 or Dw 858.4, except when q = m = 0. On the left, we notice that, since dS = b dz, we have exactly the integral of (4) if we take the charge so small that cos mo• and sin (nirz/L) have the constant values one and sin (n7rc/L), respectively, and may be taken out of the integral. We then obtain, by (4), if m 0, g sin TT, = -1Cc,„Lrief,',(17-r , d, a) 2 e L L
(6)
If m = 0, the is omitted on the right since f02'610 = 2r. Solving for Cm„ and substituting in (2) and (3), we have " (2 — .5?„)M,,(7, a, b)/??,45, d, p)
V' = —
sinmirc
nrz sin —
L
R;)„(,= 17 7d, a) L
n-1 m=0
cos m(d) —
(ror V" reL LJLJ n-1 rn=0
sin
R°(
L
(7)
p)
d, a)
nrc
L
sin
nrz
L
cos m(4, — 00) (8) where 3?„ = 0 if m 0 and S;,', = 1 if m = 0. For the cylindrical box, solved in terms of Bessel functions in 5.299, we have d = 0 and the above potentials become
V=
P (2— 31)R1(17r, a, b)Im( L)
q
reL tz....1 m=0
rr,n(nra\ L
sin in s
in - sbin
-r-
cos m( 4,
—
4,o) (9)
nrb) (nr 0
a, ". T' P sin nre sin nrz T (nra)
(2 sggm
V" =
-17 (--
n=1
A
771
R
--
cos m(4) — 00) (10) Inspection of V' shows that it satisfies all the boundary conditions, being finite, but not zero, on the axis. If the charge is on the axis, we use (10), dropping the summation with respect to m and setting m = 0.
202
THREE—DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.331
5.33. Modified Bessel Functions of Zero Order.—When n = 0, 5.32 (2) becomes Io(v) = 1 +
V2
V4
24(2o2
4_
V6
(1)
26(31)2 + • •
Thus Io(v) is real but has no real roots and and
/0(0) = 1
(2)
/o( co) = 00
When n = 0, 5.32 (5) becomes v2
Ko(v) = —Io(v) In
22 -V
v4(1 +2 24(21) 2
+
V6 (1 ± +
(3)
26(302
The derivatives of Io(v) and Ko(v) are given by 5.321 (2) and (5) to be A = I,
and
(4)
K'o = —K1
From 5.321 (7) and (8), we have EvIo(v) dv = vIi(v)
(5)
EvK o(v) dv = —vK1(v) + 1
(6)
5.331. Definite Integrals for the Modified Bessel Function of the Second Kind. Value at Infinity.—Substitute R° = f 'e—rs..h dcp in the left side of 5.32 (1), setting n = 0, and it becomes
Jo
cosh2cke—"".ho
1 —— v0
cosh (ke—v cosh
—
e—v.ho
Combining first and third terms by Dw 650.01 or Pc 657 and integrating the result by parts give (sinh 4))[e—v 4.4h d(cosh 0)] = f u dv — sinh cpe_ 60sh m + 1 cosh ci5e—v °°°1'm dy6 v o vo which cancels the second term so that the integral satisfies 5.32 (1). Since any solution of 5.32 (1) must be of the form R° = AI o(v) BK 0(v) and since the integral is zero when v = 00, it cannot contain Io(v) and must be of the form BKo(v). In 5.35, we shall show that to agree with 5.33 (3) we must take B = 1, giving Ko (v)
= f'`°e—a°°ahmd 4
=
(iv) =
We see that v„,dnKo(v)
— ( 1).f yin coshn q5e—v webs' dq5
(1)
203
DEFINITE INTEGRALS
§5.34
is also zero when v = 00 because v appears in the integrand in the form vine-at' and a > 1. From the recurrence formulas 5.321 (5) and (6), we see that K1 = so that K1(00) = 0 Thus we get from 5.321 (6) if„( 00) = Kn_2(00) = • • • = K,(00) = 0 where p = 1 if n is odd and p = 0 if 71, is even. It follows now from 5.321 (4) that Kin (00) = 0, etc. Another definite integral for Ko(v) is Ko(v) = f 'e cos (v sinh q5) cicb
(2)
Prof. H. Bateman proves this relation as follows: Consider the function W = fo e -xec.ho cos (y sinh cP) cict, where x = r cos 0 and y = r sin O.
Differentiate,
aw_ aw aw ao = x ay — y ax
=
GO
— fo
e—xe°'h o[x sinh 0 sin (y sinh q5) — y cosh cb cos (y sinh
1
dcl, C
= .10'°.-,--ctje-z ...h 0sin (y sinh OA di() = — e-xc.°1' 0 sin (y sinh ck) 0
=0
Therefore W is independent of O. Putting 0 = 0 gives x = r, y = 0 so that W = Ko(r) by (1), and putting 0 = jr gives x = 0, y = r so that W is the integral of (2). Thus (2) is proved. To get an integral useful in slit diffraction theory, we insert (1) in 5.33 (6), replace the 0 to 00 integral by half the — 00 to 00 integral, and interchange the integration order. This gives, by Pc 402 or Dw 567.1, 2vKi(v) = rviii2)(—jv) = f
e9 OOS"(1
v cosh 0) sech2 ¢d4)
(3)
5.34. Definite Integrals for Bessel Functions of Zero Order.—Substituting v/j for v in 5.331 (1), we have K 0 0 = f eiv "th dcA = f cos (v cosh 4)) dc/)
ji sin (v cosh cp) dcp
Making the same substitution in 5.32 (4) and equating real and imaginary parts gives for Jo(v) and Yo(v), if we use Watson's notation, see 5.293, Jo(v) = 2f - x'sin (v cosh (k)
rr co
7 0
Yo(v) =
7
0
cos (v cosh
(1)
¢))
(2)
204
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.351
5.35. Inverse Distance in Terms of Modified Bessel Functions.—We now derive an expression for the inverse distance between two points of which the cylindrical coordinates are po, 4,o, zo and p, (45, z. This could be done by the method of 5.304, but it is shorter to start with the expression for the potential of a point charge midway between two plates at a distance L apart. To obtain this potential, let a = 00 in 5.324 (9) and shift the origin to L/2. Only odd values of n survive so that, when p < po, (2- 31)Km(NP0)Im(Np) cos Nz cos Nzo cos m(cb - 4,o) (1)
V =-1!-fa n=0 m=0
where N = (2n + 1)71- /L. this becomes (2
7.2e
—
Now let 2nr/L = un and 2,7r/L = Au, and
451) cos m(0 - 0o)
mo
n=0
K„,[
-T Alpo]
im[(Un t--1,9] cos
Run +T Au)z] Au
Next let L co so that Au -> 0, and from the basic definition of an integral the n-summation, 0 < n < 0 , becomes the u-integral, 0 < u < co . Thus we have V-
q 4v€R 47T-E[z2
p2+ PO
2PPo cos (4) — OOP
Km(upo)/.(up) cos uz du] cos m(¢) - 4,o) (2) —q 2, (2 = 27. m-o If p > po interchange p and po. At po = 0 the summation disappears so that r-1= (p2
z 2)-i =
f: cos kzKo(kp) dk
(3)
With z and 40 zero, write R for p in (3) and compare with (2). Thus we get when p < po CO
(2 - 32.K.(p0)1.(p) cos mcp (4) m-o To verify the choice of constant in 5.331 (1) substitute it for Ko(kp) in (3), integrate first with respect to k using Dw 577.2 or Pc 415, and then with respect to q> using Dw 120.01 or Pc 99, and so get the left side. 5.351. Cylindrical Dielectric Boundaries.—The integral for the reciprocal distance just derived may be used to solve problems involving cylindrical boundaries when the fields extend to infinity. In such cases, Ko[(p2
pg - 2ppo cos 95)i] =
§5.351
CYLINDRICAL DIELECTRIC BOUNDARIES
205
expansion in a series of Bessel functions is not feasible since the integrals of 5.297 are infinite at a = co as can be seen by inspection of the asymptotic values in 5.295. In such cases, we can usually obtain the solution as a definite integral which can be evaluated numerically by graphical integration.
5.351.—Equipotentials o point charge at center of cylindrical hole in dielectric for which K = 5. Calculated from 5.351 (1) and (2) by Dr. A. E. Harrison.
As a specific example, let us calculate the field of a point charge q placed on the axis of an infinite cylindrical hole of radius a in an infinite block of dielectric of coefficient K. Let V be the potential in the hole and VK that outside. We use a method similar to that used in 5.303. We consider V to consist of two parts: first, that part due to the charge alone, given by 5.35 (3) and second, that due to the polarization of the dielectric, which must be finite on the axis. Thus V is of the form V = 2,2,v q 0
0 (kp)
*(k)Io(kp)] cos kz dk
(1)
In the dielectric the potential must be finite at infinity and can have only the form
206
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS VK =
q 0 24r2ev
§5.36
(2)
(13(10 K 0(k p) cos kz dk
The boundary condition V = V K when p = a requires that
K (k a) + (k) I (k a) = c13(k)K (k a)
(3)
The boundary condition OV/ap = KOV E/ap at p = a gives
Kaka)
(4)
*(k)I0(ka) = K
Eliminating T(k) from (3) and (4) and simplifying by the use of 5.32 (7), we obtain 1 (k) — (5) 1 — ka(K — 1)Io(ka)Kaka) Substituting (5) in (3) and solving for '(k) give ka(K — 1)Ko(ka)K,;(ka)
(k) =
1 — ka(K
(6)
1)I o(ka)Kaka)
The field when K = 5, obtained by plotting the integrands in (1) and (2) and integrating with a planimeter, is shown in Fig. 5.351. 5.36. Potential inside Hollow Cylindrical Ring.—The most important electrical applications of the modified Bessel functions occur in connection with alternating currents in cylindrical conductors. There are, however, a few applications in electrostatics. Let us, as an example, consider the potential in the region bounded by p = a, p = b, z = 0 and z = c. Let the potential be zero on all these boundaries except the first. When p = a, let V = f(z). Let us expand f(z) as a Fourier's series in the interval 0 < z < c. Since f(z) = 0 when z = 0 or z = c, we have the well-known expansion, Pc 815, Ak COS (
f(z) =
kz)
,
k=1
Ak =
2f'
f(z)
c os erz .
sin
dZ
(1)
Consulting 5.292 (2), 5.292 (3), and 5.32 (6), we see that a solution of Laplace's equation satisfying all boundary conditions is
coserllo(krp/c) Ko(krp/c)1 I o(krb/c) Ko(krb/c) j sin c (2) V= Ko(kira/c) Io(kralc) k=1 I o(krbl c) Ko(krb/c) If we wish the potential inside the cylinder p = a, on which the potential is V = f(z), and between the two earthed planes z = 0 and z = c, this expression becomes .,
A k sin k=1
k z Io (krp/c) c I o(kra/ c)
(3)
§5.37
MODIFIED BESSEL FUNCTIONS OF NONINTEGRAL ORDER
207
If we wish the potential outside the cylinder p = a, on which the potential is V = f(z), and between the two earthed planes z = 0 and z = c, we have Ak sin
V=
k z Ko(krp/c) c Ko(kira/c)
(4)
k-
5.37. Modified Bessel Functions of Nonintegral Order.—From 5.31 (1), putting in(v) = j-nJa(jv), we see that a general solution of Bessel's modified equation, when n is not an integer, is given by kii(v) = AI n(v) BI-n(v) (1) Kn(v) is a linear combination of /n(v) and /,(v). The recurrence formulas are equally valid for these values. The Eqs. 5.31 (2), (3), (4), (5), and (6) lead to ( 2y sinh v 1.4 (v) = Iry
2y - cosh v Iry 2)1 1 . 1.1(v) = — (cosh v - -sinh v v (Iry A 1 .1-f (V) sinh v - - cosh v) 3 )1 3) . (v) (. [(1 ± - sinh v- -cosh v] Iry v2 v 3 . 2) 3 I- (V) (- [(1 - cosh v - - sinh 2)] 72) v) v The general formula is simplest in exponential form I-1 (V) =
(
C
1 -r± (n+i ) (v) — (27r2))1
[( —1)sev T- — 1)ne-.1(n s)!(2v). s!(n
s)!
(8)
=o where the upper or lower sign is used throughout. Both /„±i(v) and /,_i(v) are infinite when v = 00. A solution that is zero at co is K,+I(v) as defined in 5.32 (4) when n replaces n. By 5.31 (8), eliminate Y n±i( jv) then write in terms of /±(n+i)(v) by 5.32 (2) and substitute (8). The positive exponential disappears and leaves Kn+I(V) =
1)n [I
\
— Ind-i(11)] = VI);■ •J S
=o
! (nn
+S ;e(2vS)3 (9) )
)
The simplest cases are K3 (v) = GT) e-v
r y ( 1 K I (v) = (- 1 + 2v
(10)
208
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.38
If v = 0 in (9), the s = n term of the summation dominates so that Kn+i(v)
y(2n —n 1)!! v 0 2v
(11)
The spherical Bessel function kn(v) may be defined by n
2
kn(v) = (.7 ) ) Kn±i(v) = s
=o
s)!e—v (n s!(n — s)!(2v)8
(12)
With the aid of 5.32 (4) and 5.31 (8), we obtain (13)
kn(jv) = — j—n[jn(v) — jnn(v)]
From this equation and 5.31 (11) kn(jv)j'i(v) — jk'n (jv)jn(v) = — j— n±lv-2
(14)
The relations with Schelkunoff's g„(v) and Stratton's ki)(v) are kn(v) = v--'1?-„(v) = —jnV)(jv) = —jaljn(jv)
jnn( jv)]
(15)
From (12) and 5.321, the recurrence relations are — (2n + 1)v—lkn = kn_i — kn+1,
— kn = k„T 1 ± (n
4±
i)v—lkn (16)
5.38. Wedge Functions.—The functions of 5.291 and 5.292 are inadequate when the potential vanishes on the cylinders p = pl, p = p2 and on the planes z = z1, z = z2 and has arbitrary values over the planes 4, = 01, = 4) 2 . This problem requires functions orthogonal in p and z obtained by substitution of jv for n and jk for k in 5.291 (1), (2), and (3). Thus, 43(14) = C cosh vct) D sinh vck = C' evo Z (kz) = E sin kz F cos kz R9 (kp) = AF „(kp)
D'e—
BG,,(kp)
(1)
(2) (3)
where R8 (v) satisfies the differential equation
d2R ± 1 dR dz2
The function F ,.(v)
.0 as v
F „(v) = csch
The function G8(v)
v dv
v2 v2
(1 — — )R = 0
(4)
co and is defined by
f:58cc's Bcosh P 0 dO — f
e—v cosh 8 sin vt dt
(5)
0 as v --> co and is defined by G8 (v) = fo'e—v cc.sh cos vt dt = K;8(v)
(6)
Below a certain value of v, which increases with v, both functions are periodic and oscillate as sin [v ln (iv)] and cos [v ln (iv)] as v approaches
209
MIXED BOUNDARIES
§5.39
zero. The integral with respect to v of v—' times a product vanishes unless the factors have the same v so that an arbitrary function of v suitable for expansion in Fourier series can be expanded in terms of R;(kp). The three-dimensional dielectric wedge problem is solved by methods similar to those used for the two-dimensional wedge in Art. 4.07. 5.39. Mixed Boundaries. Charged Right Circular Cylinder. Many three-dimensional boundaries met in practice do not fit any separable coordinate system; thus the methods so far used are not applicable. For example, if a charged circular disk lies inside a coaxial circular cylinder, a system of coordinates fitting the boundaries can be generated by rotating the last part of Fig. 4.22b about the y-axis. Laplace's equation, 4.11 (6), in this system cannot be separated, and the surfaces, from 5.00 (2), are not equipotentials. Clearly the boundaries fit into the cylindrical system, but at y = 0, V is a constant only over the disk while between the disk and the cylinder a V/ay is zero. Thus this becomes a mixed boundary value problem. It can be solved (see Art. 5.42). There are many cases where V or a V/an is known on part of a coordinate surface and nothing over the remainder. One of these is the charged solid right circular cylinder. A possible approach to this is to express the charge density as a series having arbitrary constants adjustable by 3.15 (3) to give a uniform potential inside the cylinder. When magnified the edge looks like a right-angle wedge for which 4.18 (1) gives dz/dW = WI so that W = zi and dW/dz = z-1. Thus the series will diverge at the edge and converge slowly elsewhere unless the infinity appears explicitly in a. A series of Gegenbauer or Jacobi polynomials with the proper weighting factor has some advantages but complicates integrations and numerical work. If the cylinder is bounded by z = ±b and p = a, then simple forms for the charge densities on the end 0-, and on the side a3 are —
N
M
A n(1 —
Qe =
b-2.Z2) 71-1
(re =
n=0
B„,(1 — a-2p2)m--1
(1
)
M=0
Note that at the edge where z = b and p = a only the Ao and Bo terms survive. Thus the matching of cre and a, there requires that (2)
IAA o = alBo
There are two ways in which 3.15 (3) may be used to get An and B„,. One is to calculate 32PV(z,0)/8z2P on the axis, set z = 0, and require that [a2PV(z,0) 4:92pz
= z=0
Op V 0
(3)
210
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.39
where bop = 0 if p 0 and 150„ = 1 if p = 0. This and (2) yield p ± 2 equations which may be solved for N M + 2 of the low-order terms in (1). Alternatively the potential in the plane z = 0 due to (1) may be expanded in even powers 2p of p and coefficients of all but the p = 0 term equated to zero. For a given b/a ratio the number of significant digits carried fixes the best values of M and N, which are those that give potentials nearest Vo at pole and equator where maximum deviations occur. Fewer a coefficients can be found for the more remote point than for the nearer one. If too big values are chosen for M and N, large terms alternating in sign appear, which reduce the number of significant digits in the sum. All the values of An and B„, found from (1) must be used in subsequent calculations as omission of one may change its neighbors so they are no longer correct. If the polynomials already mentioned replace (1) (see Taylor, J. Research, Natl. Bur. Std., Vol. 64B, pages 142 and 143), An and B,,„ decrease rapidly as n and m increase, so that dropping the last terms has little effect on the rest. The chief labor is finding the derivatives in (3). The axial values of the expressions for the potential of a charged ring in the text and problems of this chapter must be integrated over the charge distribution of (1). It is usually necessary to expand some of the functions in series in order to get fractional integrals of the form tabulated in Vol. II, Chap. XIII, of "Integral Transforms" (see bibliography). A hypergeometric series usually results. By summing these and using recursion formulas, the simultaneous equations of (3) are set up and then solved for A„ and B,„. Numerical values of b/a have been calculated (J. Appl. Phys., Vol. 33, page 2966) for 0.125, 0.25, 0.50, 1, 2, 4, 8, which show that for the extreme values 0.125 and 8 only one coefficient can be found for the more remote surface. The coefficients for b = a appear in the table below. Clearly for practical purposes the solution is exact when b = a, as the potentials V„ and V, and the surface deviations Ab and Da at pole and equator show. Charge density coefficients for unit potential when b = a.
A0 = Al = A2 =
A5 = A4 =
A5 = A5 = Be, = B, = By
=
0.55941519 0.24032463 —0.46123818 0.71795706 —0.67534061 0.34357563 —0.07271528 0.55941519 —0.35462716 1.4624910
= —3.9209295 = 6.2012014 B5 = —5.6962826 B6 = 2.8184609 B7 = —0.5816914 B3 B4
V9 = 1.0000002 V. = 1.0000001 Surface deviations Lb = —0.0000004b Act = —0.0000002a
The potential outside the sphere passing through the cylinder edges is found by the second integral in 1.06 (6) to be
211
CYLINDRICAL BOUNDARIES
§5.40
V=
aVo 26
s=0 p=0
( —1)P(Ce CS) (2p± 2s) !a2Pb23 P2s+2p(COS 0) 4P(p!) 2(2s)!r 28+2p+1
(4)
Ce = aB(p + 1, m ± -3-)B,„,
4)A n (5) Cs = bB(s +, n n =0 where B(x,y) is a beta function. Note that at a great distance (4) takes the form V = Q/ (4rer). The capacitance is then Q/V o. We get an empirical interpolation formula for C by adding a correction term containing two adjustable constants to the exact known capacitance of a circular disk given in 5.03 (2). This curve, which passes through the seven calculated points with an error of 0.2 per cent or less, is 7.76] farad (6) C = [0.708 0.615( X 10-10a m=o
a
It is clear that this method applies to other geometrical forms such as the spindle formed by two finite circular cones fitted base to base. It also works when such objects are placed in uniform fields as seen in the foregoing Taylor reference. 5.40. Mixed Spherical, Spheroidal, and Cylindrical Boundaries. Often different boundaries of an electric field belong to different coordinate systems. No method given so far handles this case. The formulas needed to relate the four systems, oblate, prolate, spherical, and cylindrical, naturally split into two groups: those relating the exterior harmonics, which satisfy Laplace's equation and vanish at infinity, and those relating the interior harmonics, solutions that are finite at the origin. The former depend on a set of Fourier transforms due to J. C. Cooke (see Monatshefte fur Math. Vol. 60, page 322, 1956; and Zeit. fur Math. and Mech. Vol. 42, page 305, 1962). These transforms have been modified to fit the notation of this chapter. Note that Hobson, HTF, HMF, and others insert (-1)m in the right side of 5.23 (6) and (7), so in their notation the sign of m in the exponent of j in the following formulas should be reversed. To shorten the formulas Rmn is written for (n — m) !/(n m)!. —
10 rilf„,(tp)In+I(tc) cos (tz) dt = j'-m(27rc-')IR,„„Q7(j)P7( ) 10 't--11C„,(tp)/7,±i(tc) sin (tz) dt = j 2-'n(i7.c-91RmnQ7,','(,g)P7(E)
(2)
1:
tnK„.(tp) cos (tz) dt = j"-ri(47)(n — m)!r-71-1137(cos 0)
(3)
10
talc„,(tp) sin (tz) dt = jn-'n-i(ir)(n — m)!r-n-iP7(cos 0)
(4)
fo-t —IK„.(tp)J.+1(tc) cos (tz) dt = jn+m(firc-1) 11?..Q7(n)PT() Jo y
(1)
t-ilc„,(tp)J,,+I(tc) sin (tz) dt =
(5) (6)
212
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.40
m is odd, if m is even and in (2), (4), and (6) n In (1), (3), and (5) n —1 < < +1 and 0 < < 00 in (1) and (2), but the reverse is true when 0 < < 1 and — 00 < < 00. The interfocal distance in the spheroids is 2c as in Arts. 5.27 to 5.281. Additional formulas needed for combining spherical and spheroidal forms are found by substitution of the power series 5.32 (1) and 5.293 (2) for /„±i(t) and J.±1(t) in (1) and (5) or (2) and (6), giving the integrals the form of (3) or (4). Thus CO
\ 28+1
(2'nn(i0P7(E)
)
=
P'2ns(cos 0)
(7)
s= in 28+1
(8)
P;s(cos
QZ(n)P„(0 = s= in B
( —1)"(2s — m)!(n m)! —
(it + 2s+ 1)!!(n — m)!(2s — n)!!
(9)
In (3) or (4), in may be written as a Neumann series involving in+I(t) and J.+#(t) by Watson (1), page 138, and HMF 9.1.87, page 364. The resultant integrals are those in (1) and (5) or (2) and (6). Thus n+1
P',,'"(cos 0) =(-1)"'"1-8 jC.Q;n8(jilP;n8 a)
r
(10)
s=in CO
= a .r n
Cmns—
—1)nCnoisQ2s(n)P78(t)
(2s + 1)(n + 2s — 1)!!(2s — m)! m)! (n — m) !(2s — n)!!(2s
(11)
(12)
From these a set of formulas for combining oblate and prolate boundaries may be derived. Let the interfocal distances and the E coordinate in the oblate case be 2c1and Eland in the prolate case 2c2 and E2. In (7) write ci for c and Eifor E, and in (11), (c2/ci)n±1 (c1/r)n+1for (c/r)n-", p for s, and then 2s for n. Elimination of spherical harmonics gives \ 28+1
27(3.0Ptnn(1) =
—1)m+P+ni-1emnspei)
(
8
Cmnsp
—
(2772(0) P 273(6)
(13)
= in P=8
(n
(2p + 1)(n 2p — 1) !!(2p — m)! m)!(2s 2s + 1)!!(n — m)!(2s — n)!!(2p — 2s)!!(2p
m)! (14)
§5.40
213
CYLINDRICAL BOUNDARIES
A similar operation on (8) and (10) gives 28+1
(21.:(77)P7(E2) =
Q'2;(3.0P'2n,(Ei)
(15)
8 = in P
In the same way for nonconfocal coaxial spheroids of the same type 23+1 1) n+'+Pemnsp0 V2np(A-2) P''p(E2)
QZUNFnnal) =
(16)
s= in P=s 23+1
(711)PV El) =
Cinnsp(7 C1)
V2np(n2)P7p(E2)
(17)
s=in P=s
The interior harmonics, like (1) to (6), depend on Bessel function formulas due to Cooke. In Hobson's notation the right sides of (18) to (23) should be multiplied by (-1)m. The relations are CO
\4- 1.1-m(2n+ 1)(n — ViC) (n m)! 7r
cos (tz)I,,(tp)
jn+I(cOPT(Z)P'nn(i0 (18)
nm
=
.ji'±nv(tr)n[(n
m)!]-1P7(cos 0)
(19)
n=m
l) (n—
( —1)m(2nn -F 2rtc = (—) n=m
( -F m) !
m)!
Jn±i(ct)PZ(OPT(n) (20)
sin (tz)/„,(tp)
(—j)m-E1(2n 1)(n — m)!
611
(n
m)!
n=m+1 in+1(c0P7(
(tr)"[(n
m)!]-1PZ(cos 0)
OPT, (3.0
(21) (22)
n=m+1
r Vtc
LJ
— 1)'n(2n + 1)(n — ) (n m) ! m !Jn+I(ct)P7(E)P7(n)
n=m+1
(23) In (18), (19), and (20) m n is even, and in (21), (22), and (23) m n is odd. The interior equations analogous to (7) to (12) are now polynomials instead of infinite series. They may be derived by writing out
214
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.40
the left sides in products of powers of its coordinates, then changing over to the coordinates of the right side by 5.27 (4) and (5) or 5.28 (4) and (5) and regrouping to form harmonics. Alternatively write in±i(ct) and Jn+1(ct) in (18) and (20) or in (21) and (23) as power series in ct by 5.32 (2) and 5.293 (3), replace (On by (r/c)"(tc)' in (19), and equate coefficients of (tc)n. Thus we get Tin
2a c) PZ(cos 0) (71.
P7.(E)Pinn(i0 =
(24)
s S
in 28
— 1) in—aDnins() P;(COS 0)
PZMPZ(n) =
(25)
s=S
Dmns —
m)! (n 2s — 1)!!(n (2s + m)!(n — m)!(n — 2s)!!
(26)
in
- PT(cos 0) (cY r
:7--nG.n8P2. ()P;(jr)
(27)
(
(28)
8=8
in
=
) P78(17) 1)1n—sGmn.P72n3(t
3=8 limns
(2s
(4s + 1)(2s — m)!(n m)! m)!(n + 2s + 1)!!(n — 2s)!!
P9)
Here S equals im or 4 (m + 1) and is an integer if n is even and an integer plus if n is odd, and s increases by unit steps. Equations (30) to (33) derive from (24) to (29) in the same way that (13) to (15) derive from (7) to (12). Thus in
\28
P7(3.0P7`(E1) =
1 ) 8—PFninsP ci) P2p(n)P2p(E2)
(30)
s=S p=P in
3
28
P7(n)P7(Ei) =
(-1)InF,anspC-1 (
8=8 p=p in
Cl
)
P2np(ii-)P7p(E2)
\28 .in( -1)8Fmnsp() ci P2np(.7?.2)P79(E2)
P'„n(in)PZ(Ei) =
(31)
(32)
s=S p=P
in
3
P',7(n)P7(E1) =
/ 1 ) 1n—PF mn8 P
2s Cl C1
P2np(712) P7p(E2)
(33)
s=8 p=P
F„,s, —
(4p + 1)(n + 2s — 1)!!(2p — m)!(n m)! (2s + 2p + 1) !!(2p m)!(n — m)!(n — 2s) !!(2s — 2p) !! (34)
§5.41 SPHERE IN CONCENTRIC, COAXIAL, EARTHED SPHEROID
215
Here S and P equal km or 4(m + 1) and are integer if n is even and integer plus 2if n is odd and s and p increase by unit steps. There are two more formulas that will prove useful in working problems. The first is IT II (6), page 29, and the second is derivable from HTF (25), page 159, by expansion of the integrand binomial and integration of the real part. In both P'.(cos 0) has been multiplied by ( —1).8 to change it from Hobson's notation to that of 5.23 (6) and (10). Thus
:e-gq,a(tp)tn dt = (n — m)!r-.-43,7,i(cos 0) I ( c)
(-1)i(8--no (n — s)!(s m)!!(s — m)!
P',7(cos 0) = (n m)!(.c)n 8
(35) P)
(36)
= In
where S is n or n — 1 and s m is even so s jumps by two in successive terms of the summation. 5.41. Sphere in Concentric, Coaxial, Earthed Spheroid. Suppose that the potential inside and on the surface of a sphere of radius a is —
)2n
P2Th(cos 0)
V =
(1)
n=0
and that outside it there is a concentric coaxial spheroid 1 o (or no) with interfocal distance 2c, at potential zero. It is desired to find the potential between them and the charge densities. Let the interior and exterior potentials of the charge distribution on the sphere in the presence of the spheroid be, using 5.40 (27) and 5.40 (10) for the oblate case, N
An(-)2n P 2.(cos f
Vin =
a
)
n=0
N
n
( — ).(4s± 1) ( 2n) !(a/ c 2s 1) ! ! (2n — 2 s)2)1;!P2s(k)P28q)
An ( 2n 1 ±
(2)
n=0 8 =0
=
N
ayn+1 P 2n(COS
V ex n=0
N =
+ 1)(2n + 2s — 1)!!(c)2 n+1 ( — 1)8(2n) !(2s — 2n) !!
a
3) Q28 00P28(t) (
n=0 s= n
In the prolate case 5.40 (28) and 5.40 (11) are used. The potential Vin of the charge density on the spheroid 1 = N (or n = no) must exactly cancel Vex there and thus is given by (3) when —Q28(.go)P2800/P28(..go)
216
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.41
replaces Q28(j0. Thus N
Vin
—n)11), !!(02''+ip )n (2)n1(2Fs 2s2 .(3 0 ) p2s(A.)p2s(E) Q22:( A„i((2s±os1(2 -
= n -= 0 s = n
(4)
V:„ must equal The sum of the coefficients of P2s(j0P2(E) due to V,„„ its original coefficient as given by (1), which is (2) with B. written for An. Thus we have N 1 simultaneous equations, one for each value of s from 0 to N inclusive, to be solved for A 0 , • • • , AN, which are 0
A0000 + A iCoi + _2 A C_ 02 + • • + ANC6N = 7 . BnCOn
1
A 0C +
om n=0
A2C12 + • • • + ANCON =
B.0
(5)
n =1
tAT
A CN ± A1CN1
A2CN2 ± • • • ± ANCIVN —
BNCNN
It is evident from (2) and (4) that when s < n only the sphere's charge, and when s n only the spheroidal charge, contributes to the coefficients of P2e(i0P2.(E) so that (ays 1)(2n)! ( —1)n(4s (6) Csn = (2n + 2s +1)!!(2n — 2s)!! c (2n)! a )2n + (2n + 1)(4n — 1) !!(c)2n+1(22.(A-0) — 1) "[ (2n — i)! 2c j(2n)!! a P2n(3To) (7) c\2n+1 1)(2n ± 2s — !!(22sW-0) j(2s ± s n (8) ( —1)8±1(2n)!(2s — 2n)!!P2s go) ct) s
Cnn
=
(
For the prolate case write (-1)s for (— 1)nin (6) and —1 for j in (7), omit j and write (-1)" for (-1)s+1in (8), and replace ji-o by no in (7) and (8). The potential between sphere and spheroid, V ex +Vin, is N AnCsn(-0 2".+1[Q P22:(( j °0)Q28(i0
-
P2.(i01132s(E)
(9)
n=0 s = n
Use of the Wronskian, 5.214 (5), will simplify the charge density formula for the spheroid. For the sphere (9) should be written in spherical coordinates by 5.40 (7) and (24) before the differentiation. Where the sphere is an equipotential, Bo = Vo and B1, B2, • • BN are zero; therefore the right sides of all but the first equation in (5) are
CHARGED SPHERE IN CYLINDER
§5.42
217
zero. The charge on the sphere is the coefficient of 1/r in (3) multiplied by 47-E, so the capacitance between sphere and spheroid is C
Q Vo
= =
zlreA oa Vo
(10)
The Cpn factors are very easy to calculate and arrange by a digital computer if recursion formulas, 5.233, are used for the Legendre functions. If the capacitance is wanted to five significant figures, then N should be taken just large enough so that using N 1 instead of N equations in (5) leaves the fifth digit in C unchanged. If the sphere with the fixed potential distribution is outside the spheroid, exactly similar steps are taken starting with N 2n-F1
n) ./37,('="
V =
P2,(cos 0)
n=0
valid on and outside the sphere but with the roles of internal and external potentials reversed. In the more general case V will depend on r, 0, and 95, so double summations involving Amn coefficients will occur. If the spheroids are concentric but the axis of the fixed potential on one spheroid does not coincide with that of the second spheroid, it may be made to do so by writing it in spherical coordinates and using 5.24. 5.42. Charged Sphere in Cylinder.—As another application of the formulas of 5.40, consider a sphere of radius c charged to potential Vo, which lies on the axis of an infinite cylinder of radius a at zero potential. The internal and external potentials of the charge on the sphere are, using 5.40 (3), 2n An(7-') P 2n(COS 0)
Vin =
(1)
n=0
N Vex =
LJ
)2n+1 An( P
2,,(cos 0)
n =o
2 = —
( —1)71A nc2n+If t2'1(.0(tp) cos (tz) dt
(2)
n=0
The induced charges on the cylinder must exactly cancel Ve x at p = a, so that its internal potential V',„ is formed by replacing Ko(tp) by —Io(tp)Ko(ta)/Io(ta) in (2). Elimination of I o(tp) cos (tz) by 5.40 (19),
218
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.43
where n is replaced by 2s and t by u/a, gives N V in
2s)An( cV n-1-23+1028 ( —1)n+8-"/(2n P 2s(COS (2n + 2s + 1)(2s)! a) c)
= n=o
'u2n4-28
I(2n
(3)
s=0
2s) =
o
du
/g(u)
(2n + 2s + 1) 50
u2.+28K 0( "du
(4)
/o(u)
The third form in (4) was integrated by parts to give the second. A table of numerical values of I(2n 2s) is given in J. Math. Phys., Vol. 4, page 834, 1963. These are needed in all cylindrical applications of the formulas of 5.40. Inside the sphere -17. 7, Vin must equal Vo, so that the coefficient of (r/c)2nP2„(cos 0) appearing in (1) and (3) must vanish if n 0 and be Vo when n = 0. Thus there is a set of simultaneous equations like 5.41 (5) to be solved where Bo = Vo and B. = 0 if n 0. From (1) and (3) n
2p)(c) 2n-1-2p+1 71-(2n + 2p + 1)(2n) ! a 2/(4n) (c)4n-1-1 Cnn = 1 7(4n + 1) (2n) ! a
p C. =
n=p
2( —1).+P±1/(2n
(5) (6)
The potential between sphere and cylinder is Vex VL. An idea of the number of coefficients N needed for different values of the radii of the sphere (c) and of the cylinder (a) can be found by looking at the data taken from the reference cited, where a comparison of (1) and (2) of this section with (3) and (4) of the reference shows that An = (An)rei/(4n + 1). The potential is taken to be 1. c/a
AO
Al
A2
A3
A4
A5
0.1 0.3 0.5
1.095373 1.353616 1.773104
—0.000226 —0.007532 —0.046118
0.000001 0.000181 0.002161
—0.000004 —0.000216
0.000014
—0.000001
In the reference cited numerical values are given for a wide range of c/a and for prolate and oblate spheroids including the disk. The reference formulas could be somewhat simplified by using more relations from 5.40, some of which were not available when it appeared. As in the last article the capacitance is Q 4reA oc C = — (7) V Vo 5.43. Cylinder in Hole in Sheet. The formulas of 5.40 also apply to the case of an infinite circular conducting cylinder of radius a at potential —
CYLINDER IN HOLE IN SHEET
§5.43
219
Vo which passes through the center of a circular hole of radius c in an infinite thin conducting sheet normal to its axis as shown in Fig. 5.43a. With no cylinder the problem is solved in 5.272 using oblate spheroidal harmonics where — < < co and 0 < < 1. In the present case with a cylindrical boundary, 5.40 (1) applies, but now Pn(E) is always even as regards z and Qn(j0 is even only when n is odd from 5.213 (3). Thus a solution of Laplace's equation giving V = 0 on the plane outside the hole is V =
(1)
.4,.(22n-F1(3T)P 2n+1(E) n=0
The problem is to choose Anso that the cylinder p = a is at potential Vo. An equation derivable from 5.40 (1) facilitates the analysis. Let (la)
L LJ 2c (a)
(b)
FIG. 5.43a, b.
designate 5.40 (1) or 5.40 (2) where n + 2 replaces n. Differentiate (la) with respect to c, and eliminate /„/±1by 5.321 (1) and then one integral by (la) to get Eq. (lb). Differentiate this with respect to c, and eliminate ./..„'+1by 5.321 (1) and then one integral by (lb). Note that the negative of the final integral is that produced by differentiation of 5.40 (1) or (2) twice with respect to z. Thus
a2[(2 n(Mipt`MI = — c-2n(n az2
2)Qn+2(i0P.+2(E)
(2)
Evidently repetitions of this operation will yield ,92p[Qnuopn()1 "'
az2"
_ ( 1)P+1
+ 2p — 2) !!(n 2p)!!Q ±2pt , pn+2,(E) (n, — 2)!!n!! C 2" n v (3)
The boundary condition applied to (1) is that V = Vo and a2PV/oz2P = 0 when z = 0 and p = a or when = 0 and c = (c2— a2)i. Equation 1 simultaneous equations, one 5.214 (4) gives 0 2n+I(j0), so that the N
220
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.44
for each value of p from 0 to N inclusive, to be solved for A n are N
2p)!!(2n + 2p - 1)!! A( - 1)"(2n P2n+2,±1[(c2 - a2)1c-1] (2n + 1)!!(2n - 1)!!
VoSop n=0
(4) where Sop is zero unless p = 0. The larger the value of c/a the smaller the N needed in (1) for a given accuracy. The equipotentials of (1) will somewhat resemble hyperboloids of one sheet. If the solution of the problem of a cylinder of radius a passing axially through the hole in the hyperboloid = Eo is desired as in Fig. 5.43b, then (1) is replaced by N
V
=
An(22n+I(i0[P2n+ia)
Q2.4-1(E)P2.+1(E0)1 (22.+1(E0)
(5)
n=0
The manipulation is similar to that given except that another formula, which appears on page 326 of the first Cooke reference of 5.40, is used for the new term. It is like 5.40 (1) but with P„m(E) replaced by - (2/71-)QT(E). 5.44. Off-axis Charged Sphere in Cylinder.—In Art. 5.40 formulas for the solution of mixed coordinate problems where the two boundaries are concentric but not necessarily coaxial were given. These are inadequate if the surfaces are not concentric, so a different approach is needed. One is to express the external fields in terms of their multipole sources. A good example of this method is the solution for the potential between a cylinder at potential zero and an off-axis sphere at potential Vo inside it. Two coordinate systems are needed. In the first p, ¢, z system the cylinder wall lies at p = b. The center of the sphere at z = 0, p = a, 1,t. = 0 is the origin of the second system r, 9, 0 whose 9 = 0 axis parallels the z-axis and whose = 0 plane lies in the 11, = 0 plane. The potential of the sphere of radius c due to its own charge density in the presence of the cylinder is Vs =
il„„,r--'-IP'„"(cos 9) cos mcp
r> c
n=0 tn=0 n
=
A„„re-2"—'P'(COS 0) cos m4
r< c
(1)
n=0 m=0
This must have even symmetry in both z and cp. The cos mcb term takes care of qs and the choice of even m n takes care of z. The potential of the induced charge on the cylinder must be added to this and if the sphere
221
OFF-AXIS CHARGED SPHERE IN CYLINDER
§5.44
is conducting the sum must have the constant value Vo when r < c. The steps in the solution are: 1. Write the external harmonics r-n-1/37, (cos 0) cos m4) in terms of x-, y-, z-derivatives of r-l. 2. Express the reciprocal distance r--' in terms of p, z, and obtain by differentiation the potentials of the multipoles of the system. This gives spherical harmonics r-n-1P„(cos 0) cos mck in terms of p, z. 3. Write down the interior potential V, of the induced charge density on the cylinder p = b by noting that V3must cancel Vs there. 4. Express this V, in terms of r, 0, O. 5. Add Vs when r < c to V3, and equate the coefficients of the sum of the harmonics containing rk to Vo when k = 0 and to zero when k 0. 6. Solve the resultant linear simultaneous algebraic equations for A„,„. 7. Put these values into Vs with r > c plus V 3. The sum is V. The multipole expansion for r-n-1P„(cos 0) cos m4 is given in complex form in HTF II (26), page 251, or Magnus 10.3.45. In the notation of 5.23 (6) the real part becomes, when 02(r-1)/8y2 is replaced by -32(r-1)/ ax2 -32(7.-1)/az', for field point insert ( - 1)non right side of (2) and (3). S
r—n-1P,n,(cos
cos mg) = s =0
(2
61),n_1ni ( n
s
1)!
(n - m) !4$(m - 2s) !s!
an(r1) axe 23 az'o'—'"-28
(2) Similarly the imaginary part becomes r-n-lP„m(cos 0) sin m(t) S
2m-1(m s - 1)! 28-1 azg—m-1-2s (n - m)!48(m - 2s - 1)!s! ay, ax,rr-1) -
(3)
3=0
The limit S is int or i(m - 1), whichever is an integer, and the source coordinates are xo, yo, zo. To avoid ambiguity in some cases, put in the s value and simplify before insertion of m. For step 2 use 5.35 (2), which gives the reciprocal distance from the source point at p = a, z = zo, ¢ = 0 to the field point p, z, 4, to be r1 = 2r1 Z
Epf: Kp(tp)Ip(ta)
cos t(z - zo) dt cos pl,t,
(4)
P= 0
where 2 - Sp° has been replaced by cp. Substitute (4) in (2) writing X,(m,n) for the product of (-1)1"-1"t+s by the factors preceding the derivative in (2) and /„("'-")(u) for am-28/p(u)/aum-28, and obtain for
222
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.44
r-n--1r 7,nm (cos 0) cos mq5 the expression xsern,n,
Ep cos p1,1/
p=0
47,
)
8=0
f tnK p(tp)I p(m-28)(ta) cos tz dt 0
(5)
Step 3 consists of replacement of I p(tp) by — A„,,,Kp(tb)Ip(tp)/ I p(tb), which yields a solution of Laplace's equation that cancels iln„,r-n-1P7(cos 0) cos m4) on the cylinder p = b and is valid inside it. For step 4 note that p2 is r2 sin2 0 + a a2— 2ar sin 0 cos cp and that r sin 0 is the p of 5.40 (19), so that from HTF II (36), page 102, or HMF 9.1.79 there results
Ip(tp) cos 14 +. (-1)qp±q(ta),IQ (tr sin 0) cos q4, q= — co
=
ZI(-1)q[I p+,(td) + (1 — 5;),_01-7,a(ta)1}1,2 (tr sin 0) cos qct,
(6)
q=0
The second form follows from the first because in(v) = /_„(v) by Art. 5.32. Multiplication of (6) by cos tz, application of 5.40 (19) to the terms with r and z, replacement of the brace in (6) by f (ta) and of m and n by q and w, where w q is even by symmetry, give
+ t .fT ((twa)w
I p(tp) cos tz cos pkt. =
I riv/1,(cos 0) cos qct. (7)
q=0 w=q
This equation and (5) as modified by step 3 show that the coefficient Cr: of rw/I, (cos 0) cos qcb in V, due to A„r-n-1P',:'(cos 0) cos m0 in V, is
nm Cwq
20s(m,n)An. rfpg(ta)twi-nic,(0)4(.-20(t)
=_
I p(tb)
— 1)w(w + P = 0 s=0
dt
(8)
By 5.32 (2) f „(ta) is written in powers of ta, and differentiation of 5.32 (2) gives /p("'-28) (ta) in powers of ta. Thus the integrals needed are
f
p(t) dt
.10 4(0
1
n
f tw dt 1 0 rip(t)12
(9)
This integral is tabulated only for p = 0 [see 5.42 (4)] and for p = 1 (see
§5.44
223
PROBLEMS
6.18). Step 5 requires that Vo =
cro')A00
ca420 + • • + cooA 11 + cVoA oi +
0=
cu A00 +
o=
48 A00 + (c-5 + 48)A20 + • • •
cHA20 + • • + (c-3 + cH)Aii
+ cflA oi +
(10)
+ cNA.11 + 44A3i + These equations are solved for A„„„ which is then inserted in (1) with r > c to get Va and in (4) as modified by step 3 to get V,. Then V = Vs ± V,
(11)
Another form for the potential may be more convenient. Once the A„'s are known, the sphere can be replaced by a set of multipoles of known magnitude by (2). In the Green's function 5.298 (4), z — zo is written for z, b for a, a for b, and 0 for ctn. Then, by differentiation with respect to zo and b, the potential in an earthed cylinder of each multipole may be found, and the sum of these, weighted according to their magnitude, gives V for a sphere centered at z = 0, p = b, ¢ = 0. If the sphere is replaced by a spheroid, the problem can be set up in the same way. By 5.40 (7) or (8) the spheroidal harmonics are expressed as a sum of spherical harmonics, which are then treated as in this article up to (7), at which point the spherical are converted back into spheroidal by 5.40 (10) or (11), and so forth. Before the advent of digital computers the numerical work involved in such methods would have been too laborious to be practical, but with these devices, solutions of any desired accuracy can be found. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Two similar charges are placed at a distance 2b apart. Show that a grounded conducting sphere placed midway between them must have a radius of approximately b/8 to neutralize their mutual repulsion. 2. The radius vector from the center of a conducting sphere of radius a, carrying a charge Q, to a point charge q is c. When the system is placed in a uniform field E along the direction of c show that, in order that no force act on q, Q must have the value qa3 2c' — ' — c2[471-€E(1 2(c)a) — c (co — a c 3. An infinite conducting plane forms one boundary of a uniform field of unknown strength X. A known charge q, when placed at a distance r from the plane, experiences an unknown force F. It is found that if a hemispherical piece of conducting
224
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
material, of radius a, is placed flat against the plane opposite the charge the force is unchanged. Show that q2 [ 2r2 1 qr6 X — F— 4r2 tre (r 4 — a4) 2 2/re(r,— 0 4. A hollow conducting sphere of radius a is half filled with a dielectric. On the axis of symmetry at a distance a/3 from the plane dielectric surface, a charge experiences no image force. Show that the capacitivity is 1.541e„. 5C. An infinite conducting plane at zero potential is under the influence of a charge of electricity at a point 0. Show that the charge on any area of the plane is proportional to the angle it subtends at 0. 6C. If two infinite plane uninsulated conductors meet at an angle of 60° and there is a charge e at a point equidistant from each and distant r from the line of intersection, find the electrification at any point of the planes. Show that at a point in a principal plane through the charged point at a distance r31 from the line of intersection, the surface density is
( 4/1-r2 4
) 71
7C. What is the least positive charge that must be given to a spherical conductor, insulated and influenced by an external point charge e at distance r from its center, in order that the surface density may be everywhere positive? 8C. An uninsulated conducting sphere is under the influence of an external electric charge. Find the ratio in which the induced charge is divided between the part of its surface in direct view of the external charge and the remaining part. 9C. A point charge e is brought near to a spherical conductor of radius a having a charge E. Show that the particle will be repelled by the sphere, unless its distance from the nearest point of its surface is less than -la (e /E)1, approximately (take e E). 10C. A hollow conductor has the form of a quarter of a sphere bounded by two perpendicular diametral planes. Find the images of a charge placed at any point inside. 11C. A conducting surface consists of two infinite planes which meet at right angles and a quarter of a sphere of radius a fitted into the right angle. If the conductor is at zero potential and a point charge e is symmetrically placed with regard to the planes and the spherical surface at a great distance f from the center, show that the charge induced on the spherical portion is approximately —5ea3/(7rf 3). 12C. A thin plane conducting lamina of any shape and size is under the influence of a fixed electrical distribution on one side of it. If of is the density of the induced charge at a point P on the side of the lamina facing the fixed distribution and o2 that at the corresponding point on the other side, prove that al — 02 = co where oo is the density at P of the distribution induced on an infinite plane conductor coinciding with the lamina. 13C. A conducting plane has a hemispherical boss of radius a, and at a distance f from the center of the boss and along its axis there is a point charge e. If the plane and the boss are kept at zero potential, prove that the charge induced on the boss is —e(1
a2
a 2)1
14C. Electricity is induced on an uninsulated spherical conductor of radius a by a uniform surface distribution, density a, over an external concentric nonconducting spherical segment of radius c. Prove that the surface density at the point A of the
PROBLEMS
225
conductor at the nearer end of the axis of the segment is —crc 2a2 (c a)(1
AB)
where B is the point of the segment on its axis and D is any point on its edge. 15C. Two conducting disks of radii a, a' are fixed at right angles to the line that joins their centers, the length of this line being r large compared with a. If the first has potential V and the second is uninsulated, prove that the charge on the first is 872Ear 2 V (r2r2 — 4aa') 16C. A spherical conductor of diameter a is kept at zero potential in the presence of a fine uniform wire in the form of a circle of radius c in a tangent plane to the sphere with its center at the point of contact, which has a charge Q of electricity. Prove that the electrical density induced on the sphere at a point whose direction from the center of the ring makes an angle v, with the normal to the plane is
—c2Q sec3 477-2a
2 .1.
r
(a2
c2 sec2 G — 2ac tan cos 0)-1 do
17. A stationary ion with charge q and mass m is formed in a highly evacuated conducting bulb of radius a at a distance c from its center. Show that the time for this ion to reach the wall is
t = q-'1c(47rmEa 3) 1(K — E)
where V = (a2 — c2)a-2
K and E are complete elliptic integrals of modulus k. 18. An earthed conducting sphere of radius a has its center on the axis of a charged circular ring, any radius vector c from its center to the ring making an angle a with the axis. Show that the force sucking the sphere into the ring is
Q2E(c2 — a2)k 3 cos a 1672Ec2a2 sin3 a(1 — k2)
where k2 —
4a2c2 sine a
a4
c 4 — 2a2c2 cos 2a
E is the complete elliptic integral of modulus k. 19C. If a particle charged with a quantity e of electricity is placed at the middle point of the line joining the centers of two equal spherical conductors kept at zero potential, show that the charge induced on each sphere is —2em(1 — m + m2 — 3m3 4m4) neglecting higher powers of m, which is the ratio of the radius to the distance between the centers of the spheres. 20C. Two insulated conducting spheres of radii a, b, the distance c of whose centers is large compared with a and b, have charges Qi, Q2, respectively. Show that the potential energy is approximately (87rE)-'[(a-' — b 3c-4)(2!
2c-,(21(22
(b-,— a3c-4)(a
21C. Show that the force between two insulated spherical conductors of radius a placed in an electric field of uniform intensity E perpendicular to their line of centers is 127rEE2a6c-4(1 — 2a3c-3— 8a 5c-5 + • • • ), c being the distance between their centers. 22C. Two uncharged insulated spheres, radii a, b, are placed in a uniform field of force so that their line of centers is parallel to the lines of force, the distance c between
226
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
their centers being great compared with a and b. Prove that the surface density at the point at which the line of centers cuts the first sphere a is approximately eE(3
6b3c-3
15abac-4
28a2b3c-5
57a3b3c-6 + •
•)
23C. Two equal spheres each of radius a are in contact. Show that the capacity of the conductor so formed is 87rect In 2. 24. Two spheres of radii a and b are in contact. Show that the capacitance of the conductor so formed is 47reab[
C-
a +b
2C' -I-
a (a
b
b)
4'(a
b)]
where C' is Euler's constant 0.5772 and gz) = r'(z)/r(z). 25C. A conducting sphere of radius a is in contact with an infinite conducting plane. Show that if a unit point charge is placed beyond the sphere and on the diameter through the point of contact at distance c from that point, the charges induced on the plane and sphere are Ira (a) - — cot —
and
7ra — cot (— c c
-1
26C. Prove that if the centers of two equal uninsulated spherical conductors of radius a are at a distance 2c apart, the charge induced on each by a unit charge at a point midway between them is
Z
(-1)' sech na
where c = a cosh a
27. Two equal spheres, of radius a, with their centers a distance c apart are connected by a thin wire. Show that the capacitance of the system is 8rea sinh
Z(- 1)n+l csch n13 n=1
where cosh )3 = c/a. 28. Two equal spheres, of radius a, with their centers a distance c apart, are charged to the same potential V. Show that the repulsion between them is 277-€V2 Z(-1)n+I(coth p - n coth n13) csch n0 n=1
where cosh 13 = 2c/a. 29C. An insulated conducting sphere of radius a is placed midway between two parallel infinite uninsulated planes at a great distance 2c apart. If (a/c) 2 is neglected, show that the capacity of the sphere is approximately 47rEa[1 + (a/c) log 2]. 30C. Two spheres of radii r1, r2 touch each other, and their capacities in this position are c,, c2. Show that c1= 47T-er2(f2Zn-2 f3 Zn-3 f4Zn-4 + • • •) where f = r, (r1
227
PROBLEMS
31C. A point charge e is placed between two parallel uninsulated infinite conducting planes, at distances a and b from them, respectively. Show that the potential at a point between the planes which is at a distance z from the charge and is on the line through the charge perpendicular to the planes is z Sre(a
b)[
2a
2b
,v (2a — z 2a 2b
2b +z 2a 2b
*(2a 2b — z)] 2a 2b
where 4/(z) = [r'(z)]/[r(z)]. 32. An electron, charge e, mass m, traveling horizontally in a high vacuum with a velocity v, must pass over an uncharged horizontal dielectric plate of length d. Show that, if it comes within a distance a of the front edge of the plate, it will be drawn into it by image forces before clearing the back edge, where, if edge effects are neglected, a' —
(K — 1)e 2d 2
2m2rae„(K
1)v'
33C. A point charge is placed in front of an infinite slab of dielectric, bounded by a plane face. The angle between a line of force in the dielectric and the normal to the face of the slab is a; the angle between the same two lines in the immediate neighbourhood of the charge is 0. Prove that a, 13 are connected by the relation
sin i:13 =
( 2K y sin 1 K
Za
34C. Two dielectrics of inductive capacities K, and K2 are separated by an infinite plane face. Charges e,, e2 are placed at points on a line at right angles to the plane, each at a distance a from the plane. Find the forces on the two charges, and explain why they are unequal. 35C. Two conductors of capacities CI, c2 in air are on the same normal to the plane boundary between two dielectrics with coefficients k,, k,, at great distances a, b from the boundary. They are connected by a thin wire and charged. Prove that the charge is distributed between them approximately in the ratio ki
[4,7re,,
c2
k, — k 2
2b(ki
kz)
2k 2 (k,
k2)(a
421-E„ k 2[ ci b)1
k, — kz
2a(ki +kz)
2k,
(ki -Fk2)(a b)]
36C. A conducting sphere of radius a is placed in air, with its center at a distance c from the plane face of an infinite dielectric. Show that its capacity is zlireoa sinh
aZ
(
K — 1 yi cosech na K +1
where cosh a = c/a. 37. Show that the charge induced on an earthed conductor consisting of two similar spheres of radius a, intersecting orthogonally, by a charge q in their plane of intersection, at a distance 3a2-1 from the axis of symmetry is (1 /3 — 2/51)q. 38. A charge q is at a distance b from the center of a circular hole of radius a in an infinite earthed flat conducting sheet and lies in the same plane. Show that the charge density induced on the sheet at a distance r from the charge and c from the center of the hole is —q(a 2— b2)-1/[27-2r 2(c2— 39. Show that, if in the preceding problem the charge is distributed around a ring of radius b concentric with the hole and coplanar with the sheet, the induced charge density at P is —q(a2— b2)1/[221-2(c2 — b2)(c2— a2)1].
228
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
40. Invert the preceding problem, apply Green's reciprocation theorem, and show that the potential at a point P on a sphere, part of whose surface is occupied by a thin conducting bowl at potential Vo, is (2V0/7) sin-,(cos a/cos 0) where 0 is the angle, measured at the bowl end of the rotational symmetry axis, between this axis and P and a that between this axis and the edge of the bowl. 41. Starting with the preceding result find the first term in the spherical harmonic expansion for the potential of a freely charged spherical howl all elements of which are at a distance a from the origin where its radius subtends an angle /3. From this show that its capacitance is 4ea(0 + sin (3). 42. Invert the result of problem 39 and find the charge density induced on a spherical earthed conducting bowl by a uniform charge distribution —0-1 over the remainder of the sphere. By superimposing a uniform spherical surface charge of density al show that the charge density at P on the inside or outside, respectively, of a freely charged conducting spherical bowl at potential Vo is
=
EV 0
sin a
.
sin a sin 0
Zra [ (sin2 0 — sin 2 a)i
=
eV o
where the angles are defined as in problem 40. 43C. A conductor is formed by the outer surfaces of two equal spheres, the angle between their radii at a point of intersection being 27/3. Show that the capacity of the conductor so formed is 2rae(5 — 3-14), where a is the radius of either sphere. 44C. An uncharged insulated conductor formed of two equal spheres of radius a, cutting one another at right angles, is placed in a uniform field of force of intensity E, with the line joining the centers parallel to the lines of force. Prove that the charges induced on the two spheres are +1171-ea2E/2. 45C. A conductor is bounded by the larger portions of two equal spheres of radius a cutting at an angle 7/3 and of a third sphere of radius c cutting the two former orthogonally. Show that the capacity of the conductor is 47re 1c + a (I — -131) — ac[2(a2
c2)-1 — 2(a2
3c2)-+
(a2
40)-111
46C. A spherical shell of radius a with a little hole in it is freely electrified to potential V. Prove that the charge on its inner surface is less than ieVS/a, where S is the area of the hole. 47C. A thin spherical conducting shell from which any portions have been removed is freely electrified. Prove that the difference of densities inside and outside at any point is constant. 48C. Prove that the capacity of an elliptic plate of small eccentricity e and area A is approximately Ay( 1 e4 e 6) Se(-— 64
64
49C. An ellipsoidal conductor differs but little from a sphere. Its volume is equal to that of a sphere of radius r, its axes are 2r(1 a), + 2r(1 0), 2r(1 y). Show that neglecting cubes of a, 0, -y, its capacity is 4orer[1 + ( A) (a2 +32 + ,2)]. 50. An oblate conducting spheroid with semiaxes a and b has a charge Q. Show that the repulsion between the two halves into which it is divided by its diametral plane is (12[1671-e(b 2 — a2)]-1In (b/a), where b > a. 51C. A prolate conducting spheroid, semiaxes a, b, has a charge Q of electricity. Show that repulsion between the two halves into which it is divided by its diametral plane is Q2[167e(a 2 — b2)]-' In (a/b) where a > b. Determine the value of the force in the case of a sphere.
229
PROBLEMS
52C. A thin circular disk of radius a is electrified with charge Q and surrounded by a spheroidal conductor with charge Qi, placed so that the edge of the disk is the locus of the focus S of the generating ellipse. Show that the energy of the system is [Q 2 L BSC + (Q Qi) 2L SBC]/ (87-ea), B being an extremity of the polar axis of the spheroid and C the center. 53C. If the two surfaces of a condenser are concentric and coaxial oblate spheroids of small ellipticities S and S' and polar axes 2c and 2c', prove that the capacity is 471-Ecc'(c' — c)-2[c' — c ;(Sc' — o'c)], neglecting squares of the ellipticities. Find the distribution of electricity on each surface to the same order of approximation. 54C. An accumulator is formed of two confocal prolate spheroids, and the specific inductive capacity of the dielectric is K1/6.), where (.7) is the distance of any point from the axis. Prove that the capacity of the accumulator is 2T-2E„K1 In
b, a+b
(al +
where a, b and al, b, are the semiaxes of the generating ellipses. 55C. A thin spherical bowl is made by the part of the sphere x 2 +y2 + z2 = cz bounded by and lying within the cone (x/a)' + (y/b)2 = (z/c)' and is put in connection with the earth by a fine wire. 0 is the origin, and C, diametrically opposite to 0, is the vertex of the bowl; Q is any point on the rim, and P is any point on the great circle arc CQ. Show that the surface density induced at P by a charge E placed at 0 is
Ec CQ 47rabI OP 2(0P2 — 0Q2) 1'
where I =
do f Jo .r (a2 sin2 0
b 2cos' 0)1
56. A charge q is placed at a distance c from the center of a spherical hollow of radius a in an infinite dielectric of relative capacitivity K. Show that the force acting on the charge is (K — 1)q' 4,rf„c 2
n=0
n
n(n ± 1) (,)"+1 K(n 1) a
57. An earthed conducting sphere of radius a has its center on the axis of a charged circular ring, any radius vector c from this center to the ring making an angle a with the axis. Show that the force sucking the sphere into the ring is Q2
4n ec 2
a)2n+1 Z(n + 1)P, Ei (cos a)P„(cos a)( 0
58. The spherical coordinates of a circular ring are a, a. A sphere of relative capacitivity K and radius b has its center at the origin. If the ring carries a charge of linear density 7, show that the potential between the sphere and the ring is
—ZP„(cos a) sin a l
2f„
re = 0
n(K — 1)b 2n±' a
an(n(K + 1) + 1)r"÷,
Pn (00s 0)
59. That portion of a sphere of radius a, lying between 0 = a and 0 = it — a is uniformly electrified with a surface density a. Show that the potential at an external
230
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
point is 2n-FI
cur {a
cosa + Z
—
e
7'
,, a
4n + 1
[P2n+i(COS a) — P2n_i(COS ah
P2n(P)
60. A circular ring of radius a and charge Q is used as a collector at the center of an earthed spherical photoelectric cell of radius b. Show that the field strength at the spherical surface is 00
Q
•
4re
1)n
n=0
3 • • • (2n — 1) 4n + 1()2n P 2o(CoS b2 b 2 • 4 • • • 2n
61. A point charge q is placed at a distance b from the center of two concentric earthed conducting spheres of radii a and c, where a < b < c. Show that when a < r < b, the potential is =
V
b2n+1
q
41T-EZ bo+1(a 2n-1-1
a270-1)
c2n -1-1 c2n+1)
r"
1.1-1
P ,t (cOs 0)
n=0
62. An infinite flat conducting plate is influenced by a charge q at a distance b from it, and there is a solid dielectric hemisphere of radius a flat against the plate opposite the charge. Show that the surface density induced on that part of the plate lying outside the dielectric is
b (b2
K) (2n + 1)(2n + 2) 1• 3 • • • (2n +1) 1 + (2n + 1)(K + 1) 2 • 4 • • • (2n + 2)
(-1)n+1(1
7.2)1
n=0
4n+3 a b2'+2r 2'±3
63. Show that the potential at any point due to a circular disk of radius c raised to a potential V1 is (2V1 /71-)Z( —1)”(2n + (c/r) 2"+1P2n(cos 8) where r > c. Find the value for r < c. 64. Two similar rings of radius a lie opposite each other in parallel planes, so spaced that the radius of one ring subtends an angle a at the center of the other. Show that when they carry charges Q and Q', the repulsion between them is
QQ' z 4rea2
n=0
un l • 3 5 • • • (2n + 1) (sin (si a) 2"±2P2.÷1 (cos a) 2 • 4 • 6 • • • 2n
65. The inside of a conducting sphere of inner radius a is coated with a uniform layer of dielectric of inner radius b. Show that the force on a point charge q, in the cavity at a distance c from the center, is CO
(72
4re,,
n=0
nc2"-11(K — 1)(n + 1)a2"+' [(K + 1)n + 1]b2n+1 1 b 2n+1 [ RK + 1)n + K]a2"+' (K — 1)702n+11
66. A point charge q is placed at a distance c from the center of an earthed con-
ducting sphere of radius a on which is a dielectric layer of outer radius b and relative capacitivity K. Show that the potential in this layer is Nk
47ro,,c
n=0
c"
(2n + 1)b2n+1(r" — Pn(cos 6) + 1)n + l]b2"+' (n + 1)(K — 1)a2n+11
PROBLEMS
231
67C. A conducting spherical shell of radius a is placed, insulated and without charge, in a uniform field of electric force of intensity F. Show that, if the sphere is cut into two hemispheres by a plane perpendicular to the field, these hemispheres tend to separate and require forces equal to tirca'F' to keep them together. 68C. A spherical conductor of internal radius b, which is uncharged and insulated, surrounds a spherical conductor of radius a, the distance between their centers being c, which is small. The charge on the inner conductor is Q. Find the potential function for points between the conductors, and show that the surface density at a point P on the inner conductor is 3c cos 0) (1 a' b3 — a3
Q
where 0 is the angle that the radius through P makes with the line of centers and terms in c2 are neglected. 69C. The equation of the surface of a conductor is r = a(1 SP„), where is very small, and the conductor is placed in a uniform field of force F parallel to the axis of harmonics. Show that the surface density of the induced charge at any point is greater than it would be if the surface were perfectly spherical by the amount
(5
[ 3neF.5 [(n 1)Pn+1 2n +1
(n — 2)P„-”
SP.) is 70C. A conductor at potential V whose surface is of the form r = a(1 surrounded by a dielectric (K) whose boundary is the surface r = b (1 + nP„), and outside this the dielectric is air. Show that the potential in the air at a distance r from the origin is
KabV
(2n
{1 b r
(K — 1)a
+1)3a.b2.+1 (1
(n 1)a2n+,] P.} (K — 1)nb'[nb 2n+1 (K — 1)(n + 1)a2n+3 nK)b 2n+3
n
where squares and higher powers of S and n are neglected. 71C. The surface of a conductor is nearly spherical, its equation being r = a(1
nSa)
where n is small. Show that if the conductor is uninsulated, the charge induced on it by a unit charge at a distance f from the origin and of angular coordinates 0, 4 is approximately
fL
+ \f/ (--Ynsn(o, )1 95
72C. A uniform circular wire of radius a charged with electricity of line density e surrounds an uninsulated concentric spherical conductor of radius c. Prove that the electrical density at any point of the surface of the conductor is L[i
102P2 + 1 . 304P4 52 a
2c
92 • 4 a
. 3 . 506P6 132 2• 4 • 6 a
73C. A &electric sphere is surrounded by a thin circular wire of larger radius b carrying a charge Q. Prove that the potential within the sphere is [ 4r€,,b
1 ± Z(— 1)n 1
1 + 4n 1 + 2n(1
K)
1 • 3 • 5 • • (2n— 1)(r)2np 2n b 2 4 6 • • 2n
232
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
74C. A conducting sphere of radius a is embedded in a dielectric (K) whose outer boundary is a concentric sphere of radius 2a. Show that, if the system is placed in a uniform field of force F, equal quantities of positive and negative electricity are separated of amount 367a2c„FK/(5K + 7). 75C. A spherical conductor of radius a is surrounded by a uniform dielectric K, which is bounded by a sphere of radius b having its center at a small distance y from the center of the conductor. Prove that, if the potential of the conductor is V and there are no other conductors in the field, the surface density at a point where the radius makes an angle 0 with the line of centers is, approximately, E,KVb
a[(K — 1)a
b][
1±
6(K — 1) ya2 cos 2)b' 2(K — 1)a3 (K -
76C. A shell of glass of inductive capacity K, which is bounded by concentric spherical surfaces of radii a, b, (a < b), surrounds an electrified particle with charge q which is at a point Q at a small distance c from 0, the center of the spheres. Show that the potential at a point P outside the shell at a distance r from Q is approximately
q
1
2c(b 3 — a3)(K — 1) 2
471- E
r
2a3(K — 1) 2 — b 3(K + 2)(2K + 1)
cos 0 r2
where B is the angle which QP makes with OQ produced. 77. Show that the attraction between a sphere of radius a and relative capacitivity K, and a point charge q at a distance c from its center is (1 — K)q 2 n(n + 471-E„c2 .4 1 (K n=1
1)
(a)2'.+1 1)n c
78. A conducting sphere of radius a is supported by an orthogonal conducting cone whose exterior half angle is a. Show that, if this system is charged, the potential is A (r" — on+ir-.-1)P.(cos 0) where 0 < n < 1 if I-7r < a and P,,(cos a) = 0. 79. Two conducting coaxial cones whose half angles, measured from the positive axis, are a and 13 intersect a conducting sphere of radius a orthogonally. If the system is charged, show that the potential outside the sphere between the cones is
A (r" — a 2n+ir-n-1)[Q„(cos (3)Pn (cos 0) — P.(cos ()Q„(cos 8)] where n is the smallest number for which Q,,(cos 0)Pn (cos a) — P„(cos 13)(2n(cos a) = 0 80. An oblate dielectric spheroid whose surface is 3•0 is placed in a uniform electric field E parallel to the axis C = 1. Show that the resultant field inside is uniform and that its strength is —E/ l(K — 1ko[(1 31) cot-' 10 — 1-0] — 81. The spheroid of the last problem is placed in a uniform field E normal to the axis C = 1. Show that the electrical intensity inside is uniform and of magnitude 2E 2 + (K
—
1
)3'0[(1
cot-230
—
"()]
82. A prolate dielectric spheroid whose surface is no is placed in a uniform electric field E parallel to the axis C = 1. Show that the resultant field inside is uniform and that its strength is —E/1(K — 1)no[(1 — no) coth -1no + no] —
PROBLEMS
233
83. The spheroid of the last problem is placed in a uniform field E normal to the axis t = 1. Show that the electrical intensity inside is uniform and of magnitude 2E 2 + (K — 1)no[( 1—
coth-i n o + nol
84. A spheroid of relative capacitivity K is placed in an electric field E, its axis of revolution making an angle a with the field. Show that the torque acting on it is iirEo(K — 1)m2nE(E — E2) sin 2a where n is the semiaxis in the direction of the axis of revolution and m is the semiaxis normal to it. For an oblate spheroid, El and B2 are the results of problems 80 and 81, respectively, where 3 0 = n(m2 — n2)-1. For the prolate spheroid, El and B2 are the results of problems 82 and 83, respectively, where no = n(n2 — 85. An earthed conducting disk of radius a whose equation is 3.= 0 is influenced by a point charge q at 1, 3-0, 0. Show that the potential due to the induced charge on the disk is CO
=
q
27r2ea
(4n + 1)Q2o.W-o)Q2.(i0P2n(t) n =0
86. Show, using 5.275 and Green's reciprocation theorem, that the potential due to a charged ring whose equations are 3- = 3 0, t = to is CO
= i Q 4 2n + i)Pn(7o)Pnctoui?-)Pn(z) , (
47rea
n=0
when 3' > 3 o and the charge on the ring is Q. Write the expression valid when t- < 87. An earthed conducting disk of radius a is influenced by a ring carrying a charge Q whose equation is 3- = 3 0, t = to. Show that the potential due to the induced charge is
V; =
27r 2ea
(4n + 1)Q2nU3"o)P2n(E0)(22n(j0P2n(t) n=0
88. The upper and lower halves of an oblate spheroidal shell = 3 o are insulated from each other and charged to potentials + Vo and —V0, respectively. Show that the potential at any external point is V = VO
1 • 3 • • (2n + 1) (4n + 3)Q2,,+I(/3-) 2 • 4 • • • [2(n + 1)1 (2n + 1)(2211-1(3?-0)
(-1)n2
n
1,\ n41 \
Write down the potential for the region inside. 89. Show, by the method used in 5.274, that if the density of electric charge on a prolate spheroid n = no is Gr(t, 95) then the potential inside is
V =ZZM „,„F,;(n)P7,(t) cos m(4,
—
Om)
n=0 7n=0
where M.t. is (
-
1)m(2 — 47reC2
(2n + 1)[ (n — m)12
+1 /21-
PN t) cos m(4
Q"'(no) f
(n
m)!
—1 0
—
O„,)hih3 dt
234
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
where hiha are given by 5.28 (6) and (7). Write the analogous formula for the potential outside. 90. Using the results of the last problem, show that the potential due to a point charge q at Ea, no, 00 is given by
4n-qcC2
I TP',;( t)(2,7(n)PZ(no)P,7((to) cos m(95 — tfro)
(2 — 5?„)( 1)'n(2n
(n —m)I (n +m)
n=0m=0
when n > no. Write out the result when n < no. 91. Starting with the potential inside an earthed cylindrical box of radius a due to a point charge q on the axis and using Green's reciprocation theorem, find the potential on the axis of a ring of radius b coaxial with and inside the box. Hence, show that the potential anywhere inside the box due to this ring is
q2 Irea
sinh µ kz sinh ihk(L — c)Jo(w.b)Jo(iikP) sinh µ k L p.k[Ji(p.ka)] 2
k=1
where z < c and the coordinates of the bottom and top of the box and the plane of the ring are, respectively, z = 0, z = L, and z = c. Take Mk so that Jo(Aka) = 0. Show by 3.08 (2) that the force toward the bottom of the box is q2
21rea2
k =1
sinh ihk(L — 2c)[Jo(Akb)] 2 J 1(Aka) sinh I.Lk L
92. The walls of a conducting box are given by z = ±c, p = a. The halves above and below the plane z = 0 are insulated from each other and charged to potentials -I- Voand — Vo, respectively. Show that the potential anywhere inside is given by V = +V 0{1
2 a
sinh [Atk(c— 1z1)]Jo(AkP)} µk sinh AtkcJi(Aka)
k =1
where Jo(µka) = 0 and the sign is that of z, and also by v = v o[z + 2 xk /o(nirp/c) sin (nirz)] I../ n/o(nra/c) c c n=1
93. The walls of a conducting box, which are given by z = ±c, p = a, are earthed with the exception of two disk-shaped areas at the top and bottom bounded by p = b, which are charged to potentials + Voand — Vo, respectively. Show that the potential inside is given by 2bV0x.k sinh 1(Ihkb)J 0(1.4kP) a2 ;Lk sinh tikc[J1(Aika)] 2 k=1
where Jo(Aka) = 0, and also by 2bV0 ,--■ c
(
n=1
1
Ko(mra/c)10(nrp/c) — .10(thra/c)Ko(nrp/c) (nr . Torz —b sin —
)'
lo(nra /c)
235
PROBLEMS when p > b and when p < b by Vo — [z 2b
(-1)'
K o(n7ra/c)Ii(nlrb/c)
/o (nra/c)Ki(mrb/c) (or
Io — p
/0(Thra/C
n=1
Torz]
sin — c
94. A semi-infinite conducting cylindrical shell of radius a is closed at one end by a plane conducting plate normal to its axis and at the same potential. Show that the image force on a charge q placed on the axis at a distance b from the plate is c's
q2 22rea 2
k= 1
e-Akb
2
where .1-o(Pka) = 0
[ Ji(Aka)
95. The infinite conducting cylinder p = a is filled below the z = 0 plane with a dielectric of relative capacitivity K. There is a charge q on the axis at z = b. Show that the potential above the dielectric is K
e-Akk-b1 271-c,,a 2
K
k=1
—
1
Jo(Ahp)
le-mk(z+"AdJi(1.4ka)J2
where Jo(mka) = 0. Find the potential in the dielectric. 96. The surface of a conductor is generated by rotating a circle of diameter a about one of its tangents. Show that the capacitance of the conductor is
87rea Z [J I(Aka)]-1.10. e -Ak..l.h dct, = 8rea k=1
80, 0 (pka)[JIGika)]-1
k=1
where .1-o(Aka) = 0 and So.o(uka) is a Lommel function. 97. The conductor in the last problem is placed, uncharged, in a uniform field E parallel to its axis of rotational symmetry. Show that, in polar coordinates, the potential at any point is e 2Ea2 7 1 =k
8 j0( Ak
a 2r
1 sin 0)
coo
(Aka)1 2
where Jo(uka) = 0. 98. The conductor in the last problem is rotated to make E normal to its axis. Show that, if J o(pka) = 0, the potential at any point is 2Ea2 r
e Aka
k=1
..Ji(Aika 2r-1sin 0) [J2(ihka)] 2
COO v
COS cb
99. Show that the potential due to a ring of radius a carrying a charge Q is Q — f K o(ka)I0(kp) cos kz dk
if p < a
Q Io(ka)Ko(kp) cos kz dk 272e o
if p > a
272€ 0 or
236
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
100. A ring of radius a carrying a charge Q is coaxial with an infinite dielectric cylinder of radius b. Show that the potential in the dielectric is Q
2 -„ 7 0NY(k)/0(kp)K o(ka)
cos kz dk
where 4,(k) = [1 kb(K - 1)0kb)Ko(kb)]-1• 101. The earthed toroidal ring generated by rotating a circle of radius a centered at p = b about the coplanar z-axis lies halfway between the plane z = c at potential Ec and the plane z = -c at -Ec. Observe that ring multipoles at p = b, z = 0, which give fields normal to the planes and zero potential halfway between them, have potentials a"Vi /aer where V is the potential of a ring dipole of strength M given by
2,c2
nit
nrb
nrb) (mrp) mrz
r
-
nr0(7)K0(1rP) sin 717z
sin — or
n=1
n=1
according as p < b or p > b. Hence show that a potential function that gives V = 0 b) 2 = a 2 is on the ring at 2m lines p = p,, z = +z, where 0 < s < m and (p, -
V = Ez -
(A in + A 2n 3 +
mrb) (nir . mrz / 0 -- sin c c c
+ A.„,n2'n+2 )K0(
• • •
72=1 This holds if p < b.
A, = E
If p > b, interchange p and b.
VII
•
"
V-r-1,1Z1Vr+1,1
"
V12
•
•
V,-1,2Z2Vr+1,2
•
•
•
The coefficient A, is given by
•
Vml
V11 V21
•
•
Vm2
V12 V22
'
" •
VInl
V m2
V1„,1 2,o • • • V„.„,
Ps
/orb It)n'11-P,, /sin n2'+'Ko(
V,, =n=1
If p, > b, interchange p, and b. Using only A i fitted at p = b, z = ± a gives excellent results when a << c and a << b. This is a typical approximate method. 102. Referring to 5.03 (3), 5.302 (7), and 5.302 (9), show that the potential due to a disk of radius a, carrying a charge Q, is jo(kp)sin
LIT-Qat/0'
ka
dk
103. If Vo is the potential due to an infinite plane conducting sheet having a uniform charge density go on each side, show, by referring to 5.272 and 5.302, that the potential due to the same sheet when it has a hole of radius a in it is
2cro
Vo —f e-klz1 J 0(kp) 7f 0
(a c ka sin ka)
dk
2
104. A cylinder p = a makes contact with the earthed plane z = 0. There is a uniform potential gradient along the cylinder which has the potential Vo where it passes through a second earthed plane z = c, from which it is insulated. Show that
237
PROBLEMS the potential between the planes outside the cylinder is V0 (_ 2 V =— 71- A-11 n =1
(7)7rz/c) K o (n7rp/c) K o(nra/c)
105. The potential on the walls of a hollow ring bounded by p = a, p = b, z = 0, z = c is given by fi(z), f2(z), f 3(p), f 4(p), respectively. Show that the potential at any point inside is given by the superposition of four potentials, two of the type of 5.304 and two of the type of 5.36. 106. A point charge q is placed at z = zo, p = b, 4) = Q. The planes cp = 0 and = a and the cylinder p = a are at zero potential where 0 <13 < a and 0 < b < a. Show that the potential is given by (3
J8:(1,b) sin —
2q V =
Sr(/' „(pr p) sin — a a
a
eaa2 A/2as, (Ara)12 —+1 s =1 r =1
where J,,(a,a) = 0. 107. A charged semi-infinite conducting sheet has an oblate spheroidal boss whose axis coincides with its edge and whose equation is ;- = ro. Show that the potential is V = C f (1 +
I
(1 +31)1[(1 + 0)1 - t.1}(1 - Ot cos -10
(1 + 1-2)i[(1
31)1 -
2
108. A solid dielectric sphere of radius a has a sector removed so that it fits the edge of an infinite conducting wedge of external angle a, its surface meeting the wedge faces orthogonally. If the wedge is charged, show that the potentials inside and outside the sphere are a + 27r irq5 .V-C Cr sin Or cos — a a + r(1 ± K) 7.14-1 V , = C[r:
ir(IC 1)a' a + 7r(1 + K) rr)
1(sin '; cos '7' 15 (sin 05 a ]
109. The conducting wedge whose faces are 4, = +a and 4, = -a and the conducting plane, z = 0, which intersects it orthogonally, are charged. Show that the potential in the space -a < 4, < +a and 0 < z is given by Cze cos mo, where m is the smallest number for which cos ma = 0. 110. Three orthogonal conducting surfaces, the wedge 4) = +a and 4, = — a, the plane z = 0, and the cylinder p = a, are charged. Show that in the space a < 4, < +a, z > 0 and p > a the potential is given by Cz(pm - a273p-m) cos mo, where m is the smallest number for which cos ma = 0. 111. Three orthogonal conducting surfaces, the wedge 4' = +a and 4. = -a, the plane 0 = Zr, and the sphere r = a, are charged. Show that when -a < 4, < +a, z > 0 and r > a the potential is given by —
C(rm+1 - a'"'-1-3r-m-2) cos 0 sinm 0 cos m4) where m is the smallest number for which cos ma = 0.
238
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
112. A charged conducting body has a deep rectangular hole whose boundaries are given by x = 0, x = a, y = 0, y = b, and z = 0. Show that, far from the opening, the potential is given by b2 P
V = C sinsin sinh (a2 b a
ab
113. The walls of an earthed rectangular conducting tube of infinite length are given by x = 0, x = a and y = 0, y = b. A point charge is placed at x = xo, y = yo, and z = zo inside it. Show that the potential is given by (9nzaz TrE
zbe)1.-lz - z01 ab
(m2a2 n2b2)le
V=
sin
mix°
n=1 m =1
a
sin
nirx a
sin
mryo b
miry
sin — b
114. The walls of an earthed conducting box are given by x = 0, x = a, y = 0, y = b, and z = 0, z = c. A point charge q is placed at xo, yo, zo. Show that the potential inside the box is given by miry sinh A„,„(c - 20) sinh A mnz nrxo nrx miry° sin sin sin sin a A„,„ sinh A „„tc a
4q V =— eab
n = 1 m =1
(m2a2
n2b2)17,.
where A,„„ -
and z < z0. If z > z0, we interchange z and z0. Show
ab
that the z-component of the force on the charge is
F, =
eab
nrxo miry° csch A„,„c sinh Amn(c - 2z0) sine a sine n=1 m=1
For F„, substitute c for a, a for c, x0 for z0 and z0 for x0 in F, and Am,,. For F y, substitute c for b, b for c, Yo for z0 and z0 for yo in F, and 115. Show that the potential of a dipole M parallel to the 0-axis at b, a 0 is ,
M
n
4reb 2 4./ n=0 m =0
when r < b.
(2 (51)(m
-
n
If r > b, write (m
1) (n - OnFn÷, (cos a)P„m(cos 0) cos m4' (n m) b n)(b/r)n+i l,"_1(cos a) for
(m - n - 1)(1)P"'+1(cos a)
b
116. Show that the torque exerted by a sphere of radius a and relative capacitivity
K at the origin on a dipole M at r is, if the angle between M and r is a, T = 111(r sin a)-1C,,,,,[(n
, (cos a) + 1)P;
-
n - 1) cos a/37,, (cos a)]
where C„,n is a summation operator given by
C„,„
(K -1)M 4re,,r2
n+1
z (2
n =0 m
0° )a2n-1-2 (m - n - 1)(n - m) !n /7+,(cos a) m)! n 1)94"+2(n
-
(nK
239
PROBLEMS Show that the radial force is
F = — Mr-2(n + 1)C„,,,[cos a PZ(cos a) + (n — m
1)137_0 (cos a)]
117. Rotate the transformation of Art. 4.13 about the x-axis letting ui = U and u2 = V and set up 5.11 (6) for this case. Substitute (cosh /hi — cos u2)1Ui(ui) U2(22) for U in 5.11 (6) and show that the differential equations for U2 and U1 are d 2U2
du; — —n
d2 U2
2U2,
du?
coth uidUl
(n2 1 4
duo
m2 ) Ul = 0
sinh 2 ui
The latter is identical with 5.23 (3) if n — is written for n and cosh u1 for z. 118. The torus generated by the circle of radius b given by ui = uo in the last problem is charged to a potential Vo. Show that the potential outside it is 2Vo(cosh u1— cos u2)2 n=0
1)Pn_i(coth uo )Pn _1(cosh ui)cos nu2 ( —2), (2n (2n + 1)!!(sinh u0)1P.-1(cosh u0)
where cosh uo is c/b, c being the distance from the center of uo to the line ul = 0. 119. Find the total charge on the torus of the last problem by comparing the potential at u1 = 0, u2 = 0 with q/(42ror) and thus show that its capacitance is 871-€21b(sinh uo)1 n =o
1)Pn_1(coth uo) (2n + 1)!!P,,_1(cosh uo)
( —2)'(2n
120. The torus of the last problem lies in a uniform field E parallel to its axis of rotational symmetry. Show that the potential of the field is
V = Ex + (cosh ul— cos u2)1
A „P„_1(cosh ui) sin nu2
n=1
where, if a is defined by 4.13 (3), U = uo,
A,, —
4(2n + 1) ( —2)nnaEoPn_1(coth uo) (2n + 1)!!(sinhuo)1P,.-1(cosh uo)
121. The torus of problem 118 is placed in a uniform electric field E normal to its axis of rotational symmetry. Show that the potential outside it is 00
E cos 514/3+ (cosh Ul — cos U2)1 Z , A.Pnl_i(cosh ui) cos nu2] n=0
where
An =
(2n — 1)(2n + 1)( —2)n+lb(sinh uo)1Pn1(coth uo) (2n + 1)!!P;;_1(cosh uo)
122. Rotate the transformation of 4.13 about the y-axis letting u1 = U and u2 = V, and set up 5.11 (6) for this case. Substitute (cosh ui— cos u2)4/1(ui)U2(u2) for U in 5.11 (6) and show that the differential equations for U2 and Ui are d'Ui = (u + 1 2 ui, duo 2
d 2U2 dU2 du; ± cot u2 ±[n(n + 1) du2
The latter is identical with 5.14 (2) with u2 in place of O.
m2
sin2 u2
]U2 = 0
240
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
123. The two spheres u1 = uo and ui = —uo of radius b generated in the last problem are aligned with a uniform field of potential Ez so that their potentials are Vo and — Vo, respectively. Show that the potential outside them is CO
Ez + (cosh u1— cos u2)1
A n sinh (n DuiPn(cos u2 ) =0
A n = 21[V0 — (2n + 1)Eb sinh uo](e(2n"uo — 1)-1 where cosh uo is c/b, c being the distance between the centers of the spheres. 124. If the spheres in the last problem are uncharged, show that 2n + 1 e(2.1-1),40 — 1 [
V 0 = Eb sinh uo n=0
1 n=0
e(2'+1)u,
—
1
125. Show that the force between the spheres in the last problem is
F = 7c
— (n + 1)A”+1 — 21aE]
A n[(n +
n=0
where An is found by substitution of Vo from problem 124 into A n in problem 123. 126. The potential on the surface of a sphere of radius a due to external sources is f(0), where 0 is the polar angle. Show by expanding as in 5.156 and summing the harmonic series under the integral by comparison with 5.16 that, when r < a, 1 ,r
- -
Cr
ar/ ad¢ rkf(a) sin a da d V(r, 0) = -7 ar _„ sin a sin 0 cos 014 [a 2 +r 2— 2ar(cos a cos 0
Jo
127. Solve the problem of a dipole normal to and halfway between two parallel earthed conducting planes. Now by inversion show that if two conducting spheres of radius a are in contact in a uniform electric field E parallel to their line of centers, the charge density is
cr
=
271- 2EE
sec' a Z(-1)"n2K0 (nr tan a) n=1
where a is measured about the point of contact from a diameter passing through this point. 128. The surface of an oblate spheroid is maintained at a potential
V=
BnP2n(t) Q2a0) n =0
Inside it a concentric sphere of radius a is at zero potential. Show that in the region between the two the potential is N V
n
= n=0s=-0
[02. 028
2a+1
(-1)nA nC,„()
a
r
a
a
—
+1 ]P2.(COS 0)
PROBLEMS
241
where A nis given by 5.41 (5) with the following values of C„, s
( —1)'(2n + 2s — 1)!!(a)4e -"Q2.(j3-o) (2s)!(2n — 2s' ! c P2.(i3-10
C,. =
(2n)! . 3(4n + 1)!!
C„„ = ( s
>n
c8'
(4n — 1)!!(ar iQ2n(iro) (2n)! \c P2.(i3"(1)
(-1)"(2s)! 2s + 1)!!(2s — 2n)!!
j(2n
8
BnC3n.
and with the right side given by n=0
129. The potential V of a concentric sphere of radius r is maintained at the value BnP2n (COS 0) n=0
Inside it is an oblate spheroid 3 = 3-0at potential zero with an interfocal distance 2c. Show that the potential between spheroid and concentric sphere is N
n
V =
A.C3.[ P23(j! °)Q23(i0
P23(j3)
Q2.0r0)
n=03=0
where An is found from 5.41 (5) with s
( C)2n Cg.r, =
en.
n> 5
— I ) n (48 + 1)(2n)!
P2g(ko)
(2n + 2s + 1)!!(2n — 2s)!! Q2,4 ji-0)
a
= (-1)8
C`,, =
(
(T+Ij(2n
+1)(4n — 1)!!
c
(2n)!
(c)2 n (2n)! P2.(ko) ] a (4n — 1)!! Q2n(73'o)
(ay-EV( —1)'(2s + 1)(2n + 2s — 1)!! c
(2n)!(2s — 2n)!! 8
and the right side of 5.41 (5) is
C.nBn. n=0
130. In the last problem if the sphere is at potential V0, then Bo = Vo and Bn = 0 if n 0. If the concentric spheroid inside it is a disk of radius c, then •o = 0, so P2.(i3-0)/Q23(i3-0) is 2j/7r by 5.214 (3). Show with the aid of 5.27 (6), 5.214 (5), and 5.214 (3) that the charge density on the disk is N
2€ —
prct
n
r r (-1)'(2s)!!A.C..P2,(E)
n=03=0
(2s — 1)!!
131. The surface of an oblate spheroid is maintained at a potential
V
=
ZBnP2n(t) n=0
242
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Outside it a concentric sphere of radius a is at zero potential. Show that in the region between the two the potential is N
a
V= n=0 s=n
'
1
1P2,(cos 0)
A„C,,„[0 2' — a
where A nis found by solving 5.41 (5) with the following values of C..: s < n,
Can =
C„,„ s > n,
Can
( — 1)n(2n + 2s (2s)!(2n
— I)!! 2s)!!
(2n)!(c/a)2n+' P2n( jr0) [ (4n — 1)!! (2n.) ! 3(4n + 1)!! ( —1)'(2s)!j(c/a)2°+' P2.(sii-o) (2n + 2s + 1)!!(2s — 2n)!! (22.(31•o)
( —
=
—
1)n
and with the right side given by n=s
132. The upper and lower halves of a conducting sphere of radius a are insulated
and charged to potentials + Vo and —V0 . It is surrounded by a concentric coaxial oblate spheroid of interfocal distance 2c at potential zero. Without the spheroid the external potential of the sphere is a)2n-I-2
=
Bn(
P2n+1(COS
(4n + 3)(2n (2n + 1)(2n
B„ — ( 1)'
f
n=0
1)!! 2)!!
Show that with the spheroid in place the potential between them is P2s (i A.G.{
V2 =
n=08=n+1
0
)P2.(ir)
Qu(N-o)
—
(22.(ii-432(t)
where Anis found by solving 5.41 (5) and Can is given by s
+
4
,
C,,,t = ( —
s > n + 4,
c.n =
(a)2n+1 ( —1)nj(4s + 1)(2n + 1)! (2n + 2s + 2)!!(2n — 2s + 1)!! c
2n + 2 c 2n+2 Q2s (jr0) (2n + 1)! a Pu(ii-o) ( —1)°(2s +1)(2n + 2s)!! (c)2'+2 (22.W-o) C.. = . P2.(71- 0) 3(2n ± 1)!(2s — 2n — 1)!! a
On (4n ± 1)!![ j02ft+1 c
Note that s is an integer plus zso right side of 5.41 (5) involves Co1„, C1.5n, etc. 133. An uncharged conducting sphere of radius a lies at a distance b from an infinite line charge of uniform density q where b > a. Take the origin at the center of the sphere, and write the potential of the line charge in circular harmonics by 4.02 (2) using gs for 0, p for r, On = 0, and V = 0 when p = 0. Express this in spherical coordinates by writing r sin 8 for p, so it has the form Zr.S.(0,0) where n 0. Add to this the induced charge potential —M(a 2n+,/rn+98.(0,0), which cancels it at r = a. Sum
243
PROBLEMS
by 4.02 and show that the total resultant potential is 192(19 2r 2 +a' sin' 9 — 2a2br sin 8 cos 0)“fr V = q In (br) 2afr(b 2 r 2sine B — 2br sin B cos 0)
Show that the attractive force on the sphere is F=
q2a2 sin sm-' re,,b(b 2 — a 2)
134. If the conducting sphere in the preceding problem is replaced by a dielectric one e, show that the total potential outside it is (€ — e0a2n-f-1
Vo =
2re,,
n
n=1
[no
(n
sin" 0
1)€,]rn+2
bn
cos no
and write down the potential inside. Show that the force on the sphere is F=
— q2
(2n
—
2)!!n(e — e„)
LI (2n — 1)!![ne
n=1
(n
(ay-E1
1)e,] b
135. A sphere of radius a at potential Vo lies at the origin between two infinite parallel earthed planes spaced a distance b apart, and its center is a distance c from the nearest plane. This is one section of the field produced by an infinite set of pairs of V o and — V o spheres at a distance 2c apart similarly aligned along the z-axis with pair centers spaced at intervals 2b. The potential between sphere and planes is
V=
Z
(ap;Ov+IPp(cos On)
CP
p=0 n=0
where rn is the distance from the field point to the center of the nth sphere and 8„ is the angle at the center of the nth sphere between r„ and the z-axis. With the aid of 5.44(2) and 5.298 (6) write down a'V/az' on the axis for all pairs except the pair that includes the origin. Sum this for all but the center pair. Add to the sum the a3v /az. term inside the sphere at z = 0 due to it and its nearest image. Thus as in Art. 5.41 show that the P equations, one for each value of s from 0 to P, to be solved for C2, are, after dividing out s!ci-s, +.+1 3 08 V 0
= P=0
f(v) =
CP[ a"
1 ao-,4-1 (1 2 avP+'+' v2
;19 r cot nv)
a P+*+1 f C (2b) (4b -) [1 — ( -1)9+13-(P 4- s + 1)
where ;In) is the Riemann zeta function tabulated in HMF, page 811. Note that as b 00 only the first two terms in the bracket survive, so the result applies to a single sphere at a distance c from a plane. The capacitance is 47reaCo/Vo.
136. A sphere of radius a at potential Vo lies halfway between two earthed parallel planes at a distance 2b apart. Observe that the potential in the region 0 < r < b
244
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
which is Vo on the sphere is
a)2n+1 V = Vo -1-Ze n =0
r
02n —
iP2„(cos 8)
a
where the (r/a) 2" term arises from the induced charge on the planes. Expand V at z = b when p is small in the form ZD,p2., so that there are N simultaneous equations D, = —Vohs, one for each value of s from 0 to N — 1, to be solved for C.. With the aid of MO, page 50, 5.298 (6), and HMF 22.10.12 show that
D„ —
(-1)'
[(2s)!112
N [ x-A (2n + 2s) !a2"+' n= 0
(2n) !b2"÷2'+'
Cn n
(2n) !b2"-". (2n — 2s) !a2"
Show that the capacitance is 4ireCoa/Vo. 137. Obtain an expression for the potential in the last problem when r > b by writing down the Green's function V i(p,z) for a charge between two planes by 5.324 (10), shifting the origin to halfway between the planes, taking a 1/"/az2" as r —> 0, and comparing this with 82"(1/r)/az2" in MO, page 50, as r 0. Thus N 21-2n (a)2n-l-l
V— V o n=0
(2n)! b
)2n1(.0
en s=0
[(2S ± 1)7 p]
2b
cos
(2s
± 1)7z 2b
Note that this expression takes the indeterminate form 0 In p as p —> 0 except at z = 0 where it is infinite as it should be. 138. The potential at a spherical-surface of radius c due to a concentric cylinder inside it having the charge density of 5.39 (1) is given by 5.39 (4) when r = c. If the spherical surface conducts, its charge density must cancel this potential and give a potential V1 when r < c like 5.39 (4) when r-( 2"+2P+0is replaced by —r2n-1-2p/c4n+4p-1-1. Show that the coefficients an and bn can be found by solving equations 5.39 (2) and 5.39 (3) provided V V 1is used in place of V. Thus show that the capacitance is still found from 5.39 (4) with s = p = 0 but with the new C, and C, values.
References The following references will be found useful in connection with this chapter. A. ELECTRICAL BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Elaborate discussions of spherical caps, toroids, and many other cases. DURAND, E.: "Electrostatique et Magnetostatique," Vol. I, 1953; Vol. II, 1966, Masson et Cie. Well-illustrated treatment of solutions in separable coordinate systems and of graphical and numerical methods. FLUGGE, S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. G. Wendt gives a clear discussion of three-dimensional potential problems, including numerical and graphical methods. GEIGER-SCHEEL.: "Handbuch der Physik," Vol. XII, Berlin, 1927. GRAY, A.: "Absolute Measurements in Electricity and Magnetism," Vol. I, Macmillan, 1888. Extended treatment of images and inversion. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extended treatment but omits Bessel functions.
PROBLEMS
245
J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Treats images, inversion, confocal surfaces, and spherical harmonics. MORSE, P. M., and H. FESHBACH: "Methods of Theoretical Physics," 2 vols., McGrawHill, 1953. Discusses the functions and methods of this chapter and tabulates solutions of Laplace's equation in separable coordinate systems. PLANCK, M. K. E. L.: "Theory of Electricity and Magnetism," Macmillan, 1932. Gives excellent treatment of confocal coordinates. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives considerable material on potential theory. THOMSON, W.: "Papers on Electrostatics and Magnetism," Macmillan, 1884. Solves many problems and gives numerical tables on spheres and spherical segments. THOMSON, W., and P. G. TAIT: "Treatise on Natural Philosophy," Cambridge, 1912. Treats spherical harmonics. VAN BLADEL, J.: "Electromagnetic Fields," McGraw-Hill, 1964. Treats approximately dielectric cubes and similar problems not in other books. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and literature references. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. MAXWELL,
B.
MATHEMATICAL
H.: "Electrical and Optical Wave Motion," Cambridge, 1915. Applies the functions of this chapter. BATEMAN, H.: "Partial Differential Equations," Dover, 1944. Important applications of toroidal and other special coordinate systems. BRITISH ASSOCIATION: "Bessel Functions, Part I," Cambridge, 1937. Extensive numerical tables of Jo, Ji, Yo, Y1, Jo, I,, K o, and K,. BRITISH ASSOCIATION: "Bessel Functions, Part II," Cambridge, 1952. Numerical tables of J,,, Y,,, I,,, K„ for 2 < n < 20. CARSLAW, H. S.: "Mathematical Theory of the Conduction of Heat in Solids," Macmillan, 1921. Bessel functions and spherical harmonies are applied to problems in heat conduction having boundary conditions identical with electrical problems. COPSON, E. F.: "Theory of Functions," Oxford, 1935. ERDELYI, A., W. MAGNUS, F. OBERHETTINGER, and F. G. TRICOMI: "Higher Transcendental Functions," 3 vols., McGraw-Hill, 1953. Very complete and concise descriptions of functions with formulas. (HTF) ERDELYI, A., et al.: "Integral Transforms," 2 vols. McGraw-Hill, 1953. A large collection of all types of integral transforms. (IT) FLETCHER, A., J. C. P. MILLER, L. ROSENHEAD, and L. J. COMRIE: "An Index of Mathematical Tables," 2 vols., Blackwell, Oxford, 1962. GEIGER-SCHEEL: "Handbuch der Physik," Vol. III, Berlin, 1928. GRAY, A., G. B. MATHEWS, and F. M. MACROBERT: "Bessel Functions," Macmillan, 1922. Includes physical applications, numerical tables, and problems, as well as mathematical developments. HARVARD UNIVERSITY COMPUTATION LABORATORY: "Tables of Bessel Functions of the First Kind of Orders 0 to 78," 11 vols., Harvard, 1946 to 1950. HossoN, E. W.: "Spherical and Ellipsoidal Harmonics," Chelsea, 1955. Includes almost everything known about the subject. JAHNKE, E., F. EMDE, and F. Loser': "Tables of Higher Functions," McGraw-Hill, 1960. Formulas and numerical tables with outstanding illustrations.
BATEMAN,
246
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
F. M.: "Spherical Harmonics," Dutton, 1927. Includes physical applications and a chapter on Bessel functions. MAGNUS, W., and F. OBERHETTINGER: "Formulas and Theorems for the Special Functions of Mathematical Physics," Chelsea, 1949. A fine collection of formulas for the functions of this chapter. (MO) MCLACHLAN, N. W.: "Bessel Functions for Engineers," Oxford, 1934. Many applications. MORGAN, S. P.: "Table of Bessel Functions of Imaginary Order and Imaginary Argument," California Institute of Technology, Pasadena, 1941. NATIONAL BUREAU OF STANDARDS: "Handbook of Mathematical Functions," U.S. Bureau of Commerce, 1964. The best collection of formulas and numerical tables for the functions of this chapter. (HMF) RIEMANN-WEBER: "Differentialgleichungen der Physik," Vieweg, Braunsweig, 1925. RYSHIK, I. M., and I. S. GRADSTEIN: "Tafeln," Deutscher Verlag der Wissenschaften, 1957. A large collection of useful formulas. WATSON, G. N.: "Theory of Bessel Functions," Cambridge, 1922. The standard treatise on the subject. It includes some numerical tables. WEBSTER, A. G.: "Partial Differential Equations of Mathematical Physics," Hafner, 1950. WHITTAKER, E. T., and G. N. WATSON: "Modern Analysis," Cambridge, 1935. MACROBERT,
CHAPTER VI ELECTRIC CURRENT 6.00. Electric Current Density. Equation of Continuity.—If we touch two conductors A and B, charged to potentials VA and VB, respectively, to two points in a third conducting body, we have seen in 1.00 that electric charge flows from one to the other until the potential of A equals that of B. We observe two phenomena associated with this transfer; one is the heating of the conductor and the other is the existence of a magnetic field in the neighborhood, while the flow is taking place. The second phenomenon will be considered in the next chapter. The rate of transfer of electric charge between A and B at any instant is called the electric current, which is therefore defined, in any system of units, by
I = dQ di
(1)
When Q is measured in coulombs and t in seconds, I is measured in amperes. If, by some mechanical electrostatic means, such as a moving insulated belt, we transfer, continuously, a charge from the point of contact of B to that of A at such a rate that the difference of potential VA — VB is kept constant and at the same time cool the conductor so that its temperature has a fixed value, we find that the current and the magnetic field remain constant. The latter, therefore, need not enter into our considerations of steady currents. Electric current, at any point, is evidently a directed quantity. If, at the point P in a conducting medium, we take an element of area dS, which is normal to the direction of the electric current at P, and if the current flowing through this element of area is dI, then we define the current density at P to be
.
dl = dS
(2)
When a steady current flows, the amount of electricity entering any element of volume must equal that leaving it. Thus, the surface integral of the normal component of the current density over this elementary volume is zero. This gives, by Gauss's theorem, 3.00 (2),
fsi • n dS fvV • i dv = 0 Since this holds for all volume elements V • i = div = 0 247
(3)
248
ELECTRIC CURRENT
§6.01
This is known as the equation of continuity, and a vector which satisfies this equation everywhere is said to be solenoidal. 6.01. Electromotance.—In the experiment considered, the belt is an agent, known as an electromotance, which exerts a mechanical force on the electric charges carried by it just sufficient to overcome the electrostatic forces between A and B. The work in joules which the belt does in moving 1 coulomb of positive charge from B to A, after eliminating friction losses in the driving mechanism, etc., measures the magnitude, 6', of the electromotance in volts. If the force driving the belt is increased, it will accelerate until the additional charge conveyed increases VA — VB so that the electrostatic forces on the charged belt just balance the driving force. If the contact between A and B is broken, the belt stops, since it cannot drive against increased electrostatic forces, and work against belt friction ceases, so that the electromotance e exactly equals VA - VB. In order to maintain an electric current, at ordinary temperatures, it is necessary to have some source of electromotance in the circuit. In the present case, the location of the electromotance is perfectly clear. It is distributed along the belt between A and B. In chemical sources of electromotance, such as storage batteries and primary cells, it is located at the surfaces of the electrodes. In thermocouples, it lies in the interface between the members. When a whole metallic circuit is placed in a varying magnetic field, the induced electromotance, discussed in Chap. VIII, may be distributed over all the elements of the circuit. When a dynamo supplies power to a circuit, the electromotance is spread over its coils. In all these cases, if the circuit is opened and the difference of potential across the opening is measured by some device drawing no current, such as an electrostatic instrument or a potentiometer which is screened from induced effects, the result equals the sum of the electromotances around the circuit. We usually define the electromotance around any given path in a different way. We defined the electrostatic field intensity at a point as the force acting on a stationary unit positive charge placed there. The nature of the force was not specified, except that it was taken proportional to the charge. As we shall see in subsequent sections, this measurement is not affected by the electrical resistance of the medium, as the latter, like viscous friction, acts on moving bodies only. Thus, if we examine the force acting on a charge on the belt, we find it to be zero, since the mechanical and electrostatic forces balance. In the part of the circuit where the current is flowing, only the electrostatic forces are present. Thus the line integral of E around the circuit just equals the potential difference between A and B, which, as we have seen, equals 6'. Thus we find 6 = j'E • ds (1)
OHM'S LAW RESISTIVITY
§6.02
249
which defines the electromotance around the path of integration. It is sometimes instructive to consider the electric field intensity E, as made up of two parts, viz., the electrostatic part E', which can be derived from a scalar potential, and for which the line integral around any closed path vanishes; and a part E" due to the electromotance, which is solenoidal in nature and does not necessarily vanish when integrated around a closed path. The intensity E" helps to visualize the case of a distributed electromotance but is not so satisfactory when the electromotance is located in a surface layer, since, in the latter case, it is infinite in these layers and zero everywhere else. The value of the integral of E" will be different around different paths. This is illustrated, in the example given, by the fact that if the path is chosen entirely on the belt or entirely off it the integral is zero. In many cases, it is possible to place barriers in the path of integration so as to exclude the sources of electromotance, such as the belt in the example given. The line integral of E over all permitted closed paths is then zero so that we can use a scalar potential throughout this region. Even when the electromotance is distributed, a barrier may exist such that, for permitted paths, the integral of (1) is zero. In this case, we are unable to distinguish between E' and E" and can use the terms electromotance and potential interchangeably. This is the sense in which E appears in the next article except in Eq. (4). We may write, in this case, E = —VV = Ve (2) 6.02. Ohm's Law. Resistivity.—If, in the experiment of 6.00, all physical conditions, such as temperature and working efficiency, are kept constant and the electromotance is increased we find that when equilibrium is reached the current has increased in the same ratio. This is Ohm's law. For a perfectly efficient machine the ratio of the electromotance between A and B to the current flowing is called the electrical resistance RAB between these two points, and so we have RAB =
VA V )3
e AB
[ AB
[ AB
(1)
Let us consider an elementary cylinder at the point P in a conducting medium whose ends, of area dS, are normal to the direction of the current at P and whose walls, of length ds, are parallel to it. The electromotance between the ends is then OE as) ds = VE • ds, and the current flowing through them is i dS. Assuming that Ohm's law applies to this cylinder, we may write (1) in the form VE • ds Ivel cos a ds RP =
where
a
.
s dS
i dS
is the angle between VC and ds.
If we take ds numerically
250
§6.04
ELECTRIC CURRENT
equal to dS, we write T for Rp and call it the resistivity or specific resistance of the medium at the point P. Thus
Ivel
(2)
=. COS CC 7,
If T is to be independent of the direction of the current, then a = 0 so that the potential gradient and the current must be in the same direction. In this case, we say the conductor is isotropic. The conductivity y is defined as the reciprocal of the resistivity T, so that for an isotropic medium (2) becomes ye i= = leYre (3) T
For a closed path in a conducting medium, we have from (3) and 6.01 (1) .95'i • ds = -ye
(4)
6.03. Heating Effect of Electric Current.—We mentioned in 6.00 that the passage of an electric current heats a conductor. To keep the temperature of the conductor constant, we must remove this heat by some cooling device. If no changes are taking place in the conductor, then the energy which we thus obtain must enter the system in some way. Evidently, we do mechanical work against electrostatic forces in carrying the charges from B to A on our insulating belt. From the definition of potential, this work is W = Q( VA — Vs), so that the power supplied is P dW =
=
, dQ A — v B)(— )= dt
V Wks
(1)
Substituting from 6.02 (1), we have P = PABRAB
(2)
If I, V, and R are all in esu or if they are all in absolute emu, then P is in ergs per second. If I is in amperes, V in volts, and R in ohms, which is the practical mks system of units, then P is in watts. A complete table of these units appears in the Appendix. The heat that is generated in this manner is called Joule heat. 6.04. Steady Currents in Extended Mediums.—In 6.00 (3), we showed that if, in the steady state, no charge is to accumulate at any point in a conducting medium, the divergence of the current density i must be zero at all points. In 6.02 (3), we found that Ohm's law, in an isotropic medium, requires that i be proportional to the potential gradient and inversely proportional to the resistivity. Combining these two equa-
§6.04
STEADY CURRENTS IN EXTENDED MEDIUMS
251
tions, we have V • i = — V • T-1V V = 0
(1)
If the medium is homogeneous, T is constant throughout it, so that this becomes (2) vzV = 0 Comparing (1) and (2) with 3.02 (2) and (3), we see that these equations are identical with Laplace's equation and that the reciprocal of T, called the conductivity, plays exactly the same role as the capacitivity does in electrostatics. It follows that all mathematical technique used in electrostatics also applies here. The tubes of flow here bear the same relation to the equipotential surfaces as the tubes of force do in electrostatics. At the boundary between two conducting mediums, both the potential and the normal component of the current density must be continuous so that, from 1.17 (5) and (6), the conditions are
1 ay' = 1 av" T" an
(3)
T' an
(4)
= V"
These two equations determine completely the relations between the potential gradients on the two sides of the boundary, so that if the relative capacitivities of the two mediums also differ and are not proportional to the conductivities, another variable is required to satisfy this additional condition. By taking a charge density o- on the boundary, this can be done, giving, by (1),
av1— K2aV 2 = _(Kin an an
o- = K1 Eu
K2T2)in
(5)
Anyone observing the magnetic forces between conductors carrying currents, which is the subject of the next chapter, might be led to believe that the current distributions computed by solving (1) or (2) would be erroneous owing to the mutual displacement of current elements by magnetic interaction. Such is not the case. It is true that, in isotropic nonferromagnetic conductors, there is always an increase of resistance in a magnetic field, sometimes known as the "longitudinal Hall effect." The resistance of those portions of an extended conductor where the magnetic field is most intense is therefore increased relative to that of other portions, producing a decrease in the relative current density. Thus when a steady current flows in a cylindrical conductor, the current density is slightly larger near the axis. If this conductor is placed in an external uniform transverse magnetic field, which adds to the field of the current on one side of the cylinder and subtracts from it on the
252
ELECTRIC CURRENT
§6.05
other, we obtain a lateral displacement of the current in the same direction as the force on the conductor. This does not change the orientation of the equipotential surfaces and does not imply any transverse force on the current carriers. Although the change of resistance can be measured, the resultant current displacement is too small to observe at ordinary temperatures. The usual transverse Hall effect changes the orientation of the equipotential surfaces in either direction, depending on the material, without, in this example, affecting the current distribution. It is possible, however, to arrange a circuit so that the latter is changed. This effect can be observed only by using very strong magnetic fields and sensitive instruments. Thus, we may conclude that the distortion of the current distribution in extended conductors caused by magnetic interactions is negligible, and the results obtained by fitting solutions of (1) and (2) to the given boundary conditions may be considered rigorous, as far as such effects are concerned. Since, as we have seen, heat is generated by the passage of electric current and since, in general, the resistivity depends on the temperature, there will actually be deviations from the solutions obtained where heavy currents heat those portions of the conductor where the current density is highest. The solutions are therefore strictly true only when the temperature coefficient of resistance of the medium is small, when the current density is small, or when the current has been circulating only a short time. 6.05. General Theorems.—The general theorems derived from Green's theorem in Chap. III may now be restated in a form suitable for the present subject. 1. If the value of the potential V is given over all boundaries of a conductor as well as the size and location of all sources or sinks of current inside, then the value of V is uniquely determined at all points in the conductor. 2. If the normal component of the current density is given over all boundaries of a conductor as well as the size and location of all sources or sinks of current inside, then the value of the potential difference between any two points in the conductor is known. 3. If the resistivity of any element in a conductor is increased, then the resistance of the whole conductor will be increased or remain unaltered. 4. If the resistivity of any element in a conductor is decreased, then the resistance of the whole conductor will be decreased or remain unaltered. To these we may add another theorem which we shall then prove. This is 5. The current density in a conductor distributes itself in such a way that the generation of heat is a minimum.
§6.06
253
CURRENT FLOW IN TWO DIMENSIONS
To prove this, let us suppose that there is a deviation from the distribution given by Ohm's law [6.02 (3)], the additional current density being j. In order that there may be no accumulation of electricity, j has to satisfy the equation of continuity V • j = 0. Applying 6.03 to an element of a tube of flow in the conductor, we have, for the heat generated in it, from 6.02 (3) and 6.03 (2),
dP
=R— --11717.
dS
s =T dS N_ d
vv
j) 2 dv
where dS is the cross section of the elementary tube and ds is its length. Integrating gives P
1 (V V) 2 — 2j • VV TP]dv = fV[r
(1)
Applying Green's theorem [3.06 (3)] to the second term, where j = Vc1) and NI/ = V, we have fv j • V V dv = — f VV • j dv
f Vn • j dS
The first term on the right is zero since V • j = 0. The second term is zero since the total electrode current is fixed. Thus, the first and third terms in (1) remain. The first gives the rate of heat generation when Ohm's law is followed, and the third, being positive, shows that any deviation from this law increases the rate of generation of heat. 6.06. Current Flow in Two Dimensions.—As pointed out in 4.00, there are, strictly speaking, no two-dimensional problems in electrostatics, since all cylindrical conductors are finite in length and there is no way to terminate the electric field sharply at the end, there being no known medium whose capacitivity is zero. In the case of current flow in conductors, we have, however, many rigorously two-dimensional problems, since the current may be confined to a finite region by an insulating boundary. All cases of flow in thin plane conducting sheets are of this type. We thus have available all the methods of Chap. IV for treating these problems. The most powerful of these is that of conjugate functions. We saw in 4.09 that a solution of 6.04 (2), when the potential is a function of x and y only, is U(x, y)
or
(1)
V(x, y)
where
W = U
V = f(z) =
iY)
(2)
254
§6.07
ELECTRIC CURRENT
The current density i at any point is, from 4.11 (2) and 6.02 (2), if V is the potential function, .
=
tau — T as
(3)
l ay 1 101W =— au—— + T as T an T dz
(4)
1 dW
dz
l ay = T an
=——
If U is the potential function, this becomes
.
=
the same sign convention being used as in 4.11 (2). Thus, if the conductor is bounded by the equipotentials U1 and U2 and by the lines of force V1 and V2, the current flowing through it will be Va / = fTT'
i ds=
_11 f v'aU ds= 1 f v2aV ds _ V2 — VI T vi an T vi as
By Ohm's law, the resistance of the conductor is
R = U2 — U1lIU2 — Uil = T IV 2 — V1I
(5)
If the equipotentials U1 and U2 in (5) are closed curves, then the electrostatic capacitance between the electrodes, in vacuo, is, from 4.11,
C —
1U2 —
(6)
UiI
where [V] is the integral of V around Ui or U2. So that if C is known, we may find the resistance between U1 and U2, if the resisting medium fills the same space as the electrostatic field would, by combining (5) and (6), giving
R = T61
(7)
Thus, from 4.13 (5), we see that the resistance per unit length between two parallel cylindrical electrodes of radii R1and R2 with a distance D between their centers is
R = —cosh-1( -F D 2ir
—
(8)
2R IR 2
where T is the resistivity of the medium between them, and we use the positive sign if they are outside each other and the negative sign if one is inside the other. 6.07. Long Strip with Abrupt Change in Width. Let us apply the theory just developed by computing the current distribution in a long straight strip of conducting material of uniform thickness whose half—
§6.07
LONG STRIP WITH ABRUPT CHANGE IN WIDTH
255
width changes abruptly from h to k. The region near the junction is shown in Fig. 6.07b. Such a boundary can evidently be obtained by bending the real axis in the zrplane, the angles being 7r/2, 37r/2, 0, 37r/2, and 7r/2 at —1, —a, 0, +a, and +1, respectively. To have a finite distance between the points —a and +a after the bending, it will be
necessary to put the origin in the z1-plane at z = — j co in the z-plane. Putting these values for the a's and the u's in 4.18 (7), we have dz = dzi
(z? — a2)i
cz1
— 1)i
[(4 —
— a2)]+
ca2 z 1[(4 — 1)(z1 — a2)11
(1)
We shall evaluate the constants c and a by the method of 4.28 before performing the main integration. As in 4.28, when r1is constant, we have dzi = jrieh do t = jz1 dot. When r1is a very small constant and 61goes from 0 to 71" in the ziplane, then y is a large negative constant and x goes from +h to — h in the z-plane. Thus, substituting in (1), we have
-h
r2e2jej
dzz — 3
o
J+h
giving h = 4jc7ra. stant so that
— 1
del
= iCi a dOi
Similarly, when r1—* 00, y is a large positive con-
k
f+k
a2y
1r
dz — jc[ - .
i le2 j9i — a2dB = + jc f del ' lv e et _ 1 1ri—,.., o 0
256
§6.07
ELECTRIC CURRENT
giving k = -Ti jcir.
Solving for c and a gives and
a=
2jk
(2)
c = — 7r
a2)/(1 u2) for Putting these values into (1) and substituting (u2 zi and 0/4, in the first and second terms of (1), respectively, we may integrate by Pc 49 or Dw 120, giving
— z = -{k tan-' rzi (1 7r
a2)]1
h tan-1 [a
—
(1 — a2)i
f
No integration constants are added since this gives, when y, = 0 and x i = + a, z = +h, and when yi = 0 and xi = +1, z = +k. The electrical conditions require that we have a different line of flow on each boundary. Clearly, if we take W = In zi or zi = ew, where U is the potential function, then the line of flow V = 0 passes from xi = 00 to x, = 0 in Fig. 6.07a and the line of flow V = ir passes from xi = — 00 to xi = 0. In Fig. 6.07b, therefore, these lines follow the right and left boundaries of the strip, respectively, from top to bottom, giving a total current I = 7r/s in the strip. Substituting for zi, we have 2
z = -[k tan- 1
(w o _ a2y 1—
e2W
1 — e2 wil \ a2 )
h tan-1 a( e2w —
(3)
where the real part of z is positive when the real part of z, is positive. If s is the resistance between opposite sides of a one meter square cut from the same piece as the strip, then the resistance of the length yh of the wide part of the strip is Rh = slyh1/2k and that of the narrow part alone is Rh = Siyhi/2h. When they are joined as in Fig. 6.07b, the resistance between the ends is not Rh + Rh but there is an additional resistance OR due to the distortion of the lines of flow near the junction. When yk >> k and yh >> h, equipotentials are parallel to the axis in Fig. 6.07b. Let us compute OR in this case. On the y-axis, V = ir so that e 2W = — e2U and (3) becomes, writing j tanh-1u for tan-1ju, x = 0 and
y =
e 2U + tanh-1 ( 1 + e2U
Let U = and 3.3
h tanh-1al
+ e2a27]
(4)
co, then ew is very large and from Pc 753 or Dw 4 2
yk = 7r (k
2e 2u, tank-1 2e 2u,
+
-
2e2U1 ± 1) a2 1 — h tank 1 a2e2U1 a2
CURRENT FLOW IN THREE DIMENSIONS
§6.08
257
Using Pc 681 or Dw 702 and neglecting 1, a, and a2 compared with ew give yk = —k In ir
2k Ui
7r
4e2ui 1 — a'
1
h In
1
+ a\ —
+1(k ln 4k2 ir — h2
h ln
k+ k—h
2kU 1
+A
Let U = U2 -- — co, then e2U is very small and, as before, Yh =
=
2 r
+ e 2U2
2a2 k tanh-1
l[k In 1+a r 1 a
h In
—
=
+ k In k ir k
+ e 2U2) 1
' 2a2 + e2u2
4a 2 1 (1 — a2)e2u2]
k+h
2h U2 1(
r
a2(2
h tanh-1
a(2 + e2u2)
h
h In 24 k ) L'i2h2
2h U2 + r
B
Subtracting and solving for U1 — U2 give
ui _
-
= ryk 7yh 2k 2h
B\ ir (A Vk 2h )
By 6.06 (5), R=s
U2
2\k
B\ + sy k + l Y h i h ) 2k 2 h _ OR + Rk Rh
Substituting for A and B, solving for AR, and simplifying give [h2
OR —
k2 _ h2] k2 k h + 2 ln ln 4hk hk k—h
(5)
Other examples of this method will be found in the problems at the end of this chapter. 6.08. Current Flow in Three Dimensions. If the entire volume between two electrodes is filled with a uniform isotropic conducting medium, then we may obtain the current distribution and the resistance between the electrodes from the solution of the electrostatic problem for the capacitance between the same electrodes when the medium is insulating. In both cases, the equation to be solved is —
(1)
V2-17 = 0
In the electrostatic case, the boundary conditions on the electrode a in vacuo are, from 1.15 (1),
V
= v.
Q„ — f ev
av dSa
s an
(2)
258
§6.09
ELECTRIC CURRENT
In the current case, they are, from 6.01 (2), and 6.02 (3),
lay
V = vaIa = — f s —T on uoa
(3)
The equipotential surfaces correspond exactly in the two cases, since the boundary conditions are identical. From Ohm's law, the resistance is Vb Vat Te,, R = 117b — Val (4) = TEv C IQ.I where C is the capacitance in vacuo in the electrostatic case. If an electrostatic case can be found such that the walls of a tube of force are identical in shape with the insulating boundary of a conductor of resistivity T and the equipotential ends of the tube have the same shape as the perfectly conducting terminals of this conductor, then the resistance between these terminals can be computed by (4) from the "capacitance" of the tube of force. By the capacitance, we mean the ratio of the charge on an end to the difference of potential between the ends. 6.09. Systems of Electrodes. Two Spheres. Distant Electrodes.— When n perfectly conducting electrodes are immersed in a homogeneous isotropic conducting medium of resistivity T, we may find all the relations between the currents entering and leaving the electrodes and their potentials by the methods of Arts. 2.13 to 2.18. It is necessary only to write the current I$ from the sth electrode in place of Qs, in the electrostatic case and to multiply the capacitances by (rEs)-1and the elastances by rev.
Thus, to find the resistance between two spheres of radii a and b, internal or external to each other, having a distance c between their centers, we have, from 2.15 (1), since I1= —/2, V1 — V2 = rev(811 — 2512 + 822)/1
so that the resistance is R—
V1 — V2
Il
Tev(Sii — 2812 + 820
In 5.08, we computed by images the values of case. In terms of these, by 2.16 (2), we have R=
rev
C11 + 2 C12 + C22
C11C22 — .12
C12,
(1) and c 22 in this
(2)
If one sphere is inside the other, this formula is of little use and the method of 5.19 can be used.
§6.10
259
SOURCE AND SINK IN SOLID SPHERE
If two electrodes are at a great distance from each other in an infinite conducting medium, we have, from (1) and 2.18 (1), the result 6'
R T
Co
T(
— 27rr 1 ±Cb
(3)
where r is the distance between them and Ca and Cb are the electrostatic capacitances of a and b alone. 6.10. Source and Sink in Solid Sphere. Spherical Bubble.—The problem of the calculation of the potential when a current I flows from a source to a sink in an insulated sphere of resistivity T indicates how spherical harmonics apply to insulating boundaries. Superposition of the potential of a source I at P1and a sink —I at the center on that of a sink —I at P2 and a source I at the center gives the potential with a source at P1 and a sink at P2 and simplifies calculation. Let the source of the first pair be at 0 = 0, r = b, which is a distance r1from the field point. Add to the potential of this pair a solution Vsof Laplace's equation which makes a V/ar zero at r = a and has no poles if r < a. From 5.153, if b < r < a,
_ 1
TI (1 _1) + vs _
TINF
47r\ri
47r—J1_7-
r/
b: n 1+ (n
n =1
1)rnbn Pn (cos 0) nen-El
(1)
The Vs term may be written, with the aid of 5.153 (1), Vs
=
•
rnbn
f r rnbn dr)
,,IVi 271-1-1
0 a2n+1
n =1
r)
TI F a br/i
1
a
f ( barl
Jo
1)dri
a) r (2)
where the distance from P to the source image at 0 r c. =
= 0, r =
a2b—' is
(r2 +a4b_z— 2a2rb--' cos 0)1
(3)
Insertion of (3) in (2), integration by Dw 380.111, and substitution of the resultant expression for Vs in (1) give ,r/[
1
V 1 =7r — — 4 r
1
r—
a
1
(br'i a— In
a2 — br cos 0) 2a2
( (4)
Equation (3) gives cos 0 in terms of a, b, r, and ri. In the general case where the radius vectors to the source, sink, and field point are a1, az, and r, two potentials, one with a source at a1, sink at center, and one with sink at az and source at center, are superimposed. Equation (4) then gives T/11 r1
1+ a r2
a azr
1 (a2 + airD 2 — 0.21 In 2r2)2 — 4r2 ] a (a2 a
(5)
260
ELECTRIC CURRENT
§6.11
Here r1, rz, ri, and r2 are the distances from the field point to the source, sink, image source, and image sink, respectively. When source and sink are at opposite ends of a diameter, r1 = rz = r2, and al = az = a. Thus Eq. (5) becomes V=
T/[ 1 r1
1 rz
1 in ri(1 + cos al)] 2a r2(1 + cos az)
(6)
where al and az are the angles at which r1and rz intersect the diameter. Any tube of flow in this case is a surface of revolution whose equation can be found by equating to the current I' in the tube the integral of the normal current density over any cap bounded by it. It is simpler to use a spherical cap centered at the source for the r1terms in (6) and one centered at the sink for the rz terms. Thus
Ii = — Ti f
ri
dS = 1(1 + —)f sin 01 del 2a 0
Since the fluxes from source and sink add, integration and the addition of a similar expression for I2 give
I' = I[(1
2ria-1)(1 — cos a l) + (1 + ir2a-1)(1 — cos a2)]
(7)
If is the obtuse angle between r1and rz, this simplifies to
I' = I[1
i(ri
rz)a-1cos 0]
(8)
A solution similar to those given applies to the case of flow from a point source to a point sink in an infinite homogeneous conductor which has a spherical hollow of radius a in it. The equation analogous to (1) for a source at b and a sink at infinity has direct powers of r in the first term since a < b. The perturbation potential of the hole vanishes at infinity and so contains reciprocal powers of r. In the equation analogous to (2) the second and fourth terms cancel, so that =
riar 4/rbLri
f dr jr rr;. -
(9)
where 7-'1is still given by (3). Integration and addition of a similar term for the sink yield a result like (5) except that the last term is replaced by 1 az(1 — cos 09)(air'i — ln a ai(1 — cos 01)(azr
a' — air cos 60 a2— azr cos 02)
(10)
6.11. Solid Conducting Cylinder.—To illustrate the application of Bessel functions to insulating cylindrical boundaries, we shall compute the potential anywhere inside a solid circular conducting cylinder of length 2c, radius a, and resistivity T when the current I enters and leaves
SOLID CONDUCTING CYLINDER
§6.11
261
by thin band electrodes at a distance b on either side of the equator. We take the width of the band too small to measure physically but mathematically not zero so that the current density and potential functions are everywhere bounded. Taking the equatorial plane at potential zero, we see that a solution of the equation of continuity which is finite on the axis is, from 5.292 (3) and 5.32, V = ZAnI o(k,,P) sin k„z
(1)
Since z = c is to be an insulating boundary, we must make a V/az = 0 there. Since cos (2n + 1)471- = 0, we satisfy this boundary condition by taking (2n + 1)r kn = (2) 2c To determine A„, we differentiate (1) with respect to p by 5.33 (4), put r = a, multiply by sin k9z, and integrate from 0 to c. By Dw 435 or Pc 359, the only term left on the right is that for which k, = kn so we have, writing n for p,
cav sin knz dz = A „I i(k„a) o ap
knc
sin2 knz d(k„z)
—
0
We use Dw 430.20 or Pc 362 for the right-hand integral. On the boundary, p = a, a V/ap is zero except in the area AS, at z = b, covered by the electrode, which is so small that we may give sin knz the constant value sin knb therein. From 6.02 (3), setting 2ra dz = dS, the left-hand integral becomes sin knb
aira
T As
i dS =
2ra
sin knb
Solving for An gives A =
2TI sinknb (2n + 1)7r2a I i (knd)
(3)
The desired solution is given by (1), (2), and (3). The boundaries of the tubes of flow, in this case, are surfaces of revolution. The equation of the tube through which a current I' flows can be obtained by integrating the current density — (1/T)(0V/az) over a disk bounded by the tube and equating it to — I', giving 1 = -I- T.
2rknA„ cos knz f I o(knp)P dP n=0
262
§6.12
ELECTRIC CURRENT
Integrating by 5.33 (5), we have, substituting for A. from (3), /' =
k.b cos k„ili(ko)
(4)
1)/i(kna)
(2n
era n=0
where is given by (2). 6.12. Earth Resistance.—Geophysicists sometimes investigate the structure below the earth's surface by observing the distribution of potential on the surface when current is passed through the soil between two or more surface electrodes. Let us investigate the simplest case, in which to a depth a the resistivity is T1and below this depth it is T2. We shall use the method of 5.303 to find the distribution about a singlepoint electrode. The case of two or more electrodes may then be found by superposition. The potential due to the electrode alone may be written down from 5.303 (1), by substituting for q, from 6.08, the value 2TEL It should be remembered that 2/, in this case, corresponds to 1 in 6.08, since all the current flows in half the field. Then, we have V
2rr
=
2r
fo
~
J o(kp)ekizi dk
(1)
owing to the electrode alone. As in 5.303 in the region of T1, we shall superimpose a second potential due to the discontinuity at z = a that may contain both the e—k' and ekz terms since z is finite in this region, thus we have V1 =
1.(k)J o(kp)e-kz dk
f %11(k),10(kp)e±kz dk
f
J o(kp)e-k' did
(2)
Since the earth's surface z = 0 must be a line of flow and since the last term already meets this condition, the remaining terms must satisfy it independently so that, if 8171/az = 0 when z = 0, —4,(1c)
I'(k) = 0
(3)
The potential in the region T2 must vanish at infinity and so can have only the form
V2 = Tlif
27r o
(4)
0(k),10(kp)e-k' dk
At z = a, the boundary conditions are, from 6.04 (3) and (4), V1 = V2
and
1 avi T1
az
1 al72
az
§6.13
CURRENTS IN THIN CURVED SHEETS
263
Substituting from (2), (3), and (4) and canceling out Jo(kp) give 43(k) (eka
e—ka — T2e — k
a
,24 (k)(eka
e—ka)
e
ka\
)
e(k)e—ka = 0 rie (k)e-ka = 0
(5) (6)
Eliminating e(k) and substituting in (2), we have, on the earth's surface where z = 0 and setting (71 — 72)/(T1 + T2) = 0, the result
=
ri
oe-2ka Jo(kp) dk 27r- Jo 1 + Oe-2ka —
(7)
Expanding the denominator by Pc 755 or Dw 9.04 and interchanging the integral and summation, we have V8 = 2[ I
f
Jo(kp) dk
2nka o (k p) did
(-1)n0nf n=1
Substituting for the integral, from 5.298 (6), gives
V'
=
ri/[1
2a Lp
2N n=1
(-1)n0n
(4n2a2
p2)1
(8)
Now if we take the earth's surface as the xy-plane and the current I passes from an electrode at x = +b to one at x = — b, the potential at the surface is, since p+= [(x — b) 2 y 2]+ and p_ = [(x b) 2 y9+1, Ve =
ri/ { 1 L7 p+ — p_
+
( —0)n[(4n2a2 +
-i]} — (4n2a2 pl)
(9)
n=1
The case for any number of layers has been worked out by Stefanesco and Schlumberger, Journal de Physique, Vol. I, p. 132, 1930. 6.13. Currents in Thin Curved Sheets. If a thin curved sheet of uniform thickness can be developed into a plane, then, by so doing, the current distribution in such a sheet can be reduced to two dimensions and 6.06 applied. If, however, the surface cannot be so developed, other treatment is necessary. From 3.03 (4) and 6.04 (2), the equation o f continuity in orthogonal curvilinear coordinates is —
a (h2h2 av) aui h1 au,
a (h3hi av) au2 h2au2
a eih2 av) = 0 011,3 au3
(1)
Let us suppose that u1and u2 are a set of orthogonal curvilinear coordinates drawn on the surface and let u3be a distance measured normal to it. The surface is so thin that the current distribution on each side is the same, making a V/au3= 0. The thickness of the surface is uniform so
264
ELECTRIC CURRENT
§6.14
that h3 is independent of u1and u2. Under these conditions, ( 1 ) takes the form + a (hi av\ = 0 a th2 (2) aui \hi aui) au2\h2 au2) This is the equation that we must solve to get the current distribution. 6.14. Current Distribution on Spherical Shell.—From 3.05 ( 1 ) , letting V be independent of r, we get 6.13 (2) for a spherical shell to be sin 01(sin
ao
Ba aog
+ago gao = 0
(1)
where 0 is the colatitude angle and 95 is the longitude angle. Although a spherical shell cannot be developed in a plane, it is possible to represent every point of a spherical surface on an infinite plane in such a way that angles are preserved. This process is known as stereographic projection and closely resembles inversion. A plane is taken tangent to the sphere at one end of a diameter. A line passing through the other end 0 of this diameter and the point P (Fig. 6.14) intersects the plane at P' which is called the projection of P. Let be a solution of the equation of continuity in the plane, where, from 4.01 (2), it takes the form
a (aT) ,
r r ar ar
a24,
u
802
(2)
In projecting the system of curves, which represents, onto the sphere, the longitude angle will be unchanged. From Fig. 6.14, we see that r and 0 are connected by the relation r = 2a tan 01 = 2a tan 10 The equation that the projected curves satisfy is obtained by substituting 0 and 4) for r and 95 in (2).
. a dO a r— = r— — = sin ar
dr
a8
so that, from (2), sin 0 (in
a
ao
+
a ae
11—
adz
=0
This is identical with (1) so that the stereographic projection of a solution of the equation of continuity in a plane gives the solution of this equation for a thin uniform spherical shell.
§6.15
SURFACE OF REVOLUTION
265
We found, in 4.09, that both U and V are solutions of V' V = 0 where U + jV = f(x + jy) = f(r cos cp ± jr sin q5) if f(z) is an analytic function. It follows from the preceding that if
U + jV = f[2a tan 4e(cos 0 + j sin 0)]
(3)
then both U and V are solutions of the equation of continuity on the surface of a sphere of radius a and that if U is the potential function then V is the stream function, and vice versa. From the laws of inversion, it follows that lines in the plane project into circles through 0 on the sphere and circles project into circles. As an example, let us find the potential at any point on a spherical shell of radius a and surface resistivity s when a current I enters at 0 = a, 0 = ,}ir and leaves at 0 = «, 4 = —47. From 4.13 (2.1), writing from 6.08, e„s I for the charge which, in 4.13, is 2/7- 6„, we have for the potential U at any point on a plane sheet 2,7r u x 2 ± y2 ± b2 r2 ± b2 = coth 2by 2br sin 4, s/ But from Fig. 6.14, we have r = 2a tan 40 so that coth
1 — cos a cos 0 27U tan2 -0 + tan2 4-a = — sin a sin 0 sin ct, 2 tan 40 tan 4« sin 0 s/
(4)
In a similar fashion, the equation of the lines of flow is, from 4.13 (4), cot
27- V
s/
=
r2 — b2 cos a — cos 0 — 2br cos 4,sin a sin 0 cos 4'
(5)
6.15. Surface of Revolution.—As another example, let us solve 6.13 (2) for the surface of revolution formed by rotating the curve y = f(z), where Az) is a single-valued function, about the z-axis. Clearly, on such a surface, any point P can be located by the orthogonal coordinates z and 4), where y6 is the longitude angle about z. In 6.13, let ul, which lies along the surface, equal z numerically and let u2 = 4,. From symmetry, h1and h, are independent of .45; so 6.13 (2) becomes
h2a(h2 av) i_ 8 2v hiaz h, az F 802 = °
(1)
-
To solve this equation, let us take a new variable u that is zero when z is zo and satisfies the relation
a _ h2 a au — hi az
Or
z h1 u= f — ,.. dz zo n2
(2)
266
ELECTRIC CURRENT
§6.15
Then (1) becomes 32V
32V
au 2 + ao 2 = 0
(3)
From 4.09, a solution of this equation is known to be either U(u, 0) or V(u, 95) where U jV = F(u jcP)
(4)
To evaluate u in terms of given quantities, we must calculate h1 and h2. The equation of the surface is p = f(z)
so that
dp = (z) dz
(5)
and we have ds2 = dp2
dz2
p2dcp2 =
(z)]2 +11 dz2
{f(z)l 2d(P2
giving hi =
(z)}2
1) 1,
h2 = f(z)
(6)
FIG. 6.15.
Putting these values in (2), we obtain u=
z
fo
{[i (OP ± f (z)
1)1dz
(7)
Since an increase of 2nir in cp returns us to the same line on the surface, it is necessary, in order to have single-valued solutions for the potential, that F(u j(I)) be periodic in with a period 27. The solution is identical in form with the solution for a cylindrical surface which can
§6.16
LIMITS OF RESISTANCE
267
be unrolled into a flat strip. If one end of the surface is closed, as in Fig. 6.15, then f(z) is zero at this point and, from (7), the equivalent cylinder will extend to u = — .0. If one end is open, as in Fig. 6.15, then the equivalent cylinder will terminate at some positive value of u and the boundary conditions on the edge of the cylinder will be the same function of as on the edge of the surface illustrated in the figure. 6.16. Limits of Resistance.— The general theorems of 6.05 frequently enable us to compute limits between which the resistance of a conductor must lie, even when we cannot compute the rigorous value. To get a lower limit, we endeavor to insert into the conductor thin sheets of perfectly conducting material in such a way that they coincide as nearly as possible with the actual equipotentials but, at the same time, enable us to compute the resistance. In this case, we know, from 6.05, that the
result is equal to or less than the actual resistance. To compute the upper limit, we insert thin insulating sheets as nearly as possible along the actual lines of flow, in such a way as to enable us to carry out the resistance computation. We know, from 6.05, that this resistance equals or exceeds the actual resistance. For example, the resistance between two electrodes of a given shape lies between that of two circumscribed electrodes and that of two inscribed electrodes. As a specific example, let us compute the resistance between perfectly conducting electrodes applied at the ends A and B of the horseshoe-shaped conductor of triangular section, shown in Fig. 6.16. To get an upper limit for the resistance, we insert infinitely thin insulating layers very close together which require the current to flow only in straight lines and a semicircle. The length of such a layer, from the figure, is 2c + ir(b x). The area is x dx so that the resistance is dR. —
T[2c
ir(b x dx
x)]
268
ELECTRIC CURRENT
§6.17
All these layers are in parallel so that the upper limit of the resistance is, if Pc 29 or Du,91.1 is used, given by (1), x dx R'
y
oa 2c + id) + rx 1f (f dR 1 „) 1 (T 2c + r(a + b)]-1 — iorka — (2c +
rb)
In
2c + irb
(1)
To get a lower limit of resistance, we insert perfectly conducting sheets across each leg of the horseshoe at M and N. The current in each leg then is uniformly distributed, and so the resistance of the legs is 4cr/a2. In the semicircular part, the conductor has the shape of a triangular tube of force in the two-dimensional electrostatic field given by W = In z. This gives, for the stream function and the potential function the values, respectively, (2) and V= 0 U = In r The resistance of the strip at y in Fig. 6.16 is then, from 6.06 (5),
d n = rir(ln ba±2by dy) i The strips are in parallel so that, when Pc 426 or Dw 620 is used,
/ r \la a +b )-1 — 71-7.1?;' = U dn) = rr (2 fo Inb ± 2y 4 - bln (a + b) — blnb — a The legs and the curves are in series, so that Ri =
T
[4c a2
ir
b ln (a + b) — b ln b — a]
(3)
6.17. Currents in Nonisotropic Mediums. Earth Strata.—Taking the divergence of 6.02 (3), we have v . (-yvV) = —17 • i = 0
(1)
If the medium is not isotropic, the conductivity in different directions varies so that -y cannot be factored out. If, however, the medium is homogeneous, the conductivity in any direction remains unchanged throughout it. In this case, we may choose a set of rectangular axes as in 1.19 (7) and 3.02 (5) so that (1) becomes 7x
a2v1-+ aw 1-yzaw= 0 Ty ax2
ay2
+
8z2
(2)
CURRENTS IN NONISOTROPIC MEDIUMS
§6.17
269
Owing to the method of formation of the soil it frequently happens that the conductivity in the horizontal direction is greater than that in the vertical direction. If we take the z-axis vertical, (2) becomes
,92-v ) -rh( ax2 + ay2 +
az2 = 0
(3)
If we choose a new variable defined by (4)
U = -1 ) Z = aZ yv
this becomes
a2v ax2
a 2v ay2
a2 v
at e
=
o
(5)
We have all the methods of Chap. V available for solving this equation but we must translate the boundary conditions into the x, y, u system and fit the solution to them, or vice versa. Suppose we have a spherical electrode of radius R half-buried in the earth. The boundary condition is then that V = Vo when x2 + y2 + z2 = x2 + y2 + (u/a)2 = R2. Thus, in the x, y, u system, our boundary condition is that for a spheroid whose equation is u2 x2 ± y2 =1 R2 a2R2 From 5.02 (1) and (2), taking a2 = b2 = R2 and c2 = a2R2, the solution is
v = v 0[f
(a2
d0 0)(c2
°' 0)1]/[. 0 (a2 ± 0) (c2
Using Pc 114 or letting x = a2 + 0 and using Dw 192.11, we have, since a > 1, 0)]1 -1[(c2— a2)/(c2 V = Vo tanh tanh-1[(c2 — a2)/c2] But from 5.28 (1) and (2)
c2 + 0 = (c2
a2),12
so, on substituting for a and c, this becomes V = Vo[tanh-i (7h —
7,\1]-1
j
tanh-1
1
(11]-1 smh-i (n 2 — 1)-1 = Vo[ cosh-1 ,.yv
(6)
270
ELECTRIC CURRENT
§6.18
and from 5.28 (4) and (5) u = az = R (a2— x2
1) (1
y2 = R2(a2
2) (712
1)
Eliminating we have a 2z2
X 2 + y2
= 1 (7) 1)n2 R2 2 R2(a2— 1) (n2— 1) The smallest value of n is on the electrode where n =a/(a2 — 1)1. Hence, the denominator of the first term is always larger than that of the second so that the equipotential surfaces are nonconfocal oblate spheroids. On the earth's surface, where z = 0, the equipotentials become (a
'Yu ]4 V = V o[cosh-1 MT sinh-1 R[ 7v (x2 ± y2)
(8)
If R is very small, we have a point electrode. Writing the angle for the arc sinh and putting a constant times the current I, for remainder, (8) becomes V = 61(x2 + y2)--i (9) These curves thus have the same form as for an isotropic medium.
(a)
( b)
(c)
FIG. 6.18a, b, c.
6.18. Vector Potential. Flow around Sphere in Tube.—So far only scalar potentials have been used for current flow. With axial symmetry the vector potential whose curl is io is easier to use for it simplifies the boundary conditions and gives 22rpAo = I for the equation of a tube of flow carrying a current I where p is the radius of a circular tube section. Formulas needed appear in Chap. VII if B is replaced by the current density io. An instructive example is the calculation of the current flow in a solid circular conducting cylinder of radius a when a concentric spherical nonconducting obstacle or cavity partially obstructs the flow. All the formulas needed for solution are in Art. 5.40 except the expression for A given in 7.06 (3). The approach is illustrated by Fig. 6.18a, b, and c. The flow (a) plus the uniform flow (b) yields the flow (c) around the sphere in the tube. If the current density far from the sphere is io, then the
§6.18
VECTOR POTENTIAL. FLOW AROUND SPHERE IN TUBE
271
vector potential of this flow is -io p from 7.02 (8). The sources on the sphere surface required to maintain the circulation of Fig. 6.18a produce the internal and external vector potentials N (A0)in
r)2n-f-1
=
N
PL,+i(cos 0),
C)2n-I-2 (110) =Dn(7,lin+1(cos 0)
n=0
n=0
(1) The physical nature of the sources needed to maintain the circulation in (a) is more evident if iois replaced by B, in which case they are obviously coaxial current loops that lie in the cylinder and spherical shell. The vector potentials of (1) are single-valued even functions of z. Values of Dn must be found such that, in the presence of the cylinder, the internal current density in (a) cancels that of the uniform flow in (b). In cylindrical coordinates (A ,) ex may be written, by 5.40 (3), N (- 1)".2Dna2'1-2r
(14 - q5) =
rn! (2 )
t2n-"Ki(tp) cos (tz) dt
(2)
n=0
Evidently from Fig. 6.18a, since no net current traverses the cylinder, A is zero at both p = 0 and p = b, so the internal vector potential of the sources in the wall that cancel (2) must be, using 5.40 (19), N
(Aib )*, =
f;(ta) 'Alt(at
t n )D 2n ) n2
—
1 7 ( ( -
2n
(tp)
cos (tz) d(ta)
n= 0
N
= n=0
+s2Dnr2s + 1
(_ s=0
r(2n)!(2s + 2)!as+114„ fcos
/1 (2n + 2s + 2) 2n + 2s + 3
(3)
The integral /'(2n 2s + 2) is that of 5.42 (4) but with first- instead of zero-order Bessel functions and is tabulated (Phys. Fluids, Vol. 7, page 635, 1964) and will be written I„,. Inside the sphere only the harmonic --ijorPI (cos 0) or --iio p, which must cancel the uniform flow, can appear. So if s 0, the coefficients of r28-"Ph+1(cos 0) must vanish. Thus each value of s gives a linear relation between the coefficients Dn. So as in 5.41 a set of N 1 simultaneous equations must be solved for Do, D1 , . . . , DN. The value of the right side is zero for all but the first equation, for which it is — The values of Cs„ in 5.41 (5) are, from (3) and (1), S
n
=
1)n+'21 n. (2s + 1) !a28-1-1'
1
( —
7r (2n) !
C nn
c2n-1-1
2/nn 2)!a2n+1 r (2n) !(2n
(4)
272
ELECTRIC CURRENT
§6.19
The above procedure works well for oblate and prolate spheroids including the disk. Extensive numerical values are given in the reference cited. Some are given below to show the interrelation of various factors. The D„ of this section equals iocCu/(4n + 3) in the reference. c/a
Do
D1
D2
D3
0.1
0.0500399 0.153298 0.277673
—0.0000001 —0.000037 —0.000870
0.000001 0.000045
—0.000003
0.3 0.5
6.19. Space-charge Current. Child's Equation.—So far, we have considered currents in conductors where the net charge is zero. Let us now consider currents in which charges of only one sign are present so that the net charge is not zero. We must assume the motion of the charges sufficiently slow so that the formulas of electrostatics are valid. In this case, Poisson's equation [3.02 (3)] must be satisfied so that V'2 V = —
E„
(1)
where V is the potential and p the charge density. The charges are supposed to be similar and associated with a mass m and to acquire their energy entirely from a superimposed field. Thus, if their velocity is v and charge q, their energy at a point where the potential is V will be my' = 2q(Vo — V)
(2)
where Vo is the potential of their point of origin. The current density at any point is given by = pV
(3)
The simplest case is the one in which the charges are freed in unlimited quantity at the plane x = 0 and accelerated with a total potential Vo to a plane x = b. At the surface x = 0, charges will be freed until there is no longer an electric field to move them away so that the boundary condition there is
(an ax
0
(4)
Here, all the velocities are in the x-direction so that, eliminating p and v from (1) by means of (2) and (3), we have 3217 axe
[
m
c, 2q(V o — V)
§6.19
273
SPACE-CHARGE CURRENT. CHILD'S EQUATION
Multiplying through by dV/dx and integrating from V = Vo and dV/dx = 0 to V and dV/dx, we have (dVy 4ir m(Vo — V)11 E,L 2q
(5
)
Taking the square root of both sides, integrating from V = Vo, x = 0 to V = 0 and x = b, and solving for i, we have = 4€„(2q\iVol 9 Vrr, b 2
(6)
This is known as Child's equation. It shows that, with an unlimited supply of charges at one plate, the current between the plates varies as the three halves power of the potential. Such a current is said to be "space-charge limited." We see from (6) that the space-charge limitation is much more serious for charged atoms than for electrons, because of their greater mass. In practice, we frequently have the emitter in the form of a small circular cylinder and accelerate the charges to a larger concentric cylinder. In this case, we use cylindrical coordinates, and if I is the total current per unit length of the cylinders, (3) becomes I = 2irrpv
(7)
Writing (1) in cylindrical coordinates by 3.05 (2), eliminating (2) and (7), and writing V for Vo — V, we have m y d2V dV = dr 271-E„\2qV
r dr'
p
and v by
(8)
The direct solution of this equation in finite terms is difficult, if not impossible. We may, however, obtain a solution in series as follows: Let us see if the assumption that q, m, and V enter into this solution in the same way as into (6) leads to a solution of (8). Try the solution _1
87r€,(2q\I 9 m) r/32
(9)
and see if 132 can be determined to satisfy (8). Substituting (9) in (8) gives the equation
d20 313d-y2 where
(12 + odo + 02 _ d-y
1=0
(10)
274
§6.19
ELECTRIC CURRENT
This equation can be solved in the regular way in series which gives =
g 72 ± Twy3 44 074
..
(12)
Here a is the radius of the inner cylinder. Tables of i3 as a function of r/a have been published by Langmuir. Problems The problems marked C in the following list are taken from the Cambridge examinations as reprinted by Jeans, with the permission of the Cambridge University Press. 1. A cylindrical column of length land radius a of material of resistivity r connects normally the parallel plane faces of two semi-infinite masses of the same material. Show that, if R is the resistance between the masses, 7/
11-7
r
_ — < R< ,rat 2a
— / In (1
♦ era/l)]
Observe that, when 1 = 0, both limits give the exact value. 2C. A cylindrical cable consists of a conducting core of copper surrounded by a thin insulating sheath of material of given specific resistance. Show that, if the sectional areas of the core and sheath are given, the resistance to lateral leakage is greatest when the surfaces of the two materials are coaxial right circular cylinders. 3C (Modified). Current enters and leaves a uniform circular disk through two circular perfectly conducting wires, of radius a, whose centers are a distance d apart and which intersect the edge of the disk orthogonally. Show that the resistance between the wires is 2(shr) cosh-1(d/2a), where s is the resistivity. 4. Two electrodes each of small radius ö are situated at a distance 2a apart on a line lying midway between the edges of an infinite strip of width r and resistivity s with insulated boundaries. Show that the resistance between them is, approximately, s
ln
sinh 2a 5
5. Instead of being situated as in the last problem, the two electrodes lie symmetrically on a line normal to the edges of the strip. Show that the resistance between them is, approximately, s ln (2 tan a) 8
6. The electrodes are situated as in problem 4 but the edges of the strip are perfectly conducting. Show that the resistance is, approximately, s - ln
tanh
7C. A circular sheet of copper, of specific resistance si per unit area, is inserted in a very large sheet of tinfoil (so), and currents flow in the composite sheet, entering and leaving at electrodes. Prove that the current function in the tinfoil, corresponding to an electrode at which a current e enters the tinfoil, is the coefficient of i in the
275
PROBLEMS imaginary part of
cz so - si In cz - a2 so ± si
roe -ln (z - c) 2.71-
where a is the radius of the copper sheet, z is a complex variable with its origin at the center of the sheet, and c is the distance of the electrode from the origin, the real axis passing through the electrode. 8C. A uniform conducting sheet has the form of the catenary of revolution y2 z 2 = c2 cosh2 (x/c). Prove that the potential at any point due to an electrode at xo, yo, zo, introducing a current C, is
c,
constant - — In cosh 47r
x - xo c
zzo yyo R y2 + z2)( + zDik
9. The edges of two thin sheet electrodes are held in a circular cylinder of resisting material of length L. Both electrodes and the axis of the cylinder lie in the same plane and their edges are parallel to, and at a distance c from, this axis. Take c small compared with a, the radius of the cylinder. If r is the resistivity and I? is the resistance between electrodes, show that Tir
711-
. 2L cosh-' (a/c) < R < 2L smh-' (a/c) 10. Show from Art. 4.29 that the resistance in the last problem is given rigorously by
R-
2TK(c2a-2) LK[(1 — c4a-4)1]
11. Show from Art. 4.30 that if, in the foregoing problems, the edges penetrate to different depths, so that the distance of the cylinder axis to one edge is c1 and to the other c2, then the resistance is
R
rK(k) LK[(1 - k 2)11
where k -
a(ci a2
c2)
cic2
Here K(k) is a complete elliptic integral of modulus k. 12. A semi-infinite plate of resistivity r is bounded by x = 0, x = 1, and y = 0. Two perfectly conducting electrodes are applied to the edge, so that their contact areas are bounded by x = 0, x = 1, z = b, z = a and x 0, = 1, z = -b, z = -a where b < a. Show from Art. 4.29 that the resistance between these strips is
R=
2rK(b/a) 1K ([1 - (b/a)9+ }
13. The contact areas in the preceding problem are of unequal width. The distance between their near edges is q, far edges h, and the width of one is w. Show from 4.30 that the resistance between them is
R=
rK(k) 1K[(1 - k 2)1]
where k = [
hg (w g)(h - w)
14. A long slab of resistivity r, thickness 2b, and width l has two perfectly conducting strip electrodes of width 2a opposite each other on opposite faces, at right angles to
276
ELECTRIC CURRENT
the sides. Show from problem 61, Chap. IV, that the resistance between them is R—
2rK(e-ra/b) 1K[(1 — c 2n.191]
15. A long strip of width 2b has a hole of radius c bored on its center line far from the ends. Using the results of 4.28, show that the hole increases the resistance by an amount equivalent to adding AL to the length of the strip where AL = (1 + X)-12b7r-iln cos [-brcb-1(1 + X)] and the parameter X must be found from 4.28 (6). 16. The curved surface of a right circular cylinder of radius r, length d, and resistivity r is given a perfectly conducting coat. Two thin coplanar saw cuts are made, so that the two halves are connected only by a strip of width 2b whose center line is the cylinder axis. Show that the resistance between the two halves of the conducting coat is rlf[(1 — b 4r-4)1] R= 2dK(bzr-2) where K(k) is a complete elliptic integral of modulus k. 17. There is a dipole source of current of moment M in a sphere of radius a and resistivity r with an insulated surface. Take the sphere center as the origin and the = 0 axis parallel to the dipole which is located at r = b, B = 00, and 0 = 0. Show from problem 115, Chap. V, that the potential when b < r < a is V=
n TM V v (2 — 5!;,.)(n — m)![(b)n+1 474) 2
n=0 rn=0
m — 1)!
(n
(n +
r
157_1(tio)PZ(p)
cos mc5
where ,uo = cos 00, A = cos 0, and r > b. If r < b, replace (b/r)n+IP7_1(20) by — (r/b)n[(n — m 1)/(n m)1/37,z4.1(Ao)• 18. A solid conducting cylinder of resistivity r is bounded by the planes z = ±c and the cylinder p = a. A current I enters at p = a, 0 = 0, z = 0 and leaves at p = a, 0 = lr, z = 0. Find the potential anywhere inside the cylinder and show that, if the electrodes are small spheres•of radius r, half embedded in the surface, the resistance between them is 2rz (2 ir 2a
n=0 nt=0
en)/2m+1,[nrr(a — r)/c] n/2.,„+1 (nira/c)
19. A perfectly conducting rod of radius b is partially insulated by a semiconducting bushing of radius a and length 2c where it passes through a perfectly conducting thin sheet at z = 0. If the total leakage current is I, show that the current density on the rod when —c < z < +c is CO
I 47 2ab
.k
(2 — SD cos (nrzlc) n[I i(nrci/ c)Ko(nirb /c) + I o(nrb /c)Ki(nra/c)!
71=0
20. Current enters an infinite plane conducting sheet at some point P and leaves at infinity. A circular hole, which does not include P, is cut anywhere in the sheet. Show that the potential difference between any two points on the edge of the hole is twice what it was between the same two points before the hole was cut.
277
PROBLEMS
21. An insulator has the form of a truncated cone of height h, base radius al, and top radius a2. The base rests on a metal plate, and the top supports a concentric metal post of radius a3. If the specific surface resistance is s, show that the surface resistance between the plate and the post is s
f[h2
27r
(al — a2)211in (at) + in (a2)} ce3 a2
al — a2
22. On a spherical conducting shell, current is introduced at the point 0 = a. = 0 and taken out at the point 0 = a 4) = 7, where 0 is the colatitude angle and the longitude angle. Show that the potential on the surface is of the form ,
A In
1 — cos a cos — sin a sin 0 cos 1 — cos a cos 0 + sin a sin 0 cos
C
23. The ends of two metal rods of radius b are embedded in a sphere of insulating material of radius a whose body resistance is very high but whose surface resistance is s. If the axes of the rods were extended, they would intersect at the center of the sphere at an angle a. Show that the surface resistance between the rods is - cosh ,
a a) sin b 2
24. A current flows in a thin shell whose equation is (z/a) 2 (p/b)2 = 1, where p2 = x3 +y2 and a > b. Show that the potential V on the shell is either Ui or U2 where U7 + jU2 = f(a j0), tan rp = y/x and
a = tanh-, [a4
(a2
bz —
b2)1 sin-1
(a 2 — b2)z2] 2
[z(a2 — b 2 )1 a2
The function f(a j(p) is periodic in 4 with a period of 27, and we must also have a V/Oa = 0 at z = ±a or a = ± .0 if there are no sources or sinks of current there, otherwise av /act, = 0 there. 25. If we have an oblate spheroidal shell carrying current instead of the prolate one treated above, so that a < b, show that the preceding considerations apply except that a = tanh-1 ([a4
+
bz ) 03 — az)z91
(b 2 — a2)1 [z(b2 — a2)1] sinh-, b a2
26. Consider the surface of the earth as plane and its resistivity to be To except for a region in the shape of a semi-infinite vertical cone with its apex at the surface which has the resistivity Ti. A current I is introduced at the surface at ¢ = 0, at a distance r = a from the apex. Show that the potential at any point r < a outside the cone is given by
Vo = gar)-4-2 Am(p)
cos mOfo [Am(P)P7v_1(A) Bm(P)P,1-1( -- i4)1 cos (p In ra-0 dp m= o (ri — T2).8.(P) = TI(Ti — T2) (2 —
S =
T2S' — TIS P7;_1(0)[TI — T2 a[Prp (cos c)] / a (cos a)
a[Pz,_+( —cos a)]/a(cos a)
s =
— 1)m(TIS — T28')]
P7p_4(cos a) P72,_1(—cos a)
278
ELECTRIC CURRENT
27. A thin circular loop current source I is coaxial with, and immersed in, a solid cylinder of resisting material of radius a and length L with perfectly conducting earthed ends. Show that the potential distribution in the cylinder is, if z < c, T/ sinh Akz sinh mk(L — c)Jo(mkb)Jo(µkP) sinh pkL ra2 L-/ Pk[Jo(pka)] 2 k= 1 where Ak is so chosen that Ji(Ihka) = 0 and the loop is at p = b, z = c. 28. Two rings made of wire of small radius d are coaxial with an infinite circular cylinder of resisting material of radius a and embedded in its surface to a depth d. Show that, if mk is chosen so that Ji(Aka) = 0, the resistance between them when their centers are at a distance c apart is, approximately, — e-Ak(-2d)ie-Akd
R= as
=1 k
29. A conducting circular cylinder with inner and outer radii a and b is cut on one side longitudinally, the edges of the cut being maintained at potentials of 1V0 and —1V0, so that current flows around the cylinder. Show that the electrostatic fields, when r < a and r > b, are given, respectively, by the transformations Vo W = — ln (a ± z)
and
W
Vo
7
ln
z
V Z +
b
30. A current I is introduced at one pole of a thin conducting spherical shell of radius a and uniform area resistivity s and removed at the other. Show that the electrostatic potential at any point inside the shell is
V =
s/ 2r
n=0
r 4n + 3 (2n + 1)(2n + 2)(a
2n+1
P2.-0(14)
31. A wire of radius a and resistivity T1is coaxial with the z-axis. The medium outside the wire is insulating except for that portion lying between z = —c and z = c, which has a resistivity T2. If T2 >> rishow that the ratio of the resistance across the gap of a perfectly insulated wire to that of the actual wire is, approximately, = + 32cTI 1r 3cts 2
n=0
( — 1)° K1[1(2n+ 1)7ra/cl (2n + 1)3 Ko [i(2n 1)ra/c]
32. The corners of the wider strip in Fig. 6.07b are cut off at 45° so that the half width of the strip increases linearly from h to k. Show that the presence of the tapered section near the center of a long strip increases its resistance over the sum of the resistances of the straight sections by an amount
rhk
(h 2
k 2) tanh-1 -
(k2 — h2) tan-1 -
hk In
— 8h2k 2
33. An infinite slab of resistive material bounded by z = a and z = —a with perfectly conducting surfaces has a spherical cavity of radius 1 centered between its faces on the z-axis. From 6.18 the vector potential for unit current density far from the
279
PROBLEMS cavity is — -1.p.
Assume as in 6.18 that inside the cavity it has the form
A4,, = ZCtr20'+1/1.+1(cos Bo) + n=0
(1 — 4)rsn-2P2n+1(cos 8 = -
where r, is the distance from the sth image sphere to the field point and B, is the acute angle between r, and the z-axis. Express AO,. in cylindrical coordinates by 5.40 (35) and (36), expand in powers of p, and require that the coefficient of p2n+1be zero if n 0 and a if n = 0. Thus show that the N 1 equations to be solved for C„ are N i-(2/ + 2n + 3)(2/ ± 2n 2)! 4;, = (2n + 2)!(2/)!(2a) 2H-2n+2 l-0
where i-(x) is the Riemann zeta function. Show that the vector potential in the current region is given by N
Ao. = ZC'nr."-2PL+1(000 re = 0
O.)
a=—
For a complete discussion of this problem see Paul Michael, Phys. Fluids, Vol. 8, page 1263, 1965. References Some material on the subject matter of this chapter will be found in most of the electrical references of Chap. V.
CHAPTER VII MAGNETIC INTERACTION OF CURRENTS 7.00. Definition of the Ampere in Terms of the Magnetic Moment.— As mentioned in the last chapter, electric currents flowing in neighboring conductors exert forces on each other known as magnetic forces. This magnetic interaction was carefully investigated by Ampere. One of his experiments showed that, when two wires carrying equal currents in opposite directions lie sufficiently close together, their power of interacting magnetically with other circuits is destroyed. Let us now construct several small plane loops of wire, twisting together the thin wires by which the current is to enter and leave the loops so that any magnetic forces observed will be due to the loop alone. Keeping them at distances great compared with their dimensions, we find that, when carrying steady currents, the forces and torques that they produce on each other are in every way like those between electric dipoles, provided that we orient the normal to the plane of the loop as we would the axis of an electric dipole. Thus, using three small loops carrying fixed currents, we can, by taking them two at a time and measuring the forces and distances between them, determine by 1.071 (6) the products m1m2 = A, M 1M3 = B and M2M3 = C and so find Pr? = AB/C, 31= AC/B, and 2/3 = BC/A. If the experiment is done in vacuo and 1/E is replaced by = 47r X 10-7, then M for each loop equals the product of its area in square meters by its current in amperes and is the magnitude of its "magnetic moment." Thus an ampere may be defined as that current which, flowing in a small plane loop of wire, gives it a magnetic moment equal to its area. In a medium other than a vacuum the forces and torques are different and the factorµ which replaces is called the permeability of the medium. 7.01. Magnetic Induction and Permeability.—If a conductor, carrying a current, when placed in a certain region, experiences a magnetic force we say a magnetic field exists in that region. We shall map the magnetic field by means of a small exploring loop just as we could map the electric field by means of a small electric dipole. When free to move, the loop assumes a certain orientation. We shall define the positive direction of the magnetic field to be normal to the loop and directed as a right-handed screw moves when rotated in the sense in which the current traverses the loop. As in the electrostatic case, we can measure the strength of the magnetic field in terms of the torque acting on the loop. 280
§7.01
MAGNETIC INDUCTION AND PERMEABILITY
281
Thus, we define the magnetic induction or flux density B to be a vector in the magnetic field direction whose magnitude, in webers per square meter, is the torque in newton meters on a loop of moment one whose axis is set normal to this direction. This torque, and hence B, depends on the medium in which the experiment is performed. We define the relative magnetic permeability Km of a substance to be the ratio of the magnetic induction when all space is occupied by the substance to its value at the same point in vacuo, the configuration of all circuits and the magnitude of all currents being the same. The permeability A is the product The experiments on which the definition of A and B depend involve no theoretical difficulty in the case of liquid and gaseous mediums. It is obviously impossible, however, to manipulate a loop of wire in a solid medium. If all space outside the small region in which our loop is situated is filled with one liquid and this region is filled with another, then we find that the measured value of B is independent of the shape and size of this region only when the liquids have the same permeability. Thus, to determine B and A at a point in a homogeneous isotropic solid, we excavate a cavity at P and fill it with a liquid such that the measurement at P is independent of the shape and size of the cavity. The value of B andµ at P, so determined, is the same as before the cavity was created. In 7.24, by using the boundary conditions derived in 7.21, we shall describe an experimental method of defining B andµ which includes the case of magnetically anisotropic crystals. Suppose that we take a great number of small loops, each carrying a current I and, without changing the area, fit them together into a mesh always keeping the direction of the induction due to adjacent loops parallel. From Ampere's experiment, the magnetic effect of all loop boundaries, except those forming the outer edge of the mesh, disappears, so that the resultant magnetic effect is exactly the same as if the current I circulated around the boundary alone and, hence, is independent of the shape of the surface of the mesh. But we know that the magnetic induction due to any elementary loop of area dS is identical with the electric field due to an electric dipole of moment AEI dS. The magnetic induction due to the whole circuit is therefore the same as the electric field due to a uniform electric double layer of strength AE/, whose boundary coincides with the circuit and which we discussed in 1.12. This magnetic equivalent of a double layer is called a magnetic shell. If we look for this shell by following a magnetic line of force produced by an electric circuit, we find that we pierce the same surfaces repeatedly, without arriving at any magnetic discontinuity corresponding to the electric double layer. Thus, the magnetic shell has no physical existence and is merely a convenient device of use in computing magnetic fields. Furthermore, this
282
MAGNETIC INTERACTION OF CURRENTS
§7.02
shows that there are neither sources nor sinks of a magnetic nature corresponding to the electrostatic charge so that divB=V•B= 0
(1)
The work required to take a unit electric charge from one face of a uniform electric double layer of strength Ae/ to the other, by a path passing around its edges is i.4/, by 1.12 (3), since the difference in the solid angle subtended by the layer at the two ends of the path is 4r. Thus, if the electric intensity is E, we have 5E • ds when ds is an element of the path. As we have seen, the magnetic induction due to a current I is everywhere the same as the electric intensity due to a uniform electric double layer of strength AEI, having the circuit as a boundary, so that, in vacuo, f B • ds = Ad", and in a medium of uniform permeabilityµ we would have (2)
JCB • ds = µI
But from 6.00 (2), we have I = fi•ndS where i is the current density and S is the area inside the path s, thus we have f B • ds = Ali • n dS. We can transform the left side by Stokes's theorem [3.01 (1)], giving x B • n dS = fsi • n dS
(2.1)
It is easy to see that this holds for any path, even if it includes only part of the current; for in that case either that part of the shell due to currents not encircled will be crossed twice, once in the positive and once in the negative direction, or it will not be crossed at all. In either case, it contributes nothing to the integral. Thus, this relation holds for the integrands alone, giving in a region of uniform permeability, V x B = µt
(3)
7.02. Magnetic Vector Potential. Uniform Field.—It is well known that the divergence of a vector which is itself the curl of another vector is zero and so satisfies 7.01 (1). Thus, we may define a new vector A, which we shall call the magnetostatic vector potential as a vector whose divergence is zero and whose curl is B, thus (1)
VxA=B V•A=0 The standard formula for curl curl A now becomes V x [V x = V(V • A)
—
(V • V)A =
—
V 2A
Substituting in 7.01 (3), we have, in a region of uniform permeability, pm = (2)
§7.02
MAGNETIC VECTOR POTENTIAL. UNIFORM FIELD
283
On writing out the components, this is iv2A z N 2A, + kVA, =
iiy
kiz) (3) Thus A, Ay, and Az all satisfy Poisson's equation [3.02 (3)]. We have found a solution of this equation in 3.09 (1) to be
As =
f iz dv jy r
A y= A r dv, A 47t-
r
A= A
dv
dv jv r
(4)
Adding components gives 471- jv r
(4.1)
In the region outside the wire, i is zero and since, in a thin wire of crosssectional area dS, we have iz dv = i dS dx = I dx, we may write, summing up the vector components,
A = p, 47r
ds r
(5)
Physically, this is a more satisfactory concept than the magnetic shell since it depends only on the configuration of the circuit, the current strength, and the location of the point of measurement and involves no artificial discontinuities. It will give the correct value for the line integral of B for any path. Evidently, we can consider the vector contribution to A by any element of the closed circuit as parallel to that element. The vector potential in rectangular coordinates associated with a uniform magnetic induction B in the x-direction has only two components Ay and A, and their values are Ay = — aZ.B Az = (1 — a)yB (6) where a is an arbitrary number. Obviously, the x-component of the curl is B and the other components are zero. In spherical coordinates, the vector potential of a uniform field parallel to the 0 = 0 axis is A4, Br sin 0
(7) In cylindrical coordinates, the vector potential of a uniform field parallel to the z-axis is Ao = ABP (8) In oblate spheroidal coordinates, the vector potential of a uniform field parallel to the = 1 axis is Am ciB = 23 (9) In prolate spheroidal harmonics, it is c2B Am=2Pi()Pi(,)
(10)
284
MAGNETIC INTERACTION OF CURRENTS
§7.04
The forms of A given in the last five formulas are not, of course, the most general forms that give the required B. We could add to each the gradient of a scalar without affecting B. 7.03. Uniqueness Theorems for Magnetostatics.—In 3.10 we found just what information concerning the interior and boundary surface of a region is necessary to determine uniquely the electrostatic potential inside it. This proof applies without change to the magnetomotance discussed in 7.28, provided there are no electric currents within the region which would make it multiple valued. The vector potential requires a different proof. We shall show that, if the location and magnitude of all fixed currents inside a closed surface are specified as well as the value of the tangential component of either the vector potential or the magnetic induction over this surface, then the value of the magnetic induction everywhere inside it is uniquely determined. The contribution of internal sources given by 7.02 (1) and (5) is unique. Suppose there were two internal values B and B' having identical external sources and boundary values and derived from A and A'. If V • A = V • A' = 0, then V2A' — '7 2A = — Ai and V x V x (A — A') vanishes in v so that putting
IP =
= A — A'
in 3.06 (5) gives
fv(B - B02 dv
si(A — A ') • [(B — if) x = f =1
dS,
(1)
If, on 8,nxA=nxA'orBxn=B'xn, then the surface integral is zero so that in either case B = B' throughout v because (B — B') 2is positive. If B — B' is zero throughout the volume, then A and A' can differ therein only by the gradient of a scalar. By 3.10, if the gradient of a scalar is zero over S, it vanishes in v so that if A — A' vanishes over S it vanishes in v and A is uniquely determined therein by its value over S. 7.04. Orthogonal Expansions for Vector Potential.—In electrostatics, after finding the potential due to a fixed charge distribution, we superimposed suitable perturbing potentials to satisfy the conditions at dielectric or conducting boundaries. To use the same method with vector potentials, it is necessary that, having found by 7.02 (4.1) that part of the vector potential due to the given current distribution, we know suitable forms of perturbing potentials to superimpose to meet magnetic boundary conditions. The direct method of solving Laplace's equation for the vector potential is not easy for, where Laplace's operator is applied to a vector, it operates not only upon the magnitudes of the components of the vector, but also upon the unit vectors themselves, as illustrated in 10.05 (2) and (3). Except in rectangular coordinates, this gives rise to a set of three
§7.04
ORTHOGONAL EXPANSIONS FOR VECTOR POTENTIAL
285
simultaneous partial differential equations whose solution may be very complicated. We are led, therefore, to look for a simpler method. We have seen [3.02 (1)] that in free space where there are no electric charges the divergence of the electric field is zero and, since from 1.06 (4) it is the gradient of a scalar, its curl is also zero. Similarly, from 7.01 (1) and (3), in free space, where there are no electric currents, the divergence and curl of the magnetic induction are zero. We would expect, in such regions, to write expansions for the two fields in terms of orthogonal functions, in the same form. In Chap. V, we obtained these expansions by solving Laplace's second-order partial differential equation. This equation was broken up into three total differential equations, each involving a single coordinate, connected by two indices. Thus, each term in the expansion involves two indices and six integration constants. As just pointed out, owing to the mathematical similarity of electric and magnetic fields, we should expect that part of the vector potential which contributes to the magnetic induction to have the same number of indices and constants. This potential should be derivable from three scalar potential functions because, in rectangular coordinates, each component of the vector potential satisfies Laplace's equation. These components are not, however, independent, but are connected by the relation V • A = 0, so that only two independent scalar functions, at most, can be used. Furthermore, as just pointed out, when properly chosen, only one of these will contribute to the magnetostatic field. The general expression for the vector potential, giving zero divergence, is A =VxW
(1)
where W is a vector which should be derivable from two scalar potential functions. To fit boundary conditions, when dealing with eddy currents and electromagnetic radiation, we shall find it convenient to split W into two components, normal to each other, each of which is derivable from a different scalar potential function. Thus we shall write W = ufri u x VW2
(2)
where V2 W1 = V2 W2 = 0 and u is an arbitrary vector to be chosen so that V2A = v2(v x W) =V x (O2 w) = 0 (3) This holds when V2 W is replaced by V2 W CW unless C depends on x, y, z. Now it is easily verified, by writing out in rectangular coordinates, that v2uw1 = uv2w1or uV2 W1 217 Wi and V 2 (u x VW,) = u x V (V2 W2) (4) where u = r, i k, or r, so that these are suitable choices of u. From the similarity of B and E, we suspect that B, like E, can be obtained from a single scalar potential function. This surmise is confirmed by examining ,
286
MAGNETIC INTERACTION OF CURRENTS
§7.05
the contribution of W2 to A. Thus we have u = .i, or k, V x (u x VW2) = u(V2W2) - V(U • VW2) (5) u = r, V x (u x VW2) = u(V2W2) — V(u • VW2) — VW2 (6) Since V2W2 = 0, the part of A derived from W2 is the gradient of a scalar and contributes nothing to B in the magnetostatic case. If u1, u2 and u3are orthogonal curvilinear coordinates, if the u specified above lies in the direction of u1, and if WI is of the form U(u1)F(u2, u3), then B . A = 0. This is proved in Lass, pp. 48 and 57. In the following articles, we shall use the theory just developed to obtain solutions of the equation pm = 0 (7) in the form A = ulU11U12U13 u2 121U22U23 u3 U31U32U33 where ul, u;, and u; are unit vectors in the directions of the coordinates u,, u2, and u3, and Ursis a function of u8only. The solution should be in such form that the vector potential anywhere inside a volume containing no sources and bounded by a set of surfaces, on each of which one of the coordinates is constant, can be calculated when the value of its tangential components on these surfaces is given. at; 7.05. Vector Potential in Cylindrical Coordinates.—In Chap. V we found that the general solution of Laplace's equation in cylindrical coordinates could be built up of a sum of terms involving, except in particular cases, Bessel functions. We now wish to find analogous solutions for the vector potential which possess orthogonal properties on the surfaces of a right circular cylinder so that we can express the tangential components of the vector potential thereon as a sum of such solutions and thus determine its value at any interior point. Choosing the Bessel function solution of VW' = 0 given in 5.291 and setting u = k, the W of 7.04 (1) becomes W = kic-1(Aeks
Be--kz)[CJ„(kp)
DY„(kp)] sin (744 4- an)
(1)
The vector potential derived from this by 7.04 (1) is
A p= —(Aekz
Be-kz)[C.1,.(kp)
A4, = (Aekz Be—kz )[C4(kP)
DYn(kp)1n(kp)-1cos (nd, ± on) (2) DY (kp)] sin (nd, +on)
This is the orthogonal surface vector function defined by 5.296 (8). If, for a given z-value, either component of k x A vanishes when p = a then at this z-value k x A may be expressed as a sum of such functions. A rarely used form containing ¢ is obtained by setting 5„ = 0, letting n --> 0 and keeping the products of nA and nB finite. In order to obtain suitable solutions for expressing the tangential components of A over the curved surfaces, we use LLie forms of 5.292 (3)
§7.06
VECTOR POTENTIAL IN SPHERICAL COORDINATES
which are orthogonal in z and 4. This gives in place of (2) yk) cos (ncl, A„ = -[aIn(kP) DKn(kP)1n(kp)-1cos (kz cos 7h) sin (kz (n4i A4, = [CI:,(kP) DIV,,(kp)]
287 5.) (3)
The z-component satisfies the scalar Laplace equation and is written Aa = [C'I,,(kp) D'Kn(kp)] cos (kz 71) cos (ncb (5:,) (4) The above solutions are inadequate when k = 0 and n 0 0. For the curved surface, solutions corresponding to (3) and (4) are A,, = (Apn-1 Bp-n-9 (Cz D) sin (n4 + on) A A pn-i + Bp n 1) (UZ D) cos (n0 + an) A. = (A' pn D') cos (nci, B'p-n)(C'z (5) 0, there are no solutions orthogonal in both p and 4) suitable for When k = the end surfaces. When both k and n are zero, some forms of interest can be found from 5.291 (9) by taking Cz D) In p W = k[(Az4, (Ez F)4,] rGz (6) A,, = Ap-'z In p Bp-' In p E p-1Z F -
-
A4, = -Ap-14 - Bp-'0 A, = (GO H) In p 141
- Dp-1 -1-Gp (7)
Other solutions not often needed are found by using i, i, or r in W. 7.06. Vector Potential in Spherical Coordinates.—In Chap. V the general solution of Laplace's equation in polar spherical coordinates was built up of a sum of terms involving spherical harmonics. We need analogous solutions for the vector potential which possess orthogonal properties on the surface of a sphere so that we can express the tangential components of the vector potential thereon as a sum of such solutions and thus determine its value at any interior point. Choosing the spherical harmonic solution of V2W1= 0 given in 5.23 and setting u = T, the W and A of 7.04 become W = T(Arn Br-n-9[CP;n,(cos 0) + DQ„ 'n(cos 0)] sin (m4, 8„,) (1) A B = (Arn Br-n-1)[CP„m(cos 0) + DQ"'„ (cos 0)]m csc 0 cos (m4, + 8,n) = (Arn+Br-n-9[CP"'„' (cos 0)-F INT' (cos 0)] sin 0 sin (m4,+ am) (2) This A is the orthogonal surface vector function mentioned in 5.231. When m = 0 and 8„, = -br, this equation becomes r Br-71-1)LCPVCOS A4, = (Arn 0) + DQVCOS 0)] (3) It is often useful to have B in terms of W. If u = r we have
B = V x (V x rW) = -V x (r x VW) = (VW) + 2vW Writing out the components of this equation gives a2 (rW) a2(rW) 1 a 2(rW) B - are Bo B,,, - . r ar ae' r sin 0 ar act.
(4)
288
MAGNETIC INTERACTION OF CURRENTS
§7.08
7.07. Vector Potential in Terms of Magnetic Induction on Axis.— Magnetic lenses for focusing electron beams are usually constructed from coaxial current coils or equivalent magnets so that the vector potential has only a 0-component. To calculate the focusing properties of such lenses, it is convenient to have the vector potential expressed in terms of the magnetic induction and its derivatives along the axis. If W = kW, where W is the solution of Laplace's equation given in 3.15 (1), 7.04 (1) yields, if •If is written for 43', A = —
aw ap
f
=
2a
\If (z jp sin 0) sin 0 d0
27r r,
(1)
where xlf(z) is a real function of z. The Taylor expansion is, by Dw 39, a2*(z) (jp sin 0)2 + • • axli(z) jp sin (2) jp sin 0) = xlf (z) xlf (z 2! az2 1! az Substitution of (2) in (1) and integration from 0 to 27r give a 2n+14/ ( z)(pyn+1
— A—
n!(n + 1)!
a z,-+i
2)
(3)
n=0
The magnetic induction V x A on the axis where p = 0 is Bo(z) = [ 1 a(pA4,)
p ap
1 =0
_ .34/(z) az
(4)
This, when put in (3), gives Ao(p, z) in terms of Bo(z) and its derivatives. ea —
n!(n
—
a2nBo(Z)
1)! az2n
fp \ 2n+1
(5)
n=0
The current elements that generate fields of this type are coaxial circular current loops so the range of p and z over which (5) holds are the same as for such loops. To find the range insert Bo(z) from 7.10 (8) into (5) and compare the result with that obtained from problem 27, which holds for all values of p and z, when J1(kp) is expanded by 5.293 (3), the summation and integration are interchanged, and the integrals are replaced by derivatives of 5.298 (6). The identity of the results shows that (5) gives a unique vector potential at all values of p and z which can be reached from the axis without crossing a current sheet. The restriction is needed because the external fields of any currents inside a closed surface are unaffected if these currents are removed, provided suitable currents are set up in the surface. 7.08. Equation of Axially Symmetrical Tubes of Induction.—A magnetic field is most easily visualized by mapping the stream lines or lines of magnetic induction. When we have axial symmetry so that all
§7.09
VECTOR POTENTIAL AND FIELD OF BIFILAR CIRCUIT
289
sections of the field made by planes passing through the axis look the same and all currents are normal to these planes, we can obtain the equation of these lines very simply from the vector potential which, in this case, has only the component Ad,. When rotated about the axis, each stream line generates the surface of a tube of induction. Any particular tube may be specified by stating the flux N through it. This is obtained by integrating the normal component of the induction over any cross section S of the tube. Thus the equation of the surface of such a tube is given by N = fsB• n dS f(p, z)
(1)
By means of Stokes's theorem [3.01 (1)], this becomes N= fsV x A • n dS = A • ds
(2)
If the section is taken by a plane normal to the z-axis, then the path s is a circle on which Aois constant and its length is 27rp so that the equation of the tube of force becomes N = 27rpAo (p, z) (3) In an axially symmetrical system of orthogonal curvilinear coordinates ui, u2, and 47, the distance p will be a function of u1and u2 so that the equation of the tubes of force is N = 2lrp(ui, u2)A4,(ui, u2) (4) 7.09. Vector Potential and Field of Bifilar Circuit.—We shall first apply 7.02 (5) to finding the magnetic field due to a long straight wire with
a parallel return at a distance a from it. Since all elements of the wire lie in the x-direction in Fig. 7.09, the vector A has only the component A. Thus, we have, from 7.02 (5), A. = pl I+
— dx = — _ ri 2rfo r2 a/ In ri x , a2 — =—n i — 2r r2 x0 27r al
[(a?
+
x2 ) — i
—
(4
+
x 2) -1] dx
(1)
290
§7.10
MAGNETIC INTERACTION OF CURRENTS
This shows the surfaces of constant vector potential to be circular cylinders coinciding exactly with the equipotentials in the electrostatic case where V = —q(2.2-€)-1In (a2/a1) if the two wires carry charges +per and —pd. For p/ = 27, the equation of these cylinders is given by 4.13 (3), where A. is substituted for U and z for x. The magnetic field which is given by V x A is at right angles to the electric field given by V V but has the same numerical value. From 3.04 (1), (2), and (3), the components of the magnetic induction, are, in vacua, since h1= h2 = h3 = 1,
aAz
aA„
aAz ay
B. = — — = 0, By = az ay az Since a1= [(y — ia)2 z 2]1and az = [(y By pIzt
1 W V4
B. = — 2,.
B. = --
ia)2
z94, this gives
1) 2 az
(3) y— 1a) a2
(4)
7.10. Vector Potential and Field of Circular Loop.—Let us compute the vector potential at the point P shown in Fig. 7.10. From symmetry, we know that in spherical polar coordinates the magnitude of A is independent of 4,. Therefore, for simplicity, we choose the point P in the xz-plane where q5 = 0. We x notice that when equidistant elements of length ds at + and —95 are paired, the resultant is normal to pz. Thus, A has only the single F IG. 7.10. component A. Let dsobe the component of ds in this direction, and 7.02 (5) becomes a cos 4,c/4, 2rio (a2 p 2 +Z2 — 2ap cos 40)1 z2)i >> a, If our loop is very small, this becomes, since r. = (p2 = pI redS5
40) r
A=
era cosq5t1 apcos ¢1do_ a2pI p _ a2A/ sin 0 4r3 r?, I 4r2
27rjo ( r.
(1)
By taking, from 7.00 the magnetic moment M of the loop equal to ra2/ and directed upward, this may be written
A = p(M x r) (4z-r3)-1
(2)
If this approximation does not hold, let 4, = r + 20 so that d4, = 2 dO
§7.10
VECTOR POTENTIAL AND FIELD OF CIRCULAR LOOP
291
and cos 0 = 2 sin2 0 — 1, then this becomes (2 sin2 — 1) d0 gaIf iT 414, = — z2 — 4ap sin2 7 0 [(a + p)2 Rearrange and let z2]-1, p)2 k2 = 4apRa m = [1 — (1 — k2)i][1 -F (1 — k2)11-1 l k3( 1 43k2 7258k4 240 = 14 p [(1 — 21—10)K — E]= 32W +
(11
= LI L-1(--a—Y [K(m) — E(m)1 = mp 4
+ -83-m2 + 6145 m4 +
p
where K and E are complete elliptic integrals of the first and second kind. When z = 0, the m modulus is simply p/a or a/ p. To determine the magnetic induction, we must write, from 3.04 (1), (2), and (3), the components of the curl in cylindrical coordinates. From 3.05, this gives h i = 1, h2 = p, and h 3 = 1 so that
a 1 a 8A4, Bp= ---(pAgs) + --(A = — az paz pact. )
B# =-aa--i(A„) — 4(Az) = 0
( 5)
1a a 1 a B. = —(Ap) +— —(PA9s) = — — (P214,) P ad) P ap P aP For the derivatives of K and we use the formulas Dw 789.1 and 789.2
K OK_ E — k 2) T ak
and
aE _E K al T Tc
We also have, from (3), that
ak_ — = -az
4ap
ak = k k 3
and
Op 2p 4p
4a
Carrying out the differentiation, collecting terms, and substituting for k give
Bp = 27 °' pRa + p)z 2
E + z2P
K + a 2 ± p2 ± z2] 2E (a — p) 2 +z
2 2 22 E E K+(a a— p 4_ zz 2 .B. = 27 PI 4_ pr ± z z i[ _ p)2 [(a
(6)
—
]
(7)
Numerical values of B, and B. can be computed for any values of p and z by finding k from (3) and looking up the corresponding values of K and E in a table (Dw pp. 208-210 or Pc p. 121). On the axis p = 0 we have 0
Bp = — —> 0 o
and
B.— ba2i
(a2 + z2)1
(8)
292
MAGNETIC INTERACTION OF CURRENTS
§7.11
7.11. Field of Currents in Spherical Shell.—We define the value of the stream functionI,G, at any point P on a thin spherical shell of radius a, to be the total current flowing across any line drawn on the surface of the shell between P and a point where IP is zero. The components of the current density are therefore related to IP by the equation —1 a,p a‘p io a sin (1) 0 a cp = ae We wish to find the vector potential and magnetic field due to these currents. Since any possible i/i can be expressed as a sum of surface harmonics, it will suffice to calculate the field of the distribution i'> = 4>), for we can then, by superimposing, obtain that due to any 1,G. We shall designate the induction outside the shell by Bo and that inside by B. Let us apply 7.01 (2) to a small circuit around an element of shell of length AO, lying in a constant (IS plane. We obtain pica De = [(No — (130),]a AO Using (1) and introducing the scalar function W, of 7.06 (4), we have
a‘p a2(rW 0) ae = ar ae
a2(rWi)
ar ae
Similarly, taking a circuit in the q5 direction gives a,p = —1 0 2(rWo) 1 a2(rWi) sin 0 ar sin e ao sin 0 ar 3 4 If we multiply the first equation through by d0 and the second by sin 0 dcp and subtract, both sides are total differentials so that, if we integrate and remember that Wo = Wi when r = a, we obtain
aaw 0
1.4
—(rW — rWi) =
aw)
(2) ar — ar To give 4> the harmonic form S:(0, 4>), to make W0 = W, at r = a, and to have Wo finite at infinity and Wifinite at the origin, we must choose, from 7.06 (1), the forms n-1-1 pc (a — W0 2n ± 1 r) Sn (0' 4)), — 2n + 1(a) 877(8' (k) (3) Hence, from 7.04 (1), the vector potential has the form ia\"+1a in = 0, (2n 1) sin 0?-) . p[A3'1(°' C6)] m aNn+1 /.1 (0, 4)] "6°' — 2n + —
ar
—
("0
(A,.); = 0, '
—
i = (2n ("
sin t9(arY46[S:(°' 4>)]
— AO ae 2n + 1 a nl[S709,
(4)
ZONAL CURRENTS IN SPHERICAL SHELL
§7.12
293
The 0 terms in A gs and A Bhave exactly the form of the first and second terms of 5.231 (7), so A is an orthogonal surface vector function on the spherical surface. 7.12. Zonal Currents in Spherical Shell.—When the currents in a spherical shell flow along parallels of latitude, we have the most important practical case. In this case, axial symmetry makes all quantities independent of cp. The stream function may then be written CaP„(cos 0)
=
(1)
n=1
The current density having only a 4, component, becomes
1 84, 4= a ae —
c'inaPn(u) a ae
Ca sin0 aPn(u) a au
n=1
enPl(u) (2) — —— a
n=1
n=1
where u = cos O. The function W, if all space has the permeability it, takes, from 7.11 (3), the form . . wo = --1 —AC7, ia\n+ip r Wj = (3) W° 2n + 1 a 22n + 1W n(u)' —ACC _ n=1
n=l
The vector potential, from 7.11 (4), may be written r
r>a
A = (141
n=1
(4)
211 C+Y 1(a) P(U)
(arr
A=
2—± Cni iP(U) (5) n=1 n We may note here that (43)x = — sin q5 and (+)„ = cos 4' so that Az and Av satisfy 7.02 (3) in free space where i = 0. We also see that A has the form given in 7.06 (3). The components of the magnetic induction may be obtained from (3) or (4) and (5). Thus, by 7.06 (4), r
B= azo.wo _ J....t
7
(97.2
05
n(n + 1)CnOn 'pn(u) a.1 2n + 1 a
(6)
Ln(n + 1)C.(a)n+2Pa(u) a.1 2n + 1 r
(7)
n=1
r>a
Br —
az(rwo) 87.2
n= .1.
From (4) and (5), we have r
Be = _1 a(r240) _ r ar
(n + 1) C7( ry a.J 2n + 1 a
" )
(8)
294
MAGNETIC INTERACTION OF CURRENTS
Be =
r>a
1 8(rA.0) r Or
§7.13
nC„ (1+2 13'(u) aL.J2n + 1\ r
(9)
n=1
These equations show us not only how to find the induction due to wires wound with any density on the sphere, but also how to wind the wires to produce a given field. For a uniform field inside, we have B, = B cos 0 and Be = —B sin 0 so that, from (6) and (8), Cn = 0 except for n = 1. Thus, from (2), the density of winding is proportional to d[Pi (cos OW dO = — sin 0. By 7.08 (3), the equations of the tubes of force, if we use (4) and (5), are
N = 2rr sin Oilo
(10)
7.13. Field of Circular Loop in Spherical Harmonics.—Suppose that the current density is zero on the surface of the sphere except in a band at 0 = a whose width is too small to measure physically but is different from zero so that the current density function and vector potential are everywhere bounded. From 7.12 (2) and (3), i may be written =
1
C.P;s(u) n=1
We determine Cn as in Chap. V by multiplying by Pi(u) and integrating from u = —1 to u = 1. By 5.13 (2), all terms on the right drop out except when n = m, and by 5.231 (3) this gives _ (n — 1) ! 2n + lf ' I aiP4(u) du (n + 1)! 2 _ The current density i is zero except in a band of width As at 0 = a, which is so narrow that P.;,(cos 0) has the constant value P1,(cos a) on it. Then the integral becomes P;,(cos
a) f ia sin 0 d0 = PI,(cos a) sin af i ds = I sin aP1(cos a)
so that we have
(2n + 1)/ sin a P4(cos a) 2n(n + 1)
(1)
The vector potential becomes, from 7.12 (4), if r < a,
sin a (y n(cos a)n(cos 0) 2 Li n(n + 1) a
A = did
n=1
and the components of the induction are, from 7.12 (6) and (8),
(2)
§7.14 BLOT AND SAVART'S LAW. FIELD OF STRAIGHT WIRE CO
,u/ sin «_r
2a
295
n
a
/11 (cos a)Pn(cos 0)
(3)
n=1
Bo _ 11/ since _r 2a Gi n a
.P1(cos a)P;,(cos 0)
11-1
(4)
n=1
r is greater than a, we must substitute (a/r)n+2 for (r/a)n-1in (5) and -n(a/r)n+2/(n + 1) for (r/a)n-i in (6). By 7.08 (3), the equations of the tubes of force, using (2), are
N = 27r sin 0244
(7)
7.14. Biot and Savart's Law. Field of Straight Wire.—Taking the curl of both sides of 7.02 (5) we have
4 1 1. fp x cY. B=VxA= ± but ds 1 1 Vx— =-Vxds+V (- x ds r r r Since the coordinates of the point at which B is being computed do not appear in ds, which depends only on the circuit configuration, ds is constant as far as V is concerned and the first term on the right is zero. But V(1/r) = - (r/r3) so that
B-
reds x r r 47rY r3
(1)
If 0 is the angle measured from ds to r, this may be written
B -µ7 µ7 sin sin 00 ds ds 44
r2
(2)
B is normal to the plane containing ds and r. If 0 is clockwise, B points
296
MAGNETIC INTERACTION OF CURRENTS
§7.15
away from the observer. Biot and Savart established this law only for straight currents. In this case, if a is the distance of P from the wire (Fig. 7.14), we have = a2 +82,
sin B = —a r
so that (2) becomes, by Pc 138 or Dw 200.03, a ds _ 21 — 1(3) 27ra 27rjo (a2 + s2)i 27ra (a' + s2)i 0 7.15. Field of Helical Solenoid.—A very frequently used form of electric circuit is the solenoid. This consists of wire wound in the form of a helix, shown in Fig. 7.15. Let the equation of this helix be
B
x = a cos q5,
y = a sin 0,
z = a4 tan a
(1)
where a is the pitch of the winding so that z increases by 27ra tan a when 0 increases by 27r. The z component of B [7.14 (1)] is [ds x r] r ds — r ds B. = Y Y (2) 3 z—
47r
r
47r
The components of r and ds are, for a point on the axis, rz = —a cos 0, dsz = —a sin 0 d0, = —a sin 0, ds, = a cos 0 dcP, rz = —aq5 tan a + C dsz = a tan a 45 We shall find the field on the axis of a solenoid having an integral number N of turns, neglecting the field of the leads, which may be calculated separately. Taking the origin at the point at which B. is to be found, setting r = (x2 + y2 + Z2)1 = a (1 02 tan' a)i in (2), and choosing the limits 07 = —Nr b/(a tan a) b/(a tan a), we find, after integrating by Pc 138 or and 02 = Dw 200.03, that B. at a distance b from the center is
B. =
f 47ra
dc/3 ck (1 +42 tan'
a)i N7ra tan a b Nra tan a — b [a2 47ra tan a [a' (N7ra tan a + b)']i (aNr tan a — b) 2]4 If fi‘i and 02 are the angles between the axis of the helix and the vectors
FIELD OF HELICAL SOLENIOD
§7.15
297
drawn from b to the extreme ends of the wire, this may be written
B — I (cos 02 — cos (31) (3) z 4ratan a It should be noted that either 131 or 02 must be greater than rr if b lies inside the solenoid so that one cosine is negative. If the distance between the extreme ends of the wire in the solenoid is L, then
L = 2arNa tan a and if there are n turns per unit length, N = Ln, so we can write
Bz 1/./n/ (cos N2 — cos 01) (4) Formulas (3) and (4) are rigorous expressions for the axial component of the field on the axis. Unless a = 0, this is not, however, the only component. Proceeding as before gives B z
4a
ds„ — ry dsz _ r o2a2tan a( — 4, cos4) + sin 0) d4) r3 r3 r3 4rJoi
(5)
This is not, in general, zero.
B—
4rY
r os— a' tan a(cos ± sin 43) d4) r3
dsz — r, dsz r3
(6)
This integral is also not, in general, zero. Thus the lines of force are not straight but spiral. In the important case of an infinitely long solenoid, (3) and (4) become
B, —
2ra tan a
=
(7)
The x-component vanishes since (5) is the integral of an odd function, but (6) becomes -cos +4, sin q5 dcp B = 271-n2aiLT [(2nara)2
0213
Integrate the second term under the integral by parts, letting u = sin 4) and dv = 4)[(2nra) 2 + 02]-1do, and we have
B„ = —27rn2citLI
(2ncroas) (12' + 614'091
[(2ncroas) c26 -1" 02]i}
If we let 4) = 2nra sinh IP, the second integral becomes, by 5.331 (2), Ko(2nara) which is the modified Bessel function of the second kind treated in 5.331. Evidently, the first integral can be obtained by differentiating the second with respect to 2nra and dividing by — 2nra so that, by 5.331 (2), (8) B, = —nis/[2nraK0(2nra) K,(2nara)]
When a = 0 or n = 00 , Bi, = 0 so that the field is axial. If a = fr or
298
§7.17
MAGNETIC INTERACTION OF CURRENTS
n = 0, our solenoid becomes a straight wire parallel to and at a distance a from the z-axis. Then B. = 0 and, from 5.33, Ko —> — In (2nra) and K1 1/(2nra) so that B n--00
21ra
which agrees with 7.14 (3). 7.16. Field in Cylindrical Hole in Conducting Rod.—The computation of the magnetic field due to a certain current density distribution can frequently be simply obtained by superimposing known solutions. As
an example of this method, let us suppose conducting material with permeability occupies all the space between two infinite cylinders. We wish to find the induction inside the smaller cylinder of radius a, which lies entirely inside the larger cylinder, of radius b, the distance between axes being c (see Fig. 7.16). The current density iz is taken as uniform and in the z-direction. If there were no hole, B0 would be constant on the circle of radius r through P, and from 7.01 (2) its value would be Arr2i = fB' • ds = 27rBo
or
B' =
xT
Addition of a similar expression for a current density —i flowing only in the inner cylinder gives zero current density there so that
B = µvi x (r — r') = C = ipt.ic(k x ii) = (1) where il, i, and k are unit vectors in the x, y, and z directions. The total current I is Orb' — rct2)i so the uniform field in the hole is Aye]. By — (2) 24r (b2— a2) 7.17. Field of Rectilinear Currents in Cylindrical Conducting Shell.— Suppose currents in an infinite circular cylindrical shell flow everywhere
§7.18 FORCE ON ELECTRIC CIRCUIT IN MAGNETIC FIELD
299
parallel to the axis and that the z-directed current density i can be expressed in circular surface harmonics; thus, iE =
(C,, cos na
D„ sin na)
(1)
n==0
From 7.09, the vector potential at P due to a long straight wire carrying a current I is parallel to it and equals (itd/r) In R if the shortest distance from P to the wire is R. Thus, from 7.02 (4), the vector potential due to the cylinder is (see Fig. 7.17) A, = bor-1 027ia In R da 4.02 (2) In R
1O m
In a ni=1
a
(2)
(cos mO cos ma sin mO sin ma)
(3)
Substitute (1) and (3) in (2), and we have integrals from 0 to 2r of the products cos mx cos nx, sin mx sin nx, and sin mx cos nx. From Dw 858.1 and 858.2 or Pc 488 and 360, these are all zero unless rit = n. If m = n, the integral of sin 2ma = 2 sin ma cos ma is still zero, but that of sine ma and cost ma is 7 by Pc 489 or Dw 858.4 so that (2) becomes As= ActCo In a
2 44ina
n(Cn cos nO + D. sin nO)
(4)
7.18. Force on Electric Circuit in Magnetic Field.—We found experimentally in 7.00 that the forces acting on a small loop of wire in a magnetic field are exactly like those acting on an electric dipole in an electric field. We also saw that any electric circuit may be considered as a mesh of such loops. From 1.07 (3), we can therefore write the force on an element dS of the mesh of a circuit carrying a current I by substituting nI dS for M and B for E; thus dF = I(n • V)B dS
(1)
We have the well-known relation V(n • B) = (B • V)n + B x (V x n) + (n • V)B
n x (V x B)
(2)
The first two terms on the right are zero since V does not operate on n. Since no current flows inside the loop, V x B = µi = 0 from 7.01 (3): We may therefore write for the total force on the circuit, substituting V I, A for B from 7.02 (1),
300
MAGNETIC INTERACTION OF CURRENTS
§7.19
F =IVfsn•BdS=IVfsn • V x A dS Applying Stokes's theorem [3.01 (1)], making use of (2) again where ds rep!aces n and A replaces B, and noting that V does not operate on ds, we have F = IV .96' A • ds = If [(ds • V)A ds x (V x A)] (3) But we see that
(ds • V)A = •9r(i dAz
dA„ k dAz) = 0
since dAz, dAy, and dAz are total differentials and A, Au, and Az are single-valued functions. From (1), V x A = B so that
F = I C ds x B
I fB sin 0 ds = dF
or
(4)
where 0, which is less than r, is the angle from ds to B. If, when the line of observation is normal to ds and B, 0 appears clockwise the force is away from the observer. Ampere's experiments give no unique law of force for circuit elements because (4) and 7.14 (1) give the force between two circuits to be
,fiyds x (ds' x r) _
F =Ids x B = yll
fyds'(ds • r) — r(ds • ds') (5)
71-ra
471-r3
The line integral around the s-circuit of the first member of the last integral vanishes because r/r3 is the gradient of the scalar 1/r. Thus Ampere's force experiments give the force between circuit elements to be ds x (ds' x r)
r3
Or
r(ds • ds') r3
(6)
Comparing (4) with 7.14 (1), we see that if we have two circuits nearly in coincidence, carrying equal currents in opposite directions so that, from Ampere's experiments, the magnetic field is nearly zero, then only those contributions to the force from portions of the range of integration for which r is small are important. These are in the positive r direction. The circuits therefore tend to separate, thus creating a magnetic field. This is exactly the reverse of the electric case where the force on equal and opposite charges tends to bring them together and destroy the electric field. Thus, there is a fundamental difference in the nature of the energy in electric and magnetic fields. We shall consider this in the next chapter, in connection with Faraday's law of induction. All these formulas for forces on circuits placed in magnetic fields have been well verified by experiment. 7.19. Examples of Forces between Electric Circuits.—When we have two parallel infinite wires at a distance a apart, carrying currents I and I', we have, from 7.14 (3) and 7.18 (4), since sin 0 = 1, the result
§7.19
EXAMPLES OF FORCES BETWEEN ELECTRIC CIRCUITS
301
F. = 11177 fds' [s1 r x a] 2ra2 so that there is an apparent repulsive force per unit length of amount F1
(1)
— 27ra
—
The positive sign is taken when I and I' are in opposite directions, and the negative sign when they are in the same direction. Next, let us compute the force between two parallel coaxial loops of wire of radii a and b, carrying currents I and I', as shown in Fig. 7.19. From symmetry, the force is one of attraction only so that, from 7.18 (4), the only effective component of the induction is B„, which is the same for all elements of either circuit. Thus, since the coordinates of the I' circuit are p = b, z = c, we have for the force
F = l'13„(a, b, c).10.27b dO = 2rbI'B„(a, b, c) Substituting from 7.10 (6), we have
F
arc [(a + b)2
[
(a2 b2 c2).E1 (a — b)2 c2
K
c2P
(2)
where, from 7.10 (3), the modulus k is
k2—
4 4ab (a + b) 2
c2
The force reverses if either current reverses. We may easily get this force in series form by combining 7.18 (4) with 7.13 (3), giving
F=
sin a(
b2
a yn
F
-
C2
Pily(COS «)P„(0)
(3)
„=1
when c2 b 2 > a2, since u = 0 in the latter, r = a and f ds = 2ra. Another, more rapidly converging expression for the force can be obtained, if the two loops have different radii, by choosing the origin at the apex of the circular cone, of half angle 13, in which the two loops lie. If r is the radius vector to the smaller loop and s the radius vector to the larger, then the force between them is
F = 2irrl' sin i(B, sin + Be cos ()
302
MAGNETIC INTERACTION OF CURRENTS
§7.20
where Br and Be are given by 7.13 (3) and (4). Thus, we have
F = irALIF sin3
(ryi - /3;,(cos (3)[nPn(cos (3) — cos /3P;, (cos 13)]
r3
s n
n=2
We start the summation from n = 2 since the n = 1 term is zero. The last term may be simplified by 5.154 (2) and (4.1) so that, writing n + 1 for n, where 1 n < co, and noting that r/s = a/b if b > a, we obtain
F = ---„II, r sine ,3
1 ay-",,i(cos ,),(cos 3) n + 1(b
(4)
n=1
7.20. Vector Potential and Magnetization.—The definition of the vector potential given in 7.02 (4.1) has proved adequate for all the cases so far treated in which the whole region is occupied by a medium of a single constant permeability. If p is variable or discontinuous, it is necessary, in order to define a magnetic vector potential uniquely, to consider the nature of magnetization. In 12.06, we shall give experimental evidence that magnetization is due to circulating currents or spinning electric charges within the body. We define a vector M to be the magnetic moment per unit volume of such currents or spins and call it the magnetization. With the aid of 7.10 (2), setting A = At, and measuring r from the field point the vector potential of M becomes
Am = — Pv iM xr dv = 14 f M 4r v 4r v r3
r
(1)
We can transform this equation by using the vector relations v x pq = Vp x q + pV x q
(2)
s fvVxMdv = fnxMdS
(3)
and
where n is the unit outward normal vector to the surface S which encloses v. We prove (3) by putting A= Mxa in Gauss's theorem [3.00 (2)] where a is a constant vector so that V • (M x a) = a • V x M, giving a•f VxMdv = fMxa• ndS = a• fnxMdS
v
s
s
Since this is true for all values of a, Eq. (3) follows. Applying (2) and (3) to (1) gives 4ir A --tam— ilv
VX VX Mxn M dv I Vx (1 dv = Iv r M dv ± si r dS (4) r fv r v
When the magnetization is uniform in the interior of a body, (4) is more
§7.21
MAGNETIC BOUNDARY CONDITIONS
303
useful than (1) since the volume integral is zero. If i is the actual current density, then the total vector potential is now given by
n
A=14 f i+V xM dv-F Av 471- fs Mr dS 4ir v
(5)
In order to apply this formula in most cases where the data include only the current distributions and the permeabilities, we must express M in terms of g and B. To do this, we note that, at great distances from closed circuits of finite extent, the induction and, hence, the magnetization are inversely proportional to the square or higher powers of r so that the surface integral in (4) vanishes. If A is constant in this region, and M is due entirely to i the vector potential must also be given by 7.02 (4.1) so that, comparing it with (5), we have = Ar(i + V M)
(6)
In a medium that is isotropic but not homogeneous we defineµ so that this equation is still valid. With it we eliminate Ai from 7.01 (3), and because the resultant equation must reduce to 7.01 (3) when g is constant and it is known experimentally that M is proportional to B and in the same direction, we have
vx
= = vx
\. 11 M=
(1 1.)B Au
A
(7) (8)
Just as we can get 7.01 (2) from 7.01 (3), we get from (7) the relation /
• ds =I A
(9)
7.21. Magnetic Boundary Conditions.—In the last article, we defined the vector potential A for regions where the magnetization is not uniform and on surfaces where it is discontinuous. To find the boundary conditions that A must satisfy, we observe that each of the three components of A is defined by a scalar equation analytically identical with 1.06 (6), which may be written in vacuo f p dv vr
r dS js r
(1.06) (6)
This defines the electrostatic potential in free space due to a volume distribution of electric charge of density p and a surface distribution of density a. From electrostatic considerations, we know the value V, of these integrals just inside the surface S is the same as the value V. just outside. Furthermore, by applying Gauss's electric flux theorem
304
MAGNETIC INTERACTION OF CURRENTS
§7.21
to a small disk-shaped volume fitting closely an area dS of the surface and so thin that dv is negligible compared with dS, we have, after canceling out dS, the relation
i av ay° — =
an
an
Ev
Thus, we know the boundary conditions that apply to integrals of the form of 1.06 (6) and hence to each component of 7.20 (5). Adding the components and writing 1.4 for 1/e„, we have Ao = Ai
(1)
and with the aid of 7.20 (8),
aA0
3A, Av(M x n)
Av[(V x Ai) x n]
=
(2)
If there is a magnetization M' on one side of the boundary and M" on the other, we may find the boundary conditions by imagining a thin layer of permeability between the boundaries, writing down (1) and (2) for each boundary, referring them to the same normal, and eliminating Ao and 0Ao/an, giving A' = A" (3) and
aA" WI,
an
liv[(111/
(4)
3/1") x
To get the second boundary condition in terms of the permeability, we write n • V for a/an and use the relation n • V (A' — A") = V[n • (A' — A")] — n x [V x (A' — A")] — (A' — A") • Vn (A' — A") x (V x n) —
The last two terms drop out because of (3), so that, writing B' for V x (A' A"), by 7.02 (1), and rearranging, we have
—
B"
—
V[n • (A'
—
A")] = n x (B'
—
,uoM'
—
B"
vM")
Substituting from 7.20 (8), we have V (A',
—
AZ) = Avn
B' B")
—
From (3) A'n — AZ is equal to zero over all the boundary so that its gradient along the boundary is zero. Thus V(A'n — A'„') is a vector normal to the boundary. But the right side is a vector directed along the boundary surface so that writing out the tangential components of this equation gives B' B" nx — = 0 (5) A
§7.22
EXAMPLE OF THE USE OF
A
AND A
305
We may now write down the boundary conditions on the components of A in the orthogonal curvilinear coordinate system u„ us, and u,, treated in 3.03, 3.04, and 3.05. Let us take ur to be constant on the boundary; then, from (3), we have
AT =
,
A's = A's',
and
A; =
(6) Writing V x A for B in the left side of (5), we have, by 3.04 (1), 3.04 (2), and 3.04 (3), the two equations
Fa (hr, L aut,,
a (hag u t,,)1 = 1[a(hr. sA"r ,$) rA r,:t,c) j aut, r These equations (6) and (7) give the required boundary conditions. To obtain the boundary conditions on B, we note that (3) shows the differences in the vector potential between two points on the interface to be the same in either medium. Hence, the derivatives of the vector potential on the two sides of the interface, in the same direction parallel to it, are equal. The vector n x A lies in the interface, and so V • n x A involves only such derivatives. We have the well-known vector equation for the derivative of a cross product s)
(ht
att„,
V •nxA= When we substitute A'
—
—
n•VxA+A•Vxn
(8)
A" for A, the last term vanishes by (3) giving
n•VxA' = n•VxA"
(9)
Writing B for V x A gives n•B' = n • B"
(10)
Thus, the normal components of the induction are continuous. From (5), we have for the tangential components of the induction 1 -7
1 (n x B') = ,7 (n x B") —
(11)
It is often possible to choose two vectors A' and A", different from and simpler than A' and A", whose curl gives the same induction at all points but which, as will be shown in 8.04, satisfy, instead of (3), the boundary conditions As' and (12) A',.' A '„t = It is evident in the example considered in the next article that these vectors may be more convenient for computation than A' and A". They are not uniquely defined, however, as are A' and A", by integrals of the type of 7.20 (5). We shall refer to them as quasi-vector potentials. 7.22. Example of the Use of A and A. As an instructive example of the use of the boundary conditions in the last article, let us consider an infinite wire carrying a current I in the z-direction, positive and nega—
306
§7.23
MAGNETIC INTERACTION OF CURRENTS
tive values of z having permeabilities A' and A", respectively. Clearly, from 7.02 (5), the simplest vectors A' and A" that we can write which satisfy 7.21 (7) but not 7.21 (3) and whose curls give us the correct values of B' and B" are A' = A" = — kG-A--144" In p) A' In p) and (1) The vector defined by 7.20 (5) requires that A' = A" at the boundary. The part of the volume integral involving I in 7.20 (5) is given by 7.02 (5). We must superimpose on this a potential to meet the new boundary condition. Inspection of the solutions available in cylindrical coordinates in 7.05 (7) shows that the suitable form is cz/p. Putting this in and evaluating the constants, we obtain A' = Fr-111 - kA,(In p (2) ILOZP-1 + C211 +e A" = -kA,,(1n p C1) eRA" — C2]) (3) The constants CI and C2 are arbitrary. The second terms represent the contribution from the integral of 7.20 (1). We get the same value of B' and B" from (2) and (3) as from (1). 7.23. Current Images in Plane Face.—The similarity of the boundary conditions for B in 7.21 to those for D in 1.17 suggests that the image method of 5.05 might be used in getting the magnetic field when an electric circuit is near the plane face of a semi-infinite block of material of permeability A. Let the plane face be at z = 0, the circuit lie in the region of positive z, and the material of permeability A occupy the entire region of negative z. Let A be the vector potential due to the circuit alone. The image law of 5.05 suggests that the quasi-vector potential above the interface can be given by A + A' and that below by A", where y, z) kf 3(x, y, z) A= y, z) (1) A' = iCY1(x, y, -z) .iCLf 2(x, y, -z) kCLI3(x, y, -z) (2) A" = kCI73(x, y, z) y, z) iC'2'f2(x, y, z) (3) A'; and A.,, A'z'„ = A'x',,„ or From 7.21 (12) at z = 0 Az + A; (4) 1 + CI = 1 + = 1+ {(
—
The boundary conditions at z = 0 on the derivatives given by 7.21 (7) are satisfied if we have 14(1+ CD = A„C3 (5) u(1- CD = Pven ti(1- CD = have the same form, Solving, we obtain, since A" and A
Ci = C2 =
-
-
-r Av
-
-
2A A
,
uo
(6)
Thus, the magnetic induction outside the permeable medium appears to be caused by two circuits, the original carrying a current I, and an
u image circuit carrying a current I' -
7 u„I.
AL 'I” Ao
The direction of I' is
§7.25
TWO-DIMENSIONAL MAGNETIC FIELDS
307
such that the projections of I and I' on the interface coincide in position and direction. In the permeable medium, the magnetic induction appears to be due to the original circuit alone but carrying a current I" = 2Iµ/(µ ± Ay) instead of I. 7.24. Magnetic Induction and Permeability in Crystals.—The similarity of the boundary conditions derived in 7.21 for B to those derived in 1.17 for D suggests that the method of measuring D and E in 1.18 may be applicable also to B and µ. Let us excavate a small disk-shaped cavity in the solid, whose thickness is very small compared with its radius so that the induction in the cavity, far from the edges, will be determined entirely by the boundary conditions over the flat surfaces. If we orient the cavity so that the induction inside is normal to the flat faces, then from 7.21 (10) this B equals that in the medium. Let us now excavate a long thin cylindrical cavity so that the induction inside, far from the ends, is determined entirely by the boundary conditions over the curved walls and orient it so that this induction is parallel to the axis. We now find, from 7.21 (11), that BIliz inside equals B/A in the solid. These two measurements give us both B and A. If we perform this experiment in a magnetically anisotropic medium we shall find, in general, that B and B' are in different directions. There will be at least three orientations of the field with respect to the crystal for which B and B' have the same direction. These are the magnetic axes of the crystal, along each of whichµ may have a different value. In the more common case, B and B' have the same direction for all orientations of the field in some plane, and for the field normal to it. This subject is further discussed in 9.02 and 9.03. 7.25. Two-dimensional Magnetic Fields.—In rectangular coordinates, the boundary conditions for the tangential component of the magnetostatic vector potential are identical with those for the electrostatic scalar potential when e is replaced by 1/A. In two-dimensional problems, involving only magnetic fields parallel to the xy-plane, the currents must flow in the z-direction and so the vector potential can have only one component Az, which is, necessarily, tangential to all surfaces. When WA, becomes very large, we see, from 7.21 (7), that the surfaces on which Az is constant, i.e., the equivector potential surfaces, are normal to the boundary. Thus, at this point, Az behaves as the electrostatic stream function does at an electric equipotential boundary. We also obtain B from A. in the same way as we obtain the electric field from the stream function; for, from 3.04 (2) and (3), we see that, in rectangular coordinates, where h1= h2 = h3 = 1 we have
B. =
OA
ay
These are the same as 4.11 (1).
By =
OA.
ax
(1)
308
§7.26
MAGNETIC INTERACTION OF CURRENTS
"Where there is no current, 7.02 (3) becomes v2Az = 0
(2)
so that all the methods of Chap. IV are available for its solution. 7.26. Magnetic Shielding of Bifilar Circuit.—To illustrate the application of the results of the last article, let us compute the magnetic induction outside a cylindrical shield of permeability 1.1, with internal and external radii a and b which surrounds, symmetrically, two parallel conducting wires carrying currents in opposite directions. Evidently, the problem calls for the use of the circular harmonics of 4.01. In 7.09, the vector potential due to two wires carrying currents in opposite directions was found to be
AZ = 14"T In r2
(1) r1 By 4.02 (1), setting Oo = 0 and 00 = r a==d po = c, we have the expansions, when p > 2r
1 c ' -(- cos nO + In p n p
In r1 = — n
_1(cy( — 1)n cos nO + In p n p
In r2 n =1
When we take the difference, even terms go out so that, writing 2n + 1 for n, we get 2..-1-1 cos (2n + 1)0 (2) In = ri 2n + c\ n=o Let us now superimpose a potential inside the shell, due to its influence, which is finite at the origin, giving in this region /2n-F1
Al z = iiii
A2.+1.p2M-1
.[
+2n 1+ 1
o In the region 2, of permeability 11, have n=
A2. =
—
7i n=0
p
cos (2n + 1)0
(3)
is neither zero nor infinity so we may
_L
2n-1\)
cos (2n
1)0
(4)
309
CURRENT IMAGES IN TWO DIMENSIONS
§7.27
Outside the potential must vanish at infinity so that cos (2n + 1)0
A3z =
(5)
n=0
These equations must be satisfied for all values of 0 so that each term separately must meet the boundary conditions 7.21 (6) and (7), giving after multiplying through by a 2n-E.' or b2"-", when p = a A2.+1.64n+2
2n+1 = B2,04a41.-1-2 _L
r
(6)
2n + 1c
and p2=12„.4_1a4n+2
C2n+1
2n + 1
AvB2n+
ia4n-1-2
(7)
when p = b B 2,,-Fib4n+2
(8)
C2214-1 = D2z1-1-1
and — ev-2n-F1
(9)
= PD2n+1
Solving for D2„.4_1, we have [
A 3= = 41//
/a 4n+2 1-1 1 (K„,± 1) 2 — (K„
— 1) 2
V,)
Cyn-l-l
2n ± 1.13
cos (2n
+ 1)0
n=0
(10) The field outside is given by 3.04 (2) and (3) where, from 3.05, h1 = 1 and h2 = p to be 1 .0.A aA z B = 139 = (11) " p 80 op 7.27. Current Images in Two Dimensions. The vector potential due to a straight linear current I and the scalar potential due to a line charge q have exactly the same form. Furthermore, as we have seen, the magnetic vector potential and the electrostatic potential, in two dimensions, satisfy the same form of Laplace's equation and the boundary conditions, if 1/Kn, is substituted for K, are the same. It follows that the results of Art. 4.04 apply to a linear current I parallel to and at a distance b from a circular cylinder of permeability pi and radius a. Thus, the vector potential in the region outside the cylinder due to its presence is the same as if the cylinder were replaced by a parallel image current I' located between I and the axis, at a distance a2/b from the latter, plus a current of strength — I' along the axis. The vector potential inside the cylinder is the same as if I were replaced by a linear current I". I' and I" are given by substituting 1/K„, for K in 4.04 (7). These —
310
MAGNETIC INTERACTION OF CURRENTS
§7.28
images lead to the potentials Azi
p
A„.1`
L.1
2K,„ (pycos nJ
+1 b
(1)
n
n=1
a
{ln aP
Azo =
— 1 a2 "COS niq ± 1 by n
[ (a) m n=1
(2) At p = a when Km = co, both B, and H, vanish outside the cylinder but only Ho inside it. If the current I lies at p = b, 415 = 0 inside a hole of radius a in an infinite medium of permeability A, the potentials are
b
a< p
Azi =
Az. =
2a
{1n p —
l)" ±
L kp/ n=1
— 1(bp\nicos
K„,-1- 1 \a2 )
n 1
(3) 2Km [ ad i Km(Km— 1) 1n p ± ln p Km-I- 1 Km + 1 2r . (bycosncb ll, — (4) n n=1
In Am, an additional term I"', the first term in (4), appears at the origin. When Km= 00 , only the in p term in (3) contributes to Bo at p = a so its value ma,I / (27a) is independent of b. This is a fairly general situation and occurs again in Art. 7.31. Outside only the first term in (4) survives, so Bo is infinite but Ho has the value I /(27rp) and Hp vanishes. 7.28. Magnetomotance and Magnetic Intensity.—In dealing with the magnetic field, we are immediately struck by the many properties that the magnetic induction and the electric current have in common. In the first place, we have 6.00 (3) V • i = 0
and
7.01 (1) V • B = 0
(1)
Then, for a simple closed path, we have 6.02 (4) f
• ds
—e
and
7.20 (9)
jB • ds
—/
(2)
The magnetic vector potential was derived on the basis of 7.20 (5). It is clear that we can proceed equally well along the lines of Chap. VI
§7.28
MAGNETOMOTANCE AND MAGNETIC INTENSITY
311
and define a new scalar quantity, analogous to which we shall call the magnetomotance Q. Thus the magnetomotance in traversing once a path enclosing a current I is =/ (3) where I is in amperes and 0 in ampere-turns. As both the magnetomotance and the electromotance are multiple-valued functions, their value depends on the path chosen. Using the method applied to e in 6.01, we can often insert a suitable barrier so that the integral of (2) becomes zero for all permitted paths. 0 then takes on the character of a scalar potential. In the simplest cases, the barrier erected is known as a magnetic shell. In (2), we see that the permeabilityµ plays the role of the conductivity -y in Chap. VI, so that we may set up a relation equivalent to Ohm's law for magnetic circuits, which corresponds to 6.02 (3) and is B= (4) It is convenient to have a symbol for the gradient of the magnetomotance. We shall call the negative of this quantity the magnetic field intensity and designate it by — H, so that
H=
=
A
(5)
The mks unit for H is ampere-turns per meter. In terms of H, (2) is
fH • ds = I
(6)
This equation is independent of the permeability of the medium. From (1) and (4), we get the relation analogous to 6.04 (2) which is vg 0
(7)
Starting with 6.00 (3), 6.02 (4), and 6.04 (2), we obtained the conditions that the electromotance must satisfy when at the boundary between two mediums of different conductivity. In the same way, we would get, from 7.28 (1), (2), and (7), the conditions 0 satisfies at the boundary between two mediums of different permeability. It is simpler, however, to start with 7.21 (10) and (11) and use (5). Thus from 7.21 (10)
1.4'H' • n = /PH" • n
(8)
H' x n = H" x n
(9)
and from 7.21 (11) In terms of the magnetomotance, these are
,ace
12 — an —
,00"
an
(10)
312
MAGNETIC INTERACTION OF CURRENTS
§7.29
and
It is to be noted that the magnetomotance satisfies just the same equations (7), (10), and (11) as the electrostatic potential, 3.02 (6) and 1.17 (5) and (6). It differs from the latter, however, in being a multiple-valued function unless we erect a barrier in the field in such a way as to prevent our making a complete circuit of any currents in the field; then i1 is single-valued in the region outside the barrier and may be treated by the methods of electrostatics. When so restricted, St is called the scalar magnetic potential and we use it to obtain solutions of magnetic problems from electrostatic solutions. For example, if we wish to find the shielding effect of a spherical shell of permeabilityµ and internal and external radii a and b when placed in a uniform magnetic field of induction B, we can use 5.18 (5). This shows us that the field inside the shell is still uniform and that its induction is
Bi
9K„,B 9K„, — 2(1c,, — 1)2(01)-3— 1)
(12)
Because (10) and (11) are identical in form with 1.17 (5) and (6) the law of refraction of magnetic lines of force at the boundary between two isotropic magnetic mediums, derived from the former, will be identical with that for electric lines of force derived from the latter and expressed by 1.17 (8). Thus, at such a boundary, the magnetic lines of force or induction make an abrupt change in direction given by 1.1' cot a' =
(13)
cot a"
where a' is the angle that a line makes with the normal to the boundary in the medium of permeability pl and a" is the corresponding angle in the medium of permeability u". 7.29. The Magnetic Circuit. Anchor Ring.—In 7.28, we pointed out the close mathematical analogy between the magnetic induction and the electric current density. To use the technique of Chap. VI for the solution of magnetic problems analogous to electric current problems, it will be convenient to define several new magnetic quantities. The magnetic flux N, through a surface S, corresponds to the electric current I, the equations being 6.00 (2) I = fsi • n dS
N=
n dS
(1)
The mks unit of flux is the weber. The magnetic reluctance R' between two points in a circuit corresponds to the electric resistance, and the
AIR GAPS IN MAGNETIC CIRCUITS
§7.30
313
analogous equations are 6.02
(1) R AB
=
St A B
E AB T
AB
." AB —
Iv AB
(2)
To illustrate the computation of such circuits, let us find the flux in a torus of permeabilityµ wound with n turns of wire carrying a current I. Let the radius of a section of the torus be b and the distance from its center to the axis of the torus a. From 7.28 (3), the magnetomotance around the torus from 0 to 27 is nI. Since, by symmetry, H is independent of 0, the magnetomotance around that portion of the path between 0 and 0 is proportional to 0, which gives
=
nI0
(3)
2r
Thus, from 7.28 (5),
Ho =
au r00
=—
In tar
So, for a given value of r, H e is constant, and we may break up the torus into a set of cylindrical laminations, as shown in Fig. 7.29, without changing its reluctance. The reluctance of one of these laminas is
dR' =
length = 2irr ,LL dS area
pi
But dS = 2[b2 — (a — r)2] dr, and since the laminas are in parallel we have a+b,b2_ (a — r)2]4 dr
r
R'
j dR' Trj a — b
Integrating by Pc 187 and putting in limits give
R'
=jr[a — (a2
—
By (2), the flux through the ring is then
N= — R'
AnI[a — (a2— b2)1]
(4)
7.30. Air Gaps in Magnetic Circuits.—It is evident that if there is a continuous path of high permeability and sufficient cross-sectional area at all points, the flux in the magnetic circuit will be confined almost entirely to the permeable medium. If, however, there is a break in the
314
MAGNETIC INTERACTION OF CURRENTS
§7.31
path, the flux must cross this gap and will distribute itself in such a way that the magnetomotance satisfies 7.28 (7) everywhere and 7.28 (10) and (11) at the gap boundaries. If the medium permeability A" is very high compared with the gap permeability then a' –4 0 from 7.28 (13), and the flux leaves it normally. If the product of A' by the length of that part of the magnetic circuit lying in the permeable medium is small compared with the product of A" by the length of the air gap and if the cross section is nowhere too small, we may consider practically the entire reluctance of the circuit to lie in the gap. This is true in many electrical machines. When the faces of the gap are parallel and their area A is large compared with the distance apart d, the field in the gap may be taken as uniform and the effect of the edges neglected so that its reluctance is d/(A'A). To obtain rigorous solutions of other cases, we must use the vector potential or work with the induction, using Biot and Savart's law and the boundary conditions of 7.21. This frequently entails more mathematical labor than is justified by the accuracy required. In such cases, a sufficiently accurate result may be given by using the scalar magnetic potential which makes available the methods of Chaps. IV and V. As mentioned in 7.28, this necessitates the erection of an artificial barrier somewhere in the permeable part of the magnetic circuit. If the permeability is high, we then assign one value of the potential to that portion of the medium on one side of the barrier and another value to that portion on the other side. If the configuration is such that we can put the barrier far from the gap, the error introduced by its presence is negligible. 7.31. Field in Shell-type Transformer.—Calculation of the fields in a shell-type transformer with infinitely permeable walls when a one-way current flows in a winding as in Fig. 7.31a is difficult because the obvious simple assumption that the field is everywhere normal to the walls makes the line integral of B around the window zero. Thus by 7.01 (3) there is no net circulating current inside. For example, in Fig. 7.31b, c, and d of the second edition of this book there is a uniform reverse current sheet occupying the remainder of the cylinder on which the primary current sheet lies. A true one-way solution needs a different approach. A clue is found in Art. 7.27 where, as K. approaches infinity, the image dominates. Thus B is tangential to the wall in the exterior region regardless of the actual position of I inside the hole. This is logical because, viewed from the outside, the permeability of the hole is zero. Thus the B outside the transformer window encircles it without penetration, and the line integral of H around the window equals the current inside but is independent of its distribution. Finding the analytical solution of Laplace's equation that meets this boundary condition in Fig. 7.31a is a difficult, unsolved mixed boundary value problem, but approximate solutions by
FIELD IN SHELL-TYPE TRANSFORMER
§7.31
315
field plotting or electrolytic tank methods are possible. If solved, this gives the interior tangential H which, together with the known interior current distribution, is sufficient to specify the interior fields. The only restriction on the normal B is that the net flux through the wall is zero subject to the condition that Laplace's equation is satisfied outside the wires. A more tractable problem is the calculation of the fields in the window due to two thin symmetrically placed layers of wire coaxial with the curved
•
0
h
P•
4
3
2
(b) FIG. 7.31a, b.
walls having the same number of ampere-turns but oppositely directed currents. Let the window be bounded by p = a, p = b, z = ±h as in Fig. 7.31a and b. One layer, at p = c, extends from z = — k to z = +k and the other, at p = c', extends from z = — k' to z = +k'. Both are closely wound with n and n' turns per unit length, so that when carrying currents I and — I' they act as uniform current sheets with total currents 2nIk = —2n'I'k' = N/g. It is evident that the modified Bessel function solutions for Acb of 7.05 (3) where n = 0 and the z factor is cos (mrz/h) will meet the boundary condition that Bp = 0 at z = + h. The m = 0
316
§7.31
MAGNETIC INTERACTION OF CURRENTS
term comes entirely from I i(norp/h) since by 5.32 (5) K1(0) = co . From 5.32 (2) Ii(norp/h) becomes 2mrp/h as m approaches zero, and if there is an m in the denominator as in (11) and (12), this term has the form.Bop. From 7.02 (8) this gives a uniform z-directed field so that Bp is z at z = + h. It cannot appear in regions adjacent to p = a or p = b because there is no z-component at these walls. Suitable solutions can now be written down. Formulas are shortened by the notation M = mir/h. Thus, taking zero order terms from 7.05 (7), 0< p
Alm =
Z, A,,[11(11/1 p)K o(Ma) m=1
K1(Mp)/0(Ma)] cos Mz c < p < c'
A 20 = Z[13.1 i(M m=1
c' < p < b
Aim =
Boc2/p(1)
(2)
Cznifi(M p)1 cos Mz + Bop
D„,[Ii(Mp)K o(Mb) m=1 Ki(111 p)I 0(Mb)] cos Mz ±Boci 2 /P(3)
The brackets in (1) and (3) are chosen so that B z = 0 = [a (pAm)/aP]/P when p = a and p = b from 5.321 (1) and 5.321 (4). The line integral fitting closely the coil between z = 0 and z = z is independent of magnetization and other coils so that B • ds = foz (Biz — B2z),z dz = AnIz Differentiation with respect to z gives the boundary conditions (Blz— B2z)p=c = An1 (B1. — B2z)p=c = 0 A similar set holds at p = c' for p = c', A.20 = A395, so that 21,4.1i(Mc)Ko(Ma) D,,,[/i(Mc')Ko(Mb)
when —k < z < +k when k < Izi < +h
B2. — B3y.
At p = c, Alm =
Ki(Mc)10(Ma)] = Bm1i(Mc) K 1(Mc')/0(Mb)] = B./1(Mc')
(4) (5) A20,
and at
C„,Ki(Mc) C„,Ki(Mci)
(6) (7)
From (1) and (2) Blz— B2z is calculated and equated to (4) and (5). The summation as usual is eliminated by multiplying through by cos Mz and integrating from —h to +h. A similar operation is done for B2 — B3z at p = c', and thus two additional equations are obtained: N sin Mk = Mhk12-1„,[1.0 (M c)K o(M a) — K o(11/1c)10(Ma)] — B,J 0(Mc) CniKo(Mc)) N sin Mk' = Mhk' {.13„,I0(Mc') — C,„Ko(Mc') — 0(Mc')K0(111b) — Ko(Mc')I0(Mb)11
(8) (9)
§7.32
SLOTTED POLE PIECE. EFFECTIVE AIR GAP
317
Equations (6), (7), (8), and (9) are solved for Am B., C,„, and D., so that ,
Nc F ,,,(a,c,k) = — sin M k[IC i(M c)I o(M + I i(M c)IC o(111 a)] k Ko(Mb)F m(a,c,k) ,k') K o(M a)F Mh[I 0(Mb)K o(M a) — I o(M a)K o(Mb)] I 0(Mb)F m(a,c,k) + I 0(Ma)F,„(b,c' cm Mh[1- °WOK o(M — Io(Ma)K o (Mb)] B„,/i(Mc) C.K i(Mc) A m K 0 (M 0.1 i(Mc) /0(Ma)KI(Mc) B„,/i(Mc') C„,Ki(Mc') Dm K 0(1 a)I i(M c') ± I o(Mb)K i(M c')
(10) (11) (12) (13) (14)
The m = 0 term yields
Bo =
(15)
From 7.08 (4) the equation of a tube of force or the stream function is xIf = 27pA
(16)
Figure 7.31b shows the results of a calculation using the foregoing formulas in which h = fa k = ih, k' = 411, a = 1, b = 4, c = 2, c' = 3, and N = The values of pAo are plotted at intervals of one-half. All the B-lines that reach the walls pass from top to bottom in the right-hand window with the current directions indicated. 7.32. Slotted Pole Piece. Effective Air Gap.—Suppose that one of the plane faces of an air gap of length B has parallel slots of width 2A ,
R
S
(6)
(c
)
FIG. 7.32.
spaced at intervals 2C between centers. If C — A is sufficiently large compared with B, the field near a slot is practically undisturbed by the adjacent slots and may be computed as if they were absent. If the depth of the slot is great compared with its width, the effect of its bottom is negligible and it can be treated as infinitely deep. Thus the section RS of Fig. 7.32a can be treated as section RS of Fig. 7.32b where the openings extend to infinity. We shall consider the permeability so high that the entire magnetomotance falls across the air gap as was discussed in 7.30. The opening has the shape of a polygon with interior angles
318
MAGNETIC INTERACTION OF CURRENTS
§7.32
zero at x = ± 00 and 3w/2 at x = ± A, y = B. Thus, an inverse Schwarz transformation will unfold this boundary into the real axis in the zrplane. Let us suppose that the 3r/2 corners come to the points x1 = ± 1, then the 0 corners come to xl = ±a where a' < 1. By 4.18 (7), the z and zi planes are connected by the relation — 1)1 a2
dz = dzi
(1)
2 —
The magnetic circuit in the zrplane is shown in Fig. 7.32c which is a section of the bifilar circuit of 7.09. Since the reluctance of the part of the magnetic circuit below the xraxis is zero, the vector potential above the x raxis is twice that of 7.09 (1) and is (2)
A = my' In w ri From 4.12 (3), this is the real part of W=
14/ In a ± zi a — zi
(3)
Before we determine W in the z-plane, we must integrate (1), which may be written
z = C'
f
dZi
— 1)1
c/(a2
dzi
_ f
— a2)(zi — 1)1
For the first integral, use Pc 127 or Dw 320.01 after taking out j, and for the second, use Pc 229 or substitute z1 = au(u2 — 1)-1and use Pc 126a or Dw 200.01. After writing jC for C', the integrals may be written in several equivalent forms. In each, the signs must be checked by differentiation and comparison with (1). These forms are z =C
i zi ± C
(1 — a2)1 { 1 a' sinh-1 a (a/zi) 2—
= C sin-' z1 + 7C(1— a2)4 sin-1 z 1 z 2 a 1 1
-2-C + jC cosh-' z1
♦
jC
(1 — a2)1
_ a2
sin-'
V
(4a)
1
(4b)
I
{ 1 — a'
V
1 — (a/z1)21 1 — OP (1 — a2)1 cosh-1 = C sin-' z, + 2C"(1 +C (1 — (a/z1)2) 2a a a
(4c) (4d)
when x, > 1, x = +A = 4C7r from (4c) so that C = 2A/r, and when 1 > x1 > a, y = B = Aa-i(1 — a2)1from (4d) so that a = A (A2 + B2)-1. From (3) by Pc 681 or Dw 702, W=
AL,,/ r
In
1 + (zi/a) 1 — (z1/ a)
=
2/.1.,/ r
orW zi tank-' — , or z1 = a tanh 5) a 21.L„I (
SLOTTED POLE PIECE. EFFECTIVE AIR GAP
§7.32
319
Then, by Pc 659 or Dw 650.08, (a2
zi)= sinh rADI
This substituted in (4a) gives 2 z = -{A sin-1 [(A2 + A
tanh
2m,,/ B sinh-1 [ B (A' +
sinh 7rW
B. (6)
212,,I
We have equivalent forms for (4b), (4c), and (4d). To find the effect of the slot on the reluctance of the air gap, we must find the flux between R and S with and without the slot. From (5), we are on the real axis when W is real. For large values of U = Az on the x-axis, which will occur when C - A is considerably greater than B, we have tanh
and, since sin-1
212,/
and
1
sinh
rA z 2m,/
7rA.
ezga
2
1 A A -tan-1T3 - 2-7r - tan-1 A' (A 2 + B7)1
(6) is r.A. X =
A + I-A tan-1 B-B sinh-
A
11
[2(A2 B 2)1e2AJ if
or
Az =
214.1 7r
{2 17(x - A) In -(A2 + B2)1sinh A t ni 11} - (7) + B B 2B a A
L
and if (x - A) >> B, this becomes
Az =
r
ln (1 +
,
A2 \ + 2,o,./F A ta1 n_ B + (x - A)71 A 2B j B2) 7 LB
Since Az = 0 at the origin from (3), its value at any point x is, from 7.25 (1),
Az = fx8 44. dx = - f By dx 0 ux This is the flux leaving the lower face between x = 0 and x = C. If the same flux crossed the same interval with the same magnetomotance when there was no slot, i.e., A' = 0, then the width B' of the air gap would be B, = 1.1„1(A C) _
Az
7(A + C)B rC + 2A tan-1(B/ A) + B In (1 + (A/B)2)
(8)
320
MAGNETIC INTERACTION OF CURRENTS
§7.32
Thus a gap of length B' without slots has the same reluctance as a gap of length B with slots of width 2A spaced at intervals 2C. Problems The problems marked C in the following list are taken from the Cambridge examinations as reprinted by Jeans, with the permission of the Cambridge University Press. 1. Show that the magnetic induction at the center of a loop of wire carrying a current and shaped like a regular plane polygon of 2n sides, the distance between parallel sides being 2a, is [An//(ira)] sin (7/2n), where I is the current. 2. A wire follows the circumference of a circle of radius a except for an arc of angular length 20 across which it follows the chord. This loop is suspended from a point opposite the center of the chord so that its plane is perpendicular to a long straight wire that passes through its center. When the currents are i and i', show that the torque on the loop is (gii'ahr)(sin 4 — 4 cos 40. 3. A filament is wound in a flat open spiral whose polar equations is r = / (27rN), where N is the total number of turns and I? is the radius vector from the center to the outer end of the spiral. If this spiral carries a current I, show rigorously that the axial component of the magnetic induction at a distance z along the axis from the coil is (ii.LN I /R)(—R(R2 z 2)-1 +In [[R + (R2 + z2)1]z-11). 4. The plane of a circular loop of wire of radius b, carrying a current i, makes an angle p with a uniform magnetic field of induction B. Show that, if a sphere of radius a and permeabilityµ is placed inside the loop concentrically, the increase in torque is [(Km — 1)/(K„, 2)]27-Bia3b-1cos 0. 5. A certain lead consists of two parallel conducting strips of width A at a distance B apart, forming opposite sides of a rectangular prism. A current I goes out on one and returns on the other. If the current is uniformly distributed, show that the repulsion per unit length between strips is A7 2(7rA 2)-11A tan-1(A /B) — 1Bln [(A 2
B2)/B2]
6. An electrodynamometer consists of two square loops of wire, each side being of length a, one of which is free to turn about an axis which bisects the top and bottom sides of both loops. Assuming the loops intersect on this axis, show that the torque on one when it makes an angle 17 with the other and both carry a current I is Ai 2a{
21(31
1) + 21n
[
51(1+± 21)] + 4 cot-1(61 •2) — 2 cos-, 1(1 61) 5
7. Two circular cylinders lie with their axes at x = +c and x = —c and each carries a uniformly distributed current 1/ in the z-direction. Surrounding these is a circular cylindrical shell with its axis at x = 0, carrying a uniformly distributed current I in the reverse direction. Show that the force toward the center acting on one of the inner cylinders is m/ 2/(16irc). 8C. Show that, at any point along a line of force, the vector potential due to a current in a circle is inversely proportional to the distance between the center of the circle and the foot of the perpendicular from the point on to the plane of the circle. Hence, trace the lines of constant vector potential. 9C. A current i flows in a circuit in the shape of an ellipse of area A and length 1. Show that the field at the center is imil/A.
PROBLEMS
321
10C. A current i flows around a circle of radius a, and a current i' flows in a very long straight wire in the same plane. Show that the mutual attraction is pii'(sec a
—
1)
where 2« is the angle subtended by the circle at the nearest point of the straight wire. 11C. Two circular wires of radii a, b have a common center and are free to turn on an insulating axis which is a diameter of both. Show that when the wires carry currents i, i', a couple, of magnitude 4-n-pb2a-4[1 — Act-2b9ii', is required to hold them with their planes at right angles, it being assumed that b/a is so small that its fourth power can be neglected. 12C. Two currents i, i' flow round two squares each of side a, placed with their edges parallel to one another and at right angles to the distance c between their centers. Show that they attract with a force 2aii' {c (2a2 +c2)1 a2
c2
+1
a2 c(a2
2c2 ,2)1
13C. A current i flows in a rectangular circuit whose sides are of lengths 2a, 2b, and the circuit is free to rotate about an axis through its center parallel to the sides of length 2a. Another current i' flows in a long straight wire parallel to the axis and at a distance d from it. Prove that the couple required to keep the plane of the rectangle inclined at an angle , to the plane through its center and the straight current is [2aii'abd(b2 d 4— 2b2d2 cos 2q5)]. 2) sin 0]/[7(b4 d 14C. A circular wire of radius a is concentric with a spherical shell of soft iron of radii b and c. If a steady current I flows around the wire, show that the presence of the iron increases the number of lines of induction through the wire by — 1)(K„, 2)(c 3— b3) 2b 3[(K„, + 2)(2K. + 1)c3— 2(K. — 1)2b3] approximately, where a is small compared with b and c. 15C. A right circular cylindrical cavity is made in an infinite mass of iron of permeability A. In this cavity, a wire runs parallel to the axis of the cylinder, carrying a steady current of strength I. Prove that the wire is attracted toward the nearest part of the surface with a force p„7 2(K„, — 1)/[27-d(K„, + 1)] per unit length, where d is the distance of the wire from its electrostatic image in the cylinder. 16C. If the magnetic field within a body of permeability A is uniform, show that any spherical portion can be removed and the cavity filled up with a concentric spherical nucleus of permeability piand a concentric shell of permeability p2 without affecting the external field, provided that A lies between pi and p2 and that the ratio of the volume of the nucleus to that of the shell is properly chosen. Prove also that the field inside the nucleus is uniform and that its intensity is greater or less than that outside according as /2 is greater or less than pi. 17C. A sphere of soft iron of radius a is placed in a field of uniform magnetic force parallel to the axis of z. Show that the lines of force external to the sphere lie on surfaces of revolution, the equation of which is of the form {
1+
2(K,, — 1)(a)3} (22 +y2) = constant K„, ± 2 r
r being the distance from the center of the sphere.
322
MAGNETIC INTERACTION OF CURRENTS
18C. A sphere of soft iron of permeability A is introduced into a field of force in which the scalar potential is a homogeneous polynomial of degree n in x, y, z. Show that the scalar potential inside the sphere is changed to its original value multiplied by (2n + 1)/(nK„, n 1). 19C. If a shell of radii a, b is introduced in place of the sphere in the last question, show that the force inside the cavity is altered in the ratio (2n + 1) 21f„,:(nK„,
1)(nK„,
n
n
K„,) — n(n
1)(K„, — 1) 2(a /b)2n+1
20C. An infinitely long hollow iron cylinder of permeability the cross section being concentric circles of radii a, b, is placed in a uniform field of magnetic force the direction of which is perpendicular to the generators of the cylinder. Show that the number of lines of induction through the space occupied by the cylinder is changed by inserting the cylinder in the field, in the ratio
b 2(K. + 1) 2— a2(K„, — 1) 2 :2K„,[0(lf„, + 1) — a2(K„, — 1)] 21. There is a zonal distribution of current flowing in an oblate spheroidal shell
r = ro such that current between t = 1 and t = to is given by 1k( to) = Evaluating the integral of 7.10 by the result in 5.274, show that the vector potential due to this current, when < ro, is
— 444(1+
eo)11 ,, Cn V,(iro)P1Cg)P;,(t) n(n + 1)
22. Using the last problem, show that the vector potential due to a loop of wire at i-o, to, carrying a current I, r < ro, is -44t4I(1
30)1(1
a)+In_2 (n + 1)-2(2n + 1)Q.1 (j3'0)P(E0)n(73")1)1(
Write, by inspection, the result when > 23. If the spheroid in 21 is prolate instead of oblate, show that, when n < no,
A = — +A(n20 — 1)4Z n(nC+ 1) Q;,(no)P;,(n)P.;,(t) 24. In prolate spheroidal harmonics, the coordinates of a loop are to, no. Show that the vector potential due to such a loop carrying a current I is —431A/[(7120 — 1)(1 — tNiln-2(n + 1)-2(2n + 1)P;i(to)(2;.(no)n(n)P.;,(t) 25. In the last problem, a prolate spheroid of permeability A whose surface is given by n = n1is inserted. Show that the additional vector potential due to its presence is given by
2n+1 AL,•/ n(o)Ono)P1(ni)P.(n) ——(K„-1)(7 2 -1)1(1—EDIz (21(n)P1(t) 2 n2(n+1) 2(2;,(ni)P,,,(no) — K4n(ni)P;,(771) where n > no.
PROBLEMS
323
26. Show that the vector potential due to a plane loop of wire of radius a at z = 0 and carrying a current I is, if p > a, mahr-Ifo I1(ka)Ki(kp) cos kz dk If p < a, interchange p and a under the integral. 27. Show that the vector potential due to a plane loop of wire of radius a, carrying a current I, is A4, = +.a/ f Ji(ka)../i(kp)e-kl'l dk
28. A loop of wire of radius a is coaxial with an infinite cylinder of radius b and permeability A and carries a current I. Show that that part of the vector potential outside the cylinder due to its presence is Avahr-i f 43(k)K,.(ka)Ki(kp) cos kz dk 0 (Km — 1)kbI0 (kb),I 2(kb) ~(k)= (K„, — 1)kbK 0(kb)7 2(kb) + 1
where
Write out the vector potential for the region inside. 29. Show that the vector potential, due to a section of length 2c of a circular cylinder of radius a around which circulates a uniformly distributed current I, is, when p > a A ct, =
AzaI f 1
rc o k
i(ka)KOP)
cos kz sin kc dk
30. The plane of a circular loop of wire of radius a, carrying a current I, is parallel to and at a distance b from the face of an infinite plate of material of permeability At and thickness t. Show that the vector potentials on the loop side of the slab, in the slab, and on the other side of the slab are given, respectively, by A l = itiva/f0 Ji(kp)Ji(ka)[e-41z1
A2 =
f i(kp),12(ka)C 0
— 1)(1 — e-2k0ek('-'20 ] dk 1)e-k= —
—
14 1),.-20+ti] I die
A3 = 2/2„Kmal f Ji(kp)Ji(ka)Ce-la dk 0
where p is measured from the axis of the loop, z from the plane of the loop, and C = [(K„, + 1) 2 — (K„, — 1) 2e-2kti-1
31. An infinite tube of external radius aiand infinite permeability is coaxial with an infinite tube of the same material and internal radius a3. In the z = 0 plane between al and a2 there is a current sheet. The current density ict, in the ranges al < p < b1 and b2 < p < a2 is —I[p In (a2/a1)]-2, and in the range b1 < p < b 2 it is I[p ln (b2 /b1 )]-1— I[p ln (a2/a1)]-2 , so that the total circulating current is zero and = 0 on both walls. Show that the vector potential when al < p < a2 and z > 0 is = ZCI.e-k.zRi(k.p) n=1
324
MAGNETIC INTERACTION OF CURRENTS
where R,„(k„p) = Yo(k„cti)J.(k.P) — Jo(k.cii)17.(k.p) and k„ is chosen so that Ro(kna2) = 0 (see HMF, Table 9.7), and C. =
Air2/1R0(k.b2)— Ro(k.bi)M(k.a2)12 41n (b2/bi) [Jo(k.a1)]2— Vo(k.a2)121
32. The potential A = —Boz/p of 7.05 (6) gives only a radial magnetic field Bo/p. Thus by superimposing the potentials In (a2/a1 —F 1 when z > 0 and zt4/z[p In (a2/a1)] -1when z < 0 on the result of the last problem, wipe out the circulating current except in the strip b1 < p < b2 of the z = 0 plane where i4, becomes I[p In (b2/b1)]-'. From 6.16 (2) this is the current distribution for a steady current
in a conducting ring. 33. The permeability of the walls of the region bounded by p = ai, p = a2, and z = 0 is infinite. At z = c there is a current sheet bounded by p = b2 and p = bi in which the current density ict, is I[p In (b2 /b1)]-1. Show from the last two problems that when ai < p < a2 the values of the vector potential, A:0 when z > c and when c > z > 0, are A'm
—Iz(p In — al
+ 2 ZC.e-kn. cosh k„cRi(k„p) n =1
= —Ic(p In — ai
+ 2 ZC„.e-kn' cosh k„zRi(k,,,p) n=1
34. An infinitely permeable rectangular conduit has its corners at x = ±a, y = 0, and y = b. Wires parallel to its axis carry currents I at c jd and —I at f jg. Show with the aid of 4.29 that the vector potential in the conduit is
+ 2[(xi — ci)2 — yncg + d: [(xi — ci)2 + In 47 [(xi — fi )2 + yn2 + 2[(xi — fi)2 — yng; g: sn mx dn my sn mx dn mx sn my dn my 1 — cn2my + k sn 2 mx sn2 my my + k sn2 sn2 my
A. =
Xi =
Similar formulas hold for xi = cl or fl and x = c or f and for yi = di or gl and y = d or g. The modulus k is found from a/b = K(k)/K[(1 — k 2)+] and ma = K. 35. An infinite wire carrying a current I is situated at x = a, y = b, between two faces of infinite permeability y = 0 and y = c. Show that the vector potential between the faces is U=
4r
iry)2 rb 71-(x — a) ir(x — a) . iry In [ (cos — -- cosh cos + sinh2 sin e — c
which is the real part of W=
ir(z — a)] [ rb In cos — — cosh
Note that the field is uniform and oppositely directed when x >> a and when x << a. 36. An infinite wire carrying a current I lies at zo to the right of a semi-infinite block of material of infinite permeability whose face is the plane x = 0 except for a circular ridge or groove of radius b whose center is at x = c. Show with the aid of
325
PROBLEMS problem 49, Chap. IV, that the vector potential is the real part of
In If(z) — Azo)l[f(z) +.nz:)1}
TV = where ir( b 2
f(z) = and cos a = c/b.
—
e
2)1 [ [z
c2)i]riti
j(b2
[z
0)/pr/a] }
j(b2
— c2)1]70« — [z— j(b2 _
a {[z
For a ridge 0 < a < 7 and for a groove 7r < a < 27.
37. Using the relation — In (z — a) = f
-i[e-k(z-a) — e-k]dk, where the real part 0k of z — a is positive, show that the transformations giving the field of a wire at x = a, carrying a current I between two semi-infinite blocks of material of permeability it whose faces are at x = 0 and x = b, where 0 < a < b, are, when 0 < x < b,
'1 Jcosh[k(z +a —b)] ± fie-kb cosh [k(z — a)] e-k = — ti= 1ln (z — a) — 20 J. dk ekb — 02e-kb 27r 1 -13 o k}l 'K,n( ica i„e ± (3e —k“ e k µv7 vi when b < x, 1 — 0 dk Wz = 7r(Km + 1) f k e _ 1 — 0 2e 216 e-k ' Km[ kr/. + &kW-2k) 1.4I when x < 0, W3 = 1 _ 0 dk 02e-2kb k e 1— r(Km + 1)
WI
—
f
where U is the vector potential, A-1V the scalar potential, W = U jV and we have = (Km— 1) (ICm 38. Show that the transformations giving the field of a wire at x = d, carrying a current I between two semi-infinite blocks of material of permeability A, whose faces are at x = —c and x = +c, where —c < d < +c, are
c
x,
x < —c,
4nc2
1.7,J( z — d W = — — In 2,7r 2c
— c < x < c,
W2 =
W3 =
+ 1) A„Km1 7r(K„, + 1)
In
In
E )3' ln {1 —
[z — (-1c/12}) )" 4n( n 1)c2
n =1
z—d 2c z—d 2c
0ln [1 + n =1
On In
—2c z — )"d]) < 2(n + 1)c
[ —1 ±
n- 1
2c z — (-1)ndl) 2(n + 1)c
where 13 = (Km— 1) (K m + 1)-', U is the vector potential, triV the scalar potential, and W = U jV. This can be done directly by images or by expanding the denominator in the previous problem, integrating and writing z c for z, 2c for b, and d for a — c. To make U finite near the origin, add the infinite constant
1)]-1 ZOn In [2(n + 1)c] n=1
326
MAGNETIC INTERACTION OF CURRENTS
39. Compare the first A.0 integral of 7.10 with 5.23 (15) and thus show that the vector potential of a circular wire loop of radius a carrying a current I is
A4, =
(cosh ui — cos u2)1Pl1(cosh ui)
where, from 4.13, 2ap coth u1= r' + a' and 2az cot uz = r' — a'. 40. By reference to 4.13, 7.06 (2), 5.233 (5), problem 117 of Chap. V, etc., add terms to the A4, of the last problem such that the magnetic field is tangent to the torus, cosh ?LI = c/b, obtained by rotation of the circle of radius b about a line in its plane at a distance c from its center. Show that the sum is i.d(cosh ul— cos u2r(c/b) Q4(c/b) 2
7L =0
(2 — 15,°,)(2;,_i(c/b)Pl_i(cosh ILI) cos nui — (2n — 1)(2n + 1)P.;,_i(c/b)
41. The current density i4, at p = a of period c in z is given by
C„ cos
-=
2nrz z
n=0
C. cos Nz
n=0
Show that the vector potentials, A'4, inside and A4, " outside, are = AL,,a ZC.Ii (Np)Ki(Na) cos Nz n=0
Ao "=µa
Cnii(Na)KI(Np) cos
Nz
n=0
42. A semi-infinite conducting sheet occupies the region 0 < 4, < ir of the z = 0 plane. A current I enters the edge at the origin and leaves at p = 00 . Show that at a point P, whose coordinates are p, z, the magnetic-induction components are A/
2
p Sin utr
it
z
B4, = — (1 + - tan-1
2
r)'
PB = B
gi
' = 21-2r
Z °
1p COS
tank-1
41
r
where r, the distance from the origin to P, is (p' + z')1. 43. A solenoidal current sheet has a density id, amperes per meter. A vector of length b from its center to any point on its edge makes an angle 13 with its axis. Show that the vector potential inside the current sheet is ) 2n+1
A4, = p.,,iob sin /3Z C,(7.,-
PL+1 (Cos 0)
n=0
where 0 is measured from the axis about its center and Co = —1 cot 0,
C. = [2n(2n + 1)(2n + 2)[-P1.(cos 13)
See 10.17 and problem 34, Chap. X. Outside the current sheet the same expression holds when r < b except that Co = 1(csc — cot (3). 44. A plane loop of wire of radius a lies parallel to the plane faces of a gap of width c between the halves of an infinite block of infinite permeability at a distance b from
REFERENCES
327
one face. If it carries a current I, show that the vector potential is co 2/ia/ Tort) nra nirb A, — — Z /1—K1— cos c
C
n=0
C
C
nirz
cos C
when p < a, and write down A, when p > a. Show that the force pulling the loop toward the z = 0 plane is raI 2 A( c
Zni,i— nra ,..,.d ra 2n2b is.1— sin n=I
c
c
c
46. Show that the magnetic induction in terms of the flux function N of 7.08 (3) and the differential equation satisfied by N where no current is present are
aN — ap
27rpl3z,
aN
— —27rpBp, az
a2N a 2N aN az'
apz
—
pap
0
Note that the sign difference in the last term rules out the solutions of 5.29 (1) with n = 0 but yields instead 2rpAo as given by 7.05 (2) and 7.05 (3). Show that the boundary conditions on N at the interface between Ai and az are N1
= N2,
aNiaN 1 aN 2 aN2) A2(91- + k— ) = P1(91— ±kaz Oz ap 01, References A.
MAGNETIC
and R. BECKER: "Classical Electricity and Magnetism," Hafner, 1950. Gives simple vector treatment. ATTwoon, S. S.: "Electric and Magnetic Fields," Wiley, 1941. Gives graphical method for plotting fields. BucHHoLz, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Gives theory and many solutions of special problems. CORSON, D. R., and P. LORRAIN : "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Well-illustrated exposition of basic ideas. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Gives theory with many examples and field maps. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XV, Berlin, 1927. GRAY, A.: "Absolute Measurements in Electricity and Magnetism," Vol. II, Macmillan, 1888. Extensive calculations of fields of circuits and shells. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extensive treatment without vectors. MASON, M., and W. WEAVER,: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Fine treatment of vector potential and boundary conditions. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment with good field maps and no vector analysis. OWEN, G. E.: "Electromagnetic Theory," Allyn and Bacon, 1963. Well-illustrated exposition of basic ideas with simple examples. ABRAHAM, M.,
328
MAGNETIC INTERACTION OF CURRENTS
K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Excellent treatment of vector potential in Chap. VIII. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives complete theory of the material of this chapter. WALKER, M.:. "Schwarz-Christoffel Transformation and Its Applications," Dover, 1964. Works out important applications in detail. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and literature references. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. PANOFSKY, W.
B. See references of Chap. V.
MATHEMATICAL
CHAPTER VIII ELECTROMAGNETIC INDUCTION 8.00. Faraday's Law of Induction.—A century ago Faraday and Henry discovered, independently, that when the magnetic flux N, defined by 7.29 (1), through a closed conducting circuit changes, a current is generated in the circuit. The direction of this induced current is such as to set up a magnetic flux opposing the change. Thus if the flux through a circuit in a certain direction is increasing, the induced current sets up a flux in the opposite direction, and if the flux is decreasing, this current sets up a flux in the same direction. The induced currents always seek to maintain the status quo of the magnetic field. The induced electromotance in volts equals the negative rate of the change of flux in webers per second so that
dN = — dt
(1)
with possible limitations mentioned later. It is immaterial what means we use to change the flux. We may move the source of the flux with respect to our circuit, we may change its strength, or we may move or change the shape of our circuit. Faraday proved (1) for a closed metallic circuit, but since we know that the tangential component of ye = E is the same on either side of the wire surface, (1) must apply equally well to the electromotance around a path just outside the wire. We may therefore assert that (1) holds for any path whatever and write, from 7.29 (1), for a two-sided surface S
fE • ds =
-dt fs B*
n dS
(2)
Applying Stokes's theorem [3.01 (1)] to the left side, we have
n • V x E dS = — f — • n dS sdt B Since this must hold for any arbitrary surface, we have
dB (3) dt in webers per square meter-second. If E is due entirely to electromagnetic induction, it is a special case of the E" of 6.01 and possesses no V xE=—
329
330
ELECTROMAGNETIC INDUCTION
§8.01
sources or sinks so that its divergence is zero. If A is obtained from 7.02 (5), its divergence is also zero. Let us now substitute V x A f or B in (3) and interchange d/dt and the curl, and we see that the curls of the vectors E and —dA/dt are equal. If two vectors have everywhere the same curl and divergence, they differ at most by a constant, so that
E = dA dt
(4)
which connects the electric intensity or electromotance per meter and the vector potential whose change induces it. Equation (1) seems adequate to deal with all cases involving a change of flux in rigid circuits and also with those cases in which the flux is changed by distorting the shape of the circuit, provided that, during the process, all elements of the circuit initially adjoining, remain adjoining. Experiments can be devised, using sliding contacts, to which the application of (1) is obscure or ambiguous. These cases can be treated by focusing our attention, not upon the area surrounded by the circuit, but upon the elements of the circuit themselves, and then applying (4). Another formulation of the law of induction is worked out in Chap. XIV, where it is shown, from the principle of special relativity, that when an observer moves relative to the fixed circuits producing a magnetic induction B he will observe, in general, an electromotance. If v is the velocity of the motion and E that part of the observed electric intensity which is due to the motion, we have, from 14.13 (12), if E is volts per meter, v meters per second, and B webers per square meter,
E = [v x B]
(5) 8.01. Mutual Energy of Two Circuits.—Let us now consider the work required to bring near each other two circuits carrying steady currents I and I'. Each circuit contains a source of electromotance that supplies or absorbs energy at the proper rate to maintain the currents constant at all times. In circuit 1 we have, from Ohm's law, 6.02 (1), and 8.00 (1),
IR = 6 —dN dt where 6 is the electromotance just described. The energy consumed in circuit 1 while bringing it up from infinity in time t is then, from 6.03 (2),
I 2R f tdt=4'6dt±I dN = f te dt ± NI where N is the flux through the first circuit due to the second circuit in the final position. Evidently, the last term represents the work done in the first circuit due to the magnetic field of the second. At the same
§8.01
331
MUTUAL ENERGY OF TWO CIRCUITS
time, work of amount +NT has been done in the second circuit to maintain its current constant. Thus, the total energy expended by the electromotance sources in the two circuits to maintain the currents constant while they are brought together is
± (IN I'N') Substituting the vector potential in 7.29 (1) and applying Stokes's theorem and 7.02 (5), we have N= fB' •ndS fVxA' • ndS= A'• ds
= xx ds • ds' 4r J' Y
and similarly
Elx peds• ds' LITYY r
(1)
so that the energy expended becomes
We
=
(IN ± I'N') = +1.1II' fr erds• ds'
(2)
2r YY
Faraday's law shows that if the two circuits attract each other the induced electromotance sets up an opposing field which tries to reverse the current in each circuit so that the sources of electromotance must supply energy to maintain I and I'. If they repel each other, the reverse is true. Let us now consider the mechanical work done against magnetic forces in moving the two circuits. From 7.18 (5), the force on circuit 1 in newtons is
f
F = I ds x B'
All'xfds x (ds' x r) r3 4r Y
All'
d '(ds • r) — r(ds • ds')
r3
47r
The first term vanishes when integrated around circuit 1 since the integrand r/r3can be written as the gradient of the scalar 1/r. The mechanical work done is then, for pure translation from r = co to r = r, TY„ =
f rF • dr =
erds • ds'r • dr _
47r
YYJ.
2-3
4a
3
f ds • ds'
(3)
Comparing (2) and (3), we see that half the energy supplied by the battery in the circuit is used in doing mechanical work. Since the only difference between the initial and final states is in the magnetic field surrounding the circuits, the remainder of the energy must be in the magnetic field. Thus, when two constant-current circuits are moved with respect to each other, the mechanical work done by the circuit and the energy in the magnetic field increase or decrease together and at the same rate. This explains the apparent anomaly mentioned in 7.18. It follows that if we know the energy WB in the magnetic fields of
332
ELECTROMAGNETIC INDUCTION
§8.02
two circuits we can get the mechanical force or torque trying to increase any coordinate 0 by taking the positive derivative of this energy with respect to 0, so that
ow',
F = + ao
(4)
8.02. Energy in a Magnetic Field.—We shall now find the energy required to establish the magnetic field of a single circuit, utilizing the results obtained in the last article and assuming all space filled with a homogeneous, isotropic medium of permeability pc. Let us build up the field step by step by bringing together infinitesimal current filaments. If the final current density is nowhere infinite, the denominator in 8.01 (3) causes no difficulty because finite current filaments are a finite distance apart. When 8.01 (3) is applied to a single circuit, we must include the factor to get the correct energy since our integration includes not only the work done in bringing filament a up to b, but also that in bringing b to a. When we write in current densities in place of currents by 6.00 (2), and make this correction, 8.01 (3) becomes TYB =
dv •
dv'
(1)
871-iviv
where r is the distance between the volume elements dv and dv', i and i' are the current densities in these elements, and the integration is performed twice throughout space, where, as in 8.01, we have assumed that is constant in the region of integration. Substituting for i from 7.01 (3) and for fr (e/r) dv' from 7.02 (4), we have Wa = 1 f VxB•A dv
(2)
v
From the formula for the divergence of a crossproduct, we have the relation A•VxB=B•[VxA]—V•[AxB]=B2 —V•[AxB] which gives 1 ,,1 f B2 dv — f V • [A x B] dv v 41.1 v where the integrals are over all space. By Gauss's theorem [3.00 (2)], we may transform the second integral into a surface integral over the sphere at infinity. This integral vanishes, since, from 7.02 (4), A goes to zero as 1/r whereas B = V x A goes to zero as 1/r2, and the surface area increases only by r2, so that fvV • [A x B] dv = fs[A x B] • n dS
0
§8.03
MUTUAL INDUCTANCE
333
Thus, the final expression for the energy becomes WB = r, f B2 dv GA V
(3)
This energy can be looked upon as residing in the space occupied by the magnetic field surrounding a current, the energy density being 1B2/p. This may be compared with 2.08 (2) which gives the energy in an electrostatic field. We see that the magnetic field, just as well as the electrostatic field, can be visualized in terms of a system of stresses. 8.03. Mutual Inductance.—The coefficient of mutual inductance M12 between two circuits is defined as the flux N12 through circuit 1 produced by unit current in circuit 2. The mks mutual inductance unit is the henry. This is expressed mathematically by writing (1),(2) B2 • n d81 = 4'A2 • dsi Js, where A2 is the entire vector potential due to unit current in circuit 2. From 8.01 (1), we may write M12 =
M12 =
XXdS1 • ds2
r
4rYJ
M 21
(3)
The flux through circuit 1 due to a current /2 in circuit 2 is, from (1), N12
=
(4)
M12/2
From 8.00 (1), the electromotance in circuit 1 produced by a fluctuating current in 2 is el
d/2 = 1V1 12-
(5)
dt
From 8.01 (3), the mutual energy of the two circuits is given by W12
=
(6)
M121112
From 8.02 (3), the total energy of the two circuits is = fv(Bi+ B2) - (B1 + B2) dv = (f B? dv + 2f B1 • B2 dv 211 v
dv) B2 The first term is the energy required to establish I l alone and the last term that for /2 alone so that the remaining term gives the energy used in bringing the two into interaction. Hence, from (6), we have 1 M12/112 = -
f
B1 • B2 dv (7) i,r From 8.01 (4), the force or torque tending to increase any coordinate of
334
ELECTROMAGNETIC INDUCTION
position 0 of one circuit relative to the other P' =
is
M12 2a
§8.05
ae
(8)
8.04. Boundary Conditions on A.—In 7.21 appear the conditions that derivatives of the quasi-vector potential A satisfy at the boundary between two regions of different permeability. In 8.03 (2), we have, for the first time, a relation involving an integral of A, and we must therefore consider what boundary conditions to impose on A itself in order that this equation will be valid where such discontinuities occur. Let us take a small rectangular circuit, the long sides of which are very close together but on opposite sides of the boundary between regions of different permeability. As these sides approach each other, the area of the rectangle approaches zero, so that the flux through this circuit, due to a current /2 flowing in a second circuit, must become zero. But by 8.03 (2) and (4), this flux is N = .9642 • ds. The integral is taken around the rectangle, the ends of which are vanishingly small, so that the whole contribution must come from the sides, which are of equal length L and which are short enough so that the vector A is constant along a side. From 7.21 (10) and (11), the tangential component of A has the same direction on both sides of the boundary, so orienting our rectangle along ut and ut we obtain for the corresponding components of A, respectively, — A.,t • ds,,t = =0 Thus, we have the two equations for the tangential components of A. A's,g = A'Zt
(1)
8.05. Mutual Inductance of Simple Circuits.—As a simple example of the computation of a mutual inductance coefficient, consider the two closely wound coils A, B, shown in Fig. 8.05. Coil B contains n turns and is wound on a ring of permeability A. Coil A has m turns. All the flux goes through the ring so that, from 7.29 (4) and 8.03 (4), we have (a2 by] (1) M12 = imm[a — If a >> b, we may factor a out of the radical and expand the remainder by Pc 753 or Dw 5.3, keeping only the square terms. This then becomes FIG. 8.05. i.annb2 (2) M12 — 2a If A = rb2 and ni = n/(27ra) is the number of turns per unit length,
MUTUAL INDUCTANCE OF CIRCULAR LOOPS
§8.06
335
then this gives the mutual inductance between an infinite solenoid wound on a cylinder of area A and permeability /.4 and a coil of m turns encircling it to be M12 = µn1m A (3) This could have been written down directly from 7.15 (7).
8.06. Mutual Inductance of Circular Loops.—The coefficient of mutual inductance for two coaxial loops of wire can be written down by
Ca) FIG. 8.06.
the use of 8.03 (2). The vector potential of A is entirely in the 95-direction and so has the same value for all elements of loop B and is parallel to each element so that, from 8.03 (2), MBA
=
4AA
• dSB = 2rb AA
:
P=b
Thus, from 7.10 (4), we have M12
= 2µk-1(ab)i[(1 — 1k2)K — E]
where, from 7.10 (3), k 2= 4ab[(a
(1)
b) 2 ± c2]-1
(2)
Numerical values of K and E may be found in Pc 121 or Dw 1040 and 1041. Another expression for M12 may be written down in the same way 0 = 0, giving, by 5.157, if b2 c from 7.13 (2), letting a =r 2 < a2, 1—> .o
n
n +1
1.3.5 •
n
n • 2 • 4.6 • •• (n
b2
1)( a2
c2y"
Pl(cos 13)
(3)
odd
n+1 c2)] 2 for c2)/a2]4n. [(b2 We may compute the mutual inductance when the two loops are in any position, provided that their axes intersect, by means of 8.03 (2), 7.13 (3), and 5.24 (5). Thus if, in Fig. 8.06b, b > a, we have, taking the origin at the point of intersection and measuring 0 from the a-loop axis, 21a (4) [NI) sin 0 d0 diri) M12 = a2f o o When b2
c2 > a 2, write [a2/(b2
ELECTROMAGNETIC INDUCTION
336
§8.07
If 0' is the angle measured from the b-loop axis, then substitution of 5.24 (5) for Pa(cos 0') in 7.13 (3), where 7 = 0, a = r, and I = 1, give
[B,lb =
sin 13
a
2a .a.7..j 0 P;,(cos /3) n=1
n (2 — 62,) (91— in) Irn, (cos -y)P7n(cos 0)
cos my6
m).
m=0
When this is inserted in (4), the cb-integration eliminates all but the m = 0 terms. By 5.154 (5) the integration with respect to 0 gives ens a sin' aP'n (cos a) — sin aPli(cos a) i) n(n n(n 1) Pn(cos 0)d(cos 0) = so that the mutual inductance is M12 = rya sin a sin )(3
1 1)(b n(n ± 1)
«)P71(cos 0)P.(cos 7) (5)
n=1
8.07. Variable Mutual Inductance.—Variable mutual inductors are frequently built in the form shown in b Fig. 8.07. Each coil consists of a single . . layer of wire wound on a bobbin with a spherical surface. One coil rotates, .\ I /1/1/// / .'- . relative to the other, about a common \ / axis. Let the angle between the axes / IN I I of the coils be 7, let the coils subtend — --r i 1e angles 2A and 2B at the center, and let I\ / i /;y \\ \ the radii of the two surfaces be a and b. \\ / , \ \/ // Consider each coil as being made up of <\ plane loops of wire, the planes being 24 . uniformly spaced with a number n per unit length. The number of turns n' 2B --A FIG. 8.07. per unit angle in each of the coils will then be na = naa sin a and 74, = nab sin )3. The mutual inductance between elements of the two coils will then be, from 8.06 (5), Op i (Ub)P n(COS 7) da t (A n L.172,(n + 1) bn' n The total mutual inductance will be obtained by integrating on both coils; thus, a = + A r p- ir+B Mab = c1.114-ab JP = — B
dMa = ira2b1.4nonb sin' a sin' g
as;a — A
• SELF-INDUCTANCE
§8.08
337
Writing u for cos a or cos 0, each integral takes the form, when 5.154 (5) is used, (1 u2)dPn(u) du = n (n + 1) f[P„ i(u) 2n + 1 du -
—
Pa1(u)] +. du
(1)
Integrating by 5.154 (3) and simplifying by repeated application of 5.154 (1), this becomes sinA
f n(A) = (n — 1)(n + 2) 1[(n + 1) — (n— 1)u21Pn(u) — 2uP,i(u)
(2) sinA
When n is even, the value at both limits is the same by 5.157 so these values drop out and only odd values of n appear. Special treatment is required when n = 1. In this case, from (1), we have u2)] -sinA 2 sin A(3 — sin' A) f1(A) = [41 — (3) =
O
--
sinA
Similar expressions hold for B so that 1—* co
\n
1
Mab = ri.4a2bnanb
n(n + l
)
ab)
fn (A) f „(B)Pn (eos
y)
(4)
n odd
From (2) and (4), the coefficients depending on the fixed dimensions of the coil can be computed and the formula then takes the form MnPn(cos y)
Mab =
(5)
n odd
where y is the angle between the axes. 8.08. Self-inductance.—From 8.02 (3), we see that the magnetic energy density of a single circuit, carrying a current /1, is proportional to B2, and since B is proportional to II, we must have 1 Wg = -„
v
B2 dv =
1,T2
(1)
We define the constant of proportionality L11to be the self-inductance of the circuit. Another useful formula for self-inductance may be obtained by substituting 7.01 (3) in 8.02 (2), giving Wg =
• A dV = L11I 2
(1.1)
where the integration must be taken throughout the region in which the current flows. In the mks system the inductance unit is the henry. If the circuit consisted of infinitely thin wire carrying a finite current, its self-inductance would be infinite. This is evident because B is inversely proportional to the distance from the wire when we are close to it. Thus, for a finite
338
ELECTROMAGNETIC INDUCTION
§8.09
length of wire, the integral in (1) is logarithmically infinite since dv may be written r dr do. This means that in computing self-inductance we must consider the actual dimensions of the wire and take the current density i nowhere infinite. We may therefore treat the wire as a bundle of current filaments of infinitesimal cross-sectional area dS each carrying a current i dS. When the current is established in such a wire, the magnetic flux about each current filament cuts across neighboring filaments and, by Faraday's law [8.00 (1)], sets up in each an electromotance opposing the current, of amount e =-- —dN/dt where dN / dt is the flux cutting the circuit in one second. To maintain the current /1, the source of power must do, in each second, an amount of work equal to dN e11 = /1
(2)
dt
in addition to any work against resistance. This energy, stored in the magnetic field each second, equals dW /dt so that we have, from (1) and (2), r dNI
1
dt
TC11-1 = L11.1 1-
dt
Canceling out the /1and integrating from no current to final current, we have N11 = Lai (3) Thus we may also define the self-inductance of a circuit as the change in the flux linking the circuit when the current changes by one unit. The electromotance in the circuit caused by the changing current is
d = T I1 (4) dt dt 8.09. Computation of Self-inductance. Thin Wire.—From what has been said, it appears natural, when computing self-inductance, to treat separately the region lying outside the wire and that lying inside. The contribution of the latter is frequently negligible, and when it is not, but when the radius of the wire is small compared with other dimensions of the circuit, we may assume that the field outside is the same as if the current were concentrated at the axis of the wire and that inside the field is identical with that inside a long straight wire of the same size, carrying the same current. The lines of force near the surface are approximately circles about the axis of the wire so that all the flux outside it links the axial filament with any line drawn parallel to it on the surface of the wire. To get this part of the self-inductance we need, therefore, to find the mutual inductance between two parallel curvilinear circuits spaced a distance apart equal to the radius of the wire. Inside the wire, the magnetomotance around any circle of radius r
el =
-( §8.10
-
SELF-INDUCTANCE OF CIRCULAR LOOP
339
about the axis depends, from 7.28 (3), only on the current inside the circle. Since, from symmetry, B depends only on r, we have, from 7.14,
B = Mu = 2irr
(1)
2ira2
where I is the total current and a is the radius of the wire. The energy inside the wire is, from 8.08 (1),
„12L
1 W;--
1LIII2
/.112L
where L is the length of the wire. The internal self-inductance, per unit length, becomes then ii
=
(2)
8r
where A' is the permeability of the wire. 8.10. Self-inductance of Circular Loop.—Using the method of the last article and the result of a previous example, we shall now compute the self-inductance of a loop of radius b of wire of permeability and radius a. The length of the wire is 2rb so that that part of the self-inductance due to the interior of the wire is, from 8.09 (2),
Lu =
(1)
As shown in 8.09, we can get the rest of the self-inductance by applying the result of 8.06 (1) to two circuits, one coinciding with the axis of the wire and one with its inner edge. In this case, from 8.06 (2), setting c = 0, and taking a << b, we have a2 ( a\2 4b(b — a) 1 — k2 = 1 — (b b — a)2 (2b — a)2 \TO = k'2 So that k is approximately unity. When k 1, the E of 8.06 (1) becomes
E = r2 1 — sin2 0)i dO = r2 cos 0 de = 1 To get K, let cb =
— 0, and split the integral into two parts
r K=
r 0.
dB
f 0(1 — k2 sin2 0)1
ir
do
Jo (1 — k2 cos2 0)1
±
2 dct. f50(1 — k2 cos2 0)1
where (1 — k2) << 00 << 1. Now put sin 4) = 4) in the first integral where 4)0 is small, and then put k = 1 in both integrals, and we have r5 K=i JO (k'2 + 02) k + J Oosin 0 "
340
§8.12
ELECTROMAGNETIC INDUCTION
Integrate by Pc 126a and 291 or by Dw 200.01 and 432.10; then put cko = tan 00, and neglect k'2 compared with cbg, and we have
K =in f (Po + (cti V2)/11 k'
8b 4 1 = In — = In — In tan — 00 = In k' a 2 k' 490
Substituting in 8.06 (1), and adding the internal inductance term (1), we get 8b Li' = b[1./(ln — — a
+_ 1 41,1 j
(2)
where te is the permeability inside the wire and /.4 is that outside. 8.11. Self-inductance of Solenoid.—If we treat a solenoid as an infinitely thin cylindrical current sheet of uniform strength, all lines of flow lying in planes normal to its 21- L axis, we may compute the self1 inductance rigorously by using 8.06 (1) to find the mutual inductance between elements of the shell and integrating over the shell. I The result is complicated, so that dx where only rough values are necesFIG. 8.11. sary and the solenoid is long compared with its diameter, the following approximation is useful. We treat the field inside the solenoid as uniform across any section taken normal to the axis. To compute the self-inductance by 8.08 (3), we must know the flux linking any element of the shell and then integrate over the shell. From 7.15 (4), the flux through the n dx turns in dx is, per unit current,
dN1 =
1— x Ri _ x)2
1
+X
[ (1 + x) ± a2]1
dx
Integrating by Pc 132 or Dw 201.01 from —1 to +1, we have az)i — a] a2)i — a] = 71,a2n2{(L2 L11 = rlia2n2[(4/2
(1)
8.12. Self-inductance of Bifilar Lead.—We shall now calculate rigorously, by the method of 8.08 (1.1), the self-inductance per unit length of a bifilar lead in which the current in one direction is uniformly distributed in a wire of radius a and the return current in a wire of radius c, parallel to the first, the axes of the wires being a distance b apart. The permeability of the wire and the region surrounding them will be taken as unity. Clearly, the vector potential and the current density have z components only. If I is the total current, then the current densities are it
I a = 7-2
—1
= 2
(1)
341
SELF-INDUCTANCE OF BIFILAR LEAD
§8.12
To obtain the vector potential A' inside the wire due to its own current, we write 7.02 (2) in cylindrical coordinates with the aid of 3.05 (2) and
FIG. 8.12.
obtain, since A' is a function of p only, a (aA' P
1.11
ap P4973 =
=
ra2
Or 7C
2
Integrating from 0 to pitwice, we have 3111
hip1
a Pi =
P1
a
ra2
4ira2
Outside the wire, putting i = 0 in (2), integrating twice from the outer boundary of the wire to pi, and observing that the lower limit in each integration is obtained by putting pl =a in (3), we have 0,47
pi— -r 41
P1 1.11 v 21'1' = — — — — 1 /1 4r 2r a
A
2r
(4)
Similarly, = C2 +
47re2
= C2
tlf In P2 ± pi + Tr
(5)
To make (4) and (5) agree with 7.09 (1), at infinity we must take 27(C1 ± C2) + p,I ln a = 0 L1112 =
r
d81 fo i2(A'2
S2 1 0
(6) A1') d82
(7)
We shall calculate the first integral only and evaluate the second from symmetry. Thus we have, using PC 523 or Dw 868.4 and canceling out Cl, a C21 (b2 2bp1cos (1))] (L11/ 2)1 — '12 + 1 + In pi dpi dcfr 472a2 f 0 Jo a a2 a 3 a a b2 = ALI2 — f dpi f dpi+ —_f pi dpi) 2,71-a2 0a a2 o .o —
=
2
(1 --F 2 1n
(8)
342
§8.14
ELECTROMAGNETIC INDUCTION
Similarly, for the second wire, (L1112)2
rtr
1 + 21n 6) -
(9)
„/-2(
Adding (8) and (9), we obtain 2 ac
(10)
b Li,. = P (1 + 4 ln -) a 47
(11)
= 1/ (1 + 2 In
47
If the wires are alike, a = c, so that
8.13. Energy of n Circuits.—If we have circuits carrying currents II, 12, . . . , Inand the induction at any point due to each circuit alone is B1, B2, . . . , lin, then the total resultant induction is the vector sum of these and the total energy in the field from 8.02 (3) is n
n
1 f(B,s • W = 1 i B2 dv = —
B5 dv
21.1 v
2ii
i
( n
Zajv
(1)
i=i) (J=1) n n
./3,?
±
Bi • B) dv
j=1 J=1 i=1. oii in the double summation, both Bi • B5 and 135 -Bi occur so that, from 8.08 (1) and 8.03 (7), this may be written n n n ZZiji/j) W = 4( ELin i =1 i=lj=1 oij = -(L1/1 + 2M121112 + ing + • • • )
+
(2)
As we saw in 8.01 (4), the force or torque tending to increase any coordinate of position 0 of the ith circuit is obtained by taking the positive derivative of the field energy with respect to this coordinate; thus, n
3211
F = I (1.Li 2
ae
J-1
)
(3)
iii
8.14. Stresses in a Magnetic Field.—We saw in 8.02 (3) and 2.08 (2) that the energy densities dW /dv in magnetic and electric fields are given by dW) _ B2 (di _D2 and (1) dv 1„, 21.4 dv e 2e (
From 7.01 (1), we know that V • B = 0 is true everywhere, and from 3.02 (1) that V • D = 0 is true in the absence of electric charges. It
§8.15
ENERGY OF A PERMEABLE BODY
343
follows that a system of stresses that will give equilibrium in an electric field will do the same in a magnetic field if we replace D by B and e by /./.. Thus we see, from 1.14, that we may visualize magnetic forces as transmitted by a tension T along the lines of force and a pressure P at right angles to them. Both T and P, per unit area, are given numerically by = IPI =
(2)
If the permeability is a function of the density r, we proceed exactly as in 2.09 except that we deal with a section of permeable material in a magnetic circuit instead of with a dielectric slab in a capacitor. This leads to a conclusion, analogous to 2.09 (2), that there is required, in addition to the above stresses, a hydrostatic pressure of amount
B2 ap
Ph =
(3)
2A2
To find the force on the plane boundary between two materials of different permeabilities iu! and 12" in a magnetic field, we add up the normal components of the stresses given by (2) and (3) and in the same way in which we obtained 1.17 (7) and 2.09 (3) we get = 1 _ A/ Bp + baz\ (4) i / 2 ar
21
1.1!
)
,,
1.112
This is the normal stress directed from pi' to A". The normal and tangential components of the inductions B' and B" in the two mediums are indicated by the subscripts n and t, respectively. For a boundary between the medium /.1. and a vacuum, we set /.4" = til = AvKm, and = 0 and obtain
F„ =
B'2 (K,,, 21.12,K1
1)
T
OK„,1
a,
j
(K„, — 1) 2B? 21.4K1
(5)
where B'2 = B'n2
(6)
The formulas derived in this article hold fairly well for ordinary substances but break down for the ferromagnetic materials discussed in Chap. IX. The latter give many complicated effects, such as changes of shape without change of volume, which cannot all be included in any simple formula. 8.15. Energy of a Permeable Body in a Magnetostatic Field.—The last article shows that the force acting on a body of permeabilityµ placed in a medium of permeability 1.4 in the presence of a magnetostatic field of induction B produced by currents fixed in magnitude and position is the same numerically as that calculated in 3.13 (6) for a dielectric body E. = j.i„ placed in an electric field of displacement D = B produced by
344
ELECTROMAGNETIC INDUCTION
§8.15
fixed electric charges. It appears in 8.01 (4) that, in the magnetic case, the force or torque acting in the direction 0 is obtained from the magnetic field energy by taking its positive derivative. Thus F = -F
aw
(1)
ae
A comparison with 3.13 (5) shows that if the magnetic induction produced in a region of permeability pi, by a fixed set of currents is B inside a volume v before it is occupied by a body of permeability /I and Bi, after it is occupied then the energy required to place this body in position is given by the following integral over its volume ( 1 _ )B dv (2) \ It is assumed that the body is isotropic, thatµ is independent of Bi, that only induced magnetism is present, and thatµ and B, are the final values in dv. This becomes, in terms of magnetization, by 7.20 (9), wB
TVB =
M • B dv
(3)
Problems Problems marked C in the following list are taken from the Cambridge examination questions as reprinted by Jeans, with permission of the Cambridge University Press. 1. Show that the coefficient of mutual inductance between a long straight wire and b) In (1 + a/b) — a], a coplanar equilateral triangular loop of wire is [1.4„/(3127-)][(a where a is the altitude of the triangle and b is the distance from the straight wire to the side of the triangle parallel to and nearest it. 2. A circular loop of wire of radius a lies with its plane parallel to, and at a distance d from, the plane face of a semi-infinite block of material of relative permeability K,,,. Show that the increase in the self-inductance of the loop due to the presence of the block is, if c is written for a (a2 d 2)-1, AA
K, — [ 1 (2 + 1 c
2 c)K (c) — -E(c)
3. A sphere of radius b and very large permeability is concentric with a wire loop of radius a. Show that the additional self-inductance due to its presence is 3 • 5 • • • (2n + 1)I(2n 212(b) 4n-I-2 = 2mb 3K 2.4.6 • • • (2n ± 2) 2n ± 1 a a' a'
[1 •
µ„n0
n=0
4. Show that the coefficient of mutual inductance between two coplanar concentric loops of wire whose radii are a and b is
p•i a z [ 1 • 3 - 5 - • • -
n=0
(2n + 1)12n + 2(b)z('+') = 2maK b 2 • 4 • 6 • • - (2n + 2) 2n + 1 a a
—
E b a
5. A circuit consists of a thin conducting cylindrical shell of radius a and a return wire of radius b inside. The axis of shell and wire are parallel but not coincident. Show that the self-inductance per unit length is 1(4/7r)[1 + 41n (alb)].
PROBLEMS
345
6. A plane circuit of any form is pressed into the face of a semi-infinite block of permeability A so that it is coplanar with the face. Neglecting the field inside the wire, show that the block increases the self-inductance by a factor 2/2/(A 7C. A long straight current intersects at right angles a diameter of a circular current, and the plane of the circle makes an acute angle a with the plane through this diameter and the straight current. Show that the coefficient of mutual induction is kh[c sec a — (c2 sec' a — a2)I] or A,,c tan (;r — la), according as the straight current passes without or within the circle, a being the radius of the circle and c the distance of the straight current from its center. 8. A cylindrical wire of permeability Ai is concentric with a thick layer of insulation of permeability Ai. This wire is placed in a liquid of permeability A2 normal to a field of induction B. Show that the transverse force per unit length on the surface of the insulation is [(142 Ai)]./B and that it is 2ai.//3/(21 + As) on the metallic IU1)/(A2 surface giving a resultant total force of IB. 9. A small sphere of radius a and permeability A is placed near a circuit which, when carrying unit current, would produce a magnetic induction B at the point where the center of the sphere is placed. Show that the presence of the sphere increases the self-inductance of the circuit by approximately 47a3B2(K„, — 1)/[A„(K„. + 2)]. 10. A cylindrical shell of permeability g and internal and external radii a and b surrounds two parallel wires, carrying currents in opposite directions, which are coplanar with, parallel to, and each at a distance c from its axis. Show that the additional selfinductance per unit length of the circuit due to the shell is, if A = Ar„K„,, 212„(la
—
(a4,,+2 —
1) n -0
b4n-I-2)
(c/a) 411+2
(2n + 1)[(K„, — 1) 2a4n+2
1)2b4".+21
11C (Modified). A circular loop of radius a is concentric with a spherical shell of permeabilityµ and radii b and c. Show that the additional self-inductance of the loop due to the presence of the shell is approximately, if A = 1.4K„„
Aes-a4(Kn, — 1)(K. + 2)(0 — b3)1 2b3RK„, + 2) (2K„, 1)c2 — (K,,, — 1)22b2] where a is very small compared with b and c. 12. The coordinates of a loop of wire of radius a in the prolate spheroidal system are A prolate spheroid of permeability A is inserted so that its surface E = Ei and n = is = no. Show that the additional self-inductance of the loop is, if A =
C2
— 1) .k 2n + 1 [n(ti )]2[(4,(771)]2P;,(no)P„(no) n2(n + 1)2i21(no)P.(no) — K.Q.(no)P1(no) n=1
13. The coordinates of two loops of wire of radii al and a2 in the prolate spheroidal system are n1, Zi, and 772, El. A prolate spheroid of permeability A is inserted in the loops so that its surface is n = no. Show that the additional mutual inductance due to the spheroid is, if A = 03
irA0a ia2 (KM C2
1)
1 P,;(t1)(2,1(ni)n(E2)C772)n(fio)Pn(no) a2(a + 1)2 (2, 1,(no)P.(no) — IC4.(no)n(no) 1 2n
14. Show that the additional mutual inductance between two parallel coaxial loops of wire of radii a and c with a distance s between centers, due to the insertion, coaxially, of an infinite cylinder of radius b and permeability A, is, if is =.
ELECTROMAGNETIC INDUCTION
346
2,a,acf 43(k)K 1 (ka)Ki(kc) cos ks dk where '(k) is given in problem 28, Chap. VII. 15. Show that the additional self-inductance of a loop of wire of radius a, due to the insertion, coaxially, of an infinite cylinder of radius b and permeability AL, is 2A,a2f cP(k)[KI(ka)]2 dk 0 where 1.(k) is given in problem 28, Chap. VII. 16. Two circuits are formed by that part of the cylinder p = a1lying between z = +ci and z = —c1and that part of the cylinder p = az lying between z = +c2 and z = —c2, where az > a1. Assuming that the currents will distribute themselves uniformly on the bands, show that the mutual inductance between them is 2,ualaz Ic-2/1(kai)K1(kaz) sin kci sin kc2 dk cie2 fo 17. Using problem 30 of Chap. VII, show that if the plane of a circular loop of wire of radius a is parallel to, and at a distance b from, the face of an infinite plate of material of permeability and thickness t the self-inductance of the loop is increased by the amount AL11
= 13 1110 - (1 — 1 2)E 132("—')M. [
71-1 2A„a[( k where M„ = 1 — -=)K„ — Ed, — k. 2
A
— +
a2
Ay
1.4 , k 2—
a2 +(nt b)2
and K.
and E. are the complete elliptic integrals of the first and second kinds of modulus k„. 18. An infinite plate of thickness t and permeability At is inserted between the loops of Fig. 8.06a, with its faces parallel to them. Show that the mutual inductance between the loops is now given by 00
M = 2/.1.„(ab)1(1 — #2 ) El (1 — —E 2 n n=0 k' K. and E„ are complete elliptic integrals of modulus k., where —
4ab (a + b)2 +(c + 2nt)2
and
13
-
A
-
µ+
19. Show that the self-inductance of a solenoid of diameter d, length c, N turns, and small pitch is igN2d{ csc a[(tan2 a — 1)E + K] — tan2ce) where d/c = tan a and E and K are complete elliptic integrals of modulus sin a. 20. A wire loop of radius a is coaxial with a solenoid of radius b and n turns per unit length. If the greatest distance from a point on the loop to the near end of the solenoid is c and to the far end it is d, show from 7.10 (4) without further integration that the force between loop and solenoid when they carry currents I and I' is
2Arn//' (ab)4 fkri[(1 —
—
— 41(1 — ikZ)K2 — Ezil
where the modulus of the complete elliptic integrals is ki = 2(ab)1/c and kz = 2 (ab)i/d. 21. Two coaxial solenoids of diameter d and small pitch, with n and m turns per unit length, are placed so that the distance between near ends is b and between far ends is c. If one is of length a, show that their mutual inductance is
PROBLEMS
347
4 iihnmd3Z ( —1). sec an[(1 — tan2 ot„)E(k„) + tan2a„K(kn)]
n=1 where tan al = c/d, tan a2 = (c — a)/d, tan a3 = b/d, tan at = (a + b)/d, and k„ = cos a,,. 22. Show that, if the currents in problem 21 are I and I', the force between the coils is 4
(-1)" sin ct. sect a„[E(k„) — K(k„)] n=i 23. Show by 5.35 (4) that the mutual inductance between two loops of radii a and b with parallel axes a distance c apart and planes a distance d apart is, if c > a -I- b, 2ktd2nmIr
I i(ka)l i(kb)Ko(kc) cos kd dk
21.1abf
0
24. Show that, if a > b + c, the mutual inductance in the last problem is .. 2gabf€i
K i(ka)Ii(kb).10(kc) cos kd dk
25. Show with the aid of 5.298 (7) that the results of problems 23 and 24 may be written irpab f'J i(ka)J1(kb)J0(kc)e-ka dk 0
26. Show from problem 23 that the mutual inductance between two solenoids of radii a and b, length 2A and 2B, and m and n turns per unit length, with parallel axes a distance c apart and central loop planes a distance d apart is, when c > a + b, 2
AB Aabmnfo k-2I 1 (ka)I i(kb)Ko(kc) sin kA sin kB cos kd dk 27. If coil b of the last problem lies inside coil a, show from problem 24 that . 2 M - — Aabmnf k-21Ci(ka)l i(kb)I o(kc) sin kA sin kB cos kd dk AB o 28. From problem 25 write the mutual inductances of problems 26 and 27 in the form 1 ' -2 AB irliabmnf k Ji(ka)..1,(kb)4(kc) sinh kA sinh kB e- I'd dk 0
29. The most general position of a loop of radius a relative to one of radius b may be stated in terms of the shortest distance c between their axes, the distances A and B of the centers of each from c, and the angle 0 between their axes measured from each center toward c. Show that the mutual inductance is /lab 47, 0
22-f 22.
0 [D2 ± a2 +
(cos 95 COS 9/,' cos # ± sin 0. sin 0') dcl, de b2
+ 2ab(cos 4, cos 0' + sin 0 sin 0' cos 0) — 2f(0, 0')]i
where D = (c2 + A2 + B2— 2AB cos i3)i is the distance between their centers and
AO, 4)') = a(c cos 0 + B sin 0 sin 0) + b(c cos 4.' + A sin 0' sin 0) 30. A bifilar lead of two parallel wires with a spacing c is placed symmetrically in the gap of width a between two parallel sheets of infinite permeability with its plane normal to them. Show that the additional self-inductance per unit length due to the presence of the sheets is GOO In ([2a tan (4irc/a)]/(7rc)}.
348
ELECTROMAGNETIC INDUCTION
31. Obtain a series solution of problem 25 by first establishing, from Watson, page 148, and HMF 15.4.14 and 8.2.5, the formula
J,
Abk) =
( -1Na2— `‘.../
m=0
(in
1)!(m
+ 2)!
a' b2 k a 2 — b2 2
Substitute this in the integral of problem 25 and integrate the result by IT I (9), page 182. Thus show that the mutual inductance is 2rAtab m=0
a2 (-1)TM(2m 1)!! (a2 — b2).+Ip1 ( ÷1 a2 (2m 4)!!
b2) b2 P2„,+2 (cos a)
where the angle between r, the line joining loop centers, and either axis is a. 32. If a = b in the preceding problem, show that the mutual inductance becomes Z(2m + 2)!( — 1)m(2m + ip(a)2'n+3 P2m+2(cos r m!(m + 1)!(2m + 4)!!
2rkta
m=0
33. Apply the method of problem 31 to problem 28 and thus show that the mutual inductance is 3 r
8AB
0:1
( -1)m÷'(2m — 1)!! (a2 — b 2)m+1
AcibmnZ a=0 m=0
(2m + 2)(2m
4)!!
r2;'+'
(a2
b2
Pm ' ÷, a2
b2)P2„,(00s a,)
where if po d + A + B, pi = d + A — B, p2 = d — A — B, and p3 = d — A + B, then r: = c2 p:,c4 is the angle between r, and either coil axis, and d > A + B. 34. If a = b in the preceding problem, show that the mutual inductance becomes 3
Z th 7//a 3 nina=0 m=0
(2m + 2)!(-1)m÷s(2m — 1)!!
(2m + 2)m!(m + 1)!(2m + 4)!!
(al 2m-1-1
r,
P2,” (COS
as)
References All the references of Chap. VII contain some of the material of this chapter. In addition there is one very important reference. GROVER, F. W.: "Inductance Calculations," Dover, 1962. This contains the most
extensive tables and instructions for making all kinds of inductance calculations.
CHAPTER IX MAGNETISM 9.00. Paramagnetism and Diamagnetism.—In most substances, with some notable exceptions to be considered in Art. 9.04, the magnetic permeability depends very little on the field strength so that taking it constant, as we have done, introduces no appreciable error. Unlike the relative capacitivity, the relative permeability may be either greater or less than 1. Substances in which it is greater are called paramagnetic, and those in which it is less diamagnetic. Let us consider the forces acting on such bodies when placed in the field of a fixed circuit carrying a constant current. From 8.01 (4) and 8.03 (4) or 8.08 (3), we see that, if any infinitesimal displacement or rotation of the circuit will increase the flux through it, then there is a force or torque trying to produce this motion. If there are any bodies in the field of this circuit whose displacement or rotation will increase the flux through it then, by Newton's law of reaction, there will be corresponding forces or torques acting on these bodies. In 7.29, we saw that the magnetic induction or flux density and the permeability are related in magnetic circuits in exactly the same way as the current density and the electric conductivity in electric circuits. Thus we may state theorems 3 and 4 of 6.05 in a form applicable to magnetic circuits. If the permeability of any element in the magnetic field of an electric current is increased or decreased, the reluctance of the magnetic circuit is decreased or increased, respectively. As we have seen, these are forces acting to increase the flux and hence to decrease the reluctance. Thus, in an inhomogeneous field, there is a tendency for bodies that are more paramagnetic or less diamagnetic than the ambient medium to move toward the more intense parts of the field, and vice versa. If a paramagnetic or diamagnetic body of elongated shape is placed in a uniform field, there is a torque tending to set its axis parallel to the field. This can be seen in the case of spheroids by substituting /./. for e, pi, for e„, and H or B/A for E in problem 84, Chap. V, and observing that the torque formula contains H1 — H2 which has the factor A in it so that the torque depends on (14 — ,u,,)2 and has the same sign for A > Ae, and < µv. Quantum mechanics gives a theoretical basis for the empirical fact that the permeability of diamagnetic bodies is usually independent of temperature. For weakly paramagnetic substances, the permeability is often 349
350
MAGNETISM
§9.02
independent of temperature. In strongly paramagnetic, but not ferromagnetic, substances, the permeability usually depends on temperature, obeying the equation =µa+
Ae,C
(1)
T+ 0
where C and 0 are constants and T is the absolute temperature. This empirical relation is called Curie's law. A theoretical basis for it is found in quantum mechanics. For gases, 0 is usually zero. 9.01. Magnetic Susceptibility.—It is often convenient to use a new quantity magnetic susceptibility defined, in an isotropic medium, in terms of the quantities already treated in 7.20 and 7.28 by the equations
IcH = M =
iho
\ — 1.)
(1)
Thus susceptibility and permeability are related by the equation K=
A;1(kt
—
Fp)
(2)
= Km — 1
When a substance is placed in a magnetic field, its energy is decreased if it is paramagnetic and increased if it is diamagnetic. From 8.02 (3) and (2), the change is given by „ H2 B2 (4 µ)H2 Aw r-v — = = 2
2µ
2
2 KH
(3)
We notice that, for paramagnetic bodies, K is positive; for diamagnetic bodies, it is negative. Curie's law [9.00 (1)] may be expressed more simply in terms of the susceptibility by the empirical equation K
_ C T +
(4)
A theoretical basis for this equation has also been worked out. 9.02. Magnetic Properties of Crystals.—Many substances, especially crystals, possess different magnetic properties in different directions. It is even possible with some materials, such as graphite, to prepare specimens that are paramagnetic in one direction and diamagnetic in another. In such cases, it is found however that for any given orientation the magnetic induction B is proportional to the field intensity H and makes a constant angle a with it. This relation is similar therefore to that connecting the electric displacement D and the electric field intensity E in a crystal and can be formulated in the same way as in 1.19 by writing a set of equations analogous to 1.19 (1) and (2). We find, therefore, from 1.19 (3) that the components of B and H are con-
§9.03
CRYSTALLINE SPHERE IN UNIFORM MAGNETIC FIELD
351
netted by the relations Bz = By =
11211-1vA31tlz A22Hy + 1232Hz
12Hx
B, = 1213H.
j. 2 3H7/
(1)
+ 1133Hz
where /212
=
A21,
/113
=
A31,
1.1 23 = 1.432
(2)
Thus B and H are now connected by a quantity having nine components of which six are different. The permeability, formerly a simple ratio, has become a symmetrical tensor. By a suitable orientation of axes, (1) may be written in a form analogous to 1.19 (7), giving Bz =
By = A2Hy,
Bz = 1.13Hz
My = K2H y,
My = K3Hy
(3) When (3) holds, the coordinate axes are said to lie along the magnetic axes of the crystal. By means of 9.01 (1), we see that the corresponding equation connecting the magnetic susceptibility and intensity of magnetization are Mx = "Ix,
(4) where Arsci. = µl — mu, etc. By rotation of coordinates, we can get a set of equations, connecting M and H in a crystal, analogous to (1) and involving a tensor magnetic susceptibility. 9.03. Crystalline Sphere in Uniform Magnetic Field.—As an illustration of the manipulation of the formulas of the last article, let us find the torque acting on a crystalline sphere when placed in a uniform magnetic field of induction B. Let the angle between B and the x-axis in 9.02 (3) be a, and let 122 = 123. The boundary conditions will evidently be satisfied if we superimpose the case of a field of induction B cos a in the x-direction, acting on an isotropic sphere of permeability and the case of a field B sin a in the y-direction, acting on an isotropic sphere of permeability /12. If we take B in the xy-plane (Fig. 9.03), then problem 18C, Chap. VII, and 7.10 (2) show that the field outside the s2here is exactly the same as if we superimposed on the original field a field due to a magnetic dipole of moment Msin the x-direction and that of one of strength muin the y-direction where — µv
47a3B cos a
and
_
— At,47ra3B
sin a
112 ;Iv The field due to these two dipoles equals that due to a single dipole of -
Zuz
352
MAGNETISM
§9.04
strength M making an angle 13 with the axis, where tan = my = (A2 — Av)(At ± 2/4) tana (1) Ar. (Ai — µv)(µ2 + 2/4) = M = of! + 471-Ba3[(ili — A.) 2(12 + 214) 2 (1 — sin a) + (µ2 — 14) 2(mi + 2;4) 2 sin ali — iL.)(A2 + 2/4) (2) The angle that Al makes with the field is, from Dw 405.02 or Pc 593, tan (fi — a) =
3(A2 — ai)u„ tan
— µv) (µ2
2/4)
(g2 —
a + 214) tang a
(3)
The torque acting is then, from 7.18,
T = 31B sin 63 — a)
(4)
This torque tends to rotate the sphere so that the axis of Al lies in the direction of the field if AI > A2 and so that it lies at right angles to the field if µ l< 112. 9.04. Ferromagnetism.—There is an important group of materials whose permeability varies with the magnetizing field, depends on the previous treatment of the specimen, and is much larger than that of
ordinary substances. These materials are said to be ferromagnetic and include iron, cobalt, nickel, and the Heusler alloys and, at low temperatures, some of the rare earth metals. Let us construct, of ferromagnetic material, a simple magnetic circuit such as the anchor ring considered in 7.29 in which the magnetizing force, or the magnetomotance per meter, can be computed easily. Starting with the zero values of both the magnetic induction B and the magnetizing force H, we find that as H is increased B also increases but the ratio of B to H which is the permeability A first increases and then decreases. Typical curves of B and A as a function of H, for a ferromagnetic material, are shown in Fig. 9.04. In many ferromagnetic substances, if we measure the magnetization
§9.05
HYSTERESIS. PERMANENT MAGNETISM
353
with sufficiently sensitive instruments, we find the steeper part of the magnetization curve to have a step structure. This is called the Barkhausen effect. It suggests that a large region, involving many similarly oriented atomic magnets, changes direction as a unit. Experiments on ferromagnetic single crystals indicate that certain preferred directions are easiest for magnetization. Since most ferromagnetic substances with which we work are polycrystalline, it is believed that the major part of the magnetization is due to the magnetic units in a microcrystal which are partially aligned with the magnetizing field growing at the expense of those which are not. When all units are so oriented, the Barkhausen effect ceases, and further increases in the magnetizing field force the magnetization of these units continuously from their preferred directions toward the direction of the field. This accounts for the absence of step structure in the flatter part of the curve where the magnetization approaches saturation. A theoretical basis for this picture is found in the electron configuration of the iron atom. In the calculation of magnetic fields, inductances, magnetic forces, eddy currents, and such quantities, we have hitherto assumed that the permeability of a given substance is fixed. We see from these curves that if there is much variation in the magnetizing force in a region occupied by ferromagnetic material this assumption is not justified. In such cases, we take a mean value of i.i, which is approximately correct for this substance, for the range of H used. Our knowledge of the magnetization curve of the particular specimen involved usually does not justify higher precision. In case, however, all ferromagnetic material used is very homogeneous and has been carefully annealed, we may be able to determine from this solution and the magnetization curve the value of i.t which should be used in different regions. It is in geometrically simple cases only that this knowledge can be used to find a more rigorous analytical solution. 9.05. Hysteresis. Permanent Magnetism.—Suppose that, having increased H to I/1, thereby increasing B to B1, as described in 9.04, we now decrease H to —H1. We find that the induction B does not retrace any portion of the path shown in Fig. 9.04 but decreases less rapidly, following the upper curve in Fig. 9.05a, and, for a normal specimen, reaches the value —B1 at —H1. If we now increase H again to +H1, we follow the lower curve in Fig. 9.05 and reach the original curve at B1, H1. This lag in the induction is called hysteresis, and the closed curve in Fig. 9.05a is called a hysteresis loop. It is clear that with a given specimen we get a different hysteresis loop when we start with a different point on the magnetization curve, but if we vary H continuously from H1 to — H1 and back we repeatedly retrace the same loop.
354
§9.06
MAGNETISM
The hysteresis loops for different ferromagnetic materials may differ greatly. For magnetically "soft" specimens, the area inside the loop is very small, giving the inner curve in Fig. 9.05b, whereas in magnetically "hard" substances it is very great as shown in curve D. In the latter case, we still have a large residual induction or retentivity 13,, when the magnetizing force is zero and it actually takes a large reverse field H„ called the coercive force, to destroy the induction in the specimen. Here, we observe for the first time the presence of a magnetic field when electric currents are apparently absent. Magnetic phenomena were
(6)
(a)
FIG. 9.05.
originally discovered associated with such "permanent" magnets, and the whole theory of magnetism was developed on the basis of experiments made with them. 9.06. The Nature of Permanent Magnetism.—Hitherto, we have considered the energy in a magnetic field as essentially kinetic, it being associated with the motion of electric charges. As the magnetic fields produced by permanent magnets appear in every respect to be identical with those produced by electric currents, it is natural to seek a similar origin for them. The nature of permanent magnetism is revealed by a group of phenomena known as the gyromagnetic effects. Since no electricity is entering or leaving a permanent magnet, any motion of electricity therein must be circulatory and this circulation or spin must be about axes that are oriented, on the average, in a definite direction to produce a definite external field. If, as we have seen in 1.04, the electrical carriers possess mechanical inertia, then when circulating in closed paths or spinning, they possess angular momentum and therefore are subject to gyroscopic forces. Such forces were predicted by Maxwell but could not be detected with the experimental technique of his day.
UNIFORM MAGNETIZATION
§9.07
355
Two effects immediately suggest themselves. The first of these is magnetization by rotation. A well-known fact in mechanics is that when the supporting system of a gyroscope is rotated and its axis a is free to turn only in the plane common to it and the axis b of rotation of the system, then a tends to set itself parallel to b. Thus, if an unmagnetized body possessed circulating or spinning electricity with axes oriented at random, a rotation of this body should produce an alignment of these axes with the axis of rotation, and the body should become magnetized. Such effects have been detected and measured by Barnett in ferromagnetic substances. The second effect is the converse of the first, rotation by magnetization. From the law of conservation of angular momentum, if the random axes of rotation are aligned by a magnetic field, then the body as a whole must rotate in the reverse direction to keep the resultant angular momentum zero. This effect was first measured by Einstein and De Haas and has since been done with greater precision by other experimenters. Both effects show that there is a rotation of negative electricity in ferromagnetic bodies and that the average magnetic moment of the individual gyroscopes is slightly greater than that of a spinning electron. The excess is supposed to be due to an "orbital" motion of the electrons. Thus, we see that the magnetic fields of permanent magnets are not different from those already studied. 9.07. Uniform Magnetization. Equivalent Current Shell.—In permanent magnets, the magnetization M is, by definition, independent of applied fields. It can be defined, as in 7.20, as the magnetic moment per unit volume of the permanent circulating currents or spins. From 7.20 (1) and(4), the vector potential due to M is given by the equation
,f
pa Mxr gyfVxM iMxn dv = dv dS (1) 47 v r3 s r 47r v r where the volume integrals are throughout the volume of the magnet and the surface integral over its surface. When M is the same in magnitude and direction in every element of a given region, we say this region is uniformly magnetized. For such a magnet, the second volume integral in (1) is zero. Let us examine the remaining surface integral. Suppose that M is in the x-direction so that M = iM. Let 0 be the angle between i and n, and let ds and dsi be orthogonal vectors lying in the magnet surface, ds being normal to i and n. Then, we have M x n dS = M sin 0dsi ds = M dx ds so that
Am —
A
Prof Cm dx ds _ _47, r
(2)
356
§9.08
MAGNETISM
We see that this is identical in form with 7.02 (5) so that the whole magnet can be replaced by a current shell, coinciding with its surface, around which the currents flow in planes normal to the direction of magnetization x. The current density, in terms of x, is uniform and equals the intensity of magnetization M. Such a shell is called an equivalent current shell and is frequently very useful in treating permanent magnets. 9.08. Magnetized Sphere and Cylinder. Magnetic Poles. For a uniformly magnetized sphere, the current flowing in the equivalent current shell between 0 and 0 — dO is —
i dO M dx = Ma sin 0 dO where i is the angular current density. The stream function is then given by e. (1) o x dO = Ma(1 — cos 0) = M a[Po(u) — P1(u)]
=f
where u = cos 0. This is identical in form with 7.12 (1) ; and setting n = 0 and 1 in 7.12 (5) and (3), we see that the vector potential outside the sphere is u,,Ma3 (2) A0 sin 0 3r2 -
X
Comparing with 7.10 (1), we see that the field is identical with that of a small current loop of moment M
-
47rMa3 3
(3)
We saw in 7.00 that the magnitude and configuration of the magnetic field at a distance from such a loop are identical with that of the electric field about an electric dipole having the same moment. For a right circular cylinder magnetized parallel to its axis, the equivalent current shell is seen to be a solenoid with zero pitch, the current per unit length being M. From 7.15 (4), the magnetic induction at any point P on the axis of such a solenoid is given by Bz = tu,M.(cos 132 — cos 0)
(4)
where 132 and 01are the angles subtended by radii of the two ends at P. The assumption that B can be found from the normal magnetization M. on a magnet surface from the inverse-square law /2„M(47r-r 2)-i also gives (4). Take Mzuniform over the flat ends of this magnet, and consider a ring of radius p = c tan 0 and width dp = c sect 0 dO which subtends an angle 0 at an axial point a distance c from its center so r = c sec 0 and .1=L at c is
§9.10 SPHERICAL PERMANENT MAGNET IN UNIFORM FIELD
seen to be A,M.1 9227rp dp cos Bs = r2 47r
Jo
1.1,M
2
fs2 s sin in 0 dO =
2
357
(cos 02 — 1) (5)
Subtraction of a similar term for the opposite end gives (4). For a magnet of length 1, this is just the electric displacement given on the x-axis with electric charges qi = itora2M at x = and qz = —1.4.2ara2M Thus, experiments with magnetic needles lead naturally at x = to the hypothesis of magnetic charges or poles. The region from which the magnetic lines of force emerge is called the north pole, and the region in which they reenter the magnet is called the south pole. As we have seen, the actual existence of magnetic charges is impossible if the divergence of the magnetic induction is to be zero everywhere. 9.09. Boundary Conditions on Permanent Magnets.—As the name implies, we assume that the intensity of magnetization of a "permanent" magnet is unaffected by the fields in which it is placed. This means that if we replace the magnet with an equivalent current shell, the region inside is assumed to have a permeability /.4,. If the magnet is immersed in a region of permeability A and an external field is superimposed, it will be distorted at the surface of the magnet to meet the boundary conditions between a medium of permeability /Ivand one of permeability /./. These boundary conditions have already been stated in 7.21, 7.22, and 7.28. 9.10. Spherical Permanent Magnet in Uniform Field.—As an example, let us compute the torque on a uniformly magnetized sphere immersed in a medium of permeabilityµ in which the magnetic induction B, before the introduction of the sphere, was uniform. Let a be the angle between M and B. Obviously, the current sheet can produce no torque on itself; so we need compute only the induction due to the external field. In spherical coordinates, the vector potential of B is
Ao Orsin 0
(1)
In 7.10 (1), we have a form of vector potential in which 0 enters in the same way but which vanishes at infinity. This is therefore the logical form to choose for the additional term due to the presence of the sphere with permeability 14. Thus, outside the sphere, we have
B A0 = —(r — r2 sin 0 2
(2)
Since Aimust be finite at the origin and must include 0 in the same way, it must have the form = Dr sin 0 (3) To determine C and D, we apply the boundary conditions of 7.21 (6) and
358
MAGNETISM
§9.11
(7) at r = a. These give the equations Bil
= -
and
201
+
= Da
(4)
Solving for C and substituting in (2) give
2L
=— 1 24r
ar ) sin 0
(5)
When r = a, this has the same form as (1) and so, from (3), represents a uniform field in the directio:i of B. The torque on the current ring element lying between 01 and 01 +d01, where 01is the colatitude angle measured from the axis of magnetization, equals the product of the current in the ring by its area, by the magnetic induction, and by sin a, giving dT = ira' sin' 01B[1
2 A — Av i do1sin a 21A + /iv
Substituting for i dOifrom 9.08 and integrating by Dw 854.1 or Pc 483 give an-Ava'MB sin a T 471-1.4a3MBsin a sin' 01d01— T= (6) 2µ + If the field in which the sphere was placed were not homogeneous, we would have a force as well as a torque. This would consist of two parts, viz., the force on a sphere of permeability 14 and the force on the current shell in the field at the surface of this sphere. These forces could be calculated, using 10.06 (13) and 7.18.
9.11. Lifting Power of Horseshoe Magnet.— For lifting ferromagnetic objects, one frequently uses a permanent magnet in the form of a horseshoe. In treating this, we shall assume for simplicity that the "legs" are eliminated, leaving only half a ring as shown in Fig. 9.11. If the legs were present, we could use an approximate method FIG. 9.11 like that in 6.16. We shall assume that the half ring was magnetized by placing it in contact with another similar piece of steel to form a complete ring, winding uniformly with wire and passing a current. To a first approximation, the magnetization M will be the same function of r, the distance from the center of the ring, as the magnetomotance was, so that we have, from 7.29 (3), M=— r
(1)
§9.11
LIFTING POWER OF HORSESHOE MAGNET
359
To find the equivalent current shell, we take a thin layer of thickness dr and radius r in which M may be considered constant. Then, if i is the angular current density, we have for the current flowing between 0 and 0 — do, as in 9.08, idO=Mdx=( C )rdo=CdO
(2)
When the layer current shells are superimposed, adjacent current layers cancel and we have a constant angular current density C around any section of the ring. Now let us suppose this magnet is placed in perfect contact with a block of infinite permeability. The lines of magnetic induction inside the current shell are then semicircles since they enter and leave the block normally by 7.21 (2). The magnetomotance around the circuit is, by 7.28 (3),
St= fove
= 71-C
(3)
The reluctance of the layer dr is, as in 7.29, length — irr Ilya dr !Iv X area since the permeability inside our current shell is 1.1.. The flux in this layer is dR —
Ba dr = dR =A,Ca dr The tension across a section of this layer is, from 8.14 (2), dT =
aB2 u C2 dr — • "a dr 21.1. 2r2
The pull on the block at each contact is T—
aC2 f 14 2
b
r2
ilvaC2(c— b) 2bc
(4)
An upper limit for C can be computed roughly from the magnetization curve of the steel. If the total number of turns in the original magnetizing winding was n and the current used was im then the magnetomotance per meter was nim/(irr) ampere-turns per meter from 7.28 (3). Suppose that the wide loop in Fig. 9.05b corresponds to our specimen and that HD = ni„,Arb). Then, for all values of r greater than b, the magnetizing force will be smaller than HD and the hysteresis loops will lie inside of that shown. We have assumed that the ratio of B. to BD is the same for all these loops (let us call it P) and that the ratio of BD to HD is also a constant (let us call it 12'). Then, from (1), C B. ni.A'P M = r ,
360
MAGNETISM
§9.111
so that we have max
ni,./213
(5)
This is an upper limit for C since any mechanical shock, heat-treatment, or "softness" of the steel will diminish C, especially after the ring has been separated into two magnets. The approximation made in assuming 1.z = 00 for the block is not serious, for in most soft iron specimens A/A2, > 500. 9.111. Field of Cylindrical Magnet.— In order to produce strong steady fields over small areas, permanent magnets are frequently made in the form of rings, with small gaps, which are magnetized, as described in the last article, by winding uniformly with wire and passing a current. To simplify the calculation of the field in this case, let us suppose that the ring is so wide in the direction parallel to its axis that we may take it as a long cylinder of internal and external radii a and b which has been magnetized, before the gap is made, by winding wire uniformly over the J P walls parallel to its axis. As explained in the last article, the resultant M is inversely proportional to r. After magnetizing, a gap with radial walls is created, without disturbing M, by removing part of the metal, leaving only the portions lying be44110 tween el = +a and Oi = — a. If b is FIG. 9.111a. sufficiently small compared with the length of the cylinder, the problem far from the ends becomes a twodimensional one. In this case, the lines of magnetic induction coincide with the lines of constant vector potential as proved in 7.25 and may be most easily calculated by the method of circular harmonics considered in 4.01, 4.02, 4.03, and 7.17. As shown in the last article, the equivalent current shell consists of that portion of the original magnetizing winding not removed with the gap and carries the current I d0 between 0 and 0 + do. Since all elements of the winding are parallel to the axis and the ends are so distant that they do not affect the vector potential A, the latter is also everywhere parallel to the axis and the current elements may be considered infinite in length. Considering the contribution of the inner elements negative and that of the outer ones positive, we find that dA at the point P at r, 0 due to the elements at +01 and —01is given from 7.09 (1), by
dA
R1 — In R2 — In Rs + In R4) des
FIELD OF CYLINDRICAL MAGNET
§9.111
361
FIG. 9.111b.—Lines of magnetic induction and constant vector potential in a permanent magnet formed by magnetizing a long thick cylindrical shell and removing a sector to form the air gap. Calculated from Eqs. (1), (2), and (3) of Art. 9.111 in which C = 1, a =7a/8,a= 1, and b = 2.
Substituting for the logarithms from 4.02 (1) and writing C for Av/hr give
dA = C n =1
ni(an
; b) cos nO1cos nO dOi
Integrating this from 01 = 0 to 01 = a gives 03
r > b,
A = cn12(an r—n b)
sin na cos nO
(1)
n=1 Similarly,
b > r > a,
A = C4celn —b
n2L
•na cos n0} (2) On sin -6
362
MAGNETISM
§9.12
and a > r,
A=
C{a ln a b- M1-2[ ( a)"—
sin na cos nO}
(3)
n=1
The lines of induction when C = 1, a = 77/8, a = 1, and b = 2, calculated from (1), (2), and (3), are accurately drawn in Fig. 9.111b. 9.12. Magnetic Needles.—The most familiar form of permanent magnet is probably the magnetic needle. This consists of a thin cylinder of steel more or less uniformly magnetized lengthwise. As we saw in 9.08 (5), the magnetic induction of such a magnet approximates in form that of the electric displacement about two charges porat2M and — pora2M placed at a distance 1 apart, where a is the radius, 1 the length, and M the intensity of magnetization of the magnet. This analogy fails if the magnet is placed in a medium of permeability /./ Ai,. However, in the case of the magnetic needle whose length-to-diameter ratio is very great, another approximation can be used. Its equivalent current shell, if filled with a medium A, and immersed in a medium behaves by 7.24 and 1.18 like a long slender cavity so that the axial flux through it from an external source is the product of gi,H or i.L,B/p. by ra2 cos 0 where 0 is the angle between its axis and B. From 9.07, the current in a length dl of the equivalent current shell that links this flux is M dl where M is the magnetization. From 8.03 (6) and 7.28 (5), the energy of the shell is W = gv/Ifira2rH cos 0 dl = gyMira2 (e2 —
(1)
This has the form of 1.071 (1) so that a thin needle uniformly magnetized lengthwise acts exactly like a pair of equal and opposite electric charges of size tz,Mra2placed in an electric field of potential V = St, where Si is the magnetomotance or scalar magnetic potential. In the absence of external magnetic field sources, the flux threading the slender current shell is practically independent of the permeability of the medium surrounding the needle for the reluctance of the magnetic circuit involved lies almost entirely inside the shell where the medium does not penetrate and the permeability is p,,. Thus the magnetic flux from a sufficiently thin magnetic needle, like the electric flux from an electric dipole, is independent of the medium surrounding it. The scalar magnetic potential or magnetomotance of such a needle is, therefore, inversely proportional to the permeability of the medium around it. These forces may be computed by the formulas of 1.071 as if the magnets possessed a mutual potential energy. Thus if r is the radius vector from ke, to it4 which make angles 01and 02 with it and a with each other and, if sp is the angle between the planes intersecting in r which contain Arl and h4 then,
PROBLEMS
§9.12
363
from (1), 1.071 (4), and 1.071 (5), this potential energy is w — - (cosa — 3 cos 01 cos 02) 4./1-Ar3 . = 471 .Lr3(sm 0 sin 02cos ¢— 2 cos 01 cos 02)
(2) (3)
All forces and torques exerted by one dipole on the other can be found from these formulas by the usual method of differentiation with respect to the coordinate involved. In particular, the force of repulsion is
F= —
aW = dr
2(sin 01sin 02cos IP — 2 cos 01 cos 02)
47rAr3
(4)
The apparent potential energy of a needle in a field of induction B, from which the forces are found by differentiation, is, from 1.071 (2), —W= be • H= Ise •
B14-1
(5)
The vector potential at a distance r from a magnetic needle is given in spherical polar coordinates by 7.10 to be, when r >> 1. -
M' sin 0 47r2
(6)
which, like its total flux, is independent of the medium's permeability. It should be noted that the classical magnetic moment AT' used in this article differs dimensionally from the loop magnetic moment defined in 7.00 and 7.10 (2). The formulas of this article are rigorous only for infinitely thin magnets. Forces on or energies of permanent magnets of large cross section in mediums where cannot be found by integrating these formulas. Problems Problems in the following group marked C are taken, by permission of the Cambridge University Press, from the Cambridge examinations as reprinted by Jeans. The classical magnetic moment is used throughout these problems. 1. An iron pipe runs east and west in a certain region. To locate the pipe, measurements of the dip at 5-ft intervals, starting from a fixed point, in a northerly direction are made. At distances x ft from the point, the dip deviates from the normal 60° dip by the following amounts: x 185 210 215 220 225 230 235 200 205 190 195 AO —4.5° —5.0° —5.5° —7.5° —6.0° +0.0° +14.5° +0.0° —1.5° —1.2° —1.0° Find x for the center of the pipe and its distance below the surface. 2. A steel pendulum bob of radius a is uniformly magnetized in a vertical direction with an intensity I. The strength of the vertical component of the earth's field at a certain location is V and the acceleration of gravity g. Find the ratio of the period to that when unmagnetized if it swings in an east and west plane, its mass being m and the distance from center of mass to pivot, 1. 3. A sphere of permeability 1000 and radius 10 cm is placed 1 m northeast (mag-
364
MAGNETISM
netic) of a compass needle. Show that, within 1 per cent, the deviation of the needle from the magnetic meridian, neglecting the magnet's own image forces, is 5'. 4. Find the magnetic field outside and in the cavity of a thick spherical shell uniformly magnetized in the x-direction with an intensity M. 5C. Two small magnets float horizontally on the surface of water, one along the direction of the straight line joining their centers, and the other at right angles to it. Prove that the action of each magnet on the other reduces to a single force at right angles to the straight line joining the centers and meeting that line at one-third of its length from the longitudinal magnet. 6C. A small magnet ACB, free to turn about its center C, is acted on by a small fixed magnet PQ. Prove that in equilibrium the axis ACB lies in the plane PQC and that tan 0 = — a tan 0', where 0, 0' are the angles that the two magnets make with the line joining them. 7C. Three small magnets having their centers at the angular points of an equilateral triangle ABC, and being free to move about their centers, can rest in equilibrium with the magnet at A parallel to BC and those at B and C, respectively, at right angles to AB and AC. Prove that the magnetic moments are in the ratios 31:4: 4. 8C. The axis of a small magnet makes an angle 0 with the normal to a plane. Prove that the line from the magnet to the point in the plane where the number of lines of force crossing it per unit area is a maximum makes an angle 0 with the axis of the magnet, such that 2 tan 8 = 3 tan 2(0 — 0). 9C. Two small magnets lie in the same plane and make angles 0, 0' with the line joining their centers. Show that the line of act pri of the resultant force between them divides the line of centers in the ratio tan 0' + 2 tan 0: tan B + 2 tan 0'. 10C. Two small magnets having their centers at distances r apart make angles 8, 0' with the line joining them and an angle e with each other. Show that the longitudinal force on the first magnet is 3mm' (47rAtr4)--'(5 cos2 0 cos 0' — cos 0' — 2 cos e cos 0). Show that the couple about the line r which the magnets exert on one another is mm'(471-Ar4)--id sin e, where d is the shortest distance between their axes produced. 11C. Two magnetic needles of moments M, M' are soldered together so that their directions include an angle a. Show that when they are suspended so as to swing freely in a uniform horizontal magnetic field, their directions will make angles 0, 0' with the lines of force given by sin 0
sin 0' M
(M2 M'2 + 2MM' cos a)-i sin a
12C. Prove that if there are two magnetic molecules, of moments M and M', with their centers fixed at A and B, where AB = r, and one of the molecules swings freely, whereas the other is acted on by a given couple, so that when the system is in equilibrium this molecule makes an angle 0 with AB, the moment of the couple is 3MM' sin 20 87µr3(3 cos2 B + 1)1
where there is no external field. 13C. Two small equal magnets have their centers fixed and can turn about them in a magnetic field of uniform intensity H, whose direction is perpendicular to the line r joining the centers. Show that the position in which the magnets both point in the direction of the lines of force of the uniform field is stable only if H > 3M (4n-pr3)-i. 14C. Two magnetic particles of equal moment are fixed with their axes parallel to the axis of z, and in the same direction, and with their centers at the points + a, 0, 0. Show that if another magnetic molecule is free to turn about its center, which is fixed
365
PROBLEMS
at the point (0, y, z), its axis will rest in the plane x = 0 and will make with the axis of z the angle tan-1[3yz/(2z2 — a2 — y 2 . Examine which of the two positions of equilibrium is stable. 15C. Prove that there are four positions in which a given bar magnet may be placed so as to destroy the earth's control of a compass needle so that the needle can point indifferently in any direction (and experiences no translational force). If the bar is short compared with its distance from the needle, show that one pair of these positions is about 1/ times more distant than the other pair. 16C. Three small magnets, each of magnetic moment M, are fixed at the angular points of an equilateral triangle ABC, so that their north poles lie in the directions AC, AB, BC, respectively. Another small magnet, moment M', is placed at the center of the triangle and is free to move about its center. Prove that the period of a small oscillation is the same as that of a pendulum of length 4774t/b3g/(351)1mM', where b is the length of a side of the triangle and I the moment of inertia of the movable magnet about its center. 17C. Three magnetic particles of equal moments are placed at the corners of an equilateral triangle and can turn about those points so as to point in any direction in the plane of the triangle. Prove that there are four and only four positions of equilibrium such that the angles, measured in the same sense of rotation, between the axes of the magnets and the bisectors of the corresponding angles of the triangle are equal. Also prove that the two symmetrical positions are unstable. 18C. Four small equal magnets are placed at the corners of a square and oscillate under the actions they exert on each other. Prove that the times of vibration of the principal oscillations are )]
[ 47rAnik2 d3 27r m23(2 + /2-2)
{47-Amk2 ds 27r-
2119 —
y,
27r
[ 47rumk2 d32 (2)i
3M2
J
where M is the magnetic moment, and mkt the moment of inertia, of a magnet, and d is a side of the square. 19C. A system of magnets lies entirely in one plane, and it is found that when the axis of a small needle travels round a contour in the plane that contains no magnetic poles, the needle turns completely round. Prove that the contour contains at least one equilibrium point. 20C. Prove that the scalar potential of a body uniformly magnetized with intensity I is, at any external point, the same as that due to a complex magnetic shell coinciding with the surface of the body and of strength Ix, where x is a coordinate measured parallel to the direction of magnetization. 21C. A sphere of hard steel is magnetized uniformly in a constant direction, and a magnetic particle is held at an external point with the axis of the particle parallel to the direction of magnetization of the sphere. Find the couples acting on the sphere and on the particle. 22C. A spherical magnetic shell of radius a is normally magnetized so that its strength at any point is Si, where Si is a spherical surface harmonic of positive order i. Show that the scalar potential at a distance r from the center is [
+ 1) is,0' 2i + 1 a
.*4-1 [2i4-Fri113'0
when r < a when r > a
MAGNETISM
366
23C. If the earth was a uniformly magnetized sphere, show that the tangent of the dip at any point would be equal to twice the tangent of the magnetic latitude. 24C. Prove that if the horizontal component, in the direction of the meridian, of the earth's magnetic force was known all over its surface, all the other elements of its magnetic force might be theoretically deduced. 26C. From the principle that the line integral of the magnetic force round any circuit ordinarily vanishes, show that the horizontal components of the magnetic force at any station may be deduced approximately from the known values for three other stations which lie around it. Show that these six known elements are not independent but must satisfy one equation of condition. 26C. If the earth was a sphere and its magnetism due to two small straight bar magnets of the same strength situated at the poles, with their axes in the same direction along the earth's axis, prove that the dip S in latitude X would be given by 8 cot (5 +
= cot iX — 6 tan 4X — 3 tan3iX
27C. Assuming that the earth is a sphere of radius a and that the magnetic potential 7 is represented by ft = Si(r/a) S2(r/a)2 Si(a/r)2 S2(a/r)3, show that is completely determined by observations of horizontal intensity, declination, and dip at four stations and of dip at four more. 28C. Assuming that in the expansion of the earth's magnetic potential the fifth and higher harmonics may be neglected, show that observations of the resultant magnetic force at eight points are sufficient to determine the potential everywhere. 29C. Assuming that the earth's magnetism is entirely due to internal causes and that in latitude X the northerly component of the horizontal force is A cos X B cos 3X, prove that in this latitude the vertical component reckoned downward is 4B
6B
2(A + --) sin X — — sin3X 3 5
30C. A magnetic particle of moment M lies at a distance a in front of an infinite block of soft iron bounded by a plane face, to which the axis of the particle is perpendicular. Find the force acting on the magnet, and show that the potential energy of the system is M2G4 3271-Att,a3(A
+
31. A small magnet of moment M is held in the presence of a very large fixed mass of soft iron of permeability At with a very large plane face, the magnet is at a distance a from the plane face and makes an angle B with the shortest distance from it to the plane. Show that a certain force and a couple — A.0 M2sin B cos 32ri.tv(Ai
/12,)a3
are required to keep the magnet in position. References
S. S.: "Electric and Magnetic Fields," Wiley, 1941. Clear practical treatment of ferromagnetism and permanent magnets with diagrams and tables. BITTER, F. T.: "Introduction to Ferromagnetism," McGraw-Hill, 1937. Excellent modern treatment.
ATTWOOD,
REFERENCES
367
R. M.: "Ferromagnetism," Van Nostrand, 1951. A classic in modern magnetic experiments. EwING, J. A.: "Magnetic Induction in Iron and Other Substances," Van Nostrand, 1900. The standard work of its time. FLUGGE, S.: "Handbuch der Physik," Vol. XVIII, Springer, 1966. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Springer, 1927. GRAY, D. E.: "American Institute of Physics Handbook," 2d ed., 5-164, McGrawHill, 1963. Valuable tables and bibliography. MATTIS, D. C.: "The Theory of Magnetism," Harper and Row, 1965. A very modern and advanced treatment of magnetism and cooperative phenomena. RADO, G. T., and N. Sum.: "Magnetism," Vol. III, Academic, 1963. First of three volumes on magnetic theory and materials by many contributors. SPOONER, T.: "Properties and Testing of Magnetic Materials," McGraw-Hill, 1931. Gives detailed experimental information. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. WILLIAMS, S. R.: "Magnetic Phenomena," McGraw-Hill, 1931. Treats experimental f acts. BOZORTH,
CHAPTER X
EDDY CURRENTS 10.00. Induced Currents in Extended Conductors.—This chapter covers the laws of magnetic interaction of currents and the induction laws of Faraday discussed in Chap. VIII as they apply to extended conductors, including skin effect, induction heating, drag on moving magnets, and the magnetic shielding by thin sheets. We should note that this treatment, and all that have preceded it, involves one approximation, viz., the assumption that electric and magnetic fields are instantaneously propagated. This is equivalent to saying that Maxwell's "displacement current" is neglected. The error so introduced is perfectly negligible if the frequencies are such that the wave length is large compared with the dimensions of the apparatus. If this condition is not met, we must resort to the complete Maxwell equations treated in Chap. XI. . Faraday's law of induction states that, if the magnetic induction B in a conductor, is changing, an electric field E is produced which is given, in magnitude and direction by 8.00 (3), to be VxE=—
dB d
In terms of the magnetic vector potential A, 8.00 (4) gives A E=—d dt
(1)
(2)
Since this electric field is produced in a conductor, a current will flow according to Ohm's law. If T is the resistivity and i the current density, we may, with the aid of 6.02 (3), write (1) and (2) in the form
(v x i) = —ITBI
r
(3)
dA (4) : = — dt These currents, flowing in the conductor of permeability A, will produce magnetic fields which, by 7.01 (3) and 7.02 (2), are given by V x B = Ai (5) v2A = — pd (6) The equations that must be satisfied by i, B, and A in a conductor when a changing field is present derive from (3), (4), (5), and (6). Elimination Ti
368
§10.01 SOLUTION FOR VECTOR POTENTIAL OF EDDY CURRENTS 369
of B by (3) from the time derivative of (5) gives, by 6.00 (3),
1.2 di = _ [v x (V x i)]= v2i — v (vi = • ) dt
V2i
T
(7)
Elimination of i from (4) and (6) gives
dA = V2A
(8)
r dt Elimination of i from (3) and (5) gives, with the aid of 7.01 (1),
II. dB = _ [V x (V x B)] = V 2B — V (V • B) = V2B dt
T
(9)
Equations (7), (8), and (9) all have the form of the well-known equation of the conduction of heat, but the dependent variable is a vector instead of a scalar. In rectangular coordinates, each component, regarded as a scalar, satisfies the same equation. This is not true for the components in any other system of coordinates except in special cases. Eddy current problems fall into two classes, transient and steady state. In the latter case the current density is often written as the real part of a complex Fourier series with terms of the form A„e3n't where n is an integer, w is the angular frequency, and A nis a complex constant called a phasor. The latter is indicated by a small flat v or inverted circumflex over the symbol, and its conjugate complex by a circumflex over the symbol. Vectors, shown by boldface type, may also be phasors. If i, the current density with amplitude i0 has its maximum value when t is zero, then the instantaneous rate of dissipation of energy in an element dv is, from 6.03 (2), ,
dP = Ti2dv = rig cos' wt dv = Tii cost wt dv The power averaged over a cycle is (10)
dP = Tii dv
10.01. Solution for Vector Potential of Eddy Currents.—We can obtain solutions of 10.00 (8) when T co in just the same way we did when T = 00 in 7.04. As in 7.04 (1) and (2), let us write A in the form
A = V x (uF rid-uxVW 2)
(1)
where u = 4i, k, or r. From 7.04 (3) and (4), we have
x v(V2 W2)] V2A = V x [uV2W1 u Substituting from these two equations in 10.00 (8) gives V x [u(V2W2
d W 2) T dt
u x V(V2W2
T
dW)] =0 t
370
EDDY CURRENTS
Thus, if W1and
W2
§10.02
are solutions of the heat conduction equation, viz., dW =r dt
(2)
then, with the aid of (1), we can obtain from them solutions for the vector potential of eddy currents. We calculate B in terms of W1and W2 by making use of 7.04 (5) and (6). Thus, B=VxA=Vx(VxuWi)+Vx(uV 2W2) (3) -dd-t(uW2) — u x vWi
Vx
(4)
J
where u = t, Is k, or r. Since B and A satisfy 'deistical equations, 10.00 (8) and (9), the similar form of (1) and (4) is to be expected. We notice that now both W1 and W2 contribute to B. We can simplify this expression somewhat further by using 7.04 (6), which gives u = i, k,
B=
u=
B=
T,
d — (u x VW 2 uW 1) u • V(VW1)
T dt
ud r dt
(5)
rWi) r • V(VW1) + 2V Wi (6)
When u lies along ul, a solution of (2) of the form U(ui)F(u2, u3, t) makes B normal to A as in 7.04. 10.02. Steady-state Skin Effect.—A simple solution of 10.00 (7) applies to a medium of permeability A and resistivity T filling all positive z-space. If the phasor current density Ix is uniform, x-directed, and of angular frequency w on the surface, then in the interior 10.00 (7) becomes jWilx = j C447IX
a rz 2 = aZ 2
=
d 21x dz2
(1)
since L is a function of z only. The solution of this equation is Ix = ae-(i.wr)iz
fle(iwwy)l,
If Ix is finite when z = 00, then 13 = 0; so writing 1
(2)
j for (2j)1gives
Lem = 2oe--(.i7)izoka-(1.4)iz1
(3)
where iois the value of Ix at the surface. Taking the real part gives
ix = ioe-(iw")iz cos [cot — (icy/17)/z]
(4)
Thus the ix amplitude decrease is exponential and the phase change uniform. The net current I cos (wt + ,p) per meter width, given by expanding the cosine and using Dw 863.1 and 863.2 or Pc 506 and 507, is Jo m
ix dz = io(2wiry)-i(cos wt + sin wt) = io (co,u7)-i cos (wt —
(5)
SKIN EFFECT OF TUBULAR CONDUCTOR
§10.03
371
This would be increased by removing all conducting matter below a certain depth since, for certain values of z, the current is reversed. The power absorbed as heat, per square meter of surface, is, from 10.00 (10), with the aid of Dw 440.20 and 565.1 or Pc 363 and 401, = O.Yr
2ir /7;
„42
e-2(1'1'7)1z dz 2 0 (6) = 7/1(10447)1 = 12(76)-1 =-icokibg = where 1„ = (2)--4/ is the effective value of I. Thus the resistance equals the d-c resistance of a skin of thickness b = (coify)-1. A uniform y-directed magnetic induction Bo cos (wt - fir) at the surface would induce the eddy currents just considered. From 10.00 (7) and (9) the same differential equation describes the behavior of i and B in the conductor so that both die out exponentially. To find the power dissipated, take the line integral of B around a rectangular path normal to x whose sides lie in the z-direction and whose meter long ends lie, one just outside the surface where the induction is Bo and the other far inside the conductor where it is zero. Only the Bo end contributes to the integral so that from 7.01 (2) (or from Maxwell's first equation) P1
2arfo o
i2dt dz =
Bo = kc„/ = 2iµ„/, (7) From (6), the power dissipated by eddy currents in a surface area S in terms of the magnetic induction amplitude Bo just outside the surface is
P = f Pi dS = (2/4,76)-1f
Bg dS
(8)
The inductance per square vicontributed by the magnetic field inside the conductor is found from (4), (6), 8.08 (1.1), and 10.00 (3).
T
27r
57)
L:I! w P Ti! - , i • A dt dz dt dz = — = (9) o s 2 26) 2w& 7rJ. f = 47r f. Comparison with the resistance R: per square gives the relation 1 R” L: = — = = (10) wb way co At high frequencies g is completely negligible. 10.03. Skin Effect on Tubular Conductor.—If the frequency is so high that the depth of penetration is small compared with the curvature of the conducting surface, then the results of the last article can be applied to the usual cylindrical wire or tubular conductor. If this is not true, we must use cylindrical coordinates. In this case, if we have symmetry about the axis so that iz is a function of p only, 10.00 (7) becomes aqz , 1 al. icokir • • (1) = 3coirrtz = 37). 0 = ap2 p
ap
EDDY CURRENTS
372
§10.04
Let v = (jp)i p, and this becomes
a2E, , 1 81„ (2) 5172. This is exactly Bessel's modified Eq. 5.292 (1) of order zero, the solutions of which were found in 5.32 (6) to be "
IZ = C/o(v)
liKo(v) = 0/4(./P)IP] /51CoRiP)iPi (3) From 7.16 (2), we know that there can be no magnetic field in the cavity in this tube because of the symmetry. If the external and internal radii of the tube are a and b, respectively, we see that the boundary condition at p = b is B = 0. From 10.00 (3), this gives x i)„_b = 0
or
\op
=0
(4)
p-b
At p = a, the boundary condition is = io From (3) and (4) using 5.33 (4), we have 0 = 0/1[(jp)ib] —
(5) Ki[(ip)ib]
and from (3) and (5), io = 0/0[(jp)ia] + 13K 0[(jp)i a] From these, we obtain
10[(ip)4a]Ki[(jP)4 b] + iRip)41911(0[0p)ial I iRip)iblio .75 — . I 0[(.7P)ia]K iRi P)Ibi -I- I ai P)INK oR j p) a]
(6)
(7)
10.04. Skin Effect on Solid Cylindrical Conductor.—If we have a solid wire instead of a tube, we must set D = 0 in 10.03 (3) since Ko(x) is infinite when x = 0 by 5.33. We would then have for fa /0[(iP)1P] . (1) • 2, I 4(3P)ial instead of the result given by 10.03 (3), (6), and (7). For numerical computation, it is necessary to split /0[(j)1x] and Ko[(j)ix] into real and imaginary parts. This is done by means of the ber, bei, ker, and kei functions of Lord Kelvin, for which numerous tables exist; see, for example, Dw 1050 or HMF, pages 430 to 433. Thus /JUN] = berox jbeiox (2) Ko[(j)lx] = kerox jkeiox (3) Series for these functions are easily obtained by substituting in 5.33 (1) and (4). Using ber and bei functions in (1) multiplied by eiwt and taking
=
§10.04
SKIN EFFECT ON SOLID CYLINDRICAL CONDUCTOR
373
the real part, we have for the current density in a solid wire of radius a beiO[(P)1Plyto cos kun, , a) {berORP)1Pi beiff(p)ia] berg[(p)ia] , bero[(p)ia]beioRp)ip] — bero[(p)ip]beioRp)ia] a = tan— bero[(p)ia]bero[(p)ip] + beio[(p)ip]beio[(p)*a]
(4)
is =
(5)
We can obtain the total current in the wire at any instant from the magnetic field just inside the surface, since by 8.09 (1), they are connected by the relation B = µI/(27a). From (1), 10.00 (3), and 5.33 (4), we have, after dividing out ei't,
= — T(N7 x i). = 448ap.1 a— 6 (jP)1/O[ciP)aiTio I 0[(3p)ia]
jc0(11)0 or
= 2rab _ 2irr(jp)iala( jp)iali jco,u10[(jp)ia] °
(6)
The average power dissipated per unit length in a ring of radius p and thickness dp is, from 10.00 (10),
dPav= iTIL122vp dp = xTIZi Z p dp where L is the conjugate phasor of L. We should note here that for electrical purposes the conjugate complex of (-Fj)i = 2—i(1 + j) is ( —j)i = 2-4(1 — j) = —j(j)i. We may write
= LA( —iP)iPl = fo[ —i(iP)iP] The total power consumed per unit length in the wire is then '2 7720 Pay =
0
dPa„, =
. I0P)ictlio[ R3 — j(iP)/a]f
dp
This integral is a special case of 5.323 (1) obtained by setting n = 0, and the result written in terms of berox and beiox is
pa
Tyra
v
(P)1
beroRp)ialbeia(p)ial — bera(p)ia]beio[(p)ial, to berff(p)ict] beiff(p)ial
(7)
From (6), the rms current I, is _ 24r2T2Pa2 fa(iP)iarla(iP)Ial .2 112‘1.12 Io[(jp)ia]Io[(jp)ia]Zo 2ir2a2 bera(p)ial}2 + beia (p)ial } 2.2 p {bero[(p)ia]) 2 + beio[(p)ia]) 2°
(8)
If R = r/(ra2) is the resistance per unit length for direct currents, then the high-frequency resistance R' is R,
_ — Pavra2R _ ct(p)iber[(p)ia]beil(p)ial — ber'[(p)iajbei[(p)ra]R (9) (p)ia] } 2 71,2 — 2 ber'[(p)ia] I 2 +
374
EDDY CURRENTS
§10.05
where, as in many tables, the zero subscript is omitted and from 10.03 (1) p = r'1.10) = ypco
(10)
and a is the radius of the cylinder. One gets the magnetic field energy inside the wire as in (6), thus drrio
jo.)13 = 441 =
49.10[(jp)ia]
as I o[UP)Icti The average energy inside the wire is, from 8.02 (3),
Op .
a a ir 7.2jap ,;I) dp f13- •ap dp — 2u(o2 /oRiP)IalioR —iP)lailo 2A 0 where we have written /0 for /0[(jp)ip] and .1- , for a/o(x)/ax. Since /“x) equals /1(x) by 5.33 (4), we see that this integral is identical with 5.323 (1) when n = 1, so that its value in terms of berix and beia is
la
1-
ap-i[beri(pia)bee,(pia) — berapia)beii(pia)] Using Dw 828.1, 828.2, 829.3, and 829.4 to reduce this to zero order and remembering that this average energy equals IL.,/! where Liis the internal self-inductance per unit length and Ie is given by (8), we obtain Li =
r(p)-1ber[(p)ia]berl(p)ia] + bei[(p)iedbeil(p)ia]
fber'[(p)icd} 2 + {bei'[(p)ict) 2
2irwa
(11)
10.05. Solution in Spherical Coordinates for Axial Symmetry.—Let us assume that the magnetic field producing the eddy currents is independent of ci5 and has no 0-component. Then the vector potential has only a 0-component and may be written A = 4)A 4,(r , 0, t)
(1)
where 4is a unit vector in the .-direction given by . = — i sin . + i cos .
(2)
Apply Laplace's operator 3.05 (1) to A. = —A 4, sin ck and Au = A4, cos 4) separately. After recombining we see that 10.00 (8) becomes il
a,aA4,
T 1-
at
= V2A = 4,[v2A, — 7
sn -AA,
i e0 r2 r sin2
cl3V 2A4, + Act,V2(13
(3)
Writing out V2 in polar coordinates by 3.05 (1) and dividing out 4give AaA0 1 a (r 2aA4, 1 a (sn o i 0m o = T at r2(3r + sin o aen2 are r2 ae ) r 2A sin 0
(1_ .2)ia2R1 = 1 a (r 2aA 9!. ± (1 2 r2 ar
ar
r
u2pAoi
—
au2
(4)
where u = cos 0. We shall now consider the steady-state eddy currents when the magnetic field oscillates with an angular frequency w. As in 5.12, we shall
§10.06
CONDUCTING SPHERE IN ALTERNATING FIELD
375
seek a solution that is the product of a function of 0 by a function of r. Then we write A0 = er-iReiwt (real part) (5) Substituting in (4), multiplying through by r2, and dividing through by Or-iRem give r2 d2R R dr'
r di? h dr
1 4
jpr2 +
(1 — U2)1 d2
[(1 — U2)103] = 0
du2
(6)
where, as in 10.04 (10), we have written P = 7112W = i" (7) Proceeding as we did in solving 5.12 (1), we set the terms in (6) involving 0 equal to —n(n 1) and those involving r equal to n(n 1), thus satisfying (6) and giving, after expanding the derivatives in (6), (1—
u2)cf — 2u du
on
(8) (9)
2
The first of these is identical with the differential equation for Legendre's associated functions with m = 1 as written in 5.23 (4.1), and the second is the modified Bessel equation in x where x = (jp)ir as written in 5.32 (1). Thus, from 5.23 (5) and 5.32, Am is the real part of r--4[A„P(u)
Btia(u)l{anIn+1[(.7P)Irl
1.5.K.+1[(jP)irll el' (10)
If n is an integer, as it must be for Pl(u) and Qgu) unless conical boundaries are involved, we can use /_(,,+i) instead of Kn+1 for the second solution, by 5.37. In a region where the conductivity is zero, the left side of (4) is zero, and if we let A0em = E'Oeic", we obtain (8) as before but instead of (9) we get die\ n(n 1)R' = 0 -c/--7 r, dr ) whose solution is given by 5.12 (3) to be
R' = Arn ± ijr-n-1
(12)
and in a nonconducting region this will replace terms involving r in (10). 10.06. Conducting Sphere in Alternating Field.—Consider now the specific example of a sphere of resistivity T, permeability j4, and radius a placed in a uniform alternating z-directed magnetic field Beiwi . The phasor vector potential of this field is, when ei‘" is divided out,
A = +Or sin 0 = (10rP1(cos 0)
(1)
376
EDDY CURRENTS
§10.06
as can be verified easily by taking its curl by 3.04 (2) and (3). Thus, n = 1 in 10.05 (10) and (12); and, since the eddy current vector potential must vanish at infinity, we have, outside the sphere, a < r < co,
= 0/3(r +
1)-7.-2) sin 0
(2)
At r = 0, is finite, so 5.37 (4) and (5) show that only Ii[(jp)ir] can occur inside the sphere. Thus, setting n = 1 in 10.05 (10), we have
= obor-iIi[(jp)ir] sin 0
0 < r < a,
(3)
From 7.21 (6) and (7), the boundary conditions when r = a are
A. = At
a
-
a
-
au,,-&.(r - sin MO = ii— (r sin 0A 0) ar
and
(4)
Putting r = a in (2) and (3), we obtain with the aid of 5.321 (1), (2), and (3), after writing In for /n[(jp)ia] and v for (jp)ia, a3 + 13 = alO/I = aio[/_i — (2a3— D)/A = Ance[iIi vI Solving for C and
= Anal[(v
— I-110
15 gives 31.wal
—
(5)
— 14)2)1_1+ [An( 1 + v2) — []i* _ (21A — [1.1„(1 + v2) +21.1]Ii.a3 0.4 — [An( 1 + v2) —
(6)
This may be expressed in terms of hyperbolic functions by 5.37 or Dw 808.1 and 808.3. From 10.00 (4), the current density anywhere inside can be obtained from (3) by the equation i = —juryAi = — jpg.-1Ai
(7)
The magnetic field at any point outside is obtained from (2), 3.04 (3), and 3.04 (2) to be Boe
= —b(1 — 13) sin 0
= —r1 1
a (,s.
r sin 0 (30
0Ao) = '41 ±
-)
r3
(8)
cos 0
(9)
By a similar method (3) yields bio and Eir. Comparison of (8) or (9) with 7.10 (2) shows the eddy current field to be like that of a magnetic dipole loop of radius a carrying a current where 1.4a27 = 21315. If 0 from 11.05 the magnetic field is not alternating, then w = 0 so that p (7). From 5.37 and Dw 657.1 or 657.2, we see that Ii(x)
(-2 ) 4(x + 6 rx
and
2 (1 /_4(x) ---> (7rX ) 1
2)
377
POWER ABSORBED BY SPHERE
§10.07
Thus (5) and (6) simplify so that (2) and (3) become
A. = (1)i r
. 2(Ifn, - a31 1)2 sin 0 (K m +2)
(10)
3Kmb
(11)
- 412(Km 2) r sin
These are the correct static fields. From (7), we obtain as a first approximation for slowly alternating fields
3jco.K„,yr3 14, = — 2 ( ICm + 2)r sin 0
(12)
The same result is obtained if the resistivity is made infinite. When the frequency is made very high, we see that A.-> 0
4,-> (14/3(r - a 3r-2) sin 0
(13)
because, from 5.37, Ii(x)
( 2 yez
/3(x) --> Li(x) ot,
71-X
2.
thus, there is no magnetic field inside, and the eddy currents are confined to the surface as we would expect. To visualize the magnitudes of the numbers involved, we shall compute p by 10.05 (7) for a field alternating at 60 cycles per second (so that w = 1207r) for several substances. If r and A are in mks units, we find = 4r X 10-7henry per for copper r P-11.7 X 10-8ohm-meter, a meter, and p 28,000. For a typical specimen of iron, r 10-7, 1.1 480r X 10-7for a magnetic field intensity of 1.5 X 105/(47r) ampere8 X 10-8, turns per meter, and p = 570,000. For graphite, r A r-=, 47 X 10-7, and p = 60. Thus for this frequency, over distances of a few centimeters, (12) would apply to graphite but not to iron or copper. The initial assumption that A, is zero and Bo tangential to the surface greatly simplifies the calculation. All the results of this section, being written in phasors, give both the amplitude and phase of the quantities involved. The same forms for the electromotances and the same types of boundary conditions apply equally well to any number of thick concentric spherical shells and could be used, for example, to calculate their screening effect. The results would be much more complicated than those given here. The distribution of a given amount of material in several separated shells will increase the screening effect. There are optimum thicknesses and spacings. 10.07. Power Absorbed by Sphere in Alternating Magnetic Field.— We shall now compute the power absorbed by the sphere in the last article. From 10.00 (10), the power absorbed in the volume element dv is
dP
dv = irriir2 sin 0 dr d0
378
EDDY CURRENTS
§10.08
Substitution for i and I from 10.06 (7) and (3) and integration with respect to 0 from 0 = 0 to 0 = 7 give
P - irw32B2aef aIiRjp)irlIg —jp)irk dr
(1)
Integrate by the last form of 5.323 (1) and note from 5.37 that .4[( ±jp)ia] = Fir( ±jp)iai—i sinh [i(2p)i(1 ± j)a] I4(±jp)ia] = [lir( ±jp)ia]-1cosh [1(2p)' (1 ± j)a]
(2)
Application of Dw 651.06 to 651.09 or Pc 669 to 672 gives for the integral (wpia)-14(2pa2)1[sinh (2pa2)1+ sin (2pa2)4] — cosh (2pa2)i cos (2pa2)1 Writing out CC by (2), 10.06 (5), and 11.05 (7) gives for P, if U = µ U2[(pa2
3,a 5w2B2A2T-1[_u(s
s) — C
-
c],
+ 1)C (pat— 1)c — u(S s)] + Upzi,pa2u(S — s) mv2p2a4(c —
where u = (2p)fa, C = cosh u, c = cos u, S = sinh u, and s = sin u. 10.08. Transients in Conducting Sphere.—In the last two articles we solved the steady-state problem of a sphere of permeability pi and resistivity T in a uniform alternating magnetic field. Let us now solve a transient problem by calculating the effects when the same sphere is placed in a uniform magnetic field B which is suddenly removed at the time t = 0. Clearly, in this case, as in the last one, A will have a 0-component only. At t = 0, surface eddy currents prevent the interior field from changing, and since the vector potential must be continuous across the boundary, we have, from 10.06 (11), at t = 0,
A, = CI)
3K,J3 2)rPi(u)' 2(Km
A,, =
3„,,13 '2(K„,
a
131.(u)
2) r2
(1)
Since T = 00 outside, the subsequent behavior of A is determined from 10.00 (8) by solving the equations
v2 Ai = T dt
V2A. = 0
(2)
In 10.05, we found a solution of the first of these equations when the time entered exponentially. Let us see if we can fit the boundary conditions in the present case by using a sum of such solutions. Clearly in the present case there will be no oscillations, so we shall need to substitute a negative exponent —gat for jest. Let us write lc! = r--41.tqa =
(3) We must then substitute —k for jp throughout 10.05. This gives us jk, instead of (jp)i in 10.05 (10) and leads to ordinary Bessel functions. From 10.05 (10), writing sin 0 for Pi(u) and noting that Aimust be finite
§10.08
TRANSIENTS IN CONDUCTING SPHERE
379
at r = 0 and contain 0 in the same form as (1), we obtain Ai = 4•ZA.r—ifi(ka r) sin Be—ot
(4)
8
A. is finite at infinity and equals Ai at r = a for all values of 1. From 10.05 (12), it therefore has the form = 42/38r-2 sin 0 —g.t
(5)
a
In addition to satisfying (1) when t = 0, we must satisfy 10.06 (4). This gives at r = a, dividing out sin 0, a(rA0)
1.,° a(rA,)
A° Ai,
a
=
Therefore, at r = a for all values of t, it is necessary that
AsaiJi (k.a) = B,,
p„Asa
d [aiJi(ksa)] = da
(6)
—
Differentiating the product in the second equation, multiplying the first equation by pi, and adding and dividing out As, we obtain a)] k8 + ilvada[ji(
1
—1.1,)T(k.a) = 0 2
(7)
To meet the boundary conditions, we must choose values of k, that satisfy this equation. This determines, by (3), the values of q, that appear in (4) and (5). The values of k, can be found with the aid of 5.31 (3) and a table of trigonometric functions. From (1), (4), and (5), multiplying through by ri and setting t = 0, we have 3KmB
2(K., ± 2)
r =
(8)
This is identical with 5.297 (1), last case, where n = I and, from (7), I-. From 5.294 (7), the B of that article equals (/.0.4) a 3K Bat , VJI(k v) dv — ' vf(v)J3(k,v) dv — 3K,„B 2k, Km± 2)J o(kea) 2(Ka. ± 2)1 With the aid of (7) and 5.294 (2), we obtain ±2 Ji (kaa) = —Ji (k,a) gsa ji(kaa) = keft ji(k8a)
f
a
So that, from 5.297 (6) we have
A. —
[k!a2
31C,0Bal (IC„, — 1)(1C„,
2)]11(k.a)
(9)
380
EDDY CURRENTS
§10.09
Substituting this in (4) and (5) gives Ai and A,. With the aid of 10.00 (4), we obtain for the current density inside
411(ksr)
3Bal sin ONI
i = +
Ayr*
2)1,11(1c,a)e
L-1[14a2+ (Km— 1)(K„,
—k 1174-1t
(10)
8
10.09. Eddy Currents in Plane Sheets.—We shall now calculate the vector potential A(x, y, z, t) of the eddy currents of area density i induced in a very thin plane sheet of area resistivity s at z = 0 by a fluctuating magnetic field whose vector potential is A'(x, y, z, t). In the interior of the sheet, the electric field —a (A' + A,)/at drives to the surface charges whose electrostatic field exactly neutralizes it. The effect of the associated current is negligible, so that inside the sheet we need consider only the tangential components A, and A. Thus using 10.00 (4), we may write
d(A: A,) = si (1) dt Let the eddy currents be confined to a finite region of the sheet which may or may not extend to infinity, and let us define the stream function c13(x, y) at any point P in the sheet to be the current flowing through any cross section of the sheet extending from P to its edge. From 7.01 (2), we have, for the closed path bounding this cross section, but not including the surface, since B is symmetrical with respect to the sheet, =
(fB • ds = 2fX C Bs dx = 2f dy
(2)
where the path of integration is in the positive x- or y-direction when z is positive. Differentiating this equation, we have 2 aA _ M = a4) = 2B„ = j'y = act. = = as
ay
/.1,
ax
az
ktv az
Adding the components gives, inside the sheet
i=
2 Ay
aA, az
(3)
Substituting this in (1) gives, inside the sheet,
d(A: + As) _ 25 9A, dt Az az
(4)
Outside the sheet, A consists of two parts, a magnetic part arising from the eddy currents and an electric part, the gradient of a scalar, arising from the electric double layer produced by Ai and Az. Since A:, the tangential component of the exciting potential, is known and since A, is continuous across each face of the double layer, (4) gives the boundary condition on A, just outside either surface of the sheet. This, combined
§10.10
EDDY CURRENTS IN INFINITE PLANE SHEET
381
with the equation V 2A = 0, determines A everywhere outside the sheet. When the sheet is finite, the boundary condition in its plane, but beyond its edge, becomes i = 0 or dasIdt = 0. The right side of (4) is finite at all times, which means that if of —4 0, then 5(A; + A,) —> 0. Thus an abrupt change in A' instantaneously induces eddy currents such as will maintain A + A' and B B' unchanged in the sheet. Therefore, for a specified change in A', the initial value of A is known and, if no further changes occur, its subsequent values, as the eddy currents decay, can be determined by putting dA'/dt = 0 in (4) and solving. A second abrupt change in A' produces a second set of eddy currents, and so forth. At any instant, the actual field of the eddy currents is a superposition of these. As the magnitudes of the discontinuous changes in the external field become smaller and the intervals between them shorter, we approach, as a limit, a continuously changing magnetic field. 10.10. Eddy Currents in Infinite Plane Sheet by Image Method.— Suppose that a thin infinite plane conducting sheet at z = 0 lies in a magnetic field produced by sources in the region z > 0. At t = 0 the field is changed, the vector potential being given by A'1 = fl(x, y, z) when t < 0 and by A', = f 2(x, y, z) when 0 < t. From the last article, the eddy currents produced at the instant t = 0 keep the vector potential over the surface of the sheet the same so that, initially, the whole field on the negative side is unchanged. Thus we have, on the negative side of the sheet, from the eddy currents alone = (1) = fl(X1 y, z) — f2(x, y, z) —
This field could be produced by reversing the sign of the new source and replacing the old source. These hypothetical sources which can replace the actual eddy currents are images as in electrostatics. Since A2 is not a function of t, 10.09 (4) reduces to
dA = 2s aA µv az dt A general solution of this equation satisfying (1) at t = 0 is
(2)
A = f i(x, y, —Izi — 2stc1t) — f2(x, y, Hz' — 241,710 (3) A identical at ±z The sign of z was chosen to make as required by symmetry and to make it die out with time. Thus, the equation shows that, added to the field ig which would exist if no sheet were present, there is a decaying field due to the eddy currents which appears, from either side of the sheet, to come from a pair of images on the opposite side which recedes with a uniform velocity 2s/A,,. Maxwell gives a formula for this law of images which applies to any type of field variation. Suppose the inducing field vector potential is A' = f(t, x, y, z)
(4)
382
EDDY CURRENTS
§10.11
The change in this field in an infinitesimal interval of time dr is
s a
dT
=
a
x, y, z) dr
The initial field of the eddy currents formed in that interval must be equal and opposite to this. This field dies out, as we have seen, as if it were due to an image on the opposite side of the sheet moving away with a uniform velocity 2s/1.4,. Thus, the vector potential of the eddy currents at the present time t due to images formed in the interval dr at a positive time 7 before the present is given by
dA =
a f(t —
at
2s y, —1z1 — — r) dr
T, x,
At the time t, the total vector potential due to eddy currents is
a 2S A= —f —f(t — x, y, —1z1 — —r) dr (5) o at When z < 0, z replaces — jzi and, by Pc 863, war = —avat — (2s/g )af/az so that substitution for of/at in (5) and integration give, since f is zero at T = 00 and equals A' at T = 0, for the total resultant field 2s a f x, 2s A A' = -F -, f(t — T, y, z — r) dr (6) az o This is often simpler to integrate than (5). 10.11. Torque on Small Rotating Current Loop or Magnetic Dipole.— If a magnetic system is rotated about an axis normal to a conducting M --- sheet, the field of induced eddy currents will, in general, set up a retarding torque on the system which is, as we shall see, proportional to its = angular velocity if this is not too great. By measuring this torque, the angular velocity can be found, and this is the basic principle of some o automobile speedometers. The simplest magnetic system is a dipole or small current loop, P. 1 which rotates about its center keeping its axis 2sr parallel to the sheet. In order to visualize the /L. images, we shall not use 10.10 (5) but shall sup-. )— 2 2sr pose that at regular intervals of time T the magp. net is instantaneously rotated through an angle 3 WT. From this result, by letting the intervals approach zero, we shall obtain the result for FIG. 10.11. continuous motion. Figure 10.11 shows the images formed by the last four jumps at the instant the dipole, of moment as, arrives in the position indicated, as viewed from the upper side of the z < 0,
TORQUE ON SMALL ROTATING CURRENT LOOP
§10.11
383
sheet. The actual torque on M will be the sum of torques due to all the images. Since all image dipoles are perpendicular to the axis of rotation, ei= 82 = ir in 1.071 (4) and, from 1.071 (7) and 7.00, the torque due to a single image making an angle 1p with M is
-fry Af 2 sin ip aip a = 47— r 3
T=
(1)
The torque due to all the images is then, if p = co/4/s, et1
T=
i.c.„m2
47r
sin nwl- — sin (n 1)wr [2c + (2ns//12,)rJ3
(2)
n=0
1.4,„p 3m2
sin non- — sin nun- cos wr — cos ncor sin wr
(pc + ncor) 3
— 327r -LJ n=0
As the motion becomes continuous, r dt, nr —> t, and sin wr so that (2) passes over into the integral
T—
— cos cot
ici,cop3m2 7r 32
0
(pc + wt)3
(or —> w dt
dt
(3)
Let x = pc + wt so that cos wt = cos (x — pc), and this becomes „v,n3m2 cos x 'sin x (cos pc dx + sin pc —T— X 3 oZT fp, x 3 dx fpc Applying Dw 441.13 and 431.13 or Pc 345 and 344 gives
—T—
647r
p2c2 COS
pc
dx — sin pc
• x sin
dx
(4)
These integrals may be evaluated by the series Pc 346 and 347 or Dw 431.11 and 441.11, or they may be put in the form of the standard cosine integral Ci and sine integral Si. Thus 7r 'cos x I sin x dx = — — Si(pc) = —si(pc) (5) 2 , x dx =Ci(pc), x
fp
Numerical tables and graphs of Ci and Si are available in Jahnke and Emde and other mathematical handbooks, especially HMF, Chap. V. If pc is much greater than unity, we integrate (3) thrice by parts and obtain for — T,
1.4,P 3m1 327r
(pc + 0) 3
co sin wt 3 coswt + 12 sin wt — 60 dt (pc + (00 50 fo (pc + cot)6 (pc + (4 4
Neglecting 20w/(pc)2 compared with unity, we have 3A04.2
Pc >> 1
T=
32irpc4
(6)
384
§10.12
EDDY CURRENTS
Neglecting w(pc)2compared with unity (4) gives
T — AvP 312 6471-c2
pc << 1
(7)
For a copper sheet 0.1 mm thick s = 0.00017 so that for 100, 10,000, and 10,000,000 cycles per second we have for p, 4.67, 467, and 467,000, respectively. Thus, for these frequencies, we would use (7), (4), and (6), respectively, if c ~ 0.01 meter. 10.12. Eddy Currents from Rotating Dipole.—It is instructive actually to calculate the equation for the eddy currents produced by the rotating dipole treated in the last article. In Fig. 10.12a, we draw one of the images shown in Fig. 10.11 and indicate at P the point on the sheet whose coordinates are p, 0 at which we wish to calculate the stream function (D. From 10.09 (2), we have
f Mpdp
= 2µi l
(1)
by choosing a radial path for our integration and noting that the contribution from the lower half equals that from the upper. The figure shows the nth positive image. Only its p-component contributes to B. Using 7.10 (5) and (2) for B„ adding the contribution from the nth negative image, and summing over all the images, we have 13, = — tivm 4r
!cos nwr + 0) — cos [(n. (
1)cor
011 a(sin 0') az R2
(2)
n=0
Substituting in (1), integrating, and breaking up the second cosine by Dw 401.03 or Pc 592, we have Mp
Ni
s(nal- + 0)— cos(ncur + 0) cos WT
— Ip2
2.7r L.)
[c
+sin (nwr + 0) sin (ilT
(2nshi,)r]211
n=0
(3)
Passing over into the integral exactly as in the last article we have
(13 =
_
wmp rsin (cot ± 0) dt 2r jo {p2 [c (2s/ /2„)t] 21i
(4)
385
SHIELDING OF CIRCULAR COIL
§10.13
Choose a new variable u so that (2s/A„)t = u — c and the limits become c and 00. Then let A = 1041„/s, substitute in (4), and expand the sine by Dw 401.01. Thus (4) becomes (see also problem 32) • (0 Ac) ' cos Au du + cos (0 — Ac) .sin Au du 1 .1) = (..44111P [sin f (p2 + u2)1 f c (p2 +u2)4 i 47rs
(5)
If icol.tc/s = Ac is very small, this reduces to
du • 0 fc° .1). = wi4MP sin c (p2 +uy = Lirs
+c2) 1 ]
wi4111[ 1 — (p2 c zlirsp
sin 8
(6)
The system of eddy currents given by this equation is shown in Fig. 10.12b, the value of 471-(13//2„ being indicated on the figure. This system of stream lines rotates with the same speed as the magnet, which lies at a
FIG. 10.12b.— Stream lines of the eddy currents induced in an infinite thin plane conducting sheet as seen by an observer on a magnetic dipole directed to the right which rotates with an angular velocity w as shown in a plane 1 cm above the sheet. The values of the stream function 4> in Eq. 6, Art. 10.12, are shown when coA„M/(47rs) = 1.
1 cm distance, on the scale shown, above the plane of the paper and is parallel to the c1 = 0 line. We see at once that the currents shown will give a field at right angles to the magnet. This will produce the torque calculated in the last article.
10.13. Shielding of Circular Coil by Thin Conducting Sheet.—Let us use 10.10 (6) to find the effect of the insertion of an infinite thin conducting sheet on the flux linkage between two coaxial loops of radii a and b at
386
EDDY CURRENTS
§10.14
a distance c apart. Without a sheet the vector potential at one loop due to the other is, by 7.10 and problem 27, Chap. VII, the real part of A'4,0" = Arir—iemfoir(u2
c2)--i cos 4c/4) = Ne" Ji(ka)Ji(kb)e—kc dk (1)
where u2 = a 2 b2 tab cos 4 and N = 11.4„aI. Then writing X for 2s/A, and c Xr for — Izi — Xr in 10.10 (6) gives at the second loop A4.
= — NXf: kJ i(ka)J i(kb)e—kc f e—(xkl-mr dr dk (2)
= —N f: Xk(jco Xk)-1J i(ka)J i(kb)e—kc dk The general solution is in problem 33, but if Xk/ co < 1, this becomes CO
CO
t42a/N(2js f '
-.J\1.4,,co
n=1
2j5 \nanA'c°
knJ i(ka)cl i(kb)e—kc dk =
0
n=1
Avw)
acn
(3)
If b and c replace p and z, the k series of 7.10 (4) is a good form for 4 From 8.06 the ratio of the new to the old flux linkage is Ro —
271-b(A0 A;). 27rb(A:Orroo
(4)
If 2s/(wuv) is small, only the first term in the series is needed so that 2s B 2s 1 aA,6' Ro = — — — ac
C01.4 A0
(5)
where B„ and Ao are given by 7.10 (6) and (4), respectively, and are evaluated at p = b, z = c. If, in addition, either a or b is small compared with c, then from 7.10 (1) we get R0 = wi,„(a2
6sc b2
c2)
(6)
As we saw in 10.12 for a sheet of copper 0.1 mm thick and a frequency of a million cycles per second, couv/s is 46,700 which justifies the approximation in (5). Hence, if a = 1 cm, b = 10 cm, and c = 10 cm, then, from (6), Ro 0.0006 so that the linkage is greatly reduced by the shield. 10.14. Rotating Sheet in Magnetic Gap.—The reasoning of 10.10 applies to the scalar as well as to the vector magnetic potential and the magnetic induction. As an example consider an infinite thin horizontal sheet of area resistivity s parallel to the horizontal plane faces of a gap of height 2h in an infinite vertical circular cylinder of radius a having a uniform permanent magnetization M parallel to its axis. The sheet
ROTATING SHEET IN MAGNETIC GAP
§10.14
387
rotates with a constant angular velocity w about an axis parallel to and at a distance c from the cylinder axis, or the magnet may rotate with the sheet stationary. In any case, the system of eddy currents is fixed relative to the magnet. The magnetization is assumed to be absolutely rigid, so that the magnet acts as a cylindrical current sheet and does not perturb the magnetic field of the eddy currents as a ferromagnetic magnet would. The integrations in this article will be carried out exactly, partly to show what can be done with the mathematical references now available, especially IITF, IT, and HMF, and partly because the resultant formulas which involve many repetitive operations are easily set up and evaluated on a digital computer. From 5.16, 7.28, 9.08, and 5.298 (6) the potential on the axis of a ring of radius p' and magnetization M is, when z < h, dJ0 = Marp' dp' [4.7r[p' 2
dp' f Jo(kp')e-"h-z) dk (1)
(h2 — z2)]1 }-1 =
Here the loop definition for M is used so pi, is omitted. Integration by 5.294 (7) from zero to a gives the potential on the axis of one pole face:
Sta = f dS2a = aM f k-lJ i(ka)e-h(h-z) dk
(2)
Off the axis, in the plane z = 0 where the sheet lies, this becomes = ictMf' k-lJ o(kpi)J i(ka)e-kh dk
(3)
it, for cp to express rotation, so that Expand this by 5.298 (7), write cot when t = 0, c lies on the 4, = 0 axis (see Fig. 10.14), and obtain = aM (2 — 32).1. 0%-1J3(kp)J8(kc)Ji(ka)e—kh dk cos [s(wt 8=o
4))] (4)
This replaces 10.10 (4). Similar changes in 10.05 (5) give the eddy current Si, so write t — r for t and h (2s/A„)r for h in (4) and put 1 swilvh P. = 2 s
2s Alv
swh
(5)
ps
Now take 8/cat, set t = 0, and integrate with respect to r, so that = am
7_),,f°°,1-0(kP)J.(kc).11(ka)e_
h
k[k2
kh(k sin so
pah—' cos so) dk
(6)
8 =1
The s = 0 term drops out since Pscontains an s factor. From 10.12 (1)
388
§10.14
EDDY CURRENTS
the stream function 4 is, adding that of the lower pole,
= 4f 1-1„dp = —4f Gdp =
(7)
Now let k = p8t/h, 2R = ps p/h,2c1 = pac/h, and 2a1 = pact/h so all symbols after the summation are dimensionless, then 4) = 4aM
8(2Rt), 8(2cit)J i(2ait)e_p. t(t sin sq5 + cos sO) dt (8) t(1 t 2)
fo =1
Watson gives on page 148 a power series in t for J 8 (2tci)J 1(2tai). (-1)mcIm+3Cms(ai,c1)0'n+8+1 J8(2tc1)J1(2ta1) = a 1 m= o s)!]-1 Cms(ai,c1) = 2F1( —m, —m — s;2;aler2)[m!(m
(9)
— 4)-1] (10)
= (4c1.2— 1)".[(m + 1) !(m + s)!)--11V)[(ci
by HMF 22.5.44, where P;,!, 8)(x) is a Jacobi polynomial whose recurrence relations and coefficients appear in HMF, pages 782 and 793. In the integral 8 (2tc1)J i(2tai) is therefore replaced by t2m+3+1, so the integrals in the sine or cosine terms are .t2m+a-1-1j 8 (2tR)
1
1 + t2
.
dt
or
r Om+8,18(2tR)e_" dt 1 + t2 jo
Expand J.(2tR) in series by 5.293 (3) and write N for m
1YR2r1f t2N+10—P'td, r=0
r!(r
s)!
1 ± t2
or
s
(11)
r, so (10) is
r .t2Ne- p,t dt J o 1 + t2
(12)
If N were zero, the integrals would be HMF 5.2.13 or 5.2.12. Since N is never zero because s j 1, divide the numerator by the denominator until the remainders have this form and integrate the quotient terms by Dw 860.07, so the integrals of (12) are
(2N — 2i ± 1) ! (— 1l
—F N (P0 = i=1
(-1)N[Ci(ps) cos p8
si(p8) sin ps] (13)
N
( -1)i(2N — 2i)!
—GN(p8) =
( 1)N[Ci(p8) sin ps— si(p8) cos p8]
i=1 (14)
389
ROTATING SHEET IN MAGNETIC GAP
§10.14
See 10.11 (5) for Ci and si and see HMF, pages 238 to 244, for tables.
Thus the final value of the stream function 4) is CO
Z
—4aM ( — 1)maicrn-"Cms(ai,ci) Z ( —1)rR2r+8 r0 8=1 m= o X [r!(r s)!]-1[F7 (pz) sin scp G, (p8) cos sq5]
(15)
The torque about the axis of rotation will be calculated only in the case where h is so small compared with the radius a of the pole pieces that the fringing field is negligible. With this narrow gap the magnetic induction under the poles has the constant value µ 2 M, which is that in the equivalent current sheet solenoid, from 9.07. The torque equals the product of the normal induction Bz by the radial current density i,, and by the lever arm p integrated over the pole area. T = f pBzi, d8
where i,, = —
Ocl) P
(16)
a 4)
For the sine term in (8), i is even about 4) = 0, and for the cosine term it is odd. Thus for the latter the torque on one-half will cancel that ,
x
(a)
(b) FIG. 10.14a, b.
on the other so it need not be considered. From Fig. 10.14, the torque will be given by (8) if the J8(2tR)t sin s4, factor is replaced by af M AvSt f0 —1r cos .34) J ,(2tR) pi dpi do Let 2/i1 = poi/h.
(17)
Then from HMF 9.1.79 we have
J,(2tR) cos s4, =
J.+8(2tc1), n(2tRi) cos nO
(18)
It=
The limits — 7 to -Fir leave only n = 0. The R integral is 5.302 so (17) is —27r/4,8J8(2t,c1)4112 73:2f0U 0(2tR OR' dB]. = —27M psahm,--1,1„(2tci)J1(2tai)
390
EDDY CURRENTS
§10.15
Insert the factor 27/1/4„sa in (8), omit the 4) terms, and replace .18(2tR) in (11) by hp,7't-iJs(2tci)J1(2tai). Then use of (9) gives ‘.2 T = 271.4„a3M2 Z Zsai(-1)m+eciNC..(ai,ci)Cr.(al,ci)FN(p.) (19) 8=1
m=0 r= 0
10.15. Rotating Disk in Magnetic Gap.—Suppose that the infinite rotating sheet of the last article is replaced by a disk of radius A rotating in a magnetic gap. The primary change is that the circulation of the eddy currents is restricted to the region inside p = A. A secondary result is that the magnetic fields of the eddy currents can now pass freely around the edge of the disk without penetrating the sheet. The main problem is therefore to find a means of confining the eddy currents to the region p < A. One method to look into is that of images. In 10.14 (3) if p = a, then pibecomes (c2 A 2— 2cA cos 6)1, and if A2/c replaces c, this becomes (A/c)(c2 +A2 -2cA cos Of. Thus if a and h are replaced by Aa/c and Ah/c and the integration variable is changed from k to Ak/c, then the value of the integral in 10.14 (3) is unchanged and the new potential or magnetomotance 0' is also the same at p = A if aM is the same. In passing from 10.14 (3) to 10.14 (6), the substitution h (2s/ µ„)T was made for h in 10.14 (3), so for the new magnet (A/c)[h' (2s/ A„).7] must replace h' and thus in 10.14 (5) p.' = (c/A)ps replaces ps. Therefore if the original magnet is replaced by one centered at c' with a pole face radius a', a gap 2h', a magnetization M', and a new p87 then the value of the stream function at p = A will be unchanged provided that , A' c=
,
a
=
Aa
—,
h h' = c,
M' = A' '
p' = s A
(1)
Thus if both magnets are present, with M' reversed in the second, the circle p = A becomes the stream line 1 = 0. The stream function for a disk is then given by adding to the 43 of 10.14 (15) and (12) a function calculated by the formulas using c'1, ai h', —M', and p' instead of ci, al, h, M, and p. in 10.14 (14). If s/(A„(.0) is large compared with a2 and c2, a simpler procedure than that of 10.14 can be used. From Fig. 10.14, in the z = 0 plane (note that there is no because the loop definition is used for M) ,
mfarr =
4a 0
p' dp' da
(h2
d2
-
2
pip cos a)}
(2)
where pi is [c2 p2— 2cp cos (4) + wt)]} for a rotating field. The 10.10 (5) substitutions are now made, but because w << sh.l.„ the h 2sr/i),, which replaces h, dominates the denominator as T increases before the
ROTATING DISK IN MAGNETIC GAP
§10.15
391
co(t — r) for t substitution can take effect in the sine and cosine terms so only cot is retained. For a narrow gap h is small compared with a and may be omitted. The resultant integral is ('2,r
=
47rf 0
where pi = (c2 SZ
coc, sinyt)(pi — p'cos a)p' dp' dot dr + 2p'pi cos a] 0 [(2s/µ„ ) 2T2 + (p')2
(2.1)
p2— 2cp cos 4,)1. The T integration by Dw 200.03 yields (P1 M AL„cocp sin o f a — p' cos a)p' dp' dot 87spi 131 — 2P pi cos a o Jo (P') 2
(3)
By Dw 859.124 the a integral is 271-/ pi if pi > p' and 0 if p1 < p', so the upper limit for the p' integration is a when p1> a and p'when p1 < a. Multiplication by 4, because of 10.14 (9) and the lower pole, gives for the stream function for an infinite plane P1 > a pi < a
M yvwcpa2sin cp 2 spi MA,,cocp sin 0 4' = 2s
(1,—
(outside pole piece)
(4)
(under pole piece)
(5)
These formulas replace 10.14 (11) and (12) when Avo)/ s is small. For the disk of radius A, the stream function of the second pole found by making the substitutions for c, a, h, M, and p, indicated in (1) must be subtracted from (4) and (5). Thus clpu under the pole piece and between the pole piece and p = A become A 2a2 MA,,cocp sin y5[ (6) 1 2s c2p2 + A 4 — 2pcA2cos 4) A2 1 MiLvWCPa2 sin ck [ c2p2 ± A 4 2cpA 2cos 4) p2 ± c2 2s 2cp cos 4) 4). —
410
(7)
Figure 10.15 shows 4,plotted from the above equations for a single pole (a) and, by superposition, for two poles of opposite polarity (b), so that the currents from both poles flow in the same direction along the line between centers. In Fig. 10.15a and b, MAcos-1 is 10, c/a is 0.7, and A is 0.1 meter. The stream function in amperes has the value zero on the boundary and increases by steps of 0.1. The torque about p = 0 is calculated by 10.14 (13). The use of polar coordinates r, 0, coaxial with the pole applying the torque, simplifies integration. The needed relations between r, 0 and p, 4) appear in Fig.
392
§10.15
EDDY CURRENTS
(a)
10.15c and d.
(b)
For the single magnet these are
A' cf = — c R2 = p2 + (c')2 — 2pc' cos0 .,___ (c '— c)2 + r2 — 2r(c' — c) cos 0 p cos cp = c ± r cos 0,
p sin cp = r sin 0,
(8)
From (6) and 10.14 (16), the torque with one magnet is T = _i_m21,20.,a2s-i f Rca-2— c' li-2)p cos 4) + 2(e)2R-4p2 sine 0] dS s = _ MWwa2s lf a.1.1.[(ca-2 — CR-2)(C + r cos 0) oo + 2(c')2R-4r2 sine 0]r dr dO (9)
Integration of the R-4term by parts changes it to an R-2 term, so the 0 integrals become Dw 859.121, 859.122. The simple r integration then gives the torque for a single magnet to be T = im21,2wc2a27rs9 — A2a2(A2 — c2)-21
(10)
The two magnets of Fig. 10.15b give approximately double the torque of one. The additional AT due to the eddy currents from one pole flowing
§10.16
ZONAL EDDY CURRENTS IN SPHERICAL SHELL
393
under the other may be set up and calculated with the aid of Fig. 10.15d. In the given by (7) and 10.14 (16) for one gap, 4) is replaced by it — cto to put it in the coordinates of the other gap. The integral is put in terms of r, 0 by the geometrical relations of Fig. 10.15d. The resultant integrals, evaluated as before, give AT = IM2/.12wc2a2rs-1[1a2c-2— A2a2(A 2
c2)-2]
(11)
The total torque for Fig. 10.15b where magnets have opposite polarity is T' = 2(T + AT)
(12)
When the poles have the same polarity it is T" = 2(T — AT)
(13)
If the magnet gaps 2h and 2h' of the M and M' magnets are small compared with the distance of their edges from the boundary p = A, the primary eddy currents induced by each will be unaffected by the field of the other. There is, however, a slight distortion of the eddy currents on one side of the boundary by the magnetic fields of the eddy currents on the other side. This should have little effect on 43. However, the magnetic field of the disk eddy currents, except near its surface, cannot be found from the scalar potentials 10.14 (6) and 10.15 (3) as modified to include the image magnets. They may be found by 7.02 (5) or 7.14 (1), where the integrals are taken over the regions p < A and the ip and i4, used are derived from the stream function 10.14 (14) and its image or from 10.15 (6) and (7). Note that Eqs. (2) to (13) apply to small rotation speeds or highresistivity disks. This means one made of material of poor conductivity or one made very thin of good conductivity. Since the resistivity T of copper is 1.75 X 10_8 ohm-meters and Ay = 1.25 X 10-6, for a sheet 0.1 mm in thickness, 2644,, is 280 and w must be considerably less than this for (2.1) to hold accurately. The formulas of the last two articles should be applied with caution to electromagnets if the eddy currents are strong. These cannot be replaced by an equivalent current sheet as a permanent magnet can. The permeable core tends to short-circuit the magnetic fields of the eddy currents, which then produce a demagnetizing force that in extreme cases can greatly reduce the torque. 10.16. Zonal Eddy Currents in Spherical Shell.—Let us now consider the eddy currents in a thin spherical conducting shell when we have axial symmetry so that all the eddy currents flow in coaxial circles. Let the total vector potential be A' + A, that part due to the eddy currents being A. The electromotance e producing the current in the ring of width a de at 0 is induced by the change in the total flux through the ring. By 8.00 (1) and 7.08, we can write E in terms of A' + A and then relate it to
394
EDDY CURRENTS
§10.16
the current by Ohm's law 6.02 (1), giving = —
dN = dt
d [2ra sin 8(A',, ± A0)] — dt
lirassin a d0
Zia d0
so that, if the vector potentials are zero at 0 = 0, (1)
i4, -7t(A +114') = s
where i4, is the current density and s the area resistivity. Let the eddy currents in the shell be expressed as a series of zonal harmonics, the nth term being in. In 7.12 (2) and (5), we found a simple relation between in and the nth term in the expansion of the vector potential produced by these currents. The relation is
mua
(2)
(A 's")' = 2n + .
If A'4, is also expanded in spherical harmonics, we see by substituting (2) in (1) that the expansions of A'4, and A95 are related at the surface of the shell r = a by the equation
d IA, dt‘ n
+ An)_ '12n ± 1Ans
(3)
n
n
If, after the time t = 0, the inducing field A'd, is constant and if, at the time t = 0, the field of the eddy currents is known to be A4,
=
(4)
2CnAn
then, clearly, a solution of (3), showing how the eddy currents decay is, 2n+ lst
Ad, = ZGAne µ
(5)
n
Suppose that the vector potential of the inducing field can be written, on the surface of the sphere, in the form
Aye = ZCn An.(t)
(6)
The change in this field in the infinitesimal interval of time dr, at a time t, is given by
r before the present time
0A0, dr
at
n
aAns(t — r)
at
dr
§10.16
ZONAL EDDY CURRENTS IN SPHERICAL SHELL
395
The eddy currents set up in that interval dr exactly neutralized the change at the surface of the sphere at that time but have been dying out according to (5) since, so that the present effect is 2n-I-1,7 CnaAns( 1-)e — A" dr at
dA 08 = — n
The total vector potential of the eddy currents at the surface of the sphere at the present time due to all these past changes is, then, A08 = —
C11.1 0 n
(t
"
r) e 2n-I-1,
—
at
dr
(7)
Replacement of aAna/at by —.9Ana/ar and integration by parts with the aid of (6) give, for the total vector potential from all sources at r = a, Ark. + A',58 = +---
(2n
1)Cni Ans(t — r)e
_ 2 n +1sT
Ava CiT
(8)
n
In the steady-state a-c case A'4„ which may result from both external and internal sources, is given at r = a by (6) where An, = Pni (cos 0) cos wt
(9)
Integration of (7), with N substituted for (2n + 1) s/(i.coa), by Dw 860.80 and 860.90, gives for the eddy currents alone at r = a CO
A le = ZCn/1,(cos 0)(01/42
N 2)-1(N sin cot — w cos cot)
n=1 ao
= — ZC.P;;,(cos 0) cos On cos (cot + On)
(10)
n=1
where tan tkn is N/co. Thus the present eddy currents are in phase with those formed at a time 1//n/co before the present. The phase lag en is The value of Ad, that vanishes at r = 00 and gives (10) at r = a is the vector potential outside the sphere due to eddy currents alone. Thus
Ao =
—
a 11+1 Cn(—) Pnl(cos 0) cos en cos (cot — r
En)
(11)
n=1
tan en —
(2n
1)s
tiv aw
(12)
Inside the shell, the eddy current vector potential As must be everywhere finite, so (r/a)n replaces (a/r)n+1. For an external source, r enters A' as
396
EDDY CURRENTS
§10.17
(r/a)n. From (6), (9), and (10), the total internal field is, by Dw 401.03,
Ai+ A' = (1:•
— C.(OnP;,(cos 0) sin en sin (cot — en)
(13)
=1
This could also have been found directly from (6), (8), and (9). The ratio of the nth term in the harmonic expansion of the new amplitude to the corresponding term in the expansion of the old amplitude is
R. — IA'
± A/I =sin en
(14)
From (12), if s is very large, this is unity and the field is unchanged, but if the shield is a good conductor, i.e., if s is small compared with izyco, Ron is very small and the shielding is high. The eddy current density in the shell as given by (2) and (10) is 2n + 1 CnPnl(cos 0) cos en cos (wt — en) /Iva
i = (1:•
(15)
n
The instantaneous energy dissipation rate in an element of area dS of the shell is its dS from 6.03 (2). In the whole shell it is, writing u for cos 0, Sf i2 dS
2 = 2sraTi2 sin 0 dO = —2sra2 f +12, 1 du '
When (15) is squared and integrated from u = —1 to u = 1, all crossproduct terms vanish by 5.13 (2) leaving only a sum of integrals of squares. Thus each harmonic component in cos (cot — en) of i behaves as an independent circuit so that the mean rate of energy dissipation from 6.03 (2) is Sf i2 dS
+1 .2 +1 2 sra2f [Zin ] du = srct'V f 1 du
Integration by 5.231 (3) gives for the energy dissipation rate in the sphere P = 271-sA,72Zn(n + 1)(2n + 1)C,2, cost en
(16)
n=1
10.17. Spherical Shell in Alternating Field Solenoid Gap.—As an application of the formulas of the last article consider a spherical shell of radius a centered in a gap of width 2d in an infinite solenoid of radius c having a current density io cos wt amperes per meter, as shown in Fig. 10.17. The potential d444, due to a ring element of width dzo at z = zo may be written down by 7.13 (2) where I is replaced by icb dzo. Superimposing the results
§10.18
GENERAL EDDY CURRENTS IN SPHERICAL SHELL
397
FIG. 10.17.
for a ring at r = a, 0 = a and one at r = a, 0 = r — a eliminates even values of n and doubles the odd values so that, writing 2n + 1 for n, sin a (2n + 1)(2n + 2) a)
dA.0 = 1.44 dzo
a)PL±i(cos 0) cos cot
n =- 0
(1)
The n = 0 integration gives
fa
a-1sine a dzo = f
C 2(C 2 4)-1
2)-1 = 1 — cos dzo = 1 — d(c2 d (2)
For s > 1 a useful formula, verifiable by differentiation, is a—. sin
a 2 = c2
zZ ,
aP!(cos a) = [(s — 1)!]-1(-1)8+1
aa+ia
az80+11 s>1 a sin a = c,
a cos a = zo,
(3) (4)
The zo integration of (3) merely reduces the derivative order by one, and (3) expresses the rtsult in terms of a, so that = Ecvnir2n+ipi n+,(cos 0) cos cot,
Co = 4ihic6(1 — cos ()
n=0 = I.liob-2n[2a(2n + 1)(2n + 2)]-' sini3PL(cos0)
Cn
(5) (6)
The applied vector potential is then 10.16 (6) where r = R, so that A„B = PL+1(cos 8) cos cat, Cl = pvioR sine 40 C 2n+i = AvioR2n-Flb-2n sini3PL(cos 0)[2n(2n + 1)(2n + 2)]-1
(7)
(8)
Thus the eddy current density i and the energy dissipation are given by 10.16 (15) and (16) provided n is replaced by 2n + 1. 10.18. General Eddy Currents in Spherical Shell.—It was shown in 7.04 and 7.06 that the vector potential whose curl gives a magnetostatic B in spherical coordinates has only 0- and 0-components given by A„,,„ = V x r(Arn
Br—n—')S:(0 45)
(1)
398
EDDY CURRENTS
§10.18
In 7.11 the vector potential of an arbitrary current distribution in a spherical shell was expressed in terms of these components. These relations provide a basis for the solution of the general eddy current problem when a thin shell is placed in a time-varying field whose wave length is much longer than the shell diameter 2a and when the skin depth is much greater than the shell thickness. First the orthogonality relations of these solutions must be found. From 7.11 their form is m [ sin ml A,,,,, = 01 + 4:111 sin OFV(cos B) PZ(cos 0) sin m 4) sin 0 cos mcb
ml
—
(2)
The 4) integration of Anin • Apq over the unit sphere gives integrals of sin m4, sin p4, or cos m(11 cos p4, from 0 to 2r. From Dw 858.516 and 858.517 these vanish unless m = p in which case the result is 7. Thus
r
+1 _i r 2,r Am?, • Am, sin 0 d0 d4, = r f PZ(cos 0) • PQ (cos 0)sin 0 de 1. J i J.
(3)
From 5.232 (7) and 5.232 (8) this is zero if n / q, and if n = q, it is +1
J-1
2n
o
Amn • IL, sin 0 de (14) =
2rn(n + 1)(n + m)!
(2n + 1)(n — m)!
(4)
Let the total vector potential be A' + A, the eddy current part being A. Then from 10.00 (4), if i is surface current density and s is area resistivity, —d(A' + A) _ si dt
(5)
The relation between the current i and the vector potential it produces at r = a is found by a procedure like that in 7.11 and 7.12 to be
A.. =
Ala
im,. 2n + 1
(6)
The reasoning of Art. 10.16 starting with 10.16 (1) and (2) applies without change to the present case, if C,nn, Am, and A replace Cn, An, and A0, so that if the applied vector potential at the surface is
ZC,,,n A,,,,,,s(t)
A8 = 71.
(7)
m
then the total vector potential at the surface of the sphere at the present time due to all past fluctuations is
,
1-, mn
- a Amns(t — T)e_ [ ( 2n + 1) / (m,,a) 1 sr dT at
(8)
TORQUE ON SPINNING SPHERICAL SHELL
§10.19
399
10.19. Torque on Spinning Spherical Shell between Magnet Poles.— The theory of the last article yields the solution for the retarding torque on a spinning thin spherical conducting shell of radius a centered in a gap between the coaxial halves of an infinite cylindrical bar magnet having a uniform permanent magnetization M parallel to its axis. The edges of the plane magnet faces are at a distance b from the sphere center, where any face radius subtends an angle 13 as shown in Fig. 10.19. The vector potential for the equivalent current shell (see 9.07) is given by 10.17 (5) if M replaces Operations are simplified by having the harmonic axis and the rotation axis coincide. To accomplish this, observe from 7.06 (1) and (2) that the A' of 10.17 is given by
cb
A' =VxrW =VxrZC;r 2n-"P2n+i(COS
(1)
n=0
where the prime indicates the 10.17 coordinate. By 5.24 (5), taking 0 to be br, the expression for W referred to the new axis is 00
n
2(2n — 2m — 1)!Ir2n+17,9,--1 C'n (2n ± 2m ± 2) !! (cos 0) cos (2m + 1)4) (2) n=0 m=0
Now taking V x rW by 7.06 (1) and (2) gives Eqs. (4) and (5) for the vector potential referred to the rotation axis, in terms of C. = 2(2n — 2m — 1)!![(2n + 2m + 2)!!]-1C',
A; =
(2m + 1)C.r2n+1csc OPrn nV(cos 0) sin (2m + 1)cp
(3) (4)
n=0 m=0 n
A.; = —
c.r2.+1 sin 0(fninV)'(cos 0) cos (2m + 1)0
(5)
n=0 m=0
Substitution of O. + wt for puts in rotation. The Aos of the eddy currents is now found from 10.18 (8). Substitute t — T for r, differentiate with respect to t, put t = 0, and perform the T integration. Writing S for sin (2m + 1)0 and C for cos (2m + 1)0, the last factors in (4) and (5) are (2m
+ 1)cofo lsCcos (2m + 1)cor
S .
—c
sin (2m + l)cor i-[(2.-1-1) /AA= dr (6)
400
EDDY CURRENTS
§10.20
where the upper symbols refer to A 9and the lower to Ao. Integration by Dw 860.90 and Dw 860.80 gives the results for (A.v.8)0 and (A.,,z)0: 1)95 + 1//mn] (7) emn cos ikm.nr2n-"(2m + 1) csc OPM-N(cos 0) sin [(2m _Gan Cos4,mnr2n1-1 sin 0(P2V)'(cos 0) cos [(2m + 1)0 + 1,Gznn] (8)
where tan tkinn is (2n + 1)s/[(2m 1)Anaco)]. Clearly Am., with these components has the same orthogonal properties as 10.18 (2) since from 0 to 27 the integration of the ck factor products is independent of'/,nn. The torque on the sphere is found by using the i of 10.18 (6). The rate of energy dissipation is constant, as in 6.03 (2), so that T = Pco—' = co—'si • i dS = 27rws n=-0 m=0 a4n-I-4(2n
1)(2n + 2)(4n + 3)(2m + 1)2(2n + 2m + 2) Cl,„ [(2n 1)2s2 + (2m + 1)2/4o/201(2n — 2m) !
(9)
10.20. Eddy Currents in Thin Cylindrical Shell. When all the eddy currents in a thin infinite conducting cylinder of radius a parallel the axis, the vector potential of the field producing them also parallels the axis and the magnetic field is a two-dimensional one. In this case, as we saw in 7.25, the vector potential behaves as an electrostatic stream function so its value at any point represents the flux between this point and some fixed point. Let the total vector potential he A; + A,, that part due to the eddy currents being A,. Then the electromotance per unit length induced in a strip of width a dO at a, 0 by the change in total flux N linking unit length of this strip gives, by Ohm's law, 6.02 (1), —
dN E = -- = dt
d(A;
A,) dt
=
s . zza dB a dB
(1)
where iz is the current density and s the area resistivity. If i is written as a series of circular harmonics, then from 7.17 (1) and (4), there is a simple relation between the nth term of this expansion and the nth term of the vector potential produced by it. At the surface of the shell this relation is (2) (A n) p=a = 2µva1Z 2n If both A; and A, are expanded in circular harmonics, we see, by putting (2) in (1), that these expansions satisfy the relation
dt n
2ns An Aza
An ) = n
(3)
§10.20
EDDY CURRENTS IN THIN CYLINDRICAL SHELL
401
Except for a constant factor, this equation is identical with 10.16 (3), so that if the inducing field at p = a is given, as in 10.16 (6), by AL = ZCnAns(t) n=
(4)
then the eddy current potential is given, as in 10.16 (7), by CO
Azs =
a Ans(t — at
n
r)
e Ava
(5)
n=1
Replacement of
aAns/at
by — aAnda, and integration by parts give 4-
flC nf Ans(t — r)e
A;s = (92c
A=s
– 2n — A'a7dr
(6)
n=1
For a sinusoidal inducing field we have, as in 10.16 (9), A ns = cos (nO + on) cos cot
(7)
As in 10.16 (11) and 10.16 (12), the eddy currents alone give, when p > a, A0 =
—
kEenanp—n cos (nO + an) cos encos (wt — En)
(8)
n=1
tan
En = — 2nS(LJA,a)-1
(9)
When p < a, (p/a)n replaces (a/p)n and for external sources (see 10.14), Ai + A' = —kZen pna-n cos (nO
bn) sin En sin (wt
—
En)
(10)
n =1
The ratio of the nth harmonic with the shell to that without the shell is
±AEI
IA' I
= sin En
The same remarks apply to this as to 10.16 (14). For internal excitation the new and old amplitude ratio outside the shell is still given by (11). The eddy current density in the shell is, from (2) and (8), i = — k2(i.1„a)-1
nCn cos (nO + an)
n=1 power
cos En cos (wt
—
En)
(12)
EDDY CURRENTS
402
§10.21
From 10.02 (6), the average power consumption per unit length is
=
2r
o
si 2 do
When (12) is squared, the integration from 0 to 2r eliminates all the cross-product terms and leaves only the sum of integrals of cos' (nO (5„) which equal 7r by Dw 854.1. Thus each n term behaves as an independent circuit, and the rate of energy dissipation per unit length is P = 27r s(1.4,0- 1n2C7, cost
(13)
n=1
10.21. Eddy Currents in Rotating Finite Cylindrical Shell.—A device often used for the damping of torsional oscillations consists of a thin z
conducting cylinder of area resistivity s suspended in the gap between one or more transverse bar magnets and a soft iron external yoke which serves as a return path for the magnetic flux between poles as shown in Fig. 10.21. The treatment is rigorous, subject to the conditions of 10.00 and 10.09, for an infinite thin cylinder with a periodic arrangement of transverse uniformly magnetized bars so that the circles at z = 0, ± b, ± 2b, etc., are eddy current flow lines. It is nearly rigorous for a single section of length b as shown if the magnet poles are not too near the edge. It is pointed out in the next article that the damping can be worked out as soon as the solution for the torque on a uniformly rotating cylinder is known. It is convenient to use the scalar potential Sl throughout, and the procedure involves the same steps used in the preceding articles.
ROTATING FINITE CYLINDRICAL SHELL
§10.21
403
Suppose the stream function is expanded in the harmonic form =
C,n„ sin n— Irz sin (m4 n
6„,)
(1)
m
so that 43 = 0 at z = 0, z = b. From 7.01 (2), the line integral of H around a path from (00)p on the outer face around the edge to the point (0i)p on the inner face is 00 — 0, which equals 43, the total current flowing between P and the edge. One can see by inspection that the potentials gi and 0° are (see 5.292 and 5.324), Sti = rab-1ZZnC„,n./.(nrpb-1)K'„(nrab-1) sin (nrzb-1) sin (m4) +
nm (2) 120 = rab-1ZZnC„,„1:„(nrab-1)K„.(nrpb-1) sin (nrzb-1) sin (nick + Sm) nm (3) because if the difference Sti— 00 at p = a is taken and the modified functions eliminated by the Wronskian 5.32 (7), the result is (1). The rate of decay of these eddy currents is found from 10.00 (3), applied at p = a.
= -pv
ao
ap
_L H) ‘1)az
r a 24,
a COO'
Vx
=6'1 a' ao2
024,
al)
aZ2
k a az
(4)
= -p,,n272ab-2/:„(nrab-1)K:„(nrab-9C...
sin (nrzb-1) sin (m4,
3,„) (5)
As noted at the start of 10.09, B, has negligible effect on a thin shell, so substitution of (5) for B and (4) for V x i in 10.00 (3) gives
_aBrnn at
[1 + )2 ][i n (n7rb a) K4nrb a)]-'13nin Ilya nr a
(6)
The same law of decay applies to B and 0, so that [2.]t = Pm,, = [1 + m2b2(nra) ][p„a1:„(nrab-1)K:„(nrab-91-1
(7 )
(8)
Suppose the inducing potential at the cylinder surface is =
(9) n
m
Then the change in this potential in an interval dr at a time T before the present was compensated by the eddy currents induced at that time. The present set of eddy currents is the result of all past changes, each change having died out according to (7) so that the present potential of
404
EDDY CURRENTS
§10.21
the eddy currents is ''' aD.,nn(P,d),z,t— T) e_,,,,,,,,
at
dr
(10)
The form of Dm„(p,ck,z,t) for Fig. 10.21 must now be found. The bar is uniformly magnetized with intensity M (loop definition), but the ends are rounded so the normal M is M cos cpo where —a < 44 < +a. The pole strength in an area dS is then cM cos 4)0 dzo d4)0. The magnetic field is normal to the permeable boundary at p = d and to the plane z = b so St' vanishes there. Thus the problem is exactly that of a point charge in an earthed box which was solved in 5.324 (1) and (10). The following substitutions must be made: cM cos 4o for —q/E and zo, b, c, d for c, L, b , a. There is another positive element at —44, which by Dw 401.06 changes cos m(4) — q50) to 2 cos mq5 cos m4 0. There are negative elements at it — rpo and 7 + 00, which cancel the positive element terms when m is even and double them when m is odd. For the four elements the last term is then replaced by 4 cos (2m + 1)4) cos (2m + 1)45 o and the integration range is 0 to a. Thus from 5.324 (10), writing s for 2m + 1, c/12' —
8cM cos cti o dzod4)0 rb
n = 1 m =o
/3(nircb-1)M(mirb-1,d,p) I $(Tordb—')
sin (nrzob-1) sin (rorzb-1) cos sck cos s4)0 (11) li2(nrb—',d,p) = K3(nirdb-913(nrpb-1) — I s(nirdb-1)Ks(nirpb—') (12) Integrations with respect to zo and 4)0 give
foacos s4)0 cos 4)0 d40 = (4m)-1sin 2ma + (4m + 4)-1sin (2m + 2)a(13) f: sin (rorzob-1) dzo = b(mr)-1[cos (nirkb-1) — cos (nirhb-1)]
(14)
Thus, if s = 2m + 1, (9) may be written St' = 2cMir-2ZZF,,,,,R2(nrb-1,d,p) sin (nirzb-1) cos s(4) + wt)
(15)
n m
Fm„ --- I s
rock (nirc)( mrh\ cos — cos b b b) (m + 1) sin 2ma + m sin (2m + 2)a nm(m + 1)/s(nrdb-1)
(16)
The potential of the eddy currents is now given by (10). Substitute t — 7 for t, differentiate with respect to t, and put t = 0, so that the eddy
§10.23
TRANSIENT SHIELDING BY A THICK CYLINDRICAL SHELL 405
currents are given at the instant that cp = 0 splits the pole piece. The integral involved is then, if tan is written for P„,„ss-'co-1, scof sin (scb — swr)e -Pm•sIdT = — cos %Gm cos (s0
11.„0
(17)
This replaces cos 8(:15 + wt) in (15) for p > a to give S20 at t = 0. Note that for a perfectly conducting cylinder or for a high frequency, 1,1/„, is zero, so that S2 completely neutralizes St' outside the cylinder. At p = a on the inside surface C2, = —S20, so 4,is 2C20 at the surface and i is given by (4). The rate of energy dissipation is constant and equals
P= sfi•idS=
b 4safj oo():=‘ [t
as2 \2 dz dI
(18)
The integrations are straightforward and the torque is given by P/co. Thus writing out cos' &,,, gives finally
T
8c2.1112cos ir'ab
.EL,,,[14m+i(nirb-1,d,a)]
2[n2,2a2+ (2m + 1)2b2]
P;„,,s2 ± (2m + 1)2(.02
(19)
n=1 ,a=0
where F„,„ is given by (16), RL+1by (12), and P„,„ by (8). 10.22. Eddy Current Damping. Numerous calculations of eddy current torques and forces have been made in this chapter, especially in the case of conducting shells whose thickness is small compared with the skin depth at the frequencies involved. In every case where the linear velocity v or the angular velocity w is small, the forces and torques are proportional to v and co. There is nearly always, however, as in the last two articles, a term involving ii,v2w2a2 plus a s2 member in the denominator somewhere which, if i.c,coa is large enough compared with s, will eventually take charge. Only the case where co is small or s large gives a simple damped solution. Then the differential equation is that of an R, L, C circuit where, in the angular case, R is the coefficient of co in the torque expression such as 10.19 (10) or 10.20 (19), L is the moment of inertia I, and C-1is the torque constant of the supporting wire or spring. Formulas for the various types of damping can be found in any circuit theory text. 10.23. Transient Shielding by a Thick Cylindrical Shell. Sinusoidal longitudinal eddy currents in a thick cylindrical shell and arbitrarily varying longitudinal currents in a thin shell were treated in Arts. 10.03 and 10.20, respectively. As a relatively simple example involving both cross and longitudinal components, consider an infinite cylinder of permeability u, conductivity y, internal radius a, and external radius b which makes an angle 47r — a with a uniform field of induction B. Find the field inside at a time t after the external field is removed or estab—
—
406
EDDY CURRENTS
§10.23
lished. Examination of 7.05 (5) shows that suitable solutions for the static vector potential when p > b, b > p > a, and a > p are, respectively, (1) E[4:14 sin a(p — Cop') + k cos a(p — qp-') sin q5] (2) sin sin 0] k cos a(D' o p-1) o p — E' a(Dop — EoP-1) (A2)0 = B[4 B[4 sin k cos aF/ aF o p + (243)o = 013 sin ct] (3)
(A1)0
=
These give the prescribed field at p = 00 , and the field is finite at p = 0. The values of the constants needed to match Al and (B x p)/ p. at p = a and at p = b are, if 13 is written for (I.L — p,,)/(p ki„) and G for b2/(b2 — 02a2), Co = µT 1(4 — ilv)(b2 — a2), Do = Eo = — A0b2, Fo = 1 (4) )G, E', 3(a2 — b2)G, D', = (1 + = 3 = 0(1 + 13)a2G, = (1 — 13 2)G (5)
Ca
After t = 0, the same solutions of 10.00 (8) are valid where y = 0 except that in A lthe p-terms disappear because B = 0 at p = co . For a < p < b, solutions are needed where Az has the factor sin 0 and A4, in independent of 0. As in 10.08 we try sums of exponentially damped solutions so that A2 = BZ[(1)1?(p)e—P.'
kP(p) sin 0e-m]
(6)
For A z10.00 (8) is a scalar equation from which sin 0e-g.' divides out and which, upon substitution of v2 = iryq 8 p2, yields 5.291 (3) with n = 1. For A4, 10.00 (8) becomes 10.05 (3) which, upon inserting — 4/3-2for V24 from 10.05 (5), dividing out e-P.' and writing v2= wyp8p2 gives again 5.291 (3) with n = 1. Thus, when t > 0, (1), (2) and (3) are replaced by Al = BZ[4 sin aC,p-ie-P•g
k
cos «C;p-i sin 0e-m]
A2 = BZ{(1) sin aD3R1[(WYP8)1 p]e-P•' A3 = B[(1) sin aF,pe-P.'
k
k cos agP1Rryq8)1p] sin
(7) 0e-g.g) (8)
cos aF's p sin 4,e-9•1]
(9)
8
where R1 and P1are Bessel functions. At p = b, equate Al to A2, write k, for (ryps)i, 1, for (iryq,)i, and cancel sin 0 out of Az. Thus we get (1)C3
kC; = 4thD5R1(ksb)
kbgP 1(18 b)
(10)
Equating tangential components of A-1/3 = p 1v x A at p = b, canceling sin 4,out of the 0-component, and multiplying through by b2give —
4)11C", = kilybD,[kablVi(k.b)
Ri(k.b)] — 4)14-18b2D:P',(15b)
(11)
§10.23
TRANSIENT SHIELDING BY A THICK CYLINDRICAL SHELL 407
The k-term is zero so, from 5.294, Ro(ksb) = 0 and, from 5.293 (9), R1(k3b) = 47r(kab)-1,
Ri(ksp) = o(ksb)Ji(k.p) — o(ksb) Y i(ksp) (12)
Similarly the boundary conditions at p = a give, writing out Pi(lsa), 43Fsa
kF;a = 43D3R1(ksa)
I7].(/3a)]
kg[E'sJ i(lsa)
kW's — 444F; = kiinksDsRo(ksa) — itiiinIsD;P;(13a)
(13) (14)
Eliminating Fs, Ds, F;, and D; from these four component equations gives ksaRo(ksa) — 2 (PL,T1A)Ri(k.a) = ksaRi(kaa) + (1 — 2A,71A)R1(ksa) = 0 (15) i(lsa) AP1(/3b ) = 0 = p„lsal'i (lsa) — (16) iA3/3bPi(i3b)
Only values of p3 = 14(1.1-y) which satisfy (15) can be used. Insertion of Yo(k8b) from (15) into Ri(ksa) and application of 5.293 (9) give Ri(ksa) = 2113J0(ksb)/1--1[A3k3aJo(k3a) — 2AJI(k3a)]-1 (17) If Pi and P iare expressed in terms of Po and P2 by 5.294 (2) and (3) and if 0 replaces (p. — p,,)/(µ + A„), (16) may be rearranged to give — Es
0Yo(i3a)
Y2(/sa)
Yo(Lb)
0,10(18a)
J2(13a)
Jo(lsb)
0Y2(/,,b) 0J2(ta b)
(18)
Only values of qs = l;/(µy) satisfying the second equality can be used. Insertion of E; into Pi(lsa) and Pi(tsb) and application of the recurrence formulas of 5.294 and of 5.293 (9) give Piqsa) —
2(1 — 0) —2(1 — 0) Pi(isb) = irtaa[0Jo(La) ± .12(4a)] 71.b[Jo(lsb) 0,12(18b)]
(19)
It remains to determine Ds and D; so that, when t = 0, (8) agrees with (2) when a < p < b or so that, dividing common factors out of each component, 41)( D oP — Eop— + k (D'op — girl) = Zl(I)DaR i(ksp) kD;P ( 4)1
(20)
3 Take the scalar product of both sides by P[43Ri(knP) kPi(lnp)] dp and integrate from a to b or subtract the 0 to a from the 0 to b integral. By (11), (15), (16), and 5.296 (3) only the nth term on the right survives, and for A4, its value is given by 5.297 (6) with B = 1 — 21A;Ii.I. Integrate the left side by 5.294 (7) and (8), solve for Dn, and from (13) F„ is Fn, —
/.1. 471.13Ri(kna) k!a2g)72[Ri(kna)]2 )
kna{41.1,2, — (41.42—
(21)
408
EDDY CURRENTS
§10.23
For Az 5.297 (6) with B = kcip. gives the right side. Integrate the left side by 5.294 (7) and 5.294 (8), solve for and from (5) F',, becomes Fn
41.4.4,1)(1 [Pl(inb)]2
(11,2,1V)2+ 142—
n
13)Pi(lnb) (11:0ba 2 + A2 — AMPI(/.02
—
(22)
The total field inside at any time is given by (9), (21), (22), (17), and (19) where k,, and ln are determined by (15) and (18). If there is no field when t < 0 but at t = 0 the field B is established, then it is clear from Eqs. (3) and (9) that the vector potential A; inside is
p < a,
A', =
(A3)0 — A3
(23)
These solutions are obtained by a different method in Phil. Mag., Vol. 29, page 18, 1940. The field, when B varies arbitrarily with time, can be found from these results by methods similar to those used in preceding articles. Problems Problems marked C in the following group are taken from the Cambridge examinations as reprinted by Jeans, with the permission of the Cambridge University Press. 1C. An infinite iron plate is bounded by the parallel planes x = h, x = —h; wire is wound uniformly round the plate, the layers of wire being parallel to the axis of y. If an alternating current is sent through the wire, producing outside the plate a magnetic force Ho cos pt parallel to z, prove that H, the magnetic force in the plate at a distance x from the center, will be given by H = Ho tan —
(cosh 2mx + cos 2mx)i
cosh 2mh + cos 2mh
cos (pt /3)
— sinh m(h x) sin m(h — x) — sinh m(h — x) sin m(h x) cosh m(h x) cos m(h — x) + cosh m(h — x) cos m(h x)
where m' = hip/r. Discuss the special cases of mh small and mh large. 2. All points of a circular loop of wire are at a distance c from the center of a ball of radius a, permeability 12, and resistivity T, and any radius of the loop subtends an angle a at the center. A current I cos wt flows in the loop. Show that the vector potential inside the sphere is the real part of 2n + 1) (a y Anh,+1[(ip)ir]Pni (cos 0)eiwi n(n+1 C
aaociiI sin a 274 ti =
where An
=
—
ao(jp)ictIo_i[(jp)ia]}-iP1 (cos a) and p = aw/T
3. A sphere of radius a, permeability and resistivity r lies in a region where, at t = 0, a field is established that, but for the sphere, would be of uniform induction B.
PROBLEMS Show in the notation of 10.08 that, when t Ao =
0, the potentials are
- m120a7.z 3 _ _ -Br + B Bd. 2e q.t] sin 0 2 (AL + 20 2
[ 3/2Br 2(.1. 2bt„)
Ai =
409
ZAJ-1,11(k,,r)e-g•i] sin 0
4. An infinite circular cylinder of radius a, permeabil ty it, and resistivity 7 is wound with wire carrying an alternating current. If H0 at the surface is uniform and of magnitude Ho cos wt, show that the magnetic field at any point inside is given by 10.04 (4) and (5) with B, and Bo substituted for i„ and io. Find the density and direction of the eddy currents by 10.00 (5). 5. A circular loop of wire is located at r = a, 0 = a and a spherical conducting shell of area resistivity s at r = b. A current is built up in the loop according to the equation i = 1(1 - e-" L). If the sphere and loop do not move appreciably before the eddy currents have decayed, show that the momentum imparted is 71-4b/ 2 x.■ sin' a by-,(2n +1)01, + 2m„bR i P ( cos a)P41(cos a) 2s %-/ (n + 1)(2n + 1)(a (2n + 1) sL
n=1
6. A small loop of wire of moment M cos wt lies above an infinite thin plane sheet of area resistivity s, with its axis perpendicular to the sheet. Show that, if 1 >> orro where sro >> Atr, the magnetic induction on the far side of the sheet is
Br =
AvM[ 277.3
cos 0 cos ca ±
Avcor
4s
sin sin 0.4 ,
Be =
479.3
sin 0 cos cot
7. Show, in the preceding case, that the current density igs at a distance p from the axis and the average rate of dissipation of power are given, approximately, by
g,copM sin wt 47s(p2
c2)1
Atm 2,, 2
and
64r SC 2
8. The magnetic moment of a small loop carrying current falls linearly from the value M at t = 0 to 0 at t = T. If the loop is at a distance c above an infinite thin sheet of area resistivity s with its axis normal to it, show that the additional B on the axis at a distance z above the sheet due to its presence is AB -
/42„Mt[A,,(c z) st] 271-T(c 2st]2 z) z)2[th,o(c
when 0 < t < T and that, when t > T, it is
-
s(2t - T)] z) 4M[A,,(c z) 27[µ,,(c 2s(t - T)Pkiv(c z)
2 s tp
9. A magnetic loop dipole of moment M is parallel to and at a distance c from an infinite thin sheet of area resistivity s and is moved parallel to the sheet and normal to its axis with a uniform velocity v. Show that in a steady state the eddy current retarding force is br-lA2,,M2vs(2c)-4[(4s2 2s]-2. /.42„v 2)1 10. A loop dipole of moment M moves with a uniform velocity v in a straight line at a distance c from an infinite plane sheet of area resistivity s. If M parallels the
410
EDDY CURRENTS
sheet and makes an angle 4, with v, show that the retarding force is F = [1 +
4s cost c6
(140
4s 2)}
F
where Fo is the retarding force in the last problem. 11C. A slowly alternating current I cos wt is traversing a small circular coil whose magnetic moment for unit current is M. A thin spherical shell, of radius a and specific resistance s, has its center on the axis of the coil at a distance f from the center of the coil. Show that the currents in the shell form circles round the axis of the coil and that the strength of the current in any circle whose radius subtends an angle a at the center is, if tan c„ = — (2n + 1)5'/(i.i,,coa), M/( ztirPa )-1Z (2n + 1) (t'a).P;,(cos a) cos en cos (wt — en)
12. A thin spherical shell of area resistivity s and radius a rotates with a uniform angular velocity w about an axis normal to a uniform field of induction B. Show that the retarding torque is 67B 2,,,a4 (9,2 + ga 2,42) -1. 13. Show that the average power dissipation in the shell of the last problem when placed in the field B cos wt is 37sB2w2a4(9s2 +1,2va1/4,2)-1. 14. Show that the field inside the shell in the last problem is, if tan E = —35'/(A„coa), 3B0(902
A2va2c4,2)—i
sin (wt — e)
15. A thin circular cylinder of area resistivity s and radius a is placed in a field E = —2s / (Anwa), show that the field inside is
B cos wt normal to its axis. If tan 2sB(4s2
w2ga2)-i sin (wt — e)
16. Show that the retarding torque per unit length, when the cylinder of the last problem is spinning on its axis with an angular velocity w in a uniform magnetic field B normal to its axis, is 471-cosa3B2(4s2 17. The cylindrical coordinates of the wires of a bifilar circuit which carry the outgoing and incoming currents are, respectively, b, 0 and b, Ir. When the current in the circuit is I cos 0.4, show that the vector potential at the point p, 0 outside a thin tube p = a of area resistivity s is 2i.t,,shr-1Z[4(2n
(p- lb ) 2n+1 cos
1)25'2
(2n + 1)0 sin (cot — €,,)
where tan €,, = —2(2n + 1)s(tt,,coa)-1and the summation is from 0 to co, a > b. 18. A thin oblate spheroidal shell g- = ro has a variable thickness such that we may consider the area resistivity to be s = h2so• At the time t = 0, a uniform magnetic field of induction B is suddenly established parallel to the axis. Show that if N = 2[(1 a )NQI(jg-0)]-1the eddy current density is given by 24,
BN (1 + 31)(1 — 2 (e + 30 )1
_Ariot
e
19. A long solid cylinder of radius a = (x2 + y2)i, permeability A, and resistivity is placed in an x-directed alternating magnetic field Bei•'t (real part). Show that the z-directed vector potentials A i and A 0 inside and outside the cylinder are
T
Ai = 0/ i[(iP)1P]ei"' sin 41 (real part),
A. = B(p
n p-i)ei40,sin (real part)
PROBLEMS
411
where, writing f o and /2 for Io[(jp)1a] and /2[(iP)Ia], =
4/.4B
n - (L
and
(ip)11(m, + AO/0 - (12 -
-(A
+120/0 -
+ A')/2,2
- Arv)/2
20. Show that the mean power dissipated per unit length in the last problem is Llirulia 2B2[bero(Pla)bei2(pla) — beio(p1a)ber2(p1a)1
P— [(m
12,)bero
(IA — p.,)ber212+ [(A + ti,,)beto -I-(A — ti,)beiz12
where — (ber2x jbei2x) is written for /2(j1x). The arguments in numerator and denominator are the same. If p is small, show that this reduces to P = 1,m 2w 2a4B27( A Av)-2
21. A long solid cylinder of radius a, permeability 12, and conductivity y is rotated about its axis with a uniform angular velocity w in a magnetic field B normal to this axis. Show that the retarding torque per unit length, due to eddy currents, is 871.,a 213 2(bero(P1a)bei2(pla) — beio(p1a)ber2(0a)]
T—
+
[Cu
— I.1,,)ber212
[Cu
/4)beio
— 1.1.)bet212
The arguments in numerator and denominator are the same, and p = 7µw. 22. A loop of radius a is coaxial with and at a distance d from the bottom of a cylindrical box of radius b and height c. If the skin effect is so large that B is nearly tangent to the walls, show that the decrease in self-inductance is AL = j
M 1„
47rma2 Z[Ii(rorac-1)]2[c/i(n7rbc-1)]-1K1(nirbc-1) sin 2 (iordc-1) n=0
n= —
M = 2µa [k,7'[(1— 114)K1 — E1] — (1 — 3,!)k-'[(1 — ik 2)K —
n2c2)-I. where the modulus k1 is a[a2 (nc — d)2]-1and k is a(a2 23. In problem 22 the skin depth is .5, the conductivity y, and the current I. Show from 10.02 (8) that if k„, is chosen so that Ji(k,,,b) is zero, the power dissipation is P=
[ /1(nira/c) sin (nrd/c)]2
27/ 2{a2 -yo bcn _ i
I i(nirb/c) -2 b
=1
[ Ji(k,„a)]2[sinh2 k„,(c — d) sinh2(k„,c1)1 } [Jo(k mb) sinh (k„,c)] 2
24. A thin disk of radius a and area resistivity s is placed in a region where, before its insertion, there was a uniform field of induction Bo cos wt normal to it. Find the secondary currents from the vector potential of the primary eddy currents induced by Bo cos wt. Thus show that the total eddy current density is WB 0 {
— p sin (0t
2s
+ 6n p s[(a
p) (2 p 2 a 2)K
+
(a
+
p)(2p2 — a2 )E] cos wt + • • •
p)-and terms in (6.4La/s) 2are neglected. where the modulus of K and E is 2 (ap)1(a Note that at 60 cycles this is good for 0.1 mm copper sheet. 25. If in the last problem the frequency is so high that the skin depth 3 is much less than the disk thickness so that B can be taken tangent to its surface, show that the
412
EDDY CURRENTS
eddy current density, including both faces, at a distance p from the center is -
srBp2 ma(a2 - 1)2)1
This icb is completely rigorous for a perfectly conducting disk. 26. The eddy current density in an infinitely long cylindrical shell of radius a, area resistivity s, and thickness much less than skin depth is (i0)0 at t = 0 and i, at t = T. If (4)0 can be written as a Fourier integral, show that /95 =
(4)0 = f0 *(k) cos kz dk,
f
4/(k) cos kze-DIal,(WICI(ka)] 4r dk
27. The vector potential of the inducing field at the shell's surface in problem 26 is (A;),-a = -
1.(k, a, t) cos kz dk
Show by the reasoning of 10.16 and the last problem that, due to eddy currents, f .0 a -4,(k, a, t - r)e-Ip.ar ickoic,(kanlsr cos kz dk dr
(A0 )5=. = -
at
28. A loop of radius b carries a current I cos cot and is coaxial with a very long shell at p = a of area resistivity s and thickness much less than the skin depth. Show that the shell increases the loop's resistance and inductance by amounts
AR = 214b2wf
sin cos 0[Ii(kb)] 21(1(ka)[Ii(ka)]-1dk
0
AL = -2i.Lb 2 f
0
sine 0[Ii(kb)PKi(ka)[Ii(ka)]-' dk
if b < a. If b > a, K and I are interchanged and tan 0 = 5-1cogaIi(ka)Ki(ka). 29. From problem 40, Chap. VII, and 7.04 show that the external self-inductance of the torus formed by rotating a circle of radius b about a coplanar line c meters from its center at high frequencies where there is a strong skin effect is
llrr
L = 2pc cos a COSla[(1 - 1- k)E -
2
2
-2 n =1
(-1)nC_4(csc a)
(4n - 1)(4n + 1)(ti(cse
a)
where sin a = b/c and the modulus k of K and E is given by k 2 =cos a sect 30. A wire loop of radius a is coaxial with and outside an infinite conducting cylinder of radius b and carries an alternating current I of such high frequency that the eddy current density in the cylinder is confined to its skin. Show with the aid of problems 26 and 28 of Chap. VII that this density is f 'ICI(ka)
= bar
Ki(kb)
cos kz dk
31. In the foregoing problem show with the aid of 8.06 (1) that the presence of the cylinder decreases the self-inductance of the loop by an amount = 2pc12.10.* Ii(kb)K1(ka)Ki-1(kb) dk
PROBLEMS
413
Show that the eddy current losses increase the resistance of the loop by an amount 4a2f Ki(ka)] 2 R=— -yab 0 K1(kb) dk where -y is the conductivity and B the skin depth for the cylinder material. 32. Show from HMF 9.6.25, 12.2.3, and 12.2.4 that the integrals of 10.12 (5) are X -K i(Xp)
cos Au du
and
f (P 2 u 2)1
7rX [I i(Xp) 2p
Li(XP)!
A
csin Au du
f 0 (p 2
P
2l 2)1
Note that Ki(z) and /1(z) - Li(z) are both tabulated in HMF. If Ac < 1, then cos Au and sin Au may be expanded in series and the integrals done. If (Ac) 2 is neglected compared with 1, show that in 10.12 (5) becomes, if II. is 0 - Ac. CrlilvM {[
47rs
C
XKI(Xp)
p(p2
C2)i
sin
7rX (—[/1 (XP ) 2
LI(Xp)]
Xp (p 2
c2)1
cos Vr}
33. Solve the shielding problem of 10.13 exactly as follows: From Watson, page 148, and HMF 15.4.14 and 8.2.5 establish the formula = Ji (az)Ji (bz)
as + b2)( z)2.+2 (-1)m(a 2- b2),.+1pi ( = Lj C ,,z 2m+2 (m 1)!(m 2)! m+1 a2 - b2 2 m=0 m=0
Substitute for Ji(az)Ji(bz) in the integral of 10.13 (2), integrate the result by IT 1(11), page 135, and evaluate Ei( -jwc/X) by HTF II (4), page 145. Replace wc/X = luvwc/s by p and thus show that
Aof _
_Avai
Cm (7 p12m.+3{
2
2m+3
en' (Ci p
(s -
j si p) s =1
For Ci and si see 10.11 (5). 34. It is desired to calculate the induction heating of a solid sphere of radius a, permeability pc, and resistivity r at the center of a cylindrical current sheet of radius c and length 2d whose current density is io cos cat amperes per meter. From 7.15 (7) and 7.02 (8) the vector potential inside an infinite solenoid with opposite current density is - luior/31(cos 8) cos wt. Superimpose on this the potential of 10.17 (6) so as to wipe out the current sheet except for a length 2d. Thus show that the potential inside the sheet is the real part of
Co = - 111,,i0 cos 0,
A.95 = C,t7-2n+2P1,, +1(cos 6)0c" n =0 C. = Atsicsd-2n[2n(2n + 1)(2n + 2)1-1sin OPL, (nos 13)
where tan 0 is c/d. Now apply the theory as in 10.06, and show that the vector potential inside the sphere is, writing P for (jp) ,
A ILI 2n (15r)P2n +1 (cos 6)
A 56i = n=0
A. = Kmai(4n
3)a2n+2C4[(K„, - 1)(2n
1)/2n+4 (Pa)
(Pa/2ct+I(Pa)1-1
414
EDDY CURRENTS
3 Now show from 10.07 and 5.231 (1) that the power absorbed by the sphere is
P = irrpay-2
(2n + 1)(2n + 2)(4n + 3) -4,,An(Pi2n÷112n+1 Pl2n+112.+I)
n=0
where 1„, and pm are written for 1„,(15a) and I,,, (Pa) and K,,, is ii/tc„. 35. A solid sphere of radius a, permeability A, and resistivity r is spinning between the magnet poles as shown in Fig. 10.19. Show that the interior and exterior vector potentials are given by 10.19 (4) and (5) provided that +,,t1 and
Amnr
(C,,,,,r 2'+'
15.nr-2')02m+ + .t]
respectively, replace C„,„r2"1cos [(2m + 1)cis + cd]. Show by using the boundary conditions of 10.06 and writing K„, for pip, that ii,„„ is given by Amn =
(4n +3)K,,,a2n÷1C,,,„ (2n + 1)(K,,, — 1).12,,+1
(jP)4(1/2,,+1
where 12+1is written for 72,,+1[(jP)la]and C,,,,, is defined in 10.19 (3). Find the power dissipation and show that the retarding torque is n 7(.°Ct
n=0 m=0
(2n + 1)(2n + 2)(2n + 2m 2)! [( (4n + 3)(2n — 2m)!
.
if
,
jp),./ 2n-Fli 21,44 (7P)112n+i12n+1]
References H.: "Electrical and Optical Wave Motion," Cambridge, 1915. Discusses boundary conditions, gives solutions of wave equations and an extensive list of references. FRENKEL, J.: "Lehrbuch der Elektrodynamik," Vol II, Springer, Berlin, 1928. Gives vector potential treatment of eddy currents. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XV, Berlin, 1927. Gives surprisingly meager treatment. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Follows Maxwell. MACDONALD, H. M.: "Electromagnetism," Bell, 1934. Uses long notation but gives rigorous treatment of eddy currents in solids. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Treats current sheets. MOULLIN, E. B.: "Principles of Electromagnetism," Oxford, 1932. Gives simple approximate treatment of eddy currents. OLLENDORF, F.: "Die Grundlagen der Hochfrequenztechnik," Springer, Berlin, 1926. Treats a-c resistance of wires and coils. RUSSELL, A.: "Alternating Currents," Cambridge, 1914. Treats manipulation of ber and bei functions. SCHELKUNOFF, S. A.: "Electromagnetic Waves," Van Nostrand, 1943. Treats chiefly very high frequency eddy currents. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Treats eddy currents associated with electromagnetic waves. BATEMAN,
CHAPTER XI PLANE ELECTROMAGNETIC WAVES 11.00. Maxwell's Field Equations.—In a region whereµ and e are continuous and in which there may be an electric charge density together with electric convection or conduction currents, we shall assume that the electric and magnetic quantities are connected by the equations B . OD + pv VxH=Vx— =1-1OB VxE = — — at V • D= p V • B= 0
(1) (2)
(3) (4)
Without the last terms the first of these equations expresses Ampere's law 7.01 (3), the second generalizes Faraday's law of induction 8.00 (3), and the third and fourth are 3.02 (1) and 7.01 (1), respectively. The D term in the first equation, introduced by Maxwell, is new and, when currents and charges are absent, completes the symmetry of the group and provides a companion equation for (2). It asserts that a change in the electric flux through a closed curve in space produces a magnetomotance around the curve; just as Faraday's law of induction states that a change in the magnetic flux through a closed curve produces an electromotance around it. Maxwell apparently thought of the magnetic field as being caused by an actual displacement of electric charge, but such a picture is not needed to justify (1). These equations are justified by the far-reaching validity of the conclusions based on them. They may be only approximately true, but for the present precision of measurement we may consider them exact. If charge is conserved, the equation of continuity, 6.00 (3), holds so that
ap v•i+— =o at
(5)
In isotropic bodies, i and E are connected by Ohm's law, 6.02 (3), which is E =ri
or
i = ^yE
(6)
In anisotropic bodies, with a proper choice of axes, 6.23 (2) gives 6.17 i=
kl'sEs
415
(7 )
416
PLANE ELECTROMAGNETIC WAVES
§11.01
The relations between D and E for isotropic bodies [1.13 (1)] or for anisotropic bodies with proper choice of axes [1.19 (7)] are or
D = eE
D = iel Ez je 2E,
le€3Es
(8)
In (6), (7), and (8) µ was taken independent of field strength so these equations cannot be used for ferromagnetic materials at some frequencies. 11.01. Propagation Equation. Dynamic Potentials. Gauges. Hertz Vector.—If, in the vector formula V x V x V = V(V • V) — 02 V, B or E replaces V, 11.00 (1) or (2) replaces V x B or V x E, zero replaces V • B or V • E, and then 11.00 (2) or (1) replaces the remaining V x E or V x B and yE from 11.00 (6) replaces i, the resultant propagation equations are
art v2B =µy at +
a2B at,
or
a2E aE v2E = wY at +12' at,
(1)
These hold in charge-free space. The first term on the right gives energy dissipation by heat and a wave damping that is absent when 7 is zero. The static scalar and vector potentials may be redefined to include rapidly fluctuating fields. The vector potential whose curl is B must reduce for constant fields to the magnetostatic potential. Replacing B in 11.00 (2) by V x A and changing the differentiation order gives B= VxA
V x E = —V x
aA
at
(2)
Integration of V x E removes the curl but adds an integration constant with zero curl which must be the gradient of a scalar. Thus
aA
E = -at
— v,fr
(3)
The potentials for a specified E and B are not unique. If A is solenoidal, then V • E along with 11.00 (3) and (5) yields Poisson's equation: '7211. =
—
p/e =
Pc' • i dt/€
(4)
This depends only on the lamellar part of i. Also, when V • A is zero, writing V x A for B in 11.00 (1) and using the above formula yield V xVx
A -= +1.4,3E v2A
at =
Let E, and is be the solenoidal parts of (3) and of the source i, and let 7E8 be the current induced in the medium by E8. Then from (3)
v2A —
aA cat
—
ate = mi„
(5)
§11.01
PROPAGATION EQUATION
417
Clearly for static fields, this becomes 7.02 (2). The potential choice in which A derives from the solenoidal part of i and ,k from the lamellar part is called the Coulomb gauge. When time-dependent, only A propagates and 1G gives a localized field in the same time phase everywhere. This A comes from a scalar solution W of the propagation equation by using V x uW or V x (u x VW), a method that appeared in 10.01 and is applied later in Chaps. XII and XIII to wave guides and cavity calculations. Another and more usual choice of potential is the Lorentz gauge, which requires that both A and 1t. have the same propagation equations as E and B. In a region of conductivity y with no sources, these equations are (3A a2A v2A =µy at + PE ate
alp °2% =µY at + PE W
(6)
This gives another relation between A and ik for, if p is zero, taking the divergence of both sides of (3) gives zero for V • E by 11.00 (3), so that
av • A at
v24, —
Comparison of this with (6) shows that for V21,1/ to agree requires that
aip v • A— — A-4 — E 797
(7)
To prove this also gives VA in (6), take the gradient of both sides of (7), expand VV • A by the initial vector relation of this article, write V x B in terms of E by 11.00 (1), (6), and (8), and eliminate V1,t by (3). Cancellation of the E terms gives v2A as in (6). The whole electromagnetic field may be represented by a single vector Z, the Hertz vector, from which A and 1,1. can be obtained by the equations
az t,
A = A-yZ + ALE 6- ---
= —v • z
(8)
These equations satisfy (7) and also (3) if we take E = V (V • Z) — V 2Z = V x (V x Z)
az a2z v2z = 1.4-y— at ± E a te
The magnetic induction is given by (2) and (8) in terms of Z. B = iryV x Z
1.4€
a(v x z) at
(9) (10)
418
PLANE ELECTROMAGNETIC WAVES
§11.02
Equation (10) contains all the properties of the electromagnetic wave. In a lossless medium the fields of (9) and (11) may be written
E=VxVxZe and
H=
Ea (v
x ze) at
(12)
This Ze is the electric Hertz vector. If 11.00 (1) and (2), in a medium without charges or currents, are put in terms of H and E, then the same two equations result if E and H as well as mu and E are interchanged. Making the same interchanges in (12) gives a new Hertz vector Z„„ thus
H=VxVxZ,„
and
E—
/la (v
x Z,, )
at
(13)
This is often called the magnetic Hertz vector and is sometimes more convenient than the Coulomb gauge potentials in solving wave-guide problems. 11.02. Poynting's Vector.—Multiplication of 11.00 (1) by —E and of 11.00 (2) by B/µ and addition of the results give
B B ap B a s — • (V x E) — E • (V x— = —i • E — E • w — • — a 1.4
(1)
Integration of this over any volume v gives by Gauss's theorem, 3.00 (2),
fv
[i. •(vx
E) — E • (v x
dv = fvV • (E x — au ) dv = isn • (E x — 12 ) dS
for the left side. The right side remains a\\volumeintegral so that
— f n • (E x — B)dS = [i • E
a (D • E B)1 dv at 2 2i.L
(2)
By 6.03 (2) and Ohm's law, the first term on the right represents the electric power absorbed as heat in v. By 2.08 (2) and 8.02 (3), the second and third term represent the rate of change of the electric and magnetic field energies in v. If all these terms are positive and if the law of the conservation of energy holds, then v absorbs power from outside, and the rate of flow of energy through its surface must be given by the left side of (2). Thus the outward normal component of the vector H
=
ExB
(3)
when integrated over a closed surface, represents the rate of flow of energy outward through that surface. The vector H is called Poynting's vector We should note that a meaning has been given only to the surface integral of this vector over a closed surface. We have not shown that H can represent the rate of flow of energy through any element of surface.
§11.03
PLANE WAVES IN DIELECTRIC INSULATOR
419
11.03. Plane Waves in Homogeneous Uncharged Dielectric Insulator. An electromagnetic disturbance is a plane wave when the instantaneous values of B, E,1., A, and Z are constant in phase over any plane parallel to a fixed plane. These planes are called the wave fronts and their normal, given by the unit vector n, is the wave normal. In the dielectric where y is zero so that the first term on the right of 11.01 (11) is missing, differentiation shows that, if v = (AE)-1, a general solution is Z = f i(n • r — vt) ± f 2(n • r + vt) = fiRn • r + ye) — v(t + t')] + f2[(n • r — vt') + v(t + t')]
(1)
so that fl has the same value at the point n • r + vt' at the time t + t' as at the point n • r at the time t. Thus it represents a wave moving in the n-direction with a velocity v. Similarly, 12represents a wave of the same velocity in the — n-direction. The vacuum velocity of an electromagnetic wave is c = (1.1„E,,)-1and equals 3 X 108meters per second. The ratio of c to the velocity v in a medium is its index of refraction n. c
(2)
n= -
v
Since Eqs. (1), (2), (6), (7), and (10) of 11.01 are identical, the simplest solutions for a wave in the positive n-direction have the form
D = Do f(n • r — vt) E = Eo f(n • r — vt) B = Bo fi(n • r — vt)
(3) (4) (5)
where Do, Eo, and Bo are the vector amplitudes of D, E, and B, respectively, and f(n • r — vt) is a scalar. When p is zero, (3) and 11.00 (3) give V • D = n • DoRn • r — vt) = 0 Thus, either f' is zero, giving a static field of no interest here, or n • Do = 0
(6)
A similar substitution of (5) in 11.00 (4) gives the similar result n • Bo = 0
(7)
Thus D and B lie in the wave front. From (4), (5), and 11.00 (2) we have
V x E = n x Eo f'(n • r — vt) = — y aB t = vilo f“n • r — vt) This holds at all times so that fiis proportional to f, and we take n x Bo = vB0
or
n x E = vB
(8)
420
PLANE ELECTROMAGNETIC WAVES
Similarly, substitution of (3) and (5) in 11.00
where i is zero gives
n x B = 1.4vD
or
n x Bo = --AvDo
(1)
§11.04
(9)
The scalar product of (8) by (9) gives, since n • Bo is zero by (7), (n x E 0) • (n x B = — 1.10 Do • Bo = (n
(E0 . Bo) — (n • B0)(12 • E o) = Eo . Bo
By (8) the right side of this equation is zero, so that E0 • Bo = 0
and
Do . Bo = 0
(10)
Thus both Eo and Do are normal to Bo. From 11.02 (3), Poynting's vector is (11) H= 1 E0 x Bo[An • r — vt)] 2 = lIo[f(n • r — vt)] 2 whereas from (6) and (7) the normal to the wave front has the direction of Do x Bo. Thus the energy propagation direction H makes the same angle with the wave normal n that E makes with D. When all B vectors parallel a fixed line, then from (10) all E vectors parallel a line normal to it and the wave is plane polarized. The polarization plane is defined as that parallel to H and B in optics and as that parallel to H and E in radio. 11.04. Plane Wave Velocity in Anisotropic Mediums. We now consider the propagation of electromagnetic effects in a homogeneous anisotropic medium of zero conductivity and permeability 1.4. By 1.19 (7), in such a medium there are axes for which D and E are connected by the equations = ciEz, Dy = € 2E„, D, = € 3E, (1) —
These coordinate axes are chosen to coincide with the electric axes of the medium. By 2.08, the electrical energy density is given by
aw 1D .E = _1(elE! av — 2
2
E2.q
+ !1:
€3ED = 21
(2) The last relation is obtained by using 11.03 (9) for D, interchanging dot and cross, and substituting 11.03 (8) for n x E. It shows the equality of electric and magnetic energy FIG. 11.04. densities whose sum is twice (2). Let the direction of ray or energy propagation be n', then B, D, E, n, and n' are related as shown in Fig. 11.04. Evidently the vectors D, E, n, and n' lie in the same plane since all are normal to B. We also notice that sin a =
n• E
——
n' • D
(3)
RAY SURFACE AND POLARIZATION
§11.05
421
To find the wave front velocity v along n, we start with 11.03 (8) and (9). m 4v2D =
—
n x (n x E) = E
—
n(n • E)
(4)
We now introduce the crystal constants v1, v2, and v3defined by 1,
AvE221 =
'4603 = 1
1,
(5)
Elimination of E from the components of (4) by (1) and (5) gives mo(vi
v2)Dx = i(n • E), 14(v2 — v 2)D, = m(n ,u4(v3 — v2)Dz = n(n • E)
—
(6)
where t, nz, and n are the components or direction cosines of n. From 11.03 (6) n • D = iDx .D z = 0. Insertion of Dx, D0, and Dz from (6) gives 12 V2
—
V
n2 m2 ± V 2 — V22± V2 —
—0
(7)
This is Fresnel's equation and gives the velocity of the wave front propagation in terms of the direction cosines of its normal. In general, for every particular direction of the wave normal, there are two different values of this velocity. We can easily verify by substitution in (7) that, if vl > v2 > v3, the only two directions in which v has a single value are to =
+ vi — 4)3/4 24 — mo = 0, no = + — vg)i(v1 — v3)—i (8) (
(
and that the velocity in these directions is v2. The directions defined by 10, ?no, and no are called the optic axes of this biaxial crystal. When v1 = v2 v 3, v is single-valued in the z-direction only, for w—.c— bi 1-1 no = 1 and t o = mo = 0. If vi v2 = v3, v is single-valued in the x-direction only, for which to = 1 and mo = no = 0. In such cases there is only one optic axis and the crystal is said to be uniaxial. 11.05. Ray Surface and Polarization in Anisotropic Mediums.—The energy propagation velocity v' in any direction in a crystal is found by dividing 11.02 (3) by E2/(Av) and multiplying through by n • E giving by 11.03 (8) n• E
E2
n = vn • E
Ex B — n(n • E) E2
(n E2E)2 •
(1)
We can then solve 11.04 (4) for n(n • E), write out components, eliminate Ds, D0, and Dz by 11.04 (6), and insert in the right side of (1), giving Av
n•E rix • = [1 E2
,
—
( n • E)21 Ex,,, E2
(2)
422
PLANE ELECTROMAGNETIC WAVES
§11.05
Figure 11.04 and 11.04 (3) show that the relation between v and v' is („)2 v'
sin2 a = 1 — (n E)2 E2
= COS2 a = 1 —
(3)
Also IIz,y,z = llon'z,y.zso that we can rearrange (2) in the form E2 AvIIo(n • E)
n'.,y,z
(v/2)1,2,0 2 — (v/v')2
Let c', m', and n' be the direction cosines or components of n' and notice that, from 11.03 (11), n' • E = l'Ez m'Ey n'Ez. Then multiplying these equations through by m', and n' and adding, we obtain /
,24.
v'2 —
/2 2
in V2
n"Vi
V"— Vi
—
=0 (4)
This equation gives the ray velocity v' in the direction /', m', and n' as a function of the crystal constants v1, v2, and v3. In general for every particular direction there are two ray velocities. If, at some instant, an electromagnetic disturbance starts from the origin, then in 1 sec the wave will be given by r = v'. When v1 > v2 > v3, the form of this wave in one octant is shown in Fig. 11.05. In the direction of OP and of its image in the yz-plane, both rays have the same velocity. Substitution in (4) shows the directions of the ray and optic axes to be related by lova 2
= m 0 = 0,
nr
noVi V2
(5)
The optic axis is shown by 00' in Fig. 11.05. In each of the coordinate planes, the section of one of the sheets of the double surface is circular and of radius v1, v2, or v3as shown. If two of the quantities v1, v2, and v3 are equal, one sheet of the wave surface is a prolate or oblate spheroid whose axis is a diameter of the other sheet which is spherical. The rays that produce the spherical surface are said to be ordinary rays, and those producing the spheroidal surface are called extraordinary rays. Let us designate the two solutions of 11.04 (7) by v. and vb and the components of the electric displacement corresponding to these two normal velocities by DP),,,z and Dr,;,z. Multiply each equation of 11.04
§11.06
ENERGY, PRESSURE, AND MOMENTUM OF A PLANE WAVE 423
(6) with v = v¢by the same equation with v = vb and rearrange. This gives „ 2 j)(a) D (b)
--x,y,z--x,y,z
(n • E)2
- v!)
- vt)
(6)
Split the right side by partial fractions and add the three equations: 2D(a) D(b)
(n • E)2
n2
1
v?, -
vx,y,z
- v?,
x,y,z
n!,,,,z - v/
By 11.04 (7), each term in the sum on the right is zero so that we have Do) Do) = 0
(7)
Thus the v¢and vb rays are plane polarized at right angles to each other. 11.06. Energy, Pressure, and Momentum of a Plane Wave.—Imagine a plane wave in an isotropic medium striking a thin infinite plane absorbing sheet normally, and consider a right prism which encloses a unit area of the sheet and whose axis is normal to it. An integration of Poynting's vector H over its surface gives the radiant energy entering it. But we have already seen, from 11.03 (6) and (7), that on the sides of the prism vBn = =0 so that II,, is zero. It is also zero on that end which is on the opposite side of the absorbing sheet from the impinging wave. Thus, the only contribution to the integral is from the base on which the wave impinges. Therefore, from 11.03 (11) and (8), the instantaneous rate of energy absorption in the sheet, in watts per square meter, is n =
E x B nE2 - (n • E)E p.v
In isotropic mediums n • E = 0 and
B = (.4,€)tE from 11.03 (8), so that
EB2 B2 1 (B2 + E D) n,, - no,€)i = niloic)i - n(p,e)i 2
(1)
But, by 2.08 (2) and 8.02 (3), the energy density in the wave field is
aw = av
E.E2 B 2 —
2
21.4
(2)
The propagation velocity is (µ6)-i so that the energy per square meter of wave front equals that in a cylinder of unit area and length (AE)--i. From 1.14 (6) and 8.14 (2) there appear to be in electric and magnetic fields pressures normal to the lines of force of iEE2 and 02/A, respectively. Since in the plane wave case, such fields parallel the face of the absorbing sheet on one side but not on the other, we should expect a total pressure,
424
PLANE ELECTROMAGNETIC WAVES
§11.07
in newtons per square meter, against the sheet, of amount
+ 412-1132
P=
(3)
From (2) this equals the energy density at the surface of the sheet. Consider the absorbing sheet free to move but so heavy that P gives it a negligible velocity. Then the relation between P and the momentum p is dp P= (4) dt The mechanical law of conservation of momentum, if applied here, requires that the momentum acquired by the sheet must have been carried by the electromagnetic wave which impinged on it. A comparison of (4), (3), and (1) shows that this implies that the wave possesses a momentum gn per unit volume, in the direction of propagation n, equal to gn
= m€11„
(5)
11.07. Refraction and Reflection of a Plane Wave. Let us now consider a plane wave in a medium of permeability A and capacitivity E, impinging due on the infinite plane boundary of a second 0 medium whose constants are A" and e". This wave may give rise to two waves: one, known as the reflected wave, returning into the first medium, and the other, known as the refracted wave, entering the second medium. Let the unit vec0 tors in the direction of propagation of the incident, reflected, and refracted waves be d, d', and d", respectively. Let d and d" make angles El and 0" with the normal to the plane surface drawn into the second medium, and d' make an angle 0' with the opposite normal. The plane of incidence is perpendicular to the FIG. 11.07. interface and contains d• Let M and N be any two points in space, the vector from M to N being p. If, as before, we take n to be a unit vector normal to the wave front, then n • p/v seconds after passing through M this wave, which has a velocity v, will pass through N. In the case being considered, the same wave front passes through a given point P in the first medium twice, once before and once after reflection. A wave front which passes through a point 0 on the reflecting surface at the time t = 0 will pass through P before reflection at t = d • r/v and after reflection at t = d' • r/v, where —
§11.08
INTENSITY OF REFLECTED AND REFRACTED WAVES
425
r is the radius vector from 0 to P. If the same law of reflection holds for all parts of the surface, the interval of time between these two passages must be the same for any other point Q at the same distance from the surface as P. The vector from 0 to Q will be r s where s is a vector lying in the surface. Equating these time intervals and multiplying through by the velocity give — df • (r
d • T — d' • T = d • (T
s)
or
d • s = d' • s
(1)
Therefore d and d' make the same angle with any vector drawn in the interface which is possible only if d is the mirror image of d' in the surface. Thus both reflected and incident wave normals lie in the plane of incidence on opposite sides of the normal to the interface and make the same (acute) angle with it. To locate the refracted beam, suppose a wave front requires a certain time to pass from P in the first medium to its image point P" in the second. Let the radius vector from 0 to P be r and from 0 to P" be r". If two other points Q and its image point Q" are at the same distances from the interface as P and P", the time for the wave front to pass from Q to Q" will be the same. The radius vectors to Q and Q" are r s and r" s, where s lies in the interface. Equating the time intervals in the same way as for the reflected wave normal, we have (11"E").}(d" • r") — (1.1E)+(d • r) = (.i."E")i[d" • (r"
s)] — (p,E)i[d • (r
(m"E")1(d" • s) = (p.E)1(d • s)
s)] (2)
If s is taken perpendicular to d, which means normal to the plane of incidence, then the right side of (2) is zero, so that d" is also perpendicular to s. Thus both reflected and refracted wave normals lie in the plane of incidence. If s is taken in the plane of incidence then, if v and v" are the wave velocities in the first and second mediums, respectively, v
cos ds cos d"S
sin 6 =n sin 0"
(3)
This is Snell's law of refraction. The ratio n of sin 0 to sin 0" is called the index of refraction. The preceding reasoning is equally valid for an anisotropic medium, although v and v" may be different for each angle of incidence so that n depends on O. 11.08. Intensity of Reflected and Refracted Waves. — The law of conservation of energy requires that the energy passing into a medium through a square meter of surface must equal the difference between the incident and the reflected energies. From Fig. 11.08 and 11.03 (11), this gives (1) (II — II') cos 0 = II" cos 8"
426
PLANE ELECTROMAGNETIC WAVES
§11.08
We shall identify all vectors associated with a beam whose magnetic vector lies in the plane of incidence by the subscript 1, and those associated with a beam whose electric vector lies in this plane with the subscript 2. In finding the intensity of reflected and refracted beams, it is necessary to treat these cases separately.
FIG. 11.08.
For the beam whose magnetic vector lies in the plane of incidence, E1 is parallel to the surface; so, from 1.17 (3), we have
= E','
Ei
In this case, from 11.06 (1), setting µ = A" €1(E?
— _OM cos
(2) =
(1)
becomes
0 = ENVI' 2 cos 0"
Dividing this by (2) and using 11.07 (3), we get
E, — E' — sin ti cos 61"E" 1 sin 0" cos 0 1
(3)
Solving (2) and (3) for E', and E',' gives E;. = E',' —
sin (0— 0") Ei sin (0 + 0") 2 sin 0" cos 0 s in (0 + 0")
El
It follows from 11.03 (8) and 11.07 (3) that sin (0— 0") B' — Bi 1 sin (0 + 6") sin 20 B1 131' = sin (0 + 0") Using 11.06 (1), we get for the reflected intensity sin2 (0 e")ll sin2 (0 + 0") 1 —
(8)
Substituting this in (1), we get for the refracted intensity — 2 sin 0" cos 0 sin 20 sin2 (0 + 8")
(9)
§11.08
INTENSITY OF REFLECTED AND REFRACTED WAVES
427
For the beam whose electric vector lies in the plane of incidence, B2 is parallel to the surface so, setting 1.1 = /I" = A„, we have B2 + B2 = B2
From 11.06 (1), setting µ =
(10)
= my, (1) becomes
(B2 — B22) cos 8 = B22e"--1cos 0,,
E1
Dividing this by (10) and using 11.07 (3), we get B2 - B' — 2
sin 20" B„ sin 20 2
(11)
Solving (10) and (11) for R2and B2 gives tan (0 — 0")B B' _ B22 2 tan(0 + 0") sin 20 B/2' B2 sin (0 + 0") cos (0 — 0")
(13)
It follows, from 11.03 (8) and 11.07 (3), that E,
= tan (0 — 0") E2 tan (0 + 0") 2 sin 0" cos 0 E'2' = E2 sin (0 + 0") cos (0 — 0") 2
(15)
Using 11.06 (1), we get for the reflected intensity tan2 (0 — 0"), tan2 (0 + 0")112
11'
(16)
Substituting this in (1), we get for the refracted intensity 11'2' —
2 sin ;" cos 0 sin 20 sine (0 ± 8") cos2(0 — 0")112
(17)
At normal incidence, the cosines in (1) are unity, so that in place of (3) 6" )1 „ v (— E = -Ti E”
E — E'
V
Combining this with (2) gives E' E"=
v
v — v" E v v" 2v" E v"
(18) (19)
428
PLANE ELECTROMAGNETIC WAVES
§11.09
From 11.06 (1), the reflected intensity is H
,
tv — v"\211 + v"
(20)
4vv" (v v")211
(21)
Combining this with (1) gives _
It should be noted that the intensity of radiation given in (8), (9), (16), (17), (20), and (21) is the energy passing in 1 sec through 1 sq m of area parallel to the wave front. To get the actual radiation energy in 1 cu m divide II by the wave velocity in the medium. Evidently from (12), if 0 + 0" = ir , B2 = 0, which means that at this incident angle radiation whose electric vector lies in the plane of incidence is not reflected. This is called the polarizing angle because unpolarized radiation incident at this angle is reflected with its magnetic field in the plane of incidence. From 11.07 (3), this "Brewster's angle" is sin
sin Op —
(fir
0p)
= tan Op =
=n
(22)
11.09. Frequency, Wave Length, Elliptic Polarization.—So far we have used the general solution for a plane wave, f(n • r — vt), in deriving the laws of reflection and refraction. For discussing circular and elliptic polarization it is convenient to take a regularly periodic function which can be built up by Fourier's series from simple sinusoidal terms. Thus
f(n • r — vt) = D cos [w(t — v—'n • r)
(1)
where n is a unit vector in the propagation direction. The mathematical manipulation is considerably simplified by writing
D cos [co(t — v—'n • r)
= Dejoemt—"''o) (real part)
(2)
The term DO' is usually written as a phasor D. The angular frequency w and the cyclical frequency I, are related by =
2irp
(3)
The wave length X is the shortest distance in the direction of propagation along a "frozen" wave in which the electrical conditions repeat. It is related to the frequencies and the wave velocity v by the equations v
A = =
277-v
=
1
211-
P (AO
W (AO
2a
(4)
§11.09
FREQUENCY, WAVE LENGTH, ELLIPTIC POLARIZATION
429
Now consider the superposition of two plane electromagnetic waves of the same frequencies traveling in the z-direction. Let one be plane polarized with a y-directed magnetic vector and the other with an x-directed one so that the equations for the electrical intensities are Ex = El cos [w(t — v—'z)] Ey = E2 cos [0)(t V-1Z) + 6]
(5) (6)
Ex and Ey are shown as a function of z in Fig. 11.09. To find the locus
in the xy-plane of the end of the vector E whose components are Ex and Ey, set z = 0 and eliminate t from (5) and (6). This gives 2ExE, cos 3 = sin' S El + E2 E 2E,
(7)
This is the equation of an ellipse, shown at the left in Fig. 11.09. Such waves are called elliptically polarized. If S = mr, (7) becomes E2Ez ±
=0
(8)
which represents a straight line, so the resultant wave is plane polarized. If S = -(2n + 1)7 and, in addition, E1 = E2 = E, then (7) becomes = E2 This locus is a circle, and the resultant wave is circularly polarized.
(9)
430
§11.10
PLANE ELECTROMAGNETIC WAVES
11.10. Total Reflection.—Our first example of elliptically polarized radiation occurs in connection with the phenomenon of total reflection. If e" < E in 11.07 (3), then 0" > 0 so that 0" may equal 47r when 0 < tea. The value of 0 which gives 0" = a is called the critical angle Oc. The agreement between experimental facts and the results of the following interpretation of this phenomenon justifies its use. When 0 > 0c, let us write 11.07 (3) in the following form, when A = A", cos 0" = (1 — sin2 0")1 = 1 (—
e
E'
sin2 0
i
E
= j —,- sin2 ei
Equation 11.08 (3) must then be written El _ E1 i[sin2 0 — (e"/OPE--1, cos 0
(1)
which, combined with 11.08 (2), gives = cos 0 — j[sin2 0 — (E'/E)]1E = . ev' El El cos j[sin2 0 — (e"/e)P 1 2 cos 0[sin2 0 — (E'/E)]1 tan ik1 = cost 0 — sin20 + (E" /6)
(2) (3)
Corresponding to 11.08 (11), we have B2
sM2 0 — 1])i I3 cos 0
j{(E/e")[(E/E")
P2
which, combined with 11.08 (10), gives _ cos 0 2
tan
ik 2
—
{(e/e")[(e/e") sin2
—
(e/e") [ (E/e") sin2 — 2 cos 0{ (e/e")[(e/e") sin2 — 1]}i COS2 0 — (e/e")2 sin2 0 (6 / E" )
cos 0 +
{
_
P2—
e
2
(4)
(5)
It is evident from (2) and (4) that 1E11 = 1E11 and 1B21 = 1B21 so that the reflected and incident rays have the same intensity. Thus the law of the conservation of energy permits no energy in the refracted beam. This can be verified by computing E"," and B12' and from this II". A comparison of (2) and (4) with 11.09 (2) shows that there is a phase change for the beam whose magnetic vector lies in the plane of incidence and 4/2 for the beam whose magnetic vector lies normal to it. Thus, for plane polarized incident radiation, whose two electric vector components are E1 and E2, the reflected radiation is elliptically polarized. The phase difference S can be obtained from the ratio of (2) to (4), since B2 = (µE)lE2 and
PLANE WAVES IN ISOTROPIC CONDUCTORS
§11.12
431
B2 = (1.1,e)1E2by 11.03 (8), giving j cos 0[sin2 0 — (€"/e)]1 sine 0 — j cos 0[sin2 0 — (e" /O]1
e,5 = ejoh_4,0 = sine
Taking the ratio of the imaginary to the real part of this gives us tan 3, which, with the aid of Pc 573 or Dw 406.02, may be reduced to tan
— (66)]i 7 3 cos 0[sin2 = 2 sine 0
(6)
11.11. Electromagnetic Waves in Homogeneous Conductors.—In a medium whose conductivity is not zero, we must use all the terms in the propagation equations 11.01 (1) and (2). In deriving 11.01 (2), we set p = 0. To justify this, take the divergence of 11.00 (1) which gives Ile— (V • E) at
iryV • E = 0
Write p for eV • E by 11.00 (3) and integrate from 0 and
pt, to t' and p'. (1)
= Ploe
Thus, an electrical distribution p'0is dissipated independently of electromagnetic disturbances so that, if p; is zero initially, p' is always zero, which justifies setting p = 0. We define the relaxation time T to be T=
TE
=
(2)
-rie
Take a simple periodic disturbance so that both B and E have the form B = beic." (real part) = Boelo"-E0) (real part) Now
a/at may be replaced by jco, and 11.01 (1) and (2) become 02B = gjoyy — u.,203 — 0,20E V 2E µ ( jwy
(3) (4)
These are identical in form with 11.01 (1) for a periodic disturbance in an insulator. The equations will have identical solutions if we write E - j-y/co for E. The complex quantity (ile)1which occurs frequently may be separated into real and imaginary parts by Dw 58.2. [A(E — ic0-17)]/ = n n={
2(1+ 0)72622)1 +1 J}1'
k
ik
(5)
0)1 — 1 ]}1 (6) 2(1 ±—;
11.12. Plane Waves in Homogeneous Isotropic Conductors.—Certain properties of plane waves in conductors can now be obtained, by the
432
PLANE ELECTROMAGNETIC WAVES
§11.12
substitution suggested above, from the corresponding properties in a dielectric derived in 11.03. In isotropic mediums the nature of e does not affect (6), (7), or (8) of 11.03 so that in a homogeneous conductor B is normal to E and both lie in the wave front. From 11.03 (9) k2)+Eoeitan-1(kIn) (1) Eo = (n 4- jk)E o = (n2 Thus, the electric and magnetic vectors differ in phase by tan—' (k/n). Writing E in the form of 11.09 (2) and substituting in (1) give &awe = ,Alecolco•reaco(t—no•r)
(n2
&jot = (, x
k2)IA lecuka•re j[w(t—no•T)-1-tan -l(k/n)]
(2) (3)
where E is a unit vector along E, and 6 is a unit vector normal to the wave front. We note that both E and B decrease exponentially as the wave advances, indicating a rapid absorption if k is large. We call n the index of refraction of the conductor and k the coefficient of extinction. It is useful to know the magnitude of these quantities in nonferromagnetic conductors whose permeability is nearly equal to lie. In conductors the capacitivity is not accurately known but seems to be of the same order of magnitude as in insulators. As a typical example take copper, whose conductivity is about 5.8 X 107 mhos/m. Designating wave length in vacuo by Xo and using 11.09 (4), we have, for copper, writing K for E/Ev, 7
Xo7
WE
2rcE
Xo -„ X 3.48 X 109
Except for very short waves, this quantity is certainly large compared with unity so that, from 11.11 (6), n and k are well approximated by n ~ — k ~ (110■ 07)1 (21-c)-1
(4)
The distance an electromagnetic wave must travel in order that E and /3fall to e--2' = 0.002 times their initial values is, from (3), 2r
Xo (47X oy
wk
kc \FLIT
meters
(5)
For air wave lengths of 1 cm, 1 m, 100 m, and 10 km, this gives d the values, in copper, of 0.0025, 0.024, 0.24, and 2.4 mm, respectively. It is also of interest to compare the average electric and magnetic energies in the wave. From (3) and (4) and 11.06 (2), we have We _ E2 _
Wm
2rEc
n2 ± k2 Xo-y
(6)
On the assumption that E = 2e,, this gives, for the waves just considered, 5.7 X 10-8, 5.7 X 10-18, 5.7 X 10-12, and 5.7 X 10-14, respectively, so that nearly the entire energy is in the magnetic field.
§11.13
REFLECTION FROM CONDUCTING SURFACE
433
11.13. Reflection from Conducting Surface.—We saw in 11.11 that in a conducting medium (A"e")i is replaced by
=n
ii1"(E" —
ik
Thus, sin 0" and cos 0" become, by 11.07 (3) with the aid of 11.11 (5), sin 0" =
LE ) 4 sin n(+ jk
cos 6" = [1
At si'0 (n + j k)2
(1)
We saw in the last article that, for electromagnetic waves over 1 cm long. —k
n > 1.4 X 104
So that, from (1), 0" is a complex and extremely small angle. Thus, from 11.08 (8), when the magnetic vector lies in the plane of incidence, , sin'(0 — 8")11.1 _> II 111 = sin' (8 + 8") 0,,,o
(2)
When the electric vector lies in the plane of incidence, we get, from 11.08 (16), provided 0 is not too close to 4r, ,tan' (8 — 0")ll 112 = tan' (0 + 6") 2 v,-,o
112
(3)
Thus, for all angles of incidence and for all states of polarization, a sufficiently long electromagnetic wave is completely reflected from a conducting surface. The same results apply to longer waves at lower conductivities. From the last article, the extremely small fraction not reflected is very rapidly absorbed so that quite thin sheets of metal can be very opaque for short waves. To study the state of polarization of the reflected beam, we divide 11.08 (14) by 11.08 (4) and obtain _
cos (0 + 0")E2 _ cos (0 — 0")E1
(cos0" — sin 0" tan 0)E2 (cos 0" + sin 0" tan 0)E1
(4)
For the wave lengths just considered, we have seen that n + ,9k = n(1 — j) If µe/fin jkl 2<< 1, then cos 8" = 1 from (1) so that Ef,
[n — (i.LE)i sin 0 tan0 — jn]E2 [n (12e)1 sin 0 tan 0 — jn]Ei [n — 0201 sin 0 tan OP + AE2 [n + (1201sin 0 tan OP + n2 e E1 2n(i.‘€)1sin 0 tan 0 tan S — 2n2 — 1.46 sin' 0 tan' 0
(5)
{
(6)
434
§11.14
PLANE ELECTROMAGNETIC WAVES
Because n is a large quantity, we note that, except when 0
E; Ei
4r,
E2
E1
When tan 0 —> co , there is not only a change in the magnitudes of E; and but a phase shift ö as well. Thus, for wave lengths of a centimeter or more, we obtain elliptically polarized radiation only when plane polarized radiation is reflected near grazing incidence from the conducting surface. 11.14. Plane Waves on Cylindrical Perfect Conductors.—Plane waves can be propagated not only in free space and infinite dielectric mediums with plane faces but also in the z-direction along a set of perfect conductors, provided all sections of the set taken normal to the z-axis are identical. From 10.02, we know that on perfect conductors the currents flow in an infinitely thin surface layer with no fields inside and no energy dissipation. This problem can be set up in several ways. The results of 7.25 suggest taking the vector potential and hence, from 11.01, the Hertz vector, in the z-direction parallel to the currents. The solution of the propagation equation for the Hertz vector yields both a scalar and a vector potential. Elimination of the scalar potential by 11.01 (12) gives a vector potential normal to z. The same result can be obtained more directly by solving the scalar propagation equation
a2w 82W1 + [a2w + ay2 az2
L axe
1/€ ate J
=0
(1)
The solution obtained by setting each bracket separately equal to zero is W = Vi(x, y)fi[z — (pie)-4t]
V2(x, y)f2[z
(1.40-it]
(2)
where V1 and V2 are solutions of Laplace's equation in two dimensions so that, if W1and W2 are the complex potential functions of Chap. IV, W1 = Ui
jVi = Fi(x
If V2V is defined as i(a V/ax)
jy),
W2 = U2 + jV2 = F2(x + jy) (3)
.i(aV/ay) we see from 4.10 that
-kxV 2 V = V2U, kxV2U = V2V, V x[V2Uf(Z)] = k xV2Uf(z)
(4)
By 10.01 (1), the vector potential for a transverse electric field is A= v x kW = = V 2U1(X,
- k x V 2 -17
- (µE)-it} - k x V 2V 2f2[Z +(AE)-it] 77
- 04 0-4 0 + V 2U 2(X, Y)f2[ 2
(5)
Evidently the first and second terms represent, respectively, waves that
435
§11.14 PLANE WAVES ON CYLINDRICAL PERFECT CONDUCTORS
travel in the positive and negative z-directions. The fields are B= V x A = V2V(x, y)f'[z + (w) —it] E=
a
(6)
A = ±(110-4V2U(x, y)f'[z -T
(7)
The upper sign goes with the positive wave. From (4), (6), and (7) (AWE = k x B
(8)
Suppose that the current goes out on one set of conducting cylinders and returns on another set. It is evident from (7) that, if all members of the same set have the same potential at one value of z, then this will also be true at any other value of z. The relation between total current and charge per unit length in either set of n members can be found from (8), the magnetomotance law 7.01 (2), and Gauss's theorem 1.15 (1), thus Q = ZfE.Endsi =
i=i
GIWZfHt dsi
(9)
i=i
If L is the self-inductance per unit length, C the capacitance per unit length, and S the area outside the conductors in a plane of constant z, then with the aid of 2.07 (4), 2.19 (1), and 8.08 (1) we have
LI 2 = rB2 = E.E2 = Q2 = p dS dS 2 j,s2p. j8 2 2C 2C
(10)
Therefore, L and C are connected by the relation
LC = ALE = v-2
(11)
Thus the product LC equals the reciprocal of the square of the electromagnetic wave velocity in the medium outside the conductors. If E and B are specified in the plane z = 0 as functions of time, it is easy from (5) towrite down A for any z. Thus, 0.1,€)1z] g[t — (w)iz] A = 1V2U(x, ( z] + f[t
[t f - 0401
— g[t
(le)iz])
(12)
If U and V are given by (3), A yields the following fields at z = 0:
—(aA/a00 = —v2u(x, Bo= (V x A)o= — (moiv2v(x, y)g'(t)
E0 =
(13)
It is equally simple to write down A from (5) when E and B are given as functions of z at t = 0. Thus, ( A = 4-V2 U(x, y)/[0.16)42
t] — f[(.te)iz — t] — g[(.1,€)1z t] — g[(Ae)iz — t])
(14)
436
PLANE ELECTROMAGNETIC WAVES
§11.15
If U and V are given by (3), A yields the following fields at t = 0: E0 = — V2U(x, Y)fI(µE)/z],
Bo = — 0201V2V(x, y)gq(1.1.01z] (15)
11.15. Intrinsic Impedance of a Medium.—When the dielectric of the last article is replaced by a medium of conductivity -y, 11.14 (1) becomes [
a 2w
a21 + [a2w ay2 az2
axe
aw 7 at
Aca2w]0 = ate
(1)
When the second bracket 's equated to zero, a solution in terms of simple arbitrary functions is no longer possible so we use the Chap. V method of separation of variables which splits it into two total differential equations with a separation constant. Their solution is exponential in form and, according to whether the separation constant is real or imaginary, lead to solutions harmonic in space (transient) or to solutions harmonic in time (steady state). We shall now consider steady-state solutions in which time enters as the factor em as in 11.09. The second factor in (1) now becomes .32 W/az2 — = 0, and in the equations of 11.14 the functions f[z ± (au)—it] are now replaced by e±r'-kiwt where a=
(Los,m)[(0,2,2
= 'y2)4 —
jom) = (a + j3)2 = {(4,,,12)[(u,2,2 ± 72)i ± cod
(2) (3)
Here P is the propagation constant, a the attenuation constant, and 0 the phase constant or wave number. Thus 11.14 (6) and (7) become, for the wave in the positive or negative z-direction,
v2v(x, y)eTi'z-Fic" (real part) T V2V(x, y)eT"z[a cos (cot T (3z) — 3 sin (cot T- 0z)] jcoV 2U(x, y)eTr.±m (r.p•) = wV2U(x, y)eT"' sin (cot 0z) ft = R=3*(.4(k x E) or Pk x E = ±jc0B
B= E=
—
(4) (5) (6)
The differential equation and its solution for the z-directed line current are the same as for W, so that
a 21 az2 =
=
(7)
where em, which appears on both sides of the equation, is canceled out. It must be put back before taking the real part as in (4) and (5) to get the trigonometric form. In a length dz the current in the positive and negative line members is — (al/az) dz, which is partly displacement current and partly leakage conduction. This current traverses an impedance Z,/dz so that by Ohm's law the potential across the line is
= —2.=
7 ' az
(8)
§11.15
INTRINSIC IMPEDANCE OF A MEDIUM
437
From 6.06 (7), the product of the resistance R per unit length between conductors by the capacitance C per unit length is TE or ch. Therefore the shunt admittance Ye and the series impedance ZL are, using 11.14 (1), 12-.c = 1 R
-y + jw€
+ 3.coc
2
ALE
2L,
L jwL
E
= jw L
(9)
For transmission in the positive z-direction, so that /3 = 0 in (7), the characteristic impedance of the line is Zk == 2cr
o h=
Y
= (2c2L)1 = (ZL Ye
(10)
If B is not zero, the ratio of E to i at z = 0, which is found from (7), (8), and (10) and called the input impedance, is
2= (A— _A — (21 + 1'3) fc A ±
BZ
If the line of length / ends in a load 2/at z = 1, then from (7) and (8)
=
=2
2,
— 13"efq Berl
7'; )e__4
(12)
Elimination of A and B from (11) and (12) gives the input impedance k2 .,/,
= 2
Zk
cosh1"/ cosh Pl
Zk sinh
rl ZL sinh 1"/
(13)
Evidently the input impedances, when the line is open or short-circuited, are Zk coth 1"/ and Zk tanh respectively. Note that ZoZ, = Z. From (9) and (10), the characteristic impedance of the line is 2k
= jcop.E
jwL =
P
I
E
PC
(14)
Now consider the special case where the conductors are two parallel planes 1 m apart. An infinite 1-m square tubular section running in the positive z-direction has a capacitance per unit length of E, so that its characteristic impedance is
2k =
:MA )1 _ jcoli
+ jam
a+
) =n a-)0 E
(15)
Also V = Ez and By = Ai where i is the current density. Note that 2k depends on the properties of the medium alone and that the configuration of the fields is identical with that in a plane polarized plane wave, so it is logical to follow Schelkunoff and call Zk the intrinsic impedance of the
438
PLANE ELECTROMAGNETIC WAVES
§11.16
medium. A comparison of (15) with 11.07 (3) shows that, in optically transparent mediums where y is zero andµ is bc,„ the intrinsic impedance is proportional to the index of refraction. From (3) and (15), we may define the intrinsic propagation constant of the medium as rk
= a + j0 =
[ice11(7
+ icoe)1' = icolA2T1
(16)
We need a formula like 6.03 (2) for power absorption in an impedance Apply an electromotance 60 cos cot, so E is 0 and real, to an impedance Z producing a current I or /0cos (cot + 1,t). From Ohm's law, 6.02 (1),
e
4, 60 = IZ = I oej (X 2
R2)iej tan-, (X/R) = Io2
(17)
so IP is —j tan—' (X / R) and the instantaneous power absorbed is the real part of (18) P = Icr = 88Z-1(cos2cot cos IP — sin cot cos cot sin ')
11.16. Reflection at a Discontinuity. Matching Section.—Suppose the plane wave described in the last article, traveling in the positive z-direction, meets a plane boundary over which z is constant and beyond which lies a medium of capacitivity ez, permeability A2, and conductivity 72. The relations between the original wave, the reflected wave, and the transmitted wave and their respective currents are, from 11.15 (14),
V1 = 2111,
f2 = 2212
(1)
At the boundary, potential and current must be continuous so that
V1 + Vi = V2
and
(2)
=
The result of solving the above equations for -FT, V2, 11, and 13 is
✓i =
11
22 —
21
f2
1222
222
✓,
11
21 ± 22
V1
1121
21 ± 22
(3)
If both mediums have zero conductivity, it will be observed from 11.15 (3), 11.15 (14), and 11.07 (3) that 22/21 may be replaced by v2/vi, the ratio of wave velocities in the two mediums. The potential therefore obeys the same law for normal reflection as the electric field in 11.08 (18) and (19). By working with components one may use the transmission line formulas to derive the laws of reflection at any angle. Electromagnetic waves may be passed from one nonconducting dielectric medium 1.4 ei into a second P.3, E3 without loss by insertion of a matching section. For perfect transmission, the input impedance of the matching layer or section must equal the characteristic impedance of the first medium. Therefore, by 11.15 (15), we must substitute in 11.15 (13) Zi = (Aii/e01,
2k = (122/e2)1,
= (123/E3)',
= ./042€2)1= 27riX21
§11.17
439
COMPLEX POYNTING VECTOR
This gives the following equation to be satisfied for a match: iktiE2\1
04E2)1cos(27r1/X2)
i(A2€3)1sin (27//X2)
kil2E1j
(ii2es)1cos (27//X2)
A/4E2)i sin (27r//X2)
(4)
The real terms on the right vanish, and the equation holds if we choose / = 1(2n 1)X2 and A2q1= (A1E171)10./3E01
( 5)
Thus, the matching section should be an odd number of quarter wave lengths thick, and its intrinsic impedance should be the geometric mean of those of mediums 1 and 2. In terms of optics, the index of refraction of the matching layer must be the geometric mean of those on either side. These quarter wave layers are widely used to eliminate reflections from lenses. 11.17. Complex Poynting Vector.— An expression for the Poynting vector in phasor form is useful when the fields vary sinusoidally so that the operator a/at may be replaced by the factor jo.,. Let us write down Eq. 11.00 (2) and (1), using conjugate phasors in the latter, for which it is equally valid because they give the same two real equations. Thus V x E = —jcoh
V x E= 12(7 — jcoa
(1)
Multiply the first equation by E and the second by —È and add.
h • (v x
— • (v x h) = —[ A-rE • 2 — joi(AEB • 2 — 13 • h)]
Proceed as in 11.02, writing V • (E.x h) for the first term, integrating over the volume v, and applying Gauss's theorem. By Ohm's law, 6.02 (3), we may replace E by i/-y or Ti where z is the current density and so get —IA- ifsn • (E.x a) dS = fvri2 dv — jwfv(et •
E — A-4 • h) dv (2)
By 10.00 (10), the real term on the right gives twice the average energy dissipated per second by eddy currents in the volume of integration. Thus, if we integrate the inward pointing normal component of the vector II =
x 2) (real part) = 111—'(2 x h + E x
(3)
over the surface of the volume v, we get the rate at which energy is being absorbed from the wave. This formula can be used to find the energy flow across the plane z = zoof the transmission line considered in 11.15. Elimination of E from (3) by 11.15 (6) and the use of 11.14 (10) and 11.15 (14) show that, for a z-directed wave, the flow is the real part of —jco 20
(k
• f3) a dS = k3wf
s 2µ
dS
jcoLI2 k
2r
V/
(4)
440
PLANE ELECTROMAGNETIC WAVES
§11.18
In 11.14 (9), a positive potential was assumed on conductors in which the current is in the positive z-direction. If either current or potential is reversed but not both, then the direction of energy flow is reversed. 11.18. Nearly Plane Waves on Imperfect Conductors. Lecher Wires. If the conductors forming the transmission line considered in 11.14 and 11.15 have finite conductivity, heat is generated in their surfaces by the eddy currents. The electric field has a z-component at the surface so that its Poynting vector is inclined toward the surface and the wave is no longer truly plane. In most practical cases, the power lost per unit length is small compared with that transmitted so that the inclination of H to the z-axis is negligible. In this case, one uses the plane wave formula to calculate the fields in any plane normal to the z-axis. The energy loss varies as the field strength squared so it damps the fields exponentially with z and adds a real term Rito the series impedance ZL. The magnetic field penetration into the wire adds an internal inductance term jcoLi to Z L. From 11.15 (14), the characteristic impedance is 2k = I ZL\ I
L[Rti
YJ
jco(L 11(7
3.coe)
(1)
For many transmission lines Ri << coL, Li << L, and y << we so that in a vacuum the characteristic impedance and propagation constants are
2k = L 04E0 –i = 2.9979 X 108L (3 X 108C)-1 P2 = ZL Y = AL-1(7 ./0)0[R,, + L1)]
(2)
(3)
It will be instructive to calculate 2k and t for a Lecher wire system composed of two parallel wires of diameter d with their axes at a distance b apart. By 4.14 (2) and 11.14 (11), the external inductance is
At = L=— c
b cosh-1 —
(4)
Note that L depends only on the configuration and not on the scale. We shall assume that the conductivity and frequency are so high that S is small compared with d, and the formulas for the resistance per square R: and the inductance per square L:, given in 10.02 (10), may be used.
Ri/2 = 2ffi f ie ds .1412 = 2I: ds
(5)
where is is the current across unit arc length in the cylindrical skin and the factor 2 is inserted to include both wires. From 11.14 (6) on the U =: U1cylinder, the tangential component of B is aV / as. The magnetomotance law applied to a surface element gives /is = B. Thus,
Je ds = AL-2f(av / a 8)2 ds = 1.4-2.95.(aV / as) dV
§11.19
441
GROUP VELOCITY
From 7.09, 4.11 (2), and 4.13 (1) we have dW dz
=
av as
.
-= 7ra in
r(Ui
s
jV)
JAI
sin
2r (Ui jV)
= Au(cosh2 rUi
I
"/
ds = Ara 0
cosh'
21-U1
d = cosh 2-TU a
- cos
2irV
) dV =
V)
os2
/2 AI -2:U1 I cosh (6) 2ira
b2 - d2 = 4a2
(7)
(7) is from 4.13 (3). Putting (6) and (7) into (5) and dividing out I', 2R:b 2r'b Rz (8) ird(b2 d2) Ocl(b2 d2)1 -
2L'i b 2r'b 71-d(b 2 - d2)1 irwSd(b2 - d2)1
(9)
Now consider a special case where b/d is 1.5, and the line operates in a vacuum or air so that µ = µ = 4r X 10-7. Then by (4), L is 3.85 X 10-7 henry. The calculation of Ri and /4requires, beside d/b, the frequency and the resistivity, permeability, and one dimension of the conductors. The most important conductor is copper for which y' or 1/T' is 5.8 X 107 and µ = /.4. Consider the frequencies 3, 300, and 30,000 megacycles which correspond to the wave lengths of 100 m, 1 m, and 1 cm for which wL is 72.5, 725, and 7250 ohms, and the skin thickness is 3.82 X 10-5, 3.82 X 10-6and 3.82 X 10-7m. From (6), (7), and (3) we then have: X (m)
d (m)
Ri(ohms)
0.01 0.01 1.00 1.00 100 100
0.001 0.01 0.001 0.01 0.001 0.01
38.5 3.85 3.85 0.385 0.385 0.0385
Li (henry)
1.225 X 10-" 1.225 X 10-1' 1.225 X10-9 1.225 X 10-" 1.225 X 10-8 1.225 X 10-9
a 5.1 X 10-2 5.1 X 10-' 5.1 X10-8 5.1 X 10-4 5.1 X 10-4 5.1 X 10-'
/3 19 19 1.9 1.9 0.19 0.19
Evidently in all cases R, and coL,, are very small compared with wL so that the characteristic impedance may be obtained from (2) to be 115.5 ohms. 11.19. Group Velocity. We have seen that any of the field amplitudes for a plane wave on a transmission line may be written in the form -
A ((3) = f(u 1, u2) cos (wt - (3z)
(1)
where u1and 222 are orthogonal coordinates in a plane normal to z, and is the wave number or phase constant. When the phase velocity of a wave depends on its frequency, as in a wave guide or a dispersive medium,
442
PLANE ELECTROMAGNETIC WAVES
§11.19
then it is evident that a signal consisting of a group of waves of different frequencies will change form as it advances. Let us assume that the wave numbers of the group lie between 00— S and 130 + S and that the amplitude in the interval dr3 is A((3). The amplitude of the signal is then
S
0+ A(0)el( wt — az) d3 so— 6 r
This may be split into a carrier wave factor with the velocity w0/00 and a modulation factor with a velocity (w — 0)0)/03 — 130). Thus „ 0(w- -(P-Po)z] Po+ '°)t ) S = Cei( wot-00) , C= f da Po- 6
Here 1C/3 is the average amplitude of the wave group whose median phase velocity is vo = wo/0o• Thus we have a modulating frequency — 0)0 superimposed on the carrier frequency coo. We wish to find the velocity of a plane in which the modulation amplitude C, which constitutes the signal, is constant. By the mean value theorem for an integral, we have eiE(c.'i—wo )t — oi—#0>zi C = 28A where 00 — S < 01 (3o + 3. If the frequency range in the group is so small that 131 — )30 and col— coo are infinitesimal, the signal velocity is V8 =
wl —
Wp
01 — 00 a—.o
(2)
Problems 1. Two airplanes are flying at a distance
d apart at a height h above a plane water
surface. One sends radio signals to the other, both sending and receiving antennas being short vertical wires of length 1. Show that the ratio of intensity of the signal
received by water reflection to the direct signal, taking the capacitivity of water to be
e and taking h >> 1, d >> 1, and X >> 1, is d6
(d2
{Re
1
—
4h2)3 [(8 —
ev)d2 e,,)d 2
4eh211— 2ee,,-11.2 48h91 2eco-41t
2. A plane electromagnetic wave polarized at 45° to the plane of incidence is totally reflected in a prism which it enters and leaves normally. Show that the intensity of the emergent beam is 16n2(1 n)-4Io, where n is the refractive index. Show that the emergent beam is elliptically polarized with a phase difference q5 given by tan id" = cos e(sin 0)-2(sin 2 B — n2)1where 0 is the angle of incidence on the back of the prism and multiple reflections are neglected. 3. A plane electromagnetic wave of wave length X is incident on an infinite plane plate of dielectric of thickness a. Show that the reflected beam intensity, if the beam is polarized either normal to or in the plane of incidence, is given by 4b2 sin2(4o)[(1 — b2)2
4b2 sin2 (0)]-1/0
443
PROBLEMS
where b2 is the fraction of the incident light intensity reflected without entering the plate, S = 47rnaX-1cos 4,, 0 is the angle of refraction, and n is the index of refraction. Include multiple reflections and phase relations but assume no absorption. 4. Show that the law of refraction derived in 11.07 does not hold, in general, for Poynting's vector at the surface of an anisotropic medium. 5. Show that a plane normal to the optic axis where it pierces the ray surface in Fig. 11.05 is tangent to this surface on a circle that encloses the ray axis. 6. A plane wave is incident at an angle 0 on the plane face of a uniaxial crystal whose normal makes an angle a with the optic axis. The angle between the plane of incidence and the plane containing the normal and the optic axis is 0. If ea = e2 show that the two refracted wave normal directions 0. and 0, are „ ej sin 0
sin 0° =
M =COs 2 + 61
ea — ea
2P2— MN ± 2P(P2— M2— MN)+
sin'
Ell
N2 ±4P2
P = sin a cos a cos
N=
Pa COS2 a
M2
sin2 — Ea ) sing —
The minus sign is taken when 0 < 2,r and the plus sign when in- < 4, < 7. Verify that possible plane wave vector potentials, when B is normal to z, are Al = Ci[ij0' — k(0'2 — ODIle —(13.2-012)1.+ i(wt-0',0 k (0; — 13'2)4sin [(4 — 13'2)1x] }e2(1ut -13'') 1312)1X]
A2 = C210' cos [(IA
where 0! = co2Atif i, 1322 = w2&2E2, and 02 > 01. With these, solve the problem of a plane surface wave moving over an infinite plane perfectly conducting surface coated with a dielectric layer A2, E2, of thickness a when the infinite medium above it has the constants Ai, ei. Show that the velocity may be found from the equation €2(0'2 —
= ea(RZ — 13'2)4 tan
— 0'2)/a]
If gia = 1.00, Az2 = Ai, and e2 = elei, then 01:0':02 = 1:1.523:2, roughly, so that the amplitude has dropped to 0.1 of its value at x = a when x = 3.00a. 8. Verify that possible plane wave vector potentials, when E is normal to z, are A l = jCie-( 3'2-012)1.±i*H3'4,
A2 = jC2 sin [(14 —
13'2)4xlei(wt-13'z)
where /3 = w2 r1E1, 0; = (02/12E2, and 02 > Oa. With these, solve the problem of a plane surface wave moving over an infinite plane perfectly conducting surface coated with a dielectric layer A2, E2 of thickness a when the infinite medium above it has the constants el. Show that the velocity may be found from the equation A2( 3'2 — 0i)1
— ALM — 0'2)1cot
— 0")Ia]
If 131a = 1.00, tt2 = Ail, and e2 = 4fi, then 01:0':02 = 1:1.03:2, roughly, so that the amplitude has dropped to 0.1 of its value at x = a when x = 10.33a. 9. Show that, if the two conductors of the section of uniform line bounded by planes of zero capacitivity and permeability at z = 0 and z = / are given equal and opposite charges producing a difference of potential Vo and if at t = 0 the surfaces of the planes are made conducting so that the resistance between conductors is R, then the potential between conductors is
V. =
Vu R
rR — 2k R 2k
R
(2m + 1)1 (2m + 1)./rz 42k x-■ ( —1)' cos sin (AO/ ar A' 2m + 1
where 0 < z < 1 and s(kie)ll < t < (s
in =0
1)(Ike)ll.
444
PLANE ELECTROMAGNETIC WAVES
10. A uniform transmission line of length 1 and characteristic impedance 2k is terminated at one end by a battery of zero resistance and electromotance e in series with a switch and at the other by a resistance R. Neglecting field distortions at the ends, show that, if the switch is closed at t = 0, the current through R is .
-
Zk — R
R
Zk
4 a=0
(2n + 1)0104 / < t < (2n + 3)(1101
Zk R
11. A uniform transmission line of length 1 and characteristic impedance 2k is short-circuited at z = 1 and has a switch in series with a battery of electromotance and resistance R connected across it at z = 0. Neglecting end field distortions, show that, if the switch is closed at t = 0, the battery current is n
o') (Zk - R)° 9 (Zk R)°+1
= eZ (2 8 = (:)
2n(p.e)1/ < t < (2n + 2)0.4011
12. If the Hertz vector has only a z-component so it satisfies the scalar wave equation 11.14 (1) for a plane wave transmission line, show that the potentials are = V(x, y)f[z - 080-It]
A = k(Aik)IV(x, y)f[z - (µe) -1t]
Show that when the scalar potential is eliminated by 11.01 (12), the resultant vector potential A' is identical in form with 11.14 (5). 13. A transmission line consists of two mutually external parallel cylinders of radii a and b, at an axial distance c. Show that the characteristic impedance is given by 11.18 (1) or (2) and the propagation constant by 11.18 (3), where c2
L = — cosh-1 27r
a2
b2
= coL , -
2ab
T' (b a)[c2 (b - a)2] a2)2 - 20(b2 a2) 21rabo[(b2 -
c.91
14. A transmission line consists of two confocal elliptic cylinders with major axes 2a and 2b(a < b). If the focal distance is 2c and K is a complete elliptic integral, show that 22 is given by 11.18 (1) or (2) and I by 11.18 (3), where L =In b +(b2 c2)1Y 22r ¢+(a
Ri =
T
21-26 a
j_ K (c)] a 'b
15. A transmission line consists of two parallel flat strips of width b and resistivity whose faces are at a distance a apart where a << b. Show that the propagation constant and the characteristic impedance are, approximately, T'
b
r2 = a-(7 +
27'
b6
(1 + j) +
jwkLa l
— {c/f2T'(1 + j) + jowasiV b2(3(-y
jcue)
16. A transmission line consists of two parallel cylinders of radii a and b, one inside the other at an axial distance c. Show that the characteristic impedance is given by 11.18 (1) or (2) and the propagation constant by 11.18 (3), where L = — cosh-2 2a
a2 b2 - c2 , 2ab
R — co—L , —
T(b— ¢)[(a +b)z - c2] a 2 )2 — 2c2(a2 281-abo[(b 2 -
b2)
ell
445
PROBLEMS
17. A lossless transmission line consists of a strip of width 2b centered in a slit of width 2a in an infinite coplanar conducting plane. Show from 4.29 (10) that the characteristic impedance given by 11.18 (2) is 2k = 4nKR1 - Va-2)+1[K(ba-91-1
where n is the intrinsic impedance of 11.15 (15) and K(k) is an elliptic integral. 18. Show that a transmission line of two coplanar strips, the spacing of inner edges and outer edges being 2b and 2a, has, from 4.29 (9), the characteristic impedance 2k = nK(ba-'){ - b2a-2)1l -1 19. A transmission line consists of a strip of width 2b centered in a cylinder of radius r = albi or of a cylinder of this radius centered in a gap of width 2a in an infinite conducting plane. Show from 4.29 (10) that in either case the characteristic impedance given by 11.18 (2), if n is the intrinsic impedance, is 2k
= inK[(1 - b2a-2)1][K(ba-1)]-1
20. A strip is coplanar with a semi-infinite plane with its edges at distances a and b from the plane edge where a > b. When used as a transmission line, show that, from 4.30 (2), its characteristic impedance is 2k = I nK(b2 a- 2){1f1(1 - bet-1)N -1 where n is the intrinsic impedance and K(k) is an elliptic integral. 21. A strip of width b with one edge on the axis of a cylinder of radius r, where r > b, forms a transmission line. Show from 4.30 that 2k = 4nK[(1 - b2r-2)4][K(br-9]-1
22. A transmission line is formed of two unequal coplanar strips, the wider having a width w. The gap between the strips is q and the distance between their outer edges h. Show from 4.30 that the characteristic impedance is 2k = inK(k){K[(1
-
k 2 = hq(w q)-1(h -
where n is the intrinsic impedance of 11.15 (15) and K(k) is an elliptic integral of modulus k. 23. A strip lies on a diameter of a cylinder of radius r, one edge being at a distance c1from its center and the other at a distance c2. Show from 4.30 that it - \ 2k = 4nKR1 - k2)11[K(k)]-1,
k = r(ci
e2)(r2
cic2)'
24. An infinite cylinder of radius c lies midway between two infinite parallel planes a distance 2b apart. Show from 4.31 that the characteristic impedance of this line is 2k = 4nK(k)(KR1 -
k = cos[Zircb-,(1 + X)]
where X is given by 4.28 (6). This is accurate to less than 1 per cent if c < lb. The attenuation may also be worked out by the method of 11.18 with the aid of HMF formulas. 25. An infinite strip of width 2a lies parallel to and halfway between two infinite parallel planes at a distance 2b apart. Show from problem 61 of Chap. IV that the characteristic impedance of this "strip line" is
2k =
inK(e-T.b-1) K[(1 - e-"rab-1)4] } -1
where K(k) is a complete elliptic integral of modulus k.
446
PLANE ELECTROMAGNETIC WAVES
26. The two halves of an infinite plane are separated by a gap of width 2h. An infinite strip of width 2d lies in the gap normal to the plane so that its center line bisects the gap. Show that the characteristic impedance of this transmission line is 2k = nK[h(d,
h2)-iliK[d(c12
10)-1]1-1
where n is the intrinsic impedance of the medium defined in 11.15 (15). 27. An infinite strip of width 2c lies midway between and normal to two infinite parallel planes a distance 2d apart. Show from problem 63 of Chap. IV that the characteristic impedance of this line is, if a =
2k = inK(cos a)[K(sin cO] —' 28. The center line of an infinite strip of width 2h is coplanar with, and at a distance
b from, the edge of a semi-infinite plane. The strip is normal to the plane. Show from problem 64 of Chap. IV that the characteristic impedance of this line is 2k
- 7/11((k)1-1-KI(1
k)2)11,
-
k = b-1[(h2
b 2)1— h]
29. A transmission line consists of two infinite conductors, a cross section of which forms two separate portions of the circumference of an infinite circular cylinder as specified in problem 65 of Chap. IV. Show that for this line
2k = inK(k)(K[(1 — k 2)4]}-1 where the modulus k is defined in the problem referred to. 30. A transmission line consists of two parallel strips of width 2h in two parallel planes a distance 2d apart. Their center lines lie in a plane normal to the parallel planes. Show from problem 66 of Chap. IV that
2k = nK(ba--1)(K[(1 — b2a-2)+] where ba-' is found as directed in the problem referred to. 31. One side of a transmission line is a circular cylinder of radius c at the origin. The other consists of four hyperbolic cylinders given by ±xy b2, Show that the character-
istic impedance of this line is
2k =
inK(k)(KR1 — k 2)111 -1
where the modulus k is found as directed in problem 79 of Chap. IV. 32. One side of a transmission line is a circular cylinder of radius a. The other is a parabolic cylinder y2 = 4p(x p) whose focus is the cylinder axis. Show that the characteristic impedance is
2k = inK(k){K[(1 — k2)4]1-1 where the modulus is found as directed in problem 80 of Chap. IV. 33. One side of a transmission line is a conductor of a rectangular cross section bounded by x' = ±b, y' = ±c. The other side is two conducting planes at y = ±a. Show that the characteristic impedance is 2k 7171K
[ MI] {K[(1 — A-IB)I ] I -1
where A and B are determined as directed in problem 82 of Chap. IV. 34. A plane wave in a medium of permeabilityµ and capacitivity E strikes normally a plane perfectly conducting mirror. Show that reflection is prevented by placing a
REFERENCES
447
thin layer of material of thickness d, capacitivity e', and conductivity y' one-quarter wave length in front of it if y'd ~ (e/ it)i and y' >> References R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Uses vector notation and treats waves on wires. BATEMAN, H.: "Electrical and Optical Wave Motion," Cambridge, 1915, and Dover. Discusses boundaries and wave solutions and gives extensive references. BORN, MAX: "Optik," Springer, Berlin, 1933. A complete, illustrated treatment. COLLIN, R. E.: "Field Theory of Guided Waves," McGraw-Hill, 1960. Chapters III and IV cover subject matter of this chapter in detail. DRUDE, P.: "Theory of Optics," Longmans, 1920, and Dover. This is the classical book on this subject and uses the long notation. FLUGGE, S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. Several of the articles in this volume contain material on plane waves. FORSTERLING, K.: "Lehrbuch der Optik," Hirzel, Leipzig, 1928. A complete and elegant treatment of the subject using vector notation. FRENKEL, J.: "Lehrbuch der Elektrodynamik," 2 vols., Springer, Berlin, 1926, 1928. GEIGER-SCHEEL: "Handbuch der Physik," Vols. XII, XV, XX, Berlin, 1927, 1928. Gives elegant treatment of electromagnetic theory in Vol. XII. HARRINGTON, R. F.: "Time-harmonic Electromagnetic Fields," McGraw-Hill, 1961. Chapter II covers much of this material with circuit analysis. HERTZ, H. R.: "Electric Waves," Macmillan, 1893. Drawings of radiation fields. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapters VI and VII are pertinent. Uses Gaussian units. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extensive treatment using long notation. KING, R. W. P., H. R. MIMNO, AND A. H. WING: "Transmission Lines, Antennas, and Wave Guides," McGraw-Hill, 1945. Practical detailed treatment of transmission lines, with some field theory. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives original treatment using long notation. PANOFSKY, W. K. H., AND MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Chapter XI is pertinent to this chapter. PAPAS, C. H.: "Theory of Electromagnetic Wave Propagation," McGraw-Hill, 1965. Contains a very lucid and detailed treatment of Maxwell's equations. RAMO, S., AND J. R. WHINNERY: "Fields and Waves in Modern Radio," Wiley, 1944. Simple treatment of waves on wires. SCHELKUNOFF, S. A.: "Electromagnetic Waves," Van Nostrand, 1943. Whole theory treated by means of the impedance concept. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Extensive and rigorous mathematical treatment of the whole subject. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Chapters IX, X, and XI cover this chapter with additional material. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. ABRAHAM, M., AND
CHAPTER XII
ELECTROMAGNETIC RADIATION 12.00. The Radiation Problem.—Any system generating electromagnetic waves and incompletely enclosed by conductors loses energy because waves escape into space. If the system is not supposed to radiate, this represents a leakage loss and the only information usually required of a calculation is the power radiated. If the function of the system is to radiate, as in the case of antennas or horns, the additional information needed may include the polarization and distribution in space of the radiant energy, the nonradiative losses, the current and charge distribution in the system, and its input impedance. These will depend on the impressed frequency and electromotance, on the system's dimensions and geometrical configuration, and on the materials of which it and its surroundings are composed. When the antenna lies in empty space, a complete solution of the problem must give fields which satisfy Maxwell's equations for empty space outside the antenna and for the antenna medium inside it and which satisfy its surface boundary conditions. These fields must also lead to electric intensity and current distributions that match those of the transmission line over its junction surface with the antenna. For existing materials and configurations, such perfect solutions appear impossible. There are, however, methods of calculating one or more of the desired quantities with considerable accuracy. One can always find, by means of retarded potentials, the fields of a given charge or current distribution. Often the distant radiation is sensitive only to the major features of this distribution, of which a rough estimate is then adequate. This also gives the ohmic losses, if they are small, with sufficient accuracy so that with the aid of Poynting's theorem the real part, but not the imaginary part, of the input impedance can be found. Information as to the actual current distribution in perfectly conducting antennas of simple geometrical forms like spheres and spheroids can be obtained by a rigorous solution of the boundary value problem in terms of orthogonal functions similar to, but more complicated than, those used in electrostatics. Skin effect formulas give the ohmic losses in terms of the magnetic fields just outside the surface. With the exception of the biconical antenna, such solutions give little information on the input reactance. The radiation from orifices is often found by assuming initial field values which are then corrected by trial to fit the boundary conditions. In the articles that follow examples of most of these methods are given. 448
§12.01
449
SPHERICAL ELECTROMAGNETIC WAVES
12.01. Spherical Electromagnetic Waves. Dipole and Quadrupole Radiation.—When the strength of a multipole that appears in 1.07 or 7.10 varies in time, it generates a spherical electromagnetic wave. The electrostatic potential of the electric multipole of order n and moment q(a) may be written 1 an(r-9 4ren! axi ax; • • • axk
(1)
) n
In rectangular coordinates xi = x, x2 = y, and xs = z. (1) 4,1 = qz x 4rer3
qy(1) y
q P)z
4rer 3
4rer 3
When n = 1, (2)
This is the NI,for a dipole of moment [(qp)2 + (412 + (c))9+ with direction cosines qPq-i When n = 2, qa), ey) and qL2) are linear quadrupoles on the x-, y-, and z-axes and Q,I2y) = g1,2x),ql2z) = q L2) and q,22 = 42) are square quadrupoles in the xy-, xz-, and yz-planes. Thus the moment is a symmetrical tensor with three different square moments and two different linear moments. One linear moment was eliminated by ,32 ay2 (3 2/az2)r—i = 0. the relation (a2/3x2 When q()oscillates, the currents transferring the charges determine the moments just as well as the charges do. Equation 7.02 (4) shows that the magnetostatic Azof an infinitesimal current element io dzo is a function of r only. If the current io f(t) depends on time and the field travels with the same velocity in all directions, then the radiation Az should be a function of r only. Thus the propagation equation and its solution are
_ a2A, r-2 a (219 Az) = ea2Az V2Az = tz at 2 v2 at 2 ar ar dzo (r1 cos 0 - 0 sin 0)f(t ± v-'r) A _ µiodzo i(t ± v-lr) 4irr 4rr
(3) (4)
The plus sign gives a converging wave and the minus sign a diverging one. From 1.07 (1), if the potentials of the two charges qofi(t) of opposite sign at a distance dzo apart are superimposed, then since a/ azo= — a/az,
_
dzo alfi(t ± v-lr) - q(i) cos or 4r ad_
47rEr2 Lf1(t
1r) + rti(t
v-11 e (5)
If the current element in (4) feeds the charges in (5), then io = ago/at and q(1) f1(t) = io dzo ff(t) dt. Thus (4) and (5) satisfy 11.01 (8), and Az and NI/ are the potentials of an electric dipole in the Lorentz gauge of 11.01.
§12.01
ELECTROMAGNETIC RADIATION
450
If f(t) is cos cot, then 11.01 (1) gives the phasor fields
_n
io dzo(1 _a Az _ \r2 Ora cos Oe—or car at _L 1 = nic = aA, sin 0e— jor 3 ) Ora r at r ae aAr = 1) sin 0e-20" B _ a(rA0) 4n- \ r ' r ar r ae Er
=
(6) (7) (8)
Replacement of io dzoby jwq(1)in (6), (7), and (8) gives the fields in terms of the moment q( i) cos wt based on the charges. Note that f3 = w(i.te)i =
= were = wmn 1
=
(9
)
Far from the dipole the 7-1terms alone survive and form the radiation field, while near it the r-3terms dominate and produce the quasi-static field. The r-2terms give the induction field. In the Lorentz gauge the Hertz vector gives more compact formulas. From 11.01 (8), using jovo) for io dzo and phasor notation, (4) becomes (1)
Z = Ze"'t =
ej(wt—sr) 41-Er
(10)
The linear quadrupole, le in (1), is found from pi') as in the static case by differentiation of ZP) with respect to z and division by 2!. In the wave equation a/ az and V2 may be interchanged only when applied to scalars or rectangular components of vectors. The formulas needed for putting Z into spherical components are, for the unit vectors, j
=
i = r1sin 0 cos ck ± 0 cos 0 cos 0 — 43 sin 0 r1sin 0 sin 0 + 0 cos 0 sin 0 + 4) cos 0, k = r1cos 0
—
(11) 0 sin 0 (12)
The result in phasor form with cos 0 written for z/r is 8wer 2
= q2) (/3r — j)(ri cos 0 — 0 sin 0) cos 0e—or
(13)
Similar procedures for Zxx and i„ give 8,ffer2isz = iqP2(f3r — j) sin 0 cos ¢e— jor 8rer2E„ = jqa)(13r — j) sin 0 sin 0e—or
(14) (15)
For a square quadrupole centered at the origin in the xy-plane with sides parallel to the x- and y-axes, Z is found by superposition of z aZ11)/ay and 28Z1,1)/ax where (10) is Z 1). This gives 167rer2112y) = qg) (Or — j)(i sin 0
j cos ck) sin 0e —or
(16)
§12.01
451
SPHERICAL ELECTROMAGNETIC WAVES
Likewise for square quadrupoles in the yz- and zx-planes, we get k sin 0 sin (p)e-of k sin 0 cos 0) e—or
167E742 = ez)((r - j)(j cos 1671-er2ZT = qg) (Or - j)(i cos 0
(17) (18)
From the above tensor components the most general electric quadrupole field can be built up. From 11.01 (1) E and B satisfy the same propagation equation, so for a magnetic dipole field, from 11.03 (8) as r 00, vB should replace the E of the electric dipole field of (6), (7), and (8), and from 11.03 (11) to propagate outward, -E should replace the B. Thus writing jcoM for iodzo gives Br
= I.LM (ji3
1 ± 775) cos 0e—or
241.
B9 = 47rI — ( T d r + -j4= wi/ MC - 47r r
7,1
(19)
sin 0e—i15r
(20)
1-
(21)
j ) sin 0e-jor r2
Note that as r -> 0 the r-3terms give exactly the curl of 7.10 (2). The mean energy radiated from a multipole source is found by integration of the mean Poynting vector, 11.17 (3), over the sphere of large radius r. This may be put in terms of the Hertz vector by noting that in its curl the only derivative which does not disturb the r-1factor is r-1 a(re-or)/ar, and when r is very large this simply replaces co' by - jOe-o r. It acts only on the 0- and 0-components of Z, so that if Zt„ denotes the tangential component of Z as r 00, 11.01 (12) gives, since jo.q.Le • -ji3 is w 2v
3,
kw = co2v-3(-02,,, + 420.,)
(22)
Now from 11.03 (8), ri x Et = vBt, so that Et. = co2v-2(020. + 4,2o.) H = 1,-4 x a= ,_i0,4v_5,1(z9,028. + zoca,w)
(23) (24)
For the electric dipole, from (10) and (12), 28,0 -
II = rico4 in 0[q(I)p = 2fl
-jq(1) sin 0e-ifir, 47rEr
P = _MT dS -
co4[q(1)] 2
4,3v[q (1)12
127rev3 =
30.4
1671-2E2i.w5r2 0)2 (io dzo) 2 r(io dzo) 2 r-----127rev3 30X2
(25) (26)
For the magnetic dipole, from (19), (20), and (21), the radiated power is
0,4m2 _ 47.3m2 w4/2A 2
P = f E13 ° dS = 127rev5 2A
3euX4
127rev5 —
473/2A 2 3euX4
(27)
ELECTROMAGNETIC RADIATION
452
§12.02
In 7.10 (2) M was defined as IA for a current / in a loop of area A. For a linear electric quadrupole, from (13) and (24), w 6r,(212
R 2w 4r n(2)12 f
P Jr dS =
2,5,[q (2)12
sin3 cos2 d0 — " " 480rEv5 12872E211v' 0
(28)
1500
For a square quadrupole, from (16), (11), and (12), then (24), = Oa° cos 0 sin 20 + 4) cos 2¢] sin Oe-or
f 2'S
P = ft dS — wqq1.2'12
25672 €0 o o
(29)
cos2 20) sin30 d0 d¢
(cost 0 sins 2
= w 6r,,(2)12 L'ixy J
3207-05
,52,ral2y)12 iv
500
(30)
12.02. Retarded Potentials.—A useful method of finding solutions of Maxwell's equations given in 11.00 is by means of retarded potentials. From 11.01 (6) and (7), the vector and scalar potentials propagate in a homogeneous isotropic dielectric with a velocity (1.46)-1. The object is to set up solutions of these equations analogous to 7.02 (5) and 3.09 (1) that give the potentials A and NI,at the point P at the time t. Both the electric current i and the electric charge density p are assumed fixed in position but with magnitudes varying with time. The treatment of moving isolated charges appears in Chap. XIV. The contribution of an element dv at the point xiyizi, having traveled a distance r from dv to P with a velocity (.1E)--1, must have left dv at the time t — (i./.6)1r and thus must represent the conditions at dv at that time. Summing the effects of all elements gives the P potential, from 7.02 (5) and 3.09 (1), at time t, A(x,y,z,t) =
‘1(x,y,z,t) =
iff
i[xi,
yi, zi, t
— (i.‘€)1r]
4.irj 4re
f fP
[xi yi
zi t r
"
(1)
dx i dyi dzi
— (1./e)i r]
(2)
dxidyi dzi
The fields derive from these retarded potentials as in 11.01. Note that the integrands for Az, A,,, Az, and ‘If in (1) and (2) have exactly the form f[t — (1.1E)1]/r of the propagation equation solutions found in 11.03 (1). In connection with 12.01 (4) and (5), it was shown that if the charges in the element dv in (2) are fed by the current in dv in (1), then 11.01 (8) is satisfied. Thus A and \If for the element dv satisfy 11.01 (8), and A is i.LE 8Z/dt. From this one writes down by inspection an expression for the Hertz vector that combines (1) and (2). Thus 1rcifffi[xi, Z(x,y,z,t) = iT
yi,zi, t r
(E)irl dxi dyi dzi
(µ6)1r]
dt
(3)
§12.03
453
RADIATION FROM LINEAR ANTENNA
The usefulness of (1), (2), and (3) depends on the accuracy of the estimate of the current and charge distribution in the source. If the source is a very thin perfectly conducting wire lying on a curve s, then no component of the electric field given by 11.01 (3) can lie along s and the equation of continuity must be satisfied. Thus
E$ =
at as
aA at
,= o
ap — at
— as
0
(4)
But, since the wire is infinitely thin, we may go so close to it that the wave length and the radius of curvature of s is infinite compared with our distance r. Then, if radiation is neglected, A and ir are given by the formulas for a long straight wire •If = — 27 ln r
A=
27r
In r
(5)
Insert these values into (4), differentiate one with respect to t and the other with respect to s, and eliminate A, or NI,from the result. Thus 321 as2
321 wat2
32 p as2
320 Aeat2
(6)
If the current is harmonic in time, the solution of (6) is
/(s) = I iej('t —es)
Leic.t+so
P(s) = biel((u —m
15 . 20(''+56)
(7)
The distribution of current and charge are therefore sinusoidal. This is the current form nearly always used in antenna theory but, as will be evident in the next article, the assumption of no radiation is not the only approximation involved. 12.03. Radiation from Linear Antenna. Retarded potentials may be used to calculate the radiation field surrounding a linear antenna. We use the cylindrical coordinate system p, cP, z and write z' for the coordinate of a point on the axis of which the antenna occupies that portion between z', and 4. Its diameter to length ratio is so small that the current forms a sinusoidal standing wave. The origin lies inside or outside the antenna at a point where there is, or would be, a current node so that —
I = I o sin Oz' cos wt
(1)
We can use either the potentials or the Hertz vector to get the fields. The physical nature of the terminal conditions is somewhat clearer in terms of the potentials. Whether or not the current is zero at the end depends on the terminal connection. The linear charge density is, by 11.00 (5), (2) a- = — (11E)1I 0 cos $z' sin cot
454
ELECTROMAGNETIC RADIATION
§12.03
Note that a is everywhere finite so the solution applies only when the lumped terminal charges do not radiate. It is a very simple matter to add to the solutions, when necessary, the additional terms arising from such sources. From 12.02 (1) and (2), the retarded potentials are Az =
2'sin 0z' cos (wt — 3r) dz' A/0 z
(3)
jzi'
= 0/0 f z2 _ cos (3z' sin (wt — 3r) dz'
(4)
4s-cot zi,
u
au au — = --, az'
_sin sin (wt— Or) OIL =
at
To get Ez from 11.01 (3), we substitute
p2.
where r2 = (z' — z)2
04410 fz2' 40
z
av
v = cos Oz',
UV,
wijo f 22f v du
az
,,
= —0 sin Oz'
470
z,,
Thus from 11.01 (3), we have for Ez Ez = —
coll.! 0
uv
"' Zi
cos 04 sin (cot — Or') r1 47r0
W it
cos 04 sin (cot — On)] r2
(5)
To find E, and I I 4, without the integration of (3) and (4), we substitute u = r z' — z and v = r — z' z so that r du = u dz' and r dv = —v dz' in the integrals of (3) and (4). With the aid of Dw 401.01 or Pc 591, these may then be written, using the upper sign for (3) and the lower for (4), Su'
(wt— 0z — /3u) du 2u
f v2sin (cot ± Oz — Ov)dv 2v
(6)
Note that p appears only in the limits of the integrals and that au1,2 =
497)1,2
P
op
r1,2
Op
(7)
so that differentiation of the integrals with respect to p gives p[ 2
sin (wt — — Oz') r(r z' — z)
sin (wt — Or + 13z1"' r(r — z' z)
(8)
Putting over a common denominator, substituting p2 for r2— (z' — z)2, and combining the sines of the sum and difference of cot — Or and 13z' give coil/ cos (wt — 0r2) sin 04 — cos (wt —girl) sin 134 470p 4 z 4 z sin (wt — 13ri) cos 04] (9) sin (wt — 0r2) cos 3z2 + r1 r2
Ep = = OP
RADIATION FROM LINEAR ANTENNA
§12.03
455
aAz 0[ — sin (cot — Or 2) cos 04 — sin (cot — Ori) cos 04 = — ap = 47rp
4
z
7.2
cos (cot — /3r2) sin )34
z
z
cos (cot — Ori) sin Ozd (10)
ri
Equations (5), (9), and (10) apply to antennas whose terminal loads are not zero, provided the loads charged by them are prevented from radiating by earthed shields like the coaxial termination shown in Fig. 12.03a. The cases shown in Fig. 12.03b and c can be treated by adding to the scalar potential of (4) the terminal charge retarded potential. If the load is replaced by an antenna section extending to the nearest current node, the charge on this section equals that on the load since the same
(a)
(c)
(6) FIG. 12.03.
charging current flows into it. Thus the integrals of o from the nearest node to zi and from 4 to the nearest node give, respectively, the charge Qi on zi and Q2 on z2 so that, from (2), Qi = — co-1/0 sin 04 sin cot,
Q2 = co-1/0 sin 04 sin cot
If the terminal loads are small enough to be treated as point charges, the corresponding retarded scalar potential to be added to (4) is
I
0 sin 04 sin (cot — Or2) ANY = / 47r-weL r2
sin 0.4 sin(cot — 13r1)] r1
(11)
The contributions to Ez and E, are, respectively, A.E'z—
402
— sin 04
a [sin az
(0.4—
07.2)]
r2 ± sin 0.z i
a [sin (cot —
az
a sin (cot — 0r2)]
AE, — ``)/L1°{— sin Oz'2ap 4702
r2 ± sin Oz'la
01
r1
[sin (cot — r1
ap
(12)
}
B.
Ori)
(13)
456
ELECTROMAGNETIC RADIATION
§12.04
The magnetic field of (10) is unchanged. It is interesting to note that Bechmann's derivation of the linear oscillator field by use of the Hertz vector, which is quoted by Stratton, gives this load type at both ends. If the antenna terminals are current nodes, then from (1), Oz',. and Oz', must be mor and mor, respectively, where m1 and m2 are positive or negative integers so that the cosine become (-1)m,and (-1)T42. The fields given by (5), (9), and (10) then take the simplified form (-1)m2 sin (cot — ,3r2 )
Ez _
wad-(-1)"4 sin (wt — (3r1) ri — 470 L z)sin (wt— On)
E, —
BQ =
r2
(4 —
-
(-1)miri
47rOpl_ 1.410
z) sin (wt— 13r2) ]
(-1)m2r 2
1Y41 sin (wt — On) — (-1)m2 sin (wt — 13r 2)]
(14) (15) (16)
Here m2 — mi = n is the number of current loops in the antenna. Suppose an antenna of length 21 has end a at z' = — 1, end b at z' = 1, and is driven from the center at an arbitrary frequency so that there may or may not be a loop there, but there are nodes at the ends. Find the fields for each half separately by (5), (9), and (10) and superimpose the results. The substitutions for ends a and b are, respectively,
zi = 0, 4 = 1, ri = ra, r2 = r, z — zi = Ti cos 0“, z — z2 = r 2 cos 0 zi = 0, z2 = —1, r 1= rb, r2 = r, z — zi = r1 cos Ob, Z = r2 cos 0 The angles 0, 0z, and Ob are measured from the positive z-axis. Substitution in (5), (9), and (10) gives, writing co = cO, Ez = cido r sin (wt — (3ra)
4/r F0 =
CA I iri:[2
4
sin
(wt
ra
—
Orb)
2
rb
cos131sin(wt
—
Or)]
(17)
cos 0 cos 01 sin (wt — (3r) — cos Oa sin (wt — Or.)
Bo = ,
[2 cos 01 sin (wt — (3r) — sill (wt — ziirp
— cos Ob sin (wt — (3rb)]
(18)
ra) — sin (wt — Orb)]
(19)
(
Note that /0 is the maximum current amplitude and, if it lies on the antenna, may occur at any point except at the ends. 12.04. Distant Radiation from Linear Antenna. We shall now obtain the radiation pattern at a great distance from a linear antenna of length 21 center driven at an arbitrary frequency with nodes at the ends. In l cos 0 for ra, r — l cos 0 for rb, and r sin 0 for p, 12.03 (19), write r whence —
B, = ( AolE e —
o[eos 131
— cos
(01 cos 0)]
27i-r sin 0
sin (wt — (3r)
(1)
DISTANT RADIATION FROM LINEAR ANTENNA
§12.04
457
From 11.17 (3), the mean radiation intensity in the r-direction is II = liEei
(2)
= 4/1-1E-11Boi 2
If the antenna is driven at resonance with a loop at the center so that 201 = nlr, it is clear from these equations that, when II is plotted as the radius vector in polar coordinates as a function of 0, the resultant curve has maxima and minima corresponding in number to the standing wave loops and nodes. The dashed curves in Fig. 12.04 show such a plot for n = 3. The total power radiated is found by the integration of tI over a large sphere centered at the origin. We shall calculate only the
FIG. 12.04.
resonance case. When we substitute -inir for f31 and write the square of the numerator in (1) in terms of twice the angle by Pc 577 or Dw 404.22, we get /Ica f'1 -(-1)'cos (nor cos0) P= 0 , sin 0 d0 1 - cos' 0 or Jo -
(3)
Substitution of u for cos 0 and resolution into two integrals by partial fractions show that the integrals are equal because one changes into the other when -u replaces u as the variable of integration. Thus,
r +1. ± cos ram +1 1+cos min/ 1 ± cos nr u du du du + 1+u j _1 2(1 - u) ) 2(1 + u r+1
5-1
Let v = nif(1 + u) and (3) becomes P=
mcio2 r2-1 — cos v
Sr Jo
dv
AcI8 [C + In (2n7r) - Ci(2n7r)] 87r
(4)
458
ELECTROMAGNETIC RADIATION
§12.05
where Ci(2nir) is the cosine integral tabulated in Jahnke and Emde and C is 0.5772. Insertion of numerical values gives for the radiation resistance Rr =
P
2P
=
= 72.4 + 30 In n - 30 Ci(6.28n) (5)
When n = 1, this becomes 73.13 ohms. The radiation resistance when there is not a connected loop at the driving point is given in 12.07. The assumption of an infinitely thin antenna is only one of the defects in the foregoing treatment. To maintain the postulated standing wave, one must supply to each element the energy lost by ohmic heating and radiation. The former is greatest at the current maximum near the driving point, and the latter is greatest near the ends of the antenna as can be shown with little difficulty from 12.03 (14) and (16). Thus it is evident not only that it is impossible to maintain the assumed current from any single driving point, but also that, to supply the losses, a damped progressive wave must be present which has no nodes and radiates in quite a different fashion from the standing wave. A rough qualitative idea of the effect on the radiation pattern of the weakening of the current near the ends of the antenna is obtained, for example, by superimposing the fields of two antennas, one oscillating in the n = 1 mode and one in the n = 3 mode. Assuming equal strengths and frequencies, this gives the radiation from a three-loop oscillator when the inner loop has twice the amplitude of the outer ones. Using (1) with 131 = +wir and simplifying by Dw 403.23 give 0.401.E0 =
,u/0 cos (iir cos0)cos Or cos 0) . sin (cot - Or) irr sin 0
(6)
When substituted in (2), this gives the radiation pattern shown by the solid lines in Fig. 12.04. We note that, although the outer loops have the same amplitude as before, the outer lobes are actually cut down in magnitude, but the central lobe is quadrupled in magnitude. 12.05. Radiation from Progressive Waves.—Consider a wave of uniform amplitude that travels along a wire from z' = -41 to z' = +41 where it is absorbed without reflection, giving a current /0 cos (cot - Oz'). From 12.02 (1), the field at x, y, z is, when r = (p2 z 2)4 is large,
aA,
= (A )+E9 - -
ap
f +1a f cos [cot —
4r -Op t
+ z') E} dz ]
r'
where r' = [(z - z') 2 p2I1 r - z' cos 0 and p = r sin 0. If r' >> 1, we may neglect terms in r-2so that, using Dw 401.11 or Pc 597,
Bm -
ao sin 0 sin(cot -(3r) sin [041 - cos 61 )] 2irr 1 - cos 0
(1)
CONICAL TRANSMISSION LINES
§12.06
459
The mean power radiated then becomes, by integration of 12.04 (2),
P
C"sin2 sine[01(1— cos 0)] sin B do (1 — cos 0)2
zlirEljo
(2)
Substitution of u for (31(1 — cos 0) gives, using Dw 403.4 or Pc 576, P
=
f 21311 — COS u du -
47rEI[ j 0
1
r 2f1/
2,31j (1 -
cos u) du]
The first is given on page 3 of Jahnke and Emde in terms of Ci 201 so that the substitution of numerical values and of 271-/X for 3 give / 47/ in — — Ci
P = 30I o[
sin (47//X) 4 /X
(3)
12.06. Conical Transmission Lines.—The radiation fields of a linear antenna driven at the center are given by 12.03 (17) to (19). In 12.04 (5), we found the radiation resistance of such an antenna with a current loop at the driving point. No antenna reactance information is obtained by considering such infinitely thin wires because their inductance per unit length is infinite. If, to avoid this difficulty, we choose a radius not zero, then the infinite capacitance appearing across the infinitely narrow gap between input terminals again prevents reactance calculations. Schelkunoff eliminated both these difficulties by considering a biconical antenna composed of two cones whose apexes meet at the driving point. To simplify matters we should first discuss the biconical transmission line. The propagation equation in spherical polar coordinates for a wave of angular frequency co is, from 11.01, 3.05 (1), and 11.15 (3), 1 sin
a(sin
ao
a r2 aw a¢22 + ar ar
1 a2 W
NI
+ sin'
r 2.2W = 0 (1)
To break this into two equations, we equate the first and second pair of terms separately to zero and obtain 5.12 (6) and 12.01 (3) with P2 in place of —v2/w2. Thus from 5.12 (7) and 12.01 (4) for an expanding wave, W = r —q7(0, 4))e—'r cos (wt — Or)
(2)
where a is the attenuation constant, O the wave number, and U(0, 0) + jV(0,
= F(eio tan 20)
(3)
The vector potential for a transverse electromagnetic wave is A = V x rW r--— 71x V2Ve—ar cos (0.4 — Or) = V2 Ue—ar cos (Cot — Or) (4) where V2 has 0- and 0-components only. This is identical in form with 11.14 (5) and shows that, when an alternating potential is applied between the apexes of two or more perfectly conducting cones whose surfaces are
ELECTROMAGNETIC RADIATION
460
§12.06
generated by radius vectors, a spherical wave is transmitted that is just equivalent to a plane wave on a cylindrical transmission line for which x = cos 4) tan 20,
y = sin 4,tan IO,
z
(5)
Let us now consider the special case of two circular cones of half angle X1 and x2 whose axes intersect at an angle 21//, as shown in Fig. 12.06. Clearly, if the trace of these cones on a sphere of radius r is projected
/
/
/
..,
/ ..•••'"*. ••• / ..,..••• / I ..,--"""
14, - _ ..,_ I--- , , " I
-..
\\ \
\
\ \ ,„,_, \
.47/
...."1
\ 1
-
-'---,r-.------1'-{ -.... S . I'
I
\■
.... ....-
/
2R2
/
FIG. 12.06.
stereographically on the tangent plane, as in 6.20, the equivalent transmission line is the two cylinders of radii R1 and R2 at an axial distance D whose sections are shown in Fig. 4.13b. From Fig. 12.06,
D+
R1, 2 =
2r tan -i(//
xi,2),
4./) —
R1, 2 =
2r tan 4Ni — xi,2)
Take the ratio of these equations, apply Pc 602, and solve for 4D/R1,2.
D
_ sin 1,/, sin X1,2
2R1,2
(6)
The capacitance per unit length of the equivalent cylindrical line is, from 4.14 (2) or 4.13 (5), according to whether x =xi = X2 or xi xi,
e CY
1 sin 1,G 1 4 sine >L — sin2 — sin2 = — cosh-1 or 2.n.cosh-1 (+ or sin x 2 sin xi sin x2
(7)
The negative sign applies if one cone encloses the other, otherwise the positive sign is used. By 11.15 (10), the characteristic impedance is 2k = e
C -y
jwilE y= (a ± ji3)C jcoe
jcoli
(8)
§12.07
THE BICONICAL ANTENNA
461
where 12 is the permeability, E the capacitivity, and y the conductivity of the medium surrounding the perfectly conducting cones. When I/. = lr, simpler formulas are given by taking 0 = 0 as the axis of one of the cones so that U is independent of 4). For this we choose F(u) to be C In u in (3) so that from 5.212 (4) U voln 0 1 n ttaann : x
v 0, W/cos 0) vo cos x)
(9)
where the potential between the cones and the capacitance per unit length are, respectively, V = 2Voe— 'r cos (wt — Or)
C = rc(ln cot +x)-1
(10)
12.07. The Biconical Antenna. When both biconical transmission line cones terminate, they form a biconical antenna. If open at the ends or if closed by spherical caps, they fit into the spherical polar coordinate system to be discussed in 12.11, and the field calculation is a boundary value problem in this system. The method works best for cones whose half angles are nearly zero or 90°. In the small angle case, numerical work is simplified by using a different method. Take perfectly conducting coaxial cones of equal length whose half angle x is so small that by 12.02 (4) to (7) the current distribution is —
i = /0 cos cut sin (0/ — 13r)
(1)
The fields are then given by 12.03 (17) to (19) and the radiated power by an integration of Poynting's vector over the conical surfaces. Let xa, x, and xb be the angles subtended at z = —1, z = 0, and z = 1, respectively, by a radius p of the cone ending on its surface at a distance r from its apex. By Poynting's vector 11.02 (3) and the magnetomotance law 7.01 (2), the total instantaneous power radiation from both cones is P = 2.10.Erli-021rp dr = 2.1 0 (Ez i cos x
E,, sin x) dr
(2)
Inserting 12.03 (17) for E„ 12.03 (18) for E„, and (1) for i gives
P=
CAP cos
2ir
wtfiRcos x cos
x°)S. (COS X
rb
COS Xb
)8b1 sin 01 — )3r) dr
where 8„ and Sb have been written for sin (cot — )3r.) and sin (wt — Orb), respectively. The terms containing cos /31 cancel out because p = r sin x. r and 1 — r replace ra and The half angle x is chosen so small that 1 rb and cos xa, cos x, and cos xb equal one. If 8„ and Sb are split up so that sin wt and cos cot factor out, then with the aid of Dw 401.05 to 401.07
462
ELECTROMAGNETIC RADIATION
§12.07
or Pc 595 to 597, we obtain for 27rP(c/2/2)-1 1 sin 20.4f 1 1—) sin (L — R)]dr (sin L — sin R) ( 1 1 —r r 4 6,[ (1 r l — 1 cos 2cot f [1 — cos (L — R)]} dr (cos R cos L) r(1 — r) 2 6 { r(1 r) where R is written for 20r and L for 201. The quantities S and 5' are to be set equal to zero except where so doing makes a term infinite. In Jahnke and Emde we find the cosine- and sine-integral formulas asin Ax dx = Si(Aa), jo x
— cxos Ax d
x = C + In (A a) — Ci(A a)
where C = 0.5772. With these formulas, the power expression integrates easily and In 5 and In 5' cancel out so that all terms are finite. Thus
P
cµ/8 sin 2cot
.
{2S1(20/) — Si(40l) cos (200
In (131)] sin (MI [Ci(400 C c/2I cos2cot [C Ci(401) — 2Ci(2(3/) {2C + 2 In (200 — 2Ci(201) 47r [Si(401) — 2Si(2(3/)] sin (2001 (3) + In (131)] cos (200 —
—
Comparison with 11.15 (18), where 62 is replaced by /822, shows that the coefficient of ig sin 2wt is 22, sin 1,G or 2X,, where X, is the radiation reactance and that of Ig cos2 wt is 2, cos 1,/, or R, the radiation resistance. If there is a current loop at the driving point so that 23l = n7r where n is odd, the value of R, is identical with that of 12.04 (5). With a node at the driving point, n is even so the result is different because in (1) and (2) we assumed the current similarly directed in the two halves, whereas in 12.04 (4) they are taken in opposite directions. Although we have now established the power loss from the antenna, we still do not know where to put it in the line to give the correct input impedance. Examination of (1) shows that Ho and hence the power loss are zero at the current nodes so it cannot be put at the ends. A simple way to discover its location is to consider the case in which the perfectly conducting antenna is tuned so that the only power loss is by radiation. There is then a current loop at the driving point so that 131 = (2n + 1)r and the input current amplitude is /0. Insertion of this value in 11.15 (13) after setting the attenuation constant equal to zero gives for the input impedance and the mean power expended
2, =
P = 422'
(real part)
(4)
ANTENNA ARRAYS
§12.08
463
where 2t is the terminal impedance. But (3) gives 12,n for the mean power so for a tuned antenna Z,Z, = Zk. We try this value in 11.15 (13) for the input impedance of an untuned antenna and, to justify it, show that for thin antennas the resistive power loss agrees with (3). Thus 2, =
2k cos 0/ ± j2, sin SI Z, cos 0/ + Z k sin (3/
(5)
Now let Zk become very large, which means a very thin cone, so that the input current becomes /0 sin 0/ and the power input is, from (4),
Pi = 12,g sin' 0/ = 1(2, — ij2k sin 200/g (6) This gives a resistive power loss of 2/?,n as it should. We note that (5) is the same expression that would have been obtained from 11.15 (13) 350 300 250 v3200 E o 150
100 50 0
it
cm
IITEMB1111 III MILIMIEWAI■ 11111M111 EFIL MOM■ NII / NI !A Eli 11011111 2 3
4
5
2rrl/a
6
1
8
9
10
FIG. 12.07.—Real (solid) and imaginary (dotted) parts of ik2 /Z plotted against 271-1/X.
if we had written 0/ — 1r for pl and 2, for 2L. This is equivalent to cutting 4Xoff the line and using Z, as the terminal impedance. Thus the effective position of the radiation impedance is one-quarter wave length from the end of the line. The inverse of the radiation impedance given by Schelkunoff in Proc. I.R.E., September, 1941, pages 493-521 appears in Fig. 12.07. Clearly from (5) the resonance wave length at which Z. is real is affected by the radiation reactance, but it is still near 20/ = na if Zk »X,. 12.08. Antenna Arrays.—The radiation pattern of a linear antenna is symmetrical about its axis. To concentrate the radiation in a single direction, a "directional array" is needed in which several radiators of length 2/, usually identical in type and parallel to each other, are set in some regular pattern. These may differ in amplitude and phase of
464
ELECTROMAGNETIC RADIATION
§12.08
excitation. Let us consider radiators parallel to the z-axis and lying in the positive octant of a rectangular lattice structure, the spacing between and z-direction, respectively. adjacent ones being a, b, and c in the x-, With integral u, v, and w the distance of radiator uvw from the origin is
ruvw = iva
.ivb
(1)
kwc
Let the radius vector from the origin 0 to the very distant field point P be rr1so that the difference in travel distance of a signal from 0 to P and from radiator uvw to P is r1 • ru„.. Assume the dimensions of the array small compared with r so that all radiators can be considered as equidistant from P for field amplitude calculations. If oscillator uvw lags behind oscillator 000 by a phase angle iku,no and has a current ./uvw, then from 12.04 (1) its contribution to Bo at P is the real part of ,,[cos j31— cos (01 cos 0)] . 27rr sin 0
u
Writing
Fu. for the first factor gives for the whole array 14 0 = (11f)1E0 =
(
U
v
Or
(2)
W
For identical radiators F is constant, and this formula simplifies if, when going in the positive x-, y-, or z-direction, each radiator shows a constant phase lag n, or respectively, behind the preceding one. The triple summations in (2) then breaks down into a triple product of summations.
FZZ
ei[u (Orr ja– E) +v (Pr rjb
-Fw woe, _0]
u v w FZeiu(firi.ja-0 Zeiv(Orrjb—n)Zeiw(Prpitc—r)
(3)
Each factor is a geometrical progression which may be summed by Dw 26 and gives, when mxis the number of oscillators in the x-direction, &my n=o
sin = 1 — ejm-# 'Pe—A. (nix-1)# .— sin Ilk 1 — 04'
(4 )
From 11.17 (4) the mean Poynting vector is 1(A3E)--ito • Bo so that F2 sin2[1m.(1'r1 is — )] sin' [Inty(Rri •ib n)] sin2 [ilrl(Ori • kC 2 mitt S1112 [El3r 1 • is E)] sin2[1(Or1 • ib — 0)] sin2 [1(07.1. kc — —
(5)
465
ANTENNA ARRAYS
§12.08
As already stated, an array is used to concentrate radiation in special directions. The directivity or gain G is defined as the ratio of the maximum intensity (Im to the average intensity .130on a large sphere concentric with the array. In decibels we write it Gd. Thus G
=
c13m
G d = 10 logio
30
cI3m
(6)
—
The gain function G(0, 95) in any direction is the ratio of 4)(0, 0) to Thus for a half wave antenna, from 12.04 (2) and (4), the ratio of maxiC — Ci(27)J-1which gives a mum to average intensity is 4[1n numerical value of 1.64 for G or a gain of 2.15 decibels in the equatorial plane. The transmitting pattern is the surface. r=
G(0, 4)
)
c,(0,
)
(7)
.4)3,1
Gm
Now consider the special case of m identical, in phase, n-loop antennas, lying a half wave length apart along the x-axis. Take my = m2 = 1,
FIG. 12.08.
= 0, i3a = 113X = 7, and 3l = inir, so which becomes
- img 8,2r2,1
On • is
is r cos 4) sin 0 in (5)
cos (lnir cos 0)121-sin (Inor cos .45 sin 0) sin sin (fr cos 4) sin 0)
_IL
12
(8)
This is called a broadside array because the second factor maximum is at right angles to the plane of the array when 4) = fir. Figure 12.08 shows the relative values of II in the plane 0 = 0 when subsidiary maxima are omitted. Near this maximum the second factor sines are small so it becomes m2. We cannot assume from this that, for the same total power input, the radiation in this direction is m times that of a single oscillator because the oscillators interact. To get the actual gain, we must calculatect,o. We note from (4) that the last factor in (8) may be written, if a is 7 cos cb sin 0, —
1
sin Imaeii(m-1). 2 sin la
2
m-1
= + 2 (m — p) cos pa p =0
P=1
466
§12.09
ELECTROMAGNETIC RADIATION
Expansion of the cosine by Pc 773 or Dw 415.02, integration over 4) by Pc 483 or Dw 854.1, and combination of the terms independent of B give rn-1
2M71"
('2r (m — p)f cos (pr. cos 4) sin 0) d4)
p =1 M—1
a.
= 4m,
021
(m — P)CiPir sin 6
228-1(s!) 2 p =1 s =1
To get the power radiated, multiply this by the remaining factors in (8) and by r2 sin 0 de and integrate from 0 = 0 to 0 = 7r. The first term was integrated in 14.05. The other integrals have the form cos2 (4717r cos 0) sines-10 de = j [1 + cos (nir cos 0)] sin28-10 de
We may integrate this by Dw 854.1 or Pc 483 and 5.302 (5) which gives
—
[(s
1) U2228-2 (2s — 1) !
1) ! 2 8—i 221-8-1 n 3-1\mr1
(s —
The s-summation of the first term by Dw 442.11 or Pc 347 gives a result expressible as a cosine integral by page 3 of Jahnke and Emde. Thus pal
2 s =1
(-1)8(137)28 = —2f 2s(2s)! c■
— cos xdx = —2C — 2 In (pr) x
2Ci(pr)
Collecting terms gives for the total power radiated p
A/ C{m 2C + m 2 In (2nr) — 8r —
1
— p =1
rr
m2Ci(2nir) co
Ci(pr) — C — In (pr.) ' s =1
(-1)878±1p28 r 4ss!(2n)8-1 ''s-1(nir) (9)
From (6) and (7) the directivity in decibels is 10 log [11.4cI 2m2/ (irP)]. For n = 1, m = 2, the ratio 43mj4 0 is 3.81 which is more than twice that of a single half wave antenna. The directivity is 5.81 db. 12.09. Earth Effects.—Dielectrics or conductors near an antenna react on it or distort its field or both. The commonest such object is the earth's surface. To a nearly spherical wave one may apply the laws of reflection and refraction at a conducting boundary derived in 11.13. If, on the other hand, the antenna is so near the surface that the angle of incidence is a complicated function, this becomes laborious. Often
§12.11
467
SOLUTIONS OF THE WAVE EQUATION
the surface may be taken as flat and perfectly conducting so that the fields above it are unchanged if it is replaced by a second identical or "image" antenna located and oriented so as to make the resultant electric field normal to the former earth's surface. If all the original antennas were either normal to or parallel to the earth's surface, then evidently the resultant fields may be calculated by the formulas of the last article. 12.10. Uniqueness of Solution. Before deriving the wave equation solutions used in boundary value problems, one should find what data are needed for a unique solution. Consider a region, bounded internally by the surfaces 51to 5„ and externally by So, wherein and y are functions of position but not of field strength, and which contains no sources. Let E, B and E', B' be two solutions of Maxwell's equations which are identical throughout the region when t = 0. Poynting's theorem 11.02, and Ohm's law give, writing AE for E E', AB for B B', etc., —
E,
—
f,
(Ai) 2
{
7
a[E(AE) 2 2
at
—
1} dv =
(AB) 2 2/.4
is,AEx AB
n dSi
o
In order that the surface integral vanish, it is necessary that AE x AB • n = AB x n • AE = n x AE • AB = 0 Thus, if n x AB or n x AE is zero when I > 0, the surface integrals are zero. The energy term in brackets is zero or positive and was zero at t = 0 so, if it changes at all, it must become positive. But the first term is zero or positive, and the whole integrand is zero so that AE and AB are zero. Thus the E, B and E', B' solutions are identical and are determined by the initial values of the fields in the region together with the tangential components of either E or B over its surface when t > 0. In practice we are usually concerned with steady-state solutions of the problem, in which case the values at any time determine all previous values. 12.11. Solutions of the Wave Equation in Spherical Coordinates. In an isotropic insulating medium all waves are spherical at distances from the source great compared with its dimensions. In radiation problems, therefore, the most useful solutions of the propagation equation are in polar coordinates and have the form of a sum of products of orthogonal functions with coefficients which can be determined from given boundary conditions. We saw in 11.01 that in such cases the entire radiation field is derivable from a vector potential with zero divergence. Expressions for such a potential can be obtained from two solutions of the scalar propagation equation as indicated in 10.01. Thus from 10.01 (2), if Wte and Wem are solutions of the last two of the equations, V2A = _ mtc02A, v2w,, = v2wtm =— µew 2Wtm (1) —
468
ELECTROMAGNETIC RADIATION
§12.12
then a solution of the first equation is obtained from the formula r x VWt”.)
A= v x (r1,17t
(2)
Note that the solution derived from Wte gives a vector potential, and hence an electric field, normal to r. This is called a transverse electric wave as the subscript suggests. From 10.01 (4), the magnetic induction is r x vWte) B = —v x (Acce2rW,„, (3) Thus the magnetic field derived from W,„, is normal to r, and the wave is called transverse magnetic. We solve (1) as in 10.05 except that for more generality the factor 4)(0) is added. We therefore write ITV = r-IE(r)0(0),I3(0)
(4)
Substitution of (4) in (1) gives the differential equation of 10.05 (9) for f?, that of 5.14 (2) for 0, and d24,/d 02 = _m 2c13 for (D. The solution for W then becomes, writing ,3 for w(ME)1and u for cos 0, W = [APB(u)
BQZ(u)][0j„(13r)
bk.(j3r)] cos (met. + 3,„) (5)
This spherical Bessel function notation is that of 5.31 and 5.37 where j, ( 3r) = (2(r)-IJ„±1(Or),
k„(jOr) = (1j7r0r)-1k„±1 (jOr)
(6)
The reason for using these forms is that the first, combined with em, represents a standing wave and the second, from 5.37 (9), an expanding or contracting traveling wave according as v is positive or negative. 12.12. Polynomial Expansion for a Plane Wave.—A plane wave expansion in the form of 12.11 (5) is often needed to fit spherical boundary conditions. For simplicity consider a wave parallel to the 0 = 0 axis. It can then be referred to any other axis by 5.24. We saw in 11.15 (4) and (5) that z = r cos 0 enters plane wave formulas only in the exponent and multiplied by P. Thus we need the coefficientof P,,(cos 0) in the Legendre polynomial series expansion of err cos O. From 5.156 (3), +1
an =
2n + 1(pr)ni 2. +in ! • , 1
errna
u2\n du
(1)
Expansion of the exponential by Pc 759 or Dw 550 gives 2n + 1
an=
"K1 (fr)8 (P-r)"2 j s.
s =0
r 1
u(1 - u2)„ du
(2)
The integral vanishes when s is odd, so write 2m for s and take twice the value of the integral from 0 to 1. For the (2m)! in the denominator we may write 22mm!ir-IF(m + 4-) by Dw 850.7 and 855.1 and obtain with the
§12.13
RADIATION FROM UNIFORM CURRENT LOOP
469
aid of 5.32 (2), after writing 1'(n + 1) for n!, CO
a"'
r(nt + i)F(n + 1)
(Fr)n+,-
(2n + 1)71-1
2"-T(n + 1) L122"'rn,!r(m in=o
(m n
(3)
CO
.= (2n + 1) / -
irt!r(m
(2n + 1)G, ) in±.}(fr)
n
m=O
—
When there is no attenuation, we replace is by j0 and obtain by 5.32 (2) r P n+I(Or) = jn(2fl an = (2n + 1)jn6
1):7„03r)
(4)
The expansion is, therefore, \\ 09' ens 9 = Zjn(2n
1)jn(3r)Pn(cos 0)
(5)
n=o
For e—jo' replace /3 by —13 or cos 0 by cos (ir — 0) which is equivalent, by 5.293 (3) or 5.157, to inserting the factor (-1)" on the right. Thus
e--Jos == e--,fic cos o =
( 3) (2n + 1)jn(3r)P„(cos 0)
(6)
n=0
12.13. Radiation from Uniform Current Loop. Magnetic Dipole.— The orthogonal solutions of 12.11 usually occur in boundary value problems but may also be preferable for sources with given current distributions. For example, the radiation from a uniform current loop of radius a found, when X >> a and r >> a, by retarded potentials and given in 12.01 (19), (20), and (21) is more difficult by that method if r < a. From symmetry this source gives a transverse electric field with no free charge and hence, from 12.02 (2), no scalar potential and a vector potential with only a ct,-component. Therefore when r > a we have, from 12.11 (5), Wee =
ZA0 P0(cos 0)j.((3a)k.(jor)
(1)
n=0
If r < a, interchange r and a. By 12.11 (2), the vector potentials are r>a
r
:=1.-„P(cos 6)j„(13a)k,,(j(3r) n=o
(2)
Ao = Z;InP,I(cos 8)j„(3r)k.(j,l3a) n=o
(3)
=
470
ELECTROMAGNETIC RADIATION
§12.14
Clearly these potentials are equal at r = a, and only odd values of n can appear in them for the field must be symmetrical about the 0 = plane, so we replace n by 2m + 1. The uniform current implies from 12.02 (7) that Oa is very small so that we may write (0r)2"0-1/(4m 3)!! for j„((r) where r < a and (4m + 1)!!(jf3a)-2.=-2 for kn(ifia) from 5.31 (9) and 5.37 (11). Except for the time factor, (3) is real and identical in form with 7.13 (2) if a = 47T so that the latter, multiplied by cos cot, gives A4, when r < a. Write (-1)m(2m 1)!! for Pl„,±1(0) by 5.23 (6) and 5.157 and substitute the An, value found from (3) in (2) which gives r > a, A4,
— 0
(2m — 1)!!(0a)2'n-I-2 , 2.+1 tcos 0)k2.+I(ji3r) (4) (2m + 2)!!(4m + 1) !!/-
To get the real part of A00", we may substitute, from 5.37 (13),
[0"k2,,,±i(jOr)1,real part = — 1)m[n2m+I(Or) cos wt — i2m-1-1(37") sin cot] When r is large, all but the m = 0 term may be neglected because 13a is small so that n1 and ji are given by 5.31 (3), (4), (8), and (10). The expression for A4, then becomes = ii.LIa2r-2sin 0[cos (wt — (3r) — Or sin (wt — Or)]
(5)
The electric field —m/at agrees with 12.01 (21) provided M replaces ra2I and v x A gives 12.01 (19) and 12.01 (20). From 12.01 (27) the power radiated and the radiation resistance are
P = 153,9000/2X-4
Rr = 2P1-2 = 307,8000X-4
(6)
Thus, in terms of antenna current, the loop radiates power inversely as the fourth power of the wave length instead of the linear antenna's square. Note from (3) that when r < a there are only standing waves. 12.14. Free Oscillations of a Conducting Sphere.—The solutions of the propagation equations given in 12.11 are evidently well adapted to the study of the electromagnetic oscillations of spherical bodies. From 12.11 (2), we see that both transverse electric and transverse magnetic oscillations are possible. Examination of 12.11 (5) discloses that outside the sphere there is a standing wave if D is zero, from 5.31 (7) and a diverging wave if C is zero from 5.37 (12). The Q;" solution is excluded because it is infinite at 0 = 0 and 0 = 7r. A dielectric or imperfectly conducting sphere will have fields inside it. We then use a sum of solutions of the type of 12.11 (5) with undetermined coefficients which are different for the interior and exterior regions. The vector potentials derived from these solutions by 12.11 (2) must be matched at the boundary to satisfy the boundary conditions 7.21 (6) and (7) as was done in 10.06. A simple and interesting case is that of a perfectly conducting sphere
§12.15 OSCILLATIONS OF DIELECTRIC OR CONDUCTING SPHERE 471
where, if there are initially no internal fields, there can be none at any subsequent time and energy can be lost by radiation only. The unknown frequency appearing in 12.11 (5) must be determined to fit the boundary conditions which usually make it complex. Let us consider first the transverse magnetic oscillations such as would be set up by moving the sphere in a static electric field. The simplest of these will be that for which m = 0 and n = 1 in 12.11 (5) so that by 12.11 (2)
A =
x (r x VIkt„,) =--
=
A- p[ 2r1cos r
r.
rk „,\
a e ae(sm .
r sin
t
ae ) 0 sin 0 a
.
ki(30pr)
r
0 a2(r Wt„) r ar ae
.
ar(ricl(Ar)
(1)
p
The condition that the tangential component — aAe/at of the electric field be zero at r = a requires, from 5.37 (12), that ( -1
1 ± 1)e— joa = (1 jRa
± 31 1 13a
Oa + 2
(32a2
+
j — 31)e—jfla = 0 (2) 2
Putting these two values of Op = cop(.1e)1into 12.11 (5) yields for yyt,,e1't
_ 2 .r sin 0{il 11'4
25t010-i
+ 33/4)a e 0+31)a
[
A2k
-
—2 jr 31)a ]e
1
-2;t(µ.> - # (.2
—3i)a
(3)
With the aid of 5.37 (10) and (12) and of the substitutions 0.)
a = 44110 -1a
-
= 31a1
1,
(4)
the real part of Iktmej.t may be written, after collecting constants, Cr-2a sin 0[31(a — r) sin (wt — Or — — (a + r) cos (cot
—
— ik)]e—at±ia-ir
(5)
From (4) and 11.09 (4), we see that the wave length of the oscillation is 7.26a and that the amplitude is reduced by the factor 1/e while the wave travels a distance equal to the diameter of the sphere. The oscillation is therefore very rapidly damped and disappears in a few cycles. The transverse electric modes can be calculated in similar fashion. An analysis similar to that given, but more complicated, applies to a prolate spheroid. When the eccentricity of such a spheroid is large, it gives an excellent approximation to a straight wire of finite length. 12.15. Forced Oscillations of Dielectric or Conducting Sphere.—The formulas of 12.11 and 12.12 permit a rigorous calculation of the steadystate diffraction effects caused by a plane monochromatic electromagnetic wave with its electric vector in the x-direction and traveling in the
472
ELECTROMAGNETIC RADIATION
§12.15
z-direction, impinging on a homogeneous sphere. We shall consider only the cases where the sphere is a perfect dielectric or a perfect conductor. From 11.15 (4), 11.15 (5), and 11.01 (2), the phasor electric field, magnectic induction, and vector potential of such a wave are Ez = (AE)-11-1, = Ee-°.,
A. =
(1)
Both E. and by have r-components, so in polar coordinates both TE and TM waves are needed. From 12.11 (2) and (3) we have
Ar = AS sin 0 cos ¢ = [V x (r x Br = By sin 0 sin 4) = —[V x (r x V -Ce)],
(2)
In 12.12 (6), e-50' appears as a sum of terms like 12.11 (5). Differentiation with respect to 0 brings in the sine factor of (2), so
jI3r sin 0 e-o.
—(—j)n(2n + 1)j„((r)P!,(cos
(3)
n=1 Ar =
Br
=
—E co s w r (1)
I
(4)
(—j).(2n + 1)jn (6r)P;,(cos 0)
(5)
n=1
E sin (1)
jour
1)in(Or)P(cos 0)
( —i)n(2n
n=1
Operating on the nth term of 12.12 (5) and using 5.12 (4) give [V x (r x VC)]r = [V x LhaWn -1 ae — sin e aTkn )jr — 3'44 1 a 4asn 1 828,i1 _ r L sin 0 ae\sin ae) sin' 0 802
n(n + 1) -
W>,
(6)
Therefore those parts We'„, and Ceof W,,„ and W,e that represent the plane incident wave are found by multiplying the nth term of (4) and (5) by —r[n(n + 1)]-1. Those parts that represent the diffracted wave must be diverging waves and so contain kn(jfir). Thus we have
= E cos oN 2n + 1
[(
(03 .Jn(n + 1)
j)nj.(0r)
Ankn(j,3r)]Pl(cos
(7)
n=1
E sin ON 2n + 1 . . bnicn(jOr)]P!,(cos 0) . Jit(n ± 1) [ ( 3)ni n(00 + n=1
(8)
§12.15 OSCILLATIONS OF DIELECTRIC OR CONDUCTING SPHERE
473
Inside the sphere, fields must be finite at the origin so that E cos ON 2n ± 1 . . ( 3)'2=1„,,n(O'r)P4(cos 0) 6.0 G,J n(n 1) n =1
E
W ei
(/)
3u)
2n ± 1
j„(f3/r)P;i(cos 0)
1)`
n =1
From (6) and (2), equating normal displacements gives e,An,j,t(tVa) = e[j„(0a)
jnilnicn(j0a)]
From (6) and (2), equating normal magnetic inductions gives —bnon(13/a) — in(0a)
jntinkn(j/3a)
From (6) and 12.11 (2), the tangential components of A are
1 a2(rtivt,.)] ae ar
A, = i f 1 a-rk, sin
e ac5
r
1 a2(r ar ae + r sin 0
Equating tangential components of E or A involves only the r-derivatives of Wtin because the Wte terms are already equal by (12). The same relation is obtained from the 0 or 0-component. -
A a[ajn(O'a)]
as
_ a[ain(0a)] as
.„A- a[arcn(i0a)] 3 as
(14)
By comparison of 12.11 (2) and (3), the contribution of Wt„, to A, is identical with that of WIe to B,. Equating either 0 or 4) components of 1.4,-'B at r = a involves only the r-derivatives of W,e because from (11), (13), and 12.11 (3) the Tkt„, terms are already equal. Thus -
1 a[aj.(0/a)] — B„, as -
lia[ajn(i3a)] p as
-
• 3nBn a[ak n(i0a)1} aa
(15)
We now equate values of A,, from (11) and (14) and solve for An. Use of 5.37 (13), 5.31 (11), and 5.31 (12) gives (1 — jN)--' for In where N is Oae1in(tTa)nn-1((a) — O'ctenn.(0a)in-1(O'a) — n(Ei — e)in(O'C)nn(0a) (16) Oafijn(3'a).j.-1(0a) — OictEjn(0a)in-1(13/ a) — n(Ei e)in(0/ a)in( 3a) —
The same formula is obtained for E„ by solving (12) and (15) except that Ai andµ replace Ei and e. When the sphere is perfectly conducting, An and Bn are obtained by equating the left sides of (12) and (14) to zero. Thus An 11
nnn(0a)ri J i3ajn_1(13a) — nj.(13a) —
_
jn(3a)
in(Oa) — jn.n((3a)
(17)
474
ELECTROMAGNETIC RADIATION
§12.16
The energy scattered by the sphere in a specified direction is, from 11.17 (3), the real part of the complex Poynting vector 411-it x
a=
x (ri x E) = -iriA— Y(PeRe + to R0)
(18)
From 5.37 (12) at a great distance, neglecting terms in r-P when p > 1, k„(jOr) =
a[ric.000] =
e-or,
1 ;flr
(19)
r ar
Substitution in (7), (8), and (13) gives for the tangential component of the scattered electrical intensity at a great distance from the sphere
= _i043= jEe_or
2n + 1
{0[Ansin OP;,/(cos 0)
L-J7i(n ± 1)
n
B„P
(cos 0)] cos 95 sin 0
4[
P,' (cos 0) ssin in 0
in sin 0P.;,' (cos 0)1 sin cb} (20)
To get the total energy scattered, (18) is multiplied by r2 sin 0 de chi) and integrated over the ranges 0 < < 27r and 0 < 0 < 7r. Substitution of (20) in (18) and integration with respect to 4) bring in a factor 7r and leave, omitting the argument in the Legendre polynomials, (2n ± 1)(2m + 1) [ (A.A”, 2/.41(327.2 LJ LJn(n 1)m(m ± 1) fiirE2
ju n (sin2 )
sPinP21 0)
n =1 m
— (A.13. + nA„,)(P!' P;7,
P.,1131')]
(21)
When multiplied by r2 sin 0 c10 and integrated from 0 to 1r, the first group of Legendre polynomials yields the integral following 5.231 (7) which is zero if m n and 5.231 (8) if m = n. Integration of the first term of the second group gives the negative of the second term and cancels it. The result is real so that the total scattered power is 7rekE2-1 P = 0/37 (2n + 1)(IA.12
IBni 2)
(22)
n -= 1
Interesting cases are discussed in the books by MacDonald and by Stratton listed at the end of the chapter. 12.16. Solution of Propagation Equation in Cylindrical Coordinates.— In 11.14 we considered the special type of cylindrical wave moving in the z-direction obtained by setting those terms in the scalar wave equation containing z and t separately equal to zero. If instead we equate the first group to +0;„„ and the second to -Ti3,„2„ and assume a sinusoidal time
§12.17
EXPANSION IN CYLINDRICAL HARMONICS
475
dependence so that 132 = co2ue, we obtain the equations
± 0;„„0 = 0,
c 22 dz2
032 T- 131n) 2 = 0,
W=
02 (1)
Comparison with 5.291 (1) to (5) and 5.293 (6) shows that the form of W in the p, 4 z system is, writing k,2„„ for 32 — 131„, and k:,& for 0 2 + ,
W = (Aelkm-z = (Aepe,nnz
jje-7e„,„.)[eim (F3 nnp)
.5,) Y.(13.np)] cos (mcb 15K„,(0„,n p)] cos (mt. + om)
+
(2) (3)
When 0, /5, knin, and k„, ' „ are real, both these equations give waves propagated only in the z-direction. If 01„ > 02 so that km„ is imaginary, (2) gives a wave exponentially damped in the z-direction. If C is real and 15 is complex, we have a radial propagation component in (2). If z is absent, k,n„ and Kin are zero, and we have cylindrical wave fronts in (2) and (3). From 5.293 (4) and 5.295 (8), D = —jC gives p-directed waves. If transverse electric and transverse magnetic waves are defined as those whose electric and magnetic fields, respectively, are normal to the z-axis then, putting u = k in 10.01 (1) and (5), we obtain A = piativte B
ap ,t, a2tivte + aqkte = az p ack az
a t7 ,m)]
3214-7„,„ [ ank ap az + p ao az +k 132W
az2
(4)
8214-7)
k (02'W te
az2
13 2 014-7 el
P
,.
+ 4)02
-wfm ap
(5)
12.17. Expansion in Cylindrical Harmonics for a Plane Wave.—From 11.15, the formula for a plane sinusoidal wave traveling in the direction n is greiwe
Atti, u2)0(‘"-13n.r) = f
u2)eme--10Pco. (a-0)
(1)
where n is normal to the z-axis and makes an angle a with the plane yt. = 0 and u1and u2 are coordinates in the plane normal to n. A plane wave may be expressed in cylindrical harmonics by expanding the last exponential factor in (1) in a complex Fourier series. In order to utilize formulas already derived, it is simpler to expand real and imaginary parts separately. Let us first expand cos (x sin 4') in a Fourier series. a„ cos n4',
cos (x sin 4/) = n=0
2 —
—
27r
so
n
f —„
cos (x sin +,&) cos nik dik
(2)
476
ELECTROMAGNETIC RADIATION
§12.18
Application of Dw 401.06 or Pc 592 gives two even integrands whose integral from n to it may be replaced by twice that from 0 to r. Thus — -
an =
2
—
2a
6° n[i cos (4 — x sin IP) d¢ +
Joy cos (4
x sin1,G) did
(3)
From 5.302 (4), the integrals are Jn(x) and J„( — x) or (-1)V„(x) so that an is zero when n is odd, and we may write 2m for n which gives cos (x sin 1,t) =
J„(x) cos 4 (4)
(2 — 5,,)J2m(x) cos 2no,G = m=o
n=—
We can expand sin (x sin lk) in a sine series in just the same way and get the difference of J„(x) and J.( — x) instead of the sum so that sin (x sin ik) =
2J2,, 1(x) sin (2m + m=o
=
J,(x) sin ?AG
(5)
n=—
If we multiply this by j and add to (4), we get the exponential form. If we then substitute Op for x and fir + — a for tp, we get ejaP c°s ('—o) =
jn,/,(13p)ein(.-0) = n=—
0 (3p) + 2 Z j"J,( F6p) cos n(a —
CP )
n=1
(6) 12.18. Radiation from Apertures in Plane Conducting Screens. The rigorous calculation of the radiation passing through an aperture is a very difficult problem. The fields must satisfy not only the propagation equations in the space outside the aperture and specified conditions at its boundaries, but also they must fit smoothly onto the fields inside the aperture. These fields are usually changed by the back radiation from the screen surrounding the aperture which practically restricts rigorous solutions to the very limited number of cases where it is mathematically feasible to treat the entire radiation space as a single volume. It was proved in 12.10 that, if the initial field values throughout an empty volume are given together with the tangential components of either the electric or the magnetic field over its surface, then the fields at any subsequent time are uniquely determined. The second condition alone is sufficient in a steady state. Evidently with conducting screens one should use the electric field because its tangential component is known to be zero on the screen. The best first approximation to the unknown electric field over the aperture seems to be that its value is the same as with no screen. The treatment that follows is particularly applicable to plane conducting screens with no restrictions on the number or shape of the apertures nor on the form of the incident wave. —
§12.18
RADIATION FROM APERTURES
477
We desire a source that will give a tangential electric field E over an area S of an infinite plane and zero over the remainder. Consider a thin double current sheet with a very small distance between layers, one of which has a current density equal and opposite to the other as shown in cross section in Fig. 12.18a. For a sheet uniform in the direction of flow, all the current passes around the edge, but for one stronger in the center, like that shown in Fig. 12.18b, part will turn back before reaching the edge. If the sheet is very thin, the external magnetic field is negligible compared with that between layers so that when we apply the magnetomotance .E law to the rectangle abcd, which is normal to i and fits closely a section of the upper layer, we find B, to be mi. As the flux N = B,6 dl through the rectangle a'b'c'd' changes, the electromotance — dN/dt around the loop equals from symmetry 2E dl as 0 so that the electric field strength E just above the sheet is — 4-jco/.46i. Clearly the double current sheet can be FIG. 12.18. built up out of infinitesimal solenoids of cross-sectional area 6 dl, length dc, and magnetic moment n xi6 dl dc which, in terms of E is —2 (jm.1) —in x E dS. Evidently these solenoidal elements can be combined in such a way as to produce any desired variation in E. It is also evident from symmetry that E is normal to the plane of the sheet outside its boundaries. From 12.13 (5) we see that the vector potential at P of the radiation from a small oscillating loop is normal to the loop axis and proportional to the sine of the angle between this axis and r the radius vector from the loop to P. Thus substitution of the moment just found for ra2/ gives for the diffracted vector potential
A = AO" = 1e coi 271-co
f
s
(j
—
13r)(n xE) x ri e — jor dS 2
r
(1)
where r1is a unit vector along r. At a great distance the j-term in the integrand may be neglected compared with Or. In the most general case, E will vary in magnitude, direction, and phase over the aperture and H will not be parallel to n, the normal to the aperture plane. Let x', y', z' refer to coordinates in the aperture and x, y, z to those of the field point. Then, since r 2 is (x — x') 2 (y— y') 2 (z — z')2, _ ju,(i — (3r)r E = —jwA =
r__ v,(r—ie—asr) = _ x ti x V'cl) dS
= _ vcp (real part)
(2)
478
ELECTROMAGNETIC RADIATION
§12.18
where E' is a function of x', y', z' only. Thus writing —v4, for v'el), observing that then V x (ciln x E') is zero, and using the usual formula for V x 43v give (2) the form E = 47--'eiwtV x
fs(n x E')r —ie— jor dS
(real part)
(3)
Formulas (2) and (3) are derived differently by Jackson. From the first Maxwell equation 11.00 (1), juweg is V x B so, except for the gradient of a scalar, canceling the curl gives
Bt = j13(271-co)—ifs(n x 2)r—'e— jor dS
(4)
This integral equation gives the solenoidal part of the tangential E in the aperture in terms of the tangential B which is known, as will soon be shown. The diffracted field far from the aperture depends only on this part of n x E since B is given by the curl of the A in (1) or (2) and E and B are related by 11.03 (8). A system of sources of electromagnetic wave lies on one side of an infinite plane perfectly conducting sheet having apertures. The eddy currents in this sheet generate additional waves which combine with the original ones to produce the reflected and refracted waves. These currents, being coplanar with the apertures, can produce therein only magnetic fields normal to their plane. Thus the tangential magnetic fields in the apertures must be entirely due to the original sources unperturbed by the eddy currents. When the incident magnetic field parallels the screen and the wave length is long compared with the relevant aperture dimension, it is often possible to calculate by potential theory the exact ratio of the normal to the tangential components of the magnetic induction in the openings and from this the tangential electric field. In Fig. 12.18 a displacement of the rectangle a'b'c'd' along B. decreases 2E dl by an amount equal to the decrease in dN/dt. This in turn equals the rate of change of the flux that escapes from both faces of the shell between the two positions. Thus, for an aperture in the xy-plane when E has only a y-component, we have
—
aEi, = ax
jcoBz
(5)
Integration of this expression gives the tangential component of E to be used in (1). All static potential solutions will give B. in terms of the induction tangential to the sheet far from the apertures. This standing wave value is twice the incident wave value that appears in the aperture. Examples at the end of the chapter include Bethe's small hole results.
§12.19
DIFFRACTION FROM RECTANGULAR APERTURE
479
12.19. Diffraction from Rectangular Aperture in Conducting Plane.— The formula of the last article will now be used to find the diffracted field from a rectangular aperture in a perfectly conducting sheet lying in the xy-plane. The magnetic field parallels the x-axis, and Poynting's vector makes an angle a with the z-axis as shown in Fig. 12.19a. If x1 and yi are the coordinates of dS, r the radius vector from dS to P, and R that from 0 to P, then we have approximately, if R >> a and R >> b, r
R — xicos ck sin 0 — yi sin ck sin 0
(1)
The tangential component Et varies in phase at z = 0 so that Etel't = E cos a
(2)
e3(`"--5711 sin a)
Because n x E parallels the x-axis, (n x E) x R lies parallel to the yz-plane, normal to DP and hence to R and is proportional to sin 0'. Thus A, = — cos 0 csc 0'A, A, =- Ity sin 4) cos 0 — Az sin 0,
A, = sin 61 sin 4) csc O'A A- 4, A., cos 0, A, = 0
When 12.18 (1) is substituted in (3), csc 0' cancels the sin 0'.
(3) (4)
Neglect of
xi and yi compared with R2leaves x1 and ylterms only in the exponent of A. A,
Writing in Et from (2) and using (4), we have la rib OE cos a sin (1:1e-JaR . e jiqx, cos 4, sin 0-1-y i (sin 4, sin 0—sin a)] dx, dyi 2ircoR — lb 2E cos a sin 4) sin (Oa cos sin0)sin [pb(sin 4) sin 0 — sin a)] .0 e-3 E (5) irgo.,R cos 4) sin 0(sin 4) sin 0 — sin a) A = A9 cot 4) cos 0 (6)
f_
The only approximation involved in these formulas is the assumption of an unperturbed electric field over the aperture. Stratton and Chu (Phys. Rev., Vol. 56, page 106) derived them by a superposition or "reflection" of two solutions of Maxwell's equations, which assume unperturbed electric and magnetic fields over the opening. This superposition was necessary to eliminate the tangential electric field over the screen and, as we see from the solution just obtained and from the uniqueness theorem, this is just equivalent to discarding all terms derived from the magnetic field. These authors check (5) and (6) at various b and a values by comparing the yz-plane intensity Ei Egi with the rigorous two-dimensional solution for a slit (a = 00) of Morse and Rubenstein (Phys. Rev., Vol. 54,
480
ELECTROMAGNETIC RADIATION
§12.20
page 895). The results appear in Fig. 12.19b, c, and d. A similar comparison of the xz-plane field with that of a slit (b = co ) for a = 0 is shown in Fig. 12.19e. This fairly good agreement shows the magnitude of the errors made by assuming an unperturbed electric field over the opening. Clearly the Kirchhoff scalar diffraction theory used in optics, which requires Fig. 12.19b and e to be identical, is completely wrong when applied to apertures of wave length dimensions in conducting screens. Dashed lines show rigorous values. 12.20. Orthogonal Functions in Diffraction Problems. Coaxial Line.— The method of the last two articles does not apply to curved screens, and the integral of 12.18 (1) is frequently difficult to evaluate close to the
(6)
(d)
(e) 12.19b, c, d, e.
aperture in a plane screen. In such cases we may start with solutions of the scalar propagation equation in the form ei.tZ ZC,n. Umn(u1) V.(u2) W„.(u3),
ffun„,-vnu„-v, dui du2 = (37,7D,,,n
n m
If u3 is normal to the screen, then u1 and u2 are orthogonal curvilinear coordinates in its surface. The integration is over the screen and aperture surface, and 677 is zero unless p = n and q = m. Such solutions appear in 12.11 (5), 12.16 (2), and 12.16 (3) and others occur in waveguide problems. For a conducting screen, we evaluate Cmn from the tangential component of the electric field, which is given over the aperture. The radiation from the open end of a coaxial line can be calculated by this method. The distant field is found more easily by 12.18 (1), but the method to be used here is better near the opening. The propagation
§12.20
ORTHOGONAL FUNCTIONS IN DIFFRACTION PROBLEMS
481
space in the line is bounded internally and externally by the cylinders p = a and p = b, respectively, and terminates in the plane z = 0 which, except when a < p < b, is perfectly conducting. To find the radiation into the region of positive z, we shall assume that b << X and use electrostatic methods to calculate the field for r < b. This is found, for any zonal potential distribution in the aperture, by integration of that above a plane at zero potential except for a ring of radius p and width dp at potential V(p). At a point P on the axis, Vp due to the ring is got by Green's reciprocation theorem from the charge dq induced on a ring element of an earthed plane by a charge q at P. To get dq from 5.06 (3), we let b — a = z and a — b -f 09. If z > p expanding by Pc 757 or Dw 9.05 gives d _ — zq2irp dp
27r(p2
z2)1'
dq dVp = --V(p)
V(p)dp(--1)n
(2n ± 1)tt p2n+i
(1)
(2n) !!en+2
n=0
To get dV at r, 0, we write r for z and multiply by P2n±i(cos 0). values of the potential in the plane z = 0 are assumed to be p > b,
V(p) = 0,
The
n (p/b) ln (a/b)' V(p) = Vo (2) a > p > 0,
b > p > a,
V(p) = VoI
When these values are inserted in (1), the result may be integrated by Pc 427 or Dw 610.9 and gives b2n+2 _ a 2"+2 (2n+ 1)!! 2)!!r2n±1(cos 0) ( 1)n (2n + 2)r2'.+2 (2n
Vo V
In (b/a)
(3)
n=0
To get the radiation fields, one puts A = 1 and B=o=m= 6=0 in 12.11 (5) so that it gives an expanding wave and evaluates D„ by setting r =b and equating the coefficient of -P2n-I-1(COS 0) in the Ee derived from Wt.m to that in the Et, given by (3). From 12.11 (5) and (2), we have CO
CEO,. = dr [ jwra2(r.W`m)1 80
r=b
±1(cos 0) j[bk2 ±i isb) n,
n db
b
(
]
n=0
Since Ob is small, we may write — (2n + 1)(4n + 1)!!(j0b)-(2n+2) for the derivative in this expression by 5.37 (11) and (12). To get E5 at r = b, we write —b-1d[P2n+I(cos 0)]/d0 = b-'11,,+1 (cos 0) for P2.+1(cos 0) in (3)
ELECTROMAGNETIC RADIATION
482
§12.20
and b2r.+2 for r2 n+2. Equating coefficients and solving for Dn give
_
(2n — 1)!![(00 2n+2— (Oarn+21 jVo ( 4) 1)!! 2)(2n + 2)!!(4n 0.) In (b/ a) (2n The electric and magnetic fields when r is greater than b are then given by 12.11 (2), 12.11 (5), 12.15 (6), and 12.11 (3) to be 00
Et?
=
c [riC2n+I(jOr) ] PL+1(COS
15 -7
0)
(5)
r
n =0
+ 1)(2n
Er =
2)/c2.+1000P2.+1(cos 0)
(6)
n =0
13,15 =
02
nnk2n±l(j3r)P2n+1(COS
n
0)
(7)
=0
These equations hold if r > b, and if b « X they satisfy the boundary conditions (2) rigorously. If we neglect higher powers of Ob and Oa, only the first term of the series survives and, if the case is further simplified by considering the field only at a great distance so that by 5.37 (12) (jOr) —'e--2fir replaces ic2n+ i(jOr), then the fields become 133(b2 — a2)Vo sin 0 cos (wt — Or) (8) = (p 41E0 = 40)r In (b/ a) From 12.01 (8), this is identical with the distant radiation from a current element iodzo cos wt where io dzo is 7ro.)E(b2 — a2) Vo/ln (b/a) so that the power radiated into the upper hemisphere is half that of 12.01 (26).
p
211-5CE(b2
a2) 2 Vg
3[ln (b/a)]2X4 V22 — 3[ln (b/a)]2X4 R, — —± 47r5ce(b2 — a2)2 2p
(9) (10)
The same method applies to other than plane screens, such, for example, as an infinite conducting cone coaxial with the line. In this case we take the same steps but start with 5.261. Other applications will be found in the problems at the end of the chapter. Problems 1. A linear quadrupole consists of charges q, —2q, q at z = —a, 0, +a. Its moment q2) is a2q sin wt. Show from 12.01 (13) that if a << r the fields are 3tir cos (wt — fir) + (3 — 132r2) sin (wt — Or) 8rer4 6 r — 133r3) cos (wt — fir) + (6 — 302r2) sin (wt — Or) = Q sin 16rer4 3 02r2) cos (wt — fir) — 30r sin (wt — (3r) = (2743 sin 20(— 16/rrs
E, = Q(3 cost B
1)
PROBLEMS
483
2. A square quadrupole consists of charges q, —q, q, —q at the corners of a square of area a2 whose sides are parallel to 0 = 0, 17r, 11", 71", respectively. Its moment q(x22 is a2q sin wt. Show from 12.01 (16) if a << r that Or)]sin2 0 sin n 20 167rer4 sin 0 cos 20 Th.E0 = Q[(60r — 037- 3) cos (wt — Or) — (30 2r 2— 6) sin (cot Or)]
Er= 3(2[313r cos (cot — Or) — (3 — 02r2) sin (wt
167E9.4
B4, = —Be cos °tan 20 sin 0 cos 241 Be = OnQ[(3 — 02r2) cos (wt — Or) — 33r sin (wt — Or)]
E0
— E0 cos 0 tan 20,
Br = 0 ,
16irr 3
3. It is desired to generate as pure quadrupole radiation as possible by charging the polar caps whose diameters subtend an angle 2a at the center of a sphere of radius a to potential Vo cos cot while the equatorial zone is at zero potential. What is the least a value? Show that if pa is small, the value of E0 at a great distance is 532a2 VQ (16r)-1sin a sin 2a sin 20 cos (wt — fir) 4. Two coaxial uniform current loops, I cos wt and —I cos wt, are a distance b apart. (a) Find the smallest value of b in wave lengths such that at a great distance the radiation at 60° is zero. (b) Verify that at a great distance there will be a maximum of radiation at 60° if Ob = 5.296 or b = 0.843X, approximately. 5. An electric dipole lies in the xy-plane at the origin and rotates about the z-axis with an angular velocity w pointing in the positive x-direction when t = 0. Show that the components of the magnetic induction are
Br = 0,
Bo — cogM [r0 sin (cot — Or — c5) — cos (cot — Or — 0)1 47rr2
cotiM cos 0 = r2 [r0 cos (cot — Or — Ltir
± sin ((a — Or — 0)]
6. An antenna of length 2/ supports a standing wave of 2n + 1 loops and current amplitude I. If there are nodes at the ends, show that the mean energy radiated per second from an element dz at a distance z from the center is µc14 dz
47r(/2 — z2)
cos2
(2n +1)71-z 2/
7. Two antennas, each (2n + 1)X long, lie parallel to the z-axis with their centers at x = 0 and x = a. The one at x = a lags 90° in phase behind the one at the origin. Show that the radiation intensity of this "end-fire" array is
yen cos2[1(2n + 1)71- cos 0] cos 2t7r[1 — (4a/X) sin 0 cos 0]) 21r 2r 2 sin 2
Draw a rough sketch of the heart-shaped intensity pattern when 0 = ico and a = iX. 8. Show that, when a = IX, the directivity of the double antenna of the last problem is twice that of a single antenna resonating in the same mode. 9. A set of p end-fire in-phase pairs like that of the last problem are set at half wave intervals along the y-axis. Show that the radiation intensity pattern is =
_ sine (ipr sin 0 sin 0) sine (fir sin 0 sin 0)
ELECTROMAGNETIC RADIATION
484
where 11, is given in problem 7 with a = X. Show that the directivity is twice that of p single antennas spaced at half wave length intervals along the y-axis. 10. A number p of half wave in-phase antennas are placed end to end along the z-axis. Show that the radiation intensity at a great distance is _
n
µcI u cost (17 cos 0) sin' (ipr cos 8) 87-2r2 sin' 0 sin' (17r cos 0)
—
Note that this is the same situation as when a long linear antenna is loaded at half wave length intervals to suppress alternate phases. 11. The current in a circular loop of radius a of wire is / 0 sin no cos wt. Show by the use of retarded potentials and with the aid of the formulas of 12.16 that at r, 8, cp, when r is very large, the potential A 0 is laar-U0,/,1 (Oa sin 0) sin nq5 sin (inir
wt — Or)
12. Electromagnetic waves are scattered from a conducting sphere whose radius is small compared with a wave length. Show that, at a distance from the sphere large compared with its radius, the radiation scattered at an angle 17 with the direction of the incident beam is plane polarized. 13. A plane polarized electromagnetic wave falls upon a nonconducting sphere of capacitivity ei and permeability a, whose radius is very small compared with the wave length. Show that the scattered radiation is the same as if there were at the origin an electric dipole parallel to the incident electric vector and a magnetic loop dipole normal to it, their moments being, respectively,
M=
471-a3(c, — )€E E,
2E
cos wt
7rr2/ =
and
47-0,3(gi —
mc(mi
+
cos wt
2A)
14. A plane electromagnetic wave, whose y-directed E-vector is E cos (wt — Ox), impinges on a perfectly conducting cylinder of radius a whose axis is the z-axis. Show that the total fields are given by the real parts of EP
jw aWu
=
P
a 4,
eiwe
E0
", 47 = af= Ow', ap
B, = 0 2#1.-t.eiwt
(—j)"(2 — 3:)Vn(Ra) 17.(13P) — Jn(OP)Y. ' (Qa)] n4, cos ( Oa) — 1 i 3a) n=0
iE
r ee = wQ
15. In a medium Ac a plane electromagnetic wave whose y-directed electric vector is p'e' of radius a, whose axis is the z-axis. Show that the scattered radiation is given by the real part of
E cos (wt — Ox) impinges on a cylinder
B, = 0 21kuei",
a lke, = P
a4>
E0 = jwa
It.
ap
ei'i
117u = (4)-1EZ .13.[J„(0p) — j17„(0p)] cos ncil n=0 ( — j)"(2— ON(P/ENn(0a)4(13'a) — (1.LE'N'„(13a)Jn(O'a)] —
(ile')1 in(o'a)[J;(ga) — :717 (0a)1 — (A'E)1J".03'0[Jn(3a) — il7n(Qa)1
16. A plane electromagnetic wave, whose z-directed E-vector is E cos (wt — 13x), impinges on a perfectly conducting cylinder of radius a whose axis is the z-axis. Show
485
PROBLEMS that the total fields are given by the real parts of 0
E, = j,„02gitmeicot,
B, =
, wgme2cot,
P
= a ik
ac
ap
( — j)'(2 — (5,,)[Jn(0a)17.(13P) — J.(RP)Yn(Ra)] = E cos n4, 0.,0 2 J„( pa) — j Yn Oa) n=0 17. In a medium ILE a plane electromagnetic wave whose z-directed electric vector is E cos (wt — fix) impinges on a cylinder µ'E of radius a whose axis is the z-axis. Show that the scattered radiation is given by the real part of B,, =
E, = jc00 2Tkimeic",
02 a We,. P
14
„n
a
ei<",
a ik
B 8 = (32
ap
Ow,
= . B V.64,1„(RP) — .i17.(RP)1 cos nO 3 42
i„,„?
n=0 = Cn
( — j)"(2— g)[(ILE')V.(004(13'n) — (AL'€)14(0a),/n(O'a)1 (A'E)1 /n(010.r,,(0a) — ./17:,(0a)] — (i-LE')14(R'a)[1,,(Ra) — ilin(Ra)]
18. The portion of a perfectly conducting sphere of radius a between 0 = y and y is removed, and a potential difference 2V0 cos wt is maintained between the cones at r = a. Assuming that the potential in the gap at r = a is given by 12.06 (9) show that the radiation fields are the real parts of 0 = it —
Ee = Vor-leiwtV'n[(2n
2)1c2n-1-1(igr) — jOric2.+2(jOr)IPL 1 (cos 8)
n=0
E, = —Vor-'eic"Z(2n
1)(2n
2)0.k2n+I(j/37-)P2.41 (cos 8)
n=0 =
V oi (3 20.1-1ei‘"Ze j2,1-1-1(i PO P L+1 (cos 6)
n=0
On =
wp(4n
17-04(2n + 1)(2n
2)[(2n
3)P2,,÷1 (cos y) 2)k2n÷1(i0a) — jPatc2n+2(iRa)1
where 2,, is given by 12.06 (8) and (10) . 19. Calculate the input impedance of the arrangement of the last problem by proving its equivalence to a conical transmission line terminated at r = a by a load comprised of an infinite number of impedances 2, 21, . . . in parallel where 1
Iraj02 sin -y
2n
1.. 0
enk 2rz+1(
)PL +1 (cos T)
20. A medium Ac fills the space between an infinite dielectric cylinder of radius a and a perfectly conducting cylinder p = b. Show that a plane transverse electric wave propagates along the cylinder for each value of un that satisfies
ACTI(u,,)
il[Ti(v„)Y1(v„b/a)
Yi(v,,),11(vnb/a)]
ALRI(v,,)
u,,,J0(un)
vn[Jo(v.)Yi(v,,b/a) — Y0 ( v„),Ii(v„b/a)]
v„R0(v.„)
—
486
ELECTROMAGNETIC RADIATION
provided that co 2a2p'e' > u„2and w 2a2p.e > v„2. Show that the velocity of the nth wave is wa(w 2a2ALY — u,,2)-1and that the fields in g'€' are = wi3„ = lc„OnciknzJi(unp/a),
jwct& = unene-iknz,10(un p/a)
20, and a < b. where w 2a2bee' — u = w 2a2pe — v„2= k„ 21. If in the last problem there is no reflector at p = b, show that with the same materials and frequencies a plane wave is impossible but that otherwise the results of problem 20 hold if the Hankel functions Jo — jYo and J1— j171are substituted for Ro and RI, respectively. This requires that p = a be an equipotential. 22. An infinite medium i.L€ surrounds an infinite dielectric cylinder 12'e' of radius a. Show that a plane transverse electric wave can be propagated along the cylinder for each value of u„ that satisfies the equations ,
pev„Ko(v„)Ji(u„)
ihu„Ki(v„)Jo(u„) = 0,
w2a2mY — un = co 2a2pe
= le;,a 2
provided that w2a2AY is greater than un and 6' is greater than e. Show that the velocity of the wave is wa(w 2a 2A'E — 4,)-i and that the external fields are k„24, =
=
= 2,,,ee-ikn'Ke(vna-iP)
23. If Yi(v„b/a) is zero, verify that the conditions of problem 20 are satisfied approximately by e = 1.996' = 1.996, = wa = 109, u = 1, v = 3.46, and = show that the phase velocity is approximately 3.15 X 108 m/sec. 24. Verify that the conditions of problem 22 are satisfied approximately by = 1.726 = 1.726•, µ = µ = wa = 109, u = 2.65, v = 1, and show that the phase velocity in the z-direction is approximately 2.87 X 108 m/sec. 25. A medium I.LE fills the space between an infinite dielectric cylinder /AY of radius a and a perfectly conducting cylinder p = b. Show that a plane transverse magnetic wave propagates along the cylinder for each value of u„ that satisfies uJo(un)
vn[Jo(v.)Y o(v,,b/a) — Yo(N)Jo(v.b/a)]
v.Ro(v.)
e[Ji(vn)Yo(vnb/a) — 171(vn)Jo(v,,b/a)]
eRi(v.)
provided that w 2a2AV > 2/,,,2 and oike 2pe > v„2 . Show that the velocity of the nth wave is coa(coWe' — u!)-1and that the fields in '€' are
A', = k,,.(cop'e)-1.134, = jOcik.'.11(unP/a),
knaE, = uX„e-ik.2J0(tt.p/a)
where o., 2a2p'e' — = co2a2p,e — 14,2 = Ic:a2 and a < b. 26. If there is no reflector at p = b in the last problem, show that with the same materials and frequencies a plane wave is impossible but that otherwise the results of the last problem hold if the Hankel functions Jo— jYo and J1 — jY1 are substituted for Ro and R1, respectively. 27. An infinite medium 11E surrounds an infinite dielectric cylinder /./'€' of radius a. Show that a plane transverse magnetic wave can be propagated along the cylinder for each value of u„ that satisfies the equations v„E'Jl(u„)Ko(vn.)
unfJo(u„)Ki(v„) = 0,
w2a2AY — u;, = w2a 2/4e
v: = k!a2
provided that w2a2A1 is greater than Show that the velocity of the nth wave is eoa(0, 2a2p'e' — u„2 )-1and that the fields outside the cylinder are
= k„(cope)-113-4, =
k,,at = jvnene-ik,='Ko(vna-1P)
PROBLEMS
487
28. If Yo(v.b/a) is zero, verify that the conditions of problem 25 are satisfied approximately by E = 2E = 2€„, A = = .85 X 100, u, = 1, v, = 3, and = show that the phase velocity in the z-direction is approximately 3.21 X 10' m/sec. 29. Verify that the conditions of problem 27 are satisfied approximately by taking = 1.81€ = 1.81G A = = coa = 0.99 X 10', u. = 2.8, v. = 1 and show that the phase velocity is approximately 2.88 X 10' m/sec. 30. An infinite plane perfectly conducting surface is coated with a dielectric layer 2f2 of thickness a and underlies an infinite dielectric medium Ale'. By 12.16, show a diverging surface TM wave is possible whose vector potentials are the real parts of A l= C1[p1(0r2 A2 = C2 {ei (s:
— —
k/3' HV )(13' p)le-(0" fli0.+iwt 13 2 )111V )' (0' p') 3' 2)1 sin [( 3 — 0'2)z]1-1?)'(13'p) — k/3' cos[(/3s — 0")iz]lle)(0'p)}eiwt -
—
1
where ,(3! = (.0 2141€2, 114 = co 2A2€2, and R2 > (3' > /3,. Show that its velocity is the same as that of the plane wave of problem 7, Chap. XI. 31. An infinite plane perfectly conducting surface is coated with a dielectric layer /12€2 of thickness a and underlies an infinite dielectric medium wei. By 12.16, show a diverging surface TE wave is possible whose vector potentials are the real parts of Al = (1)Cle —(0"-012)1q/i2)(0'p)04'4,
A2 = 8C2 sin [(I3— 0'2]iz]li(12)(0'p)eic"
where Of. = = w2i.42€2, and 02 > > 01. Show that its velocity is the same as that of the plane wave of problem 8, Chap. XI. 32. An infinite perfectly conducting cylinder of radius a which is coated with a dielectric layer A2e2 of outer radius b passes through an infinite medium Atiei. Show by 12.16 that possible vector potentials for a surface TM wave are the real parts of
where 13 = co2121€1,
Al = Ci[—piji3/Ki(piP)
kpiKo(pip)]ei (C0t -P'')
A2 = C2[ — E01,j/3'Rl(P2P)
kP2R0(p2p)]ei (We-9'z)
= W2P2€2, 02 > 0' > 01, p! = 13'2 —
= Oi —13' 2, and
Ro(p2p) = Yo(p2a)Jo(p2P) — Jo(p2a)Fo(p2P) Ri(p2p) = Yo(P2a)J1(P2P) — Jo(P2a)Yi(P2P)
Show that the velocity may be found from the equation —e2piKo(plb)R1(p2b) = eip2Ki(plb)Ro(p2b) If a = 1, b = 2, gl = Al, €2 = 4€1, and /31b = 1.174, show that 01:0'432 = 1:1.048:2 and that, when p = 4.350b, then Bohas one-tenth of its value at p = b. 33. Obtain the result in 12.20 (8) by using 12.18 (1). 34. The plane of polarization of the incident wave in Fig. 12.19a is rotated so that E parallels the x-axis. Show that the diffracted vector potential at a great distance is, if Ae' is given by 12.19 (5), A9
—
— cos ¢' A0 ,
cos a sin 0
-
cos 0
..
21° _ cos a A8'
35. The opening of 12.19 is not rectangular but annular of internal and external radii a and b. Show that the diffracted vector potential at a great distance is -tan cb
Ag =
cos 8
—
A4, =
E cos a sin 4, oiRQ
[a.11(0Qa) — bJ1(13Qb)]c)9 R
where Q = (sin2 a + sin2 e — 2 sin a sin 0 sin 0)i.
488
ELECTROMAGNETIC RADIATION
36. The polarization plane of the incident beam in the last problem is rotated so that E parallels the x-axis. Show that the vector potential of the diffracted field at a great distance is
cot
A0 =
cos
— E cos
=
43.
caRQ
[b‘II(8Qb) — 11(0Qa)lci19R
37. Show that, if in 12.19 or in the last three problems, there are not one but two identical openings centered at y = c and y = — c then, if R >> c the single opening potential must be multiplied by 2 cos [0c( — sin a + y/R)]. 38. If in the last problem the openings are at x = c and x = — c, show that the multiplication factor is 2 cos ((3cx/R). 39. A spherical shell of radius a is perfectly absorbing inside and perfectly conducting outside. An electric dipole of moment M cos wt is placed at the origin pointed in the 0 = 0 direction, and the part of the shell between 0 = a' and 0 = a" is removed. Show that 12.11 gives the external diffracted vector potential where Wt,„ = M(1
—
13 2a2j0a)ZA„P„(cos 0)ic„(j,3r) n=
= e of (2n + 1)Pu(u) — (n 2)uP„_1(u) — (n — 1)uPn+1(u) u = cos a" 87riWee(??,
— 1)(n + 2)a[aL(0a)]/aa
u = cos a'
40. The dipole in the shell of the last problem is replaced by a small coaxial loop of wire of radius b carrying a current /e3.'. Show that 12.11 (2) gives the vector potential of the diffracted field outside where Ikee = ,u1b2(1
jf3a)ZA„Pn(cos 0)k.(07.) n
A„ = e
..[(n LL
2)uP„_1(u) (n — 1)uPn÷i(u) — (2n + 1)P.,,(u)t = Cos a" 8a2kn(j13a)(n — 1)(n + 2) u = cos a'
41. The boundary of a circular hole in an infinite thin plane conducting sheet is p = a. The potential of an impinging wave at the hole surface is Ao = f(p). If the wave in the hole is unperturbed by the boundary, show that the vector potential of the radiated field at a great distance where R >> X is the real part of Ad,
a = 13R-1cos Oei(4" —fi R) f pf(p)J 1(0p sin 0) dp 0
If the source is a very small loop which carries a current leiwt and is coaxial with, and at a distance c from the hole and if a <<X and c << X, show that Ad,ei.' = —/.4/32.10(16R)-1[2c — (2c2
a2)(c2
a2)-1] sin 20eim-I3R)
42. A slit of width 2a in an infinite plane conducting surface is bisected by the z-axis. The tangential electric field over its surface is jfi(yo) kf2(yo). Show with the aid of 5.331 (3) that the vector potential of the radiation field is A=
+a _ a [46(Yo)
k cos 4,720/0)1Hi2)(OR' NY0
where R' is the shortest distance from the strip of width dyo at yo to the field point and .0' is the angle between R' and the normal to the surface.
489
PROBLEMS
43. A plane wave whose electric vector is kEoeion-13') impinges normally on the slotted plane of the last problem. Solid lines in Fig. 4.23 show the magnetic field near the slit. Get E as in 12.18 (3) and then show from the last problem that if the phase is constant over the slit, the diffracted field is
=—
= 1020E0 cos 01-1?)(Op)eic"
where p is measured from, and 4, around, the z-axis, p >> a and Op is unrestricted. In 4.23 B, the standing wave amplitude, is twice the incident amplitude. 44. An electric field iE0e,‘" is bounded by the infinite plane conducting sheet x = 0 in which there is a slit of width 2a bisected lengthwise by the z-axis. The dotted lines in Fig. 4.23 show E near the slit. From Art. 4.23 and problem 42 show that if the phase is constant over the slit the radiated field is
Egy = — firOgagEo sin cbliV )%3P) where p is measured from, and (15 around, the z-axis, p >> a, and Op is unrestricted. 45. A plane wave whose magnetic vector is kBelm-0') impinges normally on the slotted plane of problem 42. This creates an electric field across the slit whose form, for slowly varying fields, is shown by the ellipses in Fig. 4.22. Find Cfi(yo) in problem 42 from 4.22 (3) and so, assuming constant phase over the slit, obtain Act, for the radiated field. Find C approximately by equating the curl of A,, when p is zero to the incident B. Thus show that, neglecting (0a) 2 terms,
—
T-Boe ,‘" HO2)(OP) 2[a — j ln (170a )]
where In y = —.11593, p is measured from the z-axis, p >> a, and Op is unrestricted. 46. A plane polarized electromagnetic wave whose wave front makes an angle a with the yz-plane impinges on the slotted plane of problem 42. If its magnetic vector parallels the slit, show from problems 44 and 45 that the diffracted intensity at a great distance R from the slit is, if Oa is so small that phase differences over the slit are negligible and —1n(Oa) >> 1, 1-1 =
{ cos a
0 2a2 sin a sin 412 _
2aR In (Oa)
Iio
4
47. The wave of the last problem has its electric vector parallel to the slit. Show that, if Oa is very small so that phase differences over the slit may be neglected, then from problem 43 the diffracted intensity is _
=
70 3a 4 cost
8R
cosg a_ Ho
48. A hole of radius a is cut in an infinite plane perfectly conducting sheet which bounds a uniform electric standing wave field Eoeiwt. By neglecting phase differences over the hole, show from 5.272 (6) and 12.18 (1) that the vector potential at a great distance from the center of the hole is 49
— jEo [sin (Oa sin 0) — Oa sin 0 cos (Oa sin 0)]e-iPR ro)(3R sing
49. Show by the methods of 5.272 that, if a uniform standing wave magnetic field of induction Boeiwt exists in the x-direction above and parallel to an infinite thin plane perfectly conducting sheet at z = 0 having a hole of radius a in it, the z-component of
ELECTROMAGNETIC RADIATION
490
B in the hole is 2B(x/ir) (a2— X2 — y2)-1. Hence by 12.18, neglecting phase differences, show that at a great distance from the center of the hole below the sheet, where the azimuth angle from the x-axis is 47 and the colatitude angle from the hole axis is 0, the vector potential is AB =
2jBo sin [sin (Oa sin 0) — 13a sin 0 cos (Oa sin 0)] e-i0R, r(32R sins 0
A-0= A8 cos 0 cot 43
50. A plane polarized wave strikes a thin plane perfectly conducting sheet with a hole of radius a in it at an angle a from the normal. If E0 parallels the sheet, show, when Oa is very small in problem 49, that the intensity of the diffracted radiation is -
=
404a6 cos2 a(1 — sin2B 7r2R2
cos2
Note that the standing wave B0 of problem 49 is twice the incident running wave induction. 51. Show that the average power radiated from the hole in the last problem is 16134a6 cost
37r
°
52. In problem 50 let the electric vector lie in the plane of incidence and show, when pa is very small in problems 48 and 49, that the intensity of the diffracted radiation is 134ct6 _2_,2_14 sin2 7r R
cos2 0 + (2 cos 4) — sin a)21110
Note that the standing wave amplitudes in problems 48 and 49 are twice those in 110. 63. Show that the average power radiated from the hole in the last problem is
p
404a6(4 + sin2a)
3r
c II
64. An electromotance e is maintained across the center of a very narrow slot bounded by x = +16, z = ±l in the infinite plane conducting face y = 0. Assuming that the field in the slot in this plane is E0 = ia-le sin [5(l - z)], show by 12.18 (1) that the electrical intensity outside the slot, where y > 0, is E=
1(1±1) (iy — ix)Z f R-3(1 ± jOR) sin [13(/ i(i 1)
where R2 = x 2 + y2 that
(z — z1)2.
zi)leim-OR) dzi
Show by comparison with the curl of 12.03 (3)
[ELia = — 2e(a/0)-1[B].„u,„„„ so that the fields are identical in form with those of an antenna driven at the center but with B and E interchanged so that from 12.03 (18)
= —1e(irp)-'[2 cos 131 sin (cot — Or) — sin (wt — Or.) — sin (wt — Orb)] where ra and rb are the distances between the ends of the slot and the field point. 55. Apply the results of the last problem to the resonant slot 1 = IX and thus show by 12.04 (4) that the radiation resistance of the slot is 363 ohms if it radiates from both faces.
REFERENCES
491
NOTE: The results of the following problems are based on the Kirchhoff theory of diffraction. They are useful in optics but should be used with caution elsewhere. 56. Let U = y, z)el"O be a component of the Hertz vector in an insulating medium so that it satisfies 11.01 (10) when r = 00. Let
V = r-lei(44-13r) = cp(x, y, z)eiwg be a similar component for a spherical wave originating at a point P. Insert V, and (/, in Green's theorem, 3.06 (4), and, taking the volume of integration to lie between a very small sphere surrounding P and some larger surface 8, enclosing the sphere, show that 470, = fs [r-,e-it3'170 — 1,t.V(r-'e-119')] • n dS where Op is the value of at P. This formula, the basis of Kirchhoff's diffraction theory, gives the effect at P in terms of its integral over a surface surrounding P. 57. Let S have apertures, and let U be a spherical wave originating at Q outside S. Assuming that U has the same value over the apertures as if S were absent and is zero over the remainder of S, show that Gr =
—
1(n•r rri r
n• ri)
r1
e-7 00'+'1) dS'
where r1and r are the radius vectors from Q and P to the apertures and both are much larger than 0-1. S' is the area of the openings. 58. Let there be a single aperture, let R and R1be the mean distances from its center 0 to P and Q, and let the coordinates of any point in it, referred to 0, be x' and y'. Expand r and r1in powers of x' and y' and, since /3 = 271-X-1, show that —
j n• R 2XRRI R
27i(R ± RI)
n• R1 R1
2rjF(x' ,y' ,1 / R,1 / RI)
f
dS'
59. Consider a plane wave of intensity /0 incident normally on a circular aperture of radius a so that RI = 00, and take R >> a and P at x = x, y = 0 so that F(x', y',
Er')
= p'
cos 0' sin a
where a is the angle subtended by x at 0. Evaluate the integral by 5.302 (3) and (2), and show that the diffracted intensity is Id = 1/0R-2cote (1a)(Ji(27raX-1sin a)]2a 2 60. Take Q and P on the axis of a circular aperture, of radius a and take R1 and R much larger than a. Show that, if the incident intensity is /0, that at P is 4.1410(R1
R)-2sine [-n-a 2(2X)-'R-1
RT1)]
References Most recent references give extensive bibliographies. S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. "Handbuch der Physik," Vols. XII, XV, XX, Berlin, 1927, 1928. Gives elegant treatment of electromagnetic theory in Vol. XII. HARRINGTON, R. F.: "Time-harmonic Electromagnetic Fields," McGraw-Hill, 1963. Discusses multipoles, Babinet's principle, and pertinent topics. FLUGGE,
GEIGER-SCHEEL:
492
ELECTROMAGNETIC RADIATION
H. R.: "Electric Waves," Macmillan, 1893. Radiation field drawings. J. D.: "Classical Electrodynamics," Wiley, 1962. Good treatment of multipoles, Babinet's principle and diffraction, and cgs units. KING, R. W. P.: "Theory of Linear Antennas," Harvard, 1956. A complete treatment with extensive bibliography. MACDONALD, H. M.: "Electromagnetism," Bell, 1934. Solves diffraction problems. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives excellent treatment of retarded potentials. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. The original treatment of the subject. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Covers multipoles and related topics. PAPAS, C. H.: "Theory of Electromagnetic Wave Propagation," McGraw-Hill, 1965. Covers multipoles and antennas with radiation patterns and applications. SCHELKUNOFF, S. A.: "Advanced Antenna Theory," Wiley, 1952. SCHELKUNOFF, S. A., and T. FRIIS: "Antennas; Theory and Practice," Wiley, 1952. Formulas for multipoles and antennas with specific cases in problems. SILVER, S.: "Microwave Antenna Theory and Design," McGraw-Hill, 1949. Treats general antenna theory and diffraction extensively. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Extensive and rigorous mathematical treatment of the whole subject. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Treats concisely much of the material of this chapter. VAN BLADEL, J.: "Electromagnetic Fields," McGraw-Hill, 1964. Expands topics of this chapter to include stratified mediums, pulses, and other cases. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. HERTZ,
JACKSON,
CHAPTER XIII WAVE GUIDES AND CAVITY RESONATORS 13.00. Waves in Hollow Cylindrical Tubes.—That plane waves may be guided by two or more mutually external perfectly conducting cylinders was shown in 11.14 to 11.18. If the propagation space is enclosed by a perfectly conducting cylinder, other unattenuated wave types will propagate at very high frequencies, and if this is the only boundary, a simple plane wave is no longer possible. Each external and internal (if any) boundary of a cylindrical wave guide is generated by moving a straight line transversely keeping it parallel to the z-axis. It is often considered closed by the z = 0 plane and extends from z = 0 to z = co In the text following 11.01 (5) it was pointed out that in the Coulomb gauge the vector potential A is derivable from a solution W of the scalar propagation equation in two ways, one giving transverse electric and one transverse magnetic waves. The differential equations are
v 2A = am - at 2
vwt. = 1/6
aw-te at2
r• 2 W
t77)
„
a2UT
(1)
We are interested in a steady-state solution of angular frequency w. In this case from 10.01 a vector potential satisfying the first equation and its curl may be obtained from solutions of the scalar equations by t„,), = —V x (o2tivt„, = v x (kikte k x kxVWte) (2) where i32 = co2AE. These equations are written out in cylindrical polar coordinates in 12.16 (4) and (5). As in 12.16, we may write W as a product of a function of z by a function of the transverse coordinates so that either scalar equation in (1) splits into two. Thus
W = u2,
22 dz2 + (o2T- 01.)2 = 0
vw ± 131„u = 0,
(3)
Here U depends only on the transverse coordinates u1and u2, and boundary and symmetry conditions fix 0.„. If n is a unit vector which like. A is normal to the curved surface, then from (2) the conditions there are 0 = n x Age
=
—n • V(kWie)
or
-aU . nte = 0
0= n x [17 (k vtiVt„,)] = n x [kV2Tikt„ — = —132n x
= soln„Wt,
t„, — (n x
k • V (VW t,n)]
tm) = Si[132Tktm
k' a2W as az 493
(4)
a2aw z2t'n]
a vw,m k
as az
494
WAVE GUIDES AND CAVITY RESONATORS
§13.00
where s is a coordinate measured around a cylindrical surface along its curve of intersection with a plane normal to the z-axis so that s1 = k x n. The conditions imposed on U at the boundary are therefore
Ut. = 0
or
Ann =0
and
—o
as
(5)
The second pair of conditions in (5) are those met by the principal waves described in 11.14 to 11.18 whose velocity is independent of frequency. These are possible only with two or more conducting cylinders. The differential equations of (3) are of the second order so each has two solutions, and the general solution is of the form W=
bef'—)
u2) r3U2(ui, u2)](ae-i'--z Ann = :702 — 13104 = amn +
(6) (7)
When > temp, the propagation constant P.„ is a pure imaginary so a = 0 and the first and second exponential terms in (6) indicate unattenuated waves in the positive and negative z-direction, respectively. If a < gm., we put C = 0 when z is positive or D = 0 when z is negative since the fields must remain finite. For the remaining term, P.m is real ' „ = 0 which gives an exponentially damped term. The cutoff so a„, frequency v.n and the cutoff wave length for the mnth mode are P.n =
gm
COmn
n
2r (kie)-1 =
= 24r gmn
(8)
To extend the concept of characteristic impedance to wave guides, the definition of 11.15 (15) is put in terms of B a id B by 11.15 (6). Thus
a
k Wee'
te
12 klte =
k x
to
12kI tm — + Alk-Et —
=+
±jCOAIV x k Weel
jCLIA
= la(V x kW te) 1k x [V x (V x kfk teAl x [V x(kxVW 00]1 T ila(V x kWim)/a21 I _ „
l Btml
(.4EIV
kfkt,„1
(9)
coeiV x kjTVbni
jPmn
(10)
WE
In terms of the cutoff frequency v.., these are
12,, J. = [1 —-±(0 lim n/
E-4
12ki t. =
— (v,../v)9*
(11)
From 11.15 (6), the upper or lower sign is to be used according as the wave is in the negative or positive z-direction, respectively. Note that both wave types give a real characteristic impedance, indicating power propagation above the cutoff frequency and a reactive one below it. Observe that this differs from the transmission line definition in 11.16.
ATTENUATION IN HOLLOW WAVE GUIDES
§13.01
495
From (6) and (7), when # > ft.., waves propagate with a phase velocity co
— /.. 032 — gm)/ [1— (P./ v)2[1 [1— (X/X..)9/ (12) where v is the velocity of a free wave in the medium filling the tube. The signal velocity is given by 11.19 (2) and (12) to be Vmn
=
ac.)
1 NI #'„,„ Ace)i ag„, ' „= . # (pt)i
1 v2 (13) AEv. = v.. Thus v.. > v > (v.)... At sufficiently high frequencies, both phase and signal velocities approach the velocity of the free wave. From (2) and (3), the electric and magnetic fields in terms of U are (v )
—
T,
(
Et. = —jcoAt. = jog, x V Utee—o'—z
ht. = v At. = + koinu„)e-w--. Et. --= —joa. = (wo„„v2ut. + ki,407„„ut„,)e-o'—. ht. = v x At. = 02k x V Utme-7t3 Ims ' ,
(14) (15) (16) (17)
where Ut, and Ut. satisfy (3) and from (2) have different dimensions. 13.01. Attenuation in Hollow Wave Guides.—Wave guides are used to transmit high-frequency power from one point to another so transmission losses are important. If an imperfect dielectric fills the tube, energy will be dissipated there, but in the common air-filled guide such losses are seldom important and their calculation will be left to the problems at the end of the chapter. The eddy current energy loss in the walls is, however, unavoidable and must be considered. In all practical cases, the wall conductivity is so high that field solutions for perfectly conducting tubes are excellent approximations and can be used to calculate P. In any case where losses in a quantity are proportional to the quantity itself, there is exponential attenuation. Equation 10.02 (8) shows that the power loss in a conducting surface is proportional to the square of the tangential magnetic field at the surface. Therefore the fields are damped exponentially as they pass down the tube and, if their attenuation factor is a, that of the Poynting vector is 2a. The periodic factors disappear in the time average P of the transmitted power leaving Pe--2az. The z-derivative of P evaluated by 10.02 (8) and divided by .1-1 is ,
2« =
1 OP = 2122Pt5 P az
B•
h ds
(1)
where aP/3z is the average power loss per unit length, h the magnetic induction next the wall, T the wall resistivity, and 8 the skin thickness. At the wall V2Ut. = siaute/as so that, from 13.00 (15) and (17), /A u \ 2
ht. • hi. = 032
—
+ NmnU20f
h.• ht. = 134v2utn, • v2um (2)
WAVE GUIDES AND CAVITY RESONATORS
496
§13.02
By 11.17 (3), the average z-directed energy passing through unit area per second is the real part of IAL-lt x E so that 13.00 (14) to (17) give =
witni(k x
Ur,,) x V2Utel, = ii.c1c0(02 - 131n)Ir2uce
• v2ute
(3) (k vu,„Olz = -1-A-10'„,,,132[V2U 43202 - 01„)}V2Ut. • V2 Ulm
(4)
To simplify the surface integral of the scalar product in (3) and (4), we expand and note that the first term of the surface integral vanishes when transformed into a line integral around the boundary because, by 13.00 (4) or (5), either U or n • V2U is zero there. Thus, using 13.00 (3), fsV2 U • V2 U dS = fs[V 2• (UV 2U) - UqU] dS = i3,2nnis U 2 dS
(5)
From (1), (2), and (3), the attenuation of transverse electric waves is ate -
T
134[1 — (v„,„/ v) 2]se (aU 3,3) 2 ds
2,avo
(vmn/ v) 2
ds
(6)
[1 - (vm„/ y) 2]iisUL dS
Here S is the skin depth (ic0/27)-i, AI the permeability of the wall, v the frequency, v„,„ the cutoff frequency, v the free wave velocity in /le, and flmn is 221-v-1vmn. From (1), (2), and (4), since V2U is normal to the wall, 3C
T
(aut,,/ an)2 ds
(7)
ahn - 4w 5 oln[i - (vm./ 09i.fsuln dS
This equation shows that the attenuation of transverse magnetic waves increases with frequency in the same way for all forms of cross section. From Eq. (6) the attenuation of transverse electric waves depends on the form of the tube cross section and increases with frequency unless the first numerator term vanishes, as in some circular pipe modes. 13.02. The Rectangular Wave Guide.—Consider a tube bounded by the planes x = 0, x = a, y = 0, y = b. By inspection, solutions U (x,y) of 13.00 (3), which satisfy boundary conditions 13.00 (4) and (5), are mirx nay mr sin (1) U t„ = Cmn cos — cos Ulm = C„,„ sin b a b a m2 n2 m2 n2) v2 2 1 2( I or (2) le n = ir2(a2 ±b2) Pnin =TV — a2 m — b2 =-X,,, 2n if m and n are integers. From 13.00 (14) and (15) the TE fields are th, = jcore(i cos
mrx a
sin
nry m b
-i- sin a
mrx nry) . a cos -v- e-lx--z
(3)
max nryl a jo'm„rr in-aisin max cosnay ± n cos a sin b b a b
t. = {
max nay 20 + kit„ cos a cos .- e- '-^2 (4) }
.
§13.02
THE RECTANGULAR WAVE GUIDE
497
From (1), (2), and 13.01 (6), the attenuation of these waves is a 7 a ± 2b(vmo/v) 2 /Iva ab[l — (v,e0/v)]i Ite,""' latei n2a){1 — (v„,„/v)211 = 27- (m2b (a ± b)2 m2b2 n 2a 2 ab[l — (pm„/v)211 I I mn toot
(5) (6)
From (2), the longest cutoff wave length for the TE waves in a guide for which a = 2 cm and b = 4 cm occurs when m = 0, n = 1, for then Xoi = 8 cm. This is independent of a. The next longest at m = 1, n = 0 is Xio= 4 cm and is independent of b. As we shall see later, thq
TEi0 E lines. AB section
NIN I=1
A
B
C
D TE B-/fines. Top surface
11111 1111 -"I ■ % TE/i Elines.Ab' section
TB B-/fines.Top surface
TM„ B-/ines. AB section
TME B-/h7es. Center secbbr.
FIG. 13.02a.
guide coupling can be arranged to excite some modes and not others. For a 5-cm wave transmitted only in the TEta mode, b 8.5 X 10-7 m in copper. Thus, from (5), the attenuation is ate 0.005 so by Pc 757 or DID 550.2, the field strength drops about -- per cent and the energy 1 per cent in 1 m of guide. From 13.00 (12), the X10 attenuation is au) = 0i° = 0.37r = 0.94 or about 200 times that of X. From 13.00 (12) and (13), the signal velocity is 78 per cent that of a free wave. Figure 13.02a shows field maps of the TE N and TE11waves. From 13.00 (16) and (17), the TM
498
WAVE GUIDES AND CAVITY RESONATORS
§13.03
wave fields are `gym = 0{ cor 0' E
i— in cos MirX m ax n a
Coswiry 1 1sin mrx " Y + 11a b b b max nag + kiwgn sin sin a b nry
m
mrx
Wiry
sin — sin mrx cos + i cos km = 11-020 (-- 1n a b a b a b
(7) (8)
From (1), (2), and 13.01 (7), the attenuation of these waves is atm. =
n2a3 m2b3 2T )[1 ab(m2b2 mv3 n2a2 — (vmn/p)2it
(9)
From (2), the longest TM cutoff wave length in a guide for which a = 2 cm and b = 4 cm occurs when m = n = 1 at Xii = 3.58 cm, and the next is A
B B-lines. R4171 face E-/ines. AB sechhn TE10-TE0,
I
1
B E-lees.l&hCa/ seckbn B-/i nes . AB section TM *TM12 13.02b.
at
X12 = 2.83 cm. A field map of the TM11 wave is shown in Fig. 13.02a. Superposition of two square wave-guide fields gives that of the isosceles right-triangular guide of Fig. 13.02b with the same cutoff frequency. 13.03. Green's Function for Rectangular Guide.—A wave guide is often excited by wire loops or stubs carrying sinusoidal currents. The resultant fields can always be found by integration if the Green's function
GREEN'S FUNCTION FOR RECTANGULAR GUIDE
§13.03
499
for a current element is known. The most general current element is 1(x,y,z) ds = if ,(x,y,z) dx
jiy(x,y,z) dy
kr ,(x,y,z) dz
(1)
It is convenient to work in the Coulomb gauge of 11.01 (4) and 11.01 (5) in which the vector potential comes entirely from the solenoidal part of and satisfies the wave equation and NI,comes from the lamellar part of and satisfies Poisson's equation being everywhere in phase. If 111 is not a function of x, y, z as in a uniform current loop, then only A is needed, but if it depends on x, y, z as in stub excitation, then both A and 11,are needed for the whole field. In all cases the propagated field as well as the magnetic field comes entirely from A. The Green's function must satisfy the boundary conditions that the electric field is normal to the guide walls and the magnetic field tangential to them. The procedure will be to find Green's function for I x, Iv, and rz and form the general case by (1). The calculations are made for elements at z = zo in a guide running from z = — co to z = 00. The function for a guide closed at z = 0 is then obtained by superimposing an image Green's function of suitable sign and orientation in which —zoreplaces zo. First consider a y-directed element. It is evident that the guide walls could be replaced by an infinite set of images in the z = zo plane, each image being parallel or antiparallel to the element and so having a magnetic field normal to the y-axis. Such a field is given by replacement of k by j in 13.00 (2), so that
A = v xi xviTT, B = v=jp2tiV W = — emn sin in 7x cos n71a
(2) (3)
Ampere's circuital law, 7.01 (2), is used to get On0n for which needed. From (2) and (3),
B
x
nay 0' frm Z Zol Om ' „CY...co sin max cos n -
= j02
E. is
(4)
a
m=1 n=0
Now integrate E across the guide at constant y in the z = zo plane except at x = xo, which the path avoids by an infinitesimal detour as shown in Fig. 13.03a. On this path, from symmetry, B, is zero except between Yo and yo dyo where it passes around the current element and equals i.t/ from 7.01 (2). Multiplication of (4) by sin (nom/ a) cos (niry/b) dx dy and integration over the ranges 0 < x < a and 0 < y < b yield ab cos nrY° = jitn021ab(1 dyo sin in7x°
(3(.1)'7,0,n(y)
(5)
500
WAVE GUIDES AND CAVITY RESONATORS
§13.03
This determines emn(y)for A (,,) which takes the form
A(y) =
.1einn(Y) [
mnir 2 maxx sin "Y b cos a ab
m=1 n=0
j(02
n 2r2) b2
max
sin
k ji3„%nnir
b
a
cos
nay
max
sinsin a
n
vi
2 ' n)2 = ,021.te _ Mir 2 _ n'Ir Om = "32 - inn T (
(6) (7)
There remains the local field due to the lamellar part of iy dyo. This is found from problem 113, Chap. V, by interchanging m and n, using the
(a)
FIG. 13.03a, b.
(b)
13.7, notation of (7), writing f„/(jw) for q, differentiation with respect to yo as in 1.07 (1), and replacement of (aV/ayo) dyo by if. Thus .0 dyo N mirx n . mrxo nay „ wry° sin cos sin sin b (8) jo.)Eab2 L-J a a Pmn m=1 n=1 The potentials of the x-component in (1) can be formed from (5), (6), and (8) by interchanging xo and yo, a and b, m and n, and x and y as well as permuting i, j, and k in Am. Thus A
m 2,2
(x) =
m=o
n=1
.Mn7r 2
ab -
(d32 -a2 ) sin "Ycos 7717x -iC ,,n(z)[i a
nry] iz-zoi cos "Y sin max a izjit,nmir sin sin MirX si --b-- e_.0,a
21z dxo N joma2b =
m mrxo nry max n' sin cos sin sin rye a a Nmn — b
n=1
(9)
(10)
GREEN'S FUNCTION FOR RECTANGULAR GUIDE
§13.03
501
As there are no orthogonal functions in the z-direction, a description of the dipole moment in terms of x and y, such as that worked out following 12.21 (8), is needed. This suggests producing a dipole field by the arrangement of Fig. 13.03b in which the z = zo plane is earthed except for the 62area at x = xo, y = yo, which is raised to potential V 00". To get Vo in terms of Izdzo, start with 12.21 (8). IZ dzo =
jcore( 2 — a2)Vo
In (b/a)
a—> b
2j COE V ora2 = 2jcoE Vo dS
(11)
Clearly the magnetic fields of iz dzo and all its images are normal to z and given by 13.02 (7) and 13.02 (8). Thus the tangential component of E in the plane z = zo is .nt = 0)71-
rt,t(z)0,rzn[
a
m=1 n-=1
cos
max .n max sin " — sin Y j cos " bY a b a (12)
Denote the bracket by [ ], take the scalar product of (12) by [ ] dx dy, and integrate over the zo-plane. The integral of [ ] • [ ] dx dy on the right gives 4abfi,n2„. On the left side the tangential E is zero except on the boundaries of the Vo square, and its integral across the boundary is Vo or — Vo depending on the direction of crossing. Therefore the integral of the left side over the vertical edges of the square is 1 1 V o 6— {cos [m —7(xo + — 3)1 — cos [ m —7-(xo— — 5)]} sin "Y° 2 2 a a a The y-integral is similar with m, n, yo, xo, a, b, replacing n, m, xo, yo, b, a. Add the two integrals, combine terms and note that 5 is small, and obtain for the left side VoPrO,„ 2 „ sin (mrxo/a) sin (nryo/b). Thus -2./z dzo amn ( z) = n
otEablini t 7,
sin
norxo
a
sin
n 71-Yo
b
(13)
If 0 in 13.02 (7) is replaced by 0,„„w and it is summed over m and n and multiplied by j/(.0, it becomes A(2) . max miry .n max nay „[i— m a cos asin b + 3 sin a cos
_= m=1
n=1
— kgn sin '71-x 7ria sin nay e—isc-11-41 (14) The scalar potential for the z-directed dipole can also be written down from problem 113, Chap. V, by calculating (a V/azo) dzo and making the
502
WAVE GUIDES AND CAVITY RESONATORS
§13.04
same substitutions as for (8). = 2.Iz dzoN / coma b
sin
max nary n yo mirxo sin sin sin e-0--1z–zol a a
(15)
m = 0 n=0
The integral of the Green's function over a wave-guide antenna plays the same role as the retarded potential integrals in free-space antennas, and both require a knowledge of the current distribution, which is difficult to calculate. A plausible sinusoidal distribution is usually assumed, and the transmitted wave is rarely very sensitive to errors in this assumption. If the wire in loop or stub is not too thick, then the fields outside it are essentially the same as for an infinitely thin antenna along its axis. In input impedance calculations the fields between axis and surface must be excluded to avoid infinite field energies. Each mode acts as an independent circuit shunted across the antenna element. A given current element excites every mode that has an electric field along it. The antenna impedance is minus the sum of the line integrals of the mode electric fields along the wire surface, in the direction of the current. Add to this the capacitive contribution of the scalar potential (if any) field integrated over the surface of the wire. It will be noted in the rigorous inductance formulas of Chap. VIII that the per unit length inductance of a wire becomes infinite if the radius becomes zero. If the input reactance of a thin wire-guide antenna is found by integration of the mode fields along its surface, no single term approaches infinity, and the convergence of the series becomes slower and vanishes at zero radius. Thus in the rectangular-guide case where the walls can be replaced by image antennas, a much better method is to add to the free-space antenna reactance, which can go to infinity and is often in closed form, the mutual reactances of the image antennas which are usually independent of the wire radius. The resultant series is then rapidly convergent. Examples of such calculations appear among the problems.
13.04. Aperture Excitation of Wave Guides. Coaxial Opening.Wave-guide fields set up by waves entering through orifices in the sides or end give many of the same calculation problems encountered with waves diffracted through apertures and discussed in 12.18, 12.19, and 12.20. Few exact solutions are possible but in many cases the assumption that the incident wave in the opening is unperturbed as in 12.19 gives useful results. If the aperture dimensions are sufficiently small compared with a wave length, quasi-static methods give correct tangential fields. Thus the double current sheet method of 12.18 yields accurate solutions.
§13.04
APERTURE EXCITATION OF WAVE GUIDES
503
An example of aperture excitation which uses the double current sheet of 12.18 and the Green's function of 13.03 is the coaxial opening shown in Fig. 13.04a centered at x = d, y = 0, z = c in the bottom of a rectangular guide closed at z = 0. The frequency is such that only the TEiomode is transmitted, and it is absorbed at some positive value of z. The value of the output resistance of the coaxial line is desired. To make E, zero at the closed end z = 0 requires two double current sheets of opposite sign, one at z = c and one at z = — c as shown in Fig. 13.04b. The electric field of the TE10 mode everywhere parallels the y-axis, so the infinitesimal thickness dyo of the toroidal double layer is exaggerated in Fig. 13.04c. The exact value of E over the opening is unknown, but the
z
7
dyo
(a) (c) Fin. 13.04a, b, c.
field of the principal TEM mode in the coaxial line which transports the power and is the only field far from the opening is inversely proportional to r. If the potential of the central conductor is V and the current density in the double layer in amperes per radian is i, then from the text below Fig. 12.18, dy o _
E r
V r In (r2/r i)
(1)
The currents encircle the entire section of the toroidal current sheet of Fig. 13.04c. Thus they are solenoidal and none of the scalar potentials in 13.03 can appear. The amplitude of the TE10 mode excited by a vertical element of width de may be written down from 13.03 (6) by placing the torus in the z = 0 plane in a guide running from z = — 00 to z = co and bounded by x = 0, x = a, y = — b, and y = b. From symmetry the y = 0 plane is then an equipotential except for the coaxial
504
WAVE GUIDES AND CAVITY RESONATORS
§13.04
aperture. Putting m = 1, n = 0, 1 = i d0; y = 0, and 2b for b in 13.03 (5) gives def lo—
dyo sin(71-x0/a) d0 2 jabOi00 2
— V sin (irx0/ a) d0 coabOVio In (r2/r1)
(2)
For the total amplitude this must be integrated over the vertical walls. It is simpler to combine the four elements at d ± r2 sin 0 and c ± r2 cos 0 shown by crosses in Fig. 13.04b so that for the outer wall the integral is 1 irr2 . — 5 cos (— sin 0) cos (310r 2 cos 0) d0 a 20
(3)
In the notation of 13.03 (7) with tan q5 for 7/(13/10a) this may be written Ifol'1cos [Or2 cos (0 — (1))] + cos [Or2 cos (0 + O)]) dB
(4)
so that the value of the Both terms in the integrand have a period of integral is independent of 0. With 4 = 0 it becomes 5.302 (3) so 7rd or2) 2
2f wcos ((3r2 cos 0) d0
2 2A jX
(5)
Addition of similar expressions for the current at the inner edge of the r2 < z, torus and for the image torus in Fig. 13.04b gives, with c A lo = io sin Clo =
ei(c"—,3'ioz)
T-7- sin rd J 0(2/7-r2/X) — Jo(2/rr IA)
jcod(3 0ab In (r2/r1)
(6) (7)
The TE10 electric field and the magnetic induction are B. =
-aA10
az
- j/3i 0A10,
ku =
(8)
The power transmitted is the mean Poynting vector
1EP
ax jiwOlo sm2 — I Ci01 2. a
(9)
Integration of this over the range 0 < x < a, 0 < y < b and setting the result equal to 1V 2/R give for the radiation resistance R=
Ac013'10ab lng (r2/ri) sing (rd/a) sing (13'10c)[Jo(airr2/X) — J0(271-ri/X)P 871-2 sing
The same result is obtained by the integration of the Poynting vector over the coaxial opening, using r1 V[r In (r2/r i)]—' for E and v x A10 for B, but the integration is more difficult.
§13.05
RECTANGULAR GUIDE FILLED BY TWO MEDIUMS
505
13.05. Rectangular Guide Filled by Two Mediums.—Suppose the guide is filled from x = 0 to x = c with a lossless medium Aici and from c to a with /22E2, where a — c = d. Tha wave velocity or wave number and the cutoff frequency are desired. To satisfy the boundary conditions at all times and all y values, the waves in both mediums must have the same velocity and y dependence. This is impossible for most waves purely TM or TE with respect to z but works out well for waves TE or Tit/with respect to x. Such waves derive from 13.00 (2) if i and Wter replace k and Wte. The (irm/a)2 term in 13.02 (2) is still valid but the (irn/b)2 term no longer holds and will be replaced by pi in AiEiand pi in 112E2. Thus the W,„ forms in mediums 1 or 2 that satisfy the boundary conditions at the guide walls are A l sin pix cos
miry .,,„pz
or
A2 sin p2(a — x) cos nwye-oc,,z
From the relations in 13.00 (12), when 13i = (42µ1E1 and i3=
(42122E2,
— (7)2— pi = 4 — (12—
(13:.p)2 =
(2)
For continuity of E, and H, at x = c, (E0 1= (Ez)2 and A2(1301 = Thus awte. since Ez = A2 sin plc = A2 sin p2d ay 1.1.2pii12
cos pic = —mip2A2 cos p2d
since Bz =
(1)
a wtex
(ax (3z)
2•
(3) (4)
The ratio of (3) to (4) gives the transcendental equation ii1232 tan plc = —A2p1 tan p2d
(5)
The same equation results from (Ey) 1/(By) 1= (Ey)2/(By)2 at z = c, so all tangential components of E and H are continuous if (5) holds. Equating the normal components of B gives (3). For a given frequency and medium, piand p2 as functions of Om% from (2) may be inserted in (5), the left and right sides plotted as a function of 13„' p, and its correct value found by their intersection. For the cutoff frequency is zero, so from (2) 2 kFlic = WcAlE1
/MIA 2
(A) c = 04A2€2
97)2 (+
(6)
Insertion of these values in (5) and solution of the resultant transcendental equation as before yield coc. Note that if p i is real in (6), then p2 is imaginary and vice versa. The cutoff frequencies will lie between
506
WAVE GUIDES AND CAVITY RESONATORS
§13.06
those of the equivalent modes in aif i and a2E2. There are an infinite number of values of pi, p2, and Eoc. The boundary conditions for this guide can also be satisfied by waves TM to x by a perfectly analogous procedure. The chief results appear in problem 18 at the end of this chapter. 13.06. Thin Iris in Rectangular Guide.—The junction of two rectangular wave guides, which are affixed normally to opposite sides of the perfectly conducting z = d plane in which holes have been cut to connect their interiors, forms a plane discontinuity. If in both guides only the TEiowave is propagated, then standing waves can be set up so that there is an electric node at z = d. There is then no transverse electric field za
zs Z=0
Z=d (a)
NP
Z=2d Z=0
Z=d
Z =2d
(b)
(c)
Yi
7
X NF.
—x— X t --x-—b —1 +1 +b (d)
(e)
(f)
FIG. 13.06a, b, c, d, e, 1.
there, and no magnetic field links any element of z = d. In a transmission line only a pure shunt element Zsmay be inserted at an electric node without upsetting the pattern. Now consider the special case in which a thin conducting iris at z = d in a rectangular guide partially blocks TEio standing waves set up with nearest nodes at z = 0 and z = 2d. From 13.02 (3), the TED)electric field when 0 < z < d is
( frx Ey = E0 sin T1 )sin Olaz cos wt
(1)
Higher mode fields must be present to cancel this field where it is tangent to the iris. By 13.02 (3), these contain the factor e-1 0'--(z-0which, when X10 is large, will confine them to a z-interval so short that all local fields are in phase and may be found by static field methods. This is called the "quasi-static" approach.
§13.06
THIN IRIS IN RECTANGULAR GUIDE
507
The wave-guide section 0 < z < 2d is equivalent to the short-circuited T section of Fig. 13.06a in which the currents are directed as shown. It is usual in treating wave guides to use "normalized" impedances which we designate by a superior °. These are found by dividing each impedance by the characteristic guide impedance which normalizes to 1. For a guide transmitting only the TE10 mode with no attenuation, 11.15 (13) is zo _ 2i _ 2% cos C/ + j sin /3of z 2, cos o'io/ jZ% sin f3 of
(2)
For the short-circuited T section, 2°,, = 0 and 1 = d, so for resonance = —42,? =
tan iTiod
(3)
Now consider the special case where all sections parallel to the x = 0 plane, and hence to the electric field, look alike. The electric fields parallel the x = 0 plane so that, in the region near z = d where phase differences may be neglected, E at any value of x must obey the equation of continuity in y and z and can be derived from a potential that satisfies Laplace's equation in y and z. Identical currents cross all electric nodes in a guide so that the total charge on top and bottom between nodes is always the same. Let the charge per unit width in the interval 0 < z < 2d at some value xo of x be Qio for the TE10 mode and Qo for all modes. Let the static discontinuity capacitance per unit width be C1and the potential across the guide at x = xo, z = d be Vd = bEd. Then from (1),
C l = Qo —Qc— 2 (fixi'6E, dz bEd bEd 0
f deE, dz)b— oco 2 o
6
cot bo 13'10d
(4)
Insertion of this value in (3) gives
e 1 Z° = . = . ' 3b0i0C1 3WC°
or
C° = —C1 WE
(5)
As a specific example, consider an iris which leaves a window of height c in a guide of height b, as shown in Fig. 13.06b and Fig. 4.22b where the equipotentials are dotted except at the boundaries U = 0 and U = 47. Insertion of a = c, W = jV , and z = jd in 4.22 (6) gives for V, as V and d approach infinity,
V = 11-rd — ln sin 47c— b
b
ln csc — b
(6)
The charge per unit length between 0 and d is E V. The last term times E equals half the additional charge on the guide due to the presence of the iris for a potential difference of fir. Therefore twice the second term multiplied by 26/a gives C1. Insertion of C1into (5) and substitution
508
WAVE GUIDES AND CAVITY RESONATORS
of 27-X9— ' for (310give for the normalized susceptance — In csc B° = wC° = 8b ,
§13.06
B°
-brc
(7)
For the symmetrical iris bounded by U = —ir and U =sir in Fig. 4.22b, the only change is the potential difference, which is now 7r instead of fir, so that (7) must be divided by 2. The least attenuated TE11 mode is now out by symmetry; thus the result is more precise. From Fig. 4.22b by images the center strip behaves exactly the same way as the symmetrical iris. The errors in the above value of B° are about 2 and 9 per cent at b/Xe values 0.1 and 0.2. These can be improved by a variational method given later. Another interesting case is when all sections parallel to the y = 0 plane, and hence to the magnetic field, look alike. Suppose the iris is at z = d and the frequency is such that only the TE10 wave propagates. All fields are independent of y, so n is zero in 13.03 (1) to (4) and the betas are
ai
2r = — X,
=
( 32
01)4)
=
co
Mr
/3 = (40
2r ),
(8)
This case, unlike the previous one, has charges only on the guide walls. Thus in the Coulomb gauge of 11.01 the entire field derives from the vector potential. Take the origin at one end of the short-circuited T section of Fig. 13.06a so that the B component amplitudes at z = d are Bz = A1 f3i cos (,3;4) sin (31x)
mAm1,8'„I cosh It3'„di sin (0mx)
(9)
m= 2 Bz = — Av3i sin (M) cos 0(31 x)
—
mAmQm sinhl0:,,di cos
(0mx) (10)
m=2
Multiply (9) by sin i31x dx and (10) by cos ,31x dx, integrate from x = 0 to x = a, eliminate A1, and substitute from (3) to give tan
Wd) = —
O'i afaBy cos ,31x dx ° 7rf
(11)
B, sin /31x dx
where the shunt impedance is shown in Fig. 13.06a. This requires the static approximation for the higher modes only at z = d. As a specific example, take the symmetrical iris with a gap of width c as shown in Fig. 13.06d. To visualize the equivalent magnetostatic case, let X >> a so the magnetic loops in Fig. 13.02 are much elongated in the z-direction. As the impermeable iris edges push in from the walls, they press the lines together as shown in Fig. 13.06c. No matter where
§13.06
THIN IRIS IN RECTANGULAR GUIDE
509
this occurs in the loop, Bzin the window is positive at one edge, negative at the other, zero at the center, and vanishes on the iris outside the window while Bs is nearly zero in the window and tangent to the iris walls. If the guide walls in Fig. 13.06d are folded outward at the iris junction to the coplanar position, Fig. 13.06e results. These two maps bear the same relation to each other as those in Fig. 4.22b. The field in (b) comes from cutting a slit 2 units wide, shown in Fig. 13.06e in an infinite sheet of permeability zero that separates two equal and opposite uniform magnetic fields. Figure 4.23 shows the result with only one field present. Superposition of a similar field rotated 180° gives Fig. 13.06e, and application of the transformation below Fig. 4.22b, but with the origin at the iris base, is Fig. 13.06d. W = C/(42 — 1)1, Eliminate zi and make z' =
zi = b1sin [:(z' — yt)]
(12)
— c) when z1 = 1 to give
W = C(cos' 7rz — sin' /71 a a
(13)
For the z = d plane of the guide, y' = 0 so this yields, by 4.11 (1), Bz jB, = ±C
a[cos2(lrx/a)— sin'(2rc/a)]1 -
(14)
Thus with respect to the guide axis, B z is an even function that is zero when i(a — c) < x < 4-(a c) and By is an odd function that is zero if c) < x < a. The integration in (4) can 0 < x < i(a — c) and if 4(a be done by parts, letting dv = B, dx and u = sin (Tx/a) in one and dv = Bz dx and u = cos (rx/a) in the other. The uv products vanish, and if sin (71-x/ a) is replaced by t cos (lirc/a) in one and cos (lrx/a) by t sin (irc/ a) in the other, they lead to Dw 350.01. Substitution in (11) gives =— a tan' L-c = jo.)./),? a X,
(15)
Thus the iris is inductive, and the normalized shunt susceptance is B° = = — — 1 =— Yk
X ire) cot' (2a a
(16)
An exact value could be found by solving an integral equation given in 13.08, but this is impractical. A better way is to improve the above solutions by a variational method. This method, also applicable to some static problems, will now be discussed, as in Borgnis and Papas.
510
WAVE GUIDES AND CAVITY RESONATORS
§13.08
13.07. A Variational Method for Improving Approximate Solutions.— Let some physical quantity such as the iris susceptance in 13.06 be expressed in the form of the integral equation
Q=
fsh(r)E(r) dS
(1)
where h(r) is known but E(r) is an unknown function, such as the electric field, of the coordinate r in the surface 8, which can be the window area in 13.06, over which the integration extends. Now consider the integral
h(r) = fs,G(r,r')E(r') dS' where r' relates to area S' and G(r,r') = G(r',r).
(2) Put (2) into (1) so
Q = fs,E(r')G(r,r')E(r) dS dS'
(2)
Division of (3) by the square of (1) yields 1
LLE(7.1)G(r,r')E(r) dS dS'"
Q
[fsh(r)E(r) dS] 2
(4)
This expression has the useful property that, if trial values of E(r) are inserted, the Q value is stationary about the correct value of E(r) so that a first-order variation in E(r) gives a second-order Q variation. To prove this multiply both sides of (4) by the denominators giving
Q fs fs,E(r')G(r,r')E(r) dS dS' = [fsh(r)E(r) dS] 2
(5)
Take the variation, which follows differential rules, thus
Qfs fs,SE(r')G(r,r')E(r) dS dS'
Q fs fs,E(r')G(r,r') SE(r) dS dS'
= — SQ[fsh(r)E(r) dS] 2 2fsh(r) SE(r) dSfsh(r)E(r) dS (6) Since G(r,r') = G(r',r), r and r' may be interchanged in one integral on the left so it adds to the other. In the last right-side term put (2) for h(r) in the first integral and the Q of (1) for the second. Double integrals cancel leaving SQ = 0. The value of (4) is independent of the E(r) units. Borgnis and Papas, Collin, Harrington, Marcuvitz, and others treat the subject fully. Schwinger first used this method in this way. 13.08. Inductive Iris Susceptance by Variational Method.—The technique of the last article, where Q is B° or (B°)-1, may be used to improve
§13.08
INDUCTIVE IRIS SUSCEPTANCE
511
the quasi-static solutions of Art. 13.06. The accuracy of the result depends on how close the trial value of E(r) is to the true value of the tangential E in the window or the tangential B (current density) on the iris. The conformal transformations of 13.06 will provide good trial values in both cases if the static E, and B, have the same directions as the actual E, and B,. To find the latter in the capacitive case, both vector and scalar potentials are needed in the Coulomb gauge because there is charge inside the walls and the divergence of the iris current is not zero. Thus it is easier to use the magnetic Hertz vector of 11.01 (13) for the fields. This shows that E, is y-directed, but B, has both x- and y-components and thus is not compatible with the quasi-static value. This case will therefore be relegated to the problems at the end of the chapter. There is no charge on the inductive iris; therefore only the vector potential of the Coulomb gauge is needed. This gives both E, and B, independent of y in both magnitude and direction and provides an ideal case for using the quasi-static fields as trial values in the variational integrals. From 13.02 (3) and 13.02 (4), after the absorption of some common factors into C., the guide fields become 00
ty(x,z) = /0)
(1)
O. sin i3„,,xe-js'-z m=1
CO
(x,z) =
Cm(—i(3;. sin 0.x + kj0. cos (3.x)e-jtv-z
(2)
m=1
where m is odd since the fields are symmetrical about x = 4a. In the iris plane the electric field is zero on the iris surface and Ea in the window. Multiplication of (1) by sin Ona dx and integration over the guide section give C. which, put into (1), gives CO
2 E(x) = a
sin (3,,,,x f Ea(x') sin 0,ax' dx' m=1
a
(3)
But in the window, as proved in the paragraph following 12.18 (4), .13(x) is that of the incident wave, which is normalized to sin @ix). From 11.00 (1), Bix(x,0) is (.0 -1ag:(x,z)/az. Thus transfer of the m = 1 term, which is the TE10 amplitude, to the left side gives at z = 0 — f Ea(x') sin Oix/c/x'i =
01 sin 4'
a
13',„ sin Ornalata(x') sin fi'mx' dx' m=3
Multiplication of both sides by Ea(x), integration over the aperture, and
512
WAVE GUIDES AND CAVITY RESONATORS
§13.08
division by O'd fEa(x) sin 131x dx]2give a 2f Ea sin Oix dx
1 = NO'„[f Ea(x) sin 0,„x dx] 2 01[ fEa(x) sin 01x dx]2
(4)
m =3
The left side of this equation is expressible in terms of the iris susceptance. Let the total potential and current be V1and I1when z < 0 and V2 and /2 when z > 0 as in Fig. 13.06f. Both the incident wave V, and the reflected wave V, contribute to V 1, so that V1 = Vi Vr,
ZkI 2 = V2
Zkii = V, — Vr,
(5)
Combine with this the relations in (6), valid at z = 0, to get (7). V1 = V2 = 2V, = VI ± ZJI = 2V2
(6)
I 1 = I 2 + IL
jCOL/L,
V2 jc,„'L
so
Zk
j coL
2 (V,
1)
2
(7)
From (3) the ratio of normalized TE10 incident to TE10 transmitted is ri 2
=
Vi V2
2 [-j (
)
„ , ]--1 sin 13 a x
(8)
Substitution of this into (4) and replacement of ji3:„ by (131 — 02)1give Zk
-B Yk
= —Bo = m =3
2031 — (32)4[ fEa(x) sin 0,ax dx] 2 Ea(x) sin Oix dx] 2
(9)
Integration of both integrals in (9) by parts replaces the sines with cosines and Ea(x) by aEa(x)/ax which, by Maxwell's equations, can be replaced by —jcoBz(x) since Ea(x) is y-directed. Writing kBa(x) for jc4E(x) and replacement of some of the betas in (9) by (1) give aB° =
(1 — 4a2m-2X-2)1[
f Ba(x) cos 13,nx dx 2 fBa(x) cos 01x dx
(10)
m =3
The imaginary part of 13.06 (14) gives an excellent trial value of Ba(x). Integration limits are x = 2(a — c) and x = c), but the integrands are even about x = yt and thus the limits za and c) may be used. Since m is odd, replace it by 2n 1 so that the sum goes from n = 1 to n = 03. In 13.06 (14) and in (10), substitute 0 for 131c and -(2n + 1) (7r- + 4)) for 13„,x. The integrals then lead to Mehler's Legendre function integrals found in MO, page 52, WW, page 315, HMF 22.5.36 and
INDUCTIVE IRIS SUSCEPTANCE
§13.08
513
22.10.11, and elsewhere. Thus 'sin 0 sin (n 1)0 d4
g [cos (n — 1 1)0 — cos (n ) 0] d0
1/0 (cos — COS 0)1 — t: 2 0 (cos (1) — cos 0)1 = C3[Pn_3(cos
— P.±1(cos 0)] = C3
2n ± 1 sin OP;,(cos 0) (11) 1)
n(n
When n = 0, the denominator integral is C3(1 — cos 0). Substitution of these values into (10) and application of Dw 406.3 give B° =
a
cot' (7c)[(2n + 1) 2 2a
—
(2111P! cos (irc/a)]2 n(n ± 1) X
(12)
n=1
The convergence of this series can be improved by taking the difference between it and a summable series having identical terms when n is large. Note that if a = 1 and a = 0 in 7.13 (2), then the r integral of A0 from 0 to 00, if r-n-' replaces rn when r > 1, is 2(u/ sin a)-'fo'Ao dr =
(2n + 1)n-2(n + 1)-2[P;,,(cos a)]2 (13) n=1
Here A is the vector potential of a circular loop of radius sin a lying on a unit sphere. Doing the same integration on the first integral in 7.10 with a = sin a, p = r sin a, and z = (r — 1) cos a gives (2n + 1)n-2(n + 1)-2[P,1(cos a)]2 = r-ircos 0 In (1 — cos .0) d¢ = 1 n=1
(14) This might have been foreseen because X = 00 was a tacit assumption used to get 13.06 (16). Thus the series in (12) may be replaced by 2
[(2n
1+
1)2 — 4a2X-91— 2n — 1 I[n-'(n
1)--1P,I(cos --Ca )]
n=1
(15) in which the series converges so rapidly that in the cases checked five terms gave three digits correct in B° and ten terms gave four digits. Jahnke and Emde tabulate dPn/d0 and IP„1 (cos = IdP,i /c181. The integrals in (12) are positive definite and are an absolute minimum for the true fields so (12) provides an upper limit for B°. A short proof of this appears in Borgnis and Papas, page 122. A lower limit is found by use of a variational expression, based on the obstacle current, which
514
WAVE GUIDES AND CAVITY RESONATORS
§13.08
bears a strong resemblance to (9), and reads
fil(131 - 0 2)1 fol(a 4)Bz(x) sin 0„,x dx Yk =
1 = m=3
B°
B
2
}(a-c)
[JO
Bx(x)
sin 131x dx
(16)
]
2
where the obstacle current density iy(x) has been replaced by Bz(x) to which it is proportional when z = 0. The integration is over the iris surface since By is zero in the window. Note that all terms of (9) except the integrals have been inverted. The real part of 13.06 (14) is a good trial value for B5(x). Put this value into the numerator integral of (16), and substitute 7 - (#) for 27x/a, 0 for ac/a, and 2n + 1 for m, which is odd from symmetry. The integral then becomes, disregarding the sign, 'sin 4) cos (n 4)cf, dc/o (cos cos cOi
Cf 0
4)0 - sin(n - cf.] d4 0 - cos 0)1 2n ± 1 sin OP,1(cos 0) (17) = C3[Pn_1(cos 0) - Pn±i(cos 0)] = C3 n(n ± 1)
C2 fe7sin (n
(cos
The second form of Mehler's integral, already referred to, was used here. When n = 0, the denominator integral is C3(1 - cos 0). Substitution of these values into (16) and use of Dw 406.2 give a(7-c (2n + 1) { PRcos(71-c/a)]) 2 a)L-J [1 + 4a2X-2(2n 1)-21in2(n + 1)2
1
(18)
/76 = - )g tan2
n=1
Exactly as with (12) the series may be put in the rapidly convergent form
Jr 1
1+ n=1
4a2 F f Pgcos (rc/a)R2 1}(2n ± 1) (2n + 1)2X2 n(n ± 1)
(19)
This provides a lower limit for the susceptance. The following table gives values of the correction factor by which 13.06 (16) should be multiCORRECTION FACTOR TABLE
X = 0.5
X = 1.0
X = 1.3
c/a 0.2,0.8
0.4,0.6
0.2,0.8
0.4,0.6
0.2,0.8
0.4,0.6
Upper limit
0.9768
0.9582
0.8995
0.8149
0.8102
0.6385
Lower limit
0.9758
0.9558
0.8745
0.7550
0.6760
0.2940
Arithmetic mean
0.9763
0.9570
0.8867
0.7849
0.7434
0.4662
Handbook
0.9765
0.9578
0.8913
0.8045
0.7594
0.5881
§13.09
THE CIRCULAR WAVE GUIDE
A
515 C
B MI mode. E-lines in AB section spaced to give equal intensity increments.
T.Eol mode. Principal section of equiflux tubes of magnetic induction.
A
rTh TEI1 mode. E-lines in AB section spaced to give equal intensity increments on center line.
TEli mode. B-lines in principal section spaced to give equal induction increments on center line.
A
TMol mode. B-lines in AB section spaced to give equal induction increments.
TMol mode. Principal section of equiflux tubes of displacement.
A
C
B
D
T/14-11 mode. E-lines in principal section TM„ mode. B-lines in AB section spaced to give equal induction increments on center spaced to give equal intensity increments on line. center line. Fin. 13.09.
plied according to (12) and (18) and their average compared with the accurate value of the la formula on page 221 of the "Waveguide Handbook" over the ranges plotted on pages 222 and 223. It is clear that the variational treatment has made a great improvement in the quasi-static value and that the upper limit alone does pretty well. 13.09. The Circular Wave Guide.—Consider a wave guide of radius a. From 12.16, a solution of 13.00 (3) which is finite on the axis is 0(p,4) = J „,(0.7,p)(Om, cos m4) +
b„,„ sin m4)
(1)
516
WAVE GUIDES AND CAVITY RESONATORS
§13.10
where, by 13.00 (4) and (5), Omn must be chosen so that I aJm(amna)/aal to = 0
(2)
1,/m(Retteta)lt. = 0
The most important zeros of these functions are J'0(3.832) = J'0 (7.016) = Pi (1.841) = J'1(5.331) = 0 J0(2.405) = J0(5.520) = J1(3.832) = J1(7.016) = 0
(3) (4)
From 13.00 (14) and (15), the fields of the TE waves are [E]te = jco[pimp-I J.(3mo)(Omet sin mct) — /5„,„ cos melo) „,„. ± 43/3m,,Fett(Oette?)(C,,,„ cos m(#1 ± Dm. sin mcp)]e--)0'
(5)
[bite = 1:713:tta — ViOmetP,,,(0emP)(Ctu. cos m4) + /5,,,,t sin md›)
+ (1)mP 1,1.(0..P)(C>r sin mqS — /57.1„ cos mcb)] + kol„Jm(0.p)(Cmn cos MO + Dmn sin m et,) }e 'A'n'ne
(6)
From (1), (2), 13.01 (6), and 5.295 (6), the TE wave attenuation is ate =
T
[1
/Iva
—
Pm, 2
—4
M2 Olna2
—
m2
±
1,„, 2
I
(7)
We notice that the circularly symmetrical waves for which m = 0 have the unusual property of a decreasing attenuation with increasing frequency. The longest TE cutoff wave length A ll is, from (2), (3), and 13.00 (8), at 3.42a and the next, An, is at 1.64a. Take a = 2.34 cm so that A n is 8 cm and compare with the 2- by 4-cm guide of the last article at the same cutoff. The 5-cm free wave length attenuation is 0.0028 in-' which is considerably less than the 0.005 m--' of the rectangular guide. Field maps of some of the TE modes appear in Fig. 13.09. From 13.00 (16) and (17), the fields of the TM waves are ( ./5,„„ sin Ind>) E tm = oge,t[pAtt.T„t(t3mnP)(eett cos mcA — 4mp 1Jett(0m,tp)(C,,,„ sin mg!) — limncos mcb)] kiwnteint(t3ettitP)(Cmn cos mci) + Ann sin m(5)}e--0'—z (8) Btm = [p1o2mp-1J,.(0,..p)(c. sin m(tt — /5„,„ cos m40) 15— sin ntt)]e-30'—' (9) + (13/32t3ttta'n(t3,7t.p)(0.,, cos mcP From (1), (2), 13.01 (7), and 5.295 (6), the TM wave attenuation is )2
at
[1— „, = pava
n
I
(10)
The longest A,en values are A01= 2.61a, A11= 1.64a, and A02 = 1.14a. 13.10. Green's Function for a Circular Guide. The wave patterns set up by a wire loop or stub carrying sinusoidal currents can be found by integration of the Green's function of a current element. Instead of the x , y-, z-components used for the rectangular guide in 13.03 (1), the cur—
-
§13.10
GREEN'S FUNCTION FOR A CIRCULAR GUIDE
rent element will be resolved into p-, I(p,4),z) ds = oil „(p,4.,z) dp
517
z-components, thus
4)i i,k (p,4),z)p d4)
klz(p,4),z) dz
(1)
As in 13.03 the Coulomb guage, in which A arises entirely from the solenoidal part of I(p,4),z) and 4/ entirely from the lamellar part, is convenient. The comments of 13.03 also apply here. The calculations and results for radial p- and longitudinal z-cornponents follow, but the 0-component results, whose calculation resembles that for the p-component, are given in problem 27 at the end of the chapter. Consider a current element of strength pi./0" and of length dpo at po, 0 in the z = 0 plane. If the element is at 0 = 00 instead of at 0 = 0, — oo replaces 4) in the equation of the last article. The complete TB solution is a double sum over m and n of terms like 13.09 (5) and (6). To find the amplitude of the pq-term, one sets t = 0, z = zo, and Cm. = 0 in 13.09 (6), E, being an even function of 0, takes its scalar product by [910pq,P9(07,p) sin pct. + 4:1)Ipp-',/p(Op0 p) cos p4)11) dp d4)
and integrates from 0 = — r to 4) = r and from p = 0 to p = a. The first integration eliminates all but the pth term in the m-summation and puts r in place of sin' p4) and cos' 730. If v1is written for i3 ,,p and v2 for 13„p, the remaining p-integrals are 11-1pq0;ilf:P'a [JVV1),Pp(V2)
,2 dv .-Jp(V1)Jp(V2)11 1
v iv2
(2)
By 5.296 (11), the integral is zero unless n = q when, by 5.296 (12), its value is 1(027,0a2 — p2),Ig0„a). In the zo-plane, 13, is zero everywhere and Bois zero except over the element dpo at po, 0. But by 7.01 (2), at po f Bo p do is —1ALI . Solving for D„ gives PA/ dpo Jp(0„p o)
(3)
[/5„]te ==rj 4 ,a)]2 po0;),(13La2 — p2)[(13
For the TM modes excited by the radial element, set t = 0, z = zo, and D.. = 0 in 13.09 (9) and take its scalar product by [pipp-iJp(i3„p) sin MI + 4303,J;(13p,p) cos p4dp dp d4)
And integrate from 0 = — 7r to ¢ = 7r and from p = 0 to p = a. The integrals are identical with those found before except for a 2 — SP factor, and the integral in (1) is 1[090aJ;(0„a)]2because Jp((3„a) is zero, so
=
dpo J;(f3„p0)(2 — 5;1 ,) 0,
2r)3„[(3aJ;(f3„a)]2
e
"
(4)
518
§13.10
WAVE GUIDES AND CAVITY RESONATORS
For the longitudinal component, use the source of 13.03 (11), set z = Dmn = 0 in 13.03 (8), and take the scalar product of both sides by - (1)A4(014/3) sin PO] dP dcb = Rp0 dP
[000p0Pp(i3p0P) cos
dcb
Integrate over the ranges 0 < p < a and 0 <
Jo
271.
oFinqa2[J;(d3p,a)12e-0'9)zo E • Rp, dP d4) - r241Pge
o
(5)
The arcs c 13 and c - 13 and radial lines at 14) and -1,6,4, bound dSo. The integral over the arcs, since the integrand contains no 4), is Op, { (c
13),P,[0„(c + 13)] - (c - 13)J;,[0pq(c .5—* 0
Opq[13,c,P;(13pqc)
Vo J;(02)oc)1V o 8 AO
(6)
On the radial sides 3 is so small that we may set p = c, which gives apc-14(4qc)[sin (lp AO) - sin (- Zp AOglio = p2Ao Sc-117 04(0,4) (7) The sum of (6) and (7) gives the left side of (5). When we write dS0 for c3 AO, substitute for Vo dSo from 13.03 (11), combine the Bessel functions by Bessel's equation 5.293 (3), and solve for C„, we obtain j(2 dzo (0,c) 0, e' 27,3;„w2f[aJ;(137„a) 2
P9.0
(8)
]
It should be noted that, in (4) and (7), -,/p+1(13„a) may be substituted for J;(/3„a) by 5.294 (2) because ,I,(0„a) is zero. In the Coulomb gauge the scalar potentials xIf„ and •Ifz of the radial and longitudinal components of the current element in (1) may be derived from the point charge Green's function of 5.297 (4). For the radial component write dpo J;(0.np0) for J 8 (1.4b), where the second equation in 13.09 (2) defines 13., and I/(jw) for q. Thus the element at po, 0, zo yields jI dpo
(2
27rwEa2L-J n =1 = 0
3;,',)e-flmniz-=91J n(I3innPO Jm(OmnP) cos m4) (9) 2 s+1 (Ontn a)]
For the longitudinal element replace q by I/(jw) and the exponential factor by -13,nn dzo e-13..k-zol giving the result 4,2
dzo
27rwEa2
.1
(2 -
(3:,,,,)e-o„,„1=--z/m(l3Po)Jm(Ont.P)
n = 1 m= 0
The fields are everywhere in phase.
[Js+1(13.o.a)12
cos m4) (10)
§13.11
LOOP COUPLING WITH CIRCUL AR GUIDE
519
13.11. Loop Coupling with Circular Guide.—The formulas of the last article may be applied to a square plane loop at = 0 in a wave guide of radius a closed by a conducting wall at z = 0 where the boundary condition is satisfied by using an image loop, as shown in Fig. 13.11. The current is assumed uniform so that no scalar potentials are needed, although the integrals can be set up for any given distribution. Only the radial ends of the loops generate TE waves. Addition of the contributions to limnas given by 13.10 (3) FIG. 13.11. for elements I dpo at po, d; — I dpo at po, b; I dpo at po, —b; and —I dpo at P 0, —d, and integration from Po = c to 130 = a give th.nite
9" Jm(v) —2ma[sin (13:„„d) —sin (13'„„b)] (' v dv 7r13 n,(Olna2 M2 )[Jm(i3mna)] 2 jpm„,
(1)
where )3„,„ is given by 13.09 (2), Oa is 02 — (31„, and z is greater than d. The integral can be evaluated by using for J,„(v) the series 5.293 (3) of which only a few terms are usually needed. The TE fields are given by putting this value of Dm,, into 13.09 (5) and (6) and letting 6,„„ be zero. The last article shows that TM waves will be generated by all sides of the loop. Proceeding as for the TE waves but using 13.10 (4) instead of 13.10 (3), we find for the sides of the loop when z > d, te
= j AI 0(2 — 6:,)[sin (13:,,,nd) — sin (0:„„b)W.(0..c) r[013,,,naJ:„(0„,„a)]2
(2)
From Fig. 13.11 for the longitudinal wires, we add by 13.10 (8) effects of elements at zo and —zo, integrate from zo = b to zo = d, and get z > d,
— (2 —
o[sin (0',„„d) — sin (13:„„b)1J„,(13,nnc) (3) .7.700:,na.P.(0.012
Addition of (2) and (3) gives, since 132equals ,3:,?„ for all TM modes generated by the loop
(3;,,,, by 13.00 (7),
[Omn]tm — j(2 — 62,),u/otsin (0:„„d) — sin (g.b)]./.(0.nc)
(4)
This value, when substituted in 13.09 (8) and (9), gives the fields. For transmitted waves 13:„„ is real, so that the trigonometric terms in the numerators may be written, by Dw 401.09, sin (0:,,,,nd) — sin (13:„„b) = 2 sin [r(d — b)X'„,-7,1 ] cos [71-(d + b)V,j,1]
(5)
520
WAVE GUIDES AND CAVITY RESONATORS
§13.12
It is at once evident that if the length d b of the loop is an integral number of guide wave lengths or if the distance 1(d b) of its center from the closed end is an odd number of half wave lengths, then this mode is not excited. If the bottom of the loop is on the guide axis, only axially symmetric TM waves appear since J„,(0) = 0 unless m = 0. 13.12. Orifice Coupling with Circular Guide.—In the last article the wave-guide excitation was calculated for an assumed current distribution in the antenna or current loop. We shall now calculate it by assuming the tangential electric field component over the orifice. The propagation space of a coaxial line is bounded by the cylinders p = b and p = c where c > b and terminates in the plane z = 0 which is perfectly conducting save for the annular opening. This plane closes the end of a circular wave guide bounded by the cylinder p = a extending from z = 0 to z = 00 . From symmetry it is clear that only the TM waves independent of 4) can be excited, and from 13.09 (2) and (8), we must have Jo(13.a) = 0,
E, =
(1) n=1
As in 12.21 (2), the assumed boundary conditions at z = 0 are p < b, c < p < a,
E, = 0; b < p < C, E p = fo(pin -b) 1 (2)
Insert these values for E pinto (1), multiply both sides by pJ1((3.P) dp, integrate from p = 0 to p = a, and solve for C.: C — n
2-170[Jo(0.b) — Jo(0.01 ‘0,3',M,a 2 1n(blc)[Ji(0.a))2
(3)
A simpler way to derive (3) is to replace the opening by the double current sheet of Fig. 13.04c. By 13.04 (1) the angular current density is i dzo—
2Vo jaw In (c/b)
(4)
and in the Green's function of 13.10 (8) p = m, q = n, and I is replaced by i dybo at p = c and by —i dcpo at p = b. In 13.09 (8) and (9), when is replaced by cp — 00 and integrated around the opening, the result is zero except when m = 0 where its value is 2r. Thus 13.10 (8) gives (3). To estimate the input impedance of the guide when coupled to the coaxial line, integrate Poynting's vector over the opening. From 11.17 (3) by using the ratio of Ep to Boobtained from 13.09 (8) and (9), we have =
2/f
E,f 0 dS = „, ZWA
E22rp dp
e b
=
(5)
§13.13
THE COAXIAL WAVE GUIDE
521
The square of (1) is integrated by 5.297 (3). 4,rwE
1_ Zi
a4 1n2 (b/c)
1{,10(0.b) — 003.012{[a 0.0]2 — [b,1 10.b)] 2 1 0,i0:„[J1(13„a)] 2
(6)
n=1
Thus the guide presents a set of parallel impedances, one for each mode. For transmitted modes, 0„ ' is real giving resistive impedance; for others, ,3„ is imaginary giving reactive impedance. An exact solution requires a matching of fields across the opening, using 13.13 (3) on the coaxial side. 13.13. The Coaxial Wave Guide.—Transmission lines whose propagation space is bounded internally by p = a and externally by p = b fall in the class treated in 11.14 which have principal modes that can transmit waves at any frequency. From 11.18 (5), the resistance per unit length is RJ2 = R4ai,2 ds (
jbi2
ds) =
1\
27 VI
+ a)
r/ 2 (a + b)
271-abo
coL,I 2 (1)
The characteristic impedance is then given by 11.18 (1) to be
2k
L,)]11 L[Ri jco(L jcoe) ti('y
where L = L4- In —b 2r a
(2)
The inductance per unit length was written down from LC = Ile and 2.04 (2). If the skin inductance L, and dielectric conductivity y are small, there is a flat minimum in the attenuation when a(R,/L)/a(b/a) is zero. This value, which occurs at b = 3.6a is only 5 per cent below that at 2.5a and 5.0a. At short enough wave lengths, higher order waves of the type discussed in the last article are possible in a coaxial line. These waves appear chiefly in boundary conditions at discontinuities. The axial region is now excluded so both Bessel functions can occur. Thus from 12.17 (2), = cos m) + Ann sin mP)R,(0„„p) [Rm(0..P)]te = Jrn(0..P) 17 .(0mna) — 7m(O.P)f.(0..a) [Rm(0..P)]0. = J.(13..P) Ym(Om.a) —
(3) (4)
(5)
where, from 13.00 (4) and (5), 13„,„ must be chosen so that (3[R„i(i3,..)b]te = 0
ab
[Rm (R,„„b)],„„, = 0
(6)
From (4), (5), and (6), the boundary conditions of 13.00 (4) and (5) are satisfied at p = a and p = b. The fields, attenuations, etc., can be worked out exactly as in the last article but are of little interest.
522
WAVE GUIDES AND CAVITY RESONATORS
§13.14
13.14. Plane Discontinuities in Coaxial Lines.—Electrostatic methods yield nearly exact solutions of certain electromagnetic wave problems. To show this, write 13.00 (1) in terms of wave length with E for A. V2E = _ (.02/1,E = _4,r2x-2E
0
(1)
Thus for wave lengths very long compared with the dimensions of the region involved, the instantaneous field is indistinguishable from a solution of Laplace's equation that satisfies the same boundary conditions. Consider two coaxial lines coaxially joined to the z = 0 plane which is perfectly conducting save where holes connect the propagation spaces. The wave lengths are so long that only the principal mode is propagated. Clearly, if standing waves are set up so that there is an electric node at z = 0, the discontinuity has no effect for there is no radial electric field there and the magnetic induction links no element of this plane. In any transmission line, only a pure shunt element can be inserted at an electric node in a standing wave pattern without upsetting it. Therefore this discontinuity acts as a pure shunt element across the line at z = 0. Now take the special case of coaxial annular openings and suppose that standing waves are set up so that z = 0 lies halfway between two adjacent electric nodes. Obviously the radial field of the principal mode alone is not normal to the plane face at z = 0, so local higher mode fields of the type of 13.04 (4) or (5) are needed there. These contain the factor e-1 0'...1 as in 13.03 (5) which confines them to line sections so short compared with the wave length that all the local fields are in phase. By (1), the instantaneous electric field in this section is identical with the static field between inner and outer conductors, and the capacitance across the line at z = 0 is the excess of the actual static capacitance of the section over the sum of the capacitances given by the concentric cylindrical capacitor formula for each half of it. The simplest case occurs when the ratio of the external to the internal radius is nearly one. A small sector of a cylindrical capacitor then looks like a narrow strip of parallel plate capacitor which is solvable by a conformal transformation. The case of a parallel plate capacitor with a step in one wall, as shown by the lines V = 0 and V = 27r in Fig. 6.07b, is treated in 6.07. Half the increase of resistance per unit length is given by 6.08 (5), so by 6.08 (4) the increase in capacitance per unit length is 2e/(TAR). The discontinuity capacitance is the product of this by the circumference 27-r of the propagation space. Thus, from 6.07 (5),
(h2 k2 k h 1n + 2 In C = 2.irre = 2e hk k —h
—
4hk
h2 )
(2)
§13.15
GUIDES OF ARBITRARY SECTION. ELLIPTIC GUIDES
523
Whinnery and Jamison (Proc. I.R.E. Vol. 32, page 98) have shown empirically that, when a discontinuity is in one wall, the circumference of the other wall should be used. If the ratio of internal to external radius is 5, this rule still gives an error of only about 10 per cent. Reflections from such obstacles are calculated by 11.16 (3) where, if Zki and Zk2 are the line characteristic impedances on the two sides of the obstacle, Zl
= Zkl)
1 . 1 Z2 =Jc." - ' Zk2
27T-jC = X
1
(3)
Zk2
13.15. Guides of Arbitrary Section. Elliptic Guides.—A most important bit of information about a wave guide is its dominant mode cutoff frequency. From 13.00 (8) this lowest transmitted frequency co,„„ is Omn((.te)-1or for short 3c(ue)-1. It always refers to a TE mode and for the rectangular guide, from 13.02 (2) and (3), is TEio and for the circular guide, from 13.09 (2), (3), and (5), is TE11. From 13.03 (3) and (4), 13c and the transverse coordinates u1 and u2 are connected by V2 U(ui,u2)
02U(ui,u2) = 0,
au — 0 8n
and
on s
(1)
where s refers to the curve bounding a cross section of the guide of area S. The fields derive from U by 13.00 (14) and (15). In only a few simple sections can U be written as a product of orthogonal functions of u1 and u2 as in 13.02 (1) and 13.09 (1). However, as in 13.07, stationary variational formulas exist for Oc in terms of trial U1 values which may be improved by iteration and which satisfy the equations v2mn-F1)
pn) 1 =
0
and
f Up> dS = 0
(2)
where Un) is the nth trial function. The integral ensures zero net flux through a guide section because magnetic flux lines must close, and 13.00 (15) shows that Use is proportional to Bz. Borgnis and Papas, Chap. XIV, give an excellent detailed treatment which covers both TE and TM modes. At any z, (1) holds for all u1 and u2 values. Thus its product by U, integrated over the cross section, is f ( UV2 U i3W2) dS = 0
(3)
Now replace the first term by means of the vector formula (4) to get (5).
VO • VIG V • (070 = OV2‘t f V • (UV U) dS — f (V U) 2 dS 14f U2 dS = 0
(4) (5)
From the two-dimensional Gauss theorem 3.00 (2), in which dS, ds, and Uv U replace dV, dS, and A, the first integral is fU(.9U/an) ds, which
524
WAVE GUIDES AND CAVITY RESONATORS
§13.15
is zero by (1). Thus solving (5) for 0! gives 13!
f (V U) 2 dS f U2 dS
(6)
This has the familiar variational form of 13.07 (4). To prove that it is stationary, take the variation of the last two terms of (5) and obtain
20,60c f U2 dS 2Mf U W dS = 2fVU • v
dS
(7)
By (3) and the Gauss theorem, the right side may be written
an
2f SU V2U dS = —2J V U • v(W) dS
—2f SU — ds an
(8)
Add (7) and (8). The line integral is zero on s by (1) and two others add to give f S U(V2 U + MU) dS, also zero by (1), so 30c is zero. Thus in (6) a first-order variation in U gives no variation in 0,. It can also be shown that, if the trial U is sufficiently near the correct form for M, the right side of (6) gives a result greater than 0,2, just as it did in 13.08 (10) and (16). The square of (1), being positive, is greater than zero if an incorrect U form is used. Substitution of (6) for 0! in this square and rearrangement give
f (V2 U)2 dS 10702 ds
S(VU)2 dS
su2 ds
(9)
with correct U values both equal Now take the gradient of (1), Now square, substitute the left side of (9) for 02, and rearrange to get f [Iv(v2u)]2 ds > f (v2u)2 ds
f (v2 u) 2 ds
(10)
u) 2 ds
The same remarks apply to (9) and (10). Combine (6), (9), and (10) into f [Iv(v2 u)]2 ds( v2u)2 ds
j (v2 u) 2 ds
f (vu)2 ds
f (v u)2 ds
(11)
f u2 ds
To show how the iterative process improves a trial function, assume that a suitable function Un) for which aU / an is zero on s has been found and that a better one Ifn.1-1) is desired. Substitute Ur-") for U in (11) and then replace V2 U;n") by — U(') from (2) to give
f[Uin)]2 ds f [Iv u t(n)12 dS f [()"•0 2 ds % f [v dS ]
f[vMn-1-1)]2 dS in-1-1)12 dS
> (42
(12)
The first and third terms have the same form and the third lies closer to Therefore the function Ur") obtained by solving (2) for Ur") gives
§13.15
GUIDES OF ARBITRARY SECTION. ELLIPTIC GUIDES
525
an improved value of M. Any desired precision can be obtained by repetition of the process. The foregoing technique works much better for the dominant mode cutoff frequency of an elliptical guide than that used for the special case of a circular guide in 13.09. Take the origin on the guide axis, so that the equations for the boundary and its slope are
(
X:)2 a
+
(12 =
1,
dy' _ —a 2 y' dx' b2x'
(13)
where boundary coordinates are primed. To visualize a suitable trial function, imagine the ellipse formed by rounding the corners of the rectangular guide in the TEio E-line map in Fig. 13.02a. The E-lines bend to stay normal to the boundary, and the TE10 B-line map is little changed in the y = 0 plane, but the right and left boundary close in as Yi —o b. Thus from symmetry the Bz to which U is proportional by 13.00 (15) is oppositely directed in the right and left halves and so is odd in x. The B-line maps in the +y- and —y-plane are identical if lyI < b so B z is even in y. A simple dimensionless trial function meeting all these conditions is
Uo) = a (1 + Bx2 Cy2)
(14)
From (1) VU(0)must be tangent to the boundary ellipse, and thus from (13)
aU(')/ ox'1 + 3B(x') 2 C(0 2 _ a2y' b2x' auovayi 2Cx'y' —
(15)
Collect all the terms of the last equation on one side, put over a common denominator, factor out x', and write (x')2 in terms of (y')2 by (13). The resultant equation must hold for all y' values so the constant term and the (y')2 terms are separately equal to zero. The constant term gives B, which is then put into the coefficient of (y') 2to give C. Thus (14) is
U
=
a
(1
y2
X2
3a2
2a2
b2
(16)
The integrals of (6) then become I(VU)2 dS _ [irb(5a2 U2 dS [irab(101a4
21)9 (3a2 + b2)]/[6a(2a2 b2)2] b2)2] 80a2b2+ 17b2)]/[144(2a2
(17)
If p is written for b/a, the last inequality in (11) is 13,a <
p2)11 [24(5 + 2p2)(3 L 101 + 80p2 17p4
(18)
526
WAVE GUIDES AND CAVITY RESONATORS
§13.16
For the circle a = b, so p = 1 and OA = 1.842, which agrees with 13.09 (3) to better than 0.1 per cent. This is a more precise value than can be read from the graph on page 83 of the "Waveguide Handbook," in which r/[20caE(e)] is plotted against e = [1 — (b/a)91. On the next page appears a map of the fields of the fundamental or eHll mode for e = 0.75. A table of the values of the complete elliptic integral E(e) is given in HMF, page 609. 13.16. Cavity Resonators. Normal Modes.—When an electromagnetic disturbance occurs in a closed cavity with perfectly conducting walls so the energy cannot be dissipated, waves reflect back and forth between the walls indefinitely. Because the tangential electric field vanishes at each wall, a wave of arbitrary form must be built up of elementary standing waves of frequencies such as to give electric nodes at the walls. These standing waves constitute the normal oscillation modes of the cavity. Clearly, when two continuous waves having the same frequency, polarization, and intensity travel in opposite directions on a transmission line or a wave guide, there are transverse electric nodes at half wave length intervals. If we form a cavity by inserting at any two nodes perfectly conducting plane sheets normal to the axis of the guide, then that portion of the trapped waves between the sheets is a normal mode of the cavity. It is evident that, if the spacing of the sheets is arbitrary, then the normal frequencies will be those for which the guide wave length is an integral number of half wave lengths. This fact can be used to find the normal or natural oscillation frequencies of cavities formed in the manner just described from any section of wave guide. The two oppositely directed waves in 13.00 (6) must have equal amplitudes so C = ±D. If the cavity ends are at z = 0 and z = d, then for a nondissipative cavity the real part of 13.00 (6) becomes, for one of the pth transverse electric modes, Wte = [AUi(ui, u2)
BU2(ui, u2)be sin 7 cos (wpt + 11/p)te
(1)
and for one of the pth transverse magnetic modes, We. = [A Ui(ui, u2)
BU2(ui, u2)1t„, cos
cos (wpt
ifrp)t. (2)
where 0„' „ is chosen equal to pir/d to make the tangential electric field vanish over the ends. The boundary conditions of 13.00 (4), (5), and (7) used to determine 13mn still hold. Thus, writing 0.„„ for 0 and pir/d for „, we may solve 13.00 (7) for the normal frequency of the mnpth mode in an evacuated cavity. Wmnp =
mnp = Vf3mnp = VOL + 02.n)1 = 1101n + P 271-2(1-91 (3)
§13.18
ENERGY AND DAMPING OF NORMAL MODES
527
For rectangular wave guides 13mn is given by 13.02 (2), for circular wave guides by 13.09 (2), and for the higher modes of the coaxial wave guide by 13.13 (6). For the principal coaxial modes, 13,„„ is zero by 13.00 (5). 13.17. Independent Oscillation Modes of a Cavity.—An electromagnetic disturbance in a cavity will usually excite several modes simultaneously. We shall now prove that the instantaneous total energy is the sum of the instantaneous energies of each of the constituent modes. To prove that fv.E2 dv =f vE7 dv
and
vB? dv fvB2 dv = f
(1)
we substitute Ei or B, for Ar• and Ei or B1 for in the vector analogue, 3.06 (6), of Green's theorem, and for the double curl we write (.0211€ from 11.01 (1) or (2). Thus, we obtain — (4).fv Ei • E, dv = fs[E; x (V x Ei) — Eix (V x E,)] • n dS
(2)
But E, and E, are normal to the boundary, so both vectors in the bracket are tangential to it and their scalar product with n is zero. Thus the surface integral is zero, and we have, since E = wi
w1,
fvE, • E2 dv = fvBi • B.; dv = fvAi • Ai dv = 0
(3)
Thus, in any perfectly conducting cavity filled with a nondissipative dielectric, those oscillation modes having different frequencies are quite independent. If cavity losses are so large that the resonance curves of two modes overlap, then such conclusions are not valid because 64 — co; on the left side of (2) may be zero part of the time. The vector potential used here is always derived by taking the curl of W so that its divergence is zero. As shown in 11.01, all fields that satisfy Maxwell's equations are derivable from such potentials if charges are absent. If the cavity contains an electrode carrying an oscillating charge, then by 11.01 (1) a scalar potential satisfying Poisson's equation is needed whose electric field throughout the cavity oscillates in phase with the charge. That this field — VcI) has an energy independent of that of the field —jwA or —jwV x W is proved from 3.00 (2) by writing
fv(v.,„ x W) • VcI) dv = fvV„ • (W x V(I)) dv = fsn • (W x V43) dS = 0 (4) The surface integral is zero because V41 is parallel to n. 13.18. Energy and Damping of Normal Modes. Quality.—In a lossless, charge-free cavity, the Coulomb gauge gives, by 11.01 (2), (3), and (4), B=VrA
E=
(1)
528
WAVE GUIDES AND CAVITY RESONATORS
§13.18
The instantaneous magnetic energy Wm and the electric energy We for field magnitudes H and E are then, from 8.02 (2), 11.00 (1), and 1.14 (6), Wm =
f B2 dv = 11V x H • A dv = 1-0.),?Ef A, • A, dv
We = lEfE2 dv = 1,j2 Zco2ef Ai • Ai dv
(2) (3)
This shows that the time average Wm and We are equal so that f (4/21 _ dE21) dv = f
• fz — et • E) dv = 0
(4)
The metals of most cavity walls conduct well, so that an observable loss in amplitude appears only after many free oscillation cycles. Therefore damping is a small perturbation. Thus the skin effect formulas of 10.02 applied to the undamped field formulas give the energy losses. So, from 10.02 (8), P = (2A278)-11Bt • Bt dS (5) where 7 and 8 are the conductivity and the skin thickness of the wall and Bt is the magnetic induction tangent to the wall. Cavity resonator damping is given in terms of its "quality" or "Q" value defined by (6)
Wt = Woe—wtIQ
where Wt is the energy at a time t after Wo. Division of this equation by its derivative with respect to t gives
Q
wW —
(7)
aw/at — p
Note from (2) and (3) that the total energy swings back and forth between purely magnetic and purely electric, but the sum is constant except for the damping factor. Thus, from (2), Q_
co3/22eyof A • A dv f Bt • Bt dS
cowySf B • B dv
f Bt • Bt dS
21.1f B2 dv
f dS
(8)
where ;2 is the wall permeability. In the last article it was noted that the energy transfers from one resonant mode to another if the resonance curve overlaps, so it is well to know how this curve depends on Q. Suppose the oscillation starts at t = 0 when the energy is all magnetic. Then from (6), B = Boe—iw,tici cos copt
(9)
§13.19
529
NORMAL MODES OF A CYLINDRICAL CAVITY
where w, is the undamped resonant frequency of the pth mode. By Fourier's integral theorem, B as a function of w is found by multiplication by r—le" and integration from 0 to 00. Write the cosine in exponential form by Dw 408.02 for easy integration, and assume a very large Q to make the w + cop term negligible when w is near wp, so that
iB0/7 13(°))
1(cop/Qp) — i(co — wp)
(10)
pB 0/ (o pir ) (
and
113 (w)1 - [ 1 + 4Q co -
co )2/41
Thus IB(w)I is peaked at w = cop, and from (2) and (10), W„, and B2 have half their peak value when coi = cop ± Icop/Qp. The peak width there is the "band width" Ow. Thus
Aco
(11)
13.19. Normal Modes of a Cylindrical Cavity.—The shape and materials used in a cylindrical resonant cavity fix the relative field strengths and Q-factors. From 13.00 (2) and 13.16 (1) and (2), the fields are Eee _
_
at
w p(k x V Ute) sin 2 P7 -5 sin (wpt
Op)
(1)
prz
Bte
= (— 117 Ute cos — P" d — 143;„„Ute sin —) cos (wpt Op) d V
(2)
=
(3)
d
COp (T i VUt,.
sin 7 — kig,„(1,„, cos 7 sin (wpt
Op)
(4)
= ii;„„p(k x VU t„,) cos 7 cos (wpt + Cp)
From 13.00 (2) and 13.16 (1), the energy integral over the volume and the loss integrals over the two ends and the side walls are, for TE waves,
f
A 2 dv =
2.1Bt dS
(k x U te) 2 dSf sin21 o
-32 rz
dz -
df ( V U) 2 dS 2
= 2(7)2fk x U) 2 dS = 0)2f(v U)2 dS
dS =f[(22,;)2 - (n x V Ur
+
(6)
P z dz UddsJod sine ÷
au)2
= [d(2 -
(5)
d204. U2
aS
ds
(7)
Thus from 13.18 (8) the cavity quality Qte is 021.a3s (v u)2 dS 4p271-2 (2 — (n)148.1 (V U) 2 dS
2d/08 fRpr au/as)2
d213L
U21 ds
(8)
WAVE GUIDES AND CAVITY RESONATORS
530
§13.20
where s is the curve bounding a section S normal to z and 1.4, kei, and o„ 61 are the conductivities and skin depths of the end and side walls. From 13.00 (2) and 13.16 (2), the volume integral of ./3?„, is the same as (5) multiplied by )34 and the surface integrals and Q are
dS = 213,„„p J(k x V U)2dS = 2131„,f(v U) 2 dS
2f
dS = 01„,,[n x (k x V U)]2 ds
(9)
cos' (7) dz
au s y ds = #1„pd(2 — 5:)-1f(73---
(10)
Adf (V U)2 dS " = 4(2 — (5:)1.0e f (V U) 2 dS 2i.8a8 df (au /as)2 ds
(11)
In the articles that follow the detailed properties of the rectangular, right circular, and coaxial cylindrical cavities are worked out. Results for other simple forms appear in the problems at the end of the chapter. 13.20. Properties of a Rectangular Cavity.—From 13.16 (3) and 13.02 (2), the resonant frequencies of a rectangular cavity are V2
2 9
=
X,n2n,
v2 m2 n2 71
4 a2
b2
d2
v2 Mrir
472
p = 4,7r2
(1)
If M, N, and P stand for ma/a, nr/ b, and pr/d and Tmn p for co.„„t shnni„ then from 13.02 (1) and 13.19 (1) to (4), the fields are Ete
= comn,C[iN cos Mx sin Ny — jM sin Mx cos Ny] sin Pz sin T„,„,
(2) Bte = C{P[iM sin Mx cos Ny jN cos Mx sin Ny] cos Pz — k(M 2 + N2) cos Mx cos Ny sin Pz) cos Tmn p (3) Et. = co,n„„CIP[iM cos Mx sin Ny jN sin Mx cos Ny] sin Pz — k(M 2 + N2) sin Mx sin Ny cos Pz) sin T„,„,, (4) Bt. = (3.?„„ pC[iN sin Mx cos Ny — jM cos Mx sin Ny] cos Pz cos Tmn p (5) When m, n, or p is zero, E lies along the corresponding axis and p is then independent of its length. In this most common case, the fields become, taking p = 0 and writing Co for r2C'(m2a-2 n 2b-2), max nay sin sin (0.4 + (6) a b B = 7rC 0 (iN sin Mx cos Ny — jM cos Mx sin Ny) cos (wt + 0). (7)
Ez= — w,, Co sin
If the electric field is along any other axis, the fields are given by a cyclic permutation of x, y, z, a, b, c and m, n, p.
§13.21
PROPERTIES OF A RIGHT CIRCULAR CYLINDRICAL CAVITY
531
If none of the integers m, n, p is zero and (1) is used for co„,,,p and (3) for B, then 13.18 (4) gives for Qte at resonance Qte —
+24,75,(m2b2 p2a3 b3[n2a(a d)
n2a2)(m2b2d2 m2b(b d)]
n2d2d2
p2d2b2)I
b )(m2b2
d3(a
n2d2)2
(8)
If (1) is used for con,„, (5) for B,,n, and neither m nor n is zero, then viry or (m2 b2 m'
n2122) (m2b2d2
2[ab(2 — 2°,)(m2b2
n2a2)
n2d2d2
p2a2b2)+
2d(n2a3
(9)
m2b3)]
When a > b > d and m = n, = 1, the lowest resonance Q, at p = 0, is Qt.
vicybrd(a2 b2)I b2) 2d(a3
2[ab(a2
(10)
b 3)]
For a = b and the cube a = b = c, this Q becomes
_ virybir 243
virybird " 21(a 2d)'
where v is the free wave velocity in the dielectric of permeabilityµ in the cavity, y is the wall conductivity, and 5 is the skin depth. 13.21. Properties of a Right Circular Cylindrical Cavity. From 13.11 the resonant frequencies of this cavity are, in terms of the velocity v and wave length X„,„„ of a free wave in the medium that fills it, —
v2
mnp
V2
= X„,n, 2
V 2 (AL _1_ 472 "
22
P Ir d2
2)2 02
472PmnP =
(02 mnp
(1)
472
where,Qmn p is chosen to satisfy 13.09 (2). From 13.09 (1) and 13.19 (4) the fields are, if s(q5) stands for sin (m4 C„), c(ck) for cos (m(I) 11'„,) ,
Ete = cen,,,,,C[64(7-1)Jins(4))
a)mnp
(2)
Bte= CIP[60.1:,,,c(0) — 47)1. 8 (0)] cos (Pz) kin,„J,nc(cb) sin (Pz)) cos (wt Egm = co„,C{P[9107nnfmc(4))
a)mnp (3)
p)J.s(4))] sin (Pz) 40(1-7±
— kg„.1,nc(0) cos (Pz)} sin (wt + a), (4) Bt„, = 131,C191(71J.s(4))
4.13.a:ne(0)] cos (Pz) cos (wt
a)mnp (5)
532
§13.22
WAVE GUIDES AND CAVITY RESONATORS
For a TE wave, p 0. Obtain Q from 13.19 (8) and evaluate p-integrals by 5.296 (12) with U from 13.09 (1) and (2) and p'e S, = AO,. Thus
(oLa2 _ m2) it,„a2d 3 ) — 2a) -I- 2,81„a3] pad n _ viilySadeee,np _ d] d] 4µ'S[(2 — 5,),)a 2[a(2 vp.1,5ad31
Qte
(6)
21 p 2,2[m2 (d
(7)
—
13.22. Multiply Connected Cylindrical Cavities.—In a cavity which is multiply connected, it is possible to draw a closed curve which cannot be contracted to a point without passing outside the cavity. If the cavity is bounded externally by two parallel planes and by a cylindrical surface normal to them and internally by one or more closed cylindrical surfaces also normal to them, modes of oscillation are possible whose frequency depends only on the plane spacing. These are called the principal modes. They satisfy the second set of boundary conditions in 13.00 (5) in which 0m.n is zero and represent standing waves of the transmission-line type treated in 11.14 to 11.18. From 13.16 (3), the resonance frequencies of the principal modes of such a multiply connected cavity are =
P = P
AP
ivd3 P = co 7r 7r
1 = 02)
(1)
The fields can be written down from 11.14 (6) and (7). Thus
E = CpVU(x,y) sin Pz sin (opt,
B = C p(1.101V V (x,y) cos Pz cos (opt
(2) where U(x,y) and V(x,y) are the conjugate functions of Chap. IV and U(x,y) has the value U1 on one set of cylindrical surfaces and U2 on the other. From 4.10 (1), the line integral of V U or V V around any path of constant U is the increment [V] in V. This is usually 27 as with the elliptic cylinders (write U for V) of Fig. 4.22a. The integrals of 13.18 (4) must be found to calculate Q. The potential drop between the two surfaces is U2 — U1, so the instantaneous electric energy in a cavity length dz is, from 4.11 (4) with U and V interchanged,
We dz = -iC(U 2— U1)2 dz sin2copt = 4E[V]( U2 — U1) dz sin2 copt (3) Integration over z gives the total electric energy which equals, but is 90° out of phase with, the magnetic energy. Thus f B • B dv = pod[V]( U2 - (IOC; cost copt = IdfsoB • B dS
where So covers the two end surfaces. Since VV,
au/as, and
(4)
IdW/dziI
§13.24
NORMAL MODES OF' A SPHERICAL CAVITY
533
are identical, the surface integral of BR over side surfaces U1and U2 is 2llEdC412(V V) 2 ds cos2wpt =
/dzil dV cos2(opt
(5)
Substitution of (4) and (5) into 13.18 (4) gives Q
dz1 4;.1' (5,1_4[V](U 2 — Ui)(f 1 dW +
dzi dW
dV
dVti (6)
To evaluate the line integrals, z1is expressed as a function of W as in 4.22 (3) so that dzi/dW is a function of U and V. For the first integral U is set equal to U1, for the second to U2, and both are integrated over the range of V necessary to cover U. 13.23. Coaxial Cable Resonators. In the simplest kind of multiply connected cavity the curved walls are concentric cylinders. The principal modes are found as in the last article. From 4.12, we have —
W = ln
= ew,
U = ln p,
V=
(1)
pir.z cos wpt
(2)
The fields given by 13.22 (1) and (2) are, therefore, PrZ Cp EP = — sin copt, p sin
Be =
Cp
pv
cos
The range [V] of V is 21r so, if the outer radius is b and the inner a, (U 2 — U1) = ln , a
g.i
az1 --1 dV = e—ui (3W
27, 27r dV = 271-e—u, = — a
(3)
In similar fashion the other line integral gives 2r/b. Substitution of these results in 13.22 (10) gives for the pth principal mode Ad QP
2p/SL
(a +
b)d 1-1
+ 4ab In (b/a)]
(4)
In addition to the principal modes, there are the types considered in 13.09 (3) to (6), which involve two kinds of Bessel functions. If d is large coipared with b — a, these modes have much higher frequencies than the principal mode p = 1. The methods of 13.19 apply to these. 13.24. Normal Modes of a Spherical Cavity. The solution of the propagation equation in 12.11 involves products of spherical harmonics and spherical Bessel functions. Of the 0 and R terms given in 12.11 (5), only the P,(cos 0) and j„(/3r) terms survive because Q„(cos 0) is infinite on the axis and k„(ji3r) represents a progressive wave, whereas in the perfectly conducting oscillating cavity only standing waves can exist. The electric field and the magnetic induction are, from 12.11 (5), (2), and (3) and 12.13 (6) and (13), when we write s(0) for sin (mg5 + 'I'm), —
WAVE GUIDES AND CAVITY RESONATORS
534
§13.25
j n for j„(13„„r), and u for cos 0,
c(ck) for cos (mck +
(1) Ete= coC[Om csc OP,7(u)s(95) — 4) sin OP7,:"(u)c(0)]j„ sin (wt + a) a(rj ) Be. = [riu(n + 1)PT(u)c(0)in — 0 sin OPV(u)c(0) a; (rj n) csc 8.1:'7(u)s(ck) cos(wt + a) (2) — c
—
Et„, =
arJrn)
[rin(n + 1)P',,' "(u)c(0)j„ — 0 sin OPV (u)c(cP)
— (1)m csc 01:',7(u)s(cP)a(arJr1sin (cot + a) (3) Bt. = co2 A€Clom csc OP',7(u)s(0) — 4, sin OF",r(u)c(0)bn cos (wt + a) (4) At the cavity surface r = a, the tangential component of E vanishes, so that /3„„ must be chosen to satisfy the boundary conditions a[ain(Opna)]tm — 0
Lin(Opna)]te = 0,
(5)
as
The first form of 13.18 (8), with E for — jwil and (2u'a2) for coy from 10.02, is best for (2,e. By (1) and (2) this gives
1 1 fo27f (u, ck) du dck fad jn(13,r)]2 dr 11,3f I i fzn (u, 2/12 ef
Qte —
(6)
) du dyb ta[ajn(Opna)]/aa) 2
After the integration the u-integrals in the numerator and denominator are identical by 5.231 (8) and cancel out so that by means of 5.31 (10) and integration by 5.296 (6), which is valid for nonintegral orders by HTF II (48), page 70, there results a
2co2AE(47/(3,)f r[J„±i(i3,r)]2 dr Qte —
14' 5[47r/
(Opna)][J4i (Op.a)] 2
Use of the last form of 13.18 (8) for Qt.
Q tn,
u„2122,a —
6
0;,n
(7)
gives in analogous fashion
pail+(2n + 1)21 4/92a 2 IZOL
(8)
13.25. Normal Modes of an Imperfect Cavity.—So far all cavities have been treated as completely enclosed by perfectly conducting walls. Wall loss effects were found in 13.18 (11). In practice there must be openings in the walls through which waves enter or electrodes are inserted. These are usually made as few and small as possible so as to modify the fields only slightly. Let the vector potential and resonance frequency of the
§13.26
EFFECTS OF SMALL FOREIGN BODY
535
actual cavity be A and w and of the perfect cavity Aoand coo. Insertion of A for 'F, Aofor 43, and co2ihe for the double curl in 3.06 (6) gives Ao(c02 — cog)fvA • Ao dv = fs[A x (v x Ao) — Ao x (V x A)] • n dS (1) Let v be the volume of the perfect cavity and S its surface to which Ao is normal so that the second term in the surface integral vanishes. If the hole is small, co coo, w + coo = 2Wo, and in most of v A Ao so that
f E x Bo • n dS—jf Et xBo • ndS 2jcuo2 ilef A • A dv
=
412W„,
(2)
where W., Et, and Bo are instantaneous values of the magnetic energy in the cavity and the tangential components of the actual E and of the Bo with no hole. The part of Et due to Bo is found by solving the magnetostatic problem of an infinite plane face of zero permeability pierced by the given hole which bounds a field of induction Bo that is tangential to it and uniform except for distortion near the hole. This expresses the normal B. over the hole surface in terms of Bo. The effect of the hole is wiped out by covering it with a double current sheet like that in 12.18 whose radiation exactly cancels its absorption. The current sheet B. and Et are exactly equal and opposite to those of the hole, but the Bt adds to restore the standing wave Bo. For an aperture in the xy-plane and an x-directed Bo, (Et)0 can be obtained from B. by integration of 12.18 (5) Problem 49 of Chap. XII gives B. for a circular aperture in a thin wall. If, with no hole, the normal electric field is Eo, then its contribution to Et is given by the electrostatic solution of the problem of a plane conducting face that bounds a field of intensity Eowhich is uniform far from the hole. Article 5.272 treats the case of a circular hole in a thin wall. The hole losses must be added to the wall losses to get Q.
13.26. Effects of Small Foreign Body on Cavity Mode Frequency.—The electric field distribution in a complicated cavity is sometimes found by noting the frequency change produced when a small dielectric sphere is placed at various points therein. The perturbing effect on Eo and Ho of a small body may be found by starting with the vector identities
✓ aco Ro]
=
• (to
=
at) ag (v xilo) a°, aw a fi \ aw ow • (v x 0) — 0
A 0•
(1) (2)
The integrals of the right sides of (1) and (2) over the cavity volume are zero because those on the left side transform by Gauss's theorem, 3.00 (2), into surface integrals which vanish on S since n-• E x H is zero if n x E is. Now transform the sum of the right sides of (1) and (2) by elimination
536
§13.26
WAVE GUIDES AND CAVITY RESONATORS
of all vector products by Maxwell's equations, 11.00 (1) and (2), which apply to E and H. For the conjugates E and A, j must be replaced by —j in 11.00. The result after multiplication by j is
20 + w(f Ao ±
+
0 = f[AR
a
Replace the differential Sw
)] dv (3)
by the variation 3 and solve for Sco/co.
Thus
oEt • 20) dv J (AR • Ho + 62 • E 0) dv
SOW/ •
=
(4)
Clearly an increase inµ or e or both will decrease the resonance frequency. From 11.06 (2), the field products in (4) may be interpreted in terms of energy density, so that for small perturbations 1 r
Sco = w
(,,
dWe SAdW„,)SW
W j E dv
dv =
dv
(5)
W
If the foreign object is so small that current and charge distributions on the walls are essentially unaltered, then usually the phase differences over the volume it occupies may be neglected. Thus a static solution is valid. Even if the object greatly distorts currents and charges, but only on one wall so near that phase differences between it and adjacent wall areas are small, a static solution may work. This is especially true of the plane walls in rectangular and cylindrical cavities where image methods apply. One then replaces the wall by an image cavity and image object and works with that mode of the composite cavity which gives the same fields in the region of the actual cavity. Some situations like this appear in problems at the end of the chapter. As an example consider a small sphere of radius ro, permeability and capacitivity el inserted in an empty cavity. From 5.18 (1) and (2) with C = D = 0, and from the analogous magnetic case, the fields inside are = E0
3Ev
2E„
1 = Ho
311.
(6)
211,,
Assume a nearly uniform E0 and Ho in the volume 4irrg/3 of the sphere. With Ei— E„ for of and A i — A, for SA, (4) gives Sco zirrgf[ey(Ei — fv)/(ei — w
2E011E012
— µv)/(µl
2k4)]11/012 1
of 1E01 2 dv + A4511/012 dv (7)
For a small, perfectly conducting sphere set E1 = 00, µl = 0, so that Sco
w
41-rg(EidEol 2 — -10411/012)
E0 JEo 2 dv
AvilH012 dv
(8)
§13.28
537
STATIONARY FORMULAS FOR CAVITY RESONANCE
The values of Bo and Ho to be used in the numerators of (7) and (8) are the unperturbed values at the centers of the sphere. 13.27. Wall Deformation Effect on Mode Frequencies. Anywhere inside a conducting cavity where the magnetic field touches the wall it exerts a pressure 2AH2by 8.14 (2). Where an electric field touches the wall, there is a pull or tension of 2EE2 from 1.14 (2). Thus an inward deformation of the wall puts in energy by compression of the magnetic lines and absorbs energy from the contraction of the electric lines. If one assumes that the deformation is smooth and shallow, so the original fields are little altered in the process, then —
SW
— if Gilio • Ao
—
do • E0) dv
(1)
Insertion of this into 13.26 (5) gives
—=—
(.0101 2— ElE012) dv
(2)
fiE012) dv
fv (AlHol2
Thus an inward perturbation will raise the resonance frequency if made at a point of large H and will lower it if made at a point of large E. 13.28. Stationary Formulas for Cavity Resonance Frequency. It is useful to have a calculation method for the resonance frequency of a cavity whose boundaries do not fit a coordinate system in which the wave equation is separable. Two such formulas come from 13.18 (4) which state that the time average electric and magnetic energies are equal, or —
f(1,11112 -ElE12) dv = 0 (1) The first comes from putting (1) in terms of H by 11.00 (1) and the second by putting it in terms of E by 11.00 (2). The results are f [,,212,1H-12 — 1Y x H12] dv = 0 = f [w2AEIEI 2— to x E12] dv
f ly x H21 2 dv 4.1E = 0 = f IH,I2 dv P
fly x E,12 dv f 1E2,12 dv
(2) (3)
These may be proved stationary exactly as 13.07 (5) was by multiplying up the denominator and taking the variation of 0, and Hp in one case and of 0, and E, in the other. Clearly the integral fIH,12 dv or f IE,12 dv which multiplies Op does not vanish. After some vector manipulation, including the use of Gauss's theorem, 3.00 (2), the two other volume integrands combine to give the left side of 11.01 (1) or (2), and thus vanish. The surface integral over the boundary is zero if n x E, = 0 for H,
n x 45E, = 0 for E,
(4)
538
WAVE GUIDES AND CAVITY RESONATORS
§13.29
Thus the H integrals for 813, are stationary regardless of the choice of is zero on 8. SE but the E integral only if the trial field E As a specific example consider the spheroidal cavity
a
Cal
e
2
+
(5)
=1
From symmetry only a yb independent Hooccurs which has zero divergence and is tangent to the spheroid surface. To vanish on the z-axis the factor p is indicated and to be even in z a suitable form is
H4, = p(1 — B p2— Cz2),
V x H4, =
p 12C pz k2(1 — 2Bp2— Cz2)
(6)
The constants B and C can be adjusted to make E normal to the spheroid surface. From (6) and Maxwell's equations, Ep Czp Ez — 1 — 2Bp2— Cz2
az b2p Op a2z
(7)
Eliminate p by (5) and require that (7) hold for all z-values, so that 2a2B = 1, (a2 b 2)C =
1
(8)
Put these values into (6), carry out the integrations indicated in (3), where dv = 2273 dp dz and obtain 4b2) (3a2 2b2) 02a2 = 18(5a2 (9) 22b4 33a4 52a2b2 This result may be checked, for a = b, against the exact value for a spherical TM mode given in 13.24 (5). Thus Oa = 2.751
(9),
as = 2.744
13.24 (5)
For the general ellipsoid see Borgnis and Papas, page 180. 13.29. Complex Cavities.—Some of the most important cavities are the simply connected ones whose boundaries coincide only in part with single 26
coordinate surfaces. Usually a special type of solution is required for each case. Consider, for example, a right circular cylindrical cavity of radius b and length c, to one end of the interior of which a shorter closed right circular cylinder of radius a has been fastened coaxially leaving a gap d at the other end. This cavity, shown in section in Fig. 13.29, falls
539
COMPLEX CAVITIES
§13.29
outside the categories so far treated. We shall indicate the nature of the exact solution by Hahn's method (J. Appl. Phys., Vol. 12, page 62) then give a nearly exact method that applies in the important case where d is much less than a, b, or X, and finally present a rough approximation that uses the transmission line analogy. We desire the cavity fields and the lowest normal frequency. From 13.19 and 13.21, TM field components in the M-region which give E, zero at p = b, z = 0, and z = c are CO
EZ
=
R0(flop) ±
Ro(00a)
Ro(OmP) m =1
r-172 c
cos
mRo(Oma)
NC,7,14(13„,p)
m2,2
0/00 =
c2
n 2,2
02,z = 00 — d2
(2)
cos mrz ]
c r3mRo(Oma) m=i Ro(OoP) = J o(13ob)Yo(OoP) — o(Oob)J o(OoP) Ro(Omp) = Io(0.b)Ko(O.P) — Ko(0.b)10(3.P) — iw3oRo(Ooa)
(1)
4„2
Og = )32 = (4211E — x2
(3) (4)
In the N-region, fields finite at p = 0 giving E = 0 at z = 0 and z = d are = Aoh(130p) J 0(00a)
Alo(0.P) cos ram /0(3na)
n=1 ea
(5)
d
'1
— AEA 0P0(00p) NAniaonp)cos 7 d Onl 0(0,,a) .ico 00,10(00a)
(6)
n=1
A formula for A, in terms of all the Cm's is found by equating b o to fl'o, setting p = a, multiplying through by cos (p7rz/d) dz, and integrating from z = —d to z =d. Similarly C, in terms of all the An's is obtained by equating Ez to E; when 0 < Izi < d and to zero when d < Izi < c, multiplying by cos (qrz/c) dz, and integrating from z = —c to z = c. Elimination of the An's from these equations and division by Co give Cq/Co in terms of the Cm/Co ratios. There are an infinite number of these equations, one for each q, so that to obtain Cm/Co an infinite determinant must be solved. Insertion of these values of Co/Co into the q = 0 equation gives 1_J000coreoffo(00a) JO(Ooa)L Ro(00a)
00cCm RO(Oma) sinmrdl c j 7rd •Int13,,, Ro(0.a)
(7)
m=1
The lowest value of flo that satisfies this equation gives /3 by (4). The amplitude of excitation has been chosen to make Ao = 1.
540
WAVE GUIDES AND CAVITY RESONATORS
§13.29
The calculation just outlined is evidently a major undertaking which is exact in theory but not in practice because of the finite amount of time available. However Hahn has prepared tables that make it feasible. If d is much less than a, b, and X, then the following method gives X with greater accuracy than that with which a, b, and c are usually measurable and enables us to calculate Q. The tangential electric field between M and N at p = a is independent of a and b if d is small enough and depends only on d, c, and the potential across the gap. We may then take a and b infinite, which makes it the two-dimensional problem solved in 6.07 where h and k replace d and c. The choice of c >> d simplifies the work by deleting the bends at x1= ± 1 in Fig. 6.07a. The procedure of 6.07 gives 1 z = 2d ?[(z — 1.)1 + sin-1 zi
Ex =
ay ax
aw aZ i p. .
Vo d(zi — 1)i i.p.
(8) (9)
where z and z1are complex variables and the potential difference across the gap is 2 Vo. To get Exon the x-axis in terms of x, we find complex values of z1which give y = 0 in (8). Substitution of these values in (8) and (9) gives x and E. To within 1 part in 5000, these fit the formula
Ex = Vod-1{A
B cos (rxd-1)
C(xd-9 2
D[sec (2n-xd-1)]• 355 )
(10)
where A = .34091, B = — .00656, C = — .07785, and D = .49910. Integration of (5) from z = 0 to z = d with p = a shows that Vo = A o. Insert this value of Vointo (10), substitute z for x, equate the result to (1) when — d < z < d, and equate (1) to zero when d < 1z1 < c, multiply through by cos (prz/c) dz and integrate from z = — c to z = c. Solve the resultant equation for C, and insert in (7) which is then solved by trial for /3. Moreno (see references) gives curves for such a cavity on pages 230 to 238. It is evident that insertion of An and Cm in (2) and (6) gives the magnetic inductions so that Q can be calculated from 13.18 (4). The transmission line concept furnishes an approximate value of the resonance frequency. If d << c, the region M of the cavity resembles a section of coaxial line short-circuited at z = c whose input impedance is In (b / a)0(irec,))-' tan 3c from 11.15 (13) and 13.13 (2). The region N represents a capacitor of impedance (jcoC0)-1shunted across the line. At resonance, these impedances are equal and opposite so (3 satisfies, approximately, tan /3c =
2re Co In (b/a)
(11)
We can only estimate Co. Treatment of the N-region as a parallel plate capacitor where Co = Erce/d gives a lower limit because it neglects the
§13.30 EXCITATION OF CAVITIES. INDUCTIVE COUPLING
541
fringing field. In the special case where d << a, d << c, and a = b, Co can be found quite accurately as will be seen in problem 63. 13.30. Excitation of Cavities. Inductive Coupling. The most general cavity oscillation state is a superposition of modes already considered. Not only does each mode oscillate as an independent circuit, but also the exciting to cavity current ratio in any mode depends on the geometrical configuration only, as with linear circuits. E. U. Condon (Rev. Mod. Phys., Vol. 14, No. 4) simplified this mutual inductance calculation by breaking up the vector potential A, of the ith mode into two factors C, and A? where Ce is adjusted to make the integral of A? • A° over v, the cavity volume, equal to v. These "normalized," dimensionless, and by 13.17, orthogonal vector potentials depend only on the geometry and are related to the steady-state cavity vector potential by the equations —
A = ZO, ° f 4, • Ai dv = Ci ,4?.A2 dv = vC1:
(1)
Assume that the Q-values of the ith mode and its neighbors are so high that the resonance curves discussed in 13.17 and 13.18 do not overlap and that the time-dependence of A, is exp [(j - 2Q-1)wet]. Thus 11.01 (6) becomes (2) V2A, (4€(1 jQ-1)A = 0 Let A be the total cavity vector potential excited by the solenoidal part of the loop current density i(r,t) of frequency w. Then 11.01 (5) is OzA
+ co2peA =
= Z(V2 A,
co 2gEA,)
(3)
Elimination of V2Ae by (2) gives E[o.)2 - 4(1 -I- jQ-9]Ae =
(4)
Multiply through by A, integrate over the cavity volume, and solve for
C. For a thin wire loop i dv is i dS ds or I ds, so that -
(c02 - w2. - 3.4Q-1)Ev
where M,? = f A.• ds
(5)
Here, by analogy with 8.03 (2), .1W is the normalized mutual inductance between the ith mode and the wire. The electric field along the wire must be compensated by the electromotance 6' applied to the loop, so that f • ds = jcof A • ds
(6)
Put (5) into (1), multiply by A'„ integrate around the loop, add the IR
542
WAVE GUIDES AND CAVITY RESONATORS
§13.31
drop due to loop resistance and divide by I to get the input impedance.
Z =
/
.iu)(MD 2 ev(w2 - - jc4Q-1)
= R
(7)
Since Qi is usually a large number, the first terms in the denominator dominate unless w -,== coi. Modes for which co, > co give inductive reactance, modes for which coi < w give a capacitative reactance. This formula is useful in showing the qualitative behavior of driving loop reactance, but M2 must be modified to give quantitative results. The origin of the difficulty lies in the assumption of an infinitely thin wire near which B and hence the inductance L are infinite. In practice, if the wire is thin enough compared with wave length, the cavity fields will be the same as if the exciting current were all on its center line, and its surface will coincide with the boundary of a tube of induction. The true flux linkage can therefore by found by integrating A from one end of the loop to the other along any curve lying in its surface. The definition of M2 in (5) should be modified accordingly. One should note that the path of integration in (5) can be closed by going back to the starting point along the cavity boundary which contributes nothing to M2 because A is normal to it. The integral of B2 over the loop area would do as well. 13.31. Inductive Coupling to a Circular Cylindrical Cavity.—The theory just developed will be clarified by considering the case of a circular cylindrical cavity excited in its lowest natural frequency, the TE11mode shown in Fig. 13.09, so that (311a is 1.841. From 13.21 (1), this is vll
=v
, 1 1 0 343 1) = con cT) = 1.5 X 108( 'a, ±
(1)
From 13.21 (2) and 5.296 (8) and (12), A2, which contains all geometric factors and is normalized so that f A2 • A2 dv = v, is
_ 2a[ed1(O1ip) sin 4 + 4/311ai(010) cos 95] sin (rz/d) P(131a2 - 1)4.11(011a)
(2)
A small loop inserted through the wall at some point excites this mode. Evidently from (2), one encircles the maximum TE11flux for the closest coupling on the flat ends by linking Bp near the center or Bo near the edge or on the curved surface by linking Bonear the ends or B, near the center. Weak corner fields bar the Bo choice. The ratio of B, where p a,4 = 0, z = id to Bp at p 0,4 =0, z 0is _ 114.°), -
al (Oa) _ 0.682d ax
rc/1(0)
a
(3)
§13.32
EXCITATION OF CAVITY BY INTERNAL ELECTRODE
543
Thus for 0.682d > a, the tightest coupling for a given size loop will be near the curved wall halfway between the ends. Let us assume that the small loop of area S on the z = id plane at 4 = 0 is so near the wall that the value of Bz at the wall may be used over its whole area. To justify this, we observe that, if p is 0.95a, then J1(Op) is 0.997 J1(13a). Thus at cp = 0, z = 2d p a, the mutual inductance is, from 13.21 (3), ,
= Iv 11?IS =
21haS _ 4.39S a (.61.10 — 1)1
(4)
From 13.21 (6) and 13.30 (7), in a copper cavity where y = 5.7 X 107, the Q-factor and the additional loop resistance at resonance are 2.92a2)i 9.92(ad)1(d' d3 X 104, Q = 4.110 0.86a2d
R' =
125582Q 2.92a2)1 a3(d2
The maximum potential across the cavity, at z = mod, 2f0a.E,, dp or 2cofoaA„ dp .
(5)
= 4ir, is
The potential magnification is the ratio of
this to the loop electromotance and from 13.30 (1), (6), and (2) is
V
2coC li foa dp wC11M°
4afoap-IJI(311p) dp _ 0.812a2
3P(Oi1a2 — 1)1/10110
S
(6)
The integral, found by integration of the series of 5.293 (3), is 0.80. The M° comes from (4) and Ji((3lia) is 0.582 from Jahnke and Emde. It is of interest to take some numerical values at resonance. Let the area S of the loop be 1 cm', the radius a of the cavity 10 cm, the length d 20 cm, and the medium inside a vacuum, then
w
7.25 X 10' rad/sec, M° = 0.00039 m,
X 26.0 cm v = 1.15 X 10' cyc/sec, 1715 ohms R' Q = 36,000,
Note that M°, the normalized mutual inductance, is in meters, not henrys, from (4). If the peak loop electromotance in phase with E is 1000 volts, then eo = 1000 volts, I = 0.583 amp, V.. = 81,200 volts, and P = 292 watts. Some other excitation methods appear in the problems. 13.32. Excitation of Cavity by Internal Electrode. When a cavity is excited by a nonuniform current as when an electrode terminates inside it, the divergence of the current density i is no longer zero, and thus there is a charge density a. In such cases, by 11.01 (4), the fields may be derived in the Coulomb gauge from a vector potential and a scalar potential that satisfies Poisson's equation. The mutual inductance with each normal mode is still given by 13.31 (4) because, as proved in 13.17 (4), the lamellar part contributes nothing to the volume integral. The field of the scalar potential contributes an additional electric energy in phase —
544
WAVE GUIDES AND CAVITY RESONATORS
§13.32
with the electrode charge that represents a purely capacitive reactance. As in the antenna problem, the difficulty of finding the current and charge distribution appears. A very thin wire stub has negligible capacitance, and its inductance is fixed almost entirely by its radius, so that the 12.02 reasoning applies giving the 12.02 (7) sinusoidal distribution. A fairly accurate result is obtainable when the electrode is charged by a thin wire in which the current may be taken uniform, so that the charge is entirely on the electrode and may be calculated as in electrostatics. Equations (1) to (6) of 13.30 are unchanged, but q/C o adds to (7) where I = jwq and Co is the electrode capacitance. Thus
2= R—
jcoM? jo.q / Qi) (c4 — co2
coC 0
(1)
When w = coo, 2 is no longer real because of Co, but if co co,,, only the ith term of the sum is important. Therefore equating the reactance to zero gives v Q 2 (4
w 2)2
c4,2C0M2(22(4
,42)ov = 0
(2)
Often the coefficient of 4 — w2 is much larger than that of (4 — co2)2, so that from the last two terms alone, writing 2 ct4 for co + 0.4, we get oil 2C0/1/1Q2
wi
1 2o.u/rCoQi
(3)
As an example take a sphere of radius r supported with its center a distance 1 from the wall by a thin wire, and use this to excite the TE11 mode of the last article. For tight coupling the wire should coincide with a strong vector potential component. From 13.21 (2), this position is at z = 2d c = 4.7r. As with 13.31 (4), Ji(ka) may be used for J i(kP) for some distance from the wall. Thus 13.30 (5) and 13.31 (2) yield ,
mo = a —1
Ao • ds =
2af p-1 dp _
( 31a2 — 1)1
2a In a ($11a2— 1)1 a — 1
(4)
From (7), the resistance contribution to the coupling circuit is /i' = 109aQ(d2
2.92a2)-1[1n [a(a — 0-1]}2
(5)
If the ball is sufficiently small and near the wall, the capacitance Co in (1) approximates that of sphere and plane which from 5.082 (1) is Co = 47re[r
r2(21)-1
7.2(4/2 — r2)-1 ± • •
(6)
PROBLEMS
§13.33
545
The condition that (3) be valid is strongly satisfied for any reasonable choice of values, so that w is very close to wi. From (1) and 13.31 (6), the ratio of the maximum potential across the cavity to that across the wire due to R' and Co is V
1.664 X 1011CoQ a In a d(1 w2 R'2Cg)
(7)
In the calculation of e the reactance of the higher modes was neglected. When this is not negligible, it can be found from (1). Consider the mode of the last article, so that w, v, X, and Q have the values given there, and take 1 and r to be 1 cm and 1 mm, respectively. Co
R'
1.17 X 10—" farad, —
16,560 ohms,
wi
M° 0.136 m (.0 — 0.99 X 10-6
A peak stub potential of 1000 volts gives, neglecting higher mode terms, 27,100 volts,
I
0.0604 amp,
P
30.2 watts
13.33. Cavity Excitation through Orifice.—When a wave guide ends in a cavity, a rigorous calculation involves matching the vector potential across the junction surface. On either side this is represented by an infinite series of terms, so that the problem is like that encountered in 13.29 where Hahn's method was used. Often a fair approximation is found by assuming the field over the opening as in 13.12. Examples of such solutions will be found among the problems that follow. Problems 1. From 11.15 and 13.00 (7), show that when the dielectric conductivity in a wave guide is -y, the wave number and the dielectric attenuation are, respectively,
=
2-11[02 2-I {[(02
_ 0:02 + w 244 272}i + 0 2 _ font 02,02
+ ,21,21,21i
132
+ sl 11
Note that these formulas hold both above and below cutoff. 2. If the conductivity to capacitivity ratio is small compared with the difference between the frequency v transmitted in the mth mode and the cutoff frequency v,,,, show that the attenuation a' and the phase velocity v,c, are, approximately, 1— a' = 2Pvl
\ 1-1 h2 ]
. v = vmE 1
81(AvYv2)2
where v is the free wave velocity and v,,, the phase velocity for zero conductivity. 3. A rectangular wave guide is excited by a loop lying in the plane x = d. Show by images that all vector potential components are parallel to this plane, and thus the
WAVE GUIDES AND CAVITY RESONATORS
546
waves generated by this uniform current loop may be derived as in 13.00, where W = 0„,„ sin 124-.rX cos -re' r e-iS'mn',
A = V X ilk
a-
nay
E = - we „,n jR;,,,,., cos nay + kin-1sin . b b b [
B= = O..
{
r
sin -1n- ' r xe-519'..z a
ma nay nI2r X- e-ifrm.' - -r j rir sin 7I'b r ± + lejf<„,, cos r Y-cos-
nay io'„,0 )2 sin -irx tr cos a
b
a
b
a b
2 n. Note that this combines TE and TM waves. where (3„,„) 2 = 02 — 0„, 4. Show from 13.03 (6) and (14) that the vector potential of a small loop of wire with area dS, current le", and x-directed magnetic moment which lies in the xo-plane at y = yo and z = z0 in a rectangular wave guide is, if z > zo, CO
ab
CO
z
Aµ dS
(2 - t3;1) sin MirX°cos
nryo[ .
m=1 n=0
nry
.1 cos —
a
jr,7nry /3„,nb
max
sin — sin
a
b
CIP'mn(=-.0)
5. Show from 13.03 (6) and (9) that the vector potential of a small loop of wire of area dS carrying a current Iei"t, having a z-directed magnetic moment, and lying in the zo-plane at xo, yo in a rectangular wave guide is, if z > zo,
A = jµI dS z z2 - -
ab
rn = 0 n=0
4„
cos
mrx0 nryo sin — b a
b
sin
?try max cos
a
my max nay - j— sin — cos — a a 6. A rectangular wave guide, closed at z = 0, is driven by half a rectangular loop of wire with its legs normal to the y axis and its side, of length c, at x0, z0. If only the TEN, -
wave is transmitted, show that the radiation resistance is
2 2 o R, = Oomb sin 2 — sine 0 0zo
where 0'10
X
-
4a 2
) 4
7. A rectangular wave guide, closed at z = 0, is driven by a semicircular loop of radius c of thin wire that carries a uniform current. The loop lies in the x = d plane and its center and ends are in the z = 0 plane. The frequency is such that only the TE,0 wave is transmitted. Show from 13.03 (6) that the radiation resistance is ird 2/0.072c2 — sin2 [J1(010c)]2 Hr a Moab
where i3:c, =
X
- Xi 4 2
8. Show, by application of the method of images to the result of problem 44 and with the aid of 5.298 (7) and 5.35 (4), that the radiation reactance in problem 7 is CO
2rAwc2 X0 + a
{ 71" Yo(Rinb)[JI(Oic) sin
n=1
irdi 2
a
prdil Ko(Rpnb)[/1(13,,c) sin —
+2 p =2
a
547
PROBLEMS
The problem 45 reactance with a = r is Xo. oi = 2r 1 (
—
X
X2 i , 4a2
The loop center is at y
lin,= 2±r n,2X2 _ 1 i 4a' X
Except for a large loop, the first term is much larger than the sum of the image terms. 9. The rectangular wave guide of 13.02, closed at z = 0 by a perfectly conducting plane, is excited by a wire of radius r parallel to the y-axis at x = d, z = c that carries a uniformly distributed current j/ cos wt. If r << a, r << r << c, and r << d, show by superimposing solutions of the type in problem 32 that the reactance is coihb.{2
—— 4 ;(1n Or + C) + 2 E Yo (2n0a) — Z Yo(2(3ina — di) n= — co
n=1 co — Z
± co
(2 — CY0[20(n2a2 + c2)il + Z Yo(20[(na
—
d)2 +C 21i}
n= — co
n=0
10. If in the preceding problem c is of the same order of magnitude as r but the wire does not actually touch the end wall, show that if terms in /34c4, 060, etc., are neglected, pairs of terms may be combined to give the reactance ..:. + co 171 (20ina di)} comb c Yi(20na) ± 7,13,2 X .= — cosh-1- — ir/3c2 Z 2ina di na 2r n=1
which, if d
n = — o,
takes the form X =7 — :{cosh--1 c-2/rpc2 Z ( —1)n+1171(n1
na
n=1
11. The re,..,tangular guide of 13.02, closed at z = 0 by a perfectly conducting plane, is excited by a wire of radius r carrying a current j/ cos wt parallel to the y-axis at (w2A, 7,2m2a-2)1 are x = d, z = c. If r is small, show that the fields when t3'„,
2waI a
B= if z > c.
1
z
j-, sin
m=1
mad max a sinh ji3mc sin a
Mr21 . , mirx jmr cos — e-713,0 sin — smh ji3„,c : sin a a a — t3„,a
2A/ z mad
a
m=1
When z < c, interchange z and c in E and for B write
ji = 2µI sinMird max —i1 cosh j0„,z sin sin a lq a a m=1
., max jmr sinh 30„„ cos —ie-i0'.° a (3„,a
k,
1)v 12. In the last problem take the exciting frequency v so that nv < 2av < (n where v is the free wave velocity in the tube dielectric. Show by integration of the Poynting vector across the tube section at large z that the radiation resistance is n
, mrd 2c,nthz , R, = (3„)-1S1112 (3,,c sine — a a m=1
548
WAVE GUIDES AND CAVITY RESONATORS
13. A rectangular wave guide, a = 10 cm, b = 2 cm, and closed at z = 0, transmits waves whose free wave length is 15 cm. It is driven by a 0.5-mm radius wire across the guide at x = 5 cm, z = c. Show that, to make its resistance equal 100 ohms, c should be 2.62 cm, 14.0 cm, etc. Show that the phase and signal velocities are 1.52v and 0.66v, respectively, and that the inductive reactance is 274 ohms. 14. The wave guide of problems 11 and 12 is short-circuited at x = d2, z = c2 by a perfectly conducting wire. The impedances Z1 and Z2 of each wire alone in terms of its position and size may be calculated from 11 and 12. Show that, when both where are present, the impedance 2; of the first wire is 2; = [2i22+ 0,2(10) 2,ji,22
made 2Ab z M° = — (3„,)-1sin —sin a a a m=1
pence '
where 4, is real for transmitted modes. 15. It is desired to eliminate the reactance in problem 13. Show from 14 that this can be done without changing the resistance by insertion of a short-circuiting wire on the center line at c2 = 11.39 cm from the end, so that M° is real, provided its radius is 4.5 mm and assuming that our formulas hold for such a radius. 16. A rectangular wave guide, a = 1, b = 2, and closed at z = 0, is excited by the application of equal and opposite potentials to the ends of a rectangular loop in the x = 2 plane. The legs of the loop are at y = z and y = 4, and their length is not given. The loop is driven at the frequency of half unit length free waves. Show that only the following waves can be transmitted: T/1/12, TM32, TM's, (TE10), TE12, TE16, (TE N)), TE32; give reasons for excluding other waves. 17. A rectangular wave guide is filled with a medium 14161 when 0 < x < c and 142€2 when c < x < a. Starting with 13.00 (2) and choosing (We..)1 = A1 cos pis sin (IV t..)2 = A2 cos [p2(a
my
e-i$' p.'
—e- „„= x)] sin 1 rY
show that 13p„ may be found by solving the transcendental equation p2ei cot plc = —p1e2 cot p2(a — c) as in 13.05. Indicate how the cutoff frequency may be found. 18. If the divisions between the mediums in the wave guides in 13.05 and the last problem are at y = c, show that the equations to be solved for 0',,„ are --Aiq2 tan qic = g2q1 tan q2(b — c), —61q2 cot qic = 62q, cot q2(b — c),
for TEX for T MX
19. Show that the variational expression for the normalized susceptance of 13.06 (7) is
B° = 8b z
[focE(y) cos (nry/b) dy] 2 F
X0 n=i[n2 —
(2b/Xo) 911 E(y) dY] 2
PROBLEMS
549
Use the trial E in the quasi-static expression of 4.22, namely, cos (17ry/b) 21 cos (147) [5m 2 arc /b) - sine (try/b)]1 (cos — cos 0)1
E - . where
q5
= 7ry/b and 0 = 7rc/b.
With these coordinates show that
1 focE(y) cos (117Y) dy = -b[ P. cos b 2 b
P._, cos
1 rC bL„b 2
Using the fact that the trial Ey is exact at cutoff where X = co and B° is given by 13.06 (7), write the variational upper limit in the form 17-C 2b B° = -{41n (csc —) + b X,
,rn 2
[(70 - 4b2V,2 )-1 - n-1](/,„1 } b
n=1
Note that the static value provides a lower limit. 20. The rectangular wave guide in 13.02 is obstructed by a thin conducting iris in which a slit of width d with its center at y = c runs from x = 0 to x = a. Calculate the normalized shunt susceptance as in 13.06. Verify that the transformation W = sin-1
[- cos (7rzYb) +cos (77-c/b) cos (ird/b)] [sin (7rc/b) sin (iird/b)]
gives U = -ii- when z' = 0, 0 < y' < co and when y' = 0, 0 < x' < (c - id); V = 0 when y' = 0, (c - id) x' (c id); U = 17 when y = 0, (c id) x' b and when x' = b, 0 i y' < .0 and that when y' co, instead of 13.06 (6), the stream function V becomes Try irc 17rd) V = - - In sin - sin b b b The potential difference is
U2 — U1 =
B° =
7r. Thus as in 13.06
b irc 17rd) In sin - sin — b X,
This is the quasi-static approximation to the "Waveguide Handbook," la, page 218. 21. One-half of the iris in Fig. 13.06c is removed leaving a gap of width d. Put the right-angle bends at +1 and -b in Fig. 13.06e and take the origin at the base of the iris, so that the transformation and magnetic induction in the iris plane are i rz ( irc W = C cos — sin 2 — - sine a a a 13
a {cos (17rx/a)[sin 2 arc/a) - sin° (17rx/a)111
jBy = C1
ax
Proceed as with 13.06 (13) to show that the quasi-static normalized susceptance is B° =
X, ird -17 rd) cote — 1 ± csc° a a a
550
WAVE GUIDES AND CAVITY RESONATORS
22. Show that the upper and lower limits for the B° of the last problem found by insertion of its B. and B. into the variational formulas 13.08 (10) and 13.08 (16) are
[
B° = _ Xg a
1—
m =2
20) ± 13„,_i (COS 20)]2 2 cost 1,G
(2a)2 Ii[ P„,(cos2-y) — P„,_i(cos 2-y) 2 2 sin 2 -y X
a z [ m2 X, m= 2
1 Bo
2a 2 lir Pm (ma)
where yG is bird/a and -y is irc/a. Then assume, as proved for the symmetrical inductive iris by 13.06 (14), that when X = co, the above formulas give the quasi-static values of the last problem. Thus write the above limits in the more rapidly convergent form
B° =
X, cot 2 0[1 + csc2 a
1 + - sec' 4
2a 2 ]i — 1}[P„.(cos 20) {[ 1 — ( ) mX
m= 2
Pyn-i (cos 2012} 1 — =
°
B
a
cot 2
[
1 + sect -yr + csc4
Xp
m —
2a)2
ti
m-1}[Pm(COS 2y)
=2 — P„,_1(cos 2-y)]2}
23. The inductive iris in a rectangular wave guide is a centered strip of width c. Nate from the symmetry that identical areas on the top and bottom of the guide must charge via the strip so no current circulates around the windows, and the stream function has the same value, V = 0, on strip and wall. Show that in the plane of the strip
B. + jB. = C cost —
—K
sine —c — cos2 —
)-i
where E and K are complete elliptic integrals of modulus k = cos (iirc/a). show that the quasi-static normalized shunt susceptance is
B° =
Hence
X, E — 1(1 — k 2)K a E — (1 — 1k 2)K
24. Show that the upper and lower limits for the B° of the last problem, found by insertion of its B. and B. into the variational formulas 13.08 (10) and (16) are X, B° = -- [1
a
f(B) -
n=2
4a 2 (2n + 1)2X2
{P,,[cos(rd/a)]
.A 0),
.., —a 1 4a2 r — = — [ X2 (2n + 1)2 — — 1(0) B° Xg n =2
P„_i[cos (,rd /a)] 2[1 — 2(E/K)]Pn[cos Ord/al2 cos (rd/a) + 3 — 4(E / K)
Assume, as proved for the symmetric inductive iris by 13.06 (14), that when X = the above formulas give the quasi-static solutions of the last problem. Thus write
PROBLEMS
551
the above limits in the more rapidly convergent forms ..s — Ag{E—'1(1— k2)K ]i 4a2 B° = 1 (2n ± 1)2x2 — 1)f(0)} a E — (1 — ik 2 )K + Z ([ n= 2
1
—a— {E (1 — 1k2)K g E 1(1 k2)E +
4a21-1 x x2
1) ([ (2n + 1) 2
n =2
2n 1+ 1)f (e)}
25. The wave-guide section is a right isosceles triangle with legs of length a. Show that for the simplest TE wave the cutoff frequency v, is iv/a and that 2
1 a =
iz-yvaS[(
1
—
2
(1 + 21) + (3 ± 212)P11 — 72 P2
where v is the velocity of a free wave in the medium that fills the guide. 26. The section of a wave guide is a right triangle with two sides of length a. Show that the simplest TM mode cutoff frequency v, is 1 51v/a and that a=
(2 + 2i)7[1 — 02]-1 Ava5
where v is the velocity of a free wave in the medium that fills the guide. 27. A circular wave guide of radius a that extends from z = — 00 to z 00 is excited by a current element of strength 43I o podct,e'"` at po, 0, zo. Show that the amplitude of the TE modes in 13.09 (5) and (6) is = ..7(2 — 6,° )A / c90.,p0 J:„(0”„po) eis,. 02 273„„(3L„a2o M 2)[J„,.(13„„a)] 2
=0
—
where J'„,(13„,„a) is zero, and that for the TM modes in 13.03 =
Atio
dOOMIrrt(OmnP0) 13,
(8) and (9) it is =0
,[(3/3„,.a4,(0,..a)pe
where J.,(13m.a) is zero. 28. A circular wave guide of radius a that extends from z = — 00 to z = 00 is excited by a small current loop of magnetic moment kr dSe"' that lies in the plane z = zo at po, 0. Show that only TE modes appear whose amplitudes in 13.09 (5) and (6) are tomn},
j(2 — g„)AI dSOL„J„,(0„,„p 0)
=
71-0 (f32,„,„a 2 — 2,„„
=0
M 2)[Jm(13mna)1 2e
29. A circular wave guide of radius a that extends from z = — co to z = 00 is excited by a small current loop of magnetic moment 4)1 dSei'" that lies in the plane = 0 at z = zo, P = PO. Show that, if z > zo, the amplitudes of the TE modes in 13.09 (5) and (6) and of the TM modes in 13.09 (8) and (9) are, respectively, omni„ =
r Omnion =
— mAi dsJm ( unpo) 7,-poo2mna2 — m2 umom.012e
"* 2
jµ 1dS(2 — g„).1:0 (0,n,,p0 ) .0, e'' 27/3""'13mniaJ,n (0m,,a)p
[0-1 . = 0
[i5m,d,m
=0
552
WAVE GUIDES AND CAVITY RESONATORS
30. The plane cb = 0 in a circular cylindrical wave guide is an infinitely thin conducting sheet. Show that the lowest TE cutoff frequency is given by = 1.1656
or
tan (27rv,av-0 = 4irv,av-1
and that, if the sheet has finite conductivity, the attenuation is infinite. 31. Show that the lowest TM cutoff frequency in problem 30 is iv/a and that the attenuation is infinite if the sheet has finite conductivity. 32. A uniform current I cos wt flows in that portion of the perfectly conducting cylinder p = a which lies between two infinite parallel perfectly conducting planes normal to its surface. Show that the phasor vector potential of the outward moving electromagnetic radiation between the planes is
A%
110(Rp) — iY 0(3p)]
2rai3[Ji (Oa) — Yi(Ra)1 where R2 = w2Ale, and from 5.293 (5) Y i (Oct) approaches 2/(r0a) as Oa approaches zero. Note, from the Jahnke and Emde tables, that if a < 0.0181X then [11(0a)] < 0.01[Yi(13a)] 33. Show that in the last problem the wire's input resistance and reactance are Ri = ,r 2ga2m,l'
X. —
Aii[Jo(fici)Ji(Ra) Yo(fia)Y1(001 27raMEi
where M is [Ji(fla)12 [171(0a)]2, and 1 is the distance between the planes. 34. The planes in problem 32 are connected at p = b by a perfectly conducting cylinder. Show that the phasor vector potential between the planes is
=
arY 0(3b).100p) — .10(0b)Y 0(3p)] 277-a0[J 0(0)171(0a) — 110(0).1-1(0a)1
and calculate the input reactance. 35. Electromagnetic waves move radially between two parallel perfectly conducting planes normal to the z-axis. If the electric field has only a z-component and is independent of 95 and if the ratio of E, to Ho at p = a is 2, show with the aid of 12.16 (2), (4), and (5) that at p = b the ratio is
H.s
=
26 =
Jo(Ob)Y1(0a)1 j2.[J1(13a)Y0(0b) of.10(0a)Yo(ob) J0(00170(13a)] ev2.[Ji(fia)Y 1(0) — Ji(0)Y1(fla)1 .iVo(Oa)17 1(0b) — J1(13b)Y0(1301
36. A wave guide of radius b transmits the T/1/01 wave of Fig. 13.09 when v > POI. For this wave to pass a gap in the walls, a field must exist across the gap to transfer the induced charge. Thus show from the previous problem that the transmission of the frequency v, > volis blocked by covering the small gap with a coaxial cylindrical box whose plane ends terminate at its edges, if the box radius satisfies the relation JI[2-n-(w)i orb] Yo[271-(1.te)iv,a1 = Jo[270.1.04v,a] Y1[2/r (me)1 v,b] 37. It is desired to insulate two sections of the outer walls of radius b of a coaxial guide for d-c potentials. Show from problem 35 that this can be done without interfering with the transmission of the principal mode at frequency v, by leaving a small gap between sections, provided each terminates in a plane flange whose outer edge of
553
PROBLEMS radius a is coaxial with the line and satisfies the relation Jo[217-(.1, €)1vb] Yi[27r (µe)i vx] = Ji[27r(w)ipca)Y0[27r(w)iv,b]
Assume there is a magnetic or current node at the flange's edge. 38. If I is the coaxial line current in the last problem, t is the gap length, and v is (w)-i, show that the potential between the outer flange edges is
It
Vo — ,2 ,,,Eab[Ji(wa/v)Yi(cob/v) — JI(wb /v)111(codiv)i 39. A sectorial horn formed by the planes z = 0, z = b and q5 = 0, 4, = a' is excited by a uniform current kI cos wt in a thin wire at p = c, 4 = a. The wave is completely absorbed at p = oo. From 12.16 (2) and (4), find suitable forms for A: when p < c and A, when c < p. Set up the Fourier series by equating A: to A, at p = c, integrating Bo — B' over the surface p = c, and applying the magnetomotance law as in 7.31 (9). <
p < c,
=
[fm,roc) — jY.,,r(oc)P.,,(0p) sin a
m=1a
c < p < 03
2=1 = -7—
,
7
sin
a
— jYm,r (Op)117,,,„(0c) sin a
m 1 a
Mir a
M r4) a
,
mra mr4, sin —T a
a
40. Using the last problem, integrate Poynting's vector 11.17 (4) over a section of the horn when p is large, and thus show that the radiation resistance is
R
=copirb
.
a
r jm,(0c) 2sin 2 mra L a'
m= 1 - -.7,7
41. The horn angle in problem 39 is a' = 7r/n where n is an integer and the wire radius is a. If a << ac and a << [(7r/n) — a]c, show that the boundary conditions are satisfied by superimposed fields of the type of problem 32 according to the principle of images of Fig. 4.06. If the wave is completely absorbed at p = , show that the input impedance is, approximately, if a << X and R is the ohmic resistance of the wire, n-1
wAb{ 2j R — 1 — (ln 4
Sir sr)] [Jo(20c sin —) — jY0(20c sin —
C) 8 =1
n—1
[Jo(213c sin
(s
1)7r — na
jY 0 9Lic sin (s +1)7
na
8=0 The sum of the Yo terms will usually be negligible compared with the logarithm term. 42. Show that, if an electromagnetic wave having n electric field peaks in the z-direction is to be propagated radially in the horn bounded by the perfectly conducting planes z = 0, z = b, ¢ = 0, and 0 = a, it is necessary that v > 43. A uniform current 4ile1wt flows in a circular loop p = c that lies in the plane z = d between two infinite conducting planes z = 0 and z = a. With the aid of 12.16 show that there is no propagation of energy if X > 2a and that, in this case, the vector
554
WAVE GUIDES AND CAVITY RESONATORS
potential between the planes, when p > c, is, from 12.16 nerd nrz a sin— a I 1(0,,c)Ki(0,0) sin—
2cal z
Am
a
where /3„
n=1
X
nX 2a, — 1
44. If in the last problem a < X < 2a so that a single mode is propagated, show that, when p > c, the phasor vector potential is 77-2 ad (01W i(Ric) sin — sin — 2j a a
jca a
n=2
nird rnz} IC1(3.p)Ii(0,,c) sin — sin a a
where 1-1;,2) (01p) is the Hankel function of 5.293 (10), 01 = (27r/X)[1 — (iX/a)91, and = (2/r/X)[(inX/a)2 — 1]1. 45. A sectional horn is bounded by the planes z = 0, z = a and ct. = 0, ¢ = a and is excited by a thin curved wire at p = c, z = d carrying a uniform current 4)./ cos cot. If a < X < 2a and the radius r of the wire is much less than c, and if the radiation is completely absorbed at p = co, show from problem 44 that R. — X. =
2wac 2/.4{
a
[1 INTI(OIC)Y1631C) 2 --
+
E [ IC]
(
1
ircoaC2
Or J1(Ric) sin —
+— 2rc
3,,C)/1(0,,C)
n=2
2
ird sin2 — a sin, nird ± a 4,rc ln sin 7'j 2irnc a a a
+ In 2a err
Note that the series converges very rapidly. 46. If in the last problem ). = 1.5a, c = a, a = 18°, d = la, show that to make the reactance vanish the radius of the wire should be 0.0114a and that the radiation resistance in this case is 274.5 ohms. 47. The planes x = 0, y = 0, x + y = a, z = 0, and z = d bound a cavity. Show that the natural frequency of the simplest TE mode is v = v(a2 d 2)1/(2ad) and that pad(ai +di)
Q = µ'8[2a 3 + (1 + 21)a2d + (3 + 212)d3] where v is the velocity of a free wave in the dielectric filling the cavity. 48. The planes x = 0, = 0, x y = a, z = 0, and z = d bound a cavity. Show that the natural frequency of the simplest TM mode is v = 51v/(2a) and that
Q=
A'S[(2
pad + 21)cl + a]
where v is the velocity of a free wave in the dielectric filling the cavity. 49. A cavity is bounded by the planes x = 0 and x = d and by the prism y = 0, = b, y = z, z —y = 2b. If the electric field is entirely x-directed and if one assumes that the only periodic solution that gives A. = 0 at y = 0 and z = y is
A, = e[sin p(z — y) sin q(z
y) — sin q(z — y) sin p(z
y)]
where p and q are different arbitrary constants, show that the lowest resonance wave length of the cavity is 2b/51given by p = gr/b and q =
PROBLEMS
555
50. A check of the nodal lines shows that the lowest purely TM mode in the last problem consists of four right isosceles triangular cells. Hence show by problem 48, adding up losses on four side, two diagonal, and eight end faces, that Q = 21.4(b1.4')-ibd[(2 + 21)d + 2b]-' This exceeds 13.20 (10) and 13.21 (7) using the same lowest resonance co. 51. A cavity is formed by short-circuiting the biconical transmission line bounded by the cones B = a and 8 = 7r — a by the spherical surface r = d. Show that the wave lengths of the principal mode resonances are 4d/(2p + 1) and that Q—
A'S {C
In [(2p
2iLd sin a In cot la 1)7r] — Ci [(2p 1)7r] + 2 sin a In cot la1
where C is Euler's constant 0.5772. 52. A cavity resonator is formed by planes z = 0, z = d and two confocal elliptic cylinders orthogonal to them whose major and minor semi-axes are M 1, m1 and M2, m2. Show with the aid of 4.22 and 13.22 that for the pth principal mode of this cavity Q, =
21.ebA
d(M1K2
l
M2K1)
221-MiM2[1n (M2 + m2) — ln (Mi mi)]}
where M2 > M I and K1 and K2 are complete elliptic integrals of modulus — mi)1/1111 and (M; — m:)1/M2, respectively. 53. A cavity is bounded by the planes z = 0, z = d, and by two circular cylinders orthogonal to them of radii a and b whose axes are a distance c apart where a > b c. Show with the aid of 4.13 and 13.22 that for the pth principal mode
Q=
d(a — b)[(a b) 2 — c2] {1 + 2µ'S, 4a b[(a 2— b2)2 — 2(a2 b 2)c 2 c4I1cosh-' [ (a 2
b2 — c2)/(2ab)]}
2a is partly closed 54. A rectangular cavity of width a, height b, and length 2d at the center from y = c to y = b by a thin conducting iris and excited in the 7'E10 mode. Consider it as a transmission line 7' section short-circuited at the ends with the capacitance of 13.06 forming the leg of the T, and thus show that the lowest free space resonance wave length is given by 2aX1 (4a2
X1)1
d) 4b 1 = — In csc -II — tan tan — XI 2b X1
55. If a = 2, b = 1, c = I, and d = 2 in the last problem, show that X = 3.692. 56. Obtain an approximate solution for the fields in problem 54 as follows: Use fields TE to x from problem 3 but with a cos 13;„z cos wt dependence on z and tin 1471,„ Take z = 0 at the left end of the cavity and assume the quasi-static result of 4.22 (6) for E at z = d, so that Ey
sin (rx/a)
21V0cos (17y/b) b[cos (iry/b) — cos (71-c/b)]1
or
Ey sin (7rx/a) — 0
according as 0 < y < c or c < y < b, where Vo is the electrostatic potential across the iris. Thus get the coefficients in problem 3 by using the Mehler integral, see 13.08 (11), and so obtain for the electric field in the cavity, when 0 < z < d and
556
WAVE GUIDES AND CAVITY RESONATORS
u = cos (irc/b), sinh Ific„z1 niry cos cos wt [P.M P'-'("sinh
irx{ sin (3',0z
Ey = Vo sin b a 2 sin $,,d
ax Vo a Ey = — b sin —
n=1
nir cosh 'Oilnzl Wiry ,„dI sin — cos wt [P„(u) ♦ P„_1(u)] , nh 101 l sinh b 101 n=1
where = Ico2A€ — (nir/b) 2 — (ir/a)2I and co is given roughly by problem 54. Note that in the window Ey is small compared with Er but does not actually vanish, so that the electric field line which leaves the edge in the z = d plane strikes the y = 0 plane normally but slightly displaced from the z = d plane. The boundary condition on the iris is rigorously met, thus the result is best for small windows. This also gives one-quarter of the field in a cavity of height 2b with a centered window of width 2c and in a cavity with height 2b and a centered strip of width 2 b — 2c. 57. Use the method and notation of the last problem but assume E, — 0 sin (r x/a)
or
E, sin (irx/a)
21Vo cos /b) b[cos (re /b) + cos (iry/b)]1
depending on whether a < y < c or c < y < b. Thus get the coefficients in problem 3 by using the second Mehler integral in 13.08 and obtain sinh It3',„z1 airy 1/3,„,b1 [Pn(u) + P,-1(a)l cos—cos ca a 2 sin 1#„dI + n = 1 b cosh 113c.dI n7
V0 ax sin Viozl E, = — sin
b
Vo rx E, = — sin — a
n=1
airy , cosh I81„zI [P„(u) + P„_1(u)] sin cos cot cosh 1(4.dl
Here E, is small compared with E., on the iris but does not actually vanish as it should so the iris surface is not exactly plane. The boundary condition E, = 0 is rigorously met in the window. Thus these expressions are best for a small iris and a large window. 58. A rectangular cavity of width a, height b, and length 2d > 2a is partly closed at the center by a thin iris having a slit running from y = 0 to y = b with edges at x = 1(a — c) and x = 1(a c). Proceed as in problem 54 to show, with the aid of 13.06 (16), that the lowest resonance frequency corresponds to a free space wave length X given by X=
2aXi 2a 2rd rc , 1 — cot — tan' — (4a2 XD1 2a X1 Xi
59. If a = 2, b = 1, c = 1, and d = 2 in the last problem, show that X = 2.7488. 60. Obtain an approximate solution for the fields in problem 58 as follows: Use the fields of 13.06 (9) and (10) but with cos cot time dependence. Take z = 0 at the left end of the cavity as in Fig. 13.06c and assume the quasi-static By at z = d given by 13.06 (14). Thus obtain for the magnetic fields in the cavity, using relations in 13.08, =
By =
ax cos Wiz' , sin cos lOidl a +
m= 3
A„, cosh 1/3„%z1 . nirx , sin a cosh I13„,41
Al
71-sin Ii3;zI irx , 1,31 d. cos I0,al cos a — I
-k A „,rmr sinh max 0,mo cos — a AiI4a1 cosh
In= 3
—
PROBLEMS
557
where ,1 0'„,21 = Ice2A€ - (nor/a)21, is given by problem 58, m is odd, and [ 2n + 1 1)" n(n + 1)
A2"+1
irc [ 7iC sin —Pi cos — a n a
Note that on the iris B is small but does not actually vanish, so that the iris surface deviates slightly from a magnetic flux line. The boundary condition is rigorously met in the window. Thus this result is best for wide windows. 61. Obtain a second approximate solution of the last problem by assuming the quasistatic at z = d. Thus obtain for the magnetic fields in the cavity
By
=
-
cos r
sin Adl sin 113
sin
a
Am10„,al cosh
+ rt =3
cos
sin A di
rX
sin , Amur sinh 10„,d1
irx
A„, sinhl4z1
a
Al sinh 1(3„,d1
,
cos
max
a
mrx
ni= 3
where the notation is that of the last problem. Note that in the window the Bz is small but not actually zero, so that if the image field is drawn in the region d < z < 2d, there will be a discontinuity in the slope of the flux lines in the plane of the window. On the iris the boundary conditions are rigorously met. Thus this result is best for narrow windows. 62. Use the second form of the stationary variational formula 13.28 (3) to get the resonance frequency of the spheroid (p/a)2 (z/b) 2 = 1. Take for the components of the electric field the divergenceless form, which meets the symmetry conditions,
E, - A - Bp2 - Cz2,
Ep= 2Cpz
Determine A, B, and C so that E is normal to the spheroid, substitute into 13.28 (3), and show that [ 14(3a2 + 2b2)]OA ___). 2.788 a—+b a2(5a2 + 462) 63. If d << a, d << c, and a b in the reentrant cavity of 13.29, show with the aid of the results in problem 60, Chap. IV, that Co in 13.29 (11) is, approximately,
Co = € [ 122
±4a
In
k 4ak 4ad €12 +k2 d + tan-, + tan-, d 4kd d k k
271-ad] k
where k is b - a. If a = 0.01, b = 0.015, d = 0.001, c = 0.02, this gives # = 40.6. Moreno's graphs give # = 41.0. The accuracy improves as a approaches b. 64. A rectangular cavity of dimensions a, b, d in the x-, y-, z-directions is to be excited in the TMilo mode by a wire parallel to the z-axis at x = at, y = b1. The wire is terminated to give uniform current in the section of resistance R inside the cavity. If d is much less than a or b, show that the input impedance is 2.),(31,20d2(a2 +b2) rbi ral sin2 _ sia 2_ R' = R + b a ab(a 2 +V) + 2d(a3 + b3) Show that the electromotance multiplication is csc (7ra i /a) csc (rbi/b). 65. Show by the image method and with the aid of problem 32 that the reactance load on the driving circuit of the last problem when the radius of the wire is c is
wyd --2 7 r In c
4
-
5,°A)Yo[20(n2a2 + m2b2)1] -
120[(na - ai)2
+ m2b2]l} - Yo (20[n2a2 +(mb - bi)2]+) + Yo{2i3[(na - at) 2
(mb - bi)2111)
558
WAVE GUIDES AND CAVITY RESONATORS
Observe that the first term is much larger than the rest, if c is small, and that in the remainder the m = n = 0 terms are most important. 66. The cavity in problem 64 is of copper, y = 5.8 X 107, and its a-, b-, and d-dimensions are 10, 20, and 2 cm, respectively. Show that its TMlio resonance frequency is 1.676 X 109 cycles per second, its Q is roughly 9100, and that for an input resistance of 100 ohms the driving wire is at x = 0.0512 cm, y = 10 cm; at x = 5 cm, y = 0.1024 cm, or anywhere on the curve sin (7rai /10) sin (irbi/20) = 0.0160. Show that for 1 watt input, the rms potential across the center is 625 volts. 67. A rectangular cavity is driven by a semicircular loop of radius r in the y = bi plane centered at z = d1, x = 0. From 13.20 (2) and 13.30 (5) AL, is N-81 i
n
Rmn b
cos
max
m • 1,7rX ?ivy nay pvZ sin — — J — sin — cos — sin a
—
a
Show that the normalized mutual inductance between loop and the TE„,, mode is =
811- 3npr
M°
cos
/orb' prdi sin d ,I, (0„,„r) b
The integration is simplified by using f (V x A9 0 dS in place of f A° • ds. 68. Solve the preceding problem for the TM„,„ mode using 13.20 (4) and 13.30 (5). Show that the normalized vector potential A;inn, is, in the notation of 13.20, 8+(timnpfl,,,a)-1P(iM cos Mx sin Ny + jN sin Mx cos Ny) sin Pz
— 14132.„sin Mx sin Ny cos Pz] where tiln = M2 + N2 and I3L„ = M2 +N2 +P2. Show, using the same integrals as in problem 67, that the normalized mutual inductance between the loop and the TM„,„„ mode is M° =
8+71- 4mnpr
Omnp0.41.pad 2
cos
rorbi
b
sin
prdi
Ji(0,,,,r)
69. A spherical copper cavity of radius a is excited by a small plane loop of area S so close to the wall that the value of B at the wall may be used over S. Show that, when driven at the TMoi resonance frequency, the input resistance of the loop is 32,276(40gia2 - 9).42R + 1.854 X 105 y3S2 -
R, + R +
47ra2(4081a2 - 1)
a4
and that, if floia is 2.744 and R is the loop resistance, the potential multiplication is V
2[ Si(Ooia)
6
ROIS ii(Roia)
7.42 1 = 020,S
70. A rectangular cavity is filled with material mei when 0 < y < ib and material p.2 and ei A202 when ib < y < b. If 02, find the ratio of the = A2E2 but Ai energy stored in Alci to that in /1202 when the cavity is excited in the TE101 mode. 71. A cubical cavity is oscillating at the frequency p2 = (21.Lea 2)-1. Show that it is possible to have equal average pressures on all six walls, and state what directions the electric field at the center can have in this case. Show that the rms force on the wall is then ega2/24. 72. The surface of a rectangular cavity, given by x = 0, x = a; y = 0, y = and z = 0, z = a, is formed of material of conductivity -y and skin depth 8. It is
REFERENCES
559
excited by a straight wire carrying a uniform current /0 cos 0.4 that lies in the plane z = la and passes across the corner from y = *a to z = *a at 45°. When co is the lowest resonance frequency, show that the power dissipation is P = (wybe-Lir-2) (1 — cos tor)/20 . 73. A cavity is bounded by y = 0, y = 0.5; x = 0, x = 2 if 0 < z < 2; x = 1, x = 2 if 2 < z < 4. Show that if it oscillates with the configuration of the TE202 mode of a 2 X 0.5 X 4 rectangular cavity, then Q = irvAry651/32. References F. E., and C. H. PAPAS: "Randwertprobleme der Mikrowellenphysik," Springer, 1955. Excellent detailed discussion of wave guides and cavities including integral equation and variational methods. COLLIN, R. E.: "Field Theory of Guided Waves," McGraw-Hill, 1960. Gives complete exposition, with examples, of the mathematical techniques for wave guides of all types. CORSON, D. R., and PAUL LORRAIN: "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Illustrated introductory treatment. FLUGGE, S.: "Encyclopedia of Physics," Vol. XVI, Springer, 1958. Borgnis and Papas give a very comprehensive and readable treatment of "Electromagnetic Wave Guides and Resonators." GRAY, D. E.: "American Institute of Physics Handbook," 2d ed., McGraw-Hill, 1963. Section 5b gives useful collection of formulas and pictures. HARRINGTON, R. F.: "Time-harmonic Electromagnetic Fields," McGraw-Hill, 1961. Very complete, detailed, and practical treatment with problems. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Thirty pages on these subjects. All formulas in both cgs and mks units. JONES, D. S.: "The Theory of Electromagnetism," Macmillan, 1964. Treats cavities from a mathematician's viewpoint. JOHNSON, C. C.: "Field and Wave Electrodynamics," McGraw-Hill, 1965. Wellillustrated and readable treatment. KING, R. W. P., H. R. MIMNO, and A. H. WING: "Transmission Lines, Antennas, and Wave Guides," McGraw-Hill, 1945. Well-illustrated descriptions of wave guides and cavities. MARCUVITZ, N.: "Waveguide Handbook," McGraw-Hill, 1951. Most complete collection of formulas and graphs with some theory. MONTGOMERY, C. G., R. H. DICKE, and E. M. PURCELL: "Principles of Microwave Circuits," McGraw-Hill, 1948. Covers most of the material in this chapter. MORENO, T.: "Microwave Transmission Design Data," McGraw-Hill, 1948. Good collection of graphs and tables for microwave work with some theory. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," 2d ed., Addison-Wesley, 1962. Brief but lucid treatment. RAMO, S., and J. R. WHINNERY: "Fields and Waves in Modern Radio," 2d ed., Wiley, 1953. Clear, well-illustrated, practical discussion. SCHELKUNOFF, S. A.: "Electromagnetic Waves," Van Nostrand, 1943. Extended discussion of theory with many graphs. SLATER, J. C.: "Microwave Electronics," Van Nostrand, 1950. Complete theoretical treatment of wave guides and cavities. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Very brief but lucid discussion. VAN BLADEL, J.: "Electromagnetic Fields," McGraw-Hill, 1964. Discusses additional topics connected with wave guides and cavities. BORGNIS,
CHAPTER XIV
SPECIAL RELATIVITY AND THE MOTION OF CHARGED PARTICLES 14.00. The Postulates of Special Relativity.—In previous chapters, we considered electric and magnetic interaction of charges and currents whose magnitudes may vary with time but whose positions are fixed relative to each other and the observer. The special theory of relativity seems to furnish the most firmly established experimental basis for the discussion of the interaction of systems moving in a straight line relative to each other or to the observer. The two fundamental postulates of this theory are: 1. Physical laws, or the equations defining them, have the same form, in vacuo, in all cartesian coordinate systems that have a uniform translatory motion relative to each other. 2. The velocity of light, in vacuo, is the same for all observers, independent of the relative velocity of the source of light and the observer. A great number of observable phenomena, mechanical, optical, and electrical, predicted by these postulates have been investigated, and in all cases the results of experiments made on a terrestrial scale confirm them. 14.01. The Lorentz Transformation Equations.—Consider the two coordinate systems shown in Fig. 14.01 in which the x and x' axes coincide
and the system S' is moving with a uniform velocity v in the x-direction with respect to S. To find the equations connecting x, y, z and t with x', y', z' and t' which will satisfy the two postulates of special relativity, we shall consider a simple experiment. Suppose that at the instant t = t' = 0 the two origins, 0 and 0', are in coincidence and that at this
instant a light pulse leaves the common origin. Our postulates require 560
§14.01
THE LORENTZ TRANSFORMATION EQUATIONS
561
that S and S' each observe a spherical wave spreading out from his origin with a velocity c. Thus, at any future instant, the equation of the wave front must be for S x2 + y2 + z2 = Op
(1)
x'2 ± y'2 + z'2 = c20
(2)
For S' it must be From the symmetry of the situation as viewed by S and 5', we are led to try setting z = z' and y = y' so that, to satisfy these equations, we must have x2 — c212 = x'2
cy 2
(3)
For S, at the time t, 0' has the coordinate vt; and for S', at the time t', 0 has the coordinate —vt'. The simplest relations which will give this result are (4) x' = K(x — vt) x = K'(x' + ye) Eliminating x' from these equations gives
t ' = K[t - • (1 V
1 1
KK
(5)
Substituting for x' and t' in (3) from (4) and (5), we obtain an equation in which the variables appear as x2, xt, and t2, and since this equation must be satisfied for any positive values of t and any values of x, this requires that the coefficients of x2, xt, and t2 vanish separately. Setting these equal to zero and solving for K and le we obtain K
=
e = [1 — (12F
(6)
Thus, from (4) and (5), the transformation formulas become x' = K(x — vt) z' = z Yi = Y, xv) t' = K(t — — c2
x = K(x' + vt') z = z' Y = y', t = K(t' + c ) )
(7)
Suppose that the signals are emitted from a point fixed in the moving system S' at times t'l. and 4 and received by an observer in the S-system at times t1and t2. Then from (7) At/ = t'1 — 4,
At = ti — t2 = K(4 — 4) = KAt'
(8)
This "time dilation" has been observed experimentally with fast mesons produced by cosmic rays whose average apparent lifetime, as estimated
562
SPECIAL RELATIVITY
§14.03
from their velocity and travel distance, may be increased by as much as ten times its rest value by the high speed. 14.02. Transformation Equations for Velocity and Acceleration.—We shall designate velocities relative to the S coordinate system by ux
dx = dt
=
dy_ dz u, = dt
(1)
and those relative to the S'-system by =
_ dy' u; —
dx' dt'
,
=
dz'
(2)
By differentiating the first three equations of the first group in 14.01 (7) with respect to t' and those in the second group with respect to t and then eliminating dt/dt' and de /dt, respectively, from the right sides by differentiating the fourth equation of the opposite group, we get ,uz — v uz = 1 — uzv
c2
uy,
u'
u'x
uz —
v (3)
1 ± It'gv u„z —
C2
(4)
K(1 + C2
These are the transformation equations for velocity. It is interesting to note that even if we give the two systems the relative velocity v = c and a point in the S'-system has a velocity u; = c, the value of u given by (3) is still c. Thus, the velocity of light may be regarded as the upper limit of possible velocities. By a very similar process, differentiating (3) and (4), we obtain for the accelerations
dux _ dug,, = dt
ti
4v\l-3 lu; )dt' 2duy ' _14,2) [K 1 + u'zv c2 K 2c2 K
(5)
C2
+ u'zv c2
—3du;
(6)
We get the corresponding equation for duiz /de and duy',z/de by interchanging primed and unprimed quantities and substituting +v for —v. We note that a constant acceleration in the S'-system does not, in general, imply a constant acceleration in the S-system. 14.03. Variation of Mass with Velocity.—The first postulate requires that the laws of conservation of energy and of momentum be satisfied for all observers. To see the result of this requirement, let us consider a simple hypothetical experiment devised by Tolman and illustrated in
§14.03
563
VARIATION OF MASS WITH VELOCITY
Fig. 14.03. At the instant when the two origins are coincident, let S' project a sphere with a velocity u from B' toward 0', and let S project one from A parallel to y with a velocity u. Let OA and 0 B be chosen in such a way that the spheres collide when their centers are aligned in the Y-direction. The collisions as observed by S and S' are shown. We shall choose the mass of the spheres so that, when at rest with respect Y
s
17' S'
B
A' 2' FIG. 14.03.
to any observer, both have, for him, the mass mo. We shall assume that the mass is a function of the magnitude of the velocity so that and
m = f (u2)
mo = f(0)
(1)
By the formulas of the last articles, the initial velocities are For S
For S'
uaz = 0,
ub„ = u
Uby = V,
Ubg, =
7.4,x U
/4,y
=
,
u K
(2)
Ugy = — u
Ub y = 0,
Let us indicate velocities after the impact by bars. From the conditions of impact of smooth spheres, neither observer will see any transfer of the x-component of momentum. Thus, S observes that mbv = mbv or Vf(V 2
,72 u2 — ) = Vf(V 2 + bl
K
K
This requires that = —u
and hence
mb =
(3) The negative sign is chosen since we know it to be correct for small velocities, where K = 1. For the y-component, S observes that ub
uf (u2)
mbu = f (17d)
iftbilb
Mb
564
SPECIAL RELATIVITY
§14.04
In view of (3) and the known result when v = 0, this can be satisfied only if fi,„ = —u so that this equation becomes, by dividing out u,
f(u2) = ma =
mb
(4)
If the velocity of projection u becomes very small, then
U2 K2
nib = f(V 2
AO)
and f(0) = mo
ma = f(u2)
(5)
Thus, the mass of an object in motion with a velocity v with respect to a given observer appears to him to be increased by the factor
K = [ 1 — -Yr
(6)
over its mass when at rest. We shall find it convenient to introduce here two new symbols Ki and Ki defined by = (1 — u x ± ± (7) C2
_ (1 U,2 +
c2
(8)
where uz, u„, and uz are the components of velocity of a particle as observed by S and u;, uu, and u; the components of velocity of the same particle as observed by S'. Using the relations 14.02 (3) and (4), we obtain the relations
, , 2,11 Ki = 1C4(.1. -r- c 2 Ki
K ICi
1
(i —
l
(9)
il
— 2-
(10)
C
14.04. The Transformation Equations for Force.—The nature of the equations connecting the forces observed by S and S' depends on whether force is defined as the product of mass by acceleration or as the rate of change of momentum. We shall take the latter definition so that, from 14.03 (5), we have F=
d(mu)
dt
d(K iu)
—' n0 dt
(1)
THE TRANSFORMATION EQUATIONS FOR FORCE
§14.04
565
By writing out the components of this equation, we see that with this definition force and acceleration are not, in general, in the same direction. Carrying out the differentiation, we have
du , dm = du , u du F = m u— — dt m°Kiuc 2 dt dt dt 1du uu du] = moK1R1 — c2 dt c2 dt
(2)
If the force is applied in the direction u1 of u, then u = ulu and
du _ u1 du dt dt so that we have
,du du F1 = morciTt= mi T
(3)
If du/dt is zero, i.e., if the velocity changes in direction but not in magnitude, then (2) shows that the force is again in the direction of the acceleration which is at right angles to u, and we have
du du Ft = moKi— = mt-dt dt
(4)
The quantities mi = moK? and mt = moK i are usually called the longitudinal and transverse masses, respectively, of the particle. The forces observed by S' will be of the same form as those observed by S; thus, F' = mo
d(Klu')
dt
Let us substitute in the x-component of (1) for Klux from 14.02 (3) 14.03 (9) and for de / dt from differentiating 14.01 (7), and we have
u'zv\--1d[ic u; v)] dt' c2 v y[ d(1414) + Aid — = F mov(1 + u e2 dt' du' u;2)dicl u'z) 1[(. = F'x mov(1 — c2 c2
F=
md(Klux) ° dt
mo(i
(
But
1—z— —
1 K'u' U" + U/2 c2 + and —
ic'1 2
c2
du'x— K" iu„,duy' c2
K1"du, ' c22,de
1 dic', 142 de
566
§14.05
SPECIAL RELATIVITY
so that, making the substitution, we have
u, iK,du; + u,d14\ -1 mov r I ,duz; + ,d4) + uude A l dt"de ) i c 2 + uz'vLuY\K1dr „„ i =F'+ r F u,v; 2 c2-uzv ;-c2 +' vz
F = F' ± x
Z
() 5
Similar operations for Fy and Fz give
c2 F„ z F' K(c2 +/4) u' '
(6)
These are the transformation equations for force. 14.05. Force on Charge Moving in Magnetic Field. The first postulate of special relativity requires that the laws of electrostatics shall be identical for all observers. If we assume, in addition, that the magnitude of charges is the same for all observers, we shall see that it follows that the forces observed by S between charges moving in his system differ from electrostatic forces. We shall see from 14.16 that the invariance of charge follows from the invariance of the Maxwell equations. One might call these additional forces electrokinetic forces but, as we shall see, they are identical with those we have already called magnetic forces. Let us suppose that there are point charges q and qi fixed in the x'y'plane at x' = 0, y' = yi and x' = x', y' = 0, respectively, in the S'-systern. From Coulomb's law, the components of the force observed by S' to act on q are —
F' 2
-"ix'
47E, (X"
+y'12)1'
F' = Y
gqiYi 47rE v (X"
y'12)1'
F'=
0
(1)
To S, these charges appear to be moving in the positive x-direction with a velocity v, and the forces he observes may be obtained by substituting in 14.04 (5) from 14.01 (7) and (1). To both S and S', the force appears independent of the time; so let us take it at t = 0, t' = -x'vc-2. This gives, since u'r = uy' = u' = 0, x' = ta, etc., the result
F, -
—gglxx
4re„(K2x2
y1)1'
F=
qqiyi 47rEg((1C2x2
F z = 0 (2)
Now let us see what forces S observes if qiis replaced by an infinite line of charges uniformly spaced along the x-axis and moving with a velocity v. The charge on any element of axis is, by hypothesis, the same for both observers so that the force due to any element is given, by substituting in (2), the value qi = x dx = r' dx'
(3)
§14.05
FORCE ON CHARGE MOVING IN MAGNETIC FIELD
567
In passing, we should note that the assumption that qi is the same for both observers requires, since from 14.01 (7) dx' = K dx, that 0-
=
(4)
so that the charge densities observed by S and S' are different. By substituting (3) in (2) and integrating from x = — co to x = +co, we get for the forces on q due to the moving line charge, by Pc 138 or Dw 200.03 and 201.03, if we write 13 for v/c, F. = 0,
F. = 0,
F„ — q(1 2)a 271-e.yi
(5)
By putting /3 = 0, we find the electrostatic force observed by S to be q0-/(271-E„yi) so that the additional force, due to the motion, is seen to be AF„ =
—
qv2cr 2ire„c2yi
Azav
27ryi• qv
—
(6)
But the current measured by S is i = 0-v. Furthermore, we have seen in 7.14 (3) that S attributes to this current a magnetic induction in the z-direction at x = 0, y = yl of the amount B. = 1.14/(21-yi)• Substituting in (6), we see that the additional force on q due to its motion is AFw
—Bzqv
(7)
Therefore, we say that a point charge q, moving with a uniform velocity v at right angles to a uniform field of induction B, is acted on by a force at right angles to both v and B given by F„, = q[v x B]
(8)
In rectangular coordinates, this is F,„ = q[i(vr13,
—
v.By)
i(v.B.
—
v.B.)
k(v.B„
—
(9)
In 14.16, we shall see that (8) also follows from the invariance of Maxwell's equations. If we let vq = I ds, this is exactly the law of force for an element ds of a circuit carrying a current I in a field of magnetic induction B. This law as derived from Ampere's experiments in 7.18 was valid only when integrated around a closed circuit. We have now extended it to include isolated moving charges. We should note that (8) should apply quite accurately when B is variable as regards both time and position provided that the variation is sufficiently slow so that B may be considered uniform and constant over the region occupied by the actual charge involved. We may expect (8) to break down for any actual charged particles if we go to subatomic distances and frequencies.
568
SPECIAL RELATIVITY
§14.07
14.06. Motion of Charge in Uniform Magnetic Field.—Suppose a particle having a charge q and a mass m has an initial velocity v in a magnetic field B. Let s1 be a unit vector making an angle a with B such that v = vs1. Then, from 14.05 (8), we have d(mv) _ d(mv) dt dt
mvdsi = dt
Writing out components, we have si
d(mv) 0 dt
showing that v is constant, and
dsi mde =
q[si x B]
showing that ds1is normal to s1 and B so that we must use the transverse mass 14.04 (4) for m. Let dgi) be the angle, measured in the plane normal to B, through which s1 has been turned by the change ds1, and let p be the radius of curvature of the path corresponding to dq5. Then, carrying out the multiplication in (3) and dividing through by si sin a, we have
me ds, _ ml dckma) sin a — qB p si sin a dt dt =
(4)
Thus, the path is a helix of constant angular pitch I.7r. — a, lying on the circular cylinder whose radius is mov sin P = qB[1 — (v/c)91
a
( 5)
and whose axis is parallel to B. In the special case, where v is normal to B, a is lir so that the charge travels in a circular path. 14.07. Energy of a Charged Moving Particle.—We define the energy given to a charged particle by the action of a force to be the work done on it. In an infinitesimal interval of time dt, a particle will move a distance dr = u dt. Thus, from 14.04 (3), the work done on it is dW .= F • dr = Fi dr = moK
3du dr = moKlu du dt
i
(1
)
Substitution of 14.03 (7) for kiand integration from 0 to u give W = (mg — mo)c2= (K1 — 1)moc2
(2)
This proves that an energy increase is equivalent to a mass increase, which has become one of the most accurately and completely verified laws
§14.08
MAGNETIC CUTOFF OF THERMIONIC RECTIFIER
569
of physics. Confirmation has come chiefly from the vast amount of data on nuclear disintegrations. Thus we have excellent experimental grounds for the statement that a quantity of energy A W always has associated with it a mass Am given by the equation OW = c2 Am
(3)
From (1) and 14.05 (8) a charged particle gets no energy from the magnetostatic part of a static electromagnetic field. Thus it cannot enter regions where the electrostatic potential exceeds the sum of the starting point potential and that needed to give it its initial velocity. 14.08. Magnetic Cutoff of Thermionic Rectifier.—As an example of a calculation of the motion of a charge when both electric and magnetic fields are present, let us consider a device known as the magnetron. The space between a pair of concentric circular conducting cylinders is evacuated, and charged particles, usually electrons, are set free at the surface of the inner cylinder with negligible initial velocity. The cylinders are maintained at different potentials so that the charges are accelerated from the inner, of radius a, toward the outer, of radius b. A magnetic field is applied parallel to the axis of the cylinders. The induction B is a function only of p, the distance from the axis. We wish to find the induction B that is just sufficient to prevent particles from reaching the outer cylinder when the potential is V. If 4is the longitude angle, then the charges will just reach the plate when dp/dt = p = 0 there, so that we have b dq5/dt = vm. Setting the rate of change of angular momentum equal to the torque acting, we have
dt (mP20) = — 413B15
(1)
At the beginning of the path, y& = 0, and at the end, pc& = v, p = b, and m = K mo. Integrating (1) over this path, we have b K Imo bv = —qf pB dp a
(2)
If N is the total magnetic flux passing between the cylinders, then b N = 27rf pB dp so that this can be written a
— Kimo bv = —Nq 27r
(3)
Since the charge has acquired all the energy by falling through the potential V, we have, from 14.07 (2), Vq = (Ki — 1)moc2
(4)
570
SPECIAL RELATIVITY
§14.09
Eliminating Kiand v from (3) by means of (4) gives
N = — 27bc[ cT72 (2m° + V2)11 q c
(5)
If the potential is sufficiently small, the second term on the right can be neglected. In terms of potential, (5) becomes
V =
q
N2q2 (1 + 41.2c2b2mg
J
1
)1
(6)
14.09. Path of Cosmic Particle in Uniform Field.—As an example of the calculation of a path of a charged particle influenced by both electrical and mechanical forces, we shall consider a particle, traveling in a plane normal to a uniform field B, in a medium that opposes the motion of the particle with a force proportional to its velocity. It is found experimentally that the high-speed charged particles of cosmic origin dissipate their energy nearly uniformly along their paths so that the retarding force is that assumed. Let us write down the components of the equation of motion from 14.01 (1) and 14.05 (8), giving
(1)
(mociy) = —qBt — Ky dt d, = By q— K1 tkin°K1)
(2)
We now integrate these two equations from any point on the path to the origin, where t = y = 0 and x = y = 0 and take the ratio of the results, obtaining
9 _ dy _
—qBx — Ky dx qBy — Kx
(3)
Setting y = r sin 0 and x = r cos 0, this becomes
drK = d0 r qB
(4)
Choosing 0 = 0 when r = 1 and integrating from this point to any point on the path, we have KO
In r =
K —
qB
0
or
r=
B
(5)
Thus, the path is an equiangular spiral. For a high-speed cosmic electron, K roughly equals 3qP X 10-4 where P is the air pressure in atmospheres.
§14.10
MAGNETIC FIELD OF MOVING CHARGE
571
14.10. Magnetic Field of Moving Charge.—Let us consider again the force between two moving charges determined in 14.05 (2). In view of the conclusions of 14.05, we may say that the force that S observes to act on the charge crossing the y-axis due to the charge moving along the x-axis he ascribes to the combined electric and magnetic fields of the latter. The x-component which acts in the direction of motion of q must be considered entirely electrostatic, but the y-component includes both electric and magnetic parts. Thus, we have Fz = Fecos 0, F. = Fz tan 0
F, = Fe sin 0 — Fm (1) —
F, = —F4 — F,
(2)
We use the angle 0 in Fig. 14.10 since, as we shall see in 14.16, the aberration of a signal received by q from qi is zero. Substituting in the values of the forces from 14.05 (2), we obtain —
qqiy(K 2— 1) y2)4 471-6„K(K2x 2
qqvcv 2y 471-Eue (K2x2
y2)1
(3)
S observes that this force does not vary with time and concludes that, since q is a point charge, it is effectively moving in a uniform field so that it is legitimate to compute the magnetic field in which it moves by means of 14.05 (8) and (3). Thus, the magnetic field due to q, is B—
vgivYK 471-(K 2x 2 y2)I
(4)
By definition, the electric field acting along r is, substituting from (1), E
= Fe = _Fzr = qvcr qx 4re„(K2x2 ±y2)I q
By applying in the formulas of 11.01, B, = (V x As), = — k
aA„ Oy
dA E= E = [(- ax1)-
v a—z Ax - )2 (31) 3011
(5)
572
SPECIAL RELATIVITY
§14.11
we can easily verify that the vector and scalar potentials corresponding to (4) and (5) are Az —
AingiKv.
(6)
4r(K2x2 + y2)*
qvc —
47re,(K2x2
(7)
y2)4
It should be noted that (4), (5), (6), and (7) apply to a point charge moving with uniform velocity in a straight line and are stated in terms of the actual position of the charge at the instant of measurement. 14.11. Retarded Fields and Potentials of Moving Charge.—Since electric and magnetic fields are propagated with a finite velocity c, the field at Q at a time t when the charge that produced it is at P will actually have left qiat some previous point on its path indicated by [P]. If, after Y leaving [P], the charge had changed its motion, the field at Q at the time t would still be the same. Thus, in the case of nonuniform motion, it is better to describe the field at a time tin terms [r} of its motion at a time t — ([r]/c), called the retarded time, where [r] is X the radius vector drawn from the re[p] tarded position [P] to the point of measurement Q. Since the field at Q FIG. 14.11. is independent of the path taken by qi after leaving [P], we shall assume that its state of motion at [P] continues unchanged so that it moves in a straight line with a uniform velocity [v] to a fictitious position P'. We now calculate the field at Q for this uniform motion by the method of the last article. From Fig. 14.11, we can write down the necessary relations between the fictitious and the retarded magnitudes [0]2 — 2[$] cos [0]) [0]
r2 = [r]2(1
(1) (2)
y2 = [y]2 = [r]2 sin2
where [0] is [v]/c and [5] will be [v]/c. The binomial denominator factor of 14.10 (3) becomes, when [K] is written for (1 — K 2x2 + y2 = [K]20.2
\ [O]2y)=
[K]2[r]2(1 - [0] COS [0]) 2
(3)
Substitution of (2) and (3) into 14.10 (4) and (5) gives B and E for a moving charge in terms of retarded time. Thus kBz =
kl.qi[v][1/]
47-[K]2[r]3(1 — [0] cos [8]) 3
■ 1441[V] x
[r]
47r[K]2([r] — [r] • [5]) 3
[r] x E [r]c
(4)
§14.12
RADIATION FROM AN ACCELERATED CHARGE
573
This also applies to a slowly varying velocity provided we use the values [0] and [v] at the position [P]. The electric field of 14.10 acts along r, and Fig. 14.11 gives the vector relation r=
[r]
— [rM = [r]
[p])
—
where [ri] is a unit vector along [r]. Thus 14.10 (5) may be written in terms of the retarded values [0], [5], [ri], and [v]. E
—
41(1 — WP)a ril — [5]) _ qi(iril — [5]) 47-E„[K] 2[r] 2 [p] 3 4r€ v[r]2(1 — [3] cos OD' ] cos B = 1 — [p] = 1 — [ 3 [ri] • [P]
(5) (6)
Substitution of (1) and (2) into 14.10 (6) and (7) gives for the retarded or Lienard-Wiechert potentials due to a moving point charge
A=
Pivqi[v]
47r [r][p]
= qi[3] ilirevc[r][p]'
_
qi 47-Ev[r][p]
(7)
When [/3] is small, these expressions obviously are identical with those given by 12.02 (1) and (2) for small charge. When [/3] is large, we can obtain (7) from 12.02 (1) and (2) by taking account of the variation in the retarded time within the infinitesimal volume of the charge. This variation cannot be neglected even when the charge shrinks to a point. Mason and Weaver give a full discussion of this point. 14.12. Radiation from an Accelerated Charge. The results of the last article may be applied to the accelerated motion of an actual charge such as an electron provided that the acceleration is so small that the variation of retarded time throughout the charge may be neglected. If this is not the case, it is necessary to know the configuration of the latter in order to find the field and this, at our present state of knowledge, is not possible. Let us suppose that the charge is sufficiently small so 14.11 (7) may be used and compute the fields from these potentials. Assume that the retarded position of the charge qiis known as a function of the retarded time [t]. Thus the data given are —
=—
a[r]
a[t]
a[v] a2[r] a[t] = a[t] 2
(1)
The field of the accelerated charge will be found from 14.11 (7) by the familiar formulas 11.01 (2) and (3), so that E =
aA —
—
V 0,
B=VxA
(2)
Here the differentiations are carried out at the field point, so a/at is evaluated with [r] constant and V with [t] constant. By definition [r]
574
SPECIAL RELATIVITY
§14.12
is c(t — [1]). Therefore it follows that, using 14.11 (6), a[r] = 1
c at a =1
a[t] = 8[r] 8[t] _ r wok] at a[t] c at – –1-ru I"' at a[r] = – fril • [51c
(3)
Let V' be the operator with [t] constant. Then
a[r] cV[t] = V'[r] + v[t] — = [r1] – [rd. [5]cv[t] a [t] [ri] a fril v[t] = v = v' [p] c c[p]
yin] =
–
(4)
Application of the operators (3) and (4) to (2) gives 1 aA
E
= –FA
[r11 aq5
(5)
[n] aA B=V'xAd- c[p] a[t]
( 6)
The component of [v] normal to [r1] which is the vector difference of [v] and the component, [r1][r1] • [v], of [v] along [r1], divided by [r] is the time rate of change of [r1], so that
a[ril _ c a[t]
a (H[Pi) c
– -[T i ] •• [5] + [0]2
• [5] – [5] [r]
[r] • [6/1 ,
v'([r][pl) = v'arl – Er1 • [51) =
(7) [5] = a[51
(8)
– [5]
(9)
a[t]
Substitution for A and cb from 14.11 (7) and for the derivative from (8) and (9) in (5) gives for the electrical intensity
E=
[
crri [712
_ qi 47rE„ E = qi 4irt„
(c[1[1:11) }] – [51)(1 – [$]2) [Ti] • [611(fril – [5]) – [P][9 [r] 2[p] 3 c[r][pP — [51) x [6111} — [51) (1—[O]2) + Jr 1 [r] 2[p] 3 c[r][231 3 [r[ilrl 2;][a5]{[P]
(10) (11)
As before [r] is [rd[r], [51 is [v]/c, [p] is 1 — [r1] • [5], and [r1], 151, and [p] are dimensionless. Similar substitution in (6) gives
B
—
qi f([51x [ril)(1 — [,]2) ([Ti] — [51) x [511} (12) +[n] x ([rd x c2[r][p] 4re„( 3 c[r] 2[p] 3
§14.13
575
ACCELERATION PARALLELS VELOCITY
Note that B is normal to E and [r1] because cB = [r 1] = E
(13)
Observe that the [] term in (10) or (11) is proportional to [r]-1, so that the Poynting vector integral over an infinite sphere does not vanish and represents energy radiated by the accelerated charge. The remaining term in (11) is identical with 14.11 (5) and thus, together with the remaining term in (12), represents the fields of a charge moving with uniform velocity in a straight line.
14.13. Charge Radiation When Acceleration Parallels Velocity.—When [b] is parallel or antiparallel to [v], the radiation part of B by 14.12 (12) is x {[rd x ([7.1 ] x [P} B = q[r1] 47c2E v[r] [P]3
-q[n] x [L1 471-c2E„[r][p]3 ]
( 1)
since [g] [] = 0. In terms of polar coordinates r, 0, (1), as shown in Fig. 14.13a, with [6] taken antiparallel, this becomes
B -
-q[b] sin [0]
(2)
47e,,c3H(1 - [0] cos [0]) 3
When an electron strikes the target of an X-ray tube, [v] and [b] axe antiparallel and the target penetration depth is very small compared with 100° 90° 80° 70° 60° 110° 50° 120° 40° 130° 140°
30°
150° #4111 160° 170 0t,
2°' 10°
(b)
(a) Fin. 14.13a, b.
the observation distance. Therefore the wave is spherical, [r] r, [0] --> 0, and cB0 equals E8 . This formula will now be used to find the total radiated energy density for a given 0 when the electron is stopped. The instantaneous Poynting vector 11.02 (3) becomes, since (i./„E„)-1is c 2,
-
E xB
0[212 sin 2 0[r1 16r2c3t,r2(1 - [3]cos 0)6 ]
(3)
The total energy per pulse T(0,r) passing radially through unit area at r, 0 is found by integration of II over the transit time to, corresponding to
576
SPECIAL RELATIVITY
a deceleration time [to].
Thus from 14.12 (3)
dt = ('ion[p]d[t] =
T=
§14.14
oII(1 — [0] cos 0)d[t]
(4)
Write d[v] for [b]ci[t] and integrate by Dw 90.5. Thus T,. —
OM sin' 0 f[v] 167 2c3tyr 2 Jo
d[v] — ([v]/ c) cos 0) 5
e[b] sin' 0
1
6472c2e,r2 cos 0[ (1 — [0] cos 0) 4
1
(5)
The assumption that the deceleration of the electron is constant, used to derive (5), is certainly not rigorously true at the target of an X-ray tube, but the result agrees roughly with the intensity measurements of the properly polarized part of the continuous X-rays or bremsstrahlung. Figure 14.13 b shows a polar plot of T. as a function of 0 as calculated from (5) by Sommerfeld. Curves for three values of 0 (0.1, 0.2, and 0.3) for the same [id are shown, and the angle for maximum Ty is indicated by a dot on the curve. Another application of (3) is to the calculation of the power lost due to radiation by electrons in a linear accelerator. The power, measured at the charge, radiated per unit area is drl/d[t], and when multiplied by r2c/12 this gives the power radiated through the solid angle c/S2 at an angle 0.
d[P] _ c/S2
c,2[2.1]2 sin2 0 1672f,c3(1 — [0] cos 0) 5
(6)
Integration of (6) over the whole solid angle, using x for cos 0 so c/S2 is —27 dx and the limits are +1 and —1, gives, by Dw 90.5 and 92.5, [p] 0[8]2
61-Eye3 (1 — [3]2)3
(7)
One is usually interested in (6) when [0] is nearly 1 so sin 0 = 0 is near zero, and from 14.01 (6), (1 — [0])-1 = K 2(1 [0]) is very close to 20. Writing 1 — 102 for cos 0 gives 1 — [0] cos 0 the form Itc-2(1 + K202) so 2q2[1)12K2(K0) 2
d[P]
7.2€ C3(1 + K202)5
(8)
This has its maximum at KO = 4-, where from 14.01 (6) its value is
r d[P] 1
512q2[0 2 [$]2) 4 c/S2 jrn.[51 .1 3125r2e,c2(1 -
(9)
—
14.14. Charge Radiation with Acceleration at Right Angles to Velocity. Radiation losses may be very important when high-speed charged particles
§14.14
ACCELERATION AT RIGHT ANGLES TO VELOCITY
577
are bent in magnetic fields as is the case in cyclotrons and synchrotrons. As in the last article one first finds Poynting's vector by squaring the radiation term in 14.12 (10), noting that [5] • [] is now zero, that B = E/c, and that (µ„E„)-1 = c2. The result is
= E2
q2 {[p] 2[id 2 - 1] • [i ) 2(1 - [0 )) 1671-2Eve[r] 2[29]6 ]
]
2
(1)
Let the orbit and hence [V] and [v] lie in the xy-plane of Fig. 14.14a, and let the observer be in the yz-plane so [r1] makes an angle a with the z-axis. The component of [r1] in the xy-plane is of length sin a and parallels the y-axis, so that [r1] • V is - [0] sin a sin 4) and [r1] • [g] is [v] sin a cos 4)
(b)
(a) FIG. 14.14a, b.
where is the angle between the orbit radius to the electron and the x-axis. Then [p] is 1 - [0] sin a cos 41 from 14.11 (6), and d[P]/dO is calculated from H as 14.13 (6) is from 14.13 (3) and yields d[P] c/S2
g2[1)]2 1 167r2E„c3(1 - [13] sin a cos 0)3{
(1 - [$] 2)) sin2 a sin2 cb } (1 - [0] sin a cos 0) 2
(2)
For the radiated power integrate (2) over SZ, with 0 < a < ir and 0 < < 2, The following coordinate change facilitates integration x z' sin a cos 4, = - - = cos 0' ✓ r y x' sin a sin = = sin 0' cos (t.' ✓ r z y' - = sin 0' sin cp' cos a = ✓ r q 2N2 d[P] _[312) sin20' cost q5'} dS2 1671- 2E„c3(1 - [0 ] cos 0') 3 (1 - [0] cos 0') 2 —
(1
(3)
The integral of the second term gives 14.13 (7) multiplied by - 1(1 - [13]2), and the first term integral is 14.13 (7) multiplied by 4(1 — [op), so that g2[v]2
[1'1 - 67rE,,c3(1 - [0]2)
(4)
578
§14.14
SPECIAL RELATIVITY
It is interesting to compare the radiation loss when a force F acts on a charged particle in the direction of its velocity to that when the same force acts at right angles to it. When [b]2 is eliminated from 14.13 (7) by means of 14.04 (3), there results n2F2 F )2 =
q 2,6 Pt =
67-tvm,2,c3
67rc„c3 m0K3
(5)
When [i,]2 is eliminated from (4) by 14.04 (4), the result is q 2,4
Pt =
q2,c 2F2
F )2
67re„m,3c3
fire„c3 moK
(6)
Thus the same force applied for acceleration transverse to [v] gives a radiation loss greater by the factor K2 or (1 — [v/c]2)-1than when it is applied for acceleration parallel to [v], so that when any synchrotron or cyclotron has a fixed acceleration per revolution, there is a limiting velocity at which the radiation loss cancels this applied acceleration. This limitation does not apply to linear accelerators. As with 14.13 (8), one usually wants the radiation intensity near its peak, which in this case lies on a tangent to the circle in Fig. 14.14a. For a distant observer in the yz-plane, when [0] 1, this flash leaves the electron when a = 171-, 4 = co[t] = 0, or from (3) at 0' = 0. Putting into (3) the relations connecting 14.13 (7) and (8) gives
q2[b] 2,6
d[P]
c/St [sHi 271-2E„c3(1
K20' 2)
{ ( 1 — 2K0' cos ck' 2 1 -I- K20' 2
(7)
Evidently the intensity is a maximum when the second term in parentheses is zero at 0' = 0 and is zero when this term is 1. Thus q 2k)1 2
td[P]\ c112 )„,„„
271-2E,0(1
In the plane of the loop from (3), z = 0,
[ 3]2)3
(8)
= 0, a = fr. Thus, if
K0' = ±1,
d[P] _ 0 di2 —
and
O'„,;„ = ± (1— [M2)1= Amin = w0[t]min
(9)
From (8) and (9), at high velocities where [0] —+ 1, the beam in the forward direction becomes extremely intense and narrow, as shown in Fig. 14.14 b. This beam, like a rotating searchlight, sweeps across an observer in the plane of the orbit once per revolution. The interval between the two [t]„,;„ values bounding the peak is, from (9), A[t] —
2(1 — [0]2)1
—
d[t] At At = dt 1 — [0]
(10)
§14.15
579
CHERENKOV RADIATION
where d[t]/dt from 14.12 (3) has been inserted. For high velocities, [0] 1 and 1 — [13] is written 1(1 + [0])(1 — [0]) = 1(1 — [O]2), so that the interval A t in which the beam sweeps across the observer is At =
(1 — [OP)/
1
Wp
K 3Wp
(11)
There is a general rule that when a pulse of duration At is expressed as a Fourier integral in co, the frequency range needed is cooto coo OW where Aw At is approximately 1. Thus the highest frequency seen will be, roughly, neglecting coocompared with Aw, Wmax
OW
(1—
) 0]2 -1C00 =
[
K 304
(12)
Many different approximations may be used in the calculation of the intensity as a function of co. Much depends on the distribution of electrons in the orbit and whether they radiate coherently or not. If the spacing of N electrons around the orbit is perfectly random and they radiate incoherently, then the loss is N times that of a single electron. Julian Schwinger in Phys. Rev., Vol. 75, pages 1912 to 1925, 1949, gives a detailed treatment of this case. 14.15. Cherenkov Radiation.—When a particle moves through a medium with a velocity [v] exceeding (AE)--1, then even with [v] constant,
Fm. 14.15.
it loses energy by radiation. Figure 14.15 gives a physical picture of what occurs when [v] > (i.ct)--1, so that the particle outruns its radiation and shows a section of the half plane 41 = 0. The particle path with constant [v] is shown from —z1 to the origin 0. Seven semicircles with equidistant centers indicate the locus at t = 0 of the pulses generated at these points by the particle. It is evident from the figure that there is one direction a, normal to the cone of half angle 17 — a, in which all pulses are in phase. Each pulse can be resolved into a band of frequencies or Fourier components. The field without dispersion would be infinite at the conical sheet of thickness zero, but actually dispersion will make the fields finite and of thin, but not zero, thickness. The normal direc-
SPECIAL RELATIVITY
580
§14.15
tion of this shock wave is (1)
ttE[vP = cos' a
The potentials of 14.11 (11) can be modified to give the Cherenkov field when the particle moves in a straight line with uniform velocity [v]. A cylindrical coordinate system p, cb, z is used with the origin at the particle when t = 0. The equation for [r] is [r]2 = (µE)-1(t — [t]) 2= (z — [v][t])2
p2
(2)
Solve the last two members of this equation for [t], subtract [t] from t, and then use the first two members to get [r] —
(i.LE)i[v] (z
— [v]
±
{(z
—
[v]t)2+
P2(1 -AE[V
1—
] 2
)
I
(3)
If the velocity of light (AE)--i in the medium exceeds [v], then AE[v]2 is less than 1 and both terms under the radical are positive. Thus its value exceeds that of the first term and its sign must be positive to make [r] positive. This is the case treated in 14.11. If AE[v] 2is greater than 1, then the second term in the radical and the denominator in (3) are negative. If the square root is real, its magnitude is less than that of the first term. Therefore [r] is positive only if [v]t < z, and thus z must be negative in Fig. 14.15 where t is zero, for a field to exist. Since [r] must be real, the curve bounding the field is found by equating the radical to zero. Thus (z — [v]t) 2— p2(12E[v] 2— 1) = 0
(4)
The slope of this bounding curve is constant for dz d—p= — (p.e[v] 2— 1)1 = tan
(5)
for all values of t, since the cone moves with the particle without change of slope. Note that dz/dp = z/p when t = 0, so that when t = 0 the particle is at the origin. Also (5) agrees with (1) because 11, — = a. It can be shown that the [r][p] of 14.11 (6) and (7) becomes [r][p] = ± { (z — [0) 2 p 2(1 —
AE[v]2)
(6)
Thus there are two potentials for every point inside the cone. This is also evident from Fig. 14.15, where two wavelets intersect at every point. The absolute value of [r][p] must be used, so the total potential 4 is 2rEq5 = q{ (z — [v]t) 2
)92(1 — pce[v] 2)) —+
(7)
§14.16
TRANSFORMATION OF MAXWELL'S EQUATIONS
581
which is infinite on the cone with no dispersion. The vector potential is
Az = me[v]•,
(8)
The electric field components are given by 2.7rEz = vq(z — [v]t) {040-1— [v]2 11[r][P]i-1 = Avqp{ (1.40-1— [02 11[9 ][1AI-I
(9) (10)
These formulas show that E lies in the cone in Fig. 14.15 where t = 0. From symmetry the magnetic induction is Bo which also lies in its surface. Equation (1) shows that measurement of the angle a at which light of known wave length and velocity is emitted when a high-speed particle traverses a medium; pE gives the velocity of the particle. The weak radiation from single particles can be amplified for particle-counting purposes. A very complete discussion of the theoretical and experimental aspects of this subject is given by J. V. Jelley (see references). 14.16. Transformation of Maxwell's Equations.—The first postulate of special relativity requires that the Maxwell equations have the same form for both S and S'. We have now extended these to include moving isolated charges; thus we put pu in place of i in 11.00 (1), giving for the equations that must be satisfied in free space
V E = — as
B' „ , v.z =p ui- ev (: aB' V x E' = —
V • E = p/E, V• B=0
V • E' = p'/E, V • B' = 0
B , vx—= pu
(9E
vat
at
(1)
From 14.01 (7), we have the differential relations
az _ ",
v at' _ ax —"c2,
ax'= —Kv, at
at' _ at
K
(2)
Thus, we get the relations
a
a ax' +a at' K a _v a) ax at' ax ax' c2 at' a a ax' a at' a a \ at — ax' at + at' at = "Vt' vaz) aa a a ay=ay az=az (
ax — az'
(3)
Changing the independent variables in the first group of (1) from the S- to the S'-system by means of (3) and using the velocity relations from 14.02, we find that the substitution of E, B, and p from the following
582
§14.17
SPECIAL RELATIVITY
equations gives the second group in (1) and they are therefore the desired transformation equations.
Ey,. = K(E' ,. ± vif.,y) By,. = K(13; -T c--1134y)
E. = E.' B. = B'
p = p's(1 4 u+ )
(4)
(5) (6)
where the lower sign goes with the second subscript. These equations show that the way in which an electromagnetic field is separated into electric and magnetic parts depends on the state of motion of the observer. To get E', B', and p' in terms of E, B, and p, we need interchange only primed and unprimed quantities and change the sign of 13 and v. We can show that (6) implies the invariance of electric charge assumed in 14.05. Elimination of the last factor in Eq. (6) by 14.03 (9) gives picl = p'Ki where K1and K1 are defined by 14.03 (7) and (8). Thus, if So is an observer stationary with respect to the charge, Po = p[1. —
= (1)21' \c / i
pfl — M2T \c/ i
(7)
The volume elements seen by S and S' are contracted by (2), so that
dso = [1 — 02 ]-1ds = [1 — 021-4ds'
(8)
Combining (7) and (8) shows that S and 5' observe the same charge, for Po dso = p ds = p' ds' = q
(9)
We can also show that (4) and (5) lead to the expression for the force acting on a charge moving in a magnetic field given in 14.05 (8). Suppose that 8' observes a charge at rest in an electrostatic field in his system, giving F'x , y,. = q'E'r ,,,,, (10) Using the transformation equation for force [14.05 (5)], setting uz = v, uy = 0, and uz = 0, and using (4) and (5), we have, since q = q',
F. = qE.,
Fy,. = q (Ey, Z
IT
u.Bz,y)
Writing this in vector form gives
F = q[E + (u x B)]
(12)
which agrees with 14.05 (8).
14.17. Ground Speed of an Airplane.—One of the most difficult experimental problems in aviation is that faced by an aviator in determining his speed relative to the ground when conditions are such that
§14.18
583
CROSSED ELECTRIC AND MAGNETIC FIELDS
he can see nothing outside his airplane. Since an airplane flying horizontally is cutting the vertical component of the earth's field, an electromotive force, proportional to its speed, must be induced across any lateral conductor on the airplane. The idea has been suggested and patented that ground speed can be determined by measuring this electromotive force. Several methods of doing this suggest themselves, one of the most obvious of which is to rotate an elongated conductor in a horizontal plane about its center. For a uniform speed of rotation, the induced charges are constant if the airplane is stationary. If, however, the latter is in motion, the induced electromotive force is always, in the Northern Hemisphere, from right to left. Thus, there is a transfer of charge from one end of the conductor to the other, producing at its center an alternating current whose frequency is that of the rotation and whose magnitude is proportional to the ground speed of the airplane. This current is very small but could be measured by sensitive instruments in a well-shielded laboratory. The question arises as to whether the effect disappears if the apparatus is electrically shielded in the airplane. We know that the magnetic field will penetrate nonmagnetic metallic conductors, but we also know that the induced electromotive forces in the shields will set up electric fields tending to counteract the fields induced inside. So far, we have looked at this problem from the point of view of a stationary observer S on the ground and found it quite complicated. Let us now see how it looks to the observer S' in the airplane who is to make the measurements. Let z be the vertical direction and x the direction of motion and let the components of the earth's field be B, By, and Bz. There is also usually a vertical electric field present which we shall call Ez. These are the fields observed by S, from which we determine by 14.11 (4), the magnetic field observed by S' to be (1) B;„ = K(B,,z ± c-113Ez,y) = By,z B: = B, 0, Since the speed of an airplane is small compared with that of light, and K ••=-% 1 so that the magnetic fields observed by S' are unchanged. For the electric fields, we have Ex = 0,
Eb = —KvB, —vBz,
E; = KEz KvBy
E.
(2)
Thus, S' finds his airplane in a transverse electric field vBx. He cannot therefore use any metallic shields about his apparatus. Furthermore, this field is extremely minute compared with Ez, the normal atmospheric electric field, so that the slightest lateral tilt of the airplane will completely nullify the measurements. 14.18. Motion of Charged Particle in Crossed Electric and Magnetic Fields.—In the last article, we had an example of the simplification of a
584
SPECIAL RELATIVITY
§14.18
problem by using the transformation equations 14.16 (4), (5), and (6) to introduce a field. We shall now use these same equations to eliminate a field. Consider a charged particle starting from the origin with initial velocity components vz, vy, and vzunder the influence of a uniform electric field in the y-direction and a uniform magnetic field in the z-direction. For an observer moving along the x-axis with a uniform velocity v, the fields will appear to be, from 14.16 (4) and (5), E; = 0, B; = 0,
E; = 0 E; = KE„ — KvB,, B; = K(B, — B; = 0,
(1) (2)
Thus if Ey < cBz, we choose the velocity with which S' moves such that Ey = vB,
(3
)
so he observes only a single uniform magnetic field in the z-direction, given by (2), to be present. If the charge is released when the origins are coincident, then initially t = t' = x = x' = y = y' = z = z' = 0. For S', the components of the initial velocity are v;, v;, and v'z. From 14.06 (5), he observes the particle to move in a helical path of pitch y' on a circular cylinder of radius a, with an angular velocity co', where a=
K'i mo (v;,2
v;2)i
and
qB;
tan 7' —
vx
(v2 + vy' 2)1
If we make use of 14.03 (7), (8), and (10), 14 may be written = (1 02)-1(1 — 1C = (1 — 142 ± 2412 + C2 „vx) 1CK1(C 2B vzE)
= KK i(i — p
c
—
v2 + v2 + v2)-1 Y2 c
z (1 —
c2B
Substituting from (2) and (5) in (4), we obtain a—
Kimoc[c2(vzB — E)2 (c2B2— E2)4i q(c2B2 — E2)
(6)
The angular velocity about the axis of the cylinder is _ (vr'2
vy'2)i a
q(c2B2— E2) = Kimo(c2B — v,E)
(7)
where, looking in the positive z-direction, w' is counterclockwise if q is positive. In most cases, E > v,B so that vx is negative. In this case, a line normal to the z-axis and tangent to the cylinder at the origin makes an acute angle with the negative x'-axis, where, from 14.02 (3) and (4), tan ,p0 =
= K(v V—Y
VY C( C(E 2B— 2 x B E 2))
(
8)
§14.18
CROSSED ELECTRIC AND MAGNETIC FIELDS
585
We can write down the position of the charge at any time in the S'-system. x' = a[ — sin w't' cos 1'o+ (1 — cos 0.0') sin ,'o] = a[sin %Go— sin (w't' + 4/o)] y' = a[sin w't' sin 4'o (1 — cos w't') cos 4'o] = a[cos 1,tio — cos (w't' + %to)] Let us now introduce new symbols 4/ and b which are defined by 1,G = w't'
t h = co/K(t — — xE) c2B sto b _ v = k 1moE(c2B — vx.E) qB(c2B2— E2)
(9) (10)
Writing x and y in terms of x', y', and t' by 14.01 (7) and writing t' in terms of 4/ and b, we have
x(i
+ a sin 1,1,0 — a sin 4'
2211 = by, —
c8
y = a cos 1,1/0— a cos 1,1■
(11) (12)
These are the rigorous equations of the path of the particle in terms of the parameter 4'. We notice that when 4, = 2n,r + 4'0, the position of the particle is
= 2n. b(1 -
—
E2 c2B2
y. = 0
Thus, the paths of all particles starting from the origin with the same value of b intersect periodically at the same points along the z-axis.
a)b ezz.e) a(1)
Furthermore, we notice from (10) that if the initial velocities are small compared with that of light, the distance between these points of intersection depends only on the field strengths and the ratio of the charge to the mass of the particles. This principle is utilized in one type of mass spectrometer. With an electric field of 100,000 volts per meter and with an induction of 0.1 weber per square meter, the quantity E2/(cB) 2occurring in (11) is 1/90,000. In such a case, (11) and (12) become the parametric equations of a trochoid, which is the curve traced by a point on the radius of a
586
§14.19
SPECIAL RELATIVITY
rolling circle. To make the circle roll on the x'-axis symmetrically, we choose a new origin so that
x' = x
b‘1.0 — a sin 'o
and
y' = y — a cos %Go
b
We thus obtain the paths shown in Fig. 14.18 in which we see that b is the radius of this rolling circle and a is the distance along this radius from its center to P, the point tracing the curve. The cycloidal path, traced when a = b, was used by J. J. Thomson, in 1899, in finding e/mofor photoelectrons. 14.19. Aberration and Doppler Effect.—In 11.01, we saw that all the properties of an electromagnetic wave can be obtained from the Hertzian vector Z which, in free space, satisfies the propagation equation 11.01 (10).
V 2Z =
1 2Z dt2
(1)
A general solution of this equation for a plane wave was given in 11.03 (1). In the case of a monochromatic wave of frequency p, we may write this, using 11.09 (4), since v = c, 27ry (2) Z = A cos — (n • r — ct) c This represents the ray as observed by S. The same ray will appear to 8' to be of frequency V and to be propagated in some direction n', so that
Z' = A' cos
71/
(n' • r' — ct')
(3)
the velocity, by the second postulate of 14.00, appearing the same to S and S'. The transformation equations 14.01 (7) must transform (3) into (2), and, as we see, this requires that the argument of the cosine in (3) when expressed in terms of v, n, r, and t must be identical with that of the cosine in (2). To hold for all values of x, y, z, and t, the coefficients of each of these must be the same in both arguments. We may write (4) n • r — ct = tx + my + nz — ct Using the transformation equation 14.01 (7), we have, by rearranging terms, m'y' n' n' • r' — ct' = — ct' = K(' -1- 22)x + trey
n'Z
—
K(c
(5)
Equating the coefficients mentioned, we have tc (I'
—
n'
6
m
n
IC(C
IT)
V
(6)
§14.19
587
PROBLEMS
These equations enable us to find the effect of the motion of a source of electromagnetic radiation relative to an observer upon the measurements of its radiation made by him. The directional effect is known as aberration and the frequency effect as the Doppler effect. Let the source be at rest in the S'-system and the observer be at rest in the S-system. Then the light that appears to S to come from the /"n direction will appear to S' to travel in the direction, where the relations between /,.,n and are given by (6). Equate the first and fourth fractions in (6), put / = cos 0 and /' = cos 0', and we have cos 0 —
cosB' + $ 1 ± cos 0'
(7)
where 13 = v/c. This is the rigorous formula for aberration. Equating the last two fractions, we obtain v=
1 + 13 cos 0' ,
(8)
(1 — )32)i
This is the rigorous expression for the change in the observed frequency of electromagnetic radiation when the source is moving relative to the observer. If the source is approaching the observer, cos 0' is positive, and if it is receding, cos 0' is negative so that in the first case the frequency is apparently increased and in the second decreased. From the fundamental postulates, the aberration and Doppler effect appear only when there is a relative motion of source and observer. Thus (7) and (8) are equally valid if the observer is moving and the source is stationary. We might expect to find deviations from these formulas at astronomical distances since the special theory of relativity has been completely verified only on a terrestrial scale, but they should be precise on this scale. Problems 1. A long straight filament of radius a carries a current I and emits electrons with negligible initial velocity which are accelerated by a potential V toward a long concentric cylindrical anode of radius b. Show that the relation between I, a, b, and V such that the electrons just fail to reach the anode is V = [i.t2/ 2q/(8x2m)] ln2 (b/a), where the dependence of m on the velocity is neglected. 2. Treat the preceding problem rigorously when E p> cBoby transforming to a system of coordinates moving parallel to the axis of the cylinder with such a velocity that the magnetic field disappears. Show by this means that it is impossible for an electron, once leaving the filament, to return. 3. Treat the preceding problem rigorously when E5< cB,s by transforming to a system of axes moving parallel to the axes of the cylinder with such a velocity that the electric field disappears. Show that the relation such that electrons just fail to reach the anode is 271-moc(A1— /4) = A/q(c2— A )1(1 —
b a
In —
SPECIAL RELATIVITY
588
where A = v12 v': and v', and v', are related to the initial velocities vi and v2 in the x and p directions by the relations ,
Vi —. COI V1 =
f
. V10 1 1— — c
V2 =
v2(1
ODi
—
1—
V101
c
where vi = Ep/Bos and we neglect the initial vo. 4. A vacuum tube contains a cylindrical cathode of radius r0 which is surrounded by a coaxial anode of radius ri. This tube is placed with its axis on the axis of rotation of the confocal hyperbolic pole pieces of an electromagnet. Show that the potential which must be applied in order that electrons, in the plane of symmetry, reach the anode, if the magnetic field in this plane is given by B0b(r2 b2)-1, is rigorously
moc2({ B 2q 2b2 2 2 [(r? + moc2r q 1
b 2)1
+ b2)1 j 2
— 1)
5. A parallel beam of electrons which have been accelerated through a potential
V carries a current I. If the section of the beam is circular of radius a, show that the acceleration, normal to the beam, given to an electron at its surface due to electric and magnetic forces is
=
I
y (1+ Vq 11 (2qi[(1 + Vq ) 2mc 2 mV / mc2
6. A point charge q moves in the field of a stationary point charge Q. Using the conservation of angular momentum p and of energy and the expression 14.07 (2) for kinetic energy, show that the equation of its path is r-1 = A + B cos yq5, where (IQ )2 —1 (
4reocp 7. A line normal to two plane sheets at a distance a apart in a vacuum is concentric with a hole of diameter b in each sheet. A beam of charged particles, all having the same energy, but having a maximum divergence angle a from the normal, emerges from the first hole. Neglecting a compared with unity, find the set of values of the magnetic field which, when applied normal to the sheets, will make these particles pass through the second hole. Show that the maximum diameter of the beam between the holes is then [2a/(nir)] sin a b, where n is an integer. 8. If, in the last problem, all particles comprising the cone originate at the same point on the axis and the magnetic field includes this point, show that the minimum value of the field which will bring these particles to a focus at the image point beyond the second hole is 222-mva[q(aa b)]-1, where a is the angle subtended at the source by any radius of the first aperture. Show also that this field increases the number of ]-2 ions passing the first aperture by the factor brb/[2(aa b)]1 2(sin [47b/(aa 9. Two similar circular parallel conducting cylindrical shells carry charges +Q and —Q per unit length. Inside each, a wire is so situated that when a current I flows through them in opposite directions each shell coincides with a magnetic line of force. Show how, if cQ < I, it is possible to transform to a set of moving axes such that only a magnetic field appears. .
589
REFERENCES
10. In the last problem, show that, if a charged particle starts from rest on one cylinder, the component of its velocity parallel to the cylinder at r1, r2, is given by rir20 Q ln = / ln rir2n( r2rio r2r10
27rmoy1
here n o and r2o are the distances from the starting point to the two wires. 11. In a two-dimensional magnetostatic field where the vector potential has only a z-component, let the total velocity of an ion of charge q and rest mass mo be v and let its z-component be v,, at a point where the vector potential is A i. Show that, when it is at a point where the vector potential is A 2, the z-component of its velocity is given by v, — q(moc)-1(c 2— v2)1(A2 — A i)• 12. Show that, in any two-dimensional case in which the electric and magnetic fields are orthogonal, it is possible to choose a set of moving axes such as to eliminate either E or cB whichever is the smaller. 13. The position of a charge q is given by s = a cos wt. If wac-1 1 and a << r, show, by using 14.11 (7), that the periodic parts of E and B due to q are identical with 12.01 (6), (7), and (8) where r is the distance from its mean position to the field point, 0 is the angle between r and s, and M = qa. 14. The positions s1 and s2 of two identical charges q are Si = —s2 = 2a cos (10.4). Proceeding as in the last problem, taking al—,<< ceac-' << 1, and letting Q0 = qa 2, show that the periodic part of the field at a great distance is co sQ0 sin 20 sin a t — 871-€,,c3r c
c130 = E8 =
Compare with the linear quadrupole in problem 1, Chap. XII. 15. An electron moves in an axially symmetrical magnetostatic field such that
A,h (p, z)
0,
A2 -=
A p = 0,
Show that if its angular velocity about the axis at the point p = p1 and z = z1is (Pi and if at the point p = p2 and z = z2 it is co2, then z1) — P2ilp(P2,
— m402 =
z2)}
16. At t = to and z-directed particle of charge q, mass mo, and velocity vo enters at = 0, y = lb, z = 0 an evacuated retangular wave guide transmitting a TE01wave. Show that its equation of motion for an observer moving with the signal velocity is
d(mv') dt'
0301qEo
sin
(caoll')
where Eo is the maximum electric field intensity in the stationary coordinate system. Show that, if c, its position at any time tin this system is = qEo(c02m0)-1[0)(t — to ) cos wto — sin wt ± sin wto]
vo lt — to),
y = ib,
z= 0
References A. ELECTRICAL
ABRAHAM, M., and R. BECKER: "The Classical Theory of Electricity and Magnetism," (trans. 14th German ed.), Hafner, 1950. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. More extensive and detailed treatment than that given here.
590
SPECIAL RELATIVITY
J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives fairly extensive treatment. JELLEY, J. V.: "Cerenkov Radiation," Pergamon, 1958. Most extensive coverage of theory and experiment. JONES, D. S.: "The Theory of Electromagnetism," Macmillan, 1964. Treatment somewhat different from most in places. LORENTZ, H. A.: "The Theory of Electrons," 2d ed., Dover, 1952. Written by a participant in the classical development. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. A nonrelativistic treatment of the fields of moving charges. PANOFSRY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," 2d ed., Addison-Wesley, 1962. More detailed and extensive treatment than that given here. SOMMERFELD, A.: "Electrodynamics: Lectures on Theoretical Physics," Vol. III, Academic, 1952. A lucid treatment. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Treats the vector transformations in first chapter. TOLMAN, R. C.: "Relativity, Thermodynamics and Cosmology," Oxford, 1934. Gives clear and detailed treatment of special relativity. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Very concise and lucid treatment. ZWORYKIN, V. K., G. A. MORTON, E. G. RAMBERG, J. HILLIER, and A. W. VANCE: "Electron Paths and the Electron Microscope," Wiley, 1945. Gives details of path tracing of charged particles in electrostatic and magnetostatic fields. Extensive references to periodic literature. JEANS,
B. MATHEMATICAL
See Chap. V.
APPENDIX SYSTEMS OF ELECTRICAL UNITS There are three ways in which absolute systems of electrical units may be grouped. According to the mechanical units used, they belong either to the centimeter-gramsecond (cgs) system or to the meter-kilogram-second (mks) system. According to the magnitudes of the units, they belong to either the classical or the practical system. According to the way in which a 4,r factor enters, they belong to either the rationalized or the unrationalized system. The mks units used in this book belong to the practical rationalized mks group. The tables in this Appendix give the interrelations between these units, practical unrationalized cgs units, and classical unrationalized cgs units. Equivalent formulas in all the rationalized systems are very similar. But the way in which the 4r is inserted in defining the units in the Georgi mks system differs from the method used in the corresponding classical systems in that it is so arranged that the rationalization does not affect the common practical units such as the coulomb and volt. This is done by inserting a 47 in the capacitivity and the permeability. Since all numerical tables listing the electric and magnetic properties of materials give the relative capacitivity K = e/e, and the relative permeability Km = A/12,, it is often the practice, in the practical system, to write for these quantities Ke, and K„,11,, where e, and ti,„ are the values, for a vacuum, of the capacitivity, dielectric constant, or permittivity and of the permeability. If we are given a formula in one system of units, we can always write it in any other system by substituting in the given formula the values for the same quantities expressed in the desired units. To make this clear, let us, as an example, express Coulomb's law for the force between two charges in classical unrationalized cgs units. A given physical quantity, when expressed in cgs esu, will be written without subscripts and the same quantity, measured in Georgi mks units, with subscript 1. From Table II, qi = 3-110-9q and el = (3670-110-3e (1) From Table I, for mechanical units, we see that F1 = 10-5F and r1 = 10-2r The formula 1.01 (1) for Coulomb's law, which is in mks units, is F1 —
q iqi = glg1
47reirf
(2)
(3)
Substitution of F1, qi, el and r1from (1) and (2) and simplification give ,
F=
erg
(4)
The same procedure can, of course, be used for the conversion of magnetic formulas. An elaborate discussion of the latter is given by Kennelly in Am. Phil. Soc. Proc., Vol. 76, page 343 (1936). We notice that, if we put e equal to 1 in (3) and (4), the formulas are still different. Thus it is important, even if the result is to be applied to a vacuum, that the formula to be transposed, if in esu or emu, be written for a medium in which p and e are not unity. 591
APPENDIX
592
TABLE I
A. A formula given in cgs units is expressed in mks units by replacing each symbol by its value in the right-hand column. W joule F newton 1 meter m kilogram P watt
Energy Force Length Mass Power
10'W ergs 105F dynes 1021 centimeters 103m grams 107P ergs/sec
B. A formula given in rationalized mks units is expressed in cgs units by replacing each symbol by its value in the extreme right-hand column. Energy Force Length Mass Power
W erg F dyne 1 centimeter m gram P erg/sec
10-7W joule 10-5F newton 10-21 meter 10-3m kilogram 10-vP watt
TABLE II
A formula given in rationalized mks units is expressed in cgs esu or Gaussian units by replacing each symbol by its value in the extreme right-hand column. c 3 X 10'°. In very precise work each factor 3 is replaced by 2.9979. Quantity Capacitance Capacitivity Charge, quantity Conductance Conductivity, surface Conductivity, volume Current density, surface. Current density, volume Current Displacement Elastance Electromotance Field intensity, electric Impedance Inductance Potential, electric Reactance Resistance Resistivity, surface Resistivity, volume
Esu Emu
C
c-2C
e
c-2e
q G
c'q c-2G c 27 c-27
i I D S
C-12
c-1I c-1D c2S
E cE E Z L V X R s
cE c2Z c2L cV c2X c2R c2s
T
C2T
Practical 9-110-11C farad 9-110-11€ 47r-farad/cm (367)-110-8e farad/meter 3-110-9q coulomb 9-110-11G mho 9-110-117 mho 9-110-97 mho/meter 9-110-117 mho/cm 3-110-9i' amp/cm 3-110-7il amp/meter 3-110-5i amp/meter2 3-110-9i amp/cm2 3-110-9/ ampere 3-110-'D 47r-coul/crn2 (1270-110-5D coul/m2 9 X 10115 darafs 300E volts 300E volts/cm 30,000E volts/meter 9 X 10"Z ohms 9 X 10"L henrys 300V volts 9 X 1011X ohms 9 X 10"R ohms 9 X 1011s ohms 9 X lOgr ohm-meters 9 X 1011r ohm-cms
593
APPENDIX TABLE III
A formula given in rationalized mks units is expressed in cgs emu or Gaussian (starred) units by replacing each symbol with its value in the extreme right-hand column. c = 2.9979 X 1010 3 X 1010. Quantity Capacitance Charge, quantity Conductance Conductivity, surface . Conductivity, volume Current density, surface Current density, volume Current Elastance Electromotance Field intensity, electric Field intensity, magnetic Flux, magnetic Impedance Inductance Induction, magnetic Magnetic moment (dipole) Magnetic moment (loop) . Magnetization (dipole) Magnetization (loop) . . . Magnetomotance Permeability Pole strength, magnetic . . Potential, electric scalar.. Potential, magnetic vector Reactance Reluctance Resistance Resistivity, surface Resistivity, volume
Emu C q
G 7'
Esu
Practical
c2C* cq* c2G* c27'*
10°C farads 10q coulombs 109G mhos 1097' mhos 1097 mhos/cm 101'y mhos/meter 101' amps/cm 1031 amps/meter 101 amps/cm 2 105i amps/meter 2 101 amperes 10-9S daraf 10-86 volt 10-8E volt/cm 10-6E volt/meter H oersteds (103/470H amp-turns/m N maxwells 10-8N weber 10-9Z ohm 10-9/. henry 10-4B weber/meter 2 B gauss 47rm' maxwell-em 4710-19m' weber-m 10m ampere-cm 2 10-3M ampere-m2 47rM' maxwells/cm247r10-4M' webers/m2 10M amperes/cm 1000M amperes/m gilberts (10/47r) 12 amp-turns gauss/oersteds 4710-7/2 henry/m 47rm maxwells 47x10-8m weber 10-9V volt A gauss-cm 10-6A weber/meter 10-9X ohm R' gilb/max (109/47)R' amp-turns/web 10-9R ohm 10-9s ohm 10-ur ohm-meter 10-9T ohm-cm
c27* i'
i
I S
di * ci cI * c-2s*
c_ie * E c-1E* H* cH N* c-1N c-2Z* Z L c-2L* B* c-113 m'* c-lm' M* cM My' * c-1M' M* cM 0* c0 g* c-2g m* c-1m V c-1V* A* c-1A X c-2X* R'* c2IV R c-2R* c 2s* c 2r* E
594
APPENDIX
TABLE IV A formula given in cgs emu, cgs esu, or Gaussian units may be expressed in (a) unrationalized cgs practical units or (b) rationalized mks units by replacing each symbol with its value in the emu, esu, or starred column, respectively.
Quantity
Capacitance Capacitivity
(a) Practical cgs units (b) Rationalized mks units
C farad (a) e 4,r-farad/cm (b) e farad/meter Charge, quantity q coulomb Conductance G mho Conductivity, surface -y' mho Conductivity, volume (a) y mho/cm (b) y mho/meter Current I ampere Current density, surface (a) i' ampere/cm (b) i' amp/meter Current density, volume (a) i ampere/cm2 (b) i amp/meter' Displacement, electric (a) D 4ir-coul/cm2 (b) D coulomb /m2 Elastance S daraf Electromotance e volt Field intensity, electric (a) E volt/cm (b) E volt/meter Field intensity, magnetic (a) H oersted (b) H amp-turn/m Flux, magnetic (a) N maxwell (b) N weber Impedance Z ohm Inductance L henry Induction, magnetic (a) B gauss (b) B weber/meter2 Magnetic moment (dipole) (a) M' maxwell-cm (b) M' weber-meter Magnetic moment (loop) . . (a) M ampere-cm' (b) M amp-meter2 Magnetization (dipole) (a) M maxwell/cm2 (b) M weber/meter2 Magnetization (loop) (a) M ampere/cm (b) M amp/meter Magnetomotance (a) 0 gilbert (b) 0 ampere-turn Permeability (a) A gauss/oersted (b) /1 henry/meter Pole strength (a) m maxwell (b) m weber Potential, electric scalar V volt
Emu
10-9C 10-9e 4710-"e 10-1q 10-9G 10-9-y' 10-9-y 10-"-y 10-1/ 10-1/,' 10-3i' 10-1i 10-5i 10-1D 4r10-5D 1095 108e 108E 106E H*
Esu
9 X 10"C* 9 X 1011e* 367109e* 3 X 109q* 9 X 10"G* 9 X 10117'* 9 X 10"7* 9 X 1097* 3 X 1091* 3 X 109i' * 3 X 107i/* 3 X 109i* 3 X 106i* 3 X 109D* 127106D* 9-110-11S* (300)-1e* (300)-1E 3-110-4E 3 X 101°H
4r10-3H*
127r109H
N* 108N* 109Z 109/. B* 104B*
3-110-10N (300)--IN 9-110-11Z* 9-110-11L* 3-110-10B 3-110-6B
(47,-)-1M'* (470-11019W* 10-1M*
(1270-110-"M' (12r)-1/11' 3 X 109M
103M* (47)-1M* (47)-1104M* 10-1M* 10-3M* 0* 4710-10*
3 X 1013M (127)-110-10M (127)-110-6M 3 X 109M 3 X 107M 3 X 10190 127 X 109 0
A*
9-110-2°A
(470-1109ic*
(367)-110-18A (127)-110-19m (12007)-1m (300)-1V*
(47)-1m* (47)-1108m* 108V
595
APPENDIX TABLE
Quantity
IV.—(Continued)
(a) Practical cgs units (b) Rationalized mks units
Potential, magnetic vector (a) A gauss-cm (b) A weber/meter Reactance X ohm Reluctance (a) R' gilbert/maxwell (b) R' amp-t/weber Resistance R ohm Resistivity, surface s ohm Resistivity, volume (a) r ohm-cm (b) r ohm-meter
Emu
A* 106A* 106X
R'* 4710-6R'* 109R 109s 10'r 1011r
Esu
3-110-1°A 3-110-4A 9-110-11X* 9 X 102°R' 367 X 1011R' 9-110-11R* 9-110-11s* 9-110-11r* 9-110-9r*
596
APPENDIX TABLE V
This table gives the physical dimensions of electrical units. Length is 1, mass is m, time is t, charge is q, capacitivity is e, and permeability is A. Any column may be used in any system of units, but usually the first with e = 1 is used for esu, the second with A = 1 for emu, and the third for mks. Esu (e = 1)
Quantity Capacitance Capacitivity Charge, quantity Conductance Conductivity, surface Conductivity, volume Current Current density, surface Current density, volume Displacement, electric Elastance Electromotance Field intensity, electric Field intensity, magnetic Flux, magnetic Impedance Inductance Induction, magnetic Magnetic moment (dipole) Magnetic moment (loop) Magnetization (dipole) Magnetization (loop) Magnetomotance Permeability Pole strength, magnetic Potential, electric scalar Potential, magnetic vector Reactance Reluctance Resistance Resistivity, surface Resistivity, volume
C e q, Q G 7'
7
I i' i D S
6 E H N Z L B M'
M
Emu
(A = 1)
€1
11-11-1t 2
e eimiiit-1
/1-11-2t2
dt-I
Mks M-11-2i2q2 n2-11-3t2q2
At—imi/i A-11-1t A-11-1t A-11-2t ii-lmilit-1
q
m t-19-1
m-11-2N 2 m-11-2tq2 m-11-31q2 cl 014-2 t-1q Eimiiit-2 Arimit-it-' 1-it-lq eimi/-it-2 AA mil-it-' 1-2t-1q cimil-1t-1 1.4-4mil-1 1-2q mit-2 E-11-1 ml2t-2q-2 er-Imi/it-1 Almillt--2 ml 2t-217-1 c-i mi1-1t-1 111011e-2 mlt-2q-1 cl millt-2 A-I 701-1t-1 1-1t-1q c-1mill Ai 014-1 ml2r1q-1 e-11-1t Alt-1 ml2t-1q-2 c-11-42 m12q-2 ill E-im4 /-i I.L4m11-ir1 mt-1qi c-imiii A imiiiri ml3t-'g1 eimi lit-2 g-imi 15 1-1 12t-'q dr' et-1
M'
,-i mii-i
/26'11-1t-I
M
Eimilit-2
g-1mi1-1t-1 1-1t-1q p-imi/irl t-lq mlq-2 A
0
eiml/it-2
AL
,-11-212
m V
e-imili eAmi/it-1
s
e-imil-i e-11-1t elt-2 e-11-11 e--11-It
r
E-it
A X R' R
yimitit-1 Aimil4r2 Aimilit-' Alt-1 A-11-1 Alt-1 p4t-1 Al 2t-1
ml2r1q--1 ml2t-2q-, mlt-1q-1 ml2t-iq-2 m-11-2q 2 ml2t-1q-2 nzl2t-1q-2 ml2t-'g2
APPENDIX
597
TABLE VI
Numerical values for empty space in the rationalized mks system Velocity of light Capacitivity Permeability Intrinsic impedance. Intrinsic admittance
Give0-1= 2.99793 X 108 meters/second 8.85419 X 10-" (36r X 109)-7farad/meter 4r X 10-7= 1.25664 X 10-6 henry/meter 376.730 ohms 2.65442 X 10-3mho
Numerical values of atomic constants in the rationalized mks system * Faraday constant (016 = 16) Avogadro number (016 = 16) Planck constant Electronic charge Specific electronic charge
F = 96,487 ± 2 coulombs/gm equivalent N = (6.0225 ± 0.0002) X 10" 1/mole h = (6.6256 ± 0.0005) X 10-" joule sec e = (1.60210 ± 0.00007) X 10-13coulomb e/m = (1.75880 ± 0.00002) X 10" coulombs/kg
*ConEN, E. R., and J. W. M. DUMOND, NBS Tech. News Bul. October, 1963, page 175.
INDEX (Problem numbers are italicized and enclosed in parentheses with the number of the page on which they occur)
A
Antenna, loop, 469, 470, (11, 484) radiation resistance, 458, 462, 463, 470 Aberration, electromagnetic waves, 586, (See also Resistance) 587 slot, (59, 60, 491) Air gap, magnetic circuit, 313, 317-319 spherical, (18, 485) in ring magnet, 360-362 in wave guide, 499-502, 519 slotted pole piece, 317-319 (See also Wave guide) Airplane, ground speed, 583 Argument of complex number, 72 Ampere unit of current, 248, 280 Associated Legendre functions, 157-168, Ampere's experiment, 280-300 170-178, 211-215, 220-222 Anisotropic medium, axis electric, 22 applications, 163, 167, 168, 172-179, capacitivity, 21 220-222, (89, 233), (90, 234), (115, currents in, 268-270 116, 238), (17, 276), 283, 287, 292electromagnetic waves in, 420-423 294, (20-25, 322), (43-236), 335energy density in, 31 337, (12, 13, 345), 375, 378, 395, Laplace's equation in, 50 396-400, (2, 408), (5, 409), (11, 18, permeability of, 307, 350-352 410), (29, 412), (34, 413), 468, 472Antenna, 453-467 474, 481, (18, 19, 485), (39, 40, array, 463-466, (7-9, 483), (10, 484) 488), 513, 514, 533 biconical, 461-463, (18, 19, 485) argument imaginary, 159, 161, 170(See also Cavity resonator) 177, 211-216, 219, 220, 283, (21, 4Cavity, 543-545 22, 322) directivity, 465 large, 159, 171, 179 earth effects, 466 real, 158-178 gain, 464 biaxial harmonics, 164, 165, 399 input impedance, 448 differential equations of, 157 (See also Impedance, input) integral of product, 159, 161 linear, 453-458 Laplace's integral, 159 center driven, 456 nonintegral order, 165, 166, (118-121, charge density on, 453 239), (26, 277) current density on, 453 recurrence formulas for, 162, 163 fields, distant, 456-458 special values of, 163 near, 454, 455 tables of, 158, 159 half-wave multiple, 456, (6, 483) in terms of Bessel functions, 211, with losses, 457, 458 215 potentials of, 454 in terms of Legendre functions, 158 power radiation, 457, 459, 466 vector surface function, 161 progressive wave radiation, 458, 459 applications, 287, 534 radiation resistance, 458 Attenuation, constant of, 436 terminal load, 455 in hollow tubes, 495-498 transmitting pattern, 464, 465, (7, 8, (See also Wave guide) 483), (10, 484) 599
600
INDEX
B Band width, resonant cavity, 529 Barkhausen effect, 353 Barnett effect, 355 electrons in metals, 3 Ber and bei functions, 372-374, (20, 21, 411) Bessel equation, 180 solution of, 181, 182 Bessel functions, 180-187 addition theorem, 189 applications, 189-190, 192-195, (91-93, 234), (94-98, 235), 262, (27, 278), 286, (27, 30, 31, 323), (32, 33, 324), 378-380, 386-388, 406-408, (3, 408), 467-469, 476 (11, 484), (1416, 484), (17, 20, 485), (21-23, 486), (35, 487), (36, 42, 488), (4345, 489), (59, 491), 504, 515-521, 531, 534, 542, (7, 8, 546), (9, 10, 547), (27-29, 551), (32-37, 552), (38-41, 553), (67, 68, 558) argument infinite, 184 expansion in series of, 187 of first kind, 181 integral, definite, for, 192, 203 nonintegral order, 196 order negative, 196 order one-half, 196, 376-380, (2, 3, 408) (See also spherical, below) integrals of, 182, 184-186, 192, 193, 211, 215 recurrence formulas for, 183 of second kind, 180-184, 190, 191, 194, 195, 213 spherical, 196, 197, 208, 469-474, 481, 482, (18, 19, 485), (39, 40, 488), 534, (32-37, 552), (38-41, 553), (65, 557), (69, 558) vector surface, 186, 531 of zero order, 180-195 derivatives and integrals, 191, 192 graph of, 191 inverse distance, 189 roots and numerical values, 190 Bessel modified equation, 180, 197 solution of, 197
Bessel modified functions, 187-208 addition theorems, 204 applications, 199-201, 204-206, 217, 218, 221, 222, 261, 297, 316, 317, (26-29, 323), (14-16, 346), 372376, 403, 404, (2, 408), (19, 410), (22, 23, 411), (26-28, 30, 31, 412), (32, 34, 413), 469, 475, (22, 27, 486), (32, 487), (43-45, 554) argument complex, 199 infinite, 199 ber and bei functions, 372-374, (20, 21, 411) first kind, 197-202 integrals of, 198, 202, 204, 211 with complex argument, 199, 372396 inverse distance, 204 order, imaginary, 208 nonintegral, 207 one-half, 207, 208, 398-401 (See also spherical, below) zero, 202-206 definite integrals for, 202, 203 integrals of, 202, 204 recurrence formulas, 198 second kind, 197-199 in terms of first kind, 197, 207 spherical, 208, 468-474, 481, 482, (18, 19, 485) (See also Bessel functions, spherical; Hankel functions) Biot and Savart, law of, 295 Bipolar coordinates, (9, 61), (122, 239), (123-125, 240) Brewster's law, 428 C
Capacitance, of bowl, (41, 228) from conjugate functions, 75 (See also Conformal transformations, elliptic function) of cylinder, right circular, 209, 210 of cylinder and plane, 78 of cylinders, inside third, (33, 109) two circular, 76-78 of cylindrical capacitor, 28, 76-78 definition of, 26 of disks, circular and elliptic, 124 distributed, 27
601
INDEX Capacitance, of ellipsoid, 123 by inversion, 135 of iris in wave guide, 507, 508, (20, 549) mutual, 37-39, 129, 131, 132, 258 of parallel-plate capacitor, 29 with crystalline dielectric, 31, 32 with right-angle bend, 114 of ring-shaped conductor, (96, 235), (119, 239) self-, 37-39, 129, 131, 132, 258 in series and parallel, 26, 27 of sphere in cylinder, 218 and plane, 131 and plane dielectric, (36, 227) and two planes, (29, 226), (136, 137, 204) of spheres, intersecting, 135, (43, 45, 228) two, 27, 129, 131, 132 two connected, (27, 226) two in contact, (23, 226) of spherical capacitor, 27, 129, 132, 134, 152 per unit length, 75, 78, 101-105, (61, 62, 114), (63 65, 115), (70, 117); (79, 80, 118), (81 84, 119) (See also Conformal transformations) of variable air capacitor, edge correction, (26, 108), (35, 109) Capacitance coefficients, 37-39 (See also Self-capacitance) Capacitor, capacitance of, 26 (See also Capacitance) cylindrical, 28 off center, 76-78 definition of, 26 energy of charged, 30 guard rings, 29 parallel-plate, 29 with crystalline dielectric, 31 energy of, 31, 32 with right-angle bend, (60, 104) in series and parallel, 26, 27 spherical, 27 not concentric, 128-132, 152 variable air, edge correction, (26, 108), (35, 109) Cavity resonator, 526-545 with apertures, 535 band width, 529 -
-
Cavity resonator, biconical, (51, 555) complex, Hahn's method, 539, 540 coupling to, 541-545 circular cylindrical, 542, 543 input resistance, loop, 543 M between loop and mode, 543 M between stub and mode, 544 potential multiplication, loop, 543 potential multiplication, stub, 545 by orifice, 545 rectangular, (64, 65, 557) input impedance, loop, 543 input impedance, stub, 544 M between element and mode, 541-543 by semicircular loop, (67, 68, 558) by stub, 543-545 by transverse wire, (64 66, 557) by wire loop, 541, 542 cylindrical, general, 526-530 multiply connected, 532, 533 coaxial, 533 confocal elliptic, (52, 555) excentric circular, (53, 555) excitation, 541-545 (See also coupling to, above) fields of, 529-531, 534, 539 normal modes of, 526, 527, 529-531, 533-535 damping of, 527, 528 independence of, 527 parallelogrammatic, (49, 50, 554) quality factor Q, 528, 530-534, 540545, (47, 48, 554), (51 53, 555) rectangular, 531 with iris, (54 61, 557) reentrant, 538-540, (63, 557) resonance frequency, 526, 530-532, 534, (47, 48, 554), (54, 555) (58, 556) apertures, effect of, 534, 535 damping, effect of, 527-529 dielectric sphere, effect of, 536 foreign body, effect of, 535, 536 stationary formulas for, 537, 538 wall deformation, effect of, 537 spherical, 533, 534, (69, 70, 558) spheroidal, 538 triangular, (47, 48, 554) wave-length, resonant, 530 (See also resonance frequency, above) -
-
-
602
INDEX
Cerenkov radiation, 579-581 Charge, 2, 3 accelerated, normal to velocity, 577579 parallel to velocity, 575-576 radiation, 573-579 on capacitor, 26 (See also Capacitance) collinear, 12, 13 in conductor, 3 dissipation of, 431 on conductor, by induction, 2, 36, 55 (See also Green's function) cylindrical, 36 by Green's reciprocation theorem, 36 plane, 37 spherical, 36 conservation of, Lorentz transformation, 566, 567 distribution of, uniqueness, 25, 57 elementary, 3 mass of, 3 line, near conducting sphere, (133, 242) near dielectric sphere, (134, 243) near dielectric wedge, 70-72 image laws for, 69, 70 (See also Images) near plane with slit, 92 between planes, 84, 85 potential of, circular harmonics, 65 conjugate functions, 76 logarithmic, 63 moving, in dielectric, radiation, 579581 in magnetic field, energy, 567-570 force, 566, 567 in magnetron, 569-570 in uniform crossed electric and magnetic fields, 584-586 in uniform magnetic field, path, 568 with retarding force, 570 point, 4 (See also Green's function; Images; Inversion; Potential) positive and negative, 1 potential of, 4 (See also Potential) space, 272-274 unit of, 2 Charge density, 3 on antenna, 453
Charge density, on bowl, (42, 228) on capacitor, parallel plate, 29 spherical, off center, 152 on conductor, 18 on cylinder, right circular, 209-211 on disk, charged, 124 in longitudinal field, 172 in double layer, 14, 57, 58 on ellipsoid, 124 in equivalent stratum, 57, 58 inversion of, 135 on plane by point charge, 125 in problems, (13, 106), (14, 107), (30, 108), (45, 111), (6, 12, 14, 224), (16, 225), (39, 227), (42, 47, 228), (62, 230), (68, 69, 72, 231) on sheet with circular hole, 172 with cycloidal corrugations, 80 with slit, 92 on sphere, in cylinder, 209, 210 by point charge, 126 uniqueness of, 25 Child's equation, 272-274 Circuit, force, magnetic, between, 299, 300, 331, 332, 342 on, in magnetic field, 299, 300 image of, 306 magnetic, 312, 313 air gap in, 313, 314 anchor ring, 313 flux in, 313 magnetomotive force in, 313 reluctance of, 313 magnetic field of, 280-327 (See also Magnetic induction; Vector, potential) Circuit energy, dissipation of a minimum, 252, 253 mutual, 330, 331, 342 (See also Mutual inductance) of set of n, 342 Circular current loops, coaxial, forces between, 301, 302 magnetic field of, 290, 291, 294, 295 mutual inductance of, 335 not coaxial, 335, 336 screening by plane sheet, 485, 486 self inductance of, 341, 342 variable current radiation, 469, 470 vector potential of, 290, 291, 294
INDEX Circular harmonics, 63-69 (See also Harmonics) Clausius-Mossotti formula, 33 Coercive force, 354 Complex numbers, 72, 73 Conductivity, 249 Conductor, boundary conditions, current, 251 electrostatic, 17, 18, 35 capacitance of, 26 (See also Capacitance) charges in, 3 charges on, 17 (See also Charge; Charge density) definition of, 1 distant, capacitances of, 38 electromagnetic waves in, 431-433 (See also Electromagnetic waves) energy of system of charged, 39 extended, current distribution in, 251 (See also Current) in field, effect of insertion, 57 force on charged, 18, 34, 39, 40 stress on surface of, 18, 34 torque on charged, 39, 40 (See also Torque) Cone, Green's function for, 166-168 Confocal conicoids, 122, 123 Confocal coordinates, 95, 168-179 Confocal cylinders, elliptic, 90, 91 hyperbolic, 90, 91 parabolic, 84 Confocal ellipsoids, 123, 124 Confocal hyperboloids, 122, 123, 168-179 Confocal spheroids, oblate, 168-177 prolate, 177-179 Conformal transformations, 72-120 for airfoil, 94 applications, current flow, 253-257, (317, 274), (8-14, 275), (15, 16, 276) electric, 75-105, (14, 107 to 49, 112), (52, 112 to 84, 119) electromagnetic waves, 440, 441, (13, 14, 16, 444), (17-25, 445), (26-34, 446), 507, 509, (20, 21, 549), (23, 550) magnetic, 317-319, (35, 324), (37, 38, 325) boundaries parametric, 79, 80 for current, 253-257, 264-266
603
Conformal transformations, for cycloidal corrugated sheet, 80 for cylinder, circular to elliptic, 94 for cylinders, two, 76-78, 81, 82 for cylindrical lens, (72, 73, 117) determination of required, 80, 81 for ellipses, confocal, 90, 91, 94 elliptic function, 100-104 for bar, charged rectangular, (67, 116) between parallel plates, (82, 119) for charge, line, in rectangular prism, (68, 116) in triangular prism, (71, 117) for cylinder, centered in hyperbolas, (79, 118) centered in slit, 101, 102 off center in slit, 103, 104 at parabola focus, (80, 118) between parallel plates, 104, 105 near plane edge, 103, 104 for slit in front of slit, (69, 116) for strip, flat, 91 in cylinder, coaxial, 101, 102 in cylinder, off center, 103, 104 edge on parabola axis, (81, 119) in front of slit, (70, 116) in front of wedge, (83, 119) near plane edge, coplanar, 103 near plane edge, normal, (64, 115) between planes, normal, (63, 115) between planes, parallel, (61, 114) in slit, coaxial, coplanar, 100, 101 in slit, coaxial, normal, (62, 114) in slit, off center, coplanar, 103, 105 for strip pair, equal coplanar, 100, 101 on opposed rectangle faces, (66, 118) unequal coplanar, 103, 104 unequal curved on cylinder, (65, 115) for grating, 86, 91, 98-100, (22, 107) (See also Uniform field) for hyperbolas, confocal, 90, 91 equilateral, 84 for parabolas, confocal, 84 for parametric boundaries, 79, 80 for plane, half into whole with slit, 92
604
INDEX
Conformal transformations, for plate, with circular ridge, (49, 112) with groove, (49, 112) with sharp ridge, (57, 113) for polygon with rounded corners, 98 Schwarz transformation, 82-105 for angle negative, 86, 87, 92 for angle positive, 84 for angle zero, 85 for angles, two, 89-91 doublet, 87 inversion, 87, 88 plane, half into strip, 85, 86, 91 on Riemann surface, 93 for wedge, 84 strip, into bipolar circles, 81 flat, 91, 94 for strips, radial, (58, 114) (See also Conjugate functions) Conical boundaries, 152, 156, 165-169 Conical box, Green's function, 167, 168 Conical coordinates, (4, 61) Conicoids, confocal, 122, 123 Conjugate functions, 72-105, 254-257, 441, 509 applications in problems, (17, 107 to 84, 120), (4, 274 to 16, 276), (35, 36, 324), (37, 38, 325), (13, 14, 16, 445), (26-34, 446), 507, 509, (20, 21, 549), (23, 550) capacitance from, 75 for current flow, 254-257 in strip of variable width, 255-257 for cylinder, elliptic, 90, 94 elliptic dielectric, 95-98 for cylinders, two circular, 75, 76 determination of required, 80, 81 doublet, line charge, 87 elliptic, 100-105 for grating, of flat strips, 91 of round wires, 99-100 of small wires, 86 for inversion, 87, 88 for line charge, 76 near plane with slit, 92, 93 between planes, 86 in rectangular prism, (68, 116) in triangular prism, (71, 117) for magnetic field, 317-319 under slotted pole piece, 317-319 orthoganility of, 74
Conjugate functions, for plane with slit, 92, 93 resistance from, 254, 255 on Riemann surface, 93 for strip charged, 90, 91 for transmission line, bifilar, 441 for wedge, 84 (See also Conformal transformations) Continuity equation, 248, 263-265, 269, 415, 452, 453 in anisotropic medium, 268-270 in spherical sheet, 264 in surface of revolution, 265, 266 in thin curved sheet, 263 in three dimensions, 248-253, 257, 258, 262, 263, 268, 269 in two dimensions, 253-257 Coulomb gauge, 417 Coulomb's law, 2, 15, 16 limitations, 2 Crystal, electric properties of, 21 (See also Anisotropic medium) Curie's law, 350 Curl, 50, 52 in cylindrical coordinates, (1, 60) in orthogonal coordinates, 52 in spherical coordinates, (2, 60) Stokes' theorem, 49 vector Green's theorem, 54 Current, in anisotropic mediums, 268-270 in antenna, 453-456 (See also Antenna) boundary conditions, 251, 252, 257, 258, 262 continuity of, 248, 263-266, 268, 269 density of, 247, 250, 254 displacement, 368, 415 distribution, in extended mediums, 250-263 general theorems on, 252, 253 magnetic field effect on, 251, 252 in plate with spherical cavity, (33, 278) solid cylinder, 261 in sphere, solid, 259, 260 in spherical shell, 264, 265 in stratified earth, 262, 263 in strip of variable width, 254-257 in surface of revolution, 265-267 in thin curved sheet, 263-267 eddy, 368-414
605
INDEX Current, in extended mediums, 250-263, 368-414 (See also Eddy currents) Foucault, 368-414 (See also Eddy currents) heating effect of, 252, 253 image in permeable surface, 306, 309 magnetic field of, 280-327 (See also Magnetic induction; Vector, potential) shell, for cylinder, 356, 357 equivalent of magnet, 355, 356, 387, 393, 399, 400 for horseshoe, 358, 359 for sphere, 356 space-charge, 272, 273 three-dimensional, 250-253, 258-263, (1, 274), (17-19, 276), (21-23, 26, 277), (27-38, 278) two-dimensional, 253-257, 307-310, (2-7, 274), (9-14, 275), (15, 16, 20, 276) unit of, 228, 280 vector potential of, 283 Curvilinear coordinates, 50-52 (See also Orthogonal coordinates) Cylinder, current distribution, 261, 262 eddy currents in, 371-374 (See also Eddy currents) electromagnetic waves in, 434, 435, 440, 441 (See also Electromagnetic waves) in electrostatics, 28, 36, 63-120, 187, 195, 196, 209-211, 217-223, (95, 235) enclosing disk, 218 enclosing sphere, 217, 218, 220-233 enclosing spheroid, 218 in hyperboloid, 219, 220 solid right circular, 209-211 (See also Capacitance; Conformal transformation; Green's function; Images; Uniform field) uniformly magnetized, 356, 357 Cylindrical box, Green's function, 189, 190 Cylindrical coordinates, curl, (1, 60) Laplace's equation, 52, 53, 179 solution of, 180 (See also Bessel functions; Bessel modified functions) vector potential forms in, 286, 287
Cylindrical current shell field, 299 Cylindrical dielectric boundaries, 66-69, 95-98, 204-206 Cylindrical harmonics, 179 Cylindrical ring, potential inside, 194, 195, 206, 207
D Damping, by eddy currents, 405 of eddy currents, 368-420 Diamagnetic susceptibility, 350, 351 (See also Susceptibility) Diamagnetism, 350 Dielectric boundary conditions, 18-20, 94, 95 Dielectric constant, 2, 4, 14, 17 in crystals, 21 increase of, effect, 59, 60 stress dependence in liquid, 33 Difference equations, 130 Dipole, current source in sphere, 259, 260 electric, 5-7, 10 near dielectric sphere, (115, 116, 238) forces on, 6 interaction of, 6, 7 line, 87 oscillator, radiation from, 449-452 in wave guide, 498-502, 517, 518, (27, 551) (See also Wave guide) between planes, (127, 240) potential, 6 torque on, 6 magnetic, 280, 290, 362, 451 classical moment, definition of, 362 interactions of, 363 loop moment, definition of, 280, 290 oscillating, radiation from, 451, 469, 470 rotating, eddy currents from, 382 vector potential of, 290, 363 Disk, circular, capacitance of, 124 eddy currents in, (24, 411) and point charge, (85, 233) potential of, 124 in Bessel functions, (120, 236) and ring charge, (87, 233) rotating eddy currents, 390-393
606
INDEX
Disk, circular, in sphere, (130, 241) surface density on, 124 in uniform field, 173, 174 elliptic, capacitance of, 124 surface density on, 124 Displacement, 14, 20, 21 (See also Electric displacement) Divergence, 50 in curvilinear coordinates, 51 Divergence theorem, 48, 49 (See also Gauss's theorem) Doppler effect, 586, 587 Double current sheet, 14, 58 Double layer, electric, 14, 58 magnetic, 281, 311
E Earnshaw's theorem, 13 Earth, current, in anisotropic, 268-270 resisting cone, (26, 277) in stratified, 262, 263 effect on antenna, 466 Eddy currents, 368-414 in block, solid, plane face, 370 skin depth, 370 current density, 368, 370, 372, 378, 380, 396, 398, 400, 401, 403, 405, (7, 409), (18, 410), (24, 25, 411), (26, 30, 412) in cylinder, longitudinal ac field, (4, 409), (19, 410) (20, 411) solid, 372-374 transverse ac field (21, 411) in cylindrical box, (22, 23, 411) in cylindrical coordinates, 372 damping of motion by, 405 in disk, rotating in magnet gap, 390-393 uniform ac field, (24, 25, 411) force, (5, 9, 409) inductance effect, (22, 411), (28, 29, 31, 412) power loss, 369, 371, 373, 377, 378, 396, 400, 402, 403, 405, (13, 410), (20, 23, 411), (34, 413), (35, 414) scalar potential, 387, 388, 390, 403, 404 in sheet, thin plane, 380-390 ac current loop, 385, 386, (6, 409), (33, 413) dipole rotation, 382-385, (32, 413)
Eddy currents, in sheet, thin plane, image method, 382, 383, 390, 393 magnet rotation, 387-390 rotating in magnet gap, 387, 390 shielding effect, 381, 382, (6, 409) in sphere, solid, 374-380 short ac solenoid, (34, 413) uniform ac field, 375-378 uniform transient field, 378-380 in spherical coordinates, 374, 375 in spherical shell, thin, 393-396 in ac field general, 399, 400 in ac field zonal, 393, 396 in ac solenoid gap, 396, 397 spinning between poles, 399, 400 spinning in uniform field, (12, 13, 14, 410) in spheroidal shell, (18, 410) stream function of, 384, 388, 389, 391, 403, (32, 413) torque from, 404-406, 389, 390, 392, 393, 399, 400, 405, (12, 16, 410), (21, 411), (32, 413) in tube, thick, ac, 371, 372 transverse transient, 405-408 in tube, thin, 400-405, (26, 27, 412) by ac field rotating, (16, 410) by ac field, transverse, (15, 410) by rotating bar magnet, 402-405 shielding of bifilar lead, (17, 410) Einstein and de Haas effect, 355 Elastance, mutual, (36-39, 258) self, (36-39, 258) Electric charge, 3 dissipation of, in conductor, 431, 432 (See also Charge) Electric dipole, 5 7, 10 (See also Dipole) Electric displacement, 14, 15, 18, 20, 21, 415, 416 boundary conditions, 18, 19 in crystals, 21, 22, 420, 421 in plane wave, 419-421 Electric field, axially symmetrical, 60 in cavity resonator, 529, 531, 534 (See also Cavity resonator) in Coulomb gauge, 416, 418 dielectric body in, 58, 59 energy density in, 31 from Hertz vector, electric, 418 magnetic, 418 -
INDEX Electric field, in Lorentz gauge, 417 Lorentz transformation of, 582 of moving charge, 572, 573 accelerated, 573, 574 in dielectric, 579-581 neutral points and lines, 163 propagation equation, 416, 417 (See also Electromagnetic waves) in conductor, 431, 432 screening of, 37, 38 stream function of, 74, 75, 94 stresses in, 15-17 superposition of, 35, 69, 86, 91, 9599, 100, 105-107, 127, 149, 173, 189, 205, 215-223 in two dimensions, 63-120 uniform, 6, 28, 30, 66, 91, 92, 95 (See also Uniform field) in wave guides, 495 (See also Wave guide) Electric flux, 11, 74, 75 Electric intensity, 3, 20, 74, 75 in anisotropic medium, 21, 31, 420423 boundary conditions, 18, 20, 94, 95 curl of, 329, 368, 416, 419 from electrodynamic potentials, 416, 417 from Hertz vectors, 418 in plane wave, 418-421 in anisotropic medium, 420-423 in radiation fields, 415-418 (See also Electromagnetic waves) from vector potential, quasi-magnetostatic, 330-368 Electric multipole, 5, 6, 449-452 Electric screening, 37, 38 Electromagnetic induction, 329-347, 368-414 Electromagnetic waves, 415-459 aberration, 586, 587 in anisotropic medium, 420-423 polarization, 423 ray, ordinary and extraordinary, 422 ray surface, 422 wave surface, 421 in cavities, 526-545, (47, 554 to 73, 559) (See also Cavity resonator) from charge in motion, 571-581 (See also Electric field)
607
Electromagnetic waves, from charge in motion, accelerated, linearly, 575, 576 by magnetic field, 577-579 in dielectric, 579-581 in conductors, 431, 432 in cylindrical coordinates, 474-476 plane wave, 475, 476 between planes (32-35, 552) diffraction, 476-482 from aperture, plane, 476-478 in spherical shell, (39, 40, 488) in thin plane sheet, 476-480 annular hole, (35, 487), (36, 488) circular hole, from loop, (41, 488) slit, (42, 488),(43-46, 489) small hole, (48, 49, 489), (5053, 490) square hole, 479, 480, (34, 487) from cylinder, dielectric, (17, 485) metal, (14-16, 484) Kirchhoff's formulas, (56-60, 491) circular hole, (59, 60, 491) orthogonal functions in, 480-482 from sphere, dielectric, 471-474 metal, 471-474 small, (13, 484) from dipole, 449, 450 Doppler effect, 586, 587 double current sheet, 477, 478 elliptic polarization, 428, 429 from end of coaxial line, 481, 482 into wave guide, 502-504, 520, 521 energy, density in, 423, 424 propagation direction, 419, 420, 422 frequency, 428 group velocity, 441, 442 in hollow tubes, 493-525 (See also Wave guide) intensity of, 418 momentum of, 423 phase constant, 436 plane, 415-447 attenuation constant, 436 on bifilar line, 440, 441 characteristic impedance, 437 in conductor, 431, 432 on cylinder, conducting, 434, 435 from f(t) at z = 0, 435 from initial field at z = 0, 435
608
INDEX
Electromagnetic waves, plane, on cylinder, conducting, with lossy conductor, 440, 441 with lossy dielectric, 436, 437 dielectric, (20, 24, 445), (25, 29, 505) dielectric coated, (32, 446) in cylindrical coordinates, 474-476 impedance relations, 436-438 intrinsic impedance, 436, 437 on lecher wire system, 440-442 matching at discontinuity, 438 on plane with dielectric layer, (7, 8, 443), (30, 31, 487) polarization of, 420, 428, 429 polarizing angle, 428 propagation constant, 437, 438, (10, 444 to 33, 446) reflection of, 424-427 from conductor, 433 from dielectric plate, (3, 442) from discontinuity, 438 total, 430 refraction of, 424-428 in spherical coordinates, 467, 468 on transmission line, 434-436, 440, 441 Poynting's vector, 418, 439 pressure of, 424 propagation equation, 416, 417 in conductor, 431, 432 cylindrical solution, 474-476 for plane wave, 475, 476 spherical solution, 467, 468 for plane wave, 468, 469 from quadrupole, 450-452 signal velocity, 442 from slot antenna, (59, 60, 491) from sphere oscillating, 471, 474 spherical, 468-474, 480-482 from antennas, 453-466 (See also Antenna) from current element, 449, 450 from dipole, electric, 449, 450 magnetic, 451 rotating, (5, 483) Hertz vector, 417, 418, (12, 444), 450-452 from loop, uniform, 469, 470 from multipole, 449-452
Electromagnetic waves, spherical, potential, scalar retarded, 452 vector retarded, 452 on transmission lines, 434-436, 440, 441, (13, 444 to 33, 446) uniqueness theorem, 467 wave length, 428 wave number, 436 Electromotance, definition of, 249-251 (See also Potential) distributed, 250 in electromagnetic wave, 415 (See also Electromagnetic waves) gradient of, 250, 251 induction of, 329, 415 in mutual inductance, 333 in self inductance, 338 Electron, 3 (See also Charge) inertia, 3 in metals, 3 negative and positive, 3 space charge of, 268-270 spin of, 355 Electrostriction, 33 Ellipsoid, charged, 123, 124 Ellipsoidal coordinates, 122 (See also Potential) Elliptic cylinder coordinates, (5, 61) Energy, and capacitivity, 59 of capacitor, charged, 30 of cavity resonator, 528, 529 (See also Cavity resonator) of charge, moving electric, 568, 569 of conductors, charged, 30, 40 conservation, 425, 430, 566, 582 density, in electric field, 30, 31 in electromagnetic wave, 448 in anisotropic medium, 420 in magnetic field, 332, 333 of dielectric body in field, 58, 59 dissipated as heat, 248, 253 (See also Eddy currents) dissipation of a minimum, 253 in inductance, 333, 337, 339, 342 loss in conductor, 248, 250, 253, 369 mass relation, 568, 569 mutual, of electric dipoles, 7 of magnetic needles, 363 of n circuits, 342 of two circuits, 330-332
INDEX Energy, of permeable body in field, 343, 344 potential in circuit, 30 propagation of, 440, 525 electromagnetic waves, 418-559, 581, 586, 587 (See also Electromagnetic waves) in wave guides, 493-525 (See also Wave guide) radiated, from accelerated charge, 575, 579 from oscillator, 451, 452 (See also Electromagnetic waves) Equipotential surfaces, 9, 10, 121 maps of, 9, 10, 91, 92, 205 (See also Potential) Equivalent stratum, Green's, 57, 58
F Faraday's law of induction, 329, 330, 368, 415 in extended mediums, 368, 415 in Maxwell's equations, 415 in surfaces, 394, 398, 400 Flux, electric, 10-12, 74, 75 magnetic, 289 in air gap, 313, 314, 317-319 in aneh or ring, 312, 313 changing, 329 leakage in transformer, 314-317 and mutual inductance, 333 and self inductance, 337, 338 unit of, weber, 312 Flux density, magnetic, 281 (See also Magnetic induction) Force, on charged conductors, 39, 129 on charged spheres, 129 coercive, 354 on diamagnetic bodies, 344, 349 on dielectric bodies, 58, 59 on dipoles, 7 from eddy currents, 383, 384, (5, 9, 10, 409), (12, 16, 410), (21, 411) on electric circuits, 299-302, 331, 332, (5-7, 320), (10-13, 15, 321), (8, 345) on fibers, electric, (6, 7, 320), (12, 15, 321) from Green's function, 55
609
Force, lines of, 8 (See also Lines of force) magnetic, on boundary, 343 between magnetic needles, 363 magnetizing, 352 on moving charges, 566-568, 570, 582 on paramagnetic bodies, 349 on permanent magnets, 357-359, 363 of point charge on tangent spheres, 134 between sphere and plane, 131 transformation equations for, 565, 566 on two spheres, charged, 129 uncharged in field, (124, 125, 240) Foucault currents, 368-414 (See also Eddy currents) Fourier-Bessel integral, 193 Fourier series, 65, 206 (See also Bessel modified functions; Harmonics) in cavity fields, 526, 529-532 in wave guides, 497-502 Frequency, 428 characteristic of wave guide, 494, 495 (See also Wave guide) Doppler effect on, 586, 587 of electromagnetic waves, 428 (See also Electromagnetic waves) resonant, of cavity, 526, 530-532, 534, (47, 48, 554), (54, 555), (58, 556) (See also Cavity resonator) of sphere, metal, 470, 471 Fresnel's equation, 421
G Gauge, Coulomb and Lorentz, 417 Gaussian units, 591-595 Gauss's electric flux theorem, homogeneous mediums, 11-13, 71, 141, 188, 200 inhomogeneous mediums, 17 Gauss's theorem, 48, 49, 56, 248, 302, 332 General theorems, for currents, 252, 253 for electrostatics, 48-60 Georgi units, 591-595 Gradient, curvilinear coordinates, 51 Grating, charged, 86, 90, 91, 98-100 (See also Uniform field)
610
INDEX
Green's equivalent stratum, 57, 58 Green's function, 55 for box, conical, 168 cylindrical, 189, 207 rectangular, (114, 238) for cone, 166, 167 for cylinder, 188, 189 for disk, (85, 233) for prism, rectangular, (113, 238) for ring, hollow cylindrical, 199-201 for sphere, 126 for wave guide, 499-502, 517-518 Green's reciprocation theorem, 34, 35, 54 with dielectrics, 54 Green's theorem, 48, 53-55, 141 vector analogue, 54 Guard rings, 29 Gyromagnetic effects, 354, 355 H Hall effect, longitudinal, 251 transverse, 252 Hankel functions, applications, (21-26, 486), (30, 31, 487), (32, 552), (39, 41, 553), (44, 554) (See also Bessel functions, spherical) definite integrals for, 202 definition of, 182, 198 spherical, 208 value at infinity, 184 Harmonics, biaxial, 164, 165, 399 circular, 63-69 for circuit, bifilar, 308, 309 for current in cylindrical shell, 299 for line charge, 65 cylindrical, 63, 137 into spherical, 211, 213, 215 into spheroidal, 211, 213 (See also Bessel functions) spherical, 137, 139-179 into cylindrical, 211, 213, 215 into spheroidal, 212, 213 (See also Spherical harmonics) spheroidal, 137, 168-179 into cylindrical, 211, 213, 215 into different spheroidal, 212-214 into spherical, 212, 213 (See also Spheroidal harmonics)
Harmonics, surface, 141, 168 vector, 161, 287, 398 (See also Surface harmonics) zonal, 142 (See also Surface harmonics) Hertz vector, 417, 418, 450-452, 586 aberration, 586 Doppler effect, 586 electric, 418 magnetic, 418, 511 propagation equation, 417 retarded, 452 Horn, sectorial, (39-42, 553), (45, 46, 554) Hysteresis, 353, 354
I Images, of charges, line, in conductors, 68-70, 88, 89 in dielectrics, 68-70 point, in conductors, 124-131, 133, 134, 189, 194 in dielectrics, 126, 127 of current circuits, 306, 309, 310 in cylinder, circular, conducting, 69, 70 dielectric, 68, 69 intersecting, 89 in dielectric plate, 194 of eddy currents in plane, 381-385, 390 by inversion, 89, 134, 135 in planes, intersecting, 70, 89 in sphere, conducting, 126 in spheres, two, in contact, 134 intersecting, 135 separated, 128-131 in wave guides, 502-504, 519, (6-8, 546), (9-11, 547), (13, 14, 16, 548), (41, 553) Impedance, characteristic, 437 conical transmission line, 460 transmission line, 437, 440, (13, 444 to 33, 446) (See also Conformal transformations, elliptic function) of wave guide, 494 normalized, 507 (See also Wave guide) input, biconical antenna, 462, (19, 485)
INDEX Impedance, input, cylindrical guide, 507 from coaxial line, 521 electrode in rectangular guide, (6-8, 546), (9, 10, 547) from coaxial line, 504 parallel plate line, (33, 552) sectorial horn, (40, 41, 553), (45, 46, 554) intrinsic, of medium, 436 output, 438 phasor, 436-438 Induced charge, 2 (See also Charge) Inductance, 333-342 mutual, of cavity mode, 541 of coils, coaxial, 335, (21, 347) not coaxial, (28, 347), (33, 34, 348) definition and units of, 333 of loops, coaxial, 335 coplanar, (4, 344) not coaxial, 335, 336, (23, 25, 347), (31, 32, 348) around permeable cylinder, (14, 345) of solenoid, infinite and loop, 335 of toroidal coil linking loop, 335 variable, 336, 337 self-, 337 of bifilar lead, rigorous, 340-342 shielded, (10, 345) of circular loop, 339, 340 on permeable cylinder, (15, 346) of cylinder, solid, skin effect, 374 definition and units of, 337 high-frequency, of cylinder, 374 of solenoid, 340 of thin wire, 338, 339 from vector potential, 337, 341 Induction, coefficient of, 36-38 (See also Mutual capacitance) electric, 2 (See also Charge; Green's function) electromagnetic, in extended mediums, 368-414 in linear circuits, 329-348 Faraday's law of, 329, 338 magnetic, 281, 282 (See also Magnetic induction) Inductive coupling, 541-543
611
Insulator, 1 (See also Dielectric boundary conditions) Integral equation, 510 Intensity, electric field, 3, 20, 74, 75 in anisotropic medium, 20, 22, 31 boundary conditions, 18-20, 94, 95 in dielectric, 20 (See also Potential) magnetic field, 311 boundary conditions, 311 unit of, 311 (See also Magnetic induction) Intrinsic impedance, 436 Inverse distance, 4 (See also Potential) Inverse-square law, 2 limitations, 2 (See also Coulomb's law) Inversion, 87-89, 132-135 in three dimensions, 132 capacitance by, 135 geometrical properties, 132 of images, 132-134 of planes, intersecting, 135 of spheres, intersecting, 135 of surface, charged conducting, 134, 135 in two dimensions, 87-89 images by, 88-89 unequal coplanar strips, 103, 104 J Joule's law, 250 K Ker and kei functions, 372 diffraction theory, (56-60, 491) L Laplace's equation, 50-53, 64, 73, 136, 139, 169, 178, 179, 251, 522 current applications, 251 in cylindrical coordinates, 53, 179 for nonhomogeneous dielectric, 50-53 for nonisotropic dielectric, 50 in orthogonal coordinates, 51
612
INDEX
Laplace's equation, in rectangular coordinates, 50 for rotated conjugate functions, 136 in spherical coordinates, 52, 137, 139 in spheroidal coordinates, 137, 169, 178 oblate, 137, 169 prolate, 137, 178 in three dimensions, solutions, 137-244 in two dimensions, solutions, 63-120 Laplace's operator, 50-53 applied to vector, 282, 285-287, 368370, 374, 375, 378, 416, 417, 431, 467, 493, 592 (See also Laplace's equation) Legendre coefficient, 145 (See also Legendre polynomials) Legendre equation, 142 solution of, in series, 143 recurrence formulas for, 143, 144 Legendre functions, 152-179 associated, 158, 211-214, 220-222 (See also Associated Legendre functions) eddy currents, 395-400, (29, 412), (33, 34, 413) first kind, 158 hypergeometric series for, 157 order complex, 165 order fractional, 157 second kind, 152-156 with imaginary argument, 155, 156 potential applications, 156, 220, 223 recurrence formulas for, 153, 162, 163 series for, argument exceeds one, 153 argument less than one, 154, 155 special values for, 155 in terms of polynomials, 154, 155 Legendre polynomials, 144-152, (57, 229 to 79, 232), (123, 128, 240), (129, 130, 241), (132, 133, 242), (136, 243), 259, 289, 292-295, 301, 302, 335-337, (19, 548), (22, 24, 550) argument imaginary, 148 imaginary and infinite, 148 real and infinite, 145 derivative of, 146 expansion in terms of, 147 graph of, 148 intergrals of, 146, 147
Legendre polynomials, Mehler's integral, 512, 514 recurrence formulas for, 146 Rodriquez formula for, 145 table of, 147, 148 Lienard-Wiechert potentials, 452, 573 Lines of force, electric, between charges, 8 of charges, collinear, 12 at conducting boundary, 18 at dielectric boundary, 20 differential equation of, 7, 8 of dipole, 9, 10 at infinity, 13 maps of, 9, 10, 67, 68, 77, 81, 86, 91, 92, 96, 205 stresses in, 15-17 magnetic, 280, 281 maps of, 315, 361 (See also Magnetic induction) Logarithmic potential, 63-65, 76 Lorentz guage, 417 transformation, 560-562, 564-566, 581, 582
M Magnet, cylindrical, 357, 362, 363 eddy currents by, 386-393, 399, 402405 equivalent current shell, 355, 356 horseshoe, 358-360 needle, 357, 362, 363 permanent, 354-363, 386-393, 397, 400, 402-405 boundary conditions, 357 ring, with air gap, 360, 362 spherical, 357, 358 in permeable medium, 358 Magnetic circuit, 312, 313 Magnetic cutoff of thermionic rectifier, 569, 570 Magnetic dipole, 280, 290, 362, 363, 451, 469, 470 interactions of, 363 moment definition of, classical, 363 loop, 280, 290 in wave guide, (4, 5, 546), (28, 29, 551) (See also Wave guide) Magnetic field, energy density, 332, 333 intensity, 311, 312, 354, 355
INDEX Magnetic field, intensity, boundary conditions, 311 unit of, oersted, 311 (See also Magnetic induction) maps of, 315 potential of, scalar, 311, 312 (See also Magnetomotance) stresses in, 343 Magnetic flux, tubes of, 289 (See also Flux) Magnetic force, on boundaries, 343 on circuits, 299, 300, 334, 342 Magnetic hysteresis, 353, 354 Magnetic inductance, 333-342 (See also Inductance) Magnetic induction, 280-282 in anisotropic medium, 307, 420 of antenna, 455 (See also Antenna) of bifilar circuit, 289, 290 shielded, 308, 309 Biot and Savart law for, 294, 295 boundary conditions, 303, 305 in cavity resonators, 527 (See also Cavity resonators) curl of, 282, 368 of currents, in cylindrical shell, 299 in hole in tube, 298 in spherical shell, 292-294 definition of, 280, 281 of dipole, oscillating, electric, 450 magnetic, 451, 469, 470 divergence of, 282 of eddy currents, 368 (See also Eddy currents) in electromagnetic waves, 419, 420 (See also Electromagnetic waves) of electron, accelerated, 573-575 moving in dielectric, 579-581 force of, on moving charge, 566, 567, 582 from Hertz vectors, 417, 418, 451 line integral of, 282, 300 of linear circuit, 295 of loop circular, 290, 291, 294, 295 Lorentz transformation for, 302, 303 of magnetization, 302, 303 from moving charge, 571-573, 575, 581 normal to A, 370 under pole piece, 317-319
613
Magnetic induction, propagation of, 416 in conductor, 369, 431, 432 of solenoid, helical, 296, 297 in sphere, 374-377 transient, in solid sphere, 378-380 tubes of, equation, 289, 295 refraction of, 312 in two dimensions, 307-310 from two scalar potentials, 369, 370 uniqueness of, 284 Magnetic interaction of currents, 280327 Magnetic moment, 280, 356, 363 Magnetic needle, 356, 357, 362, 363 Magnetic permeability, 281 (See also Permeability) Magnetic pole, 357 Magnetic potential, scalar, 310, 311, 387, 388, 390, 391, 403, 404 (See also Magnetomotance) vector, 282 (See also Potential) Magnetic reluctance, 312, 313, 317-319, 359 Magnetic retentivity, 354 Magnetic shell, 281 Magnetic susceptibility, 350 (See also Susceptibility) Magnetic vector potential, 282 (See also Potential) Magnetism, induced, 282, 349 (See also Magnetization; Permeability) permanent, 353-355 (See also Magnet; Magnetization) Magnetization, 302-304, 350, 351, 355, 356 in crystal, 351 cure-e-of, ferromagnetic, 352 of horseshoe magnet, 358-360 intensity of, 302-304, 350, 355, 356 of needle, 356, 357, 362, 363 of ring magnet, 360-362 by rotation, 355 rotation by, 355 uniform, 355, 356 Magnetizing force, 352 on horseshoe magnet, 358-360 Magnetomotance, 310, 311, 358, 359 boundary conditions, 311
614
INDEX
Magnetron, 569, 570 Mass, energy relation, 568, 569 longitudinal and transverse, 565 variation of, with speed, 563, 564 Maxwell, displacement current, 368, 415 equations, 415 invariance, Lorentz transformation, 581, 582 gyromagnetic effects, 355 image method, eddy currents, 381-385 Meson, 3, 561, 562 Millikan, charge on electron, 3 Mode, of cavity resonator, 526-545 (See also Cavity resonator) of propagation, 512-525 (See also Wave guide) Modulus of complex number, 72 Momentum, of plane wave, 424 Multipole, 6, 449, 450 radiation from, 449-452 Mutual capacitance, 37-39, 128-132, 258 Mutual elastance, 36-39, 258 Mutual inductance, 333, 541-543 (See also Inductance)
N Neutral points and lines, 163 0 Ohm's law, 250, 251, 415 Optic axis of crystal, 421 Orthogonal coordinates, 50-52 curl, 52 divergence, 51 gradient, 51 Laplace's equation, 51 Poisson's equation, 51 by rotation of conjugate functions, 136 Orthogonal and equipotential curves, 121 Orthogonality of conjugate functions, 74
P Parabolic cylinder coordinates, (6, 61) Parallel-plate conductors, 28, 29 infinite set, alternately charged, edge effect, (24, 108) shaped for uniform density, (25, 108)
Parallel-plate conductors, semi-infinite, thick, between infinite, (35, 109) thin, between infinite, (26, 108) two, with cylinder, 104, 105 earthed with line charge, 85, 86 earthed with point charge, (31, 227) with thick strip, (82, 119) with thin strip, (61, 114) two, oppositely charged, 28, 29 with circular cylinder, 104, 105 one with slit, (36, 109) semi-infinite, (31, 109) edge correction, (59, 114) (See also Capacitor) Paramagnetism, 349 Permeability, 281, 349, 350 in anisotropic mediums, 307, 351 boundary conditions, 303-306 energy of field, 332, 333 of ferromagnetic bodies, 352-354 in magnetic circuit, 312, 313 from magnetization, 303 pressure dependence, 343 relative, 281 temperature dependence, 350 variable, 303 Phase constant, 436, 437 Phasor, 369 Plane, conducting, and cylinder, 77, 78 with groove or ridge, (49, 112) with hole, 171, 172 image, of line charge in, 70 of point charge in, 124, 125 semi-infinite, with slit, 92 and sphere, 131 transformation of, 84 dielectric, image, of line charge, 70 of point charge, 125, 126 Plane grating, 86, 90, 91, 98-100 (See also Uniform field) Plane waves, 415-447 (See also Electromagnetic waves) Planes, intersecting, 70, 84, 89 (See also Images) parallel, 28, 29 with dipole between, (127, 240) with sphere between, (136, 243) (See also Conformal transformations, elliptic function; Images)
INDEX Poisson's equation, 51, 55, 56, 272-274, 416 in orthogonal coordinates, 51 solution of, 55, 56 for space charge, 272-273 Polarization of electromagnetic wave, 420, 428, 429 in anisotropic medium, 420, 423 from antenna, 475-478 (See also Antenna) circular, 429 in diffraction, 471-474 (See also Electromagnetic waves) from dipole oscillator, 449-451 elliptic, 429 from quadrupole oscillator, 450-452 by reflection from conductor, 433, 434 total, 430, 431 in wave guide, 493-495 (See also Wave guide) Polarizing angle, 428 Positron, 3 Potential, of current distribution, 249 boundary conditions, 251 in cylinder, solid, 260, 261, (27, 278) in extended mediums, 250-252 in sphere, solid, 259, 260, (17, 276) enclosing dipole, (17, 276) around sphere in cylinder, 270-272 in spherical shell, 264, 265 in stratified earth, 262, 263 in strip of variable width, 254-257 in surface of revolution, 265-267 in thin curved sheet, 263, 264 in two dimensions, 253-257 (See also Conformal transformations; Electromotance) quasi-vector, 305, 306, 334 boundary conditions, 305, 334 of current normal to interface, 306 scalar electrodynamic, 416, 417, 452 of charge in motion, 572, 573 of dipole oscillator, 449 from Hertz vector, 417, (12, 444) Lienard-Wiechert, 452, 573 propagation equations, 417 of quadrupole oscillator, 450, 451 relation to vector, 417 retarded, 452
615
Potential, scalar electrostatic, 4-244 of axially symmetric field, 60 boundary conditions, 18-20, 94, 95 of conductors in uniform field, 6668, 172-174, 178, 179, (120125, 239) (See also Uniform field) of confocal conicoids, 122-124 conical boundaries, 152, 156, 165168, (78, 79, 232) (See also Uniform field) of cylinder in hyperboloid, 220 of cylinder, short solid, 209-211 differential equation of, 50, 53 (See also Laplace's equation) of dipole, 6, (115, 116, 238), (17, 276) between planes, (127, 240) of disk conductor, 124, 172-175, (63, 230), (85, 87, 233), (102, 236) in sphere, (130, 241) of double layer, 14, 58 function, 74 gradient, 5, 51 (See also Electric field) of harmonic charge density, 141 on oblate spheroid, 175, 176 on prolate spheroid, (89, 233) on sphere, 141, 142 on sphere in spheroid, 215-217 on sphere outside spheroid, (129, 130, 241) on sphere, split, in spheroid, (132, 241) on spheroid in sphere, (131, 241) in hollow cylindrical ring, 194, 195, (105, 236) by images, 69, 70, 123-129, 133, 134 by inversion, 87-89, 102-104, 133, 134 of line charge, 63, 65, 76 (See also Circular harmonics; Conjugate functions; Images) and conducting sphere, (133, 242) and dielectric sphere, (134, 243) and dielectric wedge, 70-72 logarithmic, 63, 65, 76 maps of, 9, 10, 81, 91, 92 maxima and minima of, 13 near neutral points, 163
616
INDEX
Potential, scalar electrostatic, of point charge, 4, 5, 55, 176, 177, 189, 204 in Bessel functions, 189 in Bessel modified functions, 204 in cylindrical hole in dielectric, 205, 206 near dielectric plate, 193, 194 in oblate spheroidal harmonics, 172-177 in prolate spheroidal harmonics, 179, (90, 234) in spherical harmonics, 140, 142, 149-152, 156, 163, 167, 168 (See also Green's functions; Images) of polygon, 82 (See also Conformal transformations) of quadrupole, 6 of ring charge, 149, (86, 233), (99, 235), (101, 236) in spherical shell, 149, 150 of sphere, in cylinder, 217, 218 between planes, (135, 136, 243) and stream function, 74, 75, 94 superposition of, 5, 35, 69, 86, 91, 95-97, 100, (20, 107), (28, 108), 127, 149, 150, 174, 189, 193, 205 of surface electrification, 5, 56 in three dimensions, 121-244 of torus, charged, (118, 239) uncharged, between planes, (101, 236) in two dimensions, 63-120 uniqueness of, 25, 56, 57 of volume electrification, 5, 56 of wedge with boss, 138, 139, (107, 111, 237) zero of, 4 scalar magnetostatic, 310-313, 387, 390, 391, 403, 404 (See also Magnetomotance) vector electrodynamic, 416, 417 of charge in motion, 572, 573 of conical transmission line, 459 curl of, 416 (See also Magnetic induction) of current loop, 469, 470 of dipole oscillator, electric, 449 magnetic, 451, 469, 470
Potential, vector electrodynamic, divergence of, 417 from Hertz vector, 417 normalized in cavity, 541-544 orthogonal properties of, 541 of plane waves, 416-447 (See also Electromagnetic waves) propagation equations of, 417 (See also Electromagnetic waves) of aperture radiation, 477, 479 (See also Electromagnetic waves) relation to scalar potential, 417 in resonant cavity, 527-529, 535, 541-544 (See also Cavity resonator) retarded, 452 of antenna, 454 of charge in motion, 572-575 solution for, in cylindrical coordinates, 474, 475 in spherical coordinates, 467, 468 of spherical waves, 449-474 (See also Electromagnetic waves) uniqueness of, 467 of waves in tubes, 493, 495 vector magnetostatic, 281-327 of axially symmetrical field, 288 of bifilar circuit, 289, 290 shielded, 308, 309 boundary conditions, 303-305 of circular loop, 290, 291, 294 in Bessel functions, (27, 323) in modified Bessel functions, (26, 323) in spheroidal harmonics, (22, 24, 322) oblate, (22, 322) prolate, (24, 302) of currents in cylindrical shell, 299 of dipole, 290 of linear circuit, 306 normal to interface, 306 of magnetic needle, 363 and magnetization, 302-305 and mutual inductance, 333 (See also Inductance) opposite groove, (36, 324) orthogonal expansions, 266, 277 cylindrical coordinates, 266, 277 spherical coordinates, 267
617
INDEX Potential, vector magnetostatic, under pole piece, slotted, 317-319 of ring magnet, 360-362 of shell transformer, 314-317 in solenoid, finite, (43, 326), (34, 413) in solenoid gap, 397 of spherical current shell, 292-294 inside spherical shell, 292-294 three-dimensional, 282-307, 314-317 two-dimensional, 307-310, 317-319 from two scalar potentials, 285, 286 of uniform field, 283 uniqueness of, 284 vector quasi-magnetostatic, 329-334, 337, 338, 341, 368-414 of eddy currents, by images, 381386 in solid sphere, 374-377 in spherical shell, 393-400 electric field from, 330 propagation equation in metal, 369 shielding, by thin plane sheet, 382, 385, 386 by thin spherical shell, 395 by thin tube, 400-402 transient, in solid sphere, 378-380 in spherical shell, 394 in thick cylindrical shell, 405406 (See also Eddy currents) from two scalar potentials, 369, 370 Potential coefficients, 36-39, 258 (See also Mutual elastance; Self elastance) Potential energy, of capacitor charge, 30, 39 of dipoles, electric, 7 magnetic, 363 Power, dissipation by eddy currents, 369, 371, 373, 378, 396, 400, 402, 405, (7, 409), (20, 23, 411), (35, 413) in cavity, 528 (See also Cavity resonator) in plane surface of solid, 371 in skin effect, 371, 373, 378 in solid cylinder, 373 in solid sphere, 378 in thin cylindrical shell, 402 in thin spherical shell, 396, 400 in wave guide, 495, 496
Power, loss in resistance, 250, 253 a minimum, 253 radiation from antenna, 457, 459, 462, 466 (See also Antenna) transmitted by aperture, 482 (See also Electromagnetic waves) Poynting vector, 418 from accelerated charge, 575-577 normal to velocity, 577 parallel to velocity, 575, 576 angle with wave front, 420 from antenna array, 464, 465 broadside, 465 endfire, (7, 9, 483), (10, 484) end-to-end, (50, 52, 490) from aperture, (46, 47, 489), (50, 52, 490) coaxial, 482 from biconical antenna, 461 complex, 439 from conductor, 433 in crystal, 422 definition of, 418 diffracted, from slit, (46, 47, 489) from hole, (50, 52, 490) from dipole, 451 energy density in, 423, 424 for linear antenna, 457 from multipole, 451, 452 from quadrupole, 452 reflected and refracted, 426-428 reflection from conductor, 433 in wave guide, 496 from aperture, coaxial, 520 Pressure, hydrostatic, in dielectric, 32 of plane electromagnetic wave, 424 Priestley's law, 2 (See also Coulomb's law) Propagation constant, free space, 438 Propagation equation, 416, 417 solution, for cavity, 526, 529 in cylindrical coordinates, 474, 475 for hollow tubes, 493-497, 515 in spherical coordinates, 467, 468 Proton, 3 Q Quadrupole, electric, 6, 449-452, (1, 482), (2, 3, 483)
618
INDEX
Quality of cavity, 528, 530-534, 543, (47, 48, 554), (51-53, 555) Quincke electrostriction, 32
R Radiation, electromagnetic, 415-459 (See also Electromagnetic waves) reactance, 463 resistance, 458, 463, 470, 482 Reciprocation theorem, Green's, 34, 35, 54 Refraction, of electric lines of force, 20 of magnetic lines of force, 312 optical, 424-428 (See also Electromagnetic waves) Refractive index, 425, 428 Relativity (special), 560-569, 581-587, (2, 3, 587), (12, 16, 589) applications, 566-587 conservation of charge, 566-567 invariance of Maxwell's equations, 581, 582 Lorentz transformation, 560, 561 mass, transverse and longitudinal, 565 variation with velocity, 562-564 postulates of, 560 time dilation, 561 transformation equations, 561, 562, 564-566 for acceleration, 562 for force, 564-566 for velocity, 562 Relaxation, time of, 431 Reluctance, magnetic, 312, 313, 317319, 362, 369 of air gap, 314, 317-319 with slotted pole piece, 317319 of anchor ring, 312, 313 Resistance, from capacitance, 254, 258 (See also Conformal transformations, elliptic function) definition of, 249 between electrodes, cylindrical, 254 distant, 258 on slab edge, (12, 13, 275) spherical, 258 high-frequency, of cylinder, 373 input, of biconical antenna, 463
Resistance, input, of cavity loop, 543 of cavity stub, 544 of coaxial, into cylindrical guide, 520 into rectangular guide, 504 between insulated sheet edges, (911, 275) limits of, 267, 268 radiation, 458, 463, 470, 482 of solid cylinder with bubble, 270-272 specific, 250 (See also Resistivity) of strip, axial hole, (15, 276) with width change, abrupt, 255-257 tapered, (32, 278) between strips on plate, (14, 275) Resistivity, 250 of anisotropic medium, 268, 270 Retentivity, magnetic, 354 Riemann surface, 93 Rodrigues's formula, 144 S
Schwarz transformation, 82-102, 107110, 114, 509 (See also Conformal transformations) Screening, by eddy currents, 382, 386, 396, 401, 408 in plane sheets, 382, 386 in thick shell, cylindrical, 408 in thin shell, cylindrical, 401 spherical, 396 electrical, 37, 38 magnetostatic, by permeable spherical shell, 312 by permeable tube, 308, 309 Self-capacitance, 37-39, 129, 131 Self-elastance, 37-39 Self-inductance, 337-340 (See also Inductance) Shielding, 37, 38, 308, 309, 382, 386, 396, 401, 408 (See also Screening) Skin depth, 371, 495, 496 Skin effect, 371-374 on plane surface, 371 on solid cylinder, 372-374 on solid sphere, 376-378 on tubular conductor, 371, 372 (See also Eddy currents)
619
INDEX Snell's law of refraction, 425 Solenoid, 296, 297, 396, 397 (See also Inductance; Magnetic induction) Sphere, crystalline, in magnetic field, 351, 352 with harmonic charge distribution, 141, 142, 215-218, (129, 241) images in, 126-130, 134 and orthogonal wedge, 138, 139 oscillations of, forced, 471, 474 free, 470, 471 and plane, force, 131 with second conductor, 215-218, 220223 (See also Capacitance; Charge density; Potential, scalar electrostatic) solid metal, currents in, 259, 260 eddy currents in, 375-378 transient, 378-380 uniformly magnetized, 356-358 torque on, in permeable medium, 358 Spheres, two, capacitance, 27, 129, 131, 132 in contact, induced charge, 134 dielectric, hollow in field, (50, 151) force between, charged, 129 uncharged, in field, (125, 240) images in, 128, 130, 134, 135 intersecting, capacitance, 135 Spherical coordinates, curl, (2, 60) divergence, 52 Laplace's operator, 52, 53 Spherical harmonics, 139-168 applications, current flow, 259, 260, 270-272 eddy currents, 375-379, 395-400, (11, 410), (34, 413), (35, 414) electromagnetic waves, 468-474, (18, 19, 485), 534 electrostatics, 141, 142, 149-152, 163, 215-218, 220-223, (57 61, 227), (64, 66, 230), (72, 73, 231), (79, 232), (115, 116, 238), (123, 128, 240), (129, 130, 241) magnetism, magnetostatics, 293-295, (43, 296), 335-337, (31 34, 348) permanent, 355-358 -
-
Spherical harmonics, magnetism, into spheroidal or cylindrical harmonics, 211-215 vector potential expansions, 387 (See also Surface harmonics) Spherical shell, currents in, 264, 265 eddy currents in, 293-400 transient, 394, 395 magnetic field of currents in, 292, 295 Spherical shells, several concentric, 35 Spheroidal coordinates, 168, 169, 177, 178 Spheroidal harmonics, oblate, 137, 168176, 211-215 applications, electrostatic, 171-177, 215-217, 219, 220, (80, 81, 232), (84 88, 233), (128, 240 to 132, 242) linear circuit, (22, 322) dielectric spheroid, (80, 81, 232), (83, 84, 233) inverse distance, 176, 177 potential, of disk, 172-174 of harmonic charge, 175-176 of plate with hole, 171, 180 enclosing cylinder, 219, 220 vector, of current loop, (22, 301) prolate, 177-179, (82, 232), (83, 84, 89, 233), (90, 234) applications, electrostatic, (82, 83, 232), (84 89, 233), (90, 234) linear circuits, (24, 25, 322), (12, 13, 345) dielectric spheroid, (82, 232), (83, 84, 233) inverse distance, (90, 234) permeable spheroid, (12, 13, 345) potential, of harmonic charge, (89, 233) vector, of current loop, (22, 322) self-inductance, 337-340 spheroid in uniform field, 178 Static machines, 3, 247, 248 Stereographic projection, 264, 265 Stokes's theorem, 49, 329, 331 Stream function, eddy currents, 380, 384, 385, 388, 391, 403, (32, 413) electrostatics, 74, 75, 94 boundary conditions, 94 for steady currents, 253, 254 -
-
620
INDEX
Stress on conducting surface, 18, 19, 33, 34 in dielectric, compressible, 32 liquid, 33 on dielectric boundary, 18, 33, 34 in electric field, 15-17 in magnetic field, 342, 343 Surface density, 3 (See also Charge; Potential) Surface harmonics, 140, 141, 156 differential equation for, 140 orthogonality of, 141 vector, cylindrical, 186, 286, 287 spherical, 161, 287 zonal, 142, 152 (See also Legendre polynomials) complex order, 165 nonintegral order, 156, 157 second kind, 152-156 argument imaginary, 155, 156 recurrence formulas for, 153 series for, 152, 153 special values of, 155 in terms of first kind, 154, 155, 157 (See also Associated Legendre functions; Harmonics) Susceptibility, 350, 351 of crystals, 351 temperature dependence, 350
T Tension on tubes of force, 15, 16 Tensor dielectric constant, 21 Tolman, R. C., metallic conduction, 3 relativity experiment, 562 Toroidal coordinates, (8, 61), (117, 239) Torque, on charged conductors, 40, (20, 107), 174, 175 on crystalline sphere, magnetic, 352 on current loop, magnetic, 280 on diamagnetic bodies, 344, 349 on dielectric body, 59 on dielectric elliptic cylinder, 97, 98 on dielectric spheroid, (84, 233) on dipole, electric, 7 near dielectric sphere, (116, 238) magnetic, 362, 363 rotating over sheet, 382, 384
Torque, on disk, in electric field, 172-175 rotating in magnet gap, 392 on magnetic needles, 363 on paramagnetic bodies, 344, 345, 349 on sheet rotating in magnet gap, 390 on spherical magnet, in field, uniform, 358 inhomogeneous, 358 on spherical shell rotating in magnetic field, 400, (12, 410) on strip in uniform electric field, (20, 107) on tube rotating in magnetic field, 402, 405, (16, 410) Torus, capacitance, (119, 239) charged freely, (118, 239) uncharged between planes, (101, 236) in uniform field, (120, 121, 239) coaxial, (120, 239) transverse, (121, 239) Transformations, conformal, 72-120 (See also Conformal transformations) Lorentz, 560, 561 [See also Relativity (special)] Transformer, shell type, leakage, 314317 (See also Mutual inductance) Transients, eddy current, 378-380, 394, 395, 397, 398, 401 Transmission line, bifilar, 434-436, 440, 441 imperfect conductor, 440, 441 coaxial, 521-523 break with no reflection, (36, 37, 552) plane discontinuity, 522, 523 radiation from end of, 480, 482 conical, 459-461 cylindrical, general, 434-436 characteristic impedance, 437, 438 dielectric, (20, 485), (21-27, 486), (28, 29, 487) dielectric coated, (31, 32, 487) plane, (30, 487) with elliptic integral characteristic impedance, (17-25, 445), (26-34, 446) coplanar strips, (18, 445) unequal, (22, 445) cylinder, in hyperbola, (31, 446) in parabola, (32, 446)
INDEX Transmission line, with elliptic integral characteristic impedance, cylinder, between planes (24, 445) strip, in cylinder, coaxial, (19, 445) not coaxial, (23, 445) edge on axis, (21, 445) and half plane, coplanar, (20, 445) normal, (28, 446) between planes, thick, (33, 446) thin, (25, 445) thin, normal to, (27, 446) in slot, coplanar, (17, 445) normal, (26, 446) hollow tube, 511-526 (See also Wave guide) reflection at discontinuity, 438, 439 round pair, external, (13, 444) one inside, (16, 444) sectorial horn, (39-42, 553), (45, 46, 554) Tubes, of flow, current, 251, 259, 260 of force, 14, 15 pressure in, 16 refraction, dielectric surfaces, 20 tension along, 16 of induction, 289 (See also Magnetic induction) U Uniform field, electric, charge path, 568 cycloidal metal boundary, 79, 80 cylinder, circular dielectric, in, 66, 67 circular metal, in, 66, 67 elliptic dielectric, in, 95-97 disk, metal, in, 172, 175 from grating, of coplanar strips, 91 of edgewise strips, (22, 107), (23, 108) of round wires, 98-100 of small wires, 86 from plane, with circular ridge or groove, (49, 112) with spheroidal boss, 178, 179 thick, with slit, (48, 111) thin, with circular hole, 171, 172 with slit, 92, 93 plane strip in, (19, 20, 107) relativity transformation of, 582
621
Uniform field, electric, ring conductor in, (97, 235), (120, 121, 239) spheres in, dielectric coated, (7 4) 232) imperfect, (69, 231) intersecting orthogonally, (44, 228) uncharged, (122-125, 240) spherical shell, dielectric, in, 150, 151 spheroid, oblate dielectric, in, (80, 81, 232), (84, 233) prolate conducting in, 178, 179 prolate dielectric in, (82, 232), (83, 84, 233) torus in, (120, 121, 239) transient, oscillations in sphere from, 470, 471 magnetic, eddy currents in solid sphere, steady-state, 375, 378 transient, 378, 379 magnetostatic, charge motion in, 568-570 in cylindrical hole, 298 and electric, charge path in, 583586 energy in, 332 magnetic needle in, 363 permeable tube in, (20, 322) relativity transformation of, 582 screening by permeable shell, 312 sphere, crystalline, in, 351-352 permeable in, (17, 321) uniformly magnetized in, 356, 358 spherical shell in, 312 vector potential of, 283 Uniform magnetization, 355, 356 Uniqueness theorem, for currents, 252 for electromagnetic waves, 467 for electrostatics, 25, 56, 57 for magnetostatics, 284 Units, capacitance, farad, 26 capacitivity, 2 charge, 2 conversion tables, 592-595 current, ampere, 247, 280 dimensions of, 596 elastance, farad, 26 electromagnetic, 592-595 electromotance, 248, 249 electrostatic, 592-595
622
INDEX
Units, field intensity, electric, 4 magnetic, 311 flux, density, magnetic, 281 weber, 312 Gaussian, 592-594 inductance, mutual, 333-335 self-, 337-338 magnetic induction, 281 magnetomotance, 313 permeability, 281 potential, 4 power, 250 reluctance, 313 resistance, 250 V Variational methods, 510-514, 523-525, 537, 538 cavity resonance frequency, 537, 538 spheroidal, 538 guide cutoff frequency, 523, 525 elliptic, 525 guide iris susceptance, 510-514 capacitative, (19, 548) inductive, 511-514, (22, 23, 550) Vector, analogue to Green's theorem, 54 potential, 282 (See also Potential) surface harmonics, 161, 186 Bessel, 186, 386, 387 Legendre, 161, 287
W Wave, definition of, 419 Wave equation, 416, 417 (See also Propagation equation) Wave guide, 511-526, (1, 545 to 46, 554) arbitrary section, 522-525 attenuation, 495-498, (1, 2, 545) characteristic impedance, 494 circular cylindrical, 515-521 attenuation, 516 cutoff frequencies, 516 excitation, coaxial line, 520 dipole charge, 518 dipole current, 517, 518, (27, 551) radial, 517 longitudinal, 518 transverse, (27, 551)
Wave guide, circular cylindrical, excitation, loop, 519 orifice, 520 fields, 516 transverse electric, 516 transverse magnetic, 516 Green's function, 516-518 with vane, (30, 31, 552) coaxial, 521-523 plane discontinuities, 522 radiation from end, 480-482 into circular guide, 520, 521 into rectangular guide, 501-504 cutoff wave length, 494, 495 elliptic, cutoff, 523-525 fields, general, 495 normalized impedance, 507 rectangular, 496-509, 511-514, (3, 545 to 24, 550) attenuation, 497 discontinuities, plane, 506-514 thin capacitive iris, 507, 508 variational, (19, 548), (20, 549) thin inductive iris, 507-509, 511514, (21, 549), (23, 550) variational, 511-514, (22, 24, 550) excitation, coaxial line, 502-504 dipole charge, 500, 501 dipole current, 500, 501 longitudinal, 501 transverse, 500 loop, circular, (7, 656) rectangular, (6, 546), (16, 548) small loop, (3 5, 546) longitudinal moment, (5, 546) transverse moment, (4, 546) transverse wire, (9 11, 547) fields, 496, 498 transverse electric, 496 transverse magnetic, 498 Green's function, 500-502 input reactance, (8, 546), (9, 10, 547) input resistance, (7, 546), (12, 547), (13, 548) transverse electric modes, 493 transverse magnetic modes, 493 triangular, (25, 26, 551) wave velocity, phase, 495 -
-
INDEX Wave guide, wave velocity, signal, 495 Wave length, guide, 494 cutoff, of guide, 494 Wave number, 436, 437 Waves, electromagnetic, 440-559 (See also Electromagnetic waves) velocity, group, 441, 442 phase, 441, 442 signal, 441, 442 Weber, unit of flux, 312 Wedge, conducting, 84, (83, 119) dielectric, 70-72
623
Wedge, orthogonal to rotation surface, 138, 139 (See also Conformal transformations; Potential) Wedge functions, 208 X X-rays from accelerated electron, 575, 576
Zonal harmonics, 142, 152 (See also Surface harmonics)
111111111111111111111111111111111111111111111111 PB232797
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SOLUTIONS TO PROBLEMS IN STATIC AND DYNAMIC ELECTRICITY
CALIFORNIA INST. OF TECH., PASADENA, CALIF
JUN 1974
U.S. Department of Commerce National Technical Information Service
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WRS-1893-7-5
Solutions to Problems in STATIC AND DYNAMIC ELECTRICITY by William R. Smythe
California Institute of Technology Pasadena, California 91109
June 1974
by NATIONAL TECHNICALINFORMATION SERVICE U S Department of Commerce
Reproduced
Springfield VA 22251
BIBLIOGRAPHIC DATA SHEET 4. Title and Subtitle
2.
11. Report No.
I
3 PB
WRS-1893-7-5
. ._ Prepared June, 1974
5. .n.cport
SOLUTIONS TO PROBLEMS IN STATIC AND DYNAMIC ELECTRICITY 7. Author(s)
232 797 Vint
8. Performing Organization Rept.
.
No.
William R. Smythe 9. Performing Organization Name and Address
10. Project/Task/Work Unit No.
None California Institute of Technology Pasadena, California 91109
II. Contract/Grant No.
None 13. Type of Report & Period Covered
12. Sponsoring Organization Name and Address
California Institute of Technology Pasadena, California 91109
Final ld.
15. Supplementary Notes
16. Abstracts
This report is a complete set of solutions to the 679 problems posed in the third edition of "Static and Dynamic Electricity", (McGraw-Hill, 1968). Of these problems, 581 also appeared in the second edition (1950). The solutions are keyed to both editions. This book has become a classic reference for engineers and physicists working on problems involving electrostatic and magnetostatic potential theory, boundary values, eddy currents, antennas, wave guides, cavity resonators, and the diffraction of electromagnetic radiation. A significant fraction of its reference value resides in the very large collection of problems, solutions to which are presented here. Also included is an extensive errata list for the third edition, first printing.
17. Key Words and Document Analysis.
17a. Descriptors
Magnetic induction Inductance Eddy currents Electromagnetic radiation Polarized electromagnetic radiation Antennas Waveguides Cavity resonators
Electrostatics Capacitance Electric fields Boundary value problems Potential theory Electric potential Electrical resistance Magnetic fields
171). Identifiers/Open•Ended Terms
IVES
17c. COSATI Field 'Group
20C, 20N
18. Availability Statement
Unlimited FORM N715-33 I/REV. 10.73)
ENDORSED BY ANSI AND UNESCO.
STJECT TO
CH 6I -•
... 11 19.. Security Class (This Report) UNCLASSIFIED 20. S ecurity Class (This Pa UNCLASSIFIED UNCLASSIFIED THIS FORM MAY BE REPRODUCED
. i
...J
USCOMM-DC 5265.P74
Solutions to Problems in STATIC AND DYNAMIC ELECTRICITY by William R. Smythe
INTRODUCTION Solutions of the 679 problems propounded on 114 pages of the 1968 third edition of Static and Dynamic Electricity (McGraw Hill) are given in the next 178 pages. The second edition has 581 of these, including all in the first, second, third and last chapters. When chapter, page or problem number differs, the second edition value appears in parenthesis. Some of the new problems cannot be solved by the second edition theory. The text problem statement, which lists special assumptions made, defines new symbols used, usually gives the answer and points out text formulas that apply, is not repeated herein. The use of the Handbook of Mathematical Functions and other recent works listed after Chapter V greatly shortens some of the original solutions. Beginning on page 179 is a list of errata of the first printing of the third edition.
Chapter 1.
Page 22. (22)
Page 1.
- 1wi,2 - aW2,1 = qQ1,2(a24.b2 a 2+b2 WI, 2 a w )/(02) Ans. Q1,2= 47eA/T4-b2 (4/2,2 -2-
47ew 1l 2 = qQ11 2 /a + aQ2,1AfT-Fh2
) (4rea
b
.
0
When lines due to 2 and 3 are added to the straight line from 1 to 4, the resultant passes inward between 1 and A, outward between 4 and A. So maximum flux passes through AB, inside. s 71 q _1 7 q rr By Gauss Theorem: FluxAB = 2. e•4-F-2--E(tan 2- 4 --) = TT-e-(-tan 2) =3-tan-1 2•
Total flux is
Fraction is -2-tan-1
. Ans. 4
-34r* = -27(1-cosO)q+ 2.7(1-cos0)q1 = 27(q'-q)(1- 1/J) . q'=q+4*/ (2-1i) Ans. - 4Force lines at infinity come from outer ring and approach 3inner ring fastest in the plane of the rings. 627Qx/[4rre (22 +x 2).11 = Qx/[47€ (1+x 2 ) 7], 27(1+x2 ) 2 = (4+x2)3, 3(1+x2 )=-(4+x 2) so x= 1/45. The neutral points are at x =
0, -1/J.
Ans.
- 5Unit length flux is q (p 15). CU= tan-1[y/(x-a):1- tan-l[y/(x+a)] a a . x 24.y2 .32_ lay -i[y/(x - a)]- y/ (x+a)J . x>a:y=0,U=0. 0<x
-6F lux lines from quarter of small sphere of radius 6, centered on A, reach B so that 2rr6 2 (1-cosa) 1 „o 1 4aq - 4 ,cos a = -4, a =ou . At C, = BC2 so 13C=1 =a. 4762 AC2 4q Near C 41-reV = 2N/4a2 +r 4ar cos0 N/a2 +r 2- 2arcos0 Lines are normal to x at C if 79= 0 as r approaches 0.
4
3r cosl 0 = -8qar sine + qar sin() __). qr sine [1+ 3r cosel 1 r32 a r32 a2 2a 7 37 3qr2sinecos0 - This is zero at 60 = 0, ., 7, -2-. Ans. 2a3
Chapter I. Page 22(22)
Page 2 ql
-7From 1.102, C is at center of gravity so AC:CB=q2:q1. At infinity total flux is qI+qeand appears to come from C. Flux to left of n is
q1- i(l+cosci)qi= 3(qI+52)(1-cos6) so q1(1-cosa)= (qI+q2)(1-rs6) 2qIsin2(ici) = 2(q1+92)sin2(i6).
8 = 2 sin-1[q (q1-1-q2) -Isin(ia)] Ans. -8- Page 23
From Gauss Theorem,flux through disk of radius OP is 24q(1-cos6)4(1-cosa) so 2sin28 = sin2icy. sina=gsiniu Ans. 4.4 -9By Gauss theorem all lines passing in must pass out, so,with lines drawn close enough together, two in every section are tangent. At this point equipotential is normal to surface. Line on surface connecting adjacent tangent points is closed. -10Two thirds of A's lines end on B or C. Sc (see prob.6) 2782(1-cosa) =1.4782. cosy = 1
= -3. of all
lines of force not reaching B or C, half will be in each hemisphere at infinity. Thus 2r62(1-cosal)=1-476a .
3. Ans. This tube bisects infinite sphere
cosal=14=
A 3q
-
q
and so is perpendicular to AC. -11An equipotential fixed for all V values must be a sphere centered at Va. Therefore choose origin at C. v q + Va = a p 470A 47efPB PC 411e[(f-x) +r -x qa Va r 2 -1 2 a ale = 47efL (a f -x) +r -x _ aa (47e) Lq(f2 -2fx+r2) -2-qa(d4 -2fa2 x+t r )
]
J
Va/r. For one of these surfaces to
be spherical there must be an r value for which V is independent of x, so terms a 4 P with x must vanish. Hence (i2-2fx+r2) = a (a -2fx.-+f-r-)-I. So r = a. Substitute in V to get V =V.
When V=q(f+a)[47e(f-a)21-1, equilibrium point is on sphere 1 -1, it is on sphere opposite A and B. between A and B. When V = a(f-a)[47e(f+a)2 J
Chapter I. Page 23 (22)
Page 3
-12-
-13Move a coordinate origin along a line of force keeping z tangent to E. When E is a minimum then aE/az = 0 = a2 V/Oz2 . By hypothesis: aV/ax = 0, aV/ay = 0 at origin and by LaPlace's Equation a2 V/ax2= -a* V/ay2 .
Therefore the
curvature of the surface tangent to the x-axis and tangent to the y-axis are always equal and opposite. If the x-axis is tangent to maximum curvature, then the y-axis is tangent to the minimum curvature. -14Let law be F = CQr-2-P. Work to take charge from one shell to the other is zero. Otherwise current would flow. If 2 -2 o 2raa r l snecosi3d8] FC = (4ra2 ) -1C A Jr -2-P sin8cosade - Blo2rb r -pi 2 2 b - 2arcose. r sina = asin81/ -cosa = (r - acos8)/r1/ r I =a + 0 ..Arqb (r-acoslii)sined8dr B(r(b (r-bcos8)sin8d8dr WTFdr = r 3-1-1) r 34p a 2.io)o 2)o)o 2 , A ri sin8 d8 sine d8 ' _,Ni lb (14 0= 1,13)/211) '" 21 a-2./ol(b2+r2-2brcos8) (e+r2 -2arcose )
(
-
Ar02 -141 -2arcos8)- 7'Ibill Bri(b2+r2 -2brcos0.gibjr
2
tar
'a o
iAlaar) -1E(a+r)i-P -(a-0113]l b a
2"
2br
'a o
iBi(2br)-1[(b+01-13 -(r-b)2-11:
1 0 = Aa -1rb-1(a+b)1 P -(a-b)1-Pb-I-a-I (2a)113 3-Bb-IEb-1(2b)1-P -a-1(a+b)1-P -a-1(a-b) -P 3 Expand by Dwight 550.1 and neglect higher powers of p. (A/a)[a(a+b)-pa(a+b)ln(a+b)-a(a-b)+pa(a-b)ln(a-b)-2ab+p2abln(2a)3 =(B/b)E2ab-2p2abln(2b)-b(a+b)+pb(a+b)ln(a+b)+b(a-b)-yob(a-bln(a-b)] In A coefficient underlined terms cancel. In B coefficient, since p is small,neglect underlined terms. Thus Ap-(a+b)ln(a+b)+(a-b)ln(a-b)+2b1n(2a)]= 2B(a-b). If f = a+b, g = a-b, 2b =f-g, 2a = f+g, then 2gB 7--6 pAC(f-g)ln(f+g)-flnf+glng]
Ans.
-
▪
Chapter I. Page 23. (23)
Page 4.
-15All flux from C goes to A and B since q' 5_ 2q. Line going up at C enters A at angle P, line going down at C enters A at angle v. Flux q/6 from B crosses AC from a to P. Flux entering v is -q/6-q 1 /124vq/(2r)=0. Thus v=(2q+q 1 )r/(6q). Ans. Flux entering P area is +q/6 - 5q 1 /(12)+0q/(2r). Thus 0 = (5q'-2q)r/(6q). Ans -16If there is a neutral point, ali/ar and aV/a8 are zero there. -1 -1 2 2 .Q2 2 2 2 4reV=qirce1-q2r +q3r13 , +a -2arcose,rp=r tP -2prcos where ra=r -1 -1 x 3 -1,0,4.-2 So 4reql =p ra k +r) - r + ro k ) . av/6r=0=PX3 (r 1-acos8 )(v+e)-1r 3- r--2 + v0(r +0cose 1 a
)(V+P) -1r -3
1.7/a8 = 0 = r;3- rr3(2) Substitute (2) in (1). -1 3 aPcosel + ari+aPcosel )(a + r) ra -r13= 0. X3r1 = rg, Xr1= ra, x2r2 r2 - tar toss (3) toss = [(1-e)raa+ a2 ](2gr )-1Substitute in (1). 1 1. ar or 0 = (1 - X2)1:1 + (0./a)]r1 + P2 + k2r2 = r2 (P/a):(1-x2)4 + 1. P(a+f ) aP (4) rlis real if X2>1 Ans. (X2-1)(1+Pv -I) - ASubstitute in (3). toss = (-4+V2)E2v(v0)(X2-1) 11-1 =i(a-P )(v0)-2(X2-1)' (5) so
a
r.
toss < 1 so 4(4)-I(X2 -1)<1, 4(a242)(0)-I),2<14(a4)2(0)-1 or X2<(a+13)2 /(a-151 )2Ans.
4revi q2
13X3 (a+P)kr,
1 r1
aX3 (V+0)kr
2
X r1
?`2
1 X -1 1 1 r 1 r -
3/2 ( 2 -1)
(4)42
The equation of the equipotential through the circle is (?2 -1)3/2
,
:0(r2 4a2 -2arcos0) -2 + v(r2 +$2 +2Prcose) 2 .1 (a 5) 1/2 - a4?
Ans..
-17Flux through neutral ring is 1 = g1i(1-cosel) + ga i(1-cosec ) - qsi(1 cos63) -
= T(q1+q2 - q9 ) + ' i( - - q_„cosfill- q2cos82+q3cos8,7 ) (1)
coseo.= (a-rcose2 )/r11
cos63 =(a+r2 cosec ) /r3 (2) -3.
(11
3r3 2r2 +q For neutral point 47ev = g1r2 -R
2 2
2 2
where r1.= (r +a-2ar acosErand r = (r +a +2r 2acoseil. 2 2 i/3 2,1708 = o = q1r3 - ga r; so r = (q„1/q3) r3 (3) a
1
-3
1.T/ar2= 0 = qI(r2 acos62)r1 -
-3
-3
3(r2+acos82)r3 q21.2 41
-
(continued)
(4)
5
Chapter I. Page 23 (23)
5
-17- (continued) -3
-2
Substitute (3) in (4): 0=2q1 r 2r 1 - q 2 r 2 , ri = (2%/q2 )
1/3
r2 . r3=(2q3 /q 2 )
1/3 r2
(5)
a2 ‘ . n- 2/3[ f 2c2/3 2 2 2/3 2/3 2/3 -2 2/3 By (5) r2i+r23 =2(r r a' 2 +a2' .42 t u3 )) J so r2=2a q 2 [(2q1) - 2q2 +(2q3 ) -] 1 +(2q - -3 ‘ 2/ 3 2-2/3 2 2 2/3 From (5) r3 -ri = 4ar cose = [ Qq3 ) - (2%) 3r 2 q 2 ,,, 2 2 2/3 2/2 2/3 2/3 2/3
so r cose = 2 a
(2q ) 3
-(2q1) a/3 4aq 2
2a q,
2/3
(2q1)
a[ ( 2q3) - (20 ] 2/r Bp 2/3 -2!2 +(2q3))2/3 - 4(2%) - 2q2 +(2q3) 2/3
4/3 2/3 2/3 2/3 2/3 a 2q1 - 2 q2 +2q32/3 -q2 +q1
2/3 4/3 2/3 2/3 a(3% - 2 q2 +q3 )
From (2) : case = '. 2r1 •i
[ ] cose 3 = 12/3+ 2/3 ) [ 2 ar3• (3q2/3-24/3 3 C12 (11 - 1/3 -1
1/ 3
But from (5): % /ri =q1 q 2 (2q1)
1/3
cose 1 -q2 cose 2+q3 cose3 = aq2 2 r 2 2/3 2/3 2/3 1/3 2/3 -3/2 2/3 2 2 3(q 3 -q1 )(% -2 q 2 +q3 ) -5/2
so
= i(c11 +qi -q3 ) +2
E acrrR
r dr 2EJo (a 2 +r 2 )
2
213 -1/3
4/3
•
-3
1/3 2/3 -3/3 -1
ra and q3 r = q2 q3 2 4/3 1/3 2/3
r2
2/3 2/3
3[q 3 -q1 +2 q (q1 -q3 )]
2/ 3 2/3 2/3 /3 2/3 3(q3 - % )[q1 -(4q) + q3
-182rrra dr cose 4rre (a 2+r 2 )
Intensity at P from ring is
] -1
r 2 = q2 qi 2
1/3 -4/3 -1
so =
(6)
Ans.
rcr cos e dr 2e (a 2+r 2 )
ac i 2E 1 (a2 +r2) 2121O
1 a(a2+R2)43 (1) 1 Total field at P is ia/E so -. tcr/E = i(c/E)[1-a(a 2 +R5-9 Thus (a 2+ R 2 )i = 2a
Ans.
-19, .1 cr ), E3= i-EV. - 3(3Z1)43 , Et5 =iy[1.- 2 ..3 , E 7=i-D.- 7(74)3 -20Q Q2 , (a + b)2+c 21 -*it ( 1 -k 2 sin2 e) - ide=dc- (ab)" .iQ2 1(.1‹.. 2TT
a Substitute in (1) above: E1= 2E(1-26 2 rrr a dcp
Q2 Q2
V= 2T
► a. 4r:E.10 r
2 c I-
2 2 1 2 2 2 where r = a +b + 2ab cos e+c , COS cp =1-2s in 8 , dy=2de, k .4ab[(a+b) +C ] 2
2
2
el bv
3
&I( n ck3 6s.r cQ1Q2 k _ft, E v , 0:21420 E F=-'11c• oc''1 4ab ak16n2e(ab) 3'z' ±1-k2-"= 16,-73Wrf 7.7 1
Ans
-21-
[,_
417E11 =-Q(r 2+a a -2ra cose)4 - Q(r2+a2+2ra cos e) +(2Q/r) P .2. a2 -2racose 3. 4r2 a 2cos 2e a2+2racose 3-4r 2 a2 c OS 2e 2] -F + +1 8r 2r 2 2r r 8r = Qa 2 r -3(3cos2 e -1) (1) 1rr 2 Vp=4Q(rrE) So [-(b +p 2 +z 2 -2bp cosy) -i+(c 2+p2 +z2 -2c p cos cp)4].V Loop (continued)
▪ Chapter I. .
Page 6.
Page 24. (24)
fr
-21- (continued) 2 2 2 2 2 Q [fr( - 1+1) dtp + b -2bPcoscp-c2 +2cPcosSodcp .173•4(b -C )P cos CP 0+ rr Vp.-44Trer[ o o 21T(P2 +z2 ) o 8TT(P2 +z2 )2 r 2 2 3 2 4rev=9Lb -c 3-17.4 (b2 -c2 )P2 J.Q(b2- c2 )(1Dipole (1- +-g-cos 6) 8 77'21'2 2r . Q(b2 -c2 ) (3cos2 - 1) (2) Eq. (1) and (2) are identical if b2 -c2 .4a2 4r
P/r=sine - q
a 2g a
Q.E.D.
- q
CHAPTER II.
Chapter II. Page 41 -1Charges distribute equally between 1 and 2 so after first contact° QI=Q2 2Q.
V2 =SuQ11-S2 2Q2' +s Q3 = V3=s13Q1+S23 '+s33Q3 . 23
so(su -s.21)Q1-1-(s22 -s23)Q-1-(s23 -s33)Q3=0.
cL
0 0 r
r
1
2
Since r>>a by page 38: r2 -al r2 -a n _ 2[ ara ] ) +re 4[rI(r are 1 2
[ri+r2 -r/ .+ ..r.z:21c + [..a.LE2 + azI2 3 _ 0 are P3- I [r I ( +r2 ) are J2 [art ar2 ] So Q3 - [ + 1] 4 [ra. (r 1+r2 ) (r2 -a)
Ans.
- 2After first contacts: Q1 =Q3=4g' Q2 =iQ•
sIl=s22 =s33 =4*/ S IR =S14=S23 =S.74- S13=S24 S =S At last contact, 1 to 4, V.3.=V4 so Vi-V4 =0-and Q+ = Q-Q1 . So Pq (IP. r 1 1 3. from 2.15 (1) page 36, Vi=siaQi+s 12 4Q-Q1 )-1-ss2 M-s1.14-Q. r V4 =s11(4Q-Q1)+si.4Q1+s12 4Q+s132Q. So V1-V4 = 0. This gives the relation
4
0 = sl.s. (2Q1--iQ)+ss4 (-2Q14Q)+s1.3 (-2Z-Q). Solve for Q1amd Q4. st.1-2s -s Q 4=1Q-Qi= s1.1.4s 13 2 12 13 Q Ans. Qi= Ans 8(s -s ) 11 12 S11- s 12 8 - 3-
r
r
n
-21 - 2+ :-;.-- + at After first contact: QI=Q; =iQ, Q=-Q, Q4=0. At second:4TreV.,4r r l- a r 9in n n. -1 4 1 i 1 -1 -. -2)r Q 4freV= -2±.--7 ( '-'3'+-3-1 1'. Subtract. V n. -V 1 -Q 4 )(a -r 1) + (-1+2 +-4:2 lb=0=(Q 2r 2r r a 1 1 i i iirJ1 i(r-aft2 Q[(2 2i [ ( -1)3a+2 ik 2-3)a+2 3(2V-1)42 2 =_739E I [ So 0 -0 -*Q. Add Q ] Ans. r-a • Qt+Q r-a -1 - - 4(r-a) ,
]r
Q
il r 2-r-(2 4-3)a] 2_2L 8
L
r-a
Ans.
Chapter II.
Page 41. (41)
Page 7
-4A 47e0=Q1/a+2Q/r B 47re0=Q2 /a+ Ql/r+Q/r Q1=-2aQ/r
C 4r€0=Q3/a+Q2 /r+Q1/r
Q2= 2 a2 Q/r2 - aQ/r=aQ(2a-r)/r2 Q3=-a2 Q(2a-r)/r2+2a2 Q/r2 =a2 Q(3r-2a)/r 3. -5-
From symmetry: COQ, Q3=4Q, Q4=Q/8, sli V/Q, 0=s12 7Q/8-VQ1/Q, so s12 =8VQ1/(7Q2 ) First battery contact: V/=V, V2'=1/=V4;,=s12 Q=8VQ1/(7Q). Q=c11V+C22 -3.8VQ1/(7Q) Q2=0=c118VQ11 (7Q) + c12[V+16VQI / (7Q) 3, cll.= -c12[2+7Q/ (8Q1) ], Q=ca tV[ -2 — 7Q/ (8Q1)+24Q1/(703 1 =c2 1V[(192Q21 -112QA -49Q2)/ (56QQ1) J = caly(24Qi+7Q) (8Q1-7Q)/ (56QQ1) so c
=
56Q1Q2
2117(24Q1+7Q)(8Q1-7Q)
and
—7Q2 (16Q1-1-7Q) Ans. c 11V(24Q1+7Q)(8Q1-7Q)
-6First method. Charges are, dividing out Q: 2, 4, i(4+i)= 2 i4)=31 • • • 1.-(na so as n approaches infinity, Qn -* Q/3. Ans. 3 Second method. From first contact: s11s33 (s22 ); From next, s12 =s32 ; From last no
rn
(
charge transfer so s11Qi+s13Q3+si..2Q1=s13Qi+s23Q1+s33Q3and sil(Q1-03)=s13(Q/-Q3)• But si„L does not equal si3 unless 1 surrounds 3 so Q3=Q1=Q/3. Ans. 7 V=silQ1s12 (Q1+Q2 )+s11Q 3 so Q3=Q1- (sa,2 /s11) (C22 1-Q2),but s12 /s1111-Q-2)/C21 so Q3 =Q1- (Q2i-Q1) /Q1= Q: Ans . /
-8Split sphere is infinitely thin so charge density on inside must equal that on outside for no separation. Thus capacitance ab equals equals capacitance bc. 4rrcab/(b-a) = 4Trebc/(c-b). ab(c-b) = b2 c - abc. Thus a = bc/(2c-b)
Ans.
-9Since interchanging subscripts 1 and 3 changes nothing, Q1 = Q3 . 1/1V2 =s11Q1+s2.2Q2A-SIsQ3 =512Qi.-1-S2 2Q2 i-S2 3Q3= (sii+sz3)Qi+s1,2Q2=2s22 Q3.+(s3.2.+si3-s 32 )Q2 . or (s11-sae +s1.3)Q1 = (s11-2s12 +sta)Q2 so
gi= Q2 = Q3. Ans.
-10From 2.14 (4): Qi=(V0-Vp)(Va.-Vo)
=(a
-3. -- 1 )(b -a )-lq=b(r-a)r (a-b) q
= [b(a-b) -1+abr -1(b-a) -l jq. r=a+vt. aQ1iat=aQ1 /6r ar/at=-ab(a-b) -Ivr-2 q -2 Ans. so agliat = abqv(b-a) -I(a+vt) -112 2 Q on ring at r=(a +c )4 induces -Q on sphere1r=b, so V = . 7e(a +b- )-4reb or 4rreV = Q[(a2 +b2 ) -r-b-I] Ans. 41
9
Chapter II. Page 42. (42)
Page 8
-12First method. Unit charge at z makes ring potential isle . For earthed ring 2 *1-1
0=-SI_IQI-FS12 Q SO Sa2=[417e(a2 +13 ) J
.
2
2
,
C=s1.1.=QI(s2.2 0 1-=477e(a +b ) Q1/Q.
Second method. Green's Reciprocation Theorem. V'= (11[4re(a2+b2)]-1 QV'-Q217,1 = 0 1 .V+Q I -0 so C=Q4./V,=-4re(a2+b2)iQI/Q. Ans
-Q1,0
Ans.
E) V Q.:J.
;
-134rev.Pg.I. agAE b 1-2 =
[1
6
4ree'ab
e
C= 9 V Oa + e(b-a)
Ans.
-14G.R.T. OW= q1(a-1-b-1)+,11e(Ob) 1
qn+q1y+q.•0 = q1•0+0•Vp-q1•000.
4ne'V' = qr-1 P 1 Solve for q1
q1 =-e'abqr-lre l (b-a)+ea]
Ans
-15G.R.T.
tciVta+qVi'D 2q.0 = -q l•0+0•Vp+q' •0 so Vp '=z-Va.
-
-
Thus when charged half the potential drop falls across each dielectric. V. /V2= 1 = C2/C1so .that cab e b(a+b) b+a-2a = 2Ans. 1 = b 2b-(b+a) ela(a+b) eta or el -16When T submerged bouyant force is mg so, when half in, it is 2mg - half weight and half electric force. Solid hemisphere is in equilibrium under normal hydrostatic pressure so force integral over plane face equals that over curved face and 2 1.„,_„2
by I.16 (2). From 1.17 (5), (6): evaVi/a8 = eaV2/a8 and V1= V2 are satisfied by radial field if Q 4-4rev aV and Q2= i•Li-reaV so 0.1= ev V/S 1 and a =eV/a. So -mg=i7V2(ev -e) and V = (2mg)2(re-rev) . Ans. -mg = ra
- 2-a e )
-17CIf charge sum is positive, force lines at infinity come from a positive conductor area. If this conductor has a negative area, then lines from a conductor at higher potential end there. In this way we eventually arrive at a conductor at higher potential than others near it. Since no near ones are at higher potential, no lines from them reach it, so it must everywhere be positively charged.
Page 9.
Chapter II. Page 42. (42) -18CQ t = clihc a Va +ci3V3• • •
Q2 = cia V1+C22V2+c32V31-• • • .
After joining (1 + Q.; =
(cI1tc22 +2c12 )V + (c3i+c32 )V3'+' • • . The new conductor formed by these two with charges Qo = Qi+Qa satisfies Qo
= cooV+c9 c
oo
= c
0V 4 0114 1 + • • • . Comparing coefficients gives
Ans.
4C +2B 4-c 42c12 = C1 2 11 22 -19C-
Repetition of the procedure of the last problem gives Q1 =(cI.1+c2I+c31+... )V, (c12+c224.c62+• .)V etc. Adding: Q= Qi= ( Q2 =
cij)V = (s1+2s2)V. Ans.
-20CEnergy change is i;:
. But 2EQI= 5E Qiso AW=i2:QiVi-i172:
(Vi-V) Ans .
-21CNo internal field energy change. Behave like two spheres of radius
21
b, charges q, q' . W = 2rLs11(q24112)+2s11qq
(it-[siii(q+q l ) 2+si2 i(q+q' )2= t(q-q' )2(sil-si2) = (16 re) -1(crq r -22CQ = (b-a) (4-Q)(4reab) = Q[1-4-reabc-1 (b-a) -1]. Charge on outside of b is Qb=- (b/r)Q' since s Qb+s22 Q 1 =0=[bQb/b - (1? /r2) ]/ (4re) from 2.113(1)( C..„\ 11 Thus 4reF = QbQ /r2 =-be /r3= -bC2 Q2Er9(c+4Treab(b-a)-31]1 -23C-
Result given by 2.14 (3). -24 Pia Pe 2_ EFirst contact: V=pia Q2 i2-957:: e +PI1Q1=1322 Qa+P12%-• Q2 =e, Qi=E-e so Ty-Ultimately no transfer so Q1=E, Qa =en . Eien=(1312"P22 )/ (P3.2 -P11 ) or en=eE/(E-e) A -25CFir st Contact: (1) slie i+spi (Q-ei)=S000-edi-SOlel'
(2) s 11e24-si2 e1+s01(Q-el-e2) =s00(Q-e -e )+s e -Fs02 (e +e )+s (Q-e -e -e )=s (Q-e -e -e )+s e 1 2 e1' (3) s11e 3+s3.2 01 2 1 2 01 1 2 3 oo 1 2 3 ol 3 +s Oa (e2+ e ) . (4) s/ie+s22 4 (ea-e2 +e3 )+soi (Q-el-e 23400 -e -e °-e Ia -e e-e)+s 34 oi402eat2 e+s (e +e32 2 )=S (2)- (1) : s3.2 (e2 -ez. ) -s0a.e2 +st2 e1=-s00e2 +s0I(ea -e2)+sce ei. c1e1 = cz e2 (3)- (2) : sii(e3 -e2 )-soie3+s22 e2 =-sooe3+soi(e3 -ea )+sez e2 .
c 1e2 = c2 e 3
(4)- (3) : s11(e4.-e3)-so1e4+si2 e3=-s o0 e4+so1(e4-e3)+s02e3 .
c1e3 = c2 e4
So eI/e2 =e2 /e3=eg/e• ., if e2 =ae1, e 3=ae2 =a2 el, e4 =ae3=a3e1. ei:ea :es :e4 = 1:a:a2 :a 3 . Ans.
Chapter II.
Page 43.(43)
Page 10.
-26CThere are now four potential coefficient combinations: a,b,c,d. aeo+bel = 0
(1)
-ae2-1-aeo+ce1+bez,=0 (2) -ae3-ae2+aeo+del+ce2+be3=0 (3)-a(to-e3l-e2+e0)+-ce1+de24-ce3+be+.0 (4) Eliminate e0in pairs. (2)-(1): (c-b)e1+(b-a)e2=0. (5) (3)-(2): (d-b)e1-1-(c-a)e2+(b-a)e3=0 (6) (4)-(3): (c-b)e1-I-(d-a)e2+(c-a)e3+(b-a)e4=0 (7). Q2=(b-c)Q1/(h-a) Q3=E(b-a)(b-d)-(a-c)(b-c)]Q1/(b-a)2, Q4=r(a-c)(b-a)(b-d)+(a-c)2(b-c)+(a-d)(b-c)(b-a)+0-0(b-4' /(b-a)3. (Q1-Q2)(Q1Q9-Q:)=(a-c)[-(b-a)(b-d)-(a-c)(b-c)-1-(b-c)2 ]Q1/(b-a)3, Q1(Q2 Q3Q1Q4). C(b-a)(b-d)(b-c)+(a-c)(b-c)2-(a-c)(b-a)(b-d)-(a-c)2(b-c)-(a-d)(b-c)(b-a)-(b-c)(b-a)9Q//(b-a) =[(b-a)(b-a)Eb-d-a+d-b+a1+(a-c)[(b-02-(3-a)(h-d)-(a-c)(b-c)]] so (Q1-Q2 )(Q1Q3-Q:) = Q1(Q2Q3Q1R4) Ans. -27CCharge -(Ql+co induced on sphere gives potential -(Q1+4Q2)/(4reR) = Vs so V1=s11Q4.+s12 Q-Vs^"'s11Q1+si2Q2 , V2=sLq+s2Q-Ves2..j.Q.,1+s22Q2since s11R*11,s22 :44S21 S1.2 Thus s11((1-Q1)+s22(Q-Q2).(Q1+Q2)/(4reR), s21(Q1- Q2)+s22 (Q. -Q2) = (q+Q)/4rteR). Subtract and rearrange:
(q-QI)/(9 -92) = (s22 -s..12)/(s11-s12)
12 1
Ans.
-28C-
V ,q Vc is on conductor) Thus for all points P where Q is constant, VI)is also constant. (Vc If q is a point charge at P and Q is the induced charge, then from 2.14 (1),
-29C-
e • P
(1)PutV'=1 and q=1 in 2.14 (1). Get Q=-V'. Q.E.D. (2)Ee'V=EeV' so el.0+0.Vp+eVq=e1-0+e.VI'+0.V4;. so Vq=1,71;. Thus if charge e at P gives potential V at Q,then e at Q gives potential V at P. -30C-
V= 0
1
• P e• Q
Chapter II. Page 43 (43)
Page 11.
-31C2
2 o V_= 2 >0 so B=°BBVB-E°BCvC' °=cCCVecCBVB s °CC/(cBB°CC-2°B4 eBB°CC>2°BC' c Q VBis positive. Ans. V_=-° Q C CB -B i(cBBcCC-2°2CB ) and ca<0 so V Cis positive. Ans. But ccelcBcf• Therefore VB>VC. All lines of force leaving B reach C or A and none
reach B. Ans. All force lines reaching C are from B and all leaving go to A. Ans. -32CThe initial system energy is *'e /c. The energy decreases as P moves up to the rest of the conductor under the influence of the induced charges. Upon contact the redistribution of charge further decreases the energy to ie2 /c. Mutual repulsion then moves P away, decreasing the energy. -33CGivenC=q11, C(l+m)=q22, Cn=q33, (1 =0. Since 2 encloses 1, q1 =0, 23 Ql=q3.2V1+q12 V2. Let Q1=0, then V1=V2 and qm2=-C. QI=C(Vi-V2) (1) Q1-q=CnVI (2) Q2=-CV1+C(l+m)V2(3) From (2) Vi=(Q1.-Q1)/(Cn). From (3) V2=Q2[C(l+m)]-1+(Q1-Q 1 1)En(l+m)C7-1. Substitute in (1): Q1=(Q1-q)/n - Q2/(1+m) - (Q1-Q1)/En(l+m): or (n+nm+m)Q1=mQ.1-nQ2. Thus Q1=(mQl-nQa )/(m4n+nm) = CV?-CV
(4) Q2+Q1-Q1-Q2=CnV2 (5)
Q2=-CV1+C(l+m)V; (6) From (5) 1/=(Q1+Q2 -Q1-Q;/(nC), Then from (6) (l+m)(Q1+Qz -Q1-Q,p/(nC) -Q2 /C = Vi. Substitute in (4) Q1=[(1+m)/n](Qi+Q2-Q1-QP-Q -(Q2.-1-Q2-q- QP/n or (n+m)Qz! = nQl+mQ2(n+m)Q1!= nQ1+mQ2 -(n+m)Qi=(m+n+mn)Qi+rmQ2 -(n+m)Qi. Thus Q= [(mnQi+(m+n)Q2)/(m+n)]
Ans.
-34CSubscript 0 refers to outer conductor. Let charge on the rth conductor be Er and its potential Vrwhen all others are grounded. Then Er=qrrVr(1). Now raise 0 to potential Vo. Fields are unchanged so all inside points have their potential increased by Vo. Thus Er=qrr(Vr+V(3)-1-qr0V0(2) Subtract (1) from (2) toget qrr=-11.0• Since r may have any value from 1 to r, there are n relations of this kind. Q.E.D. With the outer conductor nraised to o, the systems energy is n V 14.4[i cirrV2r+ 2 E qrs VrVs The first term gives the energy in the field inside conductor O. If it is grounded, these fields are unchanged although all potentials are decreased by. V . Hence the o , n in qrs(Vr-V0)(Vs-V0) r>s Q.E.D. first term must equal 2 -35COriginal energy was 3E2/C = E2(b-a)/(8',evab). Final is ig2 /C = E2(b l -a)/(8rev ab'). E2 ' Work is W=E2 (1-2-1+ )/(8re a)=E2(1 11(gre )---.--. b-b Ans. bb' v -8rev b' v b' b
Chapter II.
Page 44. (44)
Page 12
-36Cb'-b Ans. Let the contraction b'-a (b- a) -s) occur with charge constant, then with a battery, bring potential to its original value.
Work done:iC V2 -iC2 V2 =217C
v2[212_ ab l ].2r,. r2ia
V
a- b
The battery work will be, since E = CV = 4revVab/(b-a). Wa -W1=2revV2 a2(b l -b)/((b-a)(b'-a)] -2revV2 a2 b2(b-b')/[(b-a)2 b131 ] = 2reV2 a2[bi(b l -b)(b-a)-b(b-b l )(b l -a)Yrb l (b-a)2(bl-a)]. =2re V2 a2 E(W-b)(bb'-ab l +bb l -ab). WW(b-a)2(b 1 -a)] =2revV2 a2(W-b)51 (b-a)+b(b 1 -a)]/[2b 1 (b-a)2(b i -a)]
Ans.
-37CTake three conductors: No 1 earthed, No2 uncharged, No 3 positively charged. Then: V3 >V2>0,
E2=0, 0=s11E1+si3E3(1), V2=s11El+s23E3(2),(2)-(1) gives (si2 E1-s135)
-(sI1E1-s23E3)=V2>0. But Elis negative, callit -E0, where Eo5E3, so silEo+s23E>sI2 E0+s13 E But from 2.17 (3) s.. ...1>s12so if -we equate E0 to E3the inequality still holds Ans. Therefore s11+s29> S12-I-S13 (2)'s13gives sl3 2=s1.2 s13E1+s13s E3. (1).s23gives 0=s11s29E1+s13s23E3. Subtraction 23 gives sI1V2=(s12 s1.9-sils23)E1-(sI/s23-sla s13)E3. All symbols in this equation are positive so both sides are.
Therefore sI1s23> sl2 s13.
Q.E.D.
-38CConsider rth conductor with charge +Er. Adding new conductor lowers its potential and all other potential differences. Vr= srrEr) Vs=srsErY Vr-Ve(srr-srs)Er' Similarly 17;.-17;=(s r-s;s)Er. AVr=( srr-srs- qr+s;s)Eris positive so srr -srs>qr -qs(1) and -s' (2) etc. Multiply:s s -s (s +s )+s2 sss-srs›s' >s'rr s' ss -s'rs (s'rr+s' ss TS rs ss)+s.2 TS rr ss rs rr ss 22 ss>srs(srr+sss -srS S; s-S;,s (S;r4-S;s) (3) !_ r s' so that srrsss-sE )
Adding (1) and (2) gives srr+sss>s;r+s;s+2srs-2s;.s(4). Substitution of right side in (3) will strengthen inequality so that for s rr+s ss srrsss-srs;s>srs(s;r+s;s)+24s-2srss;s-ers+sg-qs(qr+s;s) >ers-2srss;s+srs+(srs-s;s)(s;.s+s;s). The last term is positive since >s's, therefore dropping it strengthens the inequality so that srsr 2
srrsss -s;rsLs > (srs-s;s) Q.E.D. -39CPotentials of all conductors are ITI,7,1 2., V1= i
slsE s +sInEn +sme Ec +sId E d m-3. V = E S Es+s2n En+s2 C E +s ,E, C 2caa 2 1. a
M-1
(1) (2)
+s 4s E c +snd E = Es Vn 1 nsEs nn E n nc d m-1 +s V = E s E +s c 1 cS cnEn ccEc +scd E d m-1 V = i d sdsEs+sdnEn+sdcEc+sddEd
(n) (n+1) (n+2)
These n+2 or p+q+2 equations are in general enough to determine all p+q+2 unknowns. (continued)
Page 44. (44)
Chapter II.
Page 13
-39C- (continued) However, if d is screened from c by conductor n, then, by 2.17 (3), ssn = ssd, where s has all values except d. Hence in the first n+1 equations En and Ed coefficients are the same, so .only En+Edcan be found. If the nth conductor is in the B group both En and Edare unknown. The two equations, En+Ed=Q and equation (n+2), will not determine En,Ed and Vd. Similar reasoning applies if the nth conductor is in the A group. Hence the distribution is determinate if c is not screened from d but is indeterminete if c is screened from d. Q.E.D. -40CWrite
If E =E =0, their presence VB=sABE+sBBEB+sBCEC+sBDED' A C is immaterial, so that (3) determines sBB,sim and s n• But from 2.17 when all are present s =s s =s s =s AB BB' AD DD' CB BD. and s E +s E 4.s E +s E It remains to find s and sAcsBD' VA AA CC' A AA A AB B AC C AD D' Let EB=Ec=ED=0, then VA= 1/CA3)+(1/CBB)EA=EA/CAB+sBBEA (1) sAAEA=EA/C=1:( (s'AA -s' )E E /C Remove C and D so VI=s 1 E +s' and let E =-E so that V'A B A A B A A AB' A AA A ABEB' gives s from (1). Ans. Also s =s -s +sDD and s' s' Substitution for CC CC CD CC CD EA/CAB AA is known from (2),giving all s's. The c's can be found from the s's by 2.16 (2). -41C-
After 2 insulate and transfer to 3. After 4 insulate and transfer to 1. Vi=silQi+sI2 Q6+si3Q3. In 1 and 2: scB=
sBC=sAB=sBA=13' s—= UA sAC=sAA sBB=c1' sCC=" =s =p, s =s =s s =r. Let charges between CA AC BA sAB CB BC. BB =sAA =q, CC 4 and 1 be QA,QB,Qcat cycle start and QI;i,Q313,q at end. In 2, 0=r(q+q(QA-1-Qc)+pQB,
In 3 and 4: s
s
QkQA+Qc (1). But Q 1 B--QB+q. In 4, 0=rg+qQii5+pq (2). Elimination of
gives
0=r( q B)+ q( A+Qc)+pQB (3). Q =QA+Qc, Q 1=-(q/r)(QA+Qc)+[(r-p)/r]QB, and Q. [(qa Q _ pt)/I](QA+Qc).cc(r.p)/r AQ jQB, q+01'c=[(q24pr+r2)/i2 ](QA+Qc)-[q(r-p)/r2 ]()B. In steady state: q/q=QA/Q13, q/q=QA/Qc, Qiyq=-QB/Qc and (q+q)/q=(QA+Qc)/QB. (q2-pr+r2)(QA+Qc)/QB- q(r-p) -
Q1;
QA+Qc -
-rq(QA+Qc)/QB+r(r-p)
-(q2-pr4T2)x - q(r-p)=-rqx2 -r(r-p)x
or
QA+QC - B.Let-x - -----,then Qt QB rx2-qx+p-r = 0
(4)
Ans.
When between 2 and 3 let charge on A be EA, on B be EBand on C be Ec. E /E =4, =Q", E/E = (q/r)±pE /(rE )=(-e+p)/(0)=(-p+r-02413)/(rP)=(142)/P. EC o CA A B B E /E =E E /(E E )= P2-1. Thus E :E =-P:l and E :E 42-1:l so E :E :E =1:-0:P2 -1 Ans. C B A C B A A B C B B A C -
Page 14.
Page 44 . (44)
Chapter II.
-42C-1 -1 1 -1 From 2.18 (2) : V1=QICI -FQ2 R (1) . V2 =Q2C2 +Qilt (2) where R=41'rer.. From 2.19 (3) 2 R C1X(1) - RC1C2 X(2) gives Q .R2 CIV / - RC2V2 so Ct.i..-. R2 Ci 5 c 3.2 .- RC iCa 1 R2 -C C R2 -C1C2 R2 -C1C2 R2 -C1C2 1 2 arW/ar = 47reaW /BR.
2W = (r2C1V21- 2RC1C2 ViV2 + R2C2 V: ) (R2 - C1C2 )-1.
aW jar = LirreE (R2 -C1C2) (RC1V1-CIC2VI; +RC2V: ) -R(R2Ciy1-2RC1C2Viy2+R2 C2 14 ) D (R2 -C1C2 )2 . 2 ...2 . _2 _2 4rre (R2 CIC2 VI.V2-RC1C2 V i-i-ulu2V1V2-RC14 V22 ) _ 4rrec1C2(RV2 -C:Lvi.) (RV1-C2V2 ) Ans . (R2 -CI.; )2
(R2 -C IC2 )2
-43C(I) s 11 (Q-e,)+s12 ei= S22 ei+si2(Q-e1) or (s11-2s12+s22)ei=(s 11-s12)Q
(II) .11 (Q-e1-e2 )+S12 e2 +S13e1
2
Q1-e
3
Q-ei-e2 3
6
2
1 3
0 co er
I OS 0 6 e; e1 e2 0 e1 12 -s22 )e2 +(sIA -s/.2 )Q=0 (-s.3.1.+s2 +s12 -s23 )e1- + (-s11 +2s
. = S e +s (Q-e1-ea )+s e z2 a a I 23 ll or (-s11+s12 +513 -S23 )e1 + (s11-2s12 +s22 ) (e1-e2 ) = 0
(2)
(R) s1I (Q-e;.-er )+s12 er+si3q. = s22 er+s2i(Q-e;-er )+s2aer (-sra.+s 12 +s13-s23 )e r' - (s 11-2s i2 +saa )er+(sin. -s-3_ 2 )Q = 0. Substitute from (1) and (2) 2 and cancel out s1::. -2s 12 +s22 to get (e2 -ei)q-eier+e i=0 (3) . Write from symmetry 2 2 a , +e =0 (e -e -e (4) Eliminate er to get -e1er+1+(e2 -e1) e'+e2 (5) r 1e2 = 0 e'-Fz.1 2 I )er 1-r +e21e2 = 0 (6). Let a=C(e2 -e1)/e1 r <1. Then from (5) and (6) and -e2a.e r' +(e2 -e1 )2 e' r..1_ or
e' -(1+a)e l-Fael = 0. Let e' = A'kr . Then, see 5.081, 2k=l+a*14-a so k=1 or k=a. r r -1 r +1 r „a r solution is e '=A+B: (e2 -ei) /en. j (7) At r=1, ei.=A+Ba; At r=co , e I=e so from (3) T r 1 -1 2 /(2e1 -e2 )=A since a<1. R= (ei-A)/a=e?.(2e1.-e2 ) (ei-e2 ) . Substitute in (7) er'=er=e.i e;.=e21 (2e1-e2 )-1[1+[ (ei-e2 )/e3 ]2 " 1] Ans. Substitute this value in (3) to obtain 2 .2T em -2 r er= [el. / (2e."-e2 ) [ 1- (ei-e2 ) ] Ans. -44Co) s22 (1) In 2 contact: sn.j.qn1+s12 (Qn+Qn )=saa (Qn+Qn1)+s12 gni s1 =sia , s23 =s22 , s ]
But Q -FQ I =E-q so (s11-2s12 +s22 )qn'+(si2 -s22)E=0= (ce+P ) qn' -PE. so qn1=-7--0,PE ii3 (2) Ans. n njBE _ ' 0E + -Qn . In 3 contact : s11Qn+s12(Qn _i+Qn1 )=s22 Qn _a+s33Qn'+sa2 gni Qn1=E -Qn_1- qn Qn'=E-ctii3 Qn=513
so that (s1I-2s22 +s33 )qn+(s33 -s22)Qn _ i+(si2 -s75)E=0 or (ce+Y)qn+(Y4) Qn -1=YE= (ce+Y )qn+1+(ce-13j PE (Y la) fry % (3). But Qn-ga (4) so (a" -P-a-y)cin- 41+13 ‘ -4N / qn-(a4f)cin _l 0,448 ' r 0'41 n or cln+1-(1+ciTy)qn ta.fron_a=0 As in last problem let qn=Akn so qn=B+C(a+f3)n (a+1f) (5 ) As n approaches infinity qn approaches qn or f3E/(cr-i-f3) (6) so B=8E/ (a+0) since a-FIEKce-Fy YE OE . _ce-FP from (1) . Qn _1=0 on first contact so by (3), (5), (6) ql=c),.. 7r-ce.444 1(..;(7-7, ±. Thus Cs-a(Y-P) 07_0 )2 E(7)
Q$ _1
Put (6) and (7) into (5) to get _
4rI
. ._E_ G syi\L-Il acy.43)/(cv+y.r ).
] n-1.1
a+P L
P (ck.+Y)
ChapterII.
Page 45. (45)
Page 15
-45C1 1 1 t 47'T b-t-a Capacitors in series. Before: Co= 4,72.32. After : - =- +- . --= + a CiCa C3 Cl ea(a+t) Evb(2+t) ea (a+t) ab (bev -ae)t+(b-a)ed C-C (bev-ae)t+ea(b-a) b-d • ev eab(a+t) rev 4rev ab[e(b-a)-bev+ae] 4rev (e-ey)b2 t Cm Ans. -e Co (b-a)[(bev-se)t+(b-a)eaj e(b- a)d -
-46Ce=C3.(Vi-V2 )=C2 (V1-V2 ) ' , Cl/C2 =evi e, Wm-142 4e (V1-V2 )-ie (V2. -Va ) ' =-keiVi -V2 -(cv/ e ) (171-172)3. = ie
)(e-ev) ie •
Ans.
-47C -2r(e+ev)ab Ore ab 477eab 4Tle ab CIV=.-.--17--z(b _ a)VQ3., Ca v=2 (o_ a)Q2 V, CV=Q3.+Q2 V b-a b-a Thus el=i(e+ev) Ans. -48CEnergy inside first capacitor is unchanged. outside energy is IsT1=Q12(8T7eb) -1, Q5.+Q=Q1. Qi/Q=477ebif4rre ab(b-a) -1+4rebj e(b-a) w_ Q (b-al C2 2 8rreb 8rb[ae2 +(b-a)e] a ea +(b-a) e Qt[
e (b-a)2
J b-a 2€(3-a) e2 (b-a)2(1-1) b-a b[ae +(b-a)e:]' "1-n2 8r[-eme[ae2 +(b-a) ej-e[ae2 +(b-a)e ]a a e2 +03-0 e] 2 b-a [ Q2 ae2 +2 (b-a) J Ans. 8r7bre ae2 +(b-a)e] =[8reb aez +(b-a)e]
8TTEbe[ae2+(b-a)c.,2
-49Cq / /q2 =e/e
qi=eqi(.e+e e ), q2 =e q/(e+e' ) . Normal unit area force
is f=a2 /(2e)• a =a /(T7a),
a =q2 ( 77a), f =q2 /(2eTa2 ), f2 =e2 /(2e'r2 a2 ). 6 1 Tr/2 fa cosed0=2fa. Component normal to dielectric face is j dF;--.J.fri2 2 2 2a e a eq2 e' q ' 2 e-e Fi=7--a— w 2erp az = TT 2a(e+e -- )2 Ans. a F2 = Tir +€1)2 • F=Fm-F2 -q-, -50C2 Apply Gauss theorem to sphere of radius r. ' eEdS = 4T7er2 E=q so E=q/(4T7er ). The r n-1(ear2)dr+J -1 ran-2 (e 2 -1 potential of the point Pr is then: i(q/r7)[..1 co nJ.r ) dr + • • • a an-1 •-1 r 1 1 1 ) + .... + —(_____ ) dr + S (Csr2 )-1dr3 =-9-E 1 ti a s e 'a a 4T7 en ao _i a (es-Fir2 as n.lia s+3. n-1 n-1 1 1 1 1 1 1 + 7---( —) + (- - -)] Ans. esr as a a S+1 s+1 s
+P
Page 46.(46)
Chapter II.
Page 16
-51CA/b
y I This solution is valid only if the slab is far enough between the plates so that there are two regions where the field is essentially uniform, of length y and yo-y. Let the capacitance of those regions where the field is not uniform be Cl. The energy is, since (e - ev)t = et', YP ?INT :22_L_u_ re b e b eye_htl_ Q2 2C 2[Ci+(evyb/d)+ evb(Yo-Y)/(d-t')]' F=Toy =2C2 d d- t' 2e(d-t') This is as exact as the above assumption. If edge effects can be neglected we have E (A-x13) C bx ev[A(d-t1 )+bxt' . Substitution in F gives C v d-t d d(d-t') evQ2bt' bdeti (d-t') oe (d-t')2 F Ans. 2(d-t') 4IA(d-t1 )+bxt1 ] 2ev[A(d-)+bxt'3 e 2 ]
-52CSurfaces 1,2 and 3 may be replaced by uncharged conducting sheets without disturbing the field. The capacitance 1 to 2 is A and 2 to 3 is B. If 1 to 2 is filled with dielectric e and 2 to 3 with e', the series capacitance is 1/C=ev/(eA)+ev/(e'B). The charges on the two faces of 2 at potential
V2are
equal and opposite and the
field is normal so that if it is removed the boundary at the dielectric surface, =V, and e(av1/a0.0(aV2 On)are satisfied and the capacitance 1 to 3 is unchanged. V2=V2 I C 53Cr 2 . BV avi Apply 1.15(1) to surface element dS. Divide out dS. elionm 2 011 / 2
e E1cos c -62 E2 cos C2= E1sin c= E2 sin c2, eEcot c sin c2 1 1 Ans e2 E2 cos C2= a.Thus E2= a/ [sin ca (e1cot c1- e2cot c2 ) 1 2 1 a a e mcas c,- e2Eicot c2sin ci= Q. So eicot c2= eiCOt C ) An e cot c (1 e1 E1cos ct 1 1 E1sin c1 1. -54C1 r 1 -1Tcrl, -1•---r E=Q/(4rer), e=Cril. Potential at r is P. P-A=IaEdr=1 4rC n+11.1a J 1 1 -Q n+1 nI n+1 P-A=(A-B)(bi r'-a'b , a =a I )/(a l r'-b'r') b'=b Let r =r , B-A= 4rC.T;1[1771-an+11' 1 a'b'A-a'b'B 1 a'r'A-b'r'B Pa'-b' a l -b' a'r'-b(r' r' n+1 n+1 A-B a b a'A-b'B Aan+l-Bbil+1 a'b' A-B r n I an+1-bn+1 Ans. r •a-b 1- an+i_bb+I a'-b' ]
Chapter II.
Page 46.(46)
Page 17.
-55Ce=e1+(eil -ei)x/h. From 1.15 (1): ebliThc= a, dv.-afe2+[x(e2-€4)/h]r1dx ,,h, V = halln[hel+(e2 - ei)xJ10/(eja -e.1)= ha[ln(he2 )-1n(hei)]/(ea 7e.1). V=haln(es /e1)/(e2 -el) C = c/V = (e2 -c1)/[hln(e2 /e1)]
Ans.
-56C4reaV/ax=-Qr-2 , e=94Iev, dV=-Qrdr[4rrev (c+r)r2 j-1 . 0-4re V =-Q1b [r(c+r)ridr ,vrr so Vr = -Q(4rev c) -iln[r(b+c)b--(r+c)-11 Ans. i) . =(Q/011o (o+r)/r]l r [
2 e=f-Lev /(eP p3 ),
2 2 -57CQ(eb/a ___I_rb/ae p2pdp -e) e dr V V 4rev ' b a 4tIev4a.1 8reolla 13/a2 2 -1 C = Q(V -V ) 1= 8revga(e -e) Ans. b a - 58C-
rbc3Q p 2
Spherical equipotentials satisfy all boundary conditions since the dielectric constant varies only over equipotentials along which the gradient is zero. av a rr av rr 1.15 (1). -Q=Jje dS=Jjevf(0,p)-.5 r sin0d0dcp. This integral is independent of r for uncharged dielectric so r2 aV/ar is constant. rbdr ab(V -V1) 1 1 equal to Cl .j v2 dV=Cij aT or V2 -V1=C1(17.; - -;). Thus Cm= 2 b-a Substituting: -Q=ev (V2 -11.1)[ab/(b-a)]Jjf(0,Y)sinOdOdo C = Q/(2 -VI) = abev /(b-a)Y1f(8,0sin0dOdy Ans. -59CSince the disk planes coincide with equipotentials of the z-component of the field, it can produce no charge displacement. Hence, see 1.19, e
=e31=0, D2=e33 z. 32 However, the y-component will probably not coincide with a symmetry axis of
the disk so it can produce an x-displacement. A similar argument applies to the x-field.
Hence D x = eIIX +
Use 2.09 (3). c= evE. ae' aT
ae'
dx
ax i . Tri =
T=
1Y ,
Dy= e12 X+ e22 Y. Ans. -60C-
m/x, if m is unit area mass.
but eu=ev is constant and
a(1/0) ax
=z--4.7==nEx
m ales] Z7 ] a2 [e e -1 a(1/e) c ,2 3Tc ' so F - , r-1,7x x j Ans
dal [e"-e' m ae' .3-3-c . Therefore: F = -277-777 +
T'
4
-61C-
'v'
e=ev (1+0P). From 2.11 (1) to which must be added the electrostriction term from 2.09(3)v aStl[ 1 4. 04 OT 2 2 2 431 -75-2-.19 = i01 (1-02 T2 +...)/ ev . Ans. 6e/aT. But D=a so r=e11[1+0T T Thus F=TD /e -62C2 -1 9a a 1 In polar coordinates the equation is: r - -7(r cos92 r1cos0 ) = a lo 2 2 This is obviously an equipotential about doublets -
of strength 47ev •9a/16 and a charge 47ev placed
as
a
a
shown. The potential V is 1/a and the charge inside is 4rev so that C=Q/V=41Tev a. Ans. (continued)
Chapter II.
Page 47. (47)
Page 18.
-62C- (continued) Equipotentials about this surface will have the form 9 4reV = (IT - --aq(r 2cos0 -r-2cos0 ) 2 -1 16 1 g9 r -a bV -3 Q = -ev67 = 7. 4 - TjaqLr2 rlJ). J On axis rn.= r-a, ra = r+a, t 7(r
-
li(r+a)--(r-a)Y=a-1
1 92 a or — —a r(r2-a2 )-2 =a -1 so 4(a-r) (a+r)a + 9a r =O. By inspection this is satisfied r + 4 by r = 2a.
Therefore on axis r=2a, ri=a, re3a. Substitute for a
1 11 1 9, 1 rl ) 7777 (141) a 3JJ = 17717L 16q q Ans. q = 3.167a2- 3ra2 -63CP =V /V when Q = 0 and V =0 if q s or q 0 r TS r s r q 0= C r1V1-I- C V -I-***-1-c rr Vr -I-C VS-4-•••4 c V rs rb b r2 2 If V = 0 whenqsandq 4 r then 0=c V +c V so V /V =-c /c rs s q rr r r s rs rr Ans. = -c /c So P rs rr rs 1S q_ = 4r/42
2.12r 1
8 127 0
/
Chapter III. Page 60.(60)
Page 19.
-1Substitution of h1, h2 , h3 from 3.05 into 3.04 (1), (2), (3) gives answer. -2Substitution of h1, h2, ha from 3.05 into 3.04 (1), (2), (3) gives answer. -3Multiply 5.01 (1) through by (1 + 0)(b2 + 0)(c2 + 8) to obtain x2 (b2 +EI) (c2 +6) +
y2 (a2 +0) (c2 +0) +z2 (a2 +0) (b2 +0) - (a2 +0) (b2 +6) (c2+0) = 0
(1)
If x, y, z and u1,u2 ,u3 are the coordinates of any point P, then this equation is satisfied by putting 0= ui or u2 or u3 , since the 0 coefficient is -1, it a a a must equal (u1-0) (u2 -0) (u3-0). Putting 6 = -a or 0 = -b or 0 = -c gives 2 (c2 y2 (2 +O ) (U +b2 ) (U3+132 ) -b2 ) X2 (b2- a2 ) (c2 - a2 ) = (um+a2 ) (u2 +a2 ) (u3+a2 ) or 2 2 2 2 2 2- 2 or z2 (a2 "-C ) (b •C ) = (111+C ) (U +C ) (U3 +C ) (2) Differentiate the logarithm of 2
both sides, keep us , u2 constant: 2dx/x=du1/(u2 1-a2 ) 1 2dy/y=du1(u1+b2 ),2dz/z=du1(u1+c2 ) (3) Then, going along side OA in Fig. 3.03, where u2 and u3 are constant, z x2 (ui+a2 ) -2 +y2 (ui+b2 ) +dy2 +d z2 (4) +2 (U1+C2 ) dS2 =dx2 Substitute for x2 , y2 and z2 from (2) ds =
[ (u
2 2 (u1+b ) (U3 +b )
2 (U 2 +8 ) (U 3+a2 ) +a2 ) (b2 -a2 ) (C2 -a2 )
2 (u1+c ) (U 2+C2
)
2
1
du1
(u2 2 ) (u +b2 )(u + c2)T1du2 = h2 du2 from 3.03 (1) = t (u - u2) (U - U ) [ (u + a 1 1 1 1 s 1 3 1 1 Thus 1 -4-(1.1 -U )(u -u ) i(u -u1 )(u -u2 ) 4(u I -u2 ) (U 1 -U ) 2 1 2 h2 h2 = h21 (u +a2 )(u +132 )(u +c2 )' 2 (u +a 2 )(u2 +b2 ) (U2 +C2 ) ' (u +a2 )(u +b2 )(u +c2 ) 1 1 2 7uuu Solve for x,y, z: x =
3- 2
2 2 2 .4".11 (U -b ) (U2a -b2 ) 2 2 Y b2 (b2 -c2 ) '
3
ab
Z
2
2 2 2 2 ••11 (U -C ) (u2 -C ) 1 2 a ) c2 ( 32
Differentiate the logarithm of both sides keeping u2 and us constant. dx/x= du /u_ = dy/y= dz/z. ds2 = dx2 + dy2 + dz2 = h2 d = [ ( x2 +y2 +z2 ) Rial]dul = dui so .
h2a.= 1. Ans.
Do the same keeping u and u const ant so that dx = u u_du2 /bc, 3 2 2 2 13 1) 0.4 2 2 dy =fu2.u2(113 -b )'/[b(b2 -c2) * (u2 -b ) 3jdu,dz = u 2 1u2 (c -U3 ) dua / [c 2 ) (c2 ) (c2 . ds2 = u2 (u2 -u2 ) u2 [(u2 -b so h22 = u21(4 -u23 ) [ (u2 Ans. U3 ) ") / 2
3
2
2
Do same with u and u constant. dx = u u du /(bc), c2 , 2 2,* du /[b(b2 _ca)*f_a_h2)Til dy = u1u3(u22 -b2-) * j tiz41.1 U012.-.C2)/[c(b2 ) **U ) 3jdu3 \"a 2 • ds2 = U22
2
)dU 3 3
3
/E(U3 -b2
)
(
I-L3
3
SO h2 = u2 (u2 -u2 ) /[ (b2 -u2 ) (c2 -.23 ) 3 3
Ans.
-5= c2 (i..u2 ) (u2 l• ) "z= 1 2
u 3. Thus ha =1 Ans. Keep u u constant. Solve: Solve: x= cu u 3- 2 ' 7 2" 3 (u2 _112 )02 _ 1) -1du2 0 _112 )'/ (u2 dx = cu2du1/ dy= [cu1-1)T ]du 1 , ds2 = dx2 +dy2= c , 1 1 2 2 1 2 il -1= c2 1112 -u2 N (u2 _ 1) Ans. Keep u u constant. dx=cu1 du2, dy=-cu (u -1) (1-u22)-* du h21 . 2 1 1' 3 ' 1 2' ' I. 2 2 • '' 4. 2 2 2 '1. ''' 2 2 • . 2 2 2 2 Ans. h = c (u -u )0.-u ) ds2=c (u -u_)(1-u ) du` 1 2 s 4 2 27 2
Chapter III.
Page 61. (61)
Page 20.
-6Solve: x= c (u2-u1), y2 =4c2 U1U2' z=u3, Thus h7=1. Ans. Keep u2 ,u3constant. 2 ) /u . Keep u ,U dx= -cdu2dy= cutu;* du/, ds2 = C2 RU12) /U11J du21 . So 2 = C2 1 2 3 _1 c2 (u +u constant. dx=cdu2 dy= CU u2 Vdu2 1 ds2=c2(u1+u2 )du2/u2. Thus h2= ) / U . Ans. I 2 2 -7Write in cylindrical coordinates. p= 2cu1u2, z= c(u1-u2), u3=.4..Keep ulu3constant. ds2 =dz2 +de)2 = c2 (LIti-U2 )de t./Ulthl= C (Ut+112 )* -*. Ans. Keep u17 u3 dz=cduI,c1P=cu2ui constant. dz=-cdu2, dP=cu1-2 `cu2, ds2= C2 (L -12 )dU2 ) h2 = C (U Ans. 2 -8Write in cylindrical coordinates. P= csinhut/(coshu1-cosu2), z= csinu2/(coshu1-cosu2), 0= u3. Keep u2,u3constant. dP= c(1-coshuleosu2)(coshut-cosu2)-2 du1. dz = csinu2sinhu2(coshu1-cosu2 -3: Ans. 3.=c (coshul.-cost2 ) ds2=c2(coshut.-cost)11. -2 Keep u1l u3constant. dP= -csinu2sinhu_1(coshu1-cosu2) du 2,dz=c(coshut.cosu2-1)(COShU.1-COSUET2 du ds2=c2(coshu1-cosu2)"2di4. 112=111=c(coshu1-cosu2) -1Ans. h3=P=c sinhu1(coshuI-cosu2)-IAns. 9 -1. Then P = cDsinu2, 0 = u3, z = cDsinhut.. Keep u2,u3constant. Let D = (coshuI-cosu2 ) dz= cD2 (1-COShUtCOSUdd111, dP = csinu2sinhu1D2du1, ds2= de+dz2= c2D2du1, hl=cD. Ans.
Keep ul,usconstant. dz=-csinu2sinhu1D2du2, dP=c(cosu2coshus-1)D2du2, ds = c2D2d4. lilt= 112= cD.
Thus
Ans.
113= P = c sinu2D
Ans,
-10Cr/a Ce-r/a Substitute e/ev= Ce in 3.05 (1) and divide out to obtain: r 2 (r2aV/ar)/6r - a-2?-V/r + (r2sin0)-1a(sinedV/a0Ya04(r2sin20)-la2V/202 Let V = R(r)e(0)M(0), M = eiPt(1), p is an integer then above becomes • (r2dR/dr ) /dr - r2(aR)-1dR/dr + (esin0)-1d (sinede/d0)/d0 - (p/sin8)2 let (esin0)-1d(sined8/d0)/d8 - (psine)-2= -m(m+l) so e(0) = PIP , (cos8)
= 0.
= 0.
(2)
• (r2dR/dr) /dr - r2(ar)-1dR/dr = m(m+1). Put u = r/a. then differential equation is d(u2dR/dul/du -u2dR/du - m(m+1)R = 0 = L(u). Try series of ascending u powers: Let R = F 0 an un+a, dR/du = L(u)
Ea
(n+a)un+c.-1, d2R/dr2= E an(n+v)(n+a-1)un-lc-2
+ a [c(a+1)-m(m+P] ucT n-1(n+a-1)1un-117 o (a-m)(a+m+1) + (n+a-m)(n+a+m+1) a =ao (n+a-lj] un4cr = 0 so a=m or -m-1. o n n-1
=
0
[(n+a) (n+a+1)-m(m+1)] -a
n
an = (n+a-1)(n+a-m)-1(n+a+m+1)-tan-1Need solution independent of 0 so p =0. It must involve cos() so m = 1 and to be finite at origin a = m = 1. Therefore _ ! R = a [r/(3a) + r2/(3-4a2) + ra/(3.4-5a3) + —1 (3) = 2ao(a/02 pr /a_ 1 (r/a)--(r / a) I. o • R cog! Ans. (continued)
Page 61.(61)
Chapter III.
Page 21
-10C-(continued) At r= cc, V = Ex = Er cos0. Perturbation term outside must vanish at r =00 and contain cosO (a dipole) so Vo= E(r+Ar-2)cos0, Vi=EB(a/r)2 Leria-1-(r/a)-i(r/a)2 ] (4) -b/a At r=b, Vi=Vband (e/ev)aVi/ar=Ce (aVi/a0= aVo/ar, so that eb/a..(a/b)2_(2/b)41, b+ab-2=B[(a/b)2 1-2Ab-3=BCab 2C1-2(a/b) (1+e -bia) , so 13 -1[21)-2 eb/a_ aa b-2 B=3{Cab 2[1-2(a/b)](1+e -bia)(5) -1 2 -1 -1 b/a b/a 1.3 -2Cab (1-2ab ) (1-Fe ) + 1 +ab + a b (1-e ) A = D Cab-1(1-2ab- 1)(1.1.eb/a)+1 (6) ab 1+a213-2 (1-ebia) Substitution of (5) and (6) in (4) gives answer. -11CProceeding as in 3.14 gives ald = krv6e(V11)2 dv +
YelV.pdv ( provided we J J mayinterchangebandVin3.14(1)0ThesecondtermvanishesbecauseV,the potential of the jth conductor, is constant. The third term vanishes because charges are confined to conductors so that p is zero in v. 6Td = iS(VV)2 6edv. This is positive if be is positive. Q.E.D. -12CLaPlace's equation for a field of revolution about the x-axis is, from inspection of 3.05 (2), y-lav/y + a2VRy2 + a2x0x2 = 0
(1). If the rotated V2 curves are
62v/ay2=f" (v. ) (aV2 /ay)2+fi(2 )a2 v2 /ay2 f"(V2)[(aV2 /ay)2+(aV2/ax)2]+ ft (V2 )r(a2 /ay) /y+ a2 v2tay2-1-a2 v2/a xe
at constant V then V=f(V2). aWay=f1 (VdaVa /ay, Substitute in (1).
is zero. But a2 V2/ax2+a2 V2/ay2=0. so y-laVOy[(aV2/ax)2+(aV2/ay)2 ] -1=71211722= F(V2 ) V(172 ) F(V2) = -f"(V2)/fl (V2)=-a[1n f'(V2 )]/aV2, IF(V2)dV2= -1n[f'(V2)] + C Therefore V = f(V ) = Afe-If(172 )c1/72 dV2 + B 2
Ans.
•
Chapter IV.
Page 105. (101)
Page 22.
-1The flux line going to the right along the x-axis to infinity returns on the y-axis to -2q. It is the V = 0 line. The line going directly from this +q to -2q is the V = Voline. By Gauss Theorem the flux between P and the x-axis is 2rq=2-20+y so TfV
= Vo(ce-
25+y) since both flux function and potential
satisfy the LaPlace equation. Thus V = !(tan-1 Y - 2tan-1 Z + tan-1x+a) Ans. -2Vo
As in the last problem with the charges as shown; V= —(0 rr 3.-02+ea-o4 ) and q = 2Ve so field of line charges is 2rea, normal to x-axis 2eVx(a2+b2 ) 2 1 x 1 2eVf 1 Ans. • (x2+b2)1 + x+a) r, ( x2 a2 ) (x2 +b2 ) cr= `x-a - (x2+1 37 7 -3Invert cylindrical equipotential of radius Q'P' which surrounds Q'.
--It inverts into an equipotential cylinder of radius a with a non- k
concentric charge Q inside it plus a charge -Q at 0, the inversion
f
center. OQ'/(k+a) = (k-OQ')/a from similar triangles.
k(k+a)7,7;
k(k+2a)
(")1=k+2a ' "= OQ'-k+a • Let f = OC = k+a, then QC = k+a-OQ
./
1
=a2/(k+a) = a2 /f. So cylinder is equipotential with charge -Q at distance f from axis and charge Q at distance a2 /f from axis.
Q.E.D.
-4From 4.00 (1): a =en.Ep =(2r)-1q(a-l-r-lcosit+41cos0), r2+a2-2arcos*=ta. P r2 + az .. 2 2 2ar2cos0= 2a2, so cor = (ri-a Thus cos*= (a2+2r2)/(4ar))/(2arl) , 1 1 2 2 7 l - a -2 ar cosO= a2, so ia2+ 2 arcosO = r2. r2 2 =r2 + +2 ar cosO and r + i i Substituting: r1 2 r so ooss6 = (2r2 -a2 )/ (2 2ar). Thus we obtain 2r2 -a cr . __q_ [3._ a2 +2r2 22q Ans. +(20-1(2r2-a2)/(2i2aH . - 4rfar 4ar2 p 2r a .
From 4.22(3):
W=
-5-1 1-al) sin (zi/a/), dW/dz1= (z2
On ellipse V= Vo.
U=
(dW/CIZ)m=(aiSillhVo) -1 .
so c /47 = coshVo/sinhVo= B/A from 4.22(4). M m
dzi/dW=alcosW -1 U=0, (d14/dz) =(a coshVo) . m I Ans
+q
4
Chapter IV.
Page 106. (101)
Page 23
6 charge I t.polarizationl ‘n colln0 q lnr +E Anrncosnei V2- q E4 1(1 )ncosn0 + C + Dlnr V = 2r_ iv e r ` 1 2rev Fr' n n -n so n (c/a) +An a = Bn a (1) For force need only A. At r=a: Vi i --
—
2
ev Vi/ar =e-OV2 /ar so -cna-n-1+ nAnan-1=-(e/ev )nBo a-11-1 (2) n-1[(eiev)..1]+ an-l An[(e/ev)+1]n= 0, (1)• ne/(aev)+ (2)gives cna =[q/(2rev)]E Anrncosn0 is potential of charge at An =Re-ev )/(e+ev )]cn/(na2ri ). V pol r = 2/c, 0= 0. q'= (e-ev )q(e+ev) -1. F=qq1 (2revR)-1= ce(e-ev)[(e+ev )2rev (a2 -c2 )]-1Ans. 1
2
Cl2
0
-74_0_4 a2 b2 24-
Solve for b2 : 2b (a2 b) -b ( a2/b)+b 2b2 a4- 13'). 1 *- 2] *Ans. 2b2 =-4a2 ± (16a4+4a4)2, b = a[-2±(5) ] , d= 2b= 2[(5) -8From Fig.4.13, lines of force of two line charges are circles. Superimpose V1and V2 . (See Prob.2) Vi = (Vo /r)tan-1 [2ay/(a2 -r V2 = (Vo /r) tan- [ .2ax/ ( a2 -r2)] .
V = V1 + V2 -9-
Ans. V=
The x-axis boundary condition is satisfied by the force lines of two
1
equal and opposite lines charges. Superimpose to turn these lines back
b along the circle r=b. From 4.02(1): for two line charges rev:4= qE(2n+1) 1 (a/r)2n+Icos(2n+1)0 so reVq=E[(2n+1) -ka/r)2n+1+Anran+lcos(2n-11)3 V=Vo r 2 n+1 2 zn+1 ben , /b + (2n+1)A Jcos(2n+1)0. Thus An=a /(2n+Db4n+2 At r=b, av /br = 0= EL-a n q V =f(r,O) is the equation of the force lines (dotted) in this problem. To get Vo
VLF'
equipotentials where there are now q/(2e) take q = 2evo and write U for V. Thus a2 n-F1 [ r-2 11-1+ (rib2 )2 n+1 ]cos (2n+1) 8 Ans. U = (2Vo /r)E (2n+1) -1 -10CPotentia 1 V at P due to three line charges is, 2reV=-1nr 1-1nra -1nr3 2 2 2 or 4reV =-1n(r r r ). Equation of equipotential through P is 1 3 e2 2rccos0 r2= r2+c2 -2rccos[(2r/3)-0], r21r2 r1 = r2+ 2 r2 constant. r2 = 3 cos0+ 3 sine, -2rccos [ (4r/3) -0]. cos [ (2r/3) -0 ]= r2 3 = r2 +c2 r2 r2= r'-r2 c2+c4-2r3ccosO -4r2 c2 cos2 0. cos[(4r/3)-8]= 2 3 r2 r2 r2 Ans. (3cos0-4cos30) = r6 + c6 - 2r3c3 cos30 +2r3c3 = r6 4c6 1 2 3 ,
2
-11CFrom 4.13, all force lines are c liirc il1: From 2aP y rob. l,when V0=V/-V2, :r.I_ i -V
-1
-1 y +C= x+a
17=V1 2t an 1-.-:.7-- - tan --117-
ft-a
Ti
2
vi-17
v+C-----atan ratan 3<2 +y2 _a-
To find C set r=a, -ir<8<2r. Then V2= •• 1.7.1
; 7
-12arcosO +C. ••• .22
ir + c so C = *(VI+V2)
2
-1r
Therefore V = i(VI+V2 ) + [ 071.- V2)/r]tan L2arcos0/(r2 -a2 )J
Ans.
Chapter IV.
Page 106. (102)
Page 24.
-12CFrom 4.04(7) polarization field is equivalent to charge q' at distance a2 /c from center and charge -q' at center, where q' = -q(e-ev)/(e+ev ). Thus cicl i r Ans. F 2rev c c-(a2 /c) J e+ev 2revc(c2 -a2 )
a2 q2
1 -,_!...,
-13CImages shown make each circle an equipotential. Distances of 1,2,4 to 7 are (2/3) a. Distances of 3,5,6 to 7 are a/3. Add up densities at P due to image pairs 1-5, 3-7,2-6 and 4. Due to single image pair in[( f2+p = -q _2 -2f Pcoss6) (P2+a4 f-2 -2Pa2 f-1cos0) -11 2 o . _eLT 1 _ q 2a-2f cos0 2a-222 f Icosg3 1 a -fa ao P=a 4r f2 +aa -2fa cosy!' - a2 -4-a4r2 -2a/ f -icos401-2raR2 q itrev
For 1-5 pair: f=2-a, a2 -f2.-a2, R21=r2 +2a2 /3 - 2ra (2/3)icos r(247/3)-e] For 2-6 pair: f=2 a1, a2 -f 2,__ a2 , 4=r2 +2a2 /3 - 2ra (2/3)icos [(2r/3)+01 1 4 3r2 3. 2 For 3-7 pair: f=(3/2ra,a2 -f2=-2a2, R23 =r2 +a2 /6+(2/3)2cose, cose= 2.6car 1 q [_ a2 a2 a] q a r+,„ i 2 2a2 c=-2rra II. -m-z Ft +2Rs+1 -2r -R22.112 + 2R24-1-e and since r + -3- - 2(2/3) ar cose = a2
-e
,
R2,.+R. 2 .2(r2 +
2.2
/3)-2r (2/3)k, Cos (2-r/3 -6)+cos (2r/3 +0)] = 2 [r2 -1-2(a2 /3)+(2/3) ra cosh' =3r2 +a2
8ra 22 i 2 R21.--2 R2 = (r2 +2a2 /3)2 -2ra (2/3) (r +2a2 /3) [cos (2r/3-0)+cos (211/3+0)1+ —cos (2r/3-9) cos (2r/3-4-0) -1 2a2 3r2 -a2 _ f3r2 N2 2a2 -a2 ' - 3r4 -r2 a2 +114 /3, 2R23 = 3r2so that ) 2r` a -i-k =(r2 +—)2 +(r2 + 3 3 3 9 -1 1 1 _ qa(3r2-Fa2 ) (-9r2 a2 1-9r4-3T2 a2 -Fa4 ) 2 43r a = -2qa + -1;(3r2+a 2171 3r2 a2 (9r4 -3r2 a2 +a4 ) 4 -r2 a2 +a+ /3 1 3r2 a2 .1 q ( 3r24 .92 ) (3r2 +Jar ) ar ) (3r2 -a 2 a2 -6 2 _ Ans. 6 rar2 (9r4 -3r2 a2-1-a 4) -14Use 4.22(3). W= stn-1(z1/a1), x=0,
a=
e(y2 +c2 )--2-
a=
e 6W/asz I= el(z2 -a2 A-1where c=a. On vertical plane:
Ans. On horizontal plane: y=0, x c, a = - e (x2 -a2 )
Ans.
-15From Fig.4 . 22a . [J] = 2r and V1=0 on strip. Set x=a, U=fr in 4.22 (3.5). a=aicoshVa =c cosh; . From 4.11(4) . C = EEU3/(V2
= 2re/cosh-1(a/c).
Ans.
-16 From 4.14(2). c-= recosh -1(c/a) . From 2.07(4) . W = Z.Q2 /C.
2 /[2re,c2_ ( 4a-
Ans.
-17Fig.4 22b. sinW = sinEirz/bJsinCirra/bJ, When U=0, sinhV0 sinh(illz i./b) /sin (fra/b) When U=ir, coshVi=cosh(irza /b)/sin(*ra/b) . Square and subtract to eliminate V. 1 = [cosh2(irrza /b)- sinh2 (irz,../b)3/sin2 (ira/b)=Icosh( ►za /b)- cosh(rzi/b)+ 2 ]/ [2sin2 (i ira/b)] cos2 (*rz/b) = sinhrir(zi+z2 )/b3sinhr1r(z1-z2 )/b], z1+za =2b, z1-z2 =25. Therefore cost (ira/b) =sinh rr sinhrO/b
(sinhr) ►6/b. 6=bcos2 (ira/b)/ (TT sinh r )=0.028bcos2 (ira/b) Ans.
Chapter IV.
Page 107. (103)
-18Za. C = C ln— gives the unit circle in the By 4.12. W= C ln 1 z -(1/c) 2 1. CZ -1 plane at potential U = 0, with charge q at zl=c and -q at z1 1/c. Apply the transformation z = lnzl
,
x=rI, y=01, which puts the ziplane
into a strip of width 2c and solve for U. Thus x x e2x+ca -2cexcosy e cosy- c+ je siny ez-c - C.. In w4q, W= U C 1002 erix-F1-2cexcosy (1) ce— cosy- 1+ jcex+c siny' Find c and C such that when x=b, 6u/x= 0; U=Uo . Solving for bU/ax: 2 e2 b ab b b c - ce cos a 1aU b -e t-ce cos a - 0, Solving 'for c: Claxff7=7 i' e2 -2ceucos al-c2 c2 e2 b-2cebcosa + 1 c(1-Fe2b)cos a= (l+c2 )eb or 2c coshb cos a = l+c2 which gives (2) Ans c = cosh b cos a + (cosh2 b costa - 1)V 2 b b e +e 2ce cos a Note: as a-90, eb . U0 = Cln 2 - CU b 1 Ans. c e2 b+1 -2ce cos a -
This gives C for potential difference U.,.,. For difference U0 we get U e2X +c2-2ce cos y „ ln ,„. U = --Q Ans. c- e—+1-2cexcos y tit a-
X
-19By circular harmonics, the conducting cylinder potential in a uniform field is U =
r-1) (1) . This is the real part of
W =E (z - z) (1). The cylinder transforms into a flat strip of width 2a by 4.25. Solving 4.25(1) for z1gives z1= z d (z1 -a 2 )1 /a. (2), where square root has sign of y. Put (2) into (1). Simplify to get W =+(2E_/a)(z2 -a2 ) 2 . U = 0 if z= jy or if z= x, Ix'
A Ans. /a = ±E. W = E(z2 - a2 )z.
-20Superimpose E sing on E cosa with strip at U = 0 in both. 2 2 .E r 2 2 W.1=-jESiTla Z, W2 = ±Ecosa(z -a ) ., W= W2.4W2 = E L±(z -a ) coscr-jz sina] Ans. Esinct
Charge in W2 alone is a2=Claid/Zra = - €EXCOSCY/(82 -X2 ) .
a
X
In field Esina of W1: dT= 2cExsinadx= -2eE2 cosasinax2(a2 -x2 ) - dx l- = eE2 sin 2al ax2 (a2 -x2 );id x = irrea2 E'2 s in2a Ans. T= Jrx=+aa,_ -a _ x.-a -21
%.17.:r ' . {
4-
This is solved for potentials +er and -T in 4.22(3) where a..i= c. W =sin-1(z/c). For +1 and -1, 11W =sin -1(z/c). For +Uo and -U0, W= (2U0 /r)sin -1(z/c), z= c sin(*17W/U0) Ans -222U0 -2 3.1 Rotate above so zi=jcsin(irW/U0 ), W=—FT sin 7--. Bend up at z=±d so jc dz/dz1=(z2.-d2 ) `, z= Asin-1(z1/d)„ zi= +d when z= +by so that A = 2b/n.
(1
U= U zl Pia o V= 0 )
z1= jc when z= ja so jc-J. =d sin(ijar/b)= jd sinh ra/b), Substitute in z, cancel d. U= -U0 sinh 17z/(2jb) sin fir \ z/b) = 2U0 . W = 2U Ans. • rr j sinh(irra/b) sinh(ifra/b) (
Page108. (103)
Chapter IV. z1 1- d = c sin (inW/1.1o)P C< d. z=A1241 z2.'
- 23. e z/A Z;
rz/b
Page 26.
4:: _7_.)1 21,p171, yl d
_ a__ T
.
'' t 1 f-,' ' -...% . When -. __ke''''. x , a.-..... __... ■
If zl< 0, z=jb so jb=jAr, A=b/r', z=e ±rz/b zi.=e+d, z=a; if z=d-c, z=-a:+c+d=e , so
2b
II
y z plane ....... _-ro----
.... _ _
sk
,
_______
4— 2c --) r z/b r d=cosh(ra/b), c=sinh(ra/b). W=(2U0/r)sin -1UA1-d:q= (2U0/r)sin- 11 -cosh Ycsch 7] Ans. - 24-
.
,..,
zIplane
1ly X'„4.
z plane
....,
From 4.22: zi= sin(irW/U0). z=A0nz1=Aemsin(irW/U0) ,'
1 t—Q If z=±jb, U= U0, V= 0. Thus jb = jrA, A = b/r. —121 ' r4 ., b ./ and z =-Onsin(irW/U0) Ans. .• .......
i!1:;:irinii;iiii
(ATTli;H:=1::::;;;1 .,
;1
IA,H.,",111.,....,
-25-
sin(rU/U0)-jsimh(rV/U0) so U 2U0cos(rU/U0)+cosh(rV/U0) dzI2. b2 sin`'(rU/U0) + sinhz ( rV/U0S3_____). b: „ U a. U = TU0 by Dw 650.01. From 24 dW 2Ug [cosh ('N/U0) - cos (rU/U0)]z 4U0 " rx/b sa.2-i rV erz/b= e17x/bejTrY/h= sin(ir+ijrV/U0). By Dw.409.1 and 408.16, e co b cosh777,, Luo 2k. Squa r e and d subtract,by enxibsin=2-*sinh2 y Dw. 403. 22 and 650.01: b e cos(2ry/b)= i or x=-(ib/r)kr2cos(2ry/b)] Ans. b dz. b2ucots From 24 answer and Dw.408.19: dW = cot2 _0-
26 In 24 put U=Uo . z= (b/r)0n(ir+jrV/U0)= (bill)2n cosh(iT1V/U0) . If V--+ co, b - irV/U b b I bV ax/aV--0. --- If no edge effect, °)= 2bV/U — °72 (Ie o - -4712 r / 2U x=2U Thus extra width is (b/r)R12.
,, . ,...,,,,,.. , , .1 „ ..,. , 0 -..-,.....%,,,,,,.„, 14•Priiirrtilt—Uc -;'(iiitf,,.1, . (!, /
o
Ans.
0
-27IYI
Set b = r in 4.22(6). For charge Q multiplyiby itv(70. W=---q—sin2rE
1 sinT4z _ i —2—sin-I r-i(ez1-e7z1) ] 2re sinva 2sin(a/z)
1.41. Xi
1
ziz =jz 1 ,1 y =x ,2 1 x 2 =j(jy1)=1-yi:jz1=j0nz=j0nr- 0, *zi=em(r)Z+*j0 e 2 1-=r(cost0+jtin2u), e Zzi=(cosie - jsint0)/r 2 . When z1= ja, 0=ci, Thus a=ct. Substitute in W, multiply by r/r
pr-l)cosTO- (r+l)sini01 2r 2 sin *a -28-
W= 2re
Put -jQ at origin to cancel
40
9/'
.2
Ans.
and leave trough an equipotential. Potential of -Q is
W =[jQ/(4re)]2nz = [jQ/(4re)](einr+ j0) if V is potential function so that W = —1-[2sin
4re
ie) - (r+1 sin 20 )] 2r sin(ia)
Ans.
- 29In problem 27 flux from outside is +iQ and from inside +iQ. from charge it is Cr r+a 1+(a/r) 0 a 0 Ans. • outside and • inside. R = 1-(a/r) r-a' r 2 7 2
Chapter IV.
Page 108.(104)
Page 27.
-30Differentiate W in problem 27 and set r = 1, jz/=0. Then
—2—. _
____). Q
cosie dW = jcosizi n li 2re 2Lsinga- sin2ljz/Ji r-q 4re[sinaia- sin2i0J dzI. Superimpose density from charge -*Q at origin. a0=1:4Q/(2r) ]= 4°Q/!. Then by 4.23(3), a = ..e dW ..i.__Q_ - _CL[ cos0 Ans. dz 4r 4r (cos2 je - coeice) -31-
1.0
(a)Sheets parallel, field outside: dz/dW = c
W2 a2
C 2
w .
1,1 is:..1 ,: I
z=c(iW2- a2 pAW)+C. If W=a, z = c(ia2 - $1122na)+C1= +1)0.
I
If W=-a, z=c(ia2-a20na -jra2)+C1=b0. Subtract, 2b0=jra2c.
i lo #
t
I% 011
1
I
•
W
so c=-j2b0/(ra2) and b0=-j(2bo/r)(i-lna)+ C/. Thus
W=a
C = bo[1+(j/r)-(2j/r)Ikm]. z =-(bo/r)[j(W2 /a2)-2.j012W -r-j+2jOna]. Ans. 1 2r 3 / 2 2a2 (b)Sheets 90°,field outside:3dz/dW=cW- 3/2(,7 2_ a 2),3z=o dw +w — ill") + I C 3 8— 7 2 If ra, bo=ca (T1-2)+C I =7a 2c1-C i . If W=-a,bo=ca'(-3j- 2j)+C 1 . 8 =
3
3
=ia7jc+C 1 . Subtract: 2b0=-Sa 2(1+j1C 1 .3c=7b0(1-pa 7. 1 3 •-7-17(1-j)1308 2(72 + 2a5aw7) l Ans. 5 C1=130-(1-j)130=j130. (c) Sheets 1 90°,fislds inside. dz/dW=c(W 2-a 2 W7=ccW 7-aV7) 1 z=c(2W7+-1a2W-7)+C =. io ,42+a2)w-ii±c 1, bo=icai+C1,-b0=ijcal+C1. i 3 -7 4 -7. (1+j)bo (W 2+a 2) jbo Ans. Subtract: 2b0=3ca 4(1-j), c=7bo a '(1+j). z= 2ale 20/2 -32)
(
(1) wr---ii=qE-em*,-- ( 2 ) 1 1 1 2+b 2)1) 2)+}, a=i1-.4±.(d 2{jd±j(df+b so ja=2 When zi=ja z=jd _ q gnz±(z 2..132‘.. )7 j(d±(d 2-14)2)7} Ans W= Zre From 4.23(1): z i=i(z±(z 2-b 2)i)
z 1 plane
z plane
a
d
Z±(Z LID 2)74j(d±(d 2+1,2)2-1
-33fa 2 (x 2.. y 2) _b4 + 4a 4 X 2y 2 a44.2jaaxy kJ-) u vre t a2 (o_ y 2,) _ a 4 J + 4a 4 x2y2 2re a2(x2-y2)N q Simplify: U- 4rE On a 4((x 2_ y 2_ a 2)24.4x 2y 2). If x=d and y=c,bU/ox=0.Solve for a (same order as d) q 14a2x[a 2.(x 2,3,2)_baj+8a 4xy z 4a zxr az(x 2_y a)_ a43+8a 4xv 21 = tbx1 47EL ta2(x2_y2)_baj24.4a4x2y1 x2y1 ta2(x2-y2)-alf2+4a4
W=
q
a2Z2-b4
=q
aa 2(x2..3,2) .,b4+2ja 2xy
,. .1/4
_
q
242
}
2rE Pm a 2z 2-a 4
U
i n-
. o
Neglecting a 4,c 4 and d4compared with b 4this becomes b4
d 2-c 2-a 2+2c 2 a2(d2-C 2_
a 2)2+4a 2d2c 2 —
-4 0--b
d 24.c 2-a 2
a 2[(0.. 1 o 2)2_2a 2(d2_0)+a 4)
0. Thus a 2= d 2+ c 2. 2 a2(d 2+c 2)__2a4(p_o2)4.ae=a2b4_(d4o2)134= Let U 1be fiber potential at x=d, y=c, a 2= d 2+c 2. Then b8 b8 9 _____q_on U,• aso 2 (o 2+d 2) zire 4 47Ek a 4((p_ c 2_p_ o 2)2_4o 2d (continued)
• -q
Page 28.
Page 109. (105 )
Chapter IV.
1_ 07, 2b 2-b
When z=b, W 2=27
q b' 2 . a 2 2 . a 4 - 4r e k7 Ej 0 so case potential is U2 4L2/n .b 9:.
8rE b2 n = ---4/72 Q=2q. Ur-From 4.11 (4) Cu- 7-2-Ans. u 1 -il 2- 2/7zi.b 4 / ( 4c 2 (c 2+d 2) ) 3 , 4rE - - 4c 2(c 4012), This encloses the actual electrode and and so gives an upper limit. For a lower limit
find the equipotential passing through x=d+c, y=0 and x=d-c, y=0. Substitute coordinates in (1) and equate:fa 20_02 _b4){a 20_02 _a 4}-1 =-{a 2 (d+c) 2-b 4 }(a 2 (d+c) 2-a 4 ) - 2. Keep only b4 in the numerator and equate denominators: (d-c) 2-a 2 = -(d+c) 2+a 2, so a 2 = d2+c 2 . Then b8 -b' q 0 Pin substitute for x,y,a 2 in U. U1- q .... 47E a 2{(d+c) 2-(d 2+c 2)3 -4Tre .4d 2c 2a 4 • fie 8r E b' q )2814 U = Ans. U2 -U =47E - -/n4C2d 2. CCP//2034/(4C2d2)) 2 47e a'' 2
4
3
-2b-
-34By 4.211, with no outer cylinder images are q,-q,q,-q etc at 0,±a l ±2a,±3a,etc. so 4a 2 -a 2 /(2n-1) 2 o 1-(2(2n-1):1 2 2rEV =qE072 A 2a(4a2- ( a/2n) 2) qT4-1{ 1+(4n) 2} gn 22-a) =-On(ia). 27E(V -V = q(ancot+7 -
2
Apply Dw.422.1,422.2
bn
b 2 -a 2 (2n-1) 2 + o n! = B A qi btb2 a 2 (2r) -2j. q 2'
2a
ic00A
' 1- (a/2b)2/(2n-1)2 - ^ qEgn zqva a - qpnEot (iTra/b)-20722b/a)-17 •1 0qi2/712b/(17a) . 1 -(a/2b){1- (a/2b) 2 /n2) As B increases VBbecomes circular, so C = q/(VB-VA) = 2rE/07z(2b/(rra)).
V Ans.
-35-
Y1
Use the lines of force of equal and opposite line charges as
r ,...--..... t ,_, \ equipotentials in the z1 plane. Apply a Schwarz transformation., t , •,, 1 with 37/2 bends at x = ±1 and 2r bends at x = ±a so that V= 2rr _., V 1 I 1E- 1 dz 7 C(z?_-1)7 1 C C(a2-1)-1 37/2 eJ 1 where 77m# ---je or -747-' -Tr7 1rie 4u zar '71Z.1 i TbT
z i -a
2b 4I
(
r'1 .........,. 101 1 %
27 V=1
3r/2
a -0
k-a
_ ,i e j°I and use p99 (top). dz i=jr i ejeidO i=jr tiejei de,l gi=.. a z i=r i eje i-a+r r-B'
017
Integrate for C and a. j dz i=jCj °de l so -2B=jC7. and C = 2jB/7. 1,,, B 1 1 rA j B dz i=70C/a)ca 2-1)ij .s cif9 1 , A-B=T(jC/a)(a 2-1)77= B(a 2-1)/a so that a=B/{A(2B-A)17. In top figure rW=gn ( (a+z 1 )/(a-z 10 or z i= a tanhinW,
1
7W a' -1 sinh--=(-T-1) 2 (1) z1 2
2dzi1/ 2
P(zf-1)7
z= jCj zi _ a2 dzi - Cf (z1-1)
+ C(a2-1)/
z1 =- x1 dz1
1
1 ' (zi-a 2 )(zt-1)7
I
Integrate by Dw 320.01( 387,507.10. z=q-jsin-lz i -fj(a 2-1)7/altan-261(a 2-1)1/( a(1-z 2)11 =C[-jsin-l z i - ( j (a 2-1)i/aJsin-l f z 1 (a2-1)112(a 2-z f)- q23)= B-± (2 jB/7)cosh"z 1+A-B -1 2 j((B-A)/r1cosh-2 [(B-A)zif(a 2 -z1)A(2B-A)33]. Substitute for z 1 to get
= A ± 2iBcosh-1Btanh(7W/) {A(2B-A) } / 7
.2(B-A)c osn.1(B-A )sinh(r W/2)
r (continued)
{A(2B-A) } 22
Ans.
Chapter IV.
Page 109. (105)
Page 29.
-35-(continued) V=0; tanh(i7R)= 1. Dw.650.04, cosh-1BfA(2B-A))2= tank- q(B-a)/B). erU/2(B_A)/9 2 1 B-A When U --* co, V=0, sinh(irU)r zed/ 2 so y=rr[-Btanh---B-+(B-A)cosh-1 (2AB-A 2)1 /2 2 [2fA(2B-A)1 2 Solve for U- n Bn cosh( 757 + B tanh-1BB -31. If y B-A then B-A 2(B-A) B-A v 2 (A(2B-A))1 / 2 B U= + + tanh _B-A. Without fringing would have uniform field B-A B-A r(BA) 2 y= (B-A)U so capa citance increase is Gy= - .13.400n (A(2B-A)1 1/ 2 +Btanh 1} ±41 n B-A B = (2/r)fOn((2B-A)/(B-A))-A0n1fA(2B-A)1 2/(B-A] Ans. z1 plane -36When U
--4co,
Art. 4.18. Bend real axis in z 1plane with p= 0 at x 1 = ±a and ft= 2r at x 1=±1. (a 2
V=1
V=1 Y
17W=An{(a 1+z1)/(a 1 -z 1)1=2tanh(z1 /a 1).By 4.13(1) 90°rotation, a l <1, z1=al tanh(rW/2), ciz i /dW=(ra/2)sech 2(rW/2)=r(af-zWa l. 1 1 dz/dW=dz/dz i-dz i /dW= rc (1-zf) /a 1=irc( al La i tanh 2(rW/2) ) (1) ri_ a 2 rw z= cS(ai l - a l tanh 2(rW/2)1d(rW/2), z=c!---T-1 T - a itanqj (2)
x1
4- 2a 4 z.=—] . 1,
rZic---1-1
fi-z 1=-a s'
zi=+a --;) h .f X- 2kii
When dW/dz = co„ tanh(rW/2)= 1/a 1from (1) and z =k+jh. Substitute in (2) to obtain 2ha 1-a 2 11 -cr(1-af) rlaf k+jh=c [—tanh--;-1] = ct----(tanh-l acii7)-1). So ha1 1 2a 1 al ; c - r(1-af) (3) -2hr k=7T-ttanh-1a l - a 1 /(1-ai)} (4) Ans. Solve (3) and (4) for c and a. (2) becomes rW r(U+-1V) sinhrU+jsinTIV a (5). By Dw.655.3: tanh--=tanh ] 2 2 2 2 coshTlU + cosrV r 1-af 2 sinrV sinhrU r 2a 2 r 2al • x= lit--1-- • y- ht r(1-a 1) coshrU+cosITV 1 Ans. r(1-a f) coshrU+cosiTV U3,
z= 212Elf_ - 74--tanh
1 -37Art .4.27 : dz/dz = 2C i(z + X(z f-1)71, z= C(z f+
so
1 (zi-1) 2+P.sin-
by Dw.350.01. If z 1=0, z=0;If z1=+1, x=h=CX17/2. So X = 2/7. Thus z= h(W 2+ (214/ 17)(4 2 1)T+(2j/r)sin-1W3 Ans. Ify1 4,-1(x 1<1; 1 -
x =hxf, y= (21/17)(W1-4)7+ sin' 'x i) so y=4.1(21:(1-2 1i)7+sin -1(!)-2) dy/dx=2(h-x)7/(rxT). If dy/dx= 1, x=4h/(112+4)= 0.289h, y=0.48x, r
2 (1_x 2)11 2+y 2h2. 0.734h 2. So r =0•86h• If circle r=h. Error 14°4.
-38 Equation of n-point hypocycloid is: nx=a( (n-l)cos0+cos(n-1)01,ny=a((n-l)sinP•sin (n-1)03 . By 4.16, a transformation for this surface, V=0, is nz= a( (n- 1) (cosWi+jsinW ' ) + cos (n-1)W ' -jsin(n-1)W
i
ri'
= af(n-l)eile+
[u] circling the conductor is 2r so the charge is, by 4.11, 651 = 2reU. For charge Q 2rnEjW/Q1 +e-(n-1)2rejW/ -1) - eAns. we have nz=af(n-l)erTEjW/Q +e 2=2r/n, U=Q/(ne) 410=r/n,U=Q/(2nn) 9=0,U=0
Chapter IV.
Page 110. (106)
Page 30.
-39Make the circumscribed polygon tangent to U=-Q/(4en) which is the line of force that divides the flux coming out between center and point in half. To get polygon side vertical, rotate field of last problem inn through angle n/n by inserting factor e- . Thus zi=(a/n)e
x 1 =(a/n)e
jt(21W/Q)+Tini
((n-1)+e-j27""1
-217eV/Q
n-l)cos ( (27eU/Q)+r/n3+
e27neV/Q cos LL (n-1) 2neU/Q1- (7/4
(1) Vertical line is tangent if OU/Ox= co or bx/OU= 0. Put in U= -0/ (4en) and set bx/OU=0 , 1 r r 7 27neV/Q _ tan T1 r or e 2rn Ev giving -sin(( (-7+1)7/n3=e sin(-y+ = -e 27neV/Qcos-2n
so V= (Q/(2rnE))2ntan(17/(20) Ans. Insertion of the U and V values of the tangent point _01, r r 7 n ,
into (1) gives its x-coordinate: x =(a/n)Ctan(jr/n)3 ((n-Dcos '-.; + tan--cos(--- - --)3 Y1 t 2n 2 2n 1 Tr -1 r cos (7/n). = 2( tan-iif,n COS — TT sec-3= — TT 3- rl n-2-1-1/ Side length is L=2x tan(7/n). (tan 2n n 2n 2n 2n cos (417/2) t 1 2n _ 2a r f (n-1) t an (r /1n ) + n sin (7/n) r -ii (1(2/n)sec (7/n) + 1 l7} L - 2n ttan 2n 3 - 2a( tan ) sin sin-2n cos (77/n) 1 2n 1 2 7 r r1 --or L = 2a sin—i.tan-3 n t (1 - --) sec- + 13 Ans. n 2n 2n 4.17/n)
y t=(a/n)e-2TIEV/Q[(n-1) sin( (27eU/Q)
errenV/Qsin [( (n-1)2reU/Q3 - 7/4
(2)
For y-coordinate 2f tangent point, put Vt and Utin (2) giving 1 r 1 7 r n -17 1 a r n ( (n-l)sin(-17/n) + tan(-2-7/n)sin (----)1=-( tan—) (n-1+1)sin-y = (a/n)( t tanTr-11 1 2 2n J 1 n 2n 2n' 7 1 .-7 xt_ ncos(77/rtsec(Tr/n) so 2y = 2aftan--1 nsin-- Ans. -1.056 if n=5.Error 6910 t 2n 2n xpf (n-1)+tan(77/n) )cos (7/n) 1 1 xt=ncos (27/n) - sec (77/n) - 1.008 if n=5. Error 1 0/o xs n-l- t an (47/n) -40CHyperbola equation is xy = p. Horizontal line in z 1 -plane transforms into hyperbola Y1 by z = zi, or z = z 2= r 2e2i€1.Since y = 2xy, hyperbola line is y=2p in z 1 -plane. Charge coordinates in z 1 -plane that make U = 0 when 2p 2 200 y= 2p are z i=x i+jy and z'1 = x1+ j (4p - y 1) = z 1*+4 jp, or z1= r o e s -2j00 0 =r e + 4jp = r 2e 2je° . Thus r'0 2cos2010 = rgcos200 and Wsin260 ,
=-r gsin200+ 4p. r '0 4= rt+16p 2-8pr gsin200. tan20;= 4prZ 2sec200-tan208. r 2e 2je..r ie 200 (re je.-rne jeC) (re e-Fr 0ei °) -q ":1 ,zi-zi Kin -2 je_t lo as 2jeb - 2rE 2(11 (reje-r lo ejeb)(reje+r;ejeti) W= 27e z -z" 1 27e r le
xy=p rem
and negative ones at r'oe' Thus in the z-plane there are positive charges at r oe and -/- 1)e je°. The charge density on the hyperbola requires lOW/Ozi on xy = p where z 2=z* 2+ 4 jxy so z 1 =
+ 4 jp, and so z 1- z';= z - z1* . Also r 2s in20 = 2p. (continued)
+q
Chapter IV. Page 110. (106)
Page 31.
-40C- (continued)
jaw 1
-q
f 1
1 idzi
Id a lhyp .- 27e Lz 1 - z1
z 1- z'd dz
qz 1 z_} Trc (zi-z)-(4" -zi*) -
cre Je
TIE
2j(rasin2430 -r 2s in20) z i z*-(z i zi*+ziet)+z1z1*
As in 4.23 (3), the surface density is given by cr = -E I dW/dZi SO that 1 2qr (r gsin280-r2sin20) 0- rrfr 4+4-2r (r4- 4p2 so eliminate A. 2cos29= 2rgcos2(0-00)) Ans. On hyperbola r 2qr(rgsin00 - 2p) Ans. o7( r 4+r 4 -24 cos20,2 (r 4- 4p 2) V2 -4prg sin2003 )
2
-41CY2
W=•.11(21TEOM(0:2 -Zi)(t 2 ;tt) 1]
• "'N
Zi= AP/72Z 2 +jB,
= zT.
If z = -a, z1 =ji2, 22=1, . B=a7. If z=-1?,
q or z 2 = e
1. - - jb 2, z 2 = -1,
A = (b 2 -a 2 ) /'R. z,=e (z i -jB)/A jr(IC - 1)/(,f - VS) When z = -c, U = co, so z o = e •
GC-}s5P/(A.rb--vr;).
(7_;Nra-)7/(j)-,./;) -e irr(si-N/3)/(j.-Nr;)Ke(^5-d^ii)nAN/F)-■ /";)_e-J(NE/(Vg-A7;))7
W=-(q/(27E)12/nje " But
U=
2m em+jn -e jP _ e +1-2e os(p-n) m-Fjn -jp e e 2m+1-2emcos (p+n ) -e
1
7_
1 cosh m - cos(p-n) T
Let C -
r
41; , then
cosh m - cos(p+n) _ -q
-q gn cosh(O/i7 cos-12.0)-cos(Wr-sin.18-1J1 cosh(C.;cos40)-cora(/sin4-21; ))
4"E
cosh(C.A7 cosi0)-cosfc(JEd4 -Vr .)-1 Ans. cosh(acosia+cos(C(fsini0.1-J-4))
-42CFrom 4.261(14), a= 0, Wi=(m+n)Ez/(m+Kn). x=c coshgcosn, y=c sinhgsinri. ri = 0 is major axis, 1 Ex(1 + cotha) Ez(cos]ha+siIlha) Tl= Trt is minor axis, so m=ccosha, n=c sinha,W iAns. (cosha+Ksinha) 'U '- K + cotha -43CThe transformation is that shown in Fig. 4.13 and Fig. 4.17. The problem is to find what to start with in the C = g +jri plane to end with cylinders in a uniform field E in the z-plane. Let W 1=Ez so C= 2tan-1 (W/(cE))= jOnf(jcE+W 1 )/(jcE-W 1 )) which gives ejC= (jcE+W I )/(jcE-W 1 ) so W I (1+e3C )= jcE(e jC -1). W 1=U1+jV 1 giving real and imaginary: U (1+encos§) + V i eTi sing = cE0sing (1) V I (140cos g) - Ut erising= cE (elcos §-1) 2cEensing o 2t1 ,n Solve: U1= c E-- 2n (14e +2e cosg) ( 3) zn 7 11 g 1+e +2e cos Let g= (2n+1)Tr+ E where -0D
fl--.O,
EEO.
(2)
-.+ I -:F
2n d 2rr -w-2 rr
2 TT -PI- 2n --V Expand exponent and cosine by Dw 550, Dw 415.02 so that near the i-axis and the point g= 2n+7 the potential is U1= CEbb.7/7Kr 2+E2) ib = 2CEE/ (r12+0). Thus U 1 near the g-axis arises from an infinite set of dipoles at intervals of 27. In order to make rl = and r = -p equipotentials, put images above and below the g -axis with positive dipoles at 2p(a+ 13), -2p (a+ f3) and negative at 2p (a +13)-2B and -2p(a +13), where 1 <13
g
Chapter IV.
Page 111. (106)
Page 32
-43C- (continued)
For a set at r,writing Ti'for ri-T1, U=cEAlin(l+ent/l ) +2/n (1+elli i )] = cE-E ct :- em Ili( e jm t+e- ni
m
co
-e
m(71+213-2p (a+ 13) )
)
e-m( -71-2p (a + 13))1
+
rn( -71+2C4-2p (a + 0) ) = -e
)([ enfri+ e -rral_ em CrH-2 o_em( -11+2a)3 1
n=4:I
J
=2cEE (-l)m emli sinmg. Adding all dipoles gives m(71- 2p (a +0) U = -2cE E(-1)msinmg [e
<-
4:
`.7
m -2cE (-1) sin mt
->
-)
-)
-0
,_-
4-
4-
''.. 4-
-)
-0
-4,
-1.
4•
4.
•
n=p
e- 2mp (cc+ B) . But the sum
-m (a + 0 1 P=1 e 1 1 e rap (a + 13 ) . ciLf.,x p =-1= 1 1- e- 2m (a+13) 1- 2sinhm (a +13) 1 1-x
4-
SO
- em(-714a-13)] (-1)msin mg [m(n-a-n) e + e m(-n-a-3) - em(nra+3) U= -cEE I sinh m(c+3) =2cE di (-1)mtem(ii -a)sinh m13 + e-m(11+"sinh ma lcsch m(a+p) sin mg 1 -44CBoundaries like last problem but with plus charge at origin • and opposite charge at co. The minus charges replace the plus
Ans. + •
Ti
+ .
-
,a
• -T---t-----1.---t --- T-%
dipoles in the last problem so the potential is obtained by
r1-13
omitting b/bt from U. Thus
.
+ .
+ .
+ e 4e n J illcos mg + --im 3ernr11 =-a-E:.111)" U'= . - -11-1(. 4rE 1 ' . . m ( imt m 27Te 1 ul=_S_ =Ifni cosm(g+r) =-19-11pmil' cos m§ 4. 2re ;1.:12f 1 me m 2re 1 _ , + + -q ' (2m+1)71' cos(2m+l)g U'=U 1 +UI +C = re -- o E e + C. since old m is odd.• • • :. - + 2m+1 4—Lr —.). The location of rows and their summation are exactly as in last problem so that q; e(2m+1) (n-a) U -=..-sin(2m+1)0+e-(2m+1)(7143) sinh(2m+1)a csch(2m+1)(a+p) cos(2m+l)g +C 7E o 2m+1 Rearrange numerator N where m = 2m+1,2N=e22(TI-a-113)4M(7-a-13)4e-1/(71111-Ct)-e-m(1+1341a)
q ,- e -23asinhm(rt+f3) + e 213 sinh m(a-i3) Then U +C =7Em=0 -L cos mg m sinh m(a+p)
Ans.
-45CTransformation for field about two equal positive line charges at +c and -c is ..(1
9 2 y2_c2)2 W= , 9 Pm(z 2-c 2)so potential is U=-Znt(x-+4x2,23----((x4y4z2)2_4c2x2 . 47E J -4re ' zre )
The desired surfaces
(X 2474z2)2_4c2x2=constant
are equipotentials in this field.
Substitute the values given for the constant. reu l="Bn a, 7Teu 2=-041b. a is eldWdz1 so a= (7q/r)12z/(z 2-c 2)1. Substitute for z2-c 2. ae-crix 2i.y 2/(ra 2), ab=_,A77;;Ti(rb2). If b > a then aais positive and ab negative and re(U I-U 2)=-107:(a/b)soq=re(U1-U2)/2/70/0 Thus ca
e(u1 -u2) (x 2+y 2) 4/2 a22/11(13/a)
crb-
e(UI-U 2) b 2P/n(b/a)
Ans.
Chapter IV.
Page 111. (107)
Page 33.
-46C-
n(' + :10)=On at a-t t=—co
t
x
4- t=-a
0= 0
t=-17a
0= 0
0= 1
0=1 a<1
2a
1 1 i -47C-1 i i pi 1, -7, 1,-7 Id= sn -lz i= j (1-z 2 ) -7(1-k 2 z 2 )idz . dW/dz i=k-1(z 1 -1) 2 (z i+1)7(z i- — ) kzi+—) . z 1 k k v
- z li= 1/k V0 -U0, 4.
z= 1
z i=4,1/k i
'.' CO *?P' it. zi =
+ cc 1
H-U o I ■
1
z 1=-1 z1 =+1 Consider dz/dz 1= z1/(z12-1) 1/ 2 z = (z? - 1) 1/ 2 = j(1 - z ?)1/2 yi
Yi
V0
U
•
• •
••
Vo
1 -u o -1 -k TT/2 Bends.), rr/2
if z 1= sn W, z= j(1 - sn 2W)
V= 0 +1 U 0 T7/2
i k IT/2
1/2
=j cnW so x+jy = cn(0+j4i)
z 1 -plane Y
rr/2
2 rr
rT/2
4K'
<-
/•-•
——-
z-plane
y
X
0=2K
1
z 1=0 -r
x
i\ (1-k 2 ) 1/2 k
4,F0 K.
0.K O=0 x
Chapter IV.
Page 112. (107)
-48dz/dW= E 0 (W 2 -1) 1 /2(W 2 -m 2 ) 1/2/W 2 (1) 0<m<1 since dW/dz --0 E°0 Integrate by parts: u= (W2-1)1/2(W 2 -m 2 )1/ 2, dv = dW/W 2 . z-
Page 34. MEI
lilt'1f
1fw2_01/2(w2_m 2)1/2 p(2w2_1_ t3 2)dw + j (w2_01/2(w2_,n2)1/2] (2) W=msin0 when -W
z
W1 let W=m/W 1 , dW=-mdW i /WI, z= -(mE 0 )-1r ) 1/2dW. Gives z same form as(: z.k.0 ru wm z = -E7)1 ,1(m 2-14?) 1/2(1-W ?) 1/2W7 2dWi . Use (3),O=s in-la- . (1-m 2 )F(sin-1a,m)+2E(sin-1 7 1:71 m)] +C3 (5) m -
}
1 r 2 Evaluation of constants: W=1, z=a. So from (4), C 2=a. From (5), C 3=a+-E-0 1(1-m )K(m)-2E(m)3 —
W= m, z=a+jb. From (3) C i=a - E c71 ( (1-m 2 )K(m) - 2Em))+jb.From (4) m is found by solving b. Er:71 -(14111 2) K,{ (1_m 2)1/9+2E( (i_m y/z) -49zi_±.12 zi =C1 071((z 2+1)(z 2-1)'1 } . z =pin
dz i /dz2 = Cl(z
4.13 (1)
When z 2=x 2 , -1<x 2 <+1, y1 =jV0 . Equating z 1 values gives 1V tn. ((z.ii-1)/(z 2-1)3 ° =(z+ja)/(z-ja). To find V0 and a in terms of b and c V see 4.13 (4). c=a cotVo, b=acscV0 so a=(b2 - c2)1/2, cosV0 = c/b. Solve for
Z2—
(z+ja)rT/V°+ (z-ja)Tr/V° 7/170 TT/V (z+ja) - (z-ja)
Let W= Cz 2
1.{)z
C =jEz, then C-
jarr
_i(b 2 _ 2 )1/2}77/a (b2_c2)1/2 z+i (b 2 c2 ) 1/+{ z {z+j (b 2-c 2 )1 / 2}" 1 - tz-j (b 2-c 2)1/airT ic'
Thus W =
- a TTE V0 • Ans.
where a = v0 = cos -1 (c/b) . 2V6217
020 + 671 -
-502v 0, axe = -m 2 and oy =-1 = 4m 2 . If m = 0, V = (A+Bx)(C+Dy)
2\7
-
If m 0 0, V = (A sin mx + B cos my)(C sinh my + Dcosh my) Substitution in the first equation above verifies the result.
Chapter IV.
Page 112. (107)
Page 35.
-51La Place equation solutions equal at y= 0 and zero on the walls are 0
sin tgq)sinmrx a a
a
= E A (mr/a)sinh(mrb/a)sin(mrx/a) which is zero except at x=c. Multiply through by m sin(mrrx/a)dx and integrate x=0 to x=a. sin(mrc/a)f (tIVVoy - bV"g)y)dS =(q/e)sin(mrc/a) pa A 2q1 sin(mrc/a) = mA sinh(mnb/a) Jo sin 2(mrx/a)d(mrx/a) = TrnrArn sinh(mrb/a) so r m Emsinh(mrb/d) Ans. bV" mr(b-2d) Tric Fy. q• --_ lat y=d,x=c so F=Z A m'sinh by by sin a Y F can be written down by a a m2a x Y symmetrical symbol exchange, so that 1 Mra m7(2c-a) mrd mn(2d-b) mrb mrci F= q 2 Et -csch --sinh sin 2 b+ =csch sinh 'sin 2---1 E b a a a a -52-
q
z= 2a(z i+z3 1 ), z o= Ta(z io+z31) so for three charges: W =
=+ 2rre
-4 an (z 1 -z m)(z 2TTE Z1
Ans.
r , / IC 00
xi
zlo)
- (zio + - )3 =--(n(z-zo) +41 zi z io 2rre -53-
Apply the image law of 4.04 (7) to the triple charge system of 52 in a dielectric shell bounded by p1 =1 and pi = b as shown in Fig. 4.61 (c),(d). When transformed as in 52, the three imagesleave a single charge and the slit formed from the unit Circle does not distort its field. Use the image laws of Art. 4.34 to meet the dielectric boundary conditions on the circle at p = b. As pointed out in Art. 4.261, these will still hold at the elliptical dielectric surface formed from p = b. Real charge at z10. Virtual Images, r > b. zn
size
n
size
1
2e 2E-171 q
Z1 0
2
2c2 E_E;2q
b 240
E -1 a 2E24,
3
2s 2E 2E.T3q
b 4 zi 0
4
2E4E 3E;4 q
13 6 4-0
Where E_= E 2 - El ,
Real Images r < b (Reiman Space) zn
size
zn
-2E2-1 q
0
(b 240 ) -1
-2E2E_EV q
0
2E2E_E:eq
(b4 zio r
-2E2E2 EV q
0
2E 2 E 2E-3 0,
(136zro ) -1
-2E2 E3E.7.4 q
0
* E2+ Eland if z = x +jy then z =x-jy. Inside the dielectric:
W = r-leE;_n-lOn((zi -zn)(zi -z;) ,zil), 1 < r
(
Page 112. (108)
Chapter IV.
Page 36.
-53- (continued)
-1 r e -ju o ±(v0+2nv i )_ e ju o 4:670+2nv i )1j . 2 sin{u o + j (v0+2nv 1 ) } so that z n +zn =j1. E2C ]neillt2Sin(11+iV) - 2sin[u o±j(v0+2nv i ) 1) W.=-, to ea 1 r(ci12--+c2) -1-El
Ans.
[
The outside field comes from q at z o and its images in the cylinder
-(E2 -EN/(es +EI)
at ba/z: and (e 2 -c 1 )g/(E 2+EI ) at 0 and to real charges for which r<1. By 5.05 (5) these appear to be 2 E l ge 1 +62 ) times their actual size and in the medium El. They are 1 4e i c_2 el -el )q 2 at 0: 4c1c2(e2-E1)q 4E...151 at at 0: and and - -11-E1E2q at 1,?zto (E1-1-62}3 (Ei+ E a) ( e l+ e 2P (Et + Ed 2 az 1 o (
--3 -, --
4e iE2 (62 -E)2 q (ei+e2)4
1 at —T— and b z io
to these images is
-
4e1c2(E,-e1)2q at 0„etc‘By4.261 (4) the potential due (E 1 + E 2)4
q e2 -El
n
e 1 , (1 l eo= ---2rre[ E2+ei 'gnu-
_ e j(u-
-v-v0 )
l[e2 -eilnk(i + e j(u±u 0 )-(v+vo+2nv1 ))]
4. E22 e (E1 4.E2 )2 1 e2+e l
Ans.
In both answers use upper sign for even n and lower for odd n.
rl q7
-54It is required that when the positive half of the -axis is folded against
the negative half, squeezing out negative 'l's, the fields cross the contact —) go line with no distortion. To do this a set of real charges, q1 , ql, qy,-. ch I are placed in the dielectric below the c axis. Images then appear above Ties ie, -
the dielectric to meet boundary consitions at r= Th. In the dielectric
4.
0.-1-i11 e, 2E i1/(E1 i-E 2)at-kr yn o+2101 (real), 262(C2 -E ) qi/(E i +E 2) 2 at , :,-j(710+2T11 ) (image), 2E 2( C2 -Ez )q i/(E2+Eitat
charges appear to be ql = 2E 2 q/(E 1 +c 2 ) at
-{go-j(710-271)}(real), 2e 2 (E 2 -E 2 )2 q1 /(e l +6 2 )3 at go +j(10+4711)
(image), etc.
as shown at right. So inside the dielectric we have Wi =
11(E i -FE 2 ) 0 E2 +E i
yri) 2 -t §0 +(-1)11j(Tl 01-2tIn 1 ))
, ; \
•
E2
+ qi / 4+q'
Ans
Above dielectric there is a real charge q at §011n o and image charge T(Ei=e1 )4/(E 2+EI) at
§0 -j(2711 -10). The real charges in the dielectric
appear to be 4e2E1q/(el+E2)2 at -
go + jr 0),
go
charge
Tlo
4E a t i ( E 2 -E0q/(E 2 +e i)sat -( §0 - j (110+2n i ) :1,, charge
=
4E 2 E 1 (E 2 -e02 q/(Ei+E2 )4at -{gol(no+470), etc.
Wilar
•>qt ■
Thus potential outside due to polarization alone ,is W'= -q 11.--. ° 2rre IE 1 +E 2
t-Fin-ft)-j(27-11-110)11ii -r (ei -4-E2 )i
_r___t emI t-Fin+( -1) n( §o+j Crl 0+2nri 01 Ans E2l^ 2
1
Chapter IV.
Page 113.
Page 37.
-55- (55,p109) by 4.16: z= a (cosCW) 1 /n+jb(sinCW) 1 /11 gives V-= 0 on cyjinder. At V= 0, tan° = y/x= (b/a)(tanCU) lin or U= (1/C)tan-1 {(a tane)/13}11 . As 0 goes from 0 to rr/2, Q/4 = Er4=lerr/C so C = 2re/Q, which gives charge Q on surface ,V= 0. 6U/N3>0 so Q>0 . As z .---> V 7> "30 so that jCUeCV) 1 / +i bcie 3CUe CV+ j CUe-CV - j CU -CV 3 lin , z = (12-ejajeand then (a+ jb) (1/2) i/riejCU/ne-CV/n, z r 2 = ZZ* = (1/2) 2 /n (a 2+b 2 )e-iCilinand potential is Q.E.D.
V -3 -(n/CP'nr +b 4-nQ/(2Tre)Jemr4B so Q' = nQ. -56- (56,p109) Apply zi = 2ne to problem ,35 where A= CX, B= Tr so C= z, a = T{cx(2r-a)} -1/2 and result is multiplied by -j to
x
rotate figure as shown at top right. Thus we get
[
1
Tr sinW rr -a)sinW Onz = -2j sin-1 + TT-atan-1 , Afia(2rcx Aja(2TT-a) -)-Tr2sin2W ,Ans. r -57(
Finite ridge should not change field at z = co. Y rr/ 2
rT,a 2 = 2r. Let W = Ezi (left). In 4.18 (7),Cti =aeLi z1 dz so dz2 -47:a-i, z=.,,/TF7f,when zi =0, z = ja So
2
2
xi
V=0 ic--- 2 a l —N
I1I
, . 1 d, x
V=0 W=Ezi = E.1Z-5----4-a 2 , and at z=02 dW/dz = E. Ans.
By 4.22 (3), W=sin-l zi .Bend at origin by 4.18 (1/m)-1 (7) with a= rr/m. dz/dzi =Cz i . So we have
7 m
V=0
IttO-Iiiii-f ' ■ )6‘ 1 1 i ‘l l i
- 58-
Yi -U=TT/ 2
a =a1 , z 1= ,vz 2+a 2 .
IY itiII r/IbI 1 IIIII I II/ i I
Y
-U=TT/ 2
yerim
1j=rriax i BW = sin-1 (z/ (Crqin . If [1+2m17/B and the total
charge per unit length is q, then B = 211ME/q. If there are 2m sectors and z=a if W=Trr then W = (q/(2mrrElsin(z/a)m .
U=1712 __ V=0 Ans.
-592rr
dziclz1 =(l+z i )/zi , z=(a/r)(071z1 + z i ) +1 since V z when x1 =-1, yi = 0 then x= 0, y= a. W = ✓r i Vokri v=Vo 01 a [TV , TrW/Vo +1. Ans. so U = rr rr z e
V=0 V=0
-60-
Yi
TT/2
3TT/2 431.
V=rt
V=V
a1 V=0 dz jk
dz - Ciz , t +1 dz
jrhodz_
rr Jo
l eje l+1
Je
del-)
--h as ri-) co k (1) -rr Use method page 99 for constants. As r i approaches infinity:
fir jCj o dei=iTTC=
jC
r le
1al
3 ai
jk. so that rrc—k (2), .1.;1 = k/h, C = kir (continued)
(3)
U=0
- x'
Page 114.
Chapter IV.
Page 38.
-60- (continued) -2u(1+a)du zi=-( 1+aiu2 )(u2 -1)-1 = (1+a l )(u 2 -1)-1-1-a l dzi- (U 2 - 1) 2 2Trk [11+ (dk u/h)2u 2 + r du 1 _ 2 {h t an_j_ctu _ P (1+a )u2 du ktanh-lu) + C1 . 1, z = -2Cj ,ug'" tiic41 t2 7 11..1.3. TT ( 1)(1+alu2) ] _ 47171-1. 2 k tanh 1 , k > h. z=0 when z=-1 or u=0 so C1=0. z=rr [htan-1 h-(k/h) 2 A/zi - (k/h) 2 1 u u . U Evaluate U as it approaches -co on V=r7 wall. z1 = e cosi', + je sinr = - e . ruy/_,..(hik)2e- lUI 114. 7:441 -.; ii(h2. 4, kz)k-ze-Ulhi(jk) . -1,2/1,2=chikwi7.7-1.4/T:7 i7T Let u 2= (zi +1)(zi
z --, (2/r)[-h tan-1 j(1.-i(h2+k 2 )k-2 e-lull + ktanh -1 ((jh/k) [1-i(hz+kz)k-ze-U ]}i Use Dw 507.31 and 722.1. Neglect e -U term compared with unity to get + 2Tr iktan_1 z = (2/rT) Rjh efit[ 24-0 2+k 2)k-2e- A.1} + kj tan-1(h/k)i___42,,,,a.0,244ca)k,23 4 nk Let U —*co, (zi +1) (zi -h2 k-2)-1/2=(-1..i..e-ux_i_kzh-ze-Uri/a___,[1_ ithz+kz)h-ze- } 2 ng 2+k 2)h- 2} _k -TTE . tan-It. + it As before: z---> --In t an- 1 (k/h)+22-k gn(-14-(h 2+k 2 )h -2e-Uil =
4
r
r,
Length increments are:L/11= DC
-Uo
+Ian
+_
+-3-1-1-t an -1111=
1 t 2 +ktan-111 +Itan-11`) Ans. -27T--E (12/n *
-61-
Ys U°0 4
k h 2+ka 2h h h 2+k 2 2ktan -1 h fc , 640 Ti2m—Wr + r tan r 4k a U
2k
Erk h2+k 2 , h
" uo
2b,
2bn,
2b m,..
1 n o,
z -7-- 7_,.-,il, z1 , x = --ir,ri, y= -r.i-goi , zi = e-2-
4--
14-la i --.)
,n(i.. 0
1
-Tra /b
, a1 - e 2
1
4 2e---) Tr —rah. dz , a i b1= 1. — Then by 4.29 (5): dzi= 2be 2 1• , „z i dz1 U0 U0 .. 1 r ETT(z+a)/b -17a/b ) K sn te ,e 121 3- a 1--°J. j Ans. W - K(bi/ai)sn' fz', ai a i K 0.,/(b2i-zf)(af-z?) rrUo di.,L dW dz i= Uorr 12-17z/bte -na/b_ errz/b,-ir na/b rTz/b1. Ans j te -e ' 2bK(edz dzi dz 2bK e "a fb)losh(rz/b)-2cosh(rra/b) 1 -zrra/b , r / -, From 4.29 (9): C= 2EK{ (l-e Ans. )2 j/Kte-rr -a- b j
b1 = e
- -17a/ b
-
2
-62z1 dz By 4.18 ( 7 ) .71--"V2_1,2, z=liFq;at z i =0, z= Pi= ibi uzi z i LI. c -hf. so b1 =h. When zi =a i , z=d so d=,/aFq or a1= jItr
(2„...a., _7.4, 2 2 rr r/2 -U0 -bi V=0 b Y
Substitute z 1 , al and b1 for z, a and b in 4.29 (5) and 4.29 (9); z =h{sn2 (KW/U 01 b/c i -,./7---Fh 2 )}= jh cn(KW/U0,h/,/314-h 2 ) Ans. C=
h air 2+h' /K( d / cFtl 2 ) Ans.
a U0
V=0
-uo
ib f-d-4
Chapter IV.
-63rKW b „ ,-1 3,4.29(5). W=U° - sn-i (sb1 ,11 ) or z 2 = bi snt— u o al
dz dzi N/T-aF
Page 39. IT Y1
Page 115. U0 1 -
0 -01
Uo
By Dw 768.1 dzi /dW = (b i K/Uo) cn(KW/U0) dn(KW/Uo). dzidW = (dz/dz1 )/(dz1/dW)
V1 al
b1 Ay
13 2 KW KW zi- a2 —a2rdz c--i sn 2 (—)+ 11= -aidna(—)by Dw 755.2. — cn 05 - , 1213 Do dW ja iUo "U o 1 a? Uo C r1CW, KW r KW, By Dw 785.2, 71- cnt—id—= 7-cos - tdrr-„ 3=z. When W= O, z=0 since al Uo U0 uo do 0=1 and cos0=1. When W=U0 , i dn K=(1 - (b i /a1)23 2 by }IMF 16.5.3.
-d -c -U0 V=0
U0
d >x
z = c so (Cl]) cos-1{1-(bi /a1)2}1= (C/j)sin-1-1-311 =c. When W= Uo +jVo , then by 4.29 (4) d = cos -1(dn(K+jK')3=cos -10 by IIMF 16.5.9, so Cl] = 2d/ft and bi /a i = sin(jc/C)=sinticr/d3 2d KW . cr , ERA_ 4 cuo= 4EI ____C(sinIcr iLcil r Thus z = —cos "ltdn(—,sin--)3 Ans. Capacitance is • Ans. rr Uo 2d V I - VI K(cosicr/d) -64A Y1 U=Uo z i = Bsn 2 (KW/110 ,,5) 4.30 (1). dz/dz1 = C1 (z 1 --1-(1+B)1a(1-z1)(zi -B). T i lT TT/2 V=0 a U=0 By Dw 380.011: ftz,/,./(1-zo(z i -Wdzi =-/777 + 2(1+017771-1dzi B ,t2
(11-13)/2
so z = -00/77: +C. When z1=0, z=0 so C= jC0/13. When z1 =1 or B, z = b so b = C1 =-jb/J, C= b. When z3 = (1 +B)/2, z= b+ jh
V=0
V= VI
=(jb/)(1-B)/2 + b, so ,,k= (-h ±./h 2+b 2 )/b. Take + sign SO k=j3+ (.Jhz±b 2 _11)/13. Z= (jb/k)•11 - k 2sn 2 (KW/U 0 ,k),/k 2 sn 2 (KW/Uo ,k)-ka + b. Use Dw 755.1 and 755.2 so z= b(1 + dn(KW/U0 ,k) cn(KW/U0,k)) Ans. The capacitance is C= 2EV I /Uo = 2EK( 1:17)/K(k) Ans. -65Invert about z= D- 2jR with radius of inversion 2R. Then from problem 64: z-
4R 2 z 1-D+2jR
-
4R 2 Bsn 2 (KW/Uo ,k) - D + 2jR From Figure:
D= 2Rcoty, B- D= 2RcotP, A - D= 2Rcota, B= 2R(cot Y + cot f3 ) = 2RM
U= U
U=0
A= 2R(cot y + cot cx) = 2RN k = 0/7/i =
-2jR
2R z
Msn 2 (ICW/Uo ,k) - cotP + j
Ans.
Then. from 4.30 (2): C=
Uo V1
2e' K(JITi7)•
2E' =
E
Ev
Ans.
>xi
2R
›X
Chapter IV.
Page 115.
-66r -(z 2-c 2)dz (z 2-c 2)dz xlpositive if z1--L1(z 2-a 2)(z2-b 2)41(a2-z 2)(b 2-z2) 0<x
Page 40. Ay
r 2 2i r 2 r 2 217 77/2 -a -c -
IT
Integrate: (1) x=b to a, (2) x=o to b, (3) z1 =d to zi =d+jh. U0 U=01 2 2 4-d 4. (1)Dw 781.12,781.02, 0=2, k 2=1- ;17, j•0=ta E(k)--,eaK(k)) (-j) so c2=i,TV=0 c2 b (2)Dw 781.11,781.01, 0=5, k'=a gives -d=(a--i)K'-aE'=a(10-1,r-E')=-1 1/1 Legendre Relation HIF II (15) p320. Use positive sign for (2) since xlis + and increases with x. k 2 =1- (b/a) 2 . jh= ja[E(k)- (c/a)2K(k)
(3)Dw 781.12, 781.02, 0s=
4E(5, k)- (c/a) 2F (0,11, h= a [(E (k)F (0, k) /K(k))-E (0,9] = 1.(4(E 1 (k)F (0, k) - E(0, k)K(k) . Since ra = 2Kd,
Sin0 = AJK(k)-E(k) / (10A(k)3.
z1= sinW, dzi/dW = costa = j1.7-7=
dz/dz1 = C1 TzTi2aa71. dz/dW =(dz2/dw)(dz/dz)
dz/dW= jC1,1zFTc2= C1 ik2- sin 2W. Let sinW = k sin0, dz/d0 = CIO cos 20/,A-k2sin 20. z= C1,1-(1-k 2)+1-k2 sin2 034/1-k2 sin2 dd0 = CI (E(95,k)- ( 1-0) F (0, k)3 + C2 , ksin 0= sinW. dz/dWis zero at corner U=U l „ V=0 so sin2W= k2 or 0 = 1T/2. z= a+1 so a= CA(k)-(1-k2)4c)) 7/ 2 A 2 -sin2 UdU and C2=jb. Between B and C 1> sinU>k. Integrate B to C; -jb= jfidy= j u vic Let s in 2U = k 2+ (1-k2) COS2* so cosU= piT=TcTsin*, dU = -V57:17Icos*Obik2 - (1-k2) cos2*, sin 2U - k 2= (1-k2)cos 2V. When sinU=k, cos*=0, *=r/ 2. When sinU=1„ cos*=1, 111=0. -b = CI I / 21(1-1(2)COS 201k2+(l-k2)COS
cl4i= C1::/1(1-k 2'-(1-k2Sill240 )/NA (1-k 2)s in 2*] d*
= CI(E( 1=171)- k 2K(A/I:172)3 so a/b= (E(k)- (1-k 2)K(k))/(k 2K( 1:177) -E(N5=(7))
Ans. (1)
E(0,k) - (1-k 2)F(0,k) +jb Ans (2) Charge densities are, since V=0 on bar, E(k)- (1-k 2)K(k) 1 Nrii.7. E(k)-(1-k2)K(k) = E dW — = dW dz, On a sinU < k aA2 ..sin2 u U < U/ a dz dzi dz 5.,/z21 -1 Co /;T:17
z- a
E dW
dz
=C
1
2 U-k 2
E(k) - (1-k 2)K(k) ar/l-k2-cos 2U
On b sinU > k.
Page 116
Chapter IV.
Page 41.
-68-
1
:1
dz _ cl 241 - )(10 z=ai sn-1(111)=Csn- 3k} W 2rrc z1 - x10 +jyio • dzi Ai(b21 -zi )(a12 -zf) -a1 -b1 aidxi -b+ja pb rb 0 < x1< b1 < Dw 781.01 b= J o KL-1 al 3, b=CK(k) 1A./(bf-xi )(at-4) al aidx i 0 < b1 < x1 < jb Dw 781.02 ja=i jaiK{,,/1-(bi/a1)2} ../(x?-bi)(ai-x?)
, 7,12 112 tDi ' al rb+j a
-
'q
so a = CK(k'). Let u= xo/C, v = yo /C, then (xio+Yio )/h = c+jd = sn(u+jv,k) c+jd snu cnjvdnjv+snjvcnudnu 1 - k 2 sn 2usn 2 jv
sn(u,k) dn(v,10-1- jcn(u,k)dn (u,k)sn(v,W)cn(v,k') cn 2(v,V) - k2 sn2(v,k)sn 2 (v,kl)
Since: Dw 759.1, sn(jv,k) = jsn(v,k')/cn(v,k'); Dw 759.2, cn(jv,k)= 1/cn(v,10); Dw 759.3, dn(jv,k) = dn(vI k I )/cn(v,k'). zi = bi sn(z/c,k) = bi sn(u+jv,k). -q 242 sn(z/c,k) - c - jd W Ans. 2i sn(z/c,k) - c+ jd -69(1 )w .
La /o . TT
:727-al,77 1 ,2.. z 2 -JvC1 -
. 2Vo slrin .1.0
-2V 0
vb2,2 ..., 1
"
11;r 1 7 zi ui
-a 2
9 Vo ___,. 1 —,..
2V When 0 lx il>ai,V=Vo , U= -T." sinh 1 3 1,2 ..,,2 .At xi =b1 , U.="0:..
AY %1 "J"1 4 i z.1= -C1--‘ 4.-.-Z I =Ci ...,bk......4 -2V 0 32 -az dz z2-c2 c2_132 When b i lx/1 1V=0 , U = -..°s . it-ail 1 1—=-1--.1-1--1-1. zf-br 4--zi= -131 z =-a N7)7791 1 dz i zf-b? b Vo cz b2 rzi=aVio zrb1-41 4416 h 34 By Dw 140.1: Z=ZI-C(C 2 -b 2 )/biltallh-l (Z i /b2)=Zi+-1":-1-COSh-I L 2 (2) 4-2 L-• vb i -zt ' hi
eb2 c?-bi b, jrrSc-b” r(c-b) C;+ k + jh= c i+ .--•-g----tanh-177, 1 = ci + 1', 1-tanh'l, so h 2b 2b1 1 (3) 1 1 U1 1 b k = c1 + (2h/r)tanh-1(b i /ci) (4) £= 81 + (2h/r)tanh'i(a l ib') (5) tanh_ 1 7,1 =sinh 1
Eliminate c 1 from (3), (4). k=#,/b1(131 +2h/r) + (2h/1 )sinh-1.i•r31 /h
b 2_ C I bi "1 (6). This gives
bi in terms of h and k, then c1 = N/bi (b1 -1-2h/T7) (7). (5) gives al in terms of bi ,h and L. From (1) .,./(bi-ai)/(bi-z2i) = sin{T1W/ (2jV0)} so and
zf =
b/ abi-zi = {131/1: ,./TrsintirrW/(j110)3
b i2 - (b?-af)/sin2 {4r1W/( jV0 )}. Substitute in (2) and use (3) to get z = [14
I b?-ai sin2 (rrW/(2jVo )ji
2h sin(TTW/ (2jVo) ) r cosh-1 b i VI); -at
Ana.
Page 42.
Page 11.6.
Chapter IV. -71-
y 4.29 (5) -d -a -b -c c b d a 2r -7/2 77/2 2r 2r r/2 -r/2 2r VI dzi _ dz
-U0 V-0 V1 U0 (z2 -c2 )(d2 -z 2 ) is transformation. -z2 ) (a2 -z2 )3
Numerator may be expanded in the form - (a 2-22 ) 2+(2a2 -c2 -d2 ) (a2 -z2 ) - (a2 -c 2 ) (a2 -d 2. WT:7 )(a2 (a2..z2) ) (a2 :dz22))3 d.z, z =bsn(1-1 2 )= bsnu by 4.29 (5) jo2 2a2-0-d2 dzi = C )
.
Dw 768.4: dz/du= bcnudnu; M= c/a, N =d/a, k= b/a; M<1,N> 1; 1-M 2 = A, N 2 -1= B. Thus dz I=Ca(dn 2 u- (2-11 2-N 2 )+ (1-M 2 ) (1-N 2 )/dn 2uldu= Ca{dn 2u- (A-B) - AB/dn 2u}du But S(1/dn 2u)du= jr,dn 2u/(1-k 2 )1du - k 2snucnu/( (1-k 2 )dnu) (Verify by differentiation) By Dw 787.3. Soudn 2u du = E(u) so z i = Cal [1-AB/(1-k2 )3 E(u)-(A-B)u+k 2ABsnucnu/((1-k2)dn-u) (la) HMF 16.5.3 Points W= 0 and W= U0 or u= K coincide so (1-AB(1-k2)-1 )E(k) = (A-B)K(k) (2) k2ABsn cnu ) (lb) Thus (la) is: z1 = Cal(A-B)CK(k)E(u) u (1-k2u)dnu E(k) By BMF 17.4.27: Z(u) = E(u) - uE/K (Jacobi's Zeta function) and (lb) may be written z i = Cal (A - B)Z(u)K/E+ k2AB snucnu/{(1-k 2)dnu) (1c) Note that z1 is zero if u is zero. At z= g-jh, W= U2 + jV i or u= u2 -1110 by 4.29 (4) and from HMF 16.8 sn (u 2+j10 )cn(u2 +jK' )(dn(u 2+jK I ))-1 = dnu2 (k2 snu 2cnu2 )-1, which is a real quantity. From Whittaker and Watson p520 Ex. 3.: j(K.K'-K.E'-K'E)Cal(A-B)/E.-jh=Ca1 (j11/2)(A-B)/E 2hE Thus Ca l - ro-B) and (1c) is When z = c, z1 = f, so u1 =
zi -
k 2ABE(k)snucnu 2h r tK(k)Z(u) + (A - B)(1-k2)dnu
(c/b)=snl(M/k)=
2hr_ f .
)+ r t(k)z(ua x
(1)
/0 and (1) gives
E(k)SIA(1-A) (A-1+k 2 ) 3 (A-B)(1-k2 )
Since d= b sn(u2 +jK')=b(k snu2 )-1 by BMF 16.3.1, 16.8.1 . snu2 = 1/ 1+B _ ._2_12. + E(k)As./13(1+B)(B+1-k 2) (K(k)Z(u2) 6 r (A-B) (1-k 2 )
(2)
( 3) Equations (1), (2) and (3) determine A,B and k 2 . Capacitance per unit length is C= eUo /V, = eK(k)/(K(1 4/7-k 2 ))=eK/IC
Ans.
Chapter IV.
Page 117.
Page 43.
-72-
y
2TrcW = -q0n (z i -z1o)/(01-4))) dz/dz1 = C1 (1-z f)-2/ 3 . Tr 3
Let 1-4= (1-,5p)3 = 1- J5 5p + 9p 2-NT37p 3 (1) where
Tq
r/3
zf = 6/33P(1-,N5P+P 2).=57( 1-47 -12 )(4-(2+85)u 2 )(1-K5T-72)-3 (2) Differentiate (1): -2z1 dz 1 = 3(1-4)-3 2(1)-2(-Anudu) 33/4 3 3 / 2 udu (1 -1-,,il-u2)3/2 dz 2-2-615 so (1-4) 4 2/3 (14 ,,F=7 3 2 )3 r7 .-712 33/ 4(1-j-712 ) 1/2(4_(24.„.5)0)01 2.j1_u2Nh_ou s' k Let u= sin0, z= 23 3/4C1F(0,k)= C2 sn -1 (singi,k) so sin0 = sn(Cz,k). If zi = 0,u= 0=0, z=0. sin0 = 2- 3 1/4(1+,5) -1 . Want z=1 so sn(C,k) = 2.3 1/4/(1+).
When z= 1, tang (0/2) =
This determines C to be 1.9138. Then from (2) z2 =3 3/424,/(1-cos0)(1-k 2 sin 20)/(1+cos0) 3 or z1 = 3 3 /4 2sin0J1-k2sin 20/(1+cos 20) 2 = 33/42 sn(Cz,k)dn(Cz,k)/(1+cn(Cz,k)) 2 so W =
q „ f(C,z) - f(C,z„)
2rE f(C,c) - f(C,4)
where
f(C,z) =
sn(Cz,k)dn(Cz,k) (l+cn(Cz,k))2
Ans.
-73From 4.03 (2) a potential, zero on the wedge surface inside (V 1 ) and outside (V 2 ) a cylinder of radius c whose axis is the wedge edge and which passes through the charge Q is
n Tr/a nr/a nne sinn-22 V1 = lAn CD a V =IA C-S3 np a
tiV i /Or - 6V2 /6r= 2r/(ac)EnAn sin(nr0/a) at r=c. ds=cdO. Multiply by sin(princx) and integrate 0 to a. E) sin (nri3/a) = 240sin arm duAn= nrAn
.
aB Q 1rpinr/asinrr rr--sinnTcr so V1 = Tr j p/c=r i /r by
Invert: rr I = c2, r/c= similar triangles. p< c.
a
1 p bOl ap 1 c
GA'
/5 ,
Q ap
p boi
Lci
at
= 0.
e a .r1 nr in a at
a'ds' = ads so a =(ds'/ds)a'= (r'/O& = (c/r) 2 a'. QA
=
Qc2
(P
A - apr C
) T
/acli (E)Tria )11-1 sin ricP. ''r Sum the series 1
(
a \
2 Tr/a = (r i /r) r/a .. by Dw 417.4 where a = (p/c) , 0 = TT13/a • (fr /a) -1 sin(rp/a) Qc(rr,) ria a a cos (713/a) + r 1217/a Ans. r ITT/a -2(rr i ) A a B
sin(13/a) Qc(rr I) (111a)-1" aria Ans. a r vr ia+ 2 (rr 1 ) T/acos (rrp/a) F ri
--___.---
Chapter IV.
Page 117.
-74Write the complex potential W in the z 2 plane for a line charge Q at y2 = a2 , x = 0 and y= 0 at potential zero. Bend the x-axis at x= b 2 to form a wedge with exterior angle Ct with Q on the x-axis which makes an angle a with the upper wedge face. Then 2rEW = -Q072((z2- j a2)/(z2-11a2)), dz1 /dz2 = (za -b2)(air') -1 a/r z2 =Cz 1 +b2 , 132 /r = p/a, b2 = (2c)rria cost-To a2 = (2c)Triasin-Tg.
tayi
Ay
-Q Inversion
e Circle ,„
r 2rr-a
To put Q on x-axis let c= -ei"Pia so that rr/a 2rreW = -Q0,2 ((z irr/6- (2c)Tr/a )/(zi -(2c)Icle2iTT/C6))
4
Take origin at x1 -1-c 1 . Invert about Q. (z1 -2c)(z-c)=4c 2
4 -1
r/a
(z+c) z+c __q_ so z1 = 2cz--cand W- 2,rre072(z+c)-Trila_
Ans.
(z-017-Pae23TIlha Tviae jO i rr/a - rr/a Jo, ti eje2!IA/ 2 Let z+c =r i e ,z-c=r2 e j°2 . w..1_2_, rria rria jr(0 2-1-20/a jOi -r 1 ,.re 4/4 rria r1 e e
2 Multiply numerator and denominator by conjugate complex and take square root, *= e1- e2 , rrria+r irria-2(r i r 2 ) 711C2cos(717/a) u==-1 on Ans. 2rre rria+ rra-2(rir2)7/acos[r(213-*)/al -75w ...._ 24z z 2 :z2a ___12_ 2fizr , rria j (rr-132+y2) /a 2 ) /a)3 )/( z2-bria 1 e - i (Tr-P 2+y 2rE z2 -z--2.- 0 2re 2.. %z 2 -bi e dzi.z (a/TO -1, zi= kz2a/rr, Q 13 2=rPict, y2 =7Ty/a, z real if za=raej(7-132)14a • dz 2 1 TT/a zTria i -b i ejy/a 2 -Yr-132)/a z2=-e • 2TTEW= - On rria rrict -J"( 2,7-2 p+v)/a zi -bi e • Inversion radius about 2c is 2c and 1-cl z7/d-z7ta W=-:--Q-072 2re 47 /a-(4-0 7/ae-2 j(77-1:1 )/a inverse plane origin is c so that
AYL 0. p...
(z-2c)(z-c) = 4c 2 , z l = 2c(z+c)(z-c). Put charge at zo Then W
(zt+c)TTial z+c) (z 0-c) jrria+f (z-c) o+c) 117/a +072 217E [Pn [(z+c)(zt-c)}17/a-((z-c)(zt+c)iriae-2j (7-13) /a 7 07c77T a
Omission of last constant factor merely changes origin of V. Multiply numerator and denominator of first factor by complex c onjugate, take square root and write rie jA 1=z+c, -z"ok+c, r 2o e je20=4-c and ob tain r 2 e i 92 =z-c, rioe j 0 =zo +c, r20 e =zo -c, rioe-je"r 20 )7T/Clcostr(*-*o )/al-1-(r 2120 )27/a (r1r20 )2r/a_ 2(r 1 -r 2 r- 10U= 242 Trki r 4e (r i r20,‘21-ria-4(r 1 r2 r 10 r20 ) cos (TT(2P-27T-41-4,0 ) /a)+(r2r20) 2r/a -
U =co if r 1 =r10, r2 =r 20 , *=4'0. U=0 if $=f3-TT and if 41=13-r-a .
Chapter IV.
Page 118.
Page 45.
-75a ib i zi 3r/2 b1 -a l + 1 37k2 4i4 ni -If V dZi Zl A / (Z 3 -a i) (zi +bi ) -17(zi -ai )(zi +b i ) zfi(zi -ad(zi +bi ) -b1 o 0 al (bi-aj)zi -2aibi Y z = N/(z i -al)(zebi) +(b1 -az* (:.57=--•acimliT7Fbi)-N771 1Di sin-1 (ai+bi )zi z =-1:4 Integrate first form on small circle ri from 0 to TT (see p 92) p rr _A f hdz = j j 0iNra17 1 del=-2h = -.1TirDiTT, so h= al b srr. (A) Wish z = h when At -h i =a i Fb 1 . z1 = a1 so h= (b1 -a1 )&14.717-"-Fb i - faTisin-1(-1) + C. C= -(b1 -a1)62,TaT--2h 1 z )z 12a1b1 (1) (13 al Then z = Niz i -ai )(zi +133. ) + (b -a )2424A17 +.b :b1/1.;7131sin-1 1 (a i +bi )zi 1
dz_,i(zi -al) (zi+131) — _ ,
-ai)/(bi+ai ) - h so jk = (b1 -a1)12firj = (jn/4 ) Oh -al ) b1 = fr(N/h2 +k2 + k) (3) Solve (A) and (B) for ai,b1 ; a1 = i(A/rh 2+k 2 - k) (2)
At z1 =-b1 , z = -h+jk=(bi
-76r/4 2r 3 4 dz/dzi = CJ (zz -b1 ) (z1 -a1 ) Integrals 1 1 - 1(z1 -a )7 1 checked by z7 z? - aI) 4_4_211..12..a.l ctanh-1 (4-a/)T Z = C[11( •/ 2 z12 + tan 4 differentiation -a1 31 2nr 3
(B)
2 2rr TT 4
z=0 when z1 =a1 and when zi =0 so b?=44. Thus z = 3Czt/2 (zi-ib t) 1 / 4 If x1 =0 and yi> 0, z =
g-j.ii(y2r4b?) 1 / 4 is real and negative.
If z = (1+j)b/.5 and b1 =1=z1 , z= 4C(17+)1 /4 1+. 3-1/4 so C = 2.31/4b and z= 32 /411./z 13 (z?-3)1/4= bz1/2(3z?-4)1- (1)
(2
y b (---- z1 =b1 gl'al First method of 4.28. dz — =-2(z i +X.TzT".1). -- z= -zi-N.zeliT4.+Xcosh-1 zi (1) -zi =a i dzi z1 E--- z1 = - b1 From Dw 721.1, cosh-l x i =jcosxi -1 if xi<1 : =cosh if xi>1. so that 0 < xi <1 xi < 0 xi >i yi z = -4- Xxi,ITI. ) +X cosh-lxi = -xf -j Xxi,ITTff +; ).cos-)x i = -x +j %Ix,iii-x21 +;xcos-ix, -77-
At x1 =0, z=0+jX': x1 =1, z=-1+j•0: x1 =-1, z=-1+p.17 as plotted. At x1 4, z.-+ +; xtcos-ii -4)=-4-1-jX(1: -.4330}=-0.25+j (0.5236-Q.4330)X or z = -0.25+0.0906 jX. Need (1-0.25) 2+(1-0.0906)0 2=1=1-.1812X+A08210
-1
x1 <-1
+1 "1 x1= -1 -
Ay
+0.5625. So )s. =3.735 puts xi = -1,-7, 0, +1, +1 on circle of radius 1. -78-
x
Rotate Fig. 4.22 90°and take a1 =1 so circle r1=1
1-x0/ touches Uo plane edge. Apply 4.23(1) so area between x2>1 x1 .71 ri and xi -axis fills lower half of z-plane. From 4.22(3) z i =j sin (Wi/U0) (zi + yi) . If W=U0 , zi=j . W= (2110 /n)sin-ljzi= (2U0 /17)sin-q-j( z+ z2 -a2)/a) -2jUo (±(x+ ,/71.7? ) - as y approaches 0. z2 -a <1 11=4Q)i 1dW 2U0 -. 1 ( /g ))(I TT-7 2 -a ) 2 -4../2x(x± - 72,./a2 +(z±./z-27-2 dz TT Wi2 plus on top of Uo , I dWI2 4Uir• x+ (TINixtlx2 -a2 ) a = E Id z I minus below Uo. n2 -a2) 2X(X2 dZI 71/TX‘ixa - a 2 _L€1,4 8r°' xAixdx Ans. 1=1 ' dF = at - agds. a icos-1-r2, 2 _ a2 1112 ra rr 2 '4,1 a = 2cu2 ax r 1L 2e z=
where L is length of edge.
)x
Page 118.
Chapter IV.
Page 46.
-79Fold x-axis into right angle, dz/dzi =laii z=2,51 . If zi =xi +jbi then z=2/x1 +jbl , Use Dw 58.1. -±z = '12(,./xf +b + x i}+ j'12f,Ixt +bal -xi3= x + jy. xy = 214 = b2 so b1 = +1)2 . When r a =c 1, r=c=24, c1 =-1 /c2 . Substitute in 4.31(3) y z = b,/j cos (k snu)-jXsin-l(cnu) / TT (1-00 3.
k2 = cos(4c2 (1+X)/b2 )
[V] is doubled so capacitance per unit length is doubled. From 4.31 (4) C = 8Eld.17:7 1 2} /K(k). For good accuracy c/b should be lATi or less. -80Like above but with folding angle 217 instead of Tr/2. dz/dz i=z i , z=z1/2. Line zi =
-11bi becomes 2x= xi2 -14, 2y=2b1x1 . Eliminate xi to,_ get y2 =4p(x+p)
Where b1 p, 2 c1 = 2a. Substitute in 4.31 (3) cl+T.)
z = -8p(cos-kksn-k2)+ Xsin-kcn k-2- ) / Crr(1-FX) ) 2 where 4.31(2), k=c os Uo Uo [V] is halved so from 4.31 (4): C = 2eK(J:(2 3/K(k) Ans. -81 dz/dz i =z i ,
z1 = 2z,
yl
U=0
b1 = 2b, x=.4(4-b2i )
-Pfp
y = bl x i . Line U=0 becomes y 2 =4p(x+p) where b12=-2p. If xi =ai , x=a. Substitute in Problem 61.
6-2a1-) sn -1(exp 1_417 (-1.,) /41 , Id, where k= exp (-TTr/TD) a . Flux is half that of 61 so that R(k) capacitance is C = EK(N/T:10)/ - K(k) Ans.
W
-82-
aN/T, .7;
From 4.30 (1), z1 = Bisn 2 (KW/1.10, nal /Al) .
d zi z= rkiZi ; t 4,/zi- c, iz, de pia Check by p 99: j o dz = zi (zi -B1 ) (z1-A1 )
I
Let
-Ci u
z1 B1-u 1
'
u-
) (Zi-A1 )
71.40 iVr---0 ;ITT0 vz?
-Bi dz1=B1-C1 du; z1 =0, u=c , z1 =B1 zl - C l ; (1-u)2
a• u= 0.
z1 -CI = (B1 -CI )/(1-u); z1 -b1 = (B1 -CI )u/ (1-u); z1-A1 = {(B1 -A1)+(A1 -C1 )0/(1-u) du rgjc i du __a (CI -B1 ) rBI,C 1 Nrf(7 -0 iNCI (Al -B1 ) u 0 /1(11-(31 /C1)}u(u- [(AI -B1) / (A1 -C1)] - jTVCI (A1 -B1) 0 <s i/c, <1< (A3 -131)/(Ai-C1); a= (A, -B, )/(Ai-B1); 13= B1 /Cl ; Y= 0 in Milne-Thompson page 29. j(c -a)=
) k 2=m=2 = B1 (Ac, a Cl (Al -B1) ( 1), 2a (CI -B1/47-C1
d-_ TEN/C1(A1 b-
Al -B1
2a (C1 -B ) rc°
11„/C, (A1-CI
du
(u)
lL
K1=171
1 1- = C (Ai BO - 1-k2= k' 2 (2). d = a-c so -
-Ci) _ 2a C -B
2 (Al -B1 )
(gym du 77717)
K JBI (Ai - CI 1:Ci (Ai -B1A)1 (C1C1B(A1 3) -Bi )
103 du r a du 4_ -1 (u) +.113 .,/f (u) j (continued)
(3) _
(u)
b=y
a du 13 f (u)
jc+b
jc
dz so that
Chapter IV.
Page 119.
Page 47.
-82- (continued) 2a(CI-B1) _1 b -17,/c1(Al.B1)cn (0,k,)
1 _2a(CI-B1) K JAI(C -B1) n.,./C1 (Al -B1 ) ,A 1(A1-B1)
(4). One of the numbers A, I B1 ,
or C1may be taken as 1. Ratio (3)/(4) gives: d K(ti) = bK(k) (5) which gives k. Let C1 =1 so (2) becomes k' 2=A 1(1-Bi )/(Ai -B1 ) (2). Write P= 2ak'K(k')/(rb) From (4),1-
28../(1-Bf)A,(1-131) 2a1 417:17K f k l ) =153(1 P so " b. " -1-1-1 /- rh1A71` Bl) 4W1
= (1-131)P2
(6)
A i(1-B 1)P2 Al SO At-(1+P2 )k l2 Aii-P 2k I 2= 0 Solve for Al. k' 2 -P2 P2 (A1 -1)P2+(l-Bi )P2 A1 (1(7) A1 = -irk' (1 1 P 2 )10 # 1(1+P 2 ) 2 k' 2 4P 2), C=4 EV1/00 = IdNil ($/A) 3/1(C - -
-
Ans.
83U0 V 1 Bend negative real axis of Fig. 4.30a up to an angle a with the x-axis =0 0 V= 1 x1 rya a/rr (a Y0,_ / =cz. If z1 =B1, z=b; By 4.18 (7) dz/dz1 = z1 /17) -1 z= c1 z1 , z •4--- a—) r/r r/a rA , 1= ca B1 2 . z= (Bsn2(KW/UB)3(Y 0 ,4/ If z1 =A1=1, z=a. B1 =Cb = (b/s) 1444 = 1204. U0 Vl x by 4.30 (1). When W= U0, b= (Bsn 2(K,k)lairr= Bairby Hd 141"16.5.3. so that —17 z=b sn[KW/U0,(b/a)2 /a 2a/r Ans. Range of U and V is same as in 4.30 (1),(2) with Ans.
U0 and VIinterchanged. C= 2eV 1 /U0 =2EKW1-(b/a)'rr/ti)/K[6./(b/a)17/") 84-
41.Y1
Proceed as in last problem but add r/2 bend at z1 =1. To get capacitance ,A in 4.30 (1), z1 =Bsn 2 (KW/U0 0
must be found
)-1/N7T7T1-, From 4.18 (7), if a1 = a, a2 = r/2, dz/dz1 = z l(c"
U= 0
V=0 B1 U 0 1 VI "1
.,77-71 )dzi. By HdMF 6.6.1 the integral is an z = rzi (z1(a/17)-1/1 A .4 a --> ra incomplete Beta Function, z =B ty„--2-3. When z1 =1 it is the CN.f z1 ra complete Beta Function Btr,Ti, see Dw 855.1. Thus V=0 0 ra 1 b = BBtr,73, a = B117,71, and by HdMF 6.6.2 and 26.5.1 the ratio is the function —
(Et'121 which is tabulated. Tables of the incomplete Beta Function. K.Pearson, a = B r Biometrika Office, University College, Camb. Univ. Press, Cambridge, England. 1948. Thus the modulus AJ = k of sn(KW/U0) is known. Since k < 1, HdMf 26.5.4 gives
B
1713(2 /747Ti frr —2 3 = aB(a/ 7,1/2)
13(1+(a/r),n+l t n+11
1+ E r
I BL;+(a/r),n+1
The potential difference is the same as 63 but the charge is twice as large so that C =
4Ev1= 4EK U0 K( )
Ans.
Chapter V.
Page 223. (199)
Page 48
-10_ a b ( b4+ 2a 2 b2+a4+b4_282b2+134 0 = -agI t/3 1 + 1 li 1617E132' q2 (134 -a4 )2 4TTEb - 0 2 /112Lb+(a 2 /b -Fq 1 2abb b 1 2ab(b4 +a 4 ) 1 —r . a=g• Ans. 2i --b2 9.1 _a )2 . If a<
1 1 -q3 a/c 2 2'"a2 i aq2 2 alts { -aq2 a2 ) c 2 J - Olt ( c 2 ...a ) 14rrEC (C -a 2C-1 )2 4rrEC 3 4rrec'717-7
q
-
Image force is F • -
q2(2 c 2 -a2)(a/c) 3 2a3 Total is 0=-22-+qE (1+r) so 4Treca 4rre (c 2- a2) i
Q=
2a3 a 32c3-c2 [4rr t E (1+ )- qi.--1 G
(c
Ans.
-3-
,. ,_ 2a3 ) ,
(1) 14-r a 1 -I F1 -F2= 0. -q +q --.7.a-2r 77.-1 1) -4r 2j ' + r(r-I F l = Eq -16rcr 2 ' F2= qE ' ' r 3 1 qr • 4a 2 r 1 r -1 2 Ea 3 r 2Ea 2 ■ ) 2 Ans. (r21 7753-lar = gat r 3 4TTE (r4 _ a4 )2 i so E- 2 ,rfe(cirr46.a , 0= +q 2 -1. (7=-ajr +4rre r (2) q 2 r 2r 6 11 F1 = ou 1.----="1 j Ans. 4r 2 (r4 -a ) ., -4-q as 3q r ) c - Ev 3q + + ( 4rrcv-q 0= , 1 ) 2 E + E v • (3-a ) 2 4TrEv (Ja + 7a 4TTE V (3a-a)2 l 3 2
r
, V
qa r -a . 4rre--irr7iN'17
25 \ c- cvC' EV _ 75 3 j_ f 3 "="671 1-\IOC, - 1 0 0 1 E+Ev ' EA-Ev 352'
+3 q' dA =
4 27 C = 27 7 Ev =
1.541ev
Ans.
-5C-
dS __ 2 da) . r3do) . From 5.04 (1) da = crdA = 1 2 1.3 9- --cick° 2rr cos () cos0 a
so a = - 2r Ans. - 6 CFrom 5.04 (1) a = -aq/ (2rr 3 ) on plane from charge q. r/2 -r r/2 -q 1-r 2 13/2 +(r2 +3r 2) a/ (.15r+-12-.150 2 +7 1i r 2 3 2 a= 2rr(..jr-1Elir )2 +4r
r3
÷ 2rr {2 8
77-T571
1
Ans.
iirrr s4 + 7%
- 7C(a2 -b 3 )q 3 4. aq To cancel this at 4rra(a2 + 132 -2ab cos0) 2 + 411a 2b. 2 . -(ab)q q _ 0. most negative point add positive charge such that, at e= v, (a+ -b)2 a b or Q = -aq(-ab-b 2+a 2 -2ab+b3)/(b(b-a) 2 )=[(3b-a)/(b(b-a) 2 )] a 2 q Ans. For zero charge, 5.07 (3),(4): a=
„
-8CBy similar triangles the plane dividing visible and invisible parts passes through the image point and divides its flux in half. ,2:,? A+a (b-alb-1)}/ 077 By Gauss 9f _ -12-aq/b + 4q(jbaAns =N47=1" Theorem: Qb
Page 224. (200)
Chapter V. -9C4rEF =0= Qqb-2 + aq2 b-3 - aq213-1(b-a2b-1)-2. or Q
f(b2 -a 2 ) 2-bg laq - 0,
-Ab2 + b3 0 -a 2 2 2 )
a3 o(a2+2aE) 0. b-a - E<<1 so bkla. Qf-r, a.4a 2e 2
a3 c1(2b 2 -a2 i) b(ba-a2)
Thus E2=4.9 2 q/Q,
0
Page 49.
E = VTIM
a 2g
4E 2
Ans. -10C-
Sphere x2+y2 4-z2 =a2 . Charge at x2071+4=rf. 1) +q at x2 ,371 ,z1 . 2) -aq/r1 at a2 x1 /r1, a2 y 1 /r1 , a2 z 1 /r 1 4) +aq/r1 at a2x1 /r1 , a 2y 1 /r 1 , -a2 z1 /r 1 6) +aq/r i at -a2 x /r1 , a2 y 1 /r
a 2 z /ri
3) -q at x2 ,371 ,-z1 5) -q at -x1,y1,z1 7) +q at -x1,y1 ,-z
8) -aq/r 1 at -a2 x1 /r1 , a2y1 /ri , -a2z1 /ri -11CHarmonics and Green's Reciprocation Theorem simpler than images. Planes earthed. Quadrants at V0 ,-V0,110, - V 0 . Let is= cosO, Expand V. ot3m A mnr -n-lPm(u)sinmo. Set r=a, multiply by PI:(u)sinpOdOde, integrate -1<:p4:1, 0<:95.4:2r. Then by 5.231 (3) right side is
V=1
Ds
(p+s)! 2(2-.Spr. On the left side we have as i(p-s): 2s+1 . rri2 rrr +137/ 2 IT .217 Jo Vsin podo =VoCS 0 317/2sin pc/Rio = 8Vo/p if p even, 0 if not. 0712 TT
VP;(p)sinpod0d8= 2 1f0 11,J or
s
Thus p is odd. For a<:Cr take smallest r power. p is equal or less than s so is 2. pl pl 4 5-1! 1 5a 3V 1411(04.= 314 (1-0)du= = 4. A 22 =4-V0 •8•4•Ta 3.-77 : -•17---5TT°.Then by Green's .P
Reciprocation Theorem 2.12 (2) p 34: Vo Q'-EV I ce =0, Q'=-17Pq' Vco r r 5a3 P coordinates are 0=7,0=-1-i,r=f. Thus Q'=-1.72q'. Ans. -12C(1) is the given system. (2) is the image system. (3) is (1) and (2) superimposed, so the lamina is uncharged and at zero potential. Thus the density at any point (1) is ao
- a2 .
-a. q2 •
' q1
Ans. -13C-
Apply Gauss'Electric Flux Theorem to four charge image system shown. cosa= f/,./1-f 2 , cos = a2/(ffia2+a4 /0 •) = aaTTET . Flux in cone of half angle A is 2 (1-cos0) multiplied by total charge flux. Flux passing
q
f , into hemisphere at right is: from +q and -q, Fl = 2q4(1-cosa)=q(1-47jFf2 j;
from +q', F2 =-1(1+cosp)aq/f=4.(aq/f)(1+aaTTi2 ); from -ql,F2 =-1(1-cosP)aq/f. f a2 Ft . Induced charge is -Ft . Total is Fl +F2+F2 = +f 2 f.,/a2 2 8.2 -Ef fs_ a 2 Thus Qi= -q{1- f‘,/f2+a2 } Ans.
Page 225. (201)
Chapter V.
Page 50.
-14CBy 5.06 (3) density at A from ring charge dq is dcrA=(a2 -c 2)dq/(4rraR3), dq = a2rc2 sinec de, R= ✓a +c -2ac cos @, dR= acsinedO/R so that / Ans. a 2 -c2 27Tcand _ (a 2_ c2)c a 2 -c2 (1 13 -c3, a+c( 1 21.133 daA a "/ uA2a2 %.1/0 R 2= ac-272 AD AB - a97 AD -15C-
-T-
r>>a
2 a' 20 from 5.03 (4) By 2.14 (1) a'charge is Q=-qVINT 1 =--- tan- r r 2 -a' 2) with y=z=0. Potential at a due to Q is V=(2/TT)(Q/(8ea')) (aWr 2 -a'
2a'--3 But since a'<
m
a
-c2 )cosOsin0 de r r0 r 'a2 _ c 2 )sin 2O -a 2 /4/a 2-(a 2 -c 2 )sin 2O3d0 = -a(E-K) dr= (a2 r-Jrr/2 wa2 A/a 2-(a'-c I)sin 2 0 ' where k 2=(a 2 -c 2 )/a 2 . Ans. Therefore t = (k/0/571 5 (E-K) -18Get force between ring at r=c, O= charge q and image ring at r=a2/c, e=a, q'= aq/c. (506 (1)3. dV = aq•d0/(4recRi2r), R =z 2+d 2+b 2 -2bdcos0, 0=r-213, did = -2E18, cos0= -cos20 = 2sin 2 0 -1. d=csina, b=(a 2 /c)sina, z= (c-a2/c)cosa= ((c 2 - a2 )/c)cosa, (d+b)2 = ((c 2+a 2 ) 2 /c 2 isin 2 a. 2 de 2 de by aq 2 zrir V - 2aq r iTT 817 2 EcJo AA 2+(d+b) 2 -4bdsin 20 ) F = "lbz 4r2 ectio (,/z 2+(d+b) 2 -4bdsin le)3 „3 (N I +-2(c4+24_2a2c2(sin 2a - cos 2cx )1- 4a 2 sin 2a sin 219.] 3/2 = c -t3 (c4+a 4 -2a 2c 2 cos2a)3/2(1-k2sin 2 0)
= 8a 3 lOsin 3a(1-k2 sin 2 0)3 / 2 where k2 = 4a 2c 2 sin 2a(c 4+a 4 -2a 2 c 2 cos2c)-' Ans. dK r.'/2 1 dO sin2 0d0 kdK de E (-K-Er TV, 3 so ITT/2 But ---- +Kkdk (1-k2sinz0)3/2 1-k2 k2 ,-. 0 (1-k2sin 2O)3/2 0 (1-0sin20)3/ 2- dk By Dw 789.1. Thus F-
aqz(c 2 -a 2 )cosa k3 qzE(c 2 -a 2 )10cosa 41T2Ec2 4a 3 sin 3a 1-k 2 -16r2ECIa2sin3C1(1-k2)
Ans.
Page 225. (201)
Chapter V.
Page 51.
-19C-• 2qa 2 2aq. a b b-2a2/b - b 2 -2a2 ' -2qa3 alb a a2 n :191L. at d 3 = at d2 b- 28 2 /b -13 2 -2a2 ' "1-b 2 -2a 2 b-a2b/(b2 -2a 2 ) - b3-3a 2b
On 1: Q 1 1=-aq/f = -2aq/b, at d1 = 2a2 /b; Q 2i= •-
-2qa 3 _ 2qa4 a 2 o 2 2a 2 ) a2 a d4= b-a 2b/(b 2 -2a 2 ) - b3-3a2b , Q41-b 3-3a 2 3 .b 4 -4a 2 b2+2a4at 3- a 2 (b 2 -2a 2 )/(b3 -3a 2b) a2 -2qa 4 . a 2 (33 - 3a2b) a b - a2(b2-2a2)/(133-3a2b) - b4_4 a a•c 2+2a 4 ' Q°1-13 4 -4a 2b 2+2a 4 b-a2 (b 3 -3a 2b)/(b4 -4a 2b 2+2a 4 ) -2qa 5 . ar , ab asb a4 , j _ 8 2 0 _1321334.5a4b . Total yi— b L I. ft_ 3 so that ba 2a 2 -r b 2 -3a 2-b4-4a2b2+2a4+ b4_5 a 21,24.5 a 4 •"' m2 m3 1 in m4 3 = -2qm( 1-m(1+2m a+4m4-1-• • • )4m 2 (1+3m 2+• • • ) Qi= -2qtafj..- 1-2012+1-3M2 1-41 M -77-4. 1-21I1 .f. T..n i7-.-1. F5rn (
-m3(1+4m2+: • • )+rn4 (1+• • •))= -2qm( 1-nrhn 2-3m 3+4m4 - • '•), where m=a/b.
Ans.
-20CFrom 2.19 (4): W= +(su Ce+2s12 QQ 2+s a2 Q1), sli= c22 (cnc22•42)-1, so from 5.08 (1) c11 =1417e(a+a 2 b c-2 (1+b 2 c-2+b 4 c -4+• • •)+ a 3b 2c -4 (1+2b 2 c -2+a 2 e"+• • *)-1-• • • 1 = 4re(a+a 2bc -24.a2b2c..4(a+b)4....3. c22 = 4rE{b+ab 2 c-2+a 2 b 2 c-4 (a+b)+•••}. Then by 5.08(2) c12 =4TTE{-abc -1 -a 2b 2 c -3-a 2 b 2 c -5 (a2-i-b 2 )-• • • }. c i c3 2 -cf 2=(4rre) 2 {ab+2a 2b 2c-2+a2b2c-4(a+b) 2 +a 3b 3 c-4 -a 2b 2 c-2-2a 3b 3c-4+• • • } =(ab)"l-c
. orro2 (c11 c22
= (ab)"1{1+abc-2+abc-4 (a+b)2+a 3b 3 c-4)-1
(a+b) 2 c -4+abc-4+abc -4+• • • .
4TTEs22=b- 1-a 3 c-4, 4'7c s12=c -t so W = (8TTE) -1 ((a-1 -b 3c-4 )0+2c -1Q 1 Q 2+(b-1 -a 3c-4 )Q11 Ans. -21C-
M = q6. Dipole image: q`=-aq/c, 6.=
a 26/ c 2, mi.qt8'.-a 3 q8/c 3=-(a/c) 3M.
Let M = M/(417E) . Then M'=-Ea 3 a 3c -3=-Ea 8 c -3 at a2 /c from 0'. -3 =Ea °(c 2 -a 2 )-3 at a2 c/ (c 2 -a 2 ) from 0. M"=Ea e c- 3 a 3 (c-a 2 c -1 ) M'"=-Eal 2 c -3 (c 2-2a 2 r3 at a2 {c-a 2 c/(c 2 -a 2 )
}
-1
=a 2 (c 2-a 2 )/(c 3 -2a 2c)
fran O'.Field . 4.--.--- C --4 image is M=41eEa
From 1.071 (6), F = 3E1 /(41"rEr 4 ) so force on original dipole is
a 3c'•3
a8
1
a°
1
+. • ) + (c-a 2 c-1 )4 (c 2 -a 2 ) 3 c-a2c/(c 2-a 2 ) (c 3- 2a2t)3.1.c-a2(c2_a2)/(0_2a2011 ,,3 1 1 a° 1 + a '8 First image: B= 4Tie• 3E2 21. c 3 (c-a2 /0 4 c° (c-2a 2 /c) 4 c3 (c2_a2)3 [c-1 -a 2 c/(c 2-a 2 )]4+" 3 t!, 1 ° Second image: C= 4TTE•3E 2 a 8 (-- --+ } (c 2-a 2 ) [c-a 2 c/(c 2 -a 2 )]4 c 3 (c 2 -a 2 ) 3 [c-a2 c-1 -a2c/(c.2-a2)]4
A= 4rE•3E 2 a 6 {c"
1a
--s -
Since only a ° is required, keep only first term in B and first two in A. Second A term 7 ,s 3 n3 , a3 1 a2 1284 IS 7• 4 t 1 + 4— + --T. +• • •3 = zt.t. + 4'-'-'--g + 6-`2n +• • •. F = A + B +. • • • so C c2 2c (1-a 2 /c2) c c c c a3 a5 12TTEE 2 a 6 r a5 a3 F = ---4---1.1 - =13 - 4=3- -• • • - ---T - 4" -11. - • • • c c c c c
3=
a 6 r 2a 3 8a° 12rreE 2-1- 1.1- 3 --gc c c
3
Ans.
Page 226. (202)
Chapter V.
Page 52. f---- 2b
-22CImage of dipole M= q6 is two charges -bq/(c-i6) at b2 /(c- 6) and bq/(c-46) at b 2 /(c-.146). This gives charge bq/(c+16)-bq/(c-16) = -bq6/(c2 -4-6 2 )=-bM/c 2 at b 2 /c from 0 'plus charge bM/c2 at 0', plus a dipole of strength M'=bq(c-4-6)-2(b2 (c-i6)1-b2 (c46)11=b3q6(c±216)-1(c 2 -4-6 2 )1=b 3Mc-3 at b2/c from 0'. From 1.07 (1), field on axis of dipole is -bV/Or =M/(2TTEr 3 ). E induces 3
dipoles 4TTEEa3inA and 4TTEEb3 in B. At P field from M1 is 41TEEa 3 (2rea )= 2E. From M 2 it is 4rrEEb 3 {21TE(c-a) 3 }-1= (2Eb 3c-3 )(1+3ac-1 +6a 2 c -2 +10a 3c -3 +- • •) From Miit is 4reEa 3 b 3{[c3(c-a-b 26-1) 3]2TrEr=2Ea 3b 3 c-1+ • • • . From q12 it is
4-- 2a 4TrEEab 3{ 4rE c 2 (a-a 2 c-1) 2}-1=Eb3a-lc-2(1+2ac-1 +3a 2e.2 +4a 3c -3 +584 c-4 +• • • ) . From neutralizing ql it is -4TTEEa3b(4TTEc 2 (c-a)2) 1 =-Ea3 bc-4(1+2ac- +3a 2c-2 +• • •). From neutralizing q 2I it is -4rEab 3 (4rEc 2 a 2 )-1 = -Eb3 a--2 c-2. From MT - 4'r1EEa 3b 3• a 3 ( 2rrec (c-b 2c" ) 3 )-' = 2Ea 3b 3c - 8+. • Charge MT - 4TTEEa 3b 3 .a(4rrEc 3 (c-b 2 c -1)2 Ca-a 2 (c-b 2 c-1) 41 2 3 -4 = Ea 213 2 c -5 (1 + 2ac -1 +- • • •). From neutralizing MT - -4T1EEa 3b 3• a {4TTEC 3 (C-b 2C-1)222}-1..
Ea 2 b 3 c -5(1 + 2b2 c -2 +. • • • ) = Ea 2bc-3(1+2ac"'+2b 2c-2+3a2c -2+6abc" +
From qT - 4TTEEa 3b•a(4TTEc 2 (c-b 2c -')Ca-a 2 (c-b 2 c-1)-'3
3b 4 c -4 44a 3 c -3 +—+b 2 c 2 - • • •). To neutralize qT - -4TTEEa 3b • a{4rEc 2 (c-b 2c -i)a2 }-1=-Ea 2bc-3(1+b 2c-2 From
- 4TTEEab3•b(4TTEc 2 (c-a2c-')Cc-a-b2 (c- a2c "1)-13 2 )-1=Eab4 c -5(1 + 2a 2 c -2-1-• • • • )
To neutralize
- 4TTEEab 3 -bf4TTEc 2 (c-a 2C-')(c-a) 2 ).-1 = -Eab 4 c-5(1+• • •+2ac "II- • •)
Other terms give more than c 5 in the denominator. Add above terms to get: = E + M1 +M 2 + etc. which gives :
Et]. + 2 + 2b 3 C -3 +6ab3c-4 +12a 2b 3c-5+20a 3b 3 c -8 + etc.
E{3 + 6 (b/c) 3 +15ab3 /c 4 + 28a 2b 3 /c 5 + 57a 3b 3 /c 5 + • • • } = Et . Then from 1.16 (1): a = EEt Ans -23CBy 5,102 invert planes under influence of charge
I
+.4 a-,e- .-K-4 a-.1/- 4a
• • • 4rrE. Distance of nth charge is 4na. By 5.10(1) size after the I 4,-4a n inversion is (-1)n 2a rre/(4na)= (-1)n2TTEn-1. Q = 4TTET(-1) n-1 =-4rEern2b) Dw 6.01. V= -(2a)- . C = QV -1=8TTEa2n2 Ans. -24CE-- 2c
•4
:q
I
+.q
Q1
-cq
I
-q
-q
Invert plane and charge systems shown about -q charge by 5.101. k=0P=OPI = 2a, 0q= 2b, 0Q=12-k2b4=2c=2a2 /b. On (1), Q I =Eq'i =Ekqn0Qn"'. On (2), Q2 = Eqi= Ekqn0P -1. If q= 4TTE, sphere potential is V= lc' = (2a)-1. Q1 =4refa/2c - a/(2c+2a) + a/(2c+2a+2c) - a/ (2c+2a+2c+2a)+• • -1 217E1) (1/a+ li(2a+b) - 1/ (a+b) + 1/(2a+b) - 1/ (2a+2b) + 1/(3a+2b) - 1/(3a+3b)+• • • I = 47E-1.1/a br a+b2 2 2rreb 2 + ./3a+b)+••.+ 1/(n 2 (a+b)-nb)+ • • = (a+b El/(n 2 -nci) ,where c 1 = b/(a+b). Q2 is the same )2 2rEab as Q1 with denominator a and c interchanged so Q2= = (n 2+nc2), where c2 =a/(a+b). 1 B6y Hd M F -2, ... so C= . 3 . 16 T2-7;IY(1k)+cl/zi-1, -
V
bc 2r , Ly[ -4+b
L,yr TbF
J}I Ans.
-
Page 53.
Page 226. (202)
Chapter V. -25CPlane Al inverts into sphere A and plane B l inverts into plane B. Those
q1
•
bF • - 41
• qi
charges to left of A l go into A. 4.
r 2 ch /b, tad = r2 = cb, 2b/d= a/c. Charge on A is Qs= qlr `2d-b + 2d+b •4d-b 4d+b- ) 2ac-1 1 -b2 q -2b ra „ ra chr 2 b n 2_ a 2c0-2 q(- c cot 7.) Ans. by HMF 4.3.91. 2d2 1 n2 -(b/(2d)) 2 - -aC n=14n(d 2 -b 2 ) b
q=
ra ra SO charge on plane is Q = q-Qs =q7cot-E Ans.
- 26C2m 3 2m4 2m 2 Let a cosha= c. In problem 19: 2m= 2a/b= a/c= secha. Q= -2m+ 1-2m2 1-3m2+1-4m2+2m + 1 2 1 3 _2 1 +1)-'-• • • =-(cosha)-1+(2cosh2 a-1)-1+(4cosh3a-3cosha)'1 13 - T =-2m +(172-1) -(772m2 2 2m' n +(8cosh4a - 4cosh2a +1)-1+• • • = -(coshar 4-(cosh2a)-1-(cosh3a)4-(cosh4a)i..= E (-1)sechna Ans. -27Q1= C nV i +Ci 2V 2 0 =C22V2 +C22V i , so Q= Q1-FQ2 =2(C11 +C12 )V. From 5.082 (4), (5) C= 811tasinh8! Ans. 'f(csch(2n-1)13 - csch2np). If m+n=2n, C = 8reasinh4(-1) m+lcschm8 - 28W=.12-CV 2, F= 61•1/bC4M/t13 • Op/oc = (bW/O8)/(2a sinh13)by5.082 (3). From last problem F = 2rreV2 { coth8 El (-1)m+l csclim8F(-1)m+imcschm8cottuni4=2TreVI (1)111+1 ( coth8-mcothmp)cschm8. An - 29CPrimary sphere images: iqa/c,-+qa/c • • at tia 2 /c, 1 - -4-a 2 /c • •
1 -4
t--2c-3.1 I ';`' I -'`' ! 0 F A I +4 I...q
Neglect secondary images of order (a/c) 2q. Thus Q:::141+![1-4-44+. •]3= q(l+fPin 2). All image pair potentials at sphere center of order a 2 /c 2 so V=q/(417ea). C=QATZArrea(142/722). Ans. -30CFrom Problem 24, for V= 4-/a, Q= 2frea[b/(a+b)] 2 (El/n 2 + [b/(a+b)]El/n3+[b/(a+b)]2 E1/n4 +• •1 so C1 = 4rEarf2 Z(4.3 2 + f 3E(0+ f 4 E(k 4+—] Ans. a 2,. . - I+ b2 I
-31CInfinite image set. VP = Va+ Vb+ Vc+ Vd . r 2a 2b + 4reVa. -1 + -1 -. + 2a+z 2(a+b)+2a+z 2a+z q 1 4rEVb _ 1 + +2a+2b + 1 it + 2(a+b)+z z z q 4rEV-1 2a+2b -2a-2b -1 2(a+b)+2b-z + 2b-z + q 4TreVd_ 1 + 1 +2a+2b + .. q 2a+2b-z 2(a+b)+2a+2b-z t..2a+2b-z -
-
I.b 1_1 c i. + a1 - 4..z7 4-b-> ;a
dI
21- in+ 2a+z]-1 4.13 It
'2a+2b n z -1 1 [-F[r' + ;7213] 2b-z7 -1 1.1 c° V-Cri+ 26.21) J + n1
cr'
-
)
cif +[n+ 2a+2b - zil_ 13 1
Sum by HMF 6.3.16 and by HMF 6.3.5
2a+2b
eg' • 1 1 1 cc -z 1 -z + E -1 ---+z n }-- z+ iE-7----= nnz) z 11(1+z)+ Cl= -T(z)+C'. Thus V becomes:
V
8TTE (a+b)
DV2 (2a+b)/(a+b)]+Y[iz/ (a+b)] +'1[4(2b-z)/ (a+b)3+11[12-(2a+2b-z) / (a+b)]) Ans.
i
Chapter V. Page 227. (203) -32E-E then: p = —iv
00
d2x - trreit-i'
dxA dx dx _ 16rev x2 = -Tilde' dt
laq 2
Page 54. or`
a 4rrev (2x) 2 .1., v 1 1 d Integrated ( _13q2 ) -1, = d'jv /T:la to x= O. dt = (--L-1dx and integrate from t= 0 to /t1/./ a -3150 -4 —: dt Prrevma a-x If
c+
c) ix )0a = -Ira, a 3 = pq 2d 2 /.(2r3 evmv 2 ) Ans. -,113q 2/(877 Evma)div = {-x(a-x) - 2a • 2 tan-14.(a73— - 33CBy Gauss Theorem charge on tube of radius AB is fq+q(c-cv )/(c+cv)31(1-cosa) =12-(1-cosf3) . Thus 1-cosp= 2e(1-cos ct )/(e+Ev ) or sin2 -ip = 2e sin 2lia/ (e+ev). Ans. - 34C-
el
1
el
4r- a---,.<-- a —0. 91 q
2
cit ei _ ea A Force on qi from its image at q2 is c i+e 2 16rei e Force on qi from q2 is 2e2 q 1 q 2 /((ei +e 2)(161 re 2a 2 )1. Total forces are: r 2-E1 , e.-e2 C12 91 MIS • Ans. F -„ ,, E 2 q2 -Ch. = F k E 1+E 2 ) Iona2 q q1 16rraa(e/i-E2)H-c11+2q 1 • -35C-
V1= V 2 SO SuQi+S21Q2 = S 2 2Q2+S12Q1 . E -E, E„ 1
at q1 of sii = --xE 1C1 2E 1 of sz = -2 E i -f-E 2.417E
Uhitcharge at q1 gives potential
—. Unit charge at q 2 gives potential at q1 E 2+e 1 8rel a
---&---&
2 (a+b)
-
2
Cv ----1 z , s12 4.17(e 1 +E 2)(a+b) • i+E 2 817e 2b e 2c 2 c 2c, e1-E2 2b(e1 -I-E 2 ) (e 1 +€ 2 )(a+b) Ans. E -E
(a+b). Similarly. 477(e 1 +E 2 )
c‘, e l- E 2 1 2 Aam 1 +e 2 8TTE 2 b (E1-Fc 2 )41r(a+b) _ E i c 1_ e2 c 2 e 2 E 2 4re 2 -1E+ et -C 2 1 C7 4Qz e1C1 ' e1+E 2 8•E1b (e1+E2)417(a+b)
c
C1 -
522--
1 2
2b( E 1+ E 2)
2E 2 ( E 1 -1-E 2)(a+b)
-36C1 1 1 1 4.• • •1 For conducting face 5.082(1) is: cu = 41Feva sinhCqsinha+ sinhal sinh3a+sinh4a The first term, from the original charge, is unchanged. The second term, from the first dielectric image and is changed by the factor (e2 -E1 )(e 2+E 1 )-1 . The third is from image of the image of the second in the dielectric and so is changed by the factor (E2E1)2(c2+9-2, co etc. Thus c 11 = 4rEva sinh a E f(e 2 -E)/(e 2+e )3""1 csch na Ans. -37Being the inverse of two orthogonal planes, this requires four charges. 9 Distance of q from 0 is f. f 2 =4/8 2 + ia 2 = 5a 2 so ql=aq/f= qb,r5. Distance 4
of q' from 0 (dashed) is f 1 . f?= 2a2 + 5`= 9a 2 /5. q"=(qA(5-)a/57(3a)=3q. qi= 2q'+q" =
- 2/.151
Ans.
Chapter V.
Page 227. (204)
Page 55.
-38By 5.04, invert disk at potential -(417EK)-1 with radius of inversion K to get inverse surface earthed with unit charge at the inversion center. K= a-b, K 2 = (a-b) 2=(2a'-a+b)(a+b), a' = a(a-b)/(a+b), p' 2 =K4/r 2 +(K-a1 )2 - 2 (K-a' )K 2 cose/r.. K-a,
2 11 22 11
a+b
( a-b ) 2 b2 so p' 2= (a-b)2 [-1---i--- + r2 + (a+b)2
2b(a-b)(c 2 -r2 -b2) r (a+b)2br
rat (a-b) 2 _a-b _i_ (a-b) (c 2 -b2 )1_ (a-b)3 (c 2-a x) a 2 -0 2 =(a-b)2 0+0 2 r 2 a+b ' (a+b)r2 (a+b)r2 j R3 -ALT-72 Ans From 5.03 (2), (3): cr= =--3 alDP r 2 2r 2 (a-b)(a' 2-ps 2) - 217 213.4ica-a2 -391.7 (217b)1 / a _az a= .- [1/ (2172)34 ( a 2_132)(c ){Q I j+ (b/(b +c -2bc cos 0)) de - Q a2 -b2 -477 3 c2-a2
1 221+17 _ 2 ( LE 1D-I 4fa2 -b2 -Q 7277 [t an -I -TT 2rf a(Ca -b2 ) 041 77aT c-b
Ans.
-40irqw 5 .10 (1) gives Q= 2aQI/Z7;i:--1-b' 2 =QI/151513 , -QVtan 2 8-tan 2abitan 2 e-tan g s a= K 3 a- cos3 0cosa•87T2a 2 (ta n 2 e- t an 2a)
'Q1COS ti -cos 2 f3acos 2 p-cos 28
c'
= Q' cosa ••
. let total charge induced
1 1
..0
•
A • \
.6-- a' Q'
•
V% (
b' =2atana a' =2atanfi =2atane
811 2 a(COS 2a-COS 2E) on bowl be Q1 . dS=2rtr sin e• a- 2de=8Tra2 sin ecosede, x= cos2 e, dx= -2cosesinede.
2QN/cos 2a-cos 2 B c-ose sinede p r(cos2a-cos 2 113)4/cos 2 B-cos 2 0
cos 2 8 _ / cos ' -Zwcos 2a-cos 2QICOS 2a-COS 113 -dcos 2 - x 20 _ xdx = , , , 'tan=j o rrkcos -a-x) cos rtVcos 2a-cos 2 f3 ,/cos2 a-costs o _-2 Q _, cosI3 - rr sin . Apply Green s Reciprocation Theorem, 2.12 (1). cosa 2V'co_sj Q QV' + Q V,; = O. \T' = ---IW, - —2sin-I cos cx Ans. P Q 1 -
2a ---)4
_ 2Qtan ta n
toss 4
VCOS 2CX -cos t s
'
-41-
ti From 5.156 (2) we need A0 =-2711 V (p)dp, p = cos° = cos2*. Let u= cosa/cos4', dp = -4sin*cos*d* = -41.1-2 cos4'posa du= -4u-3 cos la du. -10 7 . -8cos2 arlsin-lu du ri _ ya r8 2 sinlu 2712- + + j d& -- 2 Lcos al-From 40, by Dw 518.3: Ao = 2 .Losa u 3 r 1 1 ,177./Ti Ira 8cor s2 a[T4.r .. Trt -ace _ since] 2c0s2,-)VQ. ,„,,j Tr, (7-2a+sin2a) -l-cos2(13= -2 1' 2c os a 2coscx 2u Icosa
= Vo(13 + sini3) /TT, where (3 is the angle the bowl subtends at the sphere center. C = Q/1.70 = 4TreAo aVo = 4E3(8+ sins)
Ans.
Page 228. ( 204 )
Chapter V.
Page 56.
-422,3 dS= 8ra 2sina cosy da. Ifx=cosFal, N/cosAa -cos From 40: da= 8112 a 2(cos 2a-cos2e) 1/cos2 p-cos 2e ' aldS
dx= , 2sincx costa da, b= cos2P, and c= cos2 0. Bowl charge, including both sides a rb ,,i/x-b b dx 1 b > c so use dx dx - -cr I?) + (c-b) is aj Dw 191.01,195.02 217767E L"1 1 (X-C) 211%./h-cJ 1 x-c r 2 (b -c) • -1 ilr c sin _c tan To neutralize -as , superimpose cy= a' 12„5=T - Nrb---
2riszu L
.71C- _
T1
Nif.:T—
A.7-C
spherical charge density a1 . For sphere potential V°: vo =c1/(4res), 61..0(417a 2) so a1 =a0 43. There is no field inside sphere, so inside bowl density is unaltered.Outside densities add. sing _ eVo . a . . Lyor Ans. 1-111 c + o a of sine] 1 ra LV'sin ae-sio 2p sin-1 -43Two planes intersecting at 60° , each at a distance 2a from q will invert into the desired spheres with 71 As the inversion center and k = 2a by 5.102, 5.109. Spheres potential is -1/2a. 2a 2a q Inverse charges are, from 5.10 (1); 221, Al --a; 29.2,E2- 4,isa gy 4a .2a ---q. Total charge is (4+41. C= Q/V = 2rrea(5 - (4/N/T )3. Ans. 8a -44C-
. %
Dipole of strength 4rek3E inverts into uniform field E. M 9PQ ,„, OP OP OP 3 Di pole: M' q' PT ^- k 'OP' = k k k k 0 -OP c1 P+clQ= 1-57f - M q P2 9 Qi -kPQq 4rak 3k 3 Q41TEEa 3 Tour ,2 - 47 kMp 2 .Thus M I - 4TTEEk 3k-3 4176E83, M2-(4 - Lt.a ) 3 - 2,12 • (4a) 3 q 1 =-4rek 4 E/(N/1(4a) 2 1 = -4rea 2ENT. Same f lux from both M1 dipoles. 2 -
c
Gauss Electric Flux Theorem. From 1.07(1) flux from two Mi dipoles is ,rsin-1(1/,12) ,'7\ / 2eS1115212dS 87E a2 Ljo sinOcos Od0= - 2TTE a 2 E. 2rea 3 r m 2cose From M 2 dipole.-ej2re (a 4v-2-)3dS - -rea2 E -
/
-E
From q i and a '2' 4rea 2
4rEa 2 E1-1-COS-1417 VI 2
1-Cos +r 4ra 3 2
From uniform field: -eEr(a/N5) 2= -41 Ea2E. Q = rea2 E(2+1+2+4) = ll1Tea2 E/2
Ans.
1
I
/
',. .,
11
k=2a
. %
_q. 4afro 4a_fro 9
Page 57.
Page 228.(204)
Chapter V.
-45C%% 2a 4\ This is 43 with sphere c added which puts ai/sphere c' % 1 1 1 orthogonal to the planes. This adds six more charges ±q' 1 to be inverted. From figure (r-c) (4a+c ' ) = (r+c) (4a-c ' )=4a2. .12 -47 4c 2 -al, q' = -,+c'q/a= -ic-1(.fa 2 F4c 2 -a)q. x=-'4c' 2/a c' =2ac-'a 2a--,..
=2ac-2[a2 +2c2 -&,/a2 +4c2 ), x-4a = -ac ' /c. Distances of q' s from 0: c) ,/a 2 c' 2c-2+4 ax=24/Tx%/1+a 2 /c 2 . OP i =ac1 /c=2a‘raTac , EP 2=./(4a7-}x)2 +(2-7T-f x%/3+8 2 /c2. oP3-4riia-E—10) 21-(IN5f) 2=•,/16a2+4ax+x2 = 24x%/a 2+4c 2 /c . U4=0P1+2x = c' ( (a/c)+ 22-(e/a)) =c' (a +Wa 2+4c2 -a) /c = 24-
,ilk i x q'.
Values of q' (inverted) from 5.10 (1) are:
t %
1
-cf
Fc . 2.-2kq'/OP 2= 2.2a•c/(4x✓a 2+x 2 ) • /17Ta"14rrc= 417etAcr----1
.
2% tql ",
%.%
1. kq'/ °P I = 2a•c/(2a./Tc) • -,1-75 } 14Tre= -2r ec/a. 3. 2kq'F:)P-3 = 2.2a • ic/Cfax%/3 a 2+c 2 3
.
,... )..-, , ,
...s' ...- 4a
4
,/1
2c
•,175 ) 4417E= --4Fredisha 2+C 2 .
14c 2 . 14rre = TrtecAraT> 4.-kql/ OP 4= 2a •c4/(%/Tx%/a 2+4c 2 )• -017; Add Tre(WS-5) from problem 43. Potential is - 4/a. Total charge is
1-4c 2 ). The capacitance is then -3c 2)+(2c/477F.c 2 )-(4cAra-Firre([(4/45)-53-(2c/a)+(4c/fa-27— 2 2 -46CFrom problem 42, o=2cri+eV/a; Integrate, Q=2Qi+(4Tra-S/a)V0E. As S approaches „-------.% I zero, Q approaches 4'n'eaV so Qi approaches TeVo S/a.
Ans. I
/1
e
i
-47C-
, - ..... % ...
f
Invert, by 5.102, segments of earthed conducting plane lamina under the
-470 '. • . \
n
1,
influence of charge -417c at inversion center. By problem 12 and 5.04 (1),
0,4 .I
before inversion, o1 -o2= cro= 4ear-3. By 5.102 (1), alo =+(r/a)3 o0 =4€/a 2 . For potential Vo , c',3= 2aVo • 2e/a2 = eVo /a.
Ans.
-48CNote that e= 1-(b/c) 2 and A= TTbc= 'Rea] ..1.7;1. In 5.02 (4) let 0= b20, a2=0, factor out c and let c=(1-e2) 7 . Then C-
811/Ad0 ,,gT (1-e2) 40(140)(1+(1-e 2)0)
Use Integral Transforms Vol.I, No. 23 page 310.
Jima' r (21) 2 2F1 (4. ) 411 ; e 2 )"" rt( 1 + rife2+ e4 +2-2-5% e6 +• • ) by Handbook Math. Functions 15.1.1. By Dw 5.1, (1-e2)1 /4 = (14 2 - Ae4 - -ffice° -••) so (1-e2) 01:=TT(.1-1-[-14]e 2+[11-4-?-6- 2 ]e4 Je8 + • • 1= (1-144e4-1114- e 8- • • . c=8\FIii/E II( 3 = •17%Fr8 [ 1+ .1k4 + 1 +[ -9-8...14 .4 ee 256 )
Ans
-
Chapter V.
Page 228. (204)
Page 58.
-49CSince ellipsoid and sphere of radius r have the same volume: r 3=abc=(1+a)(1+0)(10)r 3 so a + p +11 IV-(743 + a Y + 13y ) 1 neglecting fourth order. (a+ 13 4. y) 221Clt 2 _Fp 2+y 2+2ap+2py+2ay ::::,(ap + py +ay ) 2 P-.. 0. Thus 2(Ci+ p + yyz( a 2+ ps+y2), gap+ py + ay ) p,-.,, _ (as + fit:4. y 2) . From 5.02 (4), if x= r-20 +1, r 2 dx= de, 0 < 0 < 03,1 < x < co. Let Z = C/ (8Tre) then C = {j0 (4,/[(1+ a) a r s+Eci[0.+0 ars+ p][(1 +y)2r2+63) dE31 1.1 = i,7(4,/[1+( 2a +a2 )/x ][1+(213 +13 2)/x][1+ (2Y +y2)/xj) (r4/30) dx 7(,I1+[2(a +p+y) +a a+ p 2-1-Y 2]/x + 4 (a131-py+aY) /x 2 )-1(4,0) -14 -ir Pe, ;(c°1 (4/1+(a 2+13 2+Y 2 )(2x -1 - 2x-2 ) ri(Nrx7)-1d3]-1 21471,174(a 2+13 2 ±y 2 ) (4./7b--4/27 )] d ) = r [ 2- (d 2-113 2+y 2 ) (4.- 4)] 1 =ir [1- (a2+13 2+y 2 )].
-1
Thus C = 4TTEr[ 1 + 115 (a2+13 21-Y 2) ] Ans.
-50See 5.27, 5.271, 5.214.
V= C0Q0(j C) = Ccot-1 C -f-Irc-e > C/Cc i /r = q/(4rer).
so V= qcot -1C / (4Trec 1 ). By 5.27 (2) : a 2 =cfCc21 b2 = ci (1+0), cal= 1,2 -a2 , C20 = a2 /(b 2 -a 2). -q -q e to.,r7-c ri q -1 a=
161 _
h2 bcCo- c 1A 2+c i 4reci i+cg 4rrco/40.+0)2- p2 -477c 0/134 (b2 -a2 )"1- p2 2 b4 2Ib rbja. q2k(b/a) 2.11p dp rb q2 rCY2 dS 2 F=J 2E 32TT 2 E (b2 -a2 ).10 b4 (b 2-a2)-1 - p2 ) JO r b2 -a 2 P 0 • F-16rE(b2-e2)
Ans.
-51CSee 5.28, 5.281. V= CQ0(71) = an C(11+1)/ (71-1) 3 m.: 2Cc 2 /r = q/(41Ter) so V -014--(471rre! i (r1-1) I r a2 = cpi e ) b 2 = c22(r)2-1), Ti g . a 2/(a 2_132) , ci = as -1)2 . a = (E/112)[N/all] at n .rio ! . F .rcradsz q q q -2 1 J 2c a= 8T7c a 'ni -1. h2 -41% 2,1C 2(118 - §1) (111 -1; - 4rrC 4/p + b4 / ( a 2- b 2 ) q2 b rb 217p d p q2k(a/b) F- 32r s e (a s _bs )..J0 p 2+ b4/02 _b2) .1 j'o = Pnitp 2-F a 22 b : b l b° . F . Ans. 16TTE(a 2-b 2) -52CIn 5.03(4) : r=b, y=0, z=0, b=a.r tan-145aab2-a2-4,/(b2-a2) 2-4b2a2 3 = (2 1.70 / TT) tan-1 (a/b). This is potential of •spheroid of minor axis b.
Q - 8eaVo 47ea 4rEa 4ree External spheroid Ci =--r V1 -V0 V1 -V0 Tr-tan-1(a/b) cotNa/b) angleBsc capacitance is C 2 8eaVo _ 41TE a 4rea w_ _g2_ + +Q,) 2 Q2 •angle.ags,+ CcH-Q,) • anglesBc C2 = tan-1(a/b) anglesBc' 2C1 2C2 8rrea
Ans.
-53C2 As in 5.19: z2+p 2 (1-e2 ) =c2 , ri =c1 (1-elsiniK--2:=c1 (1+gei)- ic aPa (p), r a c 2 (1 +aged- Te a, 204,,,• V=A+B/r+(Cr 2+D/rJ)P2 (p). C,D, order of e. VI=A+B(1-1e1)/c1, 04Bc1 +Cc 12+D/c? , V2=A+B (1- fei)/ca , rbV 0=3Bc2+Cct-I-D02 . B= (V 2- Vaic (1--2p 2)-c11(1- §ei ))-1 = c1 ca (Va c3 -a( ea c i -e ic a )). 0=-cj-57dS fl li P 2 (p)dp=0 so Q=4TTEB. Thus B=Q/(417E) . C=411eB/ (V a-V1 4Ttec ic 2 /(e1- c2-4(e 2 ci - e a cd) c f _L a eac - eic , _ 4rre i c 2 ci - c2 i(e2ci - cies)3 Ans. L 3 C1- C2
C1-C2
(C 1 - C 2 ) 2
Page 59.
Page 229. (205)
Chapter V.
- 54CLaPlace's Equation in cylindrical coordinates is, referring to 3.05 (2),
_L
by b 1 b 13 bp k"IP bp' ' bz
bv\ = 0, bz,
K'Kiev =--p- . .
. b 2v_
Thus bp: bz2 - u.
This is two dimensional La Place Equation.
A solution giving ellipses is 4.22 (4), z2/(c2cosh2 V) + p2/(c2 sinh2V) = 1. Let a1 = c coshV1, r2_ 4c2e2V, V-4 Onr+62(2/c)
b1 = c sinhVi, a2 = c coshVa , b2 = c sinhV2 . For charge let
Q = SKI (aV/br)dS= forr(EvKL2TTr 2 sine) (pr ) ide= Ev 21TK/Zde = TOKiev . VI = coshia/c) or. .
V i = Pin( (a l/c)-1,1(a 2 /c) 2 - 1) = Pgr((ai-Fb i ) Lc), 112 =071((a 2+b2)/C ) .
Q
2TTIKtE„
111 -‘12-2nUa 0 14/ (a a+b 2) }
Ans.
-
-55C-I r/2 -1 5.02(4): C=817e/r 0 (0(a 2+0)(b 2+0)3 7cle= 8rre/ro (cos40(a 2+a 2 tan 20)(134-Fa 2 tan20));95 a rr/2 CV = 877€/j o 02sin20+1)2cos 20)-2idszs. From55.03(2) . 03(2): c= 4rrab [14 a+ 2b • Invert elliptic disk on radius K=c. x 2 a'2 +y 213-2 =(x2+y2 )r-2= 7P-12/EV X2 v 2. .21A -i2 QVc =cQ/6-T so that 14— But cP7c= cP/OP and --61b2 = 1 0242 a 2a+-1 4. 007152_typ27z2 CO2_157,2)tip2_(Ep2_M2)75(12 c 2a5p2 -'6(12) - cQ 20P 2-cP at)t) 2 cQ 2-0 2 OQ 20P 2 OP 20Q 2 cQ 20P 2 .
qi -]
From 5.102(1) a'=
C 3 a CV C 3 cQ OP 4rab OP3 c,Arp 247:/(T) 2
-
c -cQ 8rEv a tirrabI OPaNt(572...7Q2
Q -2-2 From 5.102, if V= 4r cl inducing charge i s Q so a l =-1 4rfaoI Or -,vopcg 2 -57Q 2 e -56CO E(cnr -n-l+Anrn,, ,,,‘ v _ q co From 5.153 (1) Vi = 4r 4fre ci e 0 , J.L--n k../, 0 — E Bnr -1.1-1Pn (II) n -n-1 nv -n-1 n 2n+1 . v + Ana = Bna since potential is continous. , Bn= c +Ana At r = a, c a n -n-2 n-1 -n-1 n , / // //////// e v C-(n+l)c a 1.-e(n+1)Bna .-e(n+l)c a-n-1 -c(n+1)Ana n-1 +nAn a n-1 (n+1)(cv -E)cn n -n-2 (evn+(n+l)e)=(n+1)c a An a (ev- €) ,An— (e n+E(n+1))a 211+1 • /,, Field inside at r=c 0=0 due to dielectric displacement is cn E .----2- ci A_P—rn_ci (n+1)(Ev - E)ncn-1 a2n+1 . Force on q is qED . o cvn+e(n+1) Dbr 4rev D 4rev o _ (E-eu)q 2 ; n(n+1) r c ian+1 Ans. F411Eve 2 d Evn-FE (n+1) ' a' -57Add to 5.16 (1) the induced charge potential Vi, so total potential is -1 -n a2n+lc-n then V = (Q/(4rrec)q(rn c +Anr-n )Pn (coscl)Pri (cos 0). If An=Egb at r=c, 0=ci due to Vi is -cos ccovi /Or +r-sina bvi /be. Ez= (Q/(411Ec))o-a 2n+ic-n((n+1)4Pn (p.)-(1-1.12 )Pn'(13.)1c -n-2Pn (cosa), pg•cose.
V=0 at r=a.
...(Q/(4rrec 2)) f(a/c) 2n+1 (n+1)Pn+i (cosa) P (coga ) From 5.154 (5) Ans. Closed form in ep (n+1) (a /c) 2n+IP (cosa)P (cos a) F = QEz= -(Q/ (41TEc2)) i n n+1 Problem 18.
Chapter V.
Page 229. (205)
Page 60.
-58-
co n V o =(27/ ev)4( sin a (r/a)nPn (cos0)-- r-"-lAn )Pn(u), Vi= (.1T/ev)pnr Pn(11). n n -n n-1 An= Bb .(1) . evbV0 /OR= eOVi /Or so At r=b, Vo = Vi• so b a sin a Pn (Pcp+b n+2 n+1 n-1 -n-2 n-ln = (2) .b sin a Pn(Ace )- E y(n+1)Anb (2). (1) .neb a , = neBnb n evb
r (cos CX)=4(1 +1) gv + nelA,, so that n(e-el )bm-fi a-n sina ci, rn n 11(E-e)13 211+1 Vo = Tivan ( (n+i)EvillE)anrn+1 since Pn (cos ix ) Pn (cos())
Ans.
-59By 5.16(1) charge element is dQ = a0 211c 2 s in Cx, and its potential is
cc
dV = 277c 2cros in ada (4rrec)"i c (c/r)ri +1Pn(cos a ) Pn (cose) . Then by 5.154 (5),
Sol aPn (cosa)d(cosa) =2(Pn+1(cos 00- Pn _ 1 (cosa))/ (2n+1) so total potential is cc 1 cn+lr-n -1 pn (cos 49) (cos011)) (2n+1)(P (cosa) - P V = c e -' ao (cr-1 cosa +E n even n+1 n-1
2n+1 -2n-1 r P (cose ) Ans. P2n-1 (cosa)) (4n+1)-4c = ce-icro(cr-lcosa+ t c 21P 2n+1 (cosa)2n -602n+1 2n+111 ;.,°' , 1‘ In 1.3. •(2n-1) [- ( 2n +14117TI -11 - 2n(t) 5.17 (1) gives V so a= E 5.17 = it P 2n (P) Or r=b 4na V --''' 2.4- . . . 2n E = 2 . -. QE,-pri 1-3..(.2(1-11 4n+1(Al 2n hi P (u) Ans. Closed form from generalized solution: e 4rE o f -l) 2'4••••2n b2 - 2n ' 4ad r(s2+b 2 'W Qk Ans. E(k) - F (k)] where k2 E= 4r2 E,Aab L4ad(1-k 2) a 2+b 2+2ad Where d is distance from field point to axis General Solution. Internal ring. b, c,d, z coplanar R = b2-1-o 2 -2bo oosa+a 2 -2ac sioacos0. Ring charge Q Charge on dO is q = Qd0/ (2r).
v=
Q(8r20-1: 02 1-2dO
.
External ring. b',e,d',z' are coplanar. -t-D 4 c'2-2c1 (1kosa R' 2. (bt7z ) 2±f'2. (bLe) 27 a l 2 • +4_ 2a'd'cos0 = 27 + a's in•a cos 0 ) V'= (81.720-11 R-Ido . di= c'sina, z'= c' cosCx.
d=c sin a z= ccosa - b
Let 0=11-2*, do= -20, cos 0= 2sin 241-1. 0
Need field along c„
V = Q (4112 erirr/ t a 2+b 2+c 2+2c(asina-b cosa)- 4acsinaSin 20-12 20. Q r.a2-1-b2.+c2.482 +b2+c2 A-2c(asina -b c osa ) -4ac s in as in24/3 -Oc-4n20 61a 2-fb 2+c2+2c (as ina -bcosa)-4ac since s in 2 *
Ec
7/2
= Q(072E )-2(yo (a 2.1.1) 2+c)p
-
3
204014-1 P -01J where P= r 2+a 2+2ad-4ad sin2*
Since r 2= b2+c 2 -2bc cos a , d = c sin a. Also true for r' , b' and d ' . For internal ring,
k2
4ad r 2+a 2+2ad
Qk
[(a2+b2 +0)k2
E c- 8r2 ea 4ad (1-k2)
E(k) - F(k)
Z
'
0 Image relation
is N/(a 2+b 2 ) (a ' 2+b' 2 ) = c 2 . Then from similar triangles, a' = ac 2 /(a2+b2 ), b' =bc 2 /(a 2+b 2 ) 1A7.4...a 12.2(r 24.a2)/( a 24112) . Substitution shows that the external k equals the internal k. (continued)
Page 230. (206)
Chapter V.
Page 61.
-60- (continued) c( , - , a wa 24-b 2 ) ; (91a a k (a2+b2+0)ka E(k)-CF=(Ek'C).3 Total E or cr/E on sphere surface is E= 4r2 ad c 4ad(1-k2 ) Q'=-cQ/(a 2-1-b2 ). For the external ring,
where a is ring radius, d is distance of field point P from axis, r is distance of P from ring center and c is the sphere radius. The modulus k is k2d
4ad 4 ad r2+a2+2ad
where p is distance from
P P to farthest loop point. From diagram at left p 2 = a+c] 2+r2 r2+a2+2ad. -61
For q only: r b, Vq = zirrqeb E(b/r)n+1Pn (11)- 11=cos0 From induced charge: Vi= q(49Eb)'1E (Anrn+Bnr -11-1 )Pn (p) (1) An cn+Bn c -11-1+bn+le -n -1.,-. ID n -n-1 at .r = c and Ana +Bn a +an b -n = 0 at r= a . Equate the An values to get 2n+1 2n+1 2n+1 r. n , 2n+1 2n+1)), An .. . (a 2n+l -b2n+1)/ ( bn (c2n+l -a 2n+1)) Bn = a (b -c )/L b ( c -a a' -n 2n+1 2n+1 2n+1 2n+1 -1 n 2n+1 -n-1 r b: 411EbV = q b (b -a )(a Ans. -c ) (r -c r )13n (cos 13) -621 ce 2n+1 -2n-1 For q and image in plane: V= q(217Eb)_ ; r (0) = 0. (cos 0), since P b P 2n+1 -Fol s 0. p. Inc ' 2n+l -2n -1 -2n2fTevbV 0= 4 (r b +r An)P2n+1(P)' 217ElibVi=qEBnr n+1p2n+1(p). a2n+1b-2n-1+ -2n-2 2n+1 At r=a: V0=Vi, a An=Bn a and EvbVc,/br = ebVi ibr. Let K=E/Ev, then -2n-3 2n -2n-1 -2n-1 -4n-3 2n (2n+l)a b An and - (2n+2)a Ari = K(2n+l)a B +a Bn = b rr -2n-1 -1 -4nn-l -2n-1 r i a -6V KBn= b-(2n+2)(2n+1)a /t2n+2-1-(2n+1)Ki. - — at 0=-. An . Thus An = (K-1) (2n+1)b ev= r60 2 By 5.154(4.1),5.157: bP 2n+1(cos0)/b0= -(2n+1)P2p (0)= -(-1)n (2n+1)::/(2n)..: at 0=17/2. Thus
_ 1 -
b
i_
Et
.1-) n+1 (K-1)(2n+1)(2n+2) (2n+1):: a 4n+ 3 i (2n+1)K + (2n+2) (2n+2): ! b2n+2r2n+3
a- 2r ,i(b2 +r 2 )3 -F o -
Ans.
-63b 4EV 4EV 217 m' I., P 2 (0) r c 2n+id c By 5.03 (2), (3): 2a= _ /--1 lki coib 2 c 2 P2Th (cos0) See 5.16 (1) (1=17/2. nvb2 -p2 • v =17 470; Tir+ Let c= bsin0, dc = case bd0=.1b 2 -c2 d0. y0u2 s ion+10 de . (2-4.6-2n)/(1.3.5- • •2n+1) Dw 858.44 -2n-1 2n Thus V= (2V 1 M)E (-1)n (2n+l)b2n+l r P (cos0) Ans. -645.16(1). Get -51 S 5pat Pat 0=2 by formulas below Fig.5.157. and write 2n for m-1. Thus n(2n+1): 2n+2 be - QQ1(417Ea2)-1(-1) F=Q't,Vp/(.960)=-Q/(477Ea2)-Isinin+lapm(cosa)OPm(12) aP2n+1(a°sa) Ans sin 0=17 =2a2 /(217Ea). 20, 2 c 0)=2a 2 0) d0=-2d0, (1-cos2 2 (1-sin Closed Solution: cr 1=C 1 a2 (cot2 a+ 2-2sin2 0) = 2aa k-2 (1-k sin 2 0). 477EV=2aifori:d295 z 2 =a 2 cot 2a, z2 +c2 = + 2 617 _Q i ger zdo Q1(22 k3/ 2 acota7/2 Q i Q 2k 3/ 2 cota E(k) de 16774a 2 2r 2 cJo (z2+c 2 )3/ 2-2r 2 E 8a 4 o(1-k 2 sin 20)2 / 2 1-k 2
Use V
The integral is from Problem 18. k 2=2/(2+cot 2cx)= 2sin2 a/(1+sin 2a)
Ans.
Page 230. (206)
Chapter V.
Page 62.
-652n+11 +1 a: +An rnj P (p) V = " n 1 4rEvc r 717r1Pn(P). p=cose. V° -417e c o kr/ n+l -n-1 n -.7111- ) . + Anb = Bn (bn-a2n% V1 =0 when r=a. At r=b, Vo=V i so c b ,, ,n +1 bV, n by0 a E = E--' at r=b, so Ev F(n+1) (t3;) + nAnbri = 'Ern [nb +(n+1)-d vbr Or c n+1+nAnb2n+11622n+1 _ a2n+13 : ecnb2n+1 2n+1 ,. n+1 .. • 2n+11J—_evt - (n+i)C +An D )1. c Equate B values +(n+1)a n+1 2n+l 2n/.1 n+ltr 2n+1 a2n+13 EC nb (b + E v (n+l)c -a ) +(n+1) Solve for A - 1 • Then 2n+lcnb2n+1 n Evnb2n+1 2n+1 2n+1 +(n+1)a2n+1 (b -a ) - cb .. n 2
nA cn-1 Ans. F= 0sur ?;)— [4reqv i A nP c o n r n(0 r=c-417sv 3 n -66ce = Vo=q(4rrevcrl Ea(rnc-n+Anr -n-1)Pn (p), Vi=c1(411EvC)-1 EB (rn-a 2n+lr -n-1 )Pn (u) • _ ap os_101:1) 2: +cib on Vi is zero at r=a . At r=b, V0 =Vi and evbV0 /Or= tbVi/br so bnc -n+Anb-n-I=Bn (Y.' -n-1 n -n 2n+l -n -1 =e (nbn+(n+1) a ex,)B (nb c -(n+1)An b b n . Equate A values to obtain i . 2n+lc _n _ enb2n+1 t . ev( Bn (b2n+1?-a 2n+1 ) ..b2n+lc -n3 . evn (n+1)- b +(n+1) a 2n+1 1 (n+1)-1B 1 2n+nl+(n+1) (E-ev)a 2n+1, -I Solve for Bn=(2n+1)Evb2n+1 c-n ([(n+1)E+nEJb v 3 Put into Vifor Ans. -67Cv=E(r-a 3/-2)cos0. o=e— at r=a,a =3tEcos0=3tE,./a 3 -co 2 /a . dF=2E-1/42dS 2 so that br F = 2 EE 2 a- 21a (a 2-p 2)2TTpdp = 477eE 2 a 3 Ans. -68CTake origin at center of a. c is small. By 5.153(1) 11-' = 1-1(1 + 1)13.1 +. • ) so R P.,' b ( 1 iP1+ • • ). V= T^ ' 2 1'1(1.0+ C + DrPi(11), Q= - Ci g-dS = 4Trr 2 tA/r 2 . i +1 so A = ---247E . At r=a, V= 41,rt so C=0, B= -Da 3 . At r = R, V = Vo so that _1r g + 3 Q ca 3 Dso_bL-„1 a p). Thus Dv0_4rr 4Tfeb2 411E(b3 -a3 )' D- 4r (b3 -a 3 ) bV 2ca 3 3c cos a = [e J + --r-C Jcose) = —q-C+ b 3 -a03 3 Ans. br r=a= 417--(1 a2 + C a3 (b3 -a3 )-a 4r a° 3 (
+Q cos0 sign depends on choice of 0 = 0 line.
-69c-n-2 Let V= ErPi + Br-2P1 +Cr-nPii _ i+ Dr Pn+.1 . When r= a (1+8Pn), V=0 so, if C,D are order of 8, n-n+1 2p. By 5.154 (1), 0 = Ea (1+8Pn )Pi+ B(a-2-2a-28Pn)Pi+ Ca n Pn.1+ Da +-11Pn+1. P1Pn-2:+1Pn-1+2:11-4n-2D . 0 (2). (Ea 8i12.1:_jaB8)n +Can= 0 (3) Thus Ea + Ba-2= 0 (1) . (Ea 8- 2a-2 B6)(n+1) (2n+1)-4aFrom (1) B=-Ea3 , From (2) D=-3(n+1) (2n+1)-1 an+3E8, From (3) c= -3n (2n+1) _ian+1E 8. 81ra an+lr-np + (n+l)an+3r -n-2P so V =E{ (r - a3 r-2)P1 - 3 (2n+1)" n-1 n+1J3 Surface charge density is ce[71.7] =€,,/(bV/br) 3 +(bVirb0)2 , neglect (OV/rO0) 2 . i[r a= EEC (1+2a3 r -3 )1711 +26(2n+1)- man+lr -n - 1pn _1+(n+1) (n+2)an+3r -n-3Pn+0). Write r=(1+8Pn)a. +(n+1) (n+2)P 3). Substitute for Pi kl by 5.154 (1) . n+1 rI-1 33. Ans. = EEC3P1 + 3n8 (2n+1)-1[(n-2)Pn-1+(n+1)P n+1
= tE{ (1+2-2. 3 813n )Pi+ 3 8(2n+1)-1[n2 P
▪ Chapter V.
Page 231. (207)
Page 63.
-70CV1. A = r-1+Br-2n - 1Pn+ CrnPn+D. At r=a(1-1-8Pn ) Vi =V. Vo=r-3E + r -r1-1FPn . At r= b(l+nPn ) li. = Vo and EbVi/br = EvbV0 /6r . BI C and F are small of the order of ri, 6. (Ba-n-1-fr_a un)Pn+D., Aa-1+ D= V (1) -Aa-I 8 + Ba n-1+ Can= 0 (2) V= Ae1(1-6Pn)+ (A-E)b-1(1-TiPn)+((B- F)b n-l+Cbn)Pn+D = 0, A-E+bDJ (3) . -T1(A-E) + (B-F)b n+ Cbri+1= 0. (4) (EvE-EA)(1-271Pn )b-2+ (EvF -EB) (n+l)b-n-2Pn+necbn-lPn = O. (EvE - EA)b-2 = 0 (5) n-2 (n+l)b (evF-EB)+ necb11-1= 0 (6). From (5), A= ev EC, from (1) D= V-EvErla-ly from (3) E= EvEe-i+Vb - EvEbria-1or Vb= ERE -Ev )a + Evb} (ea)-1so E= eabV/C (E-ev) a + evb) • A = evabV/t(E-Ev)a+ evb). From (6), C= -(n+1) (evF-EB)/ (neb2n+1) . Substitute in (2) and : n+l 211+1 b solve for B na Bnb 2n+1+(n+1) a 211+1 (EZV 1(11;El e l bl :71F a: EvIT: n+1 evbV,8 EuF ev -nb (n+1) a (n+1) E•ty bV8 CEan+1 nb 2n+1+(n+l)a 2"1 [(E-Ev )a+Evb + ngb 2n+1 (n+1) an 1 ' [.] [(E-Ev)a+evb
1 -
Substitute for (A-E), B and C in (4) . n+l n+1 n n+1 n+1 n+1 n+11 b 1 eF na (n+l)a b +(n+1) a b ievbV6 F 0. -nabV(Ev -E) + (na Zn-r-1 }RE-9 + 2n+1 a ( C- Ey) ai-evb Crib2n+T+(n+1) "1717 1 E a +(n+l)a 2n+1 nb a+e b) 17.1 nb p[neb 2n+1+ [nb2n +1+ (n+i)a2n +I-1 2n+1 2n +1 evb b all C(E- 0 Ti (n+1) e a 2n+1-(n +1) Eva J+(2n+1)Ev anbn18) 0= Ebn cvb (E-E v )a (c_ev)r1tnb2n+1 anbn+1.5] EabV +(n+1) a 2n+1)+(2n+1) Fso and ev) a+Evb tne+(n+1) ev jb 2n+1 (E-Ev ) (n+1) a 2;1+1 n n+1 2n+1 +( e-nb vo _(E EabV 1 (2n+1)6Eva b + (n+1) a 21.1+1 }bn Ans. •—ITT r -cv ) a+Ev b [r In e+(n+1) Ev 3b 2n+ 1 + (E-Ev) (n+1) a an+i - 71C-
CO Cl. q so at r = co . Let V= Take conductor r = A(1+1)Sn) at potential V0 finer 4- rP+1 SP co V = q(4Tter) -1.Thus Vo =qt (1-1)Sn) (4rea)-1 + ii Anr- P -1 Sp ). Equate coefficients, 4rrEaVo = q, ")
0n , so that Vp= q(4rrer)-1[1+11anr
nSn (0,0) . Apply Green' s iirea•Vr as V Reciprocation Theorem, 2.12(1). Induced charge is q' =-Tri:l qi'3=I- - r t 1+Tlenn (0,0)) Ans . q -72Cn-1 co MT T r c 2rraT `13 r rn c2n+1 ., n—n--+ Et = ft At r=c, a= (la). (0) _ EL--+-IP 4rre Let cc = -1-17 in problem 57. V = or 2 0 a a o an anrn 1 n , 03 m (2m-1):'. 2m -2m c2n+ a (4m+l)c a (-1) ,, P - -e1' E "-) P n(0)P n(p) . Pn (0)=0 if n is odd, so a = (n+1)Th-r+-(2m).. 2m(11) • o ac
An=ann/ (4170, Ap=0 if p
„, a
3 r 14
1
(see problem 60) cr= -Lt1 - 1L.I3J 132+ 9•i•TLiJ
C 1e 73-[-V 13 6-1-• • •) P 4-13• i• T3• 5
Ans.
- 73C2m -4m-1 A2m+ a sina P 2m(0) • In problem 58 set a = yrr, Pn (0)=0 if n is odd. Let n=2m, B2m=b (- Om 1. 3. 5. - • (2m-1) (4m+1) ev - 2m (E-Ev ) .,_ ;I P2m (0) T =-g— la a2m ' 2.4.6- (2m) + 2me. 2rTa' -2m-(2m+1)Ev + 2m e ' J a 2m - (2m+1)Ev 1
q
F., + E,(-1)111(4111-F1) Ev •
V•= 14rrea t.1.-
1 (2m+1)Ev+2mE
1.3.5- • • (2m-1) . srlam_. r 2m (cos 0)] 2. 4 . 6 • • • (2m) La'
Ans.
Chapter V.
Page 232. (208)
Page 64.
-74C-
v. Vi=EB(r-air-2 )cos B. At r=2a, V0=Vi, 8a3+A=B(8a3-a3)=7Ba3, EvbV0=4_2. • br ' ev (8a3 - 2A) = eB(8a 3- 2a 3 ) = 10BEa 3 . B= 12ev /(7ev+5 E). At r=a CI= 3CEB=P017:COSe, br dS=2TTa 2 sin0d0. Q= 2E2IviV 7/2ka sin Ocosed0= 36ra 2 cevE/(7cv+5E) Ans. 7 ev+5 c 0
V 0 = E(r+Ar -2)cose,
-75CVi=A+Br -1 + (Cr-a+Dr)cose, V0=Er-1+Frkose. C, D, F are very small. bV0 By Gauss Theorem: € j bVi. --tI 1 =ev is2s- dS2 so eB = ev E . At r=a, Vi=V0 Si br S so A +13a-1 =V (1), aD +a-2C = 0 (2). Thus Vi=V+B (r-i-a-1)-1-C (r-2-ra-3)P 1 , V0= eev -1Br-1+ Fr-2P 1 . At r= (b+yPi ) : Vi =V0 . V +B(13-1 -a-1)=eev -1Bb-1 so abevV _r b3 -, B= a(e-ev) +bev (3). -13-2Y B+C (b-a - ba-3)= - Ec'elyB + b-2F or y(e-ev )B = -evc..1--3 j+Fev (4) a OVi= E ---OVA • -ec [2- ±1 }=-2e F or b3 2a 3y(e-ev)B from e---v, Fe = cC{1+ — } (5) • C= v or ' or b3 a 3 13 4 v 2a 3 e(2a 3+b 3)-2ev (a 3-b3 ) (4), (5) so C=
2a 4 bYev(e-e,)V bV B 3C a + --s a cos0. Thus fa(e-ev)+bev)(2a3(e-ev)+b3(e+2ev)1' a = - e&-lr =a so cr=--j -
beveV 6 a 2y(e-c„) cos° 3] at a(e-c.v ) + bey] .+ 2(e-ev)a3+(c+2Ev)b
Ans.
-76 CSince evls i nTi/ br dS1 =e1cOVd/Or dS2 = evls3bV0 /br dS3 ; 4revVi/q = r-1+ A +BrPi ,
4reVd /q =r -1+Cr-2P 1 + DrP i+E, 4ttevV0 /q = r-I +Fr-2 P1 . B,C,D,F are of order ofc . r = a -cPi , r' = el+ c a -2 Pi , Vi=Vd, ev011i / Or = EbVd/br. cca-2+Bac=evc a -2 + Da ev , olio ELAd 1 1 cP1 7 -c C(Ev+2E) + (1). r= b+cPi , 7. = 1-) + Tr, 170=17c1' cv& = br B = -2Ca-3+D, so Da 3 a3 (e-ev ) E c E C 1 Ev+2 E EC EF -2C so IF+ It +evbD= Ti+b -i (2), Tr +D=-2b3 F (3). (1) in (2) evC E-s+-7 --7] lDi a (E-Ev) - (E-E
by
L. + Ty EF
_C_. 3
a
r
(13 3 (tv+2E) -2a3(eE +2e 21 c -2F L a3(c _ ev) b3 (4). (1) in (3) C I—v— —- a-3 - b3 (5) or C a3 b3 (E-eV)
chi
=ca-3- 213-3F (6). b 3 ( (4) - (5)• ev ) gives 3ev C=(e-ev )c + (e+2ev)F. Substitute for C in (6) C _ 2F c[133 (ev+2E)-2a 3 (e - Eve + (e+2ev ) [133 (e+2e) - 2a3 (e-ev))F S o lve for F 3eva 3b 3 a3 b 3' 3 eva3 b3 (e-Ev) 2c (b 3-a 3 ) (e-ev)2 cose 2c (b 3 -a3 ) (e..e,)2 1 Ans. FV° ci [1 - 4rrev r + (2a3 (e-ev) 2 -b3 (c+2cv) (cv+2c)1r2 2a 3 (E-Ev ) 2 -ba (e+2ev) (ev+2e) ' -
-77an tis,n q co 4TrevV 0= e-E o t— en +1-+--1-3Pn (cos0), 4TTeV.3.=(cos0). At r=a, V0 =Vi so e-yi + n+1 -6vBnan • c EBnrnPn 0 c a n 2n+1 .23 eLi V0= ELS m a it1+115,n -n(c-ev)a 61.71 1 -(q/c)2 s., n(e- Eli)(n+1) ral2n+1 B A = - nan Ans vor Or ' el anal n' gln f n e+ (n+1 ) cv1 • F- (1.67 = 4trev V ne+ (n+l)ev InJ w rrn____Ja___ A
-78s
A (r -a
2s+l-s-1
r
1
)P s (cose) satisfies LaPlace' s Equation, and gives a
potential positive if 0 < 9 <6'.. From Fig. 5.157a, fractional value of s is needed if -IFTT< a
rr 2
Chapter V.
Page 232. (208)
Page 65.
-79s 2s+1 -s-1 r V = (r -a r )1Ms (cos13)Ps (cos0) - Ps (cos )q (cose) ) gives potential zero at
e=
and satisfies LaPlacet s Equation. if s is chosen so that Qs(cosp)Ps(cos Cc)
-Ps(cos13 )Qs (cos a ) is zero. V must remain positive between cones so smallest s-value is needed. -80Uniform field is V= Ex= Ec iCC = -jciPi(jC)Pi (g). Vo = Ex + jEcIAQi (jC)Pi (g), Vi= jEc1l3P1(jC)Pi (D. At C= Co, - Co+AiQi(iCo) = -BCD, ev( -1+AjbMiCoV000=BE or -Co+Aj(Coeot-1C0-13=-BCo ) e vE -1+Ajtcof'Co--Ct ii 33= -Be so that (e-Ev)Co+Ai (e -e,)C0 cot -1Co+ (E+ECO-evCg)( 1-402 /9= 0. Solve for A. (6.4,)S (1+Coi?
0 A= (E-E,) Co [ (1+Ce)cot' Co-
•
V1
e
B- e
(c -ev)COC (1-1-i02 ) COCIC0) (e-ev)C0((l+Co)cot'C o - C o -E
(e- ev)Co{(1-1-Go)cot'Co Co }-€
Ans.
-81Uniform field: V= Epcos0 = Ea4/7-1 i 7 /F -4-C 2cos95=
(DP 1 (iC)cos0 Ea. V0 =EaP1(0f-jP1(jC)+AQ1(jc)icosgs, Vi=-jBEaP1(§)P1(jC)cosgs. P 1
1+
.
V EgV ci. evbVo ,
57 1-0+AQI(go)= Bs./14C 2 (1) Qi(iC) =A1+T 2 (cot -1C-C(1+0)-1). At C=Coy Vi'Vo • •■ ....a3S211. ' _Silo_ [1+ AQICiColi a -2 2 r40EvitA [1.7-11V (iiig) 2] (2) D 1+0, -EvANA+c,f (1+4i)2 ,vrTO 1fra 1v (e -ev)C0.11-1-CO so (e_Eosco (i+cg). -2EvA. B=1--21:7-
(]Co), B=2ev (2e+(e svg0E(1 +CDcot"Co -0031-1An: -
-82Uniform field : V= Ec a rl;= EcaP i(rl)Pi (g)• V0=Ec 2P1(g)(P1(r))+AQ I (11)), Vi=Ec 2BPI(g)131(71) rlo+A(Tiocoth-'110- 1) = BT10, elf ( 1+A(coth-Lrlo + n.0(1-7120)-1 )}=EB• (E-Ev)no+A((e-ev)nocoth-1no - E-evT11 (1 -r1g)-'=0 • A = - (E - Ev)Tlo( 1-Tlo)/{(e - ev)TloL( 1-T1o)coth-1 T10+71o]-E} Ans. 2e, [1. (e-ev)-99((l-Tig)coth-'nn+ap) , V. =EBx ▪ e +71 0 -E] CE - COTIO t(1'TM COth-41 o +no 3-2Ev (e-ev no t(1-Tit)coth-17 0 + 710 )
-83Un iform field : V=Epcos0=Ec2 1- §2012-11cos0 = Ec 2pi(7-1) pi(g) coso, PI( g)=.11.7-V, p1(71).
Trci.(c 0 th-in +71(1-1-1 sr Vo= Ec2P1 (t) (P1(7) +AQI(T1)3cos0, Vi= Ec aBPI (g) c 0 s93, At 710,VA Q1(r PiCrio)+AQI(71 0)=BPIt11o), /site - 1+A' ,FA-71iot h-ino+7-100. gr =BV-no -1. evbV0/bri = EOVi /Or] so 2 e,A 2r EBTIn evri°11+A—SLI rl +E 0/1271. [ 2 + A E B-Ik-- -unA(x+B x)I v 1-71 (1-n 2] 011-1(ng-l) W-ni-1 NFIT=1 2EvA 2E, Bi= BEpcoso = BEx. Ans. (E.-co-no (71-1) 2cv- (E-Ev )7 )
0( (ng-i)coth-Ln o -.no) • v
Page 66.
Page 233.(209)
Chapter V. - 84-
From 3.121 energy of dielectric body in electric field is W =31v (Ev-E)X.rdv where Ei is field in volume after, and E that before, body was in place. Volume is 4rnt2n and fields are uniform. Thus W= fral 2n(Ev e)(EcosofE i cosa + EsinCLE a sinla) .' T= -bW/ba=*rman(Ev -E)E(2Eisinacosa- 2E2 sinacosa)= frm 2n(ey-c)E(E 1 -E 2) sin
2a
Ans.
cf(01+1 )=r0=n 2(C8-14)/C8, CO =n 2 /(r°2-n 2) Prolate spheroid,problems 82,83: cing=n 21 c2(718-1)=m2=n 2(18-1)/TT 11,1=n 2 /(n 2-m2)
Oblate spheroid, problems 80,81: cfCg=n 2 ,
co
- 85-
From 5.275 (1), (3): 4reVq= q/r = jqa-l E(2n+1)Qn (jC0)Pn (jC)Pn (t). Potential of induced
co
charge vanishes at
C=m.
4neVi= jqa-l iBnQn (jC)Pn(C). If C=0,
veviso
-(2n+1)Qn(iCo)Pn(j-0)
=BnQn (j•0), but Pn (j.0)=0 if n is odd. From. 5.213 (5) Bn=(2n+1)Qn (jC)2Pn (j•0)/(jval(j.0))
'OnTi2 ( 2n+1 )Qn (jC0). Thus Vind= (q/(2r 2 e0) (4n+1 )Q2n0C0 Q2n(JC)Pan (t)
Ans.
-86 qV 3+Q:0 = qVp+Q I •0. Ring potential on axis at CoC 0 is , if
C > C o,
V =-12--E(2n+1)Qn (jC)PnOCOPn(g0). If C Co V=(jq/(4rea))E( 2n+1 )Qn(iCo)Pn(to)Pn(jC)Pn(C)
)
(g)
n
C< Co
- 87Ring potential from 86 is Vine(jq/(4rEa))pn(4n+1)Q 2n (jC0)P 2n (g0)Q 2n (jC)P 2n (D Vr ing=-Vind at 6=0. Only even Vring terms.remain so that B anctan(i• °) "P 2n (i'0) E(4 +1 )(22n(iCo)P2n(§o)Q2n(jC)P2n (g) Ans. B 211=-2P2n(i•0)/[-iTTP2n(i•0)). V ind=7,-4-zr- ea 0 -88Only odd n values remain since V and g change signs together. On spheroid V is function
co 1 C o V= ..AnPn (g). Multiply by Pn (Ddg and integrate -1 to +1. 2VO4Pn (§)d§ An _211 -1-11 .n-1 21. : ‘31. = Ani:(Pn (g)) 2ds so by 5.154 (5), T 21..ifpPn ( :1:21 ) Vo,n odd. 1 -Pn-1]0 --a 2:+1. . . tim Q,n4., (jC) n 4n+3 1.3.5...(2n+1) r Thus V = Vo f,(-1) Ans. of g only. At
:
u/ Q2n+I(..iC0) 2n+1 2.4-6 ••• (2n+2) '2n+I ■ - 89-
.15 s.2i74 1;'1 °II/1i- -;TVIIs I -laii-%;),,,
IT, (1)
( -1 rn+.(nim:) 1
V° -
CI-MT, 01D)cg 01) smn
Of (n-m) ! 2_ s(n+m) . 01. 1)
(
6,
)
igl,
yi - (":.(+.: 1;M)' (rg- 1)4(710)11,.(n)siT (4) on (5) "I 'Io Sr _ zPrshi( g) cos s(0-90, and multiply thru by and cr=rE TI2 -E
where 4=CmnP1,;(g)cosm,(070m). (3) Let Amn=
(2)
PW(g)cosm(0-0m)h i h 2dgdo, 6 . = (2n+1)(n-m): r r r j! j 2aP .m( 0 cos m (0- 0m)h i h 3d drzi and integrate over 'n 0 . mn 2rc 2(n8-1) (n+m)! 1 D n Vi=itMmn P1.11 (-
0m)) (9) Vo=itNmnQT(71)4ni(g)cosm(0-0m) (10) -0m i r,v, (-1)m( 2-5 m) (2n+1) Fli.,2„9 2 (g)cosm(0-0m)h ih 2dgd0 (11) M (n )1: j 0oPm , Qm n n 0 4rec 2 L(n+m). mn
(8)
Ninh- Q P1Ini:(0110 1°)) Nmn (12)
• Chapter V.
Page 234. (210)
Page 67.
-90Take q at fa = 0 0 , t = g o . In last Problem: PT(C0),Efahih 2dtd0 = py::(g o)rsadS = q1 (C0) .6j20 [ (-1)m(2-8%) (2n+l)qrki, 'mn n (gc), Nmn4TTEC 2 (n4m) Qrn n (71°) Pui L(11410 PIR (rid PR( 4rec 2 CO CC
03 CO
Vi=E Ellinn P(TOP(DCOS M(95-00), ri<710 • n n
V0=E o EN o mnQui n (T1)Pnil(g)cosm(0-0o), Tl >"no • 00 - 91. [..11 (11r a)I. CD ... .N Axial charge from 5.298: Trea 2Vp=fsinhEpr (L-c)3 sinhprzJo(prP)(pr sinhprLrl P is point on ring. qq+0•Vi)+Q•0 = 0•Vc+q•Vp+Q' •0. Replace p and z in above formula by b and c to get potential at z=c on axis due to ring at p=b,z=c. Potentials reducing to this on axis are: If z
sinh[Pr (L-c)Jsinhpr z Jo (prp)Jo (Pri}) If Z >c, write (L-z) for z and c for (L-c). Ans. prcji oran •. TTea2sinhprL
Second solution: Set q= q'd0/(270 in 5.299 (1). Integrate from 0= 00 to 0 = 2rr. All terms but m=0 vanish, giving above result. Force downwards is the limit as z approaches c of 21qE (oV/bz)
+(bV/bz) J. Numerator of first factor is prsinh[pr (L-c)Jcoshprc minus z>c z‘ c Or sinhprc cosh[pr (L-c)J which gives prsinh pr (L-c-c) so the force is F=Eq 2 /(2rrEa 2 )4 sinhpr (L-2c)cschprL [Jo (prb)/..T 1 (pr a)J 2 Ans. -92Solution finite on axis and giving V0 on cylinder when Jo (pr a)=0 is V =Vo +E(Akcosh pkz+ Hi'c sinhpk z)Jo (pkP). To make potential V0 at z=c take V = Vo + EAk(sinhpkc cosh pkz -cosh pkc sinhpkz)J o (pkP)=Vo + EAksinh[pk(c-z)]..To (pkp). When z=0, V=0 so -Vo= TAksinh pkc Jo (pkp). Multiply thru by pJ0 (pkp) as in 5.296
4-- 2a—* and integrate. 5.296 (4) gives As= -2V0 (a 2 sinhpsc[J i (psa)J 2 3-11:pJo (pkp)dp. From 5.302 E the integral is ps-2fo'Is a xJ 0 (x)dx= p-s'aJi (ps a) so V= Vo [1 2 sinhIpk(c -z)).10(PkP)] Ans. uksinhpkc J i(pkp) A potential giving proper value at z= ±c and z= 0, and finite on axis is V= Vo C(z/c)+ EAn Io (nrp/c)sin(nrrz/c)J . Multiply by sin(nrrz/c) and integrate from -c to +c. pc
c
c
V oJ o sin (nrrz/c)dz = VoJ o (z /c) sin (nrrz /c)dz +An Io (nrra/c) j 0 sin a (nrz/c)dz An=2Vo/CnTTIo(nra/c)) . Thus V= Vo [(z/c)+ (2/1)E 10 (nrp/c)sin(nrrz/c)/(nIo (nTra/c))3
Ans.
Extension: 172 = EBkSinhpkZ,To(Pkp) (1) 14=170-1- EAksinhilk(C -040.1k0 =E4+Aksinillik(C - 4J0 (10) (2) where Ck=2Vo(pkaJi(pka)) by HTF, 7.10.4 (54). At z=b, VI=V2 and bvi /bz =6112/6z so Ck+Aksinhpk (c-b)=Bksinhpkb and -Akcoshpk (c-b)=Bkcoshpkb, (3) and (4). -2Vocoshpteb 2Vocoshpk(c-b) Eliminate Bk : AkBk-iikaJi(pka)sinhpkc iikaJi(pka)sinhpkc' VI=V0
coshukb sinhpk(c -z).4 ( 10)) t1-2E pkaJi(pka)s inhpkc
V 2=21.70 Z
coshpic(c-b) PkaJi (pka)sinhpkc sinhpkz Jo (pkp)
Ans.
t t
Ans.
2a --*
Chapter V.
Page 234. (210)
Page 68.
-93-
Solution giving V= 0 on walls and the z = 0 plane is V =TAksinhpkz.ro (PkP) where Jo (pka) = 0. When z=c and b> p> 0 we have V=Vo. Multiply thru by pJ o (pkp)dp and integrate as in 5.296. rib Ak= 2V0( a 2 LJI (pka)] 2 s inhpkc ',I 0p Jo(pkp) dp-
2bV„TiOeb) ra s inhpkcLJ I (pka Pka 2 ,
Ans.
Boundary conditions are also met by V2= Vo[(z/c)+3FBn I o (nrrp/c)sin(nrrz/c) 4-2134
if p b. At p=b,V 1=V 2 . Multiply V I (b) = V 2 (b) by sin(nrrz/b) and integrate z=0 to z=c. 4An[Ko(nria/c)I0(nrrb/c)-I0(nra/c)K0(nrrb/c)]= - (-1)n (c/(nrr)}+ iBncIo(nrrb/c)
(1)
At p=b, OV I /Op=bV 2 /6p. Multiply both sides by sin (nriz/c) and integrate z=0 to z=c. An ko (nrra/c)I;(nrrb/c) - Io(nrra/c)K8 (nrrb/c) i= Bn I8(nrrb/c)
(2)
SUbttact (2) • I o (nrrb/c) from (1) • I“nrrb/c) and use 5.32 (7) to obtain
n 2I,;n(n rrnb/c) ApI o (nra/c)No(nrrb/c)I;, (nrrb/d) - Ko (nrrb/c)I0 (ntrb/c) 1 = Ap(I o(nrra/c))/ (nrb/c) =(_1) So VI = 2(b/c)V0 E(-1)n { Ko (nrra/c) I o (nrrp/c) - Io (nrra/c)K0 (brp/c)ll I 1 (nTTb/C)/I0 (nrra/C)) Sin(1117z/C) V 2= (VoiC +2bE(-1)9 Ko (nTl'a/c)Ii(nrra/c)+ Io(nrra/c)Ki (nrrb/c) (To (nrrp/c) /Io(nra/c)isin (nrrz/c)] An: 94From 5.298(4) for lower charge : q2 Force is qE- — 2re a
e
2rea2 le -2prb { ji(pra ) _ 2. Ans. }
PrJikPrai
E
e "Pr z q oV Oz-2rrea 2 7(J 2 (p
q•
-95-
(e-ev) Use above potentials and plane face image laws. Dielectric image at -b is (e+ev) lz-b I Ans. VI = q(2rreva 2 )-1E[e (E -Ev)/ (E+Ev)3 e-Pr( z+b) ]cibi(Pra)}-2 To(PrP) i e-pr(b+ (pr o)1 -2 J0 (PrP) Ans. V= q(2rrev a 2 )-4E2tv (e+Ev )-962 ( Pr a) By 5.102 invert charge q=4Tre at 0 on cylinder axis. V= — Ee p aa J141r 2e1,0 e -Prz bV -E-513 at p=a gives a- a2Tt ji (pop 5.102 (1) gives a' = OP 3 /a 3=a/cos 3(x.
z = tancx, dS = 2r(acoske) (a/2)dB, but e= 2a so dS= 2rra 2cos 2a da. Q.2.1 2a i d_S a Q= 8rEE[Ji(i}s)}-1,17/2e-pra tan secada. Let tana =sinh0, sec aada= cosh0 do u-,. a sinh 95 d0j/Ji (pr a), where Jo (pr a) = 0. c o°e-• seta da = cosaill+tan 2ad0 = d0 . C---q v=aQ= area ELr so C = 8rrea E S o, o (Pra)/JKura) . Here S0,0(pr a) is a Lommel Fctn. HTF II (50) p 84. -97a2 p e sin 0, Dipole M= 4TTEa 3E inverts into uniform field E. Take M in cylinder p=a. rr'=a
z = rcosel= (aa/r 1 )cos13. From 1.07 and 5.297 (4), Vd= 2E4 ellkz Jo(likP)/{J i(11ka)}2 . From 5.10(2), V'=
f,V= -(.2E8 2 /r 1 )Ee-
pka 2 cos Oir
r Jo t (pka a sine)/r 1 }01 (pka)}-2
And.
Chapter V.
Page 235. (211)
Page 69.
-98From 5.298 (4), potential of +q at 0=0, p= b and -q at
0=
r, p=b is, since cos sO
(2-44)e-Pkizink- 1 Ds+1 021( 03-2 Js (pkb)J s (110)cos s0 • - cos(17-s0) =2 cos sO, V= --T 2, rgea i 1.111 where s is odd. As b --->0, Js (pkb)-",-(pkb) s (2ss!)-1 by 5.293 (3). Neglect s>1 and let 2qb=M=4rea 2E as in 97, then V= 2aE ie -Pk IzIJI(PkP)L-T2 (11ka)r acos 0. Invert as in 97, 2Ea 2 l e -pka2cos 0 /r. J 1 (pka 2r -lsin 0) Ans. cos 0 r 1 (J2(Pka)) 3 -999 By 5.35 (3) axis potential is: Va=4TrER 21.1 2 SccskzKo (ka)dk. A solution of V' -
LaPlace's equation with this axis value is.-* .1. 12efo Ko (ka)I0(kp) cos kz dk, (ka)K0(kp) cos kz dk if p> a. Ans. 2T if p
p = b, Ko(ka)10(kb)+11(k)Ko(kb)=1'(k)I0(kb) and Ev(K o (ka)I'o (kb)
Y(k)KO(kb)1= ecID(k)Vo (kb) so EvKo(ka)(I01C0-1CDIO)=4(k)(evI0100- eiliKo) =4:.(k)(ev (I 01q-IpC 0)+(ev-E)I 10 K 0 )-ev (kb)-1 K0(Ica)=4(k)C-Ev (kb) l+(cv-e)I 0I K0) Thus 4' (k) -
evi(0 (ka) Ev+(e-Ev)kbI;(kb)K o (kb)
and vi_
Ko(ka),o(kocoskz
2rt 0 Ev+(e-e)kbI(kb)Ko(kb)
•
dk
Ans.
-101Start with charged ring between two plates to which field is tangent. Then if p>13, nrrb nrrp, nrz . . nrrb nrrpn'rTz VI=EAmn1(0(— )Io( c lcos c , p
-Ez 2 = A1 V1 2 + A2 V22 4
...
AnVn2
-Ezs = AlV is + A2 V2s
"'
ArVrs + ....+ AnVns
-Ezn = A I V in + A2 V2n
"'
ArVr n +
P.> b and p4:13 points, p and b must be interchanged in Vr .
+ AnVnn
-Ez
2c
4.
Chapter V.
Page 236. (212)
Page 70.
-102From 5.03(3) c°J (kp) s in (kb )dk = -E and 5.302(9) c1=4TrbVb 2 - 0 2 4rrb 0 °
V --q—r 4rreb 0 e
kz
o ) sinks k Ans.
T
-103Removing the ± +, from 5.272 (8) gives the density on a charged plate with a hole and multiplication by two gives ±E at z= to so that a=(2€E/r) {cos-1(a/P)+ abr2 -a 2 }. From Int. T. I,p99 (4) and 5.302 (6): cr=2EE'rr1 f -17 2 -1:k-1Jo(kOsinkadk+aJo(kp)coskadk)=-4 . z e-klziJo V= Integrate: cos ka k-asin ka) dk. Arts. -'a (kP)fk -104V =0 at p=0 and z=0,c. At p= a, V= zVo /c so put x=rz/c in Dw 416.07. Then at p = a l V= 2n-1 1.10 (-1)nri'sin(nrrz/c). From 5.322 (2) and 5.292 a solution of La Place': equation quation meeting all boundary conditions is lr+i sinnrrz Ko(nro/c) Ans. V c Ko (nrra/c) 05 n 170 Vi AmntIm(nEb)Km‘c Km(ncb)im -1 )3sin4 cos (mr6 +a ) 1 i M' m V1 is zero on all boundaries but p=a. Find Amn to give f i (z,91) there. (
V= i c
c a ‘i,nTT c p ) ) sizt4z_ i nrra lK tr4E) _Km(nr E Bmn( 1in` / mk c ' m`
2 0 0
cos (mizt + ft ) m to give f 2 (z,0) there.
V 2 is zero on all boundaries but p=b. Find B mn co co V 3= E E C (Jm(kn p)Y (k a)-Y (k o)J (k a) lsinh k (c-z) cos(03+ y )
oo mn n m m n m n m n where k is chosen so that J (k b)Yin(k a)-Y (k b)J (k a) = 0. V 3 = 0 on all boundaries n m n m n m n n to give f 3 (p,0) there. but z=0. Find C mn m = n p)Ym (kn a)-Ym(knp)Jm (kn a))sinh knc cos (m0 +6rn). Here k is chosen as for V4 =0 oDmn {Jm(k n V 3 . V 4 is zero on all boundaries but z=c. Find Dmnto give f 4 (p,95 ) on z=c face.
Then V= V 1+V 2+V 3+V 4 meets all boundary conditions. -106V is zero on the wedge, the cylinder and at z = ±x' so that s ros i (g rp)sin--eig- , where Jiir(pra) = O. c1 A sr e-Pr lz-z°1Jsr/Ct V =f1 i At z=z o , -16V/oz I= P r A r s J s r r / a ( 11r P ) sin(sr0/a). Multiply thru by
f E,
a (iir
p)sin(proi/a)pdodp and integrate.
(II b)sinVedS Jpr/a r a ,2 sr ( Ct-TTSEJ sTriaPrP) =A rSp? (112-11)Sin—e - Ars4aPra2 J saTr i tlit-P )d (Pr OS sin 2s-LeS • (lira)) a drr a s +1 % -92ejsria 0 j rr/
2q 12,JsTrice (prb)sin • (Sntlia) e Sr0 -PrIZ-ZOljer(prp)Sin—a-
So V - ea 2a 1 lir Jsr
+1(,11r a)1
a
Ans.
Chapter V.
Page 237. (213)
Page71.
- 107If V=0 at 95=±17 and peaks at 0=0, then f(0)= cos 40 in 5.111 (10) so from 5.11 (13) tja __4--i_i_pa r dux , 1+ 2= coshu2 . From 5.111(9) a second solution is g i=co17:3 gl and 1. Fc C2 dC dG C . By 5.111(4),(5) V=cavl-g 4r---2 . U 2 c-.4.c-2 du2 G0 407-71K 4cos40 Ans. 1+Cz sinhu2 -J1+C 2'
WiTC2-
-
,-"a
j
1+
To get a more symmetrical form, multiply by C i 0/(0-1
[-...5.-1-Ci +411.-ra-G0NIC:7 IL - a _i_ i fI.C2 4l F4 I... 40_} 571 COLCO cE47.47 .sr— i-I-cAt 1-Ca 1; 3 1 47 Pcos40 Ans.
V=C
.
COS40
iA-vwi.+ci- co}
- 108r/ Tria r/a From 5.291(7) for wedg9ra lone : p cos (T70/a) so that cos (r0/a) = r ( s ine) Vo = Ci rria[l - A(a/r) 1+72}(sine) ria cosif, Vi= CI B(rsine) riacos(170/a)
r)Aizejta. Solve for A and B: At r=a, Vo=Vi so 1-A=B, tvbV0 /br= ebvi /Or so evG+ (1-1-1; 1:2T r(e-ev)arla c ailj1( r0 (21i+a)E v fr(e- ev) sinO)acos a Ans. A= 01-1,a)Ev-i-rE' B- (r+a) ev+TTE' Vo- C[r (rr+a) Ev+11E r v c (217+ e u(r sin 0)acos11° a- Ans. 3- (TT+ cy,)Ev +rt )
- 109From 5.291 (6.1) a suitable form is
Ans.
V = Cpillz cos m0
-110From 5.291 (6.1) V= C(pm + Dp-111)z cos m0 . To make V=0 at 0=-±a choose m so that costna is zero. Take smallest m so that V is everywhere positive between -fa and -a. For V=0 2m Ans. at p=a, take D=-a . Thus V= C(pm-a 2n1p -m)z cos mo - 111m+l From 5.291 (6.1) V= Czpmcosmro = Cr cos0(sin0)Incosm 0 is a solution of Laplace' Eq. By 5.12(5) if rin±l f(0,0) is a solution, then r-m-2f(0,0) is also. Thus -m-2 m+1 Ans. V = C(r -a 2m-F3r )cose(sinO)mcos m 0 - 1122
2V
b 2V b 2V
1 a 2X 1 6.2Y j. Oxa -w o. Let V=XYZ, -cc by 2
2Z = TO
o.
Ox
Let -5;-c2 = -m
Then a solution meeting boundary conditions at x= ±a, z rfx t ab Ans. V= Csin--sinsinh a
l
2
2
Oy -n t Y,1,70=(m2+n2)Z Oz
z= 0 is
-113x= a, y= 0, y =b is From 112 a solution giving V = 0 at z= cc nx mrrY EcoEA e - 1(ab)-Vm 2a 2+n lb 2rrz I sin-T-sin— b— so that, at z= 0, V 0 o mn by c° co (m 2 a 2+n 2b 2 )17 . nrx . mry EA sln b bz = Eo o mn ab ( continued)
Chapter V.
Page 238. ( 214)
Page 72.
-113- (continued) Multiply thru by sin(prx/a)sin(qry/b)dxdy. Integrate: x from 0 to ?, y from 0 to b. mry ry nnx rx rb nrc mrd 1 q 4/m2a 2+n 2b 2rra sin--a-sin-b---• 2*-E— ✓ jo sin2 --g—d-rj 0 sin2—E—d-5- Arun so that r2 1 _ 1gco cc nrc mrd nrx mrbry 2 2,1-n 2b 2flz/ (ab) V= SEE (m 2 a 2-Fn 2b 2 P e 46 a sin-i--sIn-B-sin--sin
Ans..
l'E 1 1
-114co m co
=E EA
nrx mry sinhpmn (c - z0or z ) sinhpmn (z or z0)sin--a--sin-b---;
1 1 nin At z=z0 (bV1 /bz-bV 2 /bz)= E E
n
mi7y --a -sin A ran pmnsinhprmc sin nnx
Ent Nim 2a 2+n ab2 n -
Z
ab
Multiply
both sides by sin(prx/a)sin(qr/b)dxdy and integrate 0 to a and 0 to b
r{___l_
J bx
1 OVI+ OV 2-1 ..tg,.1jdS= 3so Amn-4qsin(nrx0 /a)sin(mry0/b) E • F = -ici bz bz J at xo , yo , zo bz c sinhpmn Eabpmn {
note that sinhpmn(c-z0) coshpmnzo - sinhpmnz0 coshpmn(c-z0)= sinhpmn (c - 2z0 ) so 2q2 00 00 F= -----Ans csch(li c)sinhu (c-2z0 )sin2 (nrx 0/a)s n 2 (mry0 /b) mn cab EE mn -115M x' e r n 0, is From 5.16(1) potential of point chare cos0 1 ). g M at r0 ,00=0 4reR - 4rer0 r -"NV When P is at r, 0, 0 the potential is, from 5.24 (5), if r ■ (r0 , M (n )! V- 4 r0 E E , ( + )!rro InP n (cos00) P n (cosO)cosm 0. Dipole potential o 0 is bV/bz= (OV/bro)(Oro/Oz) + (bV/o80)(600 /oz)+ (ZW/orzioXboo /Oz). z = r nose, x=rsinOcos0, y=rsinOsin0, br0 /bz=cos00, b00/Oz=-41sinO, bOo /bz= 0,so VD=4 -TTerg ±
p(2_80 0(12:.T 0;e1(cose ) p(n+0[4,-,rpopran oio)+G r (1-4 )Plni(pc)] cosmo (n-Frn). n
From 5.233 (8), (1-0) en= (n+l)pOrii+(m-n-1)0n11+1 so that, if r4(ro ,
E '4
11 2 -61p--(61i. VD=V TTero o o
1 1= -np0,711+(m+n (m-n - 1) go)PIIIIII (1-10 )111(p)cosm0 Ans. By 5.233(8), (1-0)01:
so for r>r write rPr -n-I for rnr0 -11- land differentiate, eliminate Prinl i (p0 ) to obtain
il!( 2 4 -) (4=24 1Pm ( p ) Pm( c 0 s 0 ) c 0 s m0 Ans. n+m):( m+n ) q-cTIF n-1 ° n
V =7 1-, D 4rer6 oo
-1162n+1 TITI P (p)cosm0 Use 115 Ans. (Ev fore) for Vm and []. Vi=[]Amn PT(p)cosm0, V0=Vm+[ ]Bmn{E?) Amn. Bmil i01 2n+1+1, rr.. nIlBmnq) 2n+1+1, so Bmn=-n(K-1)a 2n+1hTgri-n+1] ro 211-" • LatKAmn= At r=a:
rn-1 I BmnSina Pri'll(cosa). Torque is At dipole: Er br ------AB +1)04(cos00, Ee= ri b li = [iril r ir ar ern mn MErsina+MEe cosa= M[]r31 8min since[(n+1)41+110PV . Use 5.233(8) . Torque is T =Msina[ ]rot Bum( (n+1)Pri+( 1-par 1[ (n+l)pplin+(m-n--1)1.1041+1 ]) =14 [ ham /rsinal ( (n+1)Pri (cosa) -
(
+ (m-n - 1) cosaTiT+1 (co;a) 1 Ans. Radial force is Fr=(ti.V)Er from 1.07(3). E =r1B ltifFsin+24(cose). M0=-Msinia, Mr=Mcosial Fr={-since (bEr /rb0)+cosa(bEr /br)) r=r 0 ' i'T " mn 1-0"-' ni = - (n-m+1)1141 by Fr= ME jBmn (n+l)r-02 (s in2aPinr°(Co sa) - (n+2) c °sot II(c os a). (1-11 2)Pm n '- (n+1)On 5.233(8) so
Fr=-Mc 3B=(n+l)ra 2(COSa Pn(CO sa) + (n-m+1)13;141 (cosa)1
Ans.
Page 239. ( 214)
Chapter V.
-117sinhui W dz z - a cot _ ., From From 4.13 (2); v - - coshui - cos u2' zj du
Page 73.
2_ -a a2 kdu u)i (cosh urcos u2) 2. 2 j sin 2(-ju/2) ' (d
sinhui au .t_l_ b r sinhui m 2U NJ } In 5.11 (6): V 2 U- 6 I' 6.11- coshul -cosupuis ' bu2'-coshui -cosua Ou2 sinhui (coshui - cos u 2 ) sinhu sin uaALJ a m U - 0. Oul R sinhul R4
sinh LI, i6 2U + b aU l+r coshui sinhau ,OU R 2 ' bu ? bui -3 " R2 R l ?Jul
.a
If R---VCOSh UrCOS112
let U = RW, bU/bui4R-1(sinhui)W + RbW/bul , bu/bua = -1/R-1(sin u2)W + RbW/6u2 ,
6 au r cosu2 sin2u21,..,_,_ sin u2 :?1_47 + Rb 2W o 2u r coshui _sinh 2u w+ sinhui bW +Rb 2w so that R Oui dui y.,1= I 2R 4R 3 i " -'- R Oua 64 FCITI 2R 4R3 o 2 1.1 + b au _ cosh ut+ cos u2w÷ sinhuiOW + sinuat231 +Rr o 2W + b 2W.I. Substitute in third line abov∎ ouf bu 2 R bu2 1 bu 1 6112 I 4R R Oui 172U=
.I. sinhui I. b 214 + b 2w1+ coshuiOw +r 3coshu,sinhu,4-cosu2 sinh u t sinhaui +sinhui sinaui _ m2 211 2 Rsinhu1 I R bui 1 R 1- but twil 4112
In the above all 12-3in the OW/Ouland bW/bu2 coefficients cancel out eliminating bW/bu a Simplify the W coefficient and multiply thru by R/sinh ui . Equate V 2U t 0 zero to get. 62w b 2w 2 1 bw ., 2 + 1 ---ff, + COthtliF + ( 4 . . 2 3=0 Let W= W IT,42 and v 2 (W I W2 )=0 separates. Thus ou i C.W. 2 Ui sinn ui 2 1 b 2W1 ,_ cothulOW I +1 1 baw m sii -T11u-+ w ---2 6u2 '= 0 yields two total differential equations ci:Ful ' WI tali 4 1 2 2 d21071
1
M2
dW
___i + Coth uidui du ? i
t n 2--4 1 +sinh 2 u1 3W1 -.7. T
°
(1)
2W and d du 2 --= -n 2W 2
(2)
2
-118-
0 AnPn _i (coshui ) cos nu t, 0 .1.12< 2 rr . V =^/cosh ui- cosu2 E At ui=u0 , V= Vo . Multiply by cos nu2du 2 , integrate 0 to 2/1 or let u 2= Tr -93, due= -dO, cos nut= ( -1)ncos 110.
N
US) .......-
-A/ N1
.
1....."
1
, I ► \ i ... „1—L.....t. \ I ....„ ! , N. / „ I NI, , --i -- r/ 1 ... I , Ii -... I I t -7- --1--. I ! I /` , , // I %. I ....e .• ---- 1.- -I -
Write cosh uo= zag.7. so sinhuo=1//j71. f---- Tr r 2r cos nu,du2 _ (-1)n2Voksinhuo V0 rcosn0d0 A n rip 1 (coshuo )Jo ,IcoiEuo-cosuo TTP 1 (coshuo )40./Z+(z2-1) n-n-T 2 This is Laplace's integral 5.23 (15) so that n n n a (-2)2Vo Pn_ q2(coth tici) ( 2n+1) Au _ el ) 2Vo P- 3 /, (c oth u0)/... .1 111.10 P 1 (cosh uo) (2n+1) :1. P 1 (cosh uo )4.1.4.• • (n-1) n- 2 n-T -119I I By 5 .23(15): P 1 (cosh ui )TrMp-i.1-17(coshui)11-id0 -,(coshul )n T -el. n--2-
112=3r/4
1
7
-
.
=17/2
Ans.
At u 2-4' 0, y= a sinhul / (cosh u1 -1)1.7i1,e au i / (1-iut • • • -1)---w- 2a/ui
A/cosh url -0- u1 1,11= A./1 al y so V
4/2 -i- E An= 0/ (4rtr) since y=r in ring axis plan( o I 2) n 24/2(coth uo) (2n+1) From 4.13(3): a=bsinhu,3 . Thus C= 2.--- 8rte.12sinhuo b E `Ans. Vo 0 (2n+1)V.P (cosh uo) --** 1.1 1-1. 1.1 2-v 0
n-1/2
Page 239. ( 215)
Chapter V.
Page 74.
- 120By 4.13 (2) x= a sin u2 /(coshui - cos u2 ) . V is zero i3 ui =u0 or u2 = 0 or n. (coshuo innu2 For uniform field V= -Ex= -Eoa sin uo(coshui -cosu2 )i= EAn P o n-2 Multiply by sin nu2 du 2 and integrate 0 to 217. rr 3 -Eoa sin nu,sin u2du2 Let u = sin nu„, du= n cos nu2 du 2 so that A n -rirp (cosh uo ) (cosh uo-cos u0 3/2 v = -2/4/cosh uo-c osu2,dv = sin ua du2 (cosh uo - cos u2 ) 2 r1.• 1/ 2 r2rcos nu2du2 (-2r 4nEoaP12 v_2(coth uo) (2n+1) 2nEoa Ans. (from problem 118) 11_1/2 (cosh uo)•0,/coshuo-cos u P _ (coshuo )(2n+1). 2 n This potential cancels uniform field potential at torus surface cosh uo = R/b.. -121-
co V is to be zero when ui =uo . V= Epcos0 +4/cosh ui -cos u2 oAn P1 (cosh ui )cos nu2c0s0
3
By 4.13 (2), p = a sinh u1 / (cosh urcos u2) . At ui= uo , V=0 so -Ea sinh u1 (cosh urcos u2)-7 =131 1 (coshuo ) cos u2 . Since the left side is an even function of u2 , it is clear_ T n r,2rr that sin nut terms cannot enter for jo is zero. Multiply by cos nu 2 du2 , integrate 0 to TT. rr - 2Ea sinh uo cos nu2du2 cos nu t due - Ea sinh uo r 21" An - 9rPn11 / 2(cosh u0 )1 0 (cosh u1 - cosu2) 3/2 - 17P/1.1 _ 1/ 2(cosh u o ) (sinh u0) 3/ 2brtz-,fir=I co su a )3/ 2 pri 3/2. p nin so The z-substitution is from 118. In 5.23 (15) n is -I, but from 5.22(1) n n n+1 Ea P/,(coth u]) -2 (-1) EaPv,(cothuo) /(. . 3 ... [n _ 3 3) _ (2n-1) (2n+1) (-2) A = Ans. 2 2 2 2 n ,,./sinh uo (2n+1):: P/1-1/2(COShUo) v2(cosh U0) a = b sinhuo , cosh uo =R/b. -122From 117 1 dz/du 1 2 = a2 /(coshu1-cosu 2 )2 , x= a sin u2 /(cosh ui-cosu2 ). Substitute in 5.11 (6). mzu sin u2 1 bU + O sin u2 1bU - 0 sin u2(cosh urcos u 2) bui coshui cos ua jbul Ou cosh th-cos u2 bu2 sin 21tU m 2U sin u 2U t 21 sin u sinhu bU .4. [cos u 2----1— --,2 + - 0 bUt Otq R4 bu t Re sin U 2 R2 R4 R' oul -
Let U= 4/cosh u1-cosu2 W = RW. Derivatives were evaluated in 117.
m2 , sinuar O 2W + b 21,41+cosu 2OW +(sin u2 (cosh ui+3cos u2) sin u2sinh2u1 1W = 0 R 1oul bu P R bua 4R 3 2R5 Rsinu2 big ILW In the above all R-3 terms in the Pl. and coefficients cancel out eliminating -6— 1 . I Write 1-cos 2u 2 for sin2u 2 , cosh 2u 1 -1 for sinh 2u1 and simplify so that m2 6 aw b W bw j_ x 2 +-i, —2-+ C 0 t U2'571 23W - O. Let W = U l (u i )U 2 (u 2 ) and V 2W=0 separates. Cill i UU 2 C 4 ' sin2u2 72u _
-
2
m2 1 d U 2 +cotuadU_ 2 _ 1 2u1 + 1 -- -= 0 yields two total differential equations U2 du22 U2 du 2 T sin2 U2 UL out 2 U 2+ 12 d 2U m2 CO t u (12.2 + (n-1-1 ) (1) and —13= (n 2+n++) (n + U dui 2dU2 11 ' 7.7-711172= '
(2)
Chapter V.
Page 240. (215)
Page 7-5-.
-123From 4.13 (2) z = a sinhui / (cosh ui - cosu2). To cancel field potential Ez on sphere ti1=u0 and give it value Vo write U1 0
V0=Ez + REAn sinh(n+4)uo Pn (cosuo ) where R= ,icosh ur cos u2 so that co
:
11"4 Vo - R-3aE sinhuo= EAn sinh (n+4)uo Pn (cos u2 ) (1). If we let cosh u°-l= (a-b)2 = 2 sinhaliuo , ° ±u /2 r7-14 , / r- fu /2 cosh uo+1 = (a+b) 2 = 2 coshaiu o so that a= e ° /v2 b= e - 2 /N/2. Then a/b = e- ° and i 1 i -(n+ .0 0 expansion by 5.153 (4) ields Ri= (cosh o-cosu2 ) "E= (a 2 +b2 -2ab cos u2)72-=4/Yi pn(cosu2) lc° - isinh uo(coshuo -cosu2 ) 7 = b(r -')/0u0 =-29 (n+i)e-t1÷7)u° Pn (cos u2) (3) Substitute (2) i ‘ and (3) in (1). Collect coefficients of Pn (cos u2) : E(./27V0 - (2n+1)EaJe - 'n u ' t u°-Ansinh(n+i), 3 1 r (2n+l)u 1 •Pn (cos u2 ) = O. So An = 22(v0 -(2n+1)Eathe ° -13 Ans. From 119: a = bsinhuo , coshuo=1 02
SO
V=
Ez +./cosh ui - cos u2 ; An sinh(n4)u1 Pn (cos u2)
Ans .
-124From 4.13 (2) and problem 123 p= alr2sinu2, z = ar2sinh Li" R 2 = cosh u1- cos lir In 123 V = 0 when z = 0, u1= 0 so sphere charges are equal and opposite and all flux between them crosses the u1 =0 plane. Thus for no charge 21710 (61/s /bz)pdp = 0. Vs is last term i cos u2[bvs1 bvs by tu _ O. cos u2 ) 3/2 c°(n+ 7)AnPn i in 123 answer. — -s •---1 b (cos u2 ) Oz - Oui 6 z u1 -0. 0 NO ui -3.0 a a -
-
But (pdp)ui =0 = (1-cos u 2) -2 a2sinu2du2 =-a 2 (1-pr 2 dp, where p= cos u2 . Thus -folews/Oz)pdp pi
= -L i (1-p) 2 E (n+ 4 )An Pn (p)dp = 2-faE j_ (p)f(n+-1-)AnPri (p)dp, use 5.153 (1) (a=b=1), o 1 s i 1r =2"fa fo (n+-2-)Anf i tPn (p)32dp= 2 2 a pin . by 5.155(3). Substitute for An from problem 123: E(2n+1)Ea(e(2n+1)uo _1)1. Q = 0 = EAn or E Vo (e0n+1)uo 1)-1 = By problem 119 a = bsinhuo , o ..13E[ e (2n+1)uo.1311-1 V o = Ea (2n+1) He (2n+l)u Ans. o -125-
g
Integrate stress across u1=0, z =0 plane. From 1.14(2), F = ler( (aV/t)z) 2 -E 2 )2Trpdp. By 124, -pde = (1-0-2 a adp where V is total potential of 123 and by/bz is from 124. i co rl 2Ea co 1 3. (n+-0AnPn F= TrE Li (:7 (p) + (1-p) [F3 (n-F-i) A n Pn (II) 3 941 = ItEt1 -27EaAn+(n+1 a )2Atlf_i[Pn (p)]adp pi co ,, i i -E(n+-i ) (s+ 1)AnAsj_ipPn (p)Ps (p)(p))• Since from 5.155 (1), (2) j_I Pn (p)Pm (p)dp = 0 if min. o 1 -1 but j, (P_ (p)) 2 dp = (n+ i) . Expand pPs (p) by 1.154 (1) and last term becomes pi co , 'a:. ,. „ E(n+ii+AnAs t s j iPs _ 1( p.)Pn (p)dp+ (s+1)11Ps+101)Pn (p)dp) only s=n+1 in first integral o co
r
0
and s=n-1 in second so = -2-.(n+1)AnAn+i+r3lAnAn _ i j= (n+1)AnAn+i , since 7A 0A_ I =0. Thus F = ITEE An -27Ea + (n+-10An - (n+1)An _ i ) o (
Ans.
Chapter V.
Page 240. -126-
Page 76.
(126,p216)
Potential inside sphere has form V(r,0) = iaAnrnPn (cos 0). If potential at r=a is f(p'), r.1 1(n+-ii)rn a-nPn (p')Pn (u)dp.' (1) then, by 5.156 (2) this is V(r, 0)=.1.2 f Note that co summation may be written 41'reaViQ-16(17Vp ) /Or (2), 4TTEVp=Qpn a -nPn (cos a) Pn (case), r
a/2 at a point P on a circle subtending an angle 8/2 as shown. But this V may also be found by direct integration. Thus v =
p
Qa since do i.r4TTER• 217a
do Q r'rr 41T 2 E,10 4,/aa+r2-2ar (cosacos e + sina s in 0 coo)
(3)
Since, by figure: R2= (acosa -r cos0) 2+a2 sinkt+ r2 sin2 0+ 2ar sinasinecoso. Substitution of (3) in (2) gives f (a) sinada d9s 4" Jr r TT im V (1- ' rt SZ-rr (4a2 +r 2 -2ar(cosa cos0+ sinasin0cos0)
a sin
.
-127CO
V = Z AnKo (nTrp/b)sin(nrz/b) . See 5.36 (4). on a small cylinder parallel to and enclosing a dipole M. c= EbV/bp=CEAnKi(nTro/b) (nr/b)sin(nrrz/b) •
21 --
Multiply by sin(prz/b)217pdp . a is zero except over area at z = ±6, since M= 2q6. Let u=
osin Intidu= oAnKi(nip/b)ri2TTp /b. q sin (nrr6/b)-->qn116/b = eAnK i(nrrp /b)2Trpnr Mn r na K tang \ 1-1 b 1 .1
1
An =2eb roKi(nro /b) 2E11 21- b
Mn 2eb 2
by 2.32 (5).
Invert with R= b = 2a, choosing M= 4rre (2a) 3E so by 5.102 (1) (2a)3., 1 r OP 3 = oos 3aL cozi.
4Tre(2a)3E 7 o / . _tana. a cos3a2e(2a) 3. 2a 2172 eE n so a = 3 E(-1.) n 2K 0 (nr tan a) cos a
2
nn (-1) K °(nrt a lia Ans.
-128Spheroid potential from its charge in spheres presence is, by 5.40(7), U c° 2s+1 P 2s (cose) . Vin= iAn[Q2n0C0)/P2naGo.)3P2n(iC)P2n(t) i rslAnCsn (c/r) g) = L, i V ex=AnQ2n(jC)P2 s+1 n-° (r/a) P 2s (cos()) by 5.40 (25). Csn=(-1) s (2s):/t(2n+2s+1)::(2s-2n)r.) , = j-0 s=0An Cns(n/a) Cns= (-1)n(2n+2s-1 ):'1:? 2n (jc o)(a/04s+1/1 (2s): (2n-2s )!!P2n (go)), Crin=(Csn+Cns3s=n' Sphere's Nn 2s+1 r 'AC charge potential must be V' =P 2s (cos0) to cancel Vin at r=a, so ex N n.-9 s4:1 n ns (c/r) 2s+lr , ,a , 2s _ a ,r ,2s+11 total V t between them is V = E E ( / ) !P (cos0). But ( ) ) tkrt ) t n=0sF0An Cns c a 2s r)2s+1 spheroid's potential is maintained at P 2s (cos()) BnC sn (c / V=3302n (.5°)P2n() 40 st 2s+1 The sum of (c/r) P (cos0) due to V and V' must equal 2s ex ex the original V coefficients so N 0 A 0 0 00 + A i Co 1 + A 2 CO2 +• • •+ ANCoN = jo BnC'on 1
A o C io
+
A /CI1
+ A 2C 12
+...+ AN% =nti BnC in
N
A oCN0
+
A 1 CNI
+ A 2CN2
+. • •+ ANCNN =
BNC
NN
• Chapter V.
Page 241.
Page 77.
-129Potential of sphere due to its charge in presence of spheroid is N = 2n+1 from 5.40 (10) V x=j0An(a/r) P2n(cos0) =nosEhAnCsnQ2s(jC)P2s") 2n Vin=ntoAn(r/a) P2n(cos0) =
from 5.40 (27) 0AnCnsP2s(jC)P2s(4) . 1-Al2n+1 (-1)si(2s+1)(2s+2n-V.1 c „ _/-12n (-1j1(4s+1)(2n):: CneCsn+Cns 'sn kci (2n)' (2s-2n):: ns "' (2s+2n+1)2(2m-2.5)::' at s=n To cancelVi . at C= Co , spheroid's charge potential must be N n n So Vtbetween them is qx -jOAAn Cns(1)2s (jC°)/Q2sa C°)1Q2s(-1C)P2s ( C)• N n (g). But sphere's potential is kept at Vt=nE0 s-E-0AnCnsEIP2s(iC0)/Q2s(i")3Q2s(iC) -P2s (j6)3P2s N = 5.4) (10) The sum of the Q (jC )P (C) V = B P (cos0)= ° 2s 2s j0:F411Bri CsnQ2s(j‘ 0)P2s(g) I n 2n 'xmust equal the original BnCsn coefficients so that Vex +Ve coefficients due to A0000 + A1C01 + A2CO2 +*"+ ANCON = 1 COOBO AoC i o + AICII + A2C12 +
.
•
•
+ ANC1 N =rigo C ln Bn
-
•
•
•
•
•
•
•
•
•
•
•
•
•
•
n A0CN 0 + A 1 CN1 + A2CN 2 + + ANCNN =ngoCNnBI
-130From 129 and 5.214 (3) V= When COD, a= -ehbVoc • nliOsikAnCsj(2j/11)Q2s(j4)-4)2s(jC))P2s“). a= ---4 16, C ((2j/11)Q 1(i*0)-P' (i 0))IP (g) h -oc.g by by 5 27(7). Then by 5.2142(3),(5: c i , n=0s=u n sn 2s 2s .".IT 2s . 2 0)) so that P2S(j.0)Qs(j.0)-P2S (i .°)Q2S a.°) = 1 =TSP2S(i .°)C (2j 1")qS(j.°) -PiS (.j • 2e a= je.21 __1211L A n P p2s((j) ) A C (§) Ans.5.214 (3) Trc i t n=0 s=O 1-1‘s cl r n=0 sWIn'sn 2s -() k /n sn (2s-1)” P2s -131N N n 2s For spheroid: Vin=nEcAnP 2n(jC)132nM nE0sFoAnCns(r/c) P2s(cos0),5.40(7) = co N A1-.. islAncsn(?)2s+1 _2s 5.50 P (cose) )3 Q2n(j C)P2n(D=n) Vex=nErn"E2n(j"/Q2n(j" (7) s n 2s+1 _(-1) (2n+2s-1):: at _ (-1) (2s):(c./0 = + where C ns (2s): (2n - 2s):: ' Csn j(2n+2s+1)::(2s -2n):: ' Cnn Csn Cns s= n N = To cancel Vex at r=a need V! = E En-An C sn(r/a)2s P2s (cos0). So Vit between in n=0s= N Z AnC 2s+1 2s , sphere and spheroid is V E ((air) -(r/a) jP (cos0). The internal spheroid = sn N 0s= = n= 2s t N potential remains nE0BnP2n(jC)P2n M (r/c)2sP2s(cos0) so this must be the =nE0sEnBn interior value of the potential needed to neutralize the induced charge on the sphere, 2s namely -Vin. Thus the coefficients of (r/c) P2s(cose) must be same in Vin and V'in so • • • • • • AoCoo + A1C01 + A2CO2 +...+ ANCoN = nEen Cno • -
A0C10 + A1C11 + A2C 12 +'"1• • • • ANC1N = nkBn Cni ' A OCNO + A1 CN1 + . • • • • A2CN2 + 0 + ANCNN = • •
•
•
BN CNN
Chapter V.
Page 242.
Page 78.
-132This is a special case of problem 129 and the result can be obtained by the substitution of n+1for n in that problem and rearrangement. -133cle Pin{l + (f) )2 -Icos0} Potential of line charge q at 0=0, F=b is VI =4; q °:., pn cos n0 q ''' , nn sinne cosn0. If rnf 0,0) satisfies V 2V=0, so n nbn -27rE nt'l 1pi 2rre 1114 1 r-n-lf does (0,0) . Thus is solution giving V=0 on the sphere r=a is V.
E rcy__4r ( bra 21 sinnn0 cosno = V1 + Vs. By Dw418 new term is
2rE n=11. -13
a 2a2sin0 a 4 sin2 9 cos0) . The addition of V1 gives =-1 Pm(1+ bara Vs TTE br a a2n+1 n • b2 (b2r 2 +sa sin 2e- 2a2br sin Ocos0) r V= q sin0i pin ri a 2 sin 0 s = -:— qa 41%272 (boaa/r(ba+ra si n 2 0 - 2brsin Ocos0) • At 0=0, V 217er rl4.1 2re nbnrn+1 2n+1 n 1 -q a p 1 pndz rc° F = 9 j_: By Dw 601.1: Vs=— p sdz= 3c- r 2re Z 1 nbn (p2+z 2 )n+ a • At p=b, 600 (p2+z2)n 2 re, 1 nb q2 OD ral 2n+lr rno)+ ( 1'1(112 2 co 2n+1 bp -n r.00 xn '' ldx 1 /)2) Use Dw856.11 Let z-2_ - , F-i-ra 77 E 1-13-1 o 2 1 J (1+x) 2 277E 1 nbn -57 q 2 a2 2 sa 1,a fa .12n , = 22_.2 ., a Ans. -a2 sin-1b ' reb 4'1' (2n-1):. I Ei - Tre b2-a2 tan- 4/b2 -a2 We •sr:77-l
L
i- _1
E---a Tr_ 6
-134Potentials inside Viand outside Vc, which are equal at r=a are now
1
CO r r i nsinne Sinne (4 C° rin ,(Ar 4)a 2n n c os no, . - — V • = ---1- E A,LT, J brirni-i n cosn0 1 2i 6 170— 1 C oviv bv = Evo--° For c l. at r=a give An the value An = (2n+l)ev /(ne + (n+1)ev ) so that tn. c7)ln r sinn0 °D 2n+1 V.= -qcos ng3 n r< a 1 2r 1 n{rie + (n+1)E v }L q 2n+1 sinne (e -ev)a rn cos nr6 V r> a 0 2rev n (ne +(n+i)ev irnii- —97 , r pri . q E 2n+1 n a n(E-ev) cos n0 = V +V s 2rev [pbn (ne 1 (n+l)ev ) nbn (pa-Fz 2)11+ 2
1--1-r My 1 [
'—
--
So Vs = n(E-ev)Vs /{ne+(n+1)Ev } where Vs is from problem 133. Thus the force on the wire
is from 133,
F
E
(2n-2): :n (e-Ev) rev 1 (2n-1):: { ne+(n+l)ev
ral2n+1 Ans.
Chapter V.
Page 243.
Page 79.
-15 352L
1 L
1R
From diagram Right side images are: rti+ =(4nb-z)2+p2,
2R
1 .12 1...=(4nb+2c-z)aiP
2, 0.
Left side images are: rti+=(4nb+z)2 +p2, rria_=(4A-2c+z)2+p 2, 0
Ec
V= (1) outside sphere of radius a centered at p= 0, z= O. p=0 Pn=0 {r rn Assume N images on each side gives required accuracy. The potential inside the sphere
co
Apply 5.40 (35) CP(r/a)PPP (cos0) (2) p+1 bp . _ 1)pe -kznjo ,. p . (a/rn ) P+1Pp (cosOn)= 0134-1 4!)S:kPe-kzn.J0-p' uc )dk=L)dk. p. bznINJ 0 p+1/13:) On axis where p=0 this becomes (a kpe-kz ndk. In differentiation with respect
due to its own charge is
__ r (
Vsphere=pEO
s7
to z for multipoles it is necessary, in order that planes remain equipotentials, to take charges of alternating sign for even p and all charges plus for odd p. Thus axial potential of images, using upper sign for even p and lower sign for odd p, is p+1 rwk p ;e-k(2c-z)+Ec e -k(4nb+2c-z)q.e -k(4nb+2c-z)+e -k(4nb+z)7e -k(4nb-2c+z)1 V=C a j dk Pi P 0 p+1 c a rk p4-k(2c-z)+Ec e -k(4nb+c) [ek(c-z e -k(c-z)]+e-k(4nb-c)L ek(c-z e -k(c-z)1 .11d1 = P PLJ 0 L p even . c aP'irkp17.e -k(2c-z)+E2e -k4nbc e -kc+ekcl sinh k (c-z)l dk p odd 'coshk(c-z)j n p 41 o L = c a rpk p 4e -k(2c-z)+4 coshkc sinhk(c-z) Ee -k4nb dk p even (3) p p: JO p odd cosh kc coshk(c-z) n (bsV/bzs )= 60sV o at z=0,p=0
The s equations for Cp are:
(4)
Due to the sphere itself, from (2) with Pp (cos0)=1 since 0=0 and with r=z, we get s -s s co b V /bz = E0 6 C s-ta . Due to the images, from (3), when p is even we have sphere -k4 , bsV q 13=0P1- P + s even Write as exponent: coshkcsinhkc r -2kz Ee At z=0. -'-P---CP ' s t-e +4 nb jdk 1---rkP p; o s odd n -cosh 2kc Integrate Dw 860 bzb 1. 1 = r icp+sc_e-2ck+r[e-k(4nb-2c)-e-k(4nb+2t k. (p+s): r -1 +E s even rcc J o Jo 2P+s+11-cp+s+1 n(2nb-c)P+s+ 2 (2nb+OP+s+ n k4nb+e -k(4nb-2c)+e -k(4nb+2c)Idk (p+s):Ef -2 1 1 s odd =r-kP+sEt2eo (2nb-c)P +s+ 1 (2nb+cy+s+13 n 3-Fs-FL n (2210 2P+s+1 j Due to images, from (3), when p is odd we have
p+1
bsV. a cosh akc k4nb)dk jip p+sr -kz e At z=0. ---PI-C ---- k Le +4 p p! 0 -cosh kc sinhkc n Ozs -
s even s odd
As before, if m = p+s+l
1 In 3 1 +z[ 2 m 1 + ] ' cm n (2nb)m (2n1 3-7an (2nb +c) 0p_ kp+sEE e -k(4nb-2c) -e-k(4nb+2c), 1 ] s odd r 1 _ik = 1,14 E[ JO n n (2nb-c)m (2nb+c)m (continued)
i, 221( c+E[e 4nbk 2kc+2e 4nbk+e 4nbk+2kc-zdk. p-i s even r=ricp4I-e2m -
0 0
-
-
-
Chapter V.
Page 243.
Page 80.
.
-135- (continued) p+s even, n 5.-.1 : (p+s):( (n-x) m -(n+x) 111 )=I+s ( (n- x)-1-(n+x) -1}/6xP+s 6P+s(-2x/ (n 2 X2)".1)/bXP±S =-6m (n 2 -x2)-1 /Oxm . From Jolley No. 124: c(n2 +x2 )-1 = 12-(17/x) (errx+e -rrN) /(erx e •TrX ) - 1 2 co {T7COth 17X 1I 1 _Id Let x = jy and this becomes E X X2 3 i n 2 ..y2 2tya -
y
p+s odd, n .. *-. 1. : (p+s)![(n -x) m+ (n+x)9 = 6P+s f (n x) (n+x)-13/6xP+s = -bni(n 2 -x a )-1 so for 1 1 1 bP+S+1 r TT COt rrXl a 11 p+s : (p+s ): --p+s+] = I --+----+— bxp x 1 x2 n-x) (n+x) X j cc 2 73 Thus : n -+-s7T = 2C (p+s+1) is the Reiman Zeta Function tabulated p811 Hd. M. F. p+1 ts,, CP a (p+s ):_,_ i 613+s+1 ',cot rTYll. s even, p even where Dps-(4b)p+S+1p! yp s 1 Ely.2 bzsv in = Dps{ ,p+s+i -78---+-7— Y bsv b p+s+i 1 r ]} s odd, p even = -D 12C (p+s+1) + a byp+s+i[ya -----ozsim ps Y b sv TT cot 1 g+s-F s even p odd oz s in= D --ps yP Or'"Tr+7 2 y j s j op+s+1 v. l TT cot odd p odd .871m= Combine all s and p: [y ps12 — ay Y s s 1 4:4s+ i V- _ ( -1) Cpa p+1 { _ bOyp+s.f.1{3712 rr c ot qTy]_,.. ( - 1 ) p+s (p+s-F1)} -1- yP+S+ 1 + {1- (-1)P+ y (4b)P+s+Ip: 2 -
g_____,+s+, ±
i_sqg
ir
1}
s}C
sr p+s+1 r 11 P j Thus, after dividing out s'. /a s, (4) becomes 60sV o =EoCpt5 p+ ;4)1!L4b./i f 1.2bj r o p+s+1 p+i . 1 TT cot Trvj where f (v) + (1: .1 + (1-(-0P+s 1i C(p+s+1) 2brP+s+1 [v 2 v v 1P if distance between planes is b then 80sV o = p_30 C485p +* ( 1 s [firs+ P a] p+s-1-1 r a 1 p+s-f- 1 Fn. Alternatively: 6 0sVo = pE C {8 - [f I2b = 0 P sP + L2T).1 2c p+s+1 1 fr cot 171 where f 1 (v) - 6 +(1- (-1)P+s 3C (p+s+1) 26vP+s+1 [v2 v :+ 1) s
-136ce 2 n -kz 2n 2n - 2p 2p 13.2riSs (-1)P (2n): e Jo(kp) dk by By 5.40(36) r P2n (cose)= r2n+1 (3 1 p=0 (2n-2p):{(2p)::3 z (2n): 540(35) 2p 2p 1 1 %2n+2pe-kzdk = (-1t)P (2n+2p): so that (2n): p=0 (2p):: So 13=0 (2n): (2p)::i 2 zan+2p)-1 By Dw 860.07 2p coefficient of p in the potential summation is (-pp i (2n+2p): a 2n+I r N (2n): b 2n+ 2P 0 D Ans. P (2p):: p=0 (2n): b 2n+ 2P+1 " 11 n:. p (2n-2p): a In '11 pi Since jP 2n ( P)"=jl:P0(1.1 )P201)(1)-1= 0 by 5.155(1) l if T-1 0,only P 0 (p) = 1 enters Q=E f by — dS 6r over r=a. d S= 2rra sine. adO = - 2rra 2dp so Q=2TreaCol dp=4rra eco, C=Q/v0=4rreac o /Yo Ans.
, r
2
1
[
-
Chapter V.
Page 244.
Page 81.
-137The potential for a point charge q on the axis (12=0) of a box of infinite radius (a=ce) and height 1,72b
is
given by 5.324 (10) with m=0, 10 (0)=1 (by 5.32 (2) and with
Rg(;nrqb,a,o) KnOSOIn(kP)-10(ka)Kn(kP) K °(kg)) a->oo Io(ka) 10(4n ra/b) co Thus V"= q(2rtb)-T0 (1'nrro/b) sin (iniT) sin(4nrz/b). Even n is out. Shift origin to z1 =b. .-q(211 tb)-1 K.0(0+ -i) Tr (3/ bj(-i)SS in{(s+12-)17z/b}
---->
t.
V —q(2rrsb) -1 EK0 (s+frP/b}cost(s+i)rrz/b),
1
{
r rs+112n, rs±orro] (2s+1)rrz .....„ an [ q ] q q 12E01P2dcose, (1) , cos 6 z 2n 2nEb s=0 2b 2b " 2b r-41:-) -67174rrzr - 4rre Ozan r - 4rre r In+ 1 a2n+1 - 2C c (2s+l)rrrnir (2s+1)1 (2s+1)Trz 2n+1 so Cn(?) cos P 2n (cos0)=- 11---2b (2n)! b s0[ 2b ''°[ 2b pa-12)21 ilE±117E 2n n ral2n+1 2C = n 7 13 cos Ans. E (s+1) K o (2n. [ j 2b 2b s=0
o 2nv _ q
-01n11
c°
r2
1
This field has the same multipole sources as problem 136 and is zero at z = ±b. Proof of (1) Or bz
op 1 - p 2 6 [Pn (p) = 31' Oz = r , Oz rn 1
un(P) + (1-P)Pn(P) rn+2 . Apply 5.154 (5) to give
(n+1) rn+2
4(21 (n+1)Pn41(p) = ..(n,1)EN.11 +(n+1 )Epn_ so bz L[L3-1= r. 2 141+2 rni" Thus O 2n (l/r)
Let 5.39 (4) be written:
/6z 2n
(20! (1/r 2114-1 )P
2n
n+1 P ri+1111) rn
(cos0) Q.E.D.
-138
co -2s-2p-1 (cos0). To cancel this E M (a,b) r P 2s+2p s0 p=0 sp
co V= E
at r=c, the interior potential of the sphere charge must be cc co 2n+2p -4n-4p-1 V = - E E M (ab)r c P s=0 p=0 sp ' 2s+2p(cos9) Let re ,he the radius vector to a point on the cylinder. Then the V' on the cylinder must n+2pr-4n-4p-1 (cos()) and the internal form have the external form r P2s+2p 7+2p -4n-4p-1 r r P (cos0). These may be expressed in cylindrical coordinates by 2s+2p c 5.40 (35) and 5.40 (36). This gives a correction factor to as and ce of 5.39 (1) since the tangential components must vanish on the cylinder. Thus new values of C 1 and C 2 will be needed.
Page 274. (256)
Chapter VI.
-1For a lower limit put conducting disks on column ends. Capacitance T
T
of one side to earth is, from 5.03 (2), 4ea. By 6.06(7), R1=Fra-2+1-7; Ans. For upper limit insert coaxial circular cylindrical thin shells ending in confocal paraboloids at both ends. resistance of upper half of parabolic shell whose base has radius x and width dx is 477;1 R.2 Te_ 2TeV 4rala 2-x2 2Te_Tfa---2--T 1 =%ra {..r[ 2 • + 33 2x dx 2Trxdx 2rxdx.8ea 1 0 C Q 2 xdx Ru 21Tra R.
2a 1+ Q,71 1 x dx Td o +Th/a 2 -71 T ra 1+1m .
Tr Ru - 2(ra - tan (1 + reit)) -2C-
Consider outer cylinder to lie on equipotential of freely charged cylinder of section A. By thin insulating layers isolate two tubes of flow of identical crossection, b and c. Transfer a layer of resisting material from c to b. Resistance of tubes t' t' becomes (1 ----)R and (1+-- )R respectively. Resistance of tubes t ° t 0' t ir in parallel is now T1. 1-(-) JR0 which is less than original resistance. TR° so net resistance is decreased. Removal of insulating lamina further decreases resistance. Therefore outer layer should lie on natural equipotential surrounding cyoinder of section A for maximum resistance. From geometry we have minimum area for conductor if the boundary of A is circular. Therefore concentric cylinders give maximum insulation. - 3C-
qr
This is just the field of Fig.4.13(b). Half the flux between
2a
2
cylinders passes inside orthogonal circle and half outside. Thus by Te T 2T d 2-a 2 2 6.08(4) and 4.13(5), -zR= c= -237 cosh-1 or from 4.13(2) R=-rr cosh2a 2
■
- 4z-a bil z-a 1- 'Tr +gm a..i.2.i 17 + • . . +onz - a - • rr -+ ... -• • + Image solution: w=2//2 — + z+a z+a+jr z+a+2jr z+a-j1' --- -2(z-a)2) • _ gn j(z -a)(1-17-2fj(z-a)) 2)0.- (21.r2(i (z - a)) 2) • t-a -*-- a 4 (z-a)(1+17-2 (z _a) a) (1+(LB) -072 (z+a)(1+17-2(z+a)2)(1+(2r)-2(z+a) 2 )• j(z+a)(1-11-2ij(z+03 2)(1- (21-2 1.i(z+a)1 2) • . + =Pnitsinj(z -a)/sini(z+a)3Ans. Conf. Trans. Sol. giving U=0 if r1 =1 is " W=k ((z1-a1) /(z1-ail))-Fina l 44(z1-as)/(a izi -l)1, z= z1 , x=ignri , r= 101 2 • upper half plane in strip of width 11/2. e z_e 2 aa sinh(z -a) + _pm W _art y esz+2a -1 + sinh(z+a) Y1 sinh8 , , f If x=a-±8,y=0;V=0 or rr.u1 U=0 i Pin sinh(2aR6) - 4 r/2 If x=-a±45,y=0;V=0, r. u 2=-1.71 so as 6-00 -. .+ - '+ 1t4- Xi -0 ■ % +1 ,\ taii ii ... I 1.11 -U 2 -* 22m(6/sinh2a), Lv]= 2r. So by 6.06(5) A' 1 U -U - - - -.-,--y; x al /"'4-a --3(--a-4\ wn x1 aiR=T —21- - Tn_sinh2a Ans. ..."
N]
ft
8
-
t r
i r Tr
i(
7
- + i'
Chapter VI.
Page 274.
Page 83.
-5- (46,p256)
Yi
• __1_ ,
z =Pinzi , x =Plltri , y= 01 ; zi=eia, z;=e-ja, z;=e j(r-a)- -e-ja
+ •/
. +ii,
13
zy4 = ei(r-a)=-e ja . Images will also work. 1 ,,4\ a 1 „; 1 i ) a% z ja z : tanhT(z+ja) +e ) _o (e (e -e i -z1) t ..,.. _ .442 (z 1 -z1) (z ja 1 t W= 2n (z 1-zi)(z1 -z4) . tanh1(z-ja) ft (ez-e-l a )(e 2-Cia) ; :).• a 2:A 1 When z= j(i--1-6) )1.11=124 tani(2a+ 6)cot76), if z=-j(a76)) 132--g-P1' '. - ■ _ - ''' '14712p Ans. T(Uvi-U 1 ) 11 la .[V] = 2r, By 6.06 (5) R = —r2--- SO 1.11 -1T2 2 072L4L 0 )
-
W
z i+ja i
_ on
- 6- (47,p256) sin z + sin ja_em sin(z+ja) cosl(z-ja)_ ontanT(z+ja) sin z - sin ja sin-i(z-ja) cosT(z+ja) tanT(z-ja)
zi= sin z
1 T
taonh a . When z=j (a7-6) U1=pin tanh(a36) 4 0 If z—j(a7 6 tanh.1.6 2 r T 2 tanh a Ans. 112= -U1 . [VI= 2r By 6.06 (5) R=1'114 -1.11 1 / - Tr zffl 6
a
f— Tr --0
-7C- (48,p256) By 6.04(3),(4) and 4.04
-2C0C I I -1 072 " T.T W = 1-"IiWz-0+;1)-3 )+C ° 2r o+C i cz- a2 " 1-217(Co +anz(z-c -8C- (49C ,p257)
p =c cosh = f(z) (6.15), u= ji0xN/sinh 2 (x/c)+1/(c costs) dx=. From problem 4: F(z)=(-12-TI/r)einsinhYL, (11 ups strip width to 2r) d ...2rI n_ J1 • 1 -LA 4_.3cosh----si-L-. W= -72-11:e2 sinh [1c-21:0+if7°] real Lsinh?cos° 2 2c _..TI fQ u 2,72EinhaSocos29LC2 " roo+sxhx23-2co _ ccs — 21.0+ cosh2 — x -x°sin2- 41 = 2TTT yy 2 2r 2c 2c 2 j Ieil 02c TI n _ [ ,x xo_ Ans. 7r KM cosn cos (0-950)1+C = -712/72 co s 01-6) c .,/ye+zj N/701 4r, ) - yy0+zzo Since cos (0-00 PPo -9- (50,p257) .... 2. 1
Use Fig. 4.22a. To find V value on outer boundary set x=0, y=a U=0 in 4.22 (3). ja=c sinjVi= jc sinhVi, so VA = sinh-1 -a Set y=0, x=a,U=fr/ 2 so a=c cosh Vu,Vu=cosh-1 §. From 6.06 (5), -t3=-V=0 U2 U1 since upper half has twice the total resistance R=Tand U2 -Ul=r. -
L
Thus
Tr Tr
\T1
Ans.
- 10The total resistance per unit length between parallel strips is, from 6.06 (7) and 4.29 (9), Rs = Tev /c = TK(b/a)/K(N/T:17i,T1 ) where b=c, r 2=ab=a2 and a=a 2 /c . The portion inside the circle a is in parallel with that outside, so the resistance is 2Rs . R = 2T K(c2 /a2 ) /K(%/1-c4 a-4 )
Ans.
Page 84.
Page 275.
Chapter VI.
-11From 6.06(7) and 4.30(5) the total resistance per unit length between parallel ,..strips ca, k i) where d i= aakil d o of unequal width is Rs = TE/C= -TK(k)/K(.117.7 R= a in Fig. 4.30 b. h= a 2 (ci+c2)/(c.ic2), .g= c i+c 2,w+g= (a 2+c i c 2)/o l +c c c2 a .? 2 h-w = (a2 +c1 d2)/ca , so by 4.30(3): ka=zF2 (ci +c 2)2 . a2+c 2 aa+cic 2 (a 1-c " ic2) Parts inside and outside the circle have equal resistance and are irrY-f ___L a c itca 2bc parallel so that R= 2Rs . Thus R= TK(k)/1 2 ) where k= a +c I c a -12ELK(1-1 2) From 4.29 (9) half of the capacitance between strips is 2K(b/a) )) Ans. 4 Thus by 6.06 (7) resistance is R = 211((b/a)/(iK(1-1 -13Use 4.30 (5), instead of the 4.29 (9) of the last problem , and 6.06 (7). R= TK(k)/(iK(J.:70)) Ans. where k = het (w+g)(h-w)3 (Refer to Fig. 4.30b) 14 From problem 61 page 114, capacitance per unit length is 2E1(1(1-e
-217a/b
) 2 3/Kfe
-ra/b
R = -FTK(e rra/b
3 Thus from 6.06 (7)
E ( 1-e-2T7a/b)lin )/(1K-
Ans.
-15If flow shown at right is to fit boundaries of Fig.4.28c then Fig. 4.22a (right) must appear as below: U = -r/ 2 -ai
= rr/ 2
V=0 -1
4
+1
Thus 4.22 (3) becomes zi= b i sinW + C. When zi= -1, W= 71/2, at z1=-a1 , W= -17/2 so 2z1 = (a 1 -1)sinW-(a1 +1), Substitute this in 4.28 (3) which gives I I i[(al- 1) sinW- (a1+3)r (a i -1)sinW-(a, -1) T 2b z = +X tanhi - ".] [(a l -1)sinW+(al - 1) 2 17(1+20 ftanh- (81 -1)sinW+(a i -1) 1 }
_. ... 2.,±2e-v)(i_ev)fti_44 1 fttl-
1 V. j sin (U+jV) = sinUcoshV+ jcosUsinhV ---÷ cosh1I--------0.Te a l -1 V-3- c'D •L 1 VU +1T/2 1 len 2(1-(a,+1)(a1 -1)-le v) tantri [11/2=4-k eV-1211711:n tanh [II = 2 a i -1= +V+ Tik2171 1 ' al+1 . 1 2 1 2(a1 +1)(a 1 -1)-le -V • 1 1 [ 1 p, z 1 - 2 e -V R 1 - 2e -V . tanh-liSon eV =4V. x 4 r( 211+3 x) C2 acii +eV+ xieVl 2 [1+2e-X 2 b 2411t(1 +X) x.± bi 1 &zaj...--_ 1 ta- (1+X) 2b rr = 1 Tr rp. +),. al +11 + Vi Use 4.28 (6). x-132 -
So LL
=
2b an 1c(I+X) 2b 17(1+%)
Ans.
Get X from 4.28 (6).
V
Chapter VI.
Page 276.
Page 85.
- 16In Fig. 4.29a dotted circle, radius r, is surface of cylinder and cuts are Uoand-U0 . Half of the capacitance per unit length of strip in cylinder in Fig. 4.29d is 4EK(r 2 a-2 ) 4eK(13/a) 4€1((b 2r-2 ) • r 4 1 C so by 6.06(7) TRAns. K(4/1-b 2 a-z )ICW1-1, 4 r-4 )• 4dK(b2 r-2 ) -
- 17From problem 115, Chapter V the dipole potential is, when r is greater than b, n+1 ov V P1 ; 1 (p)cosm0. When r=0, 7 t: .. must be zero. ( 13 ) d 4762 n=0m=0 r Pm n-1 4/°) -
(2-'14*1-11°!
n-1 . n+1 -n-1 n+1 n+1 -2n-1 n. Note that at, r=a, u(b r +—n—b a Vd by r )/Or= 0, so replace bn+lrbn+lr n+l -2n-1 n+1 -n-1 m lb a r11 if b
/c)cos(rerrz/c)cos(2ra+1)0. =nicOmE0I2m+1("P ' in+1 (nrpk)cos(nrz/c)cos(2nr1-1)0. From 6.08 (3), At p=a bV/bP=gnEoniroAum I2 over small area at z=0=0, TI=a1(6V/bp)d0dz. Multiply OV/Op at p=a by factor cos (nIrrz/c) cos ((2m1+1)03d0 dz,integrate over range -17/2.< 0< 17/2, -c
p=b. Thus
(2-6bcos(nrrz/c) 6V TI ; bo -4Tr 2 ab n=0 ntIVnTra/c)K o (nrrh/c)-KVn1 ra/c)I0(nrrb/c) 3 -20-
Ans.
To get field without hole invert without images. To get field with hole must include images shown. Since images are at the same distance from any point on boundary as original sources the potential at all points on the boundary is doubled. Q.E.D.
•
Chapter VI.
Page 277. -21- (52,p257)
Page 86.
Conical part : dRefds / (2ra), ds/dx -4h2+(ai -a 2)2/h, a=a2+(a1-a 2)x/h1. f h 3c dx ti pal f',/ OnCha 1- (a,- a 2)x) 0 PINai la2) R =----.1 c 217 a a ha2+(ara 2)x 217(a1 -a 2 - 2rr(a1 -a2) )
h
Top: dRp= dr/ (2rr), Rp=14f/T71012(a2 /a 3 ) so total resistance is „„ I Ans. A= Tri[Vh2+(ac a2 ) 22/71-a(lal la a 32) + 212 tl i -22- ( 53 ,1325 7) _onra+b2 -2bx . (x-a)2 +y2 -2bx Use stereographic projection 6.14. In plane: W = 20n ' U-k 2 z+a(x+a)2 +y r 2+b2 z-a
b=2atan-+z a, r=2atan-1le, x=rcosgs. U -Pin
Use tan2(e/2) + tan2(a/2) - 2tan (a/ 2) tarkSe/2) cos0 tana (0/2) + tan (a/2)+ 2tan(a/2)tan(9/2)COS0 Dw 406.2
_ on (1-cose)(1+cosa)+(l+cose) (1-cosa)-2sinesinacos0 _ Atnl-cosacod3 -sinasinecos0 + .0 Ans. 1-cosctcose+sinasinecosits (1-cosh)(1-1-cosa)+(l+cose)(1-cosa)+2sinesinacos0 -23- (54 ,p257) 1 1 Stereographic projection. See 6.14. From figure TD+ R= 2a tan -1-(e+y), 4.D-R= 2a tan 2(e-y). -..., ...- D +2R_ Dw 401.08 =sint3 + siny soD _sill = asin -.., by 4.14(2y D - 2R Dw 401.09 sine - siny 2R siny b .., ...rt 1 1 1 From 6.06(7) R=— - = —cosh-1 D —= —cosh-I a sine / ... ...C . C rrE 2R rrE b i1 •••• 2 a sine I so R= - coshAns . ... .." -rr -b i 1----- -24- (55,p258) . b 1 6 Prolate spheroid, 5.28. z= c2 7-105) "5i-= coo y,i, b2. c i (rie-1), a 2 = cirj og-opr1g=b2/a2, rig = a 2 /(a 2 -b 2 ), eng-1= b 2 /(a 2 -b 2 ), ci=a 2•-b 2 ,
a2 z2 g= z/(c 2ri o ) = z/a, b/bg=i abibz, r1 2 -g2=1 ;171- a 2.
h2 is constant.
13 i b ,. 2_1: ,2 [a4 -(a a _1) 2)z2 7- a4 ..(2a2.4) 2 )z , hs, 45-z-_,..2 ra2-z2 b2J /;+++ (a2_b2) (a 2- b2) " a2_b2.1 a -z 2 a2 a b(a 2 -z 2 )a b by .., ala 4 -(aa -b2 )z2 b2V Substitute in 6.13 (2) ; a-, i Vi = 0. Rearrange b(a 2,.. z 2) bz ' )z2 -b2 Oz an/a - - (a2 b(a 2z 2) bl b(a 2-z 2 ) 6 b2V A b f byl 6 2V r dz (1) Let a=j — 4/3 4_( a 2_b2) z 2 o z [/a v 4_ (a 2-b 2 )z 2 bz + b0 2 = "zbz"zrzJ+ Fii = 0 z Lz ( a 2_b2)(a 2_z2)_ a2b2 bz Use 3:167 387; a= sin Ans. 1 a4 -02_b2)z 2 +4/;T:72 a2 b ilz b(a2-z2Va4- (a2-b2)z2 dz= tanh-,/ zisi--.C72 a2 h l a2
oz /6z= 1/ba so (1) is b 2V/ba2 + b 2V/b0 2 =0. Solution is U or V where u+jv=f (a+jr6). Period 2' -25- (56,p258) Oblate Case (5.27). Proceed as in 24. ,,,so t, a2=cicg, b2=cF(1+;_02),X+ +-0/Co=b/ 0/C0= a „, 6 b hi ,,,/ g2+Cd a2 p, 2a-2+a2(b2,-a 2)-1 aVa4+(b2 -a 2*2 Co=p":7, b 2=q+a2 , x=as, Fra-67c, h = (1.. g2),A 40 -7 ( .32 _x2),A.Fa2/(ba_a2) b (a2- x2) Ap_a2 4(a22-b2)(a2-x2)+a2b2ilx h a p b ( by 62Vbx _x_ Ar -tanh-' )+ 0 A/a4+ _ 432)x2 b si sinkh2 xbx` -- bx Ns 2-- (1) a=J b(a2-x2).,/a4-+(b2-a2)x2 (b2 For the integration used Dw. 387 with b' = a 2, A = -1, .g= a4 and f . b2...a2.
Al
Page 277. (258)
Chapter VI.
Page 87.
-26- (5 7) Use conefunctions of 5.25. Let Mm(0)= Pm. 1(p), Nm (e) = Pm. 1(u)• Then take JP-2 P 3P-2 v1= + B.(p)Nm(0)3r—icos{pem(r/a)} dp outside cone. (1) cosmo m0 =
r,i,m(olc(e)
Va =mt0cosm0 JF:Cm (p)Nm(0)r—licos{pk(r/a)} dp inside cone (2) At e=a, VI= V2 so, by Fourier 's Integral Theorem, Am(p)M;(a) + Bm(P ) N;(a) = Cm(P) Nill(a) (3) .
p '(a ) (4) At 0=a., s1 oV1 /tn=e 2 tV2/bn: elAm(p)Mr:' (a)-Ei Bm(p)Nnpil (a)= E 2 Cm (p)Nm Multiply -bV1 /b0 at p=0 or 0=77/2 through by r-lhcos{q&n(r /a)} cosmOdrd0 and integrate • •
from 0 = -1 to 0=Fr and r=0 to r==. 6111 /6 =0 except in small charge area at 0=0, r=a. Thus SEa-1/2 ds = Q a-1/2/E.-- - arr oV/(r 60) r-1/2 cos(q/kt(ria ))cos m0 r dr d 0 = 217/ (2-6Vilos (Onr
r)ro tAm(p) Mtpn1(0)+Bm(p)111 (0)) cos (p inr)dp
7r/ (2 - elm° )31 cccos qu du To (Am(p)-(-1)m Bm (p) )11(0) cos(pkr) dp. Use Fourier (In-Kral The orem. =C 2r/ (2-6m ° )3.1i—r(Am (q) - (4)53 m 0341 (0) = Q a -1/ 2/ = I a-1 /2 . Am- (-1. Bmpm fox D (5 ) a" ) n
'r l e i=7'2 E 2 = 1. Eliminate Cm from (3) and (4). (TIMN -T2M1N)Am + (Ti-1'2 )NN'Bm=0.
Write S'=111/N I, S=M/N. Then (TI S- T2St) Am+ (ri- .T2 )Bm = 0
(6) Solve (5) and (6):
T1 (T1 -T2) (24)1 Arn (P) = 7r— T& /.2sBm(P) - rtaNr-a-psip_ 1/2(0) (71 _ T2+(_ offi(ri s _125 )3 • Cra(0 = 2
1
(S'mS) (71 -Td-IA/n(0
r.cp I/ (172.,57)} jocos m0,1 0 (Nm(p) Pnjip _ v2(0) + l4i (p)111p _ v2 (-03 cos(Pk(r/a)}dp -
(Ti-T2)B,n Ts (P) (Ti -T2)Cm (P) T I (S' - S) 2
jp- 1/2
TI(TI-T2)(2-6g) (0)(T1 - T 2 +W(TI S -T 2 V)
On the sphere r=a, the angular dependence of potential varies inversely as a and thus cannot contain the term cos(pna). -27-
(58)
Identical with problem 91 Chapter V except that J i (pka) = 0 (not Jo (pka)=0) so from 5.296 (4), (5) the (J (pka)) 2 of 91 must be replaced by Do (pka)32 and q/E by TI. Thus Z < C
T1 ' sinhpkz sinhpkilall Jo(Pkt1ILLilkf2 V=FDIkEi sinhpkL
T1 E sinhpk(L z)sinhpkc -
Z > C
V= ra2 k=1
sinhpkL
Ans.
1k:0:()Za(1)1 1 Jol(1
Ans.
Pk(Jo (Pka) )
-28Rke c TI c° In 27 let 1.,=0D, c= land a=b so, if 0 < z< c, V=r -r-a-a jie 2sinh pkz
(1-1k0)
ukJ o(uka)
c If e -Pkd -e "Pk(cm d)}. Then, as p approaches a, Let z=-2 Id , d<
(1-em as before to get R- rI E aTk= 1Pk
pk(c+2d)lepkd
Ans. Probably an upper limit.
The mean is R =,,T a2kg1 c° Tr pike-/Ike cosh pkd
Ans.
• Chapter VI.
Page 88.
Page 278. -29- (60,p258)
On cylinder boundaries V= V0 0/(217). In 4.12 (2) and 4.13 (4), W = U+ jV so that : e= 9 sine v.1.2 , r sine Vo Vo ..., y At r=a, since tan V = - tan •--- 'FT- tan' Inside: 2r - 1 + cose' ✓ a+x a+ rcose. V0 _1 br sine -1 V0 At r=b same as above so V= 11by . + b x fr , ri tan Outside: V= V0 tan x .+ 217 ✓ r2+br cose' -30- (61,p259) Set a =Tr/2 in 6.14 (4) and measure f from electrode so 217U= Itanh-1(sinesin0)= Itanh-lcosl 2n+1 01 2n+1 Expand by 5.156 (2) . u= tanh-'11, du= dli2 , dv = Pn(p ) du, v = n(n+1) ''' -1 -2n 0+1)(1-(-1)113 ' 4n+3 1-u I ' 2n+ip (p) Ans. v=0 if n even, 2 if n odd so that : U= ar 7,-n=0(2n+1)(2n42)(i) E 2n+1 p
-31-
TZ T 2 >T1 Vo
In 1. 2 with bV/Oz= 0 at z= ±c, V=0 at z=0, V finite at p=co, potential is ,l). . If T2 7• 1 wire flow is nearly "Vo V2= = Z A sin (2n+l)rrz Kor(2n+1)Trpi k L 2c 2c n uniform so at p=0 1.mtzV0ic (2). Set p=a in (1), substitute (2) for V2 , multiply through by sin(4(2n+1)/c), integrate from -c (2n+1)Tra] n 8cVo (3). Substitute in (1). 2=A 2c na°1 (2n+1) 217 (-1)n 8V0 g(2n+1)11p/c1 V sin s in 2c K1{-2-(2n+1)77a/c} (4). The leakage rr2 n=0(2n+1)2 v 2. to c: (-1)
E
current gives a slight deviation from uniformity in bV2/6z . Leakage in dz at z is,at p=a {(2n+1)Tro1 d(2n+1)17z -1 oV (2n+1)Trz bbit ---i- dz=r 7r,, 2Tradz= 2rra -T T -- _FAAnsin . The total current,IL, 2c K g_ 2c 2c 2 " ' 2 'I) (2n+1)17Z [(2n+1)1 277a . The current cos . - E A Ki leaked out between boundary at z=c and z is -Tr2c 2c 2 n0 n in Ti at z is I- II, and the potential drop is (I-I1)T1 dz/(17a 2 ) approximately, so 2178 T cci pcT rn+i)rfalfc (2n+1)17z dz -Vco 21'1 r 2c11._ (4)11 Kr(2n+1)Tral . •TTr 1 2 E A Ki Vo= j oTTI-E2 dz +--iij cos T2a n=0 (2n+1)Tr` ..1 2c 'L 2c a n=0 n o 2c 2
1
f(2n+1)17a/(2c)1 E (-1)n Kif(2n+1)11a/(2c)) R„,,z,,i + _ V 32cTiva_ K0t(2n+l)n a/(2c)} n=0 (2n+1)3 a K0[(2n+1)17a/(2c)} 3 aT 17 ' Ro 75.-a-T2 n=0 (2n+1)3 32cT1
°i (-1)n K1
Ans. AY
cr
-32- (63,p259) 2 z? _a2 a: _ u2 rTi2 - I x1 4-02dzi_cze d itiu17 z12-a2 , Let zi = 1-u 2 ' u - 4-1 ' -I -a 0 a 1.7- 2 -1)u du r u 3 i 2 dU Awa(a (a2 -1)udu dza= (B2-1)uzdu - C(a2-1)j (i_u2)(a2_u3) - cl ZidZi= • z (1-0) (a2-0) (1-u2) 2 ' zi (1-u2)(a2-u2 ) Use Erra 186.11 and 188.11 r adu r adu Cf du + r du Get constants by 6.07 p235. - 2 Vg(1-u) ..) ,179(1+u) .4,5(a-u) 0 ,1/45(a+u)
‘ ..1, U=7r....
k-->
u
317/4
xr _
+.i.c°dz= jCS: dei so k= Tij err . f i-,h.-i. i'dz = jCS17.,rada1 so h = 74jC AnIT. ritic
Thus a = h2 k-2, C=T2jk17. Put in a and C and integrate. Verify signs by direct 11({-jtanj.,,M+ tan-kg --A-tan-V differentiation. z= 2: -stp
- tan-kg/; }+ C 1 Va
= 2Frick tanh-V.i+k tan-VP -h tanh-'kJi/h- h tan -1kiP/h3+ C1 . (continued)
2h--)`
Page 89.
Page 278.
Chapter VI.
-32- (continued)
u=4,0D, z=k-h+j(k-h)+Cl :
z1 = 0, u=a= (h/k)2 , z=-jco+CI : zi= a-e, u=0, z=j.0+CI :
u=1, z=jm+k-h+Ci . Thus C1=h. If z1= e40, u2->a 2(1- (cia)2)(1+0-> a( 1-2ai-E2):
,Vt( 14(1-a 2)(e/a) 23 . yh -4 (MO fic tan 11-ksra-+k tan-l ra-- 2hk2+12-111242 (1 (1-a2)(E/a) ) - 2thr) = (2/r) (k tanh-1(h/k)+k tand(h/k)+(h/ 2)2/Ai (k4-0)/h4 3-Fhkc- -1Thr)
NIE"
When z1= R-)co, u2= (1 - a21372)/ (1-R-2), u
1+3(1- al)/R2,
y k-> ( 2 prf) t-ilain2-1kintr÷ (1-a 2)/ R -41cr-htanh-Vi-htofkli W=0nz vU=Onxv[V] =fr. At wide end Uk=e/nR=-Z-1 :+ rt U -On e-Tjh - k an1-1-11' + tan-11 .. 12n2112--14 + 4 4 8h 2 k k 2h h h-
1+4(1-a 2)/R 2 . Thus {107471141- +kkR-1-1-4-, `"-h tanhl -h t k8 4_hh4 . At narrow end
,f . Total resistance is (Uk-uhqr
6.06 (6), ( 7:
Wide part resistance: ifk-I{yk-(k-h)), Narrow part resistance: iykh-lir. Thus effective resistance of tapered part is k
1
k4 h4. h tan-1 --+27i k 4h4
d5R = R_rf.Kkih) +
2 +.0 r [1: tanh 1 h + k2
r
hk
k
hk
0•=
[hZ2 tanhl
tan-1 h +241k4 411 k 4h4
tan-1 t
Ans.
-33Abstract of paper "Ideal Flow along a Row of Spheres" by Paul Michael in "Physics of Fluids, Volume 8, July 1965, pp. 1263-1266. Current flows between parallel planes around hollow sphere(half shown by heavy line). The flow is the same as around an infinite row of spheres. Solve by images. Use the method of 6.18. Assume the source distribution on (1) which gives inside sphere j= each sphere is of the form k= E (4n+3)CnP12114.1 (cos0) n= 0 2n+1 (2) The subscript zero indicates +iLri-2n-2PI 2n+1(cos() )3 Cn Cr A li n-O the portion due to its own sources and the subscript j that due to image sources. The 2n-2 (3) Equations total external vector potential is A0e= ilo CnicLr3 P12,14.1(cosej) (2) and (3) correspond to 6.18 (1). From 7.02 (8)the vector potential at ID= co must (4) when r o <1 .
be -p /2 and Aosimust cancel this if r o<1. Thus 0=A0=-(p/2) +Aoi r-2n-2PI (35), s'e-tzji (tp)t2n+l (cos 0 ) From 5.40 dt = (2n-0): 2n+1 (6). So from (4) z =0, ro =p, prp, zj= 2 j a c1.,..2n+1
2
n=0 111.'d
2 i° (4%-2jat ji (tp) t 2n+ldt) + (2n): j=1.1 0
(5) In the plane
(7)
To solve for Cn , expand J1 (tp) in powers of p by 5.293 (3) and equate the coefficients (2 (2n+1):Can+2) 2n+1 je-2jat P E 1 (8) t d of p p in (7) to zero. By Dw 860.07, (zcomi* 2 :7 1(2a)nan1÷2 j0+ 31 = (2A+2n+2): (2/+2n+3) _I By Hd. M. F. 23. 2.1. From (7), (8) Cn+1. ,C1 (9) 0 (2n+2):(21): (2a) 2L-i- 2i1-1- 2 2
+):
This gives N+1 simultaneous equations to be solved for Co, Cl, C2 ",CN'
Page 320. (298)
Chapter VII.
Page 90.
-1irrin 24 n rirsrin2 a ds rifr a3. rincosed0 s F- a tang, ds= a seca6 dO, rcos0= a, B'= _p___ B
IP-" Tra
Ans.
2n -2-
B=EL 2rrr
0 x= r sin@ = acosOtanO sina 0 a dO , dT = iBsin0 xdx - pii'cos 2Trc.os 20 piA cosszl )111' 7 a Ans. tan2 0 de = _7_(sino -0cos0) T2Tr -0
-30 2de Ir dE) r p pl Ode 31INIR2 3 From 7.07 (2), dB (41.7 2 N2 z2+R 20 2) 2 z=7:17-d Orr 4NA51-7-7 -1- ) 2 Integrate by Dw 202.03 211'n -0 pNie 2d 0 pNI +072(6 +■ ./(2 Tfliz/R) 2+e 2)] 0 2 ) 2 - 2R [4./4172N 2 z 2 R_ 2 + 0 2 B z foo R ( 4 TT 2 N 2 z 2 411NIFC1( (i+Z 2R-2)7 +2/1 2rN (1-Ftl1+Z 2
)
-
071
( 211Nzirl))
2+R 2 )+, /n[ (R-1-firr=pNIR-1(- (12/,, ----R2 )/z]) Ans. 4.1IIR-1f-(R/A/T2+12 ) + sinh-l(R/z))
Ans.
-4 From 7 .07 (7), Ao=
2BDrsin0. At r=a, Ao=Ai, a3-F-C=a3D. iB (r+ Cr-2) sine, pio(rAo)/Or = po t (rAi)/Or so pi( 2a 3 -C) = po 2a 3 D. C = 2(pi-Po ) a 3 /(Pi+2P0) LBr= -IFBC/-73 o(sin 20) / (sin1360)=BCr-3cos0. b cos0=bsin0cosS. d (LT)=iarbsine bd0. 3 B(Pi -po )cos 2Tria 3 B(pi -po )cosS a = 2iab(2p0+ j sin 20 d0 Ans. pi) b(pi+2P0) -5 Component of force normal to surface between two elements dx1 and dx2 is dx I dx 2 rArA dx idx2 dF- 1112 2TTA 2 V(x2 -xi )2+ B 2a/(xa -x1) 2 + B 2 2Tra 2-loJo (x2 -xi ) 2+B 2
4— dx
_ _i2„,..„]
A
dx2 A-x a=Bu, dxi= -Bdu = Bdv Same integrg 1 ' xi= Bv, dxi for both tel a 1A/B pI2 42 BrA/B 1, pI2B r A Bp4A 2 +Ba] F= = [A tan-1 tan --yuy = — L y tan-1y - -Pm (1+37-) 3 Ans. o rA2 B 2 B2 rrA 2 d rrA2 -6- ( on next page)
t, laBI 2l'A r2 x1 Adx 2S 2T3 o 1 -2rA2 0A(tart-113-B21(1+tan- B ' = 2rrA bio L2B tan —
-7Outer cylinder field is zero inside so force is all due to inner ones. From 7.07 and 7.11, the force between them is
12-1. p-12-1/ (2r• 2c) = pI 2 / (161/c) .
Ans.
-8CCircular loop is axially symmetrical so force line equation is given by choosing N value in 7.024 which gives A0 = N/(2rrp)
Ans.
Chapter VII.
Page 91. Page 320. (300) -6jirra/2 hdx 3_ pIa At distance h from center of wire of length a B.n 4 -a/2 012ifi c2 ) 47hVb2+14Torque on top wire is T t= 1,N z y dy with same limits. f.a/2 E--- a ---)f• t a/2 poa dy y2 dy I a[f T tt-R 11-- Orr _ an.Vy 2+(a /2) 2 _ a /2(a2 +y 2 )/15(a/2) 2 +y21 217 {!/71(3T+572-4 ^r- / 2)237 2 a a dy 4 1 I 2 nn , (a 4 -P 2r , a , - ` ,r 4.7Eai7aa-] N 2 +yd /2) a °75(a *I -Y 2 +f0 2+54) I z 2 -a z2-a Iadu 3 1 Sides •• By-Jo ra4Trt(zPia dzi 1 2 211'-- 21 2 )2 i -z2) 2+ i a24132- 2 41702 + -02)2 2Tra (z2+1 1 Rzg-a) 2 +tia a ,..a 3112 if% a z dz 1 P ° udu it pI 21(z+1 a2\ a p_aI 2 " :1„.. 3--1 =— j 72-r-22- 2 .-, .a-T ssi..1 oB 2 2 ' 2/7 Y dz (u 2-1T.41-i ) f 2 4TT 0 (Z2 ±a ) o 1 2rr lat . pa 1 1 On one side T ts = 2. (--iaI)J Bhsinedx, sine= x/h, h= (4a 2+x 2)1. o x dx du _pa 21 2 ra2 1 at T ts - a2i2f h (-1-a 2+0 (4a 2+01 u=x 2 Dw 195.02 4r. o (4-a 2+x2 )(20 2 +x2) 8rr [4 pa I2 ( tanh-1,16 - tanh"147]= — (tan 4./6 8rr -)• a 2 rr )=— T = 2(Ttt+T -T tsPI2 + 2cot 24/6 +,55 - tanh-lsig + tanh-14 Ans. ss i+ 6
ifi
",
Liz -11 1
-
-
I ---9
rd
-
,"2-
-
)
-7- and -8C- are on last page. -9CComponent of ds normal to r is rat. x=r cose, y= r sinO, so that _ Ui 2 r dO = pi gde _ pi fria2 sin20+ 132 cos2 O de B --y— (1) Since x2 a -2 + y2 b -2 .1, r2 4773' r - 4TFY 471 ab 1 = r2 (a-2 cos2 0 +13-2 sin2 0) = r 2 a-2 b4 (b2 cos 20 +a 2 sin 2 0). A= rat) (2) Circumference is
dx 2+dy 2 = Pa2 sin2 41 + b2 cos2 * di/ = / (3) since x= a co s *,
y= b sin* is parametric equation of ellipse dx = -a sin* d*, dy = acos* d*. Substitution 1 of (2) and (3) in (1) gives Ans. B= Irpii/A •I -100acosOde dF = Bids cosO. B= pil/(27tr), ds = a dO, r = d - a cos0. F rr-- Jod-acose F=
d - 1 dO. Dw 858.524 for d-acose
d . aa 1 =piiI (secl -1) `pii 4-:—. -11C-
From 7.13 (5) Br =(-11-pI/a)t (b/a)° P?(cos0)- 2 (b/a)2 P3(cos0) + • • •) 3 b2 07/2 1. rrr/2 COS 2 dO -7;2 j0 75cos4O 3cos 20) de] T = 4111, 2.fori 2Br c °se dO = (2 pirb2 / a ) [j -
= 411'132a-1 (+17 - a2 6- 27-1 = 1111M 2 II. a-I (1-
4 (b/a )2 )
Ans.
Page 321. (300)
Chapter VII.
Page 92.
-12Cr 1Tr 2sine -----2 ,-ds 02 pi4 Force on element of 1 due to nearest side of 2 is -i1B2ds1=-iidsij e - 1 a-x -x a-x 1.2,4 ra-xi cds ; 3 -pii i 2 dusa.._1 1 ---Pili2 . ds Fx2 cz +s2 -x -417 "1 c./c2+(a-x)2 c.IC-277-14-xi (c 2+s 2 ) a - 4r 4r 2_ Let a-x= y so if x=0, y=a :if x=a, y=0. Two term integrals same , ...pj.,1.2 ra xdx F = /21114( cir- 1-x 2 ) o a= -Th' tihi-Fc 2 - c). Similarly the 2Trc a 2rc JofciTici -27Tc replusive force between far parallel sides is sin0=c/r c 7"Fa2 =va_--Pi i (N/a 2+ca+a2,1j ). Axial component is F Fai F r . r 2=c 2 +s2 Fr rx r 2 c2+a2 rcA/2a2+c2 i i a2+2c2] i_a i) F - Pii i2c W a a_ka _,..0:-21" rt-11 at. +1 - , . Total F - 4(Fa+F Ans. rx) = --1 rx 217(c +a2) c2 +a2 cA/a2+c2 -13CBy 8.03(2), 7.09(1) mutual inductance between wire and parallel side of pikr so m. 22071b2+d2-2bd cos 0 length 2a at a distance r is Ark-c1..9 =2a• 2Tr b2 -1-d2 +2bd cos 0 217 2pii'abd (141-d 2)sine A ., bli pi i'abd sin 0 f l 4.1 ., By 8.03(8): T= ii beTr +0 -2b2 d 2 cos0 'Ins. L A• ' r23 2 - 11(134 -14Cri . _ p.I [ a2 ' 0, A02= 7.13(2), r>>a: A =a [2-+C-1 sine , A 03 -jFr2 a- sin 4- D172 + Es sine. =1-14 0 1 4 r2 a b Or ) 2_ (p pD)a3= 2(p2 C -pE)b3 (2) At r=b, A1=A 2 so a3 (1-D)=b3 (E-C) (1): -(--- 1_ rAr ir PXc 3 -133)3 3 At r=c, a 3(F-D)= c3E (3), (-p2F+pD)a3=2pEc3 (4). C- (112 -11X112+2 (1 2+ 2P)(2112+11)c3- 2(113-103 TrIa 4 p(K„-1)(Km +2)(c 3 -b3 ) Additional Flux is 217a 4 4 .C.et LetKm-112 ii. • N - 3 2b ((+2)(2Kr3+1)c3-2(Kin-1) 2 b3 3
Ans.
-15CIn 7.27 (3) designate the last term, which alone contains Km and so arises from the ItiA l 7.26(11). magnetic materials, as A. Then the force on the wire at p =b, 0 = 0 is I*=--z bp n 1 . 1.2..v. L icl b Ectp] p . So at : F . Ey I2 Km-1 E b by Dw9.04, write b for p to get Kin+ln= lb p La2 2r Km+1 a 2 p= a2 • 0=0 ' n 21 F . _EvI 2 Km-1 1 This is the force between I and an image I(Kin-1)/(Km+1) at a 2r Km+1 alb-1 • distance d = a 2 b-1-B from it. -16C-
ri22,
From 7.02 (7) : A0 = 4Br sine outside, A0 2 = 4B[C1r+rgi9Sitle, Aoi = ZBC3r sine. From 7.21 (3),(7) p2 b(rA0)/br =pb(rA03)/br and A0=A0 2 . Thus b3 = C1b3+C2 (1), 2.1221, 3= 2pb3C1 - pC2 (2) Cia 3+ C 2= C 3a 3 (3), 2p1a 3C 1 - p1 C 2= 2p2a 3C 3
(4).
From p(1)+(2) and 111 (3)+(4) get C3= 111(1+ 21.12)/f p.(11 +242) 3 From 2p(1) -(2) and 2p1 (3)-(4) get C3=p113 3 (p-p2)/(Pa 3(111 -112)). Equate C3 values, 21.12.±E . b3 Since a, b, p, If p<pa then p1 p.2 then pi>p2 . (6) Ans. Since b 3> I: 2p1-2p2 p i +pp2 -ppi>2p1- 211211+41112-ppi . So p>pi . From (5), if p2 >p then p2>p.>pi .
Ans.
Chapter VII.
Page 321. (301)
Page 93.
-17C-
A B(r+Cr-2 ) sine, A0i= 1B Dr sine, a3+C=Da3, p2 (2a3-C)= 2a3D . (1124 110-31 sin28 2(p1 -p2 )a3+(2p1+p2)C =O. So 21rr sin0A00 = TrBrz +-11. is flux through circle. But r sine= Nix2 +y 2 so equation of tubes is Ans. [1 + 2112=211(411 (K 2 + y 2) = constant 112+ 21-11
A00=
-18CFrom 7.17 and 5.12 C2= Arn Sn . Magnetization Potential zero at infinity so that 2n+11.s.ca 2n+1 (1) -S).:0=pii/jCT.Ii s C.20= (Arn+Br -n-1)Sn , ni=CrnSn . At r=a, nb=ni so Aa }roa r 2n+1 I 2n+l +1 ° 2n+1 po (2n+1)Aa -po(n+1)B=pinCa 2n (2). From (1) and (2) po(2n1-1)Aa =(npi+(n+l)po)Ca (2n+l)po 0 Thus CkAns. npi+(n+1) po 19C
nj = (Arn+Br -n -l)sn , n2 = (Crn+Dr -r-1)Sn , C23 = Er Sn . Let K = pi /p. Then
2n+1 2n+1 2n+1 2n+1 +B=Cb +D (1), nAb -K(n+1)D (2), -(n+1)B=KnCb 2n+1 2n+1 2n+1 2n+1 (4) At r=a: Ca +D=Ea -K(n+1)D=nEa (3) , KnCa 2n+1 (1)and (2) give {n+K(n+1)}Ab (5) +(K-1) (n+1)B=K(2n+l)b2"4.1C 2n+1 2n+1 2n+1 2n+1 (3) and (4) give K(2n+1)a C=CK(n+1)+nla (6). (1)and (3) give Ab +B-Ea 2n+1 2n+1 2n+1 2n+1 2n+1 2n+1 )(K(n+1) +n) +(b -a -Ab =C(b -a ) (7) . (6) and (7) give B=Ea (2n+1)K Substitute in (5) for B and C: 0= (n+K(n+1)-K(n+1)+n+11b 2n+1A +(n+1) (K-1) a2n+1E (02n+1-a (n+1) (K-1) 2n+1 )fK(n+1)+nl-fK(n+1)+n3K(2n+l)b2n+1 E. This simplifies to (2n+1)K a 2n+1 r 0 = b 2n-/-1(2n+1)A+C- (n+1+nK)(K(n+1)+nlb2n+1+n(n+1) (K-1)2 JEt (2n+1)K3-' At r=b: Ab
Thus E/A = (2n+1) 2 K/[ (n+1+nK)(K(n+1)+n)-n(n+1) (K-1)2
(a /b)2n+1]
Ans.
-20CFrom 7.02 (6) : Uniform Field Az = -Bx= -Br sine. A3=-BC4 r sine, Let K=pi/p. A2=-B(C2r-i-e .)sin9, A1 =-B(r4)!ine. At r=b: b2+C1 =C2 b4C3 (1), Kb2 -C i =b2C2-C (2). At r=a: a 2 C2 +C3=a2 C4 (3), a2C2 - C3=a2 C4 K (4). (3) and (4) give (K-1)a2 Ci=-(K+1)C3 . Substitute in (1) to get 132 +01 = (b2 - Iiia2}. Substitute this in (2). Kb2 -KCI=f132+ TMa2)C2 . Write p for (K-1)/(K+1) then (K-1)(K+1)(b2 -a2 b 2(b2 +8a2 -K(b2 -Ba2))+Ci (b2+Ba2+K(b2 -Ba2 ))=0 so CI- ((44)2132_((_1)2a2 and flux ratio is 5_1 b2 (K14) 2 b2 K-1) 2 a2 RR 2AZ R Ans. e=2 , r=b. = 2A1 at 21([ (K+1)b2-(K-1)a23 )
-
(
Chapter VII.
Page 322. (301)
Page 94.
-21-
aPt., (g) _ PAM Current density due to nth term is in= COPn / (hi bg) . From 5.23(6) liCja r.1f.2cos0 OPn(t) From 7.04 at 0=0, An= 4rr j 11-1 3d t 46. From 5.27(6), (7) hrs/T-TiT = ha.TFTI. o h 1-p-J o R Also hih2dtdos = dS, PAM cosse =S11 so An = 4
JA- dS. From 5.274(6), (7) at 0=0, C j -11Cn Ans. A9in=.1-74n(n+1) 1 (1+3 Qt (i°)P11(i °Pr" = --jj484j-1--n(r;!2— F1) CA(i")PniCi C); ) -22Let current density be I/ (hi d t) if g o<
g< goi-a g
and zero elsewhere. Then
Multiply through by hiPnl(t)dt, integrate from go to to +ago .
in
aj pg (g)h, d = (to)fsidS = Cn j i(Pn( )) 2 d t (see 7.06). By 5.231 (3) this gives h i6 t I,,,-tt(2n+1)Difr 1 by 21 2n+1 /7--i.3 2. (jS 0)Pnl(to)Pi!pC)Pnl(t) Ans. ""0-2n (n+1) Erik."' if C< Co • -V= 2 n=ln2(r+1) -23Current density due to nth term is in=Crh72bPn (t)/bt. From 7.04 at 0 = 0, r co_sg a• By 5.26(6),(7) Aff:-thi=N/9i.hi An 41°H- 0 R g h3 dtdo. By 5.23(6)V- 12) pc r el dS Also hill 3d tdo= dS, PA(t) cos0 = Sn . Compare V-89, 5.274(6), (7): A =—13 n g1,J s -n h2R • n 2 _1
r1
A= E_Cn
A no
)&-T_ CnV1°)Pni(n)Prt
Write0Ni:71, 1-1 for 1 in 22.
A0 11171.r n£1-PCrtglri)+111)(Trl( -24 ../7-ti 1
n 2 2(nn++11) 2 QA (ri
Ans.
(t 0) Pr (rl)Pil( t). Ans.
-25 Magnetization potential is zero at r1 =C° and so t4 if T1> Tl i and A if rl
Prii(gG)+BnQL(71) = CnPn (1li) (1) ()Ivor!) = o (h3 AO Aildri) From 5.275 (8)
Qin(T10)PA
O(h 3(A0-1-Alp)
From 3.04 (3) and 7.13 (4) we have h3 = C2Ng.
2- 1
so that
L.,./n 1- if q,(ri o )pmg P,;(7-1 i ) +Bn ctliCrii))3/(pvor = cnatorir-ippi 3/(11Ori) which may be written from 5. 2 3(8)5 ( 9):PEri [Cnf -1)(QW7lo)Pili(S ct)
bgo+BnIfi (.91)
j= pvCnb oTii( (1112 -1)bbP pi (711)
By 5.23 (3) (n(n+1) cancels out) : 14(41 (r1o) Pn(UPn(rli) + BnQn (rii)3 = PvCnPn (T)i) Solve (1) and (2) for Bn .
Bn = 01-120 (no )Pgrli )1411 (rli)Pri (go) livPn(ni)4n(711)-114(711) 4nCri1)
(2)
Ans
- 26When p
Chapter VII.
Page 323. (302)
Page 95.
-27 Let A0 have form Ao _ra(k)e-kz J i (kP)dk• From 7.10(8), on axis, Bz=-17321:. (r a 2ke +f k 2 ):::/..2 -1-za ti(k)e-lczkdk. At p= 0 O(PA123) /(PbP) = J0(0)=1. From 5.298 (6), a(a 2 ) dk. -kz _0 so 0(k) =ipaIJI(ka) . A0 = 2paIre J (ka)J1 (kp)dk. Ans. -28pC0
From 26, A0for loop, if p
Ans.
29pa crdzo From problem 26 A0 of element dzo at zo is: dA0= ----2 I2(ka)K1(kP)cosk(z-zo* irccos k(z-z0) dzo = -k-l[sink(z-z o)lce = 2k-lcos kz sin kc so that
where 2cr=I/c
ok_hI l (ka)K i (kp)cos kz sin kc dk.
A0 =
Ans. 2a --4
-30Al =
J2 (kp)J1 (ka) o
(e-k z +1,1 e+k z)dk,
A 2 = jilivaIni(kP)Ji (ka){Y1e
-kz+1,2 e+kz
co A 3 = 2pvaIS Ji(kP)Ji (ka)8e -kzdk
}dk. At z=b AI =A 2 , so that
3 Pv
z
11
0
e
-kb... kb -kb kb -ve l e =11,1 e +1'2e
(1). At z=b+t A 2=A 3 so that
k(b+t) (2) . Since 1.11;=}4 or p"bA,poz=p l oA'931 /oz +11ae -kb kb kb -kb -k(b+t) k(b+t) (4) -pe +A e = e +pvT2 e (3) =-11‘,T1 e pie +211 11 ve (p.v _ p)e -kb±(pv+p),pi ekb.2pvyaekb (5). (2)-(4) (6) (i give give (pv+p)0 = 2pv`i1 Se
-k(b+t)
e
-k(b+t)
kb -kb , -kb (pv _ oee -k(b+t).2pvy2ek(b+ give (1-1v+11)e (7). (2)+(4) give +(Pv-P),t.ie =2pvTi e (pv _ p)e-2kb+(pv+ocp1 y Lpv _ p)e-2k(b+t) give Solve for (Di : (9 liv+11 = (Pv+P)+(Pv-11)01e 2kb -2kt 2 -2kb -2kt 2 2 -2kb ) (1-e ) (}12 -31v)e (1-e )e _ (10) Ans 1- ( 1.1v _ p) 2e , 2kt _ (pvi_02 (p -F pv)2 (p..pv ) 1 e' 2kt )
Let C = (p.+1.1,0 2 -(p-pv )2e-2 kt
(11)
From ( 7 ) 111=( (P+P•v ) -(p-Pv)Ce 2kb)/(2pv)
=2CK71 [( 11+1103- (P+12v) (11-P0 2 e-kt (P-11v) a (P+Pv) (P-Tiv) 2(11+Pv) e 2k t
=1(p+pv)141coo= 2(p/pv) (p+pv)C From (8)
Ans.
g= 2pv (p+py )-l Y =4pC. Ans.
From(9)
y2 . _ 2(pi pv ) (p_ poce -2k(b+t)
Arts.
• Page96.
Page 323.
Chapter VII.
-31- (31,p302) B= 0 at z=
for zeracirculating current, from 7.05 (2) with no 0 dependence.
At p=a1 A0 = nri CnRi (knp)e-kn z where Rj(knP)= An Ji(kn p) + BnYi(kn p) (1) and p=a 2 we must have B0 = (pA0) (Op) so that AnJo(knai)-t-BnYo (knai)=0 (2) AnJo(k na2 )+BnYo(knaa ) = 0 (3). Eliminate Bn: Ja(knairidkaa)-Jaknaa)YoGcai) (4), which determines kn . Write An=CnYo (knal) , Bn=-CnJo(knai) to satisfy 5.-& (2) and (3). 111(knP)=YoOtrr3i) Ji(k.nP)-Jo(kn ai)Yi(kn p) (5) If z>0 Bp= -76 -kn z Bp = E oknCn Iti(knp) e (6) Multiply through by pRi(knP) d(kn p) and d"..ii— lkr1.1.11n.P.L 4:1 4 integrate a l to a2 . On left side have knia aB pRi(k odp -2 4a i gn(a 2 /a i) 2 4aiefft(b 2 /131 ) ai P n 0 by k2n,16a 2 pRi (knp) do = Yo (knai)Jo(kna) - Jo(kn a l) Y 0(knad- Ya(knai)Jo(knai) +Jo (knas) Yo ( (na,i) Ro(kna2) = 0 (4) a ukn I pRi(kr,p)dp So Sa2 Bn pRi(knp)dp = - Cn r 2 0( RI Ow))) 2d (kn P) 2 101gi2(b2 / b 2 / 131) 2&z(b al 4. 1(nCnta22ERI(cna2)] 2-a? CR1(Irna1)] 2) by 5.296 (5), (7) since Ro(krra2)= RP (kra a l ) = 0. 5.293 (9) can be written Jn (v)Yn _ i (v) - Yn (v)Jn _ i (v) = 2070 -1 so (5) is Rocna2) yo(kna2)
_f aQcri a211112.na2)1 Yo(kna 2) Substitute for every ratio from (4) so that 2
Lrrkn a2 Y0(k na2 )
Jon, ao I 2 "sn Trkn a 2 Jo(kna
(nikano2()akon a 2si c) !-2)]
4,10(k„a i ) 4_ Jo((rr32)YArrao)JpOrnaL Yo(k n_a_i)Ji(kn a2)Y0(kna) _
( a ) c r2 r_Yo(kn:2) a TT 2k 1(1::{2:dik .1c na22)12 2J0(irnai) (kn n aai2)) 21:(1,T (kri j Or RiOCria2)n 2d(kn o)= C11—Fr 7• n ptR (k p),I rrkn a2Ji (knad. But Ri(knai) - m nai ai 1( x) 1 2 2 Cn= 7T 21111.31(kna232 fRa (knq Ro(kabi)3/ [42//2 ( 32/b i)(knai)] - [J1(k n a 2)] 3..] Ans.
111(k na2)
rrkna2 Jo(kna
Jo(kn a2) 2
-32Add 22-pIz(p2in(a 2 /4)-1to (1) above. Alt=luIz{Pk(a 2/a 1)}-i + SoCn Ri (kn p)a -kr z -33-
Ans.
2
4 = t1Z Cn te -k-n (z+c)+ e -kn (z-c)}- iiiII[z+c+z-cl(POn (a2/a1)1 1 (e'kn(z+c)+ e-kn(z-c)1_ A j ( a 2 /ai ) J =nE1Cn So 45=2 E Ans. e -kn zcos h knc Ri(k np) -}az(Pk(a2 /ai )) n= 1Cn /,; = 2 IA Cne -knccoshkn 2 Ri(knP) -PIC[P62 (a2 /al )) Ans.
c1
Chapter VII. ci 42 4,r/ 2 TT/2 fr/21DI _ 171
- 34-
)71
7‘
Page 324.
dz/dz1 = (k.17F-14/77ca )-1=6/7-Z14,r7F cazt
i=se'lz i
zi= sn z, sn(K+jK' )=k-1, ma=K, mb=K I , a/b = K/KI.
W=2--Pat[(z -c + id )(z -c -id )]= 21.2fizt( -0 2 +d 2)=-1-11071 ((x 1 - c i)2 -yt+d ?+2j (xi -c y i ). If ci+di =zi 1 2r 1 - 1 1 1 -1 2r z1 2r +jd,= sn (mc + jmd) c+jd=z so from Handbook Mathematical Functions 16.21 p575: sn (mc) dn (md)j sn(mc)dn(mc)sn(md)dn(md) + . U=Pert((x -q2-yf+df]a+4(xi-ci) a Yi) 4r cn2(md)+ k sna (mc) sn2(md) cn2(md) +k sn2(mc) sna(md) )
=111-247C(x1-02+y213+2[(xi-c1 ) 2-yi]di+df) Ans. Subtract a similar expression for -I1at fiand g 4Tr This -I cancels the singularity at y=b produced by +I. Then from the above reference: sn(mx) dn (my) sn (mx) dn (mx)) sn (my) dn (my) Ans. cna (my) +k sna (mc) sn2(md) Y1 - cn 2(my) +k sn 2(mx) sn2(my) -35-
-bi dz p_„_thL
w=-41.iirrign((zrjb)(zi+ Jkl)(zi-1)
+WI ) = - 21T wil z?-1 ' dzi z sinh(2x/C) +jsin(2y/C1) i is rea1 if z=jc z = C' tank' z1 , z = tanh_, -r --lif--f-I u cosh(2x/C1)+ cos (2y/C') ' z _1_ 11_( 12 ilTtz/c)+t an2(-10113/c) . rb w.pionta 2c sin--0 = 0, CI= -F Vi cr-. . If zi= jbi , z= jb so jbi= tanh -- =j tan-27. 217 tanh 2(i-Trz /c)-1
112
Let zio=f (z0), zi*0 =-f (4) so that W=V2/71([ f(z)-f(z0)3[f(z)-f(zn i ] TT/a Verification : If z=jy and y>D j factors out of f (z), .z+jD -xo+jR1 kficr---2 Fxf, [-cos01 - jsinOik N ( ,u) On arc : z-jD--xo -jR 1 R 2-jx0 - s, /Rf+x 2 cos02 -jsin02 LR:+x since el , 02 factor ctoris i(T+e 1)/ e -je2 - e i (r+°1 + 2) =eja and eirr=-1. if r a2T/a i-m 277aria Thus z1 = _a ria_T/_ +l1eic TT/CFEaT/
I]
37 kz kz 2 +4,2 (k)e -
±k(z-a) -k pvI 0 < x < b, W1=-e + 2Tr j 0 e -k ty(k)e -kz- (114. b<x, 2pv ilvex -13)jdkk
-
-
x<0,
PI
W3=71.0
J k
kz
2p e dk (p+iiv )(1-0)] k
U is vector potential. Vila is scalar potential. Bound ary conditions are -k(b-a) kb+pvci,2e -kb -kb x=b, U1 =U2 (cosky), pve +pvtie (1) -k(b-a) kb -kb -kb -e + Ole -02e =-Ye (2) Vi /1117=172N (sinky) -ka -ka Ui= U3 . uve +pvci' l+pvt. 2=pe (3) (4) V1/Pv=V3/11. e -4 1 -42 8 (continued)
Chapter VII.
Page 98.
Page 325.
-37- (continued) -ka 2kb (5) (3) - (4) (p-pv )e +(p-pv )01 01 -(p-uv )1,2 = 0 (1) + (2): - (11-11v )e +(p+pv )e )(1+1/4)e+ka - (p_ poe-k9 +f(i_ p)2e2kb _ (p_ v) 2)01 . 0. - (u+ )0Z = 0 0'11v (6 ) 02. pe -ka+oxi . -ka k( a - 2b) -ka ka 2kb e +Pe +e e Pr2. 112-11 11v". SO 01 =13e1_ 13 2e - 2kb 4'2 (8) (9 ) p 2e- 2kb ka
e _2uv
k ( a -2b) -ka ka 2pv e +pe +Pe (10) (11) 1-1-Py 1- e akb 1.1+11v 1-13 ae" akb e-kz= -kb ) /(1_ 132e -2kb) 11) 1 (k) ek2-14 2 (k) 213e - kb( cosh k(z+a-b)+13e coshk(z-a) -kb e -k ) dkl cosh k(z+a-b)+Pe cosh k(z -a) ekb - f32 e-kb wl 217 Kilz(z -a) -213J o 1-p k -ka k( a -2b) pu.„I r kz e 1-kz eka+f3e -ka e -k dk +I3e W — 1-132 e- 2kb 1-132 e-kb 1-p k 2 r(P+Pv)Jo W 33 11(11+11v )J0 e
-ka
e dk 1-13 k
-38- (Boundary conditions check) c 2_ (-11vI)"l211W 1= [PA( z -c)+ 1307 [4 c 2- (z+d) 2] +132Zn[1.6 c2 -(z-d)2] +13361[36c 2 _ ( z+d) + • • -2//z2c _ fapyle 2- 3071 24c 2- p 32/n.48c 11W2= {- ppvI/(u+ pv)}{Pnt(z-d) + 132112 (2 c +z+d ) +1322/n(4c+z-d) 1-133&t (6c+z+d) + • • -0n2c - pem4c -132k6c -13 30728c - - • TN 3= {-11Pii (}1+1.1a(2/7Z(d -z)+072(2c-z-d)+13 2k(4c-z+d)+133071(6c-z-d)+..•k2c-p0n4c-p2k6c -133243c- • • (-11v1) -12rW 2= (z-d)+PPir.[(z-d)(2c+z+d)]+13 22./Q2c+z+d)(4c+z-d)1+13 312/71[(4c+z-d)(6c+ztd)] + • • • -0n2c - pent8c 2 -822/n24C 2 - 832M 48C2 - • • • 3 ( -Pv I rl2rW3= [PAO - +Pk [(3- Z)(2C- -C) ]-432Pill R2C-Z -CD (4c - z+d )j+1330n[(4c -z -Fd)(6c-z-d)] + • • • -0112c -Pan8C2-1320n24c 2 - P3S,248C 2 - • • • • 3
At z=c+jy, 1.11 = U 2 , pV1=p,,V2
2171.11= Pvl r [''Cl) 2 +y2] +NA [(2c - c - 02 +y2 ][(2c+c+d) 2+y2 31-1322na4c -c+d )2 -Fy 2R4c+c - d) a÷ya -
+1332nza 6c -d -c) 2+y2] [(6c+c+d )2 +y2] + • • • +0= -pvl fem [(c- 2+y 2] +13Pin[(c -d) 2+y 2][(3c+d) 24y2 ] -H3 Viz [(3c+d)2+y 2] [(5c-d)2+y2] + 1331/n[(5c -d) 2 +y2][ (7C+d) 2-Fy 2] + • • • • +C . 21rij2= (Pm [(c- d) 2 + y2]+ P2 r(C- d) 2 +y2][0e+d) 2+y 2] 11;22/71 [oc+d)2+y2][(5e..d) 2+y2 ]4133pin Dc.d) 2+y 21(1.44 2.457 9.. e_ d]+• • . 2rrV1= -pvI f t any :7+ P[tan-ki.:Ya + tan-'37L +d - ]+133[t an-i-j- 4 +tan-15-Y—
2rrv2= 21'114= -pl
+1k-tan-1j.Y.Ti + tan-13--cY:cd-]+133[-tan-
- -ZEIL - 1-13 so that 11 11+1-1v 11+Pv
-Itan-5 27d ]+• •. At z=-c+jy: U1= U3,
pee4.0 2 +y2]+ VARc + d) 2+ yai(d- 3 c) 2+y 2]+ [3 2P./n[(d- 3c) 2+y2][5c +d)2+y2] +13 304(5 c+d)2 +y 2][(7e -d)2 +y2] +. +C
217113=2rrU1 asitishould . 21TV1 =-pvI{tan"1 (C
+p[tan-3-e
+tan-1 --Y-e±d ] + f3 2[tan- + tan-t Y7_71 ] +
217V3=-p.It t an -1 --Y-- +p[- tan-1 +tan -'--=Y---]-1-13 2[- tan-L-1Y— + (c+d) 3c-d c+d 3c-d This checks the V boundary condition.
-I- • • • 3 5c+d ]
1 43 .
Chapter VII.
Page 326.
Page 99.
-39a cos *d* From 7.10: Agl In 4.13(2) coordinates 2ap- 2a 2sinhu -- tirfola ' ir -F 2 - 2apcoakr 2apcoshu coshui -cos u2' i_ 2a2coshui From 4.13(2.1), U=ui , V=u2, r2 +a2 = so that sinh ul cosh ui - cos ui '11 a cos* cl* A__3.1I 4/cosh u,-c os u2 r 9g'-211 4Oolcoshui -4/COsh 2o i -1 cos* . The integral is that of Nei a 5.23 (15) with z=coshui , til, u=-1/2 and *=17-0 so that : i cr nn
Ao = -11IN/(cosh uccos us) /2 EI/ 2(cosh u2) Ans . Toroidal Coordinates. -40-
,...----
r,rg , Since A =A cos sts satisfies 72A =0 satisfies differential equation of Y 'A0 Y 0 I V - 117 with m=1. Q1 1 (cosh Ili) is out, being infinite at u1 = 0. Added A0 must be of form -T co A; = Al(cosh ul-cos u2 )/2 311 E A_P 11 (cosh ui)cosnu2 (1) and from 111-8 hi =h2=a/(coshuccosu3) 1n 113=11 2 sinh ul . By 3.04 (2), B is tangent to Ili= uo if (7xA0 )2 = Bi= 0 = b(h3 A0 )/(h2h 3 6u 2) so (coshuo -cos u2) 2 6(A0 / (cosh uo - cos u2)) /6112 =0. Let Co =coshuo, differentiate A0+A;6 , divide by N/C0 - cosh uo get : 0=isinu2 Pli+n ( ' sin u2cos nufF(Co-cos u2 )n sin nu 2 }P1 'An . Rearrange of n -7 i1-2= 1 in u2P l i +-Lyn[4nCosin nu2 - (2n -1) s in (n+1) u2- (2n+1) sin(n-l)u21 P 1 1 (2) trig. terms gives : (3,i.s n-T i " This holds for all values of u2 so coefficient of each sin mug term is zero. Thus for m=1, • 2P 1 1-1-4A 2 C0 P1-5A 2P=0 : m=2, -A 2 131+8A 2 Co P1-7A 3P1=0 and for the general case, m=p, 7 7 7 -7 7 1 DQ 1 1 - (2p-3)A ,P1 3 +4pA Co P ' 1 - (2p+3)A P1 1 =0 (3) Let A-P 1 1 =---T- (4) F p--i (2p+1)(2p- 1) p-i p-T p+1 p+T P -T P We cannot use P 1 (Co) which gives result independent of C o . Multiply by (2p+1X2p-1) /D. n-I7 (2p+1)Q 1 3(C 0 -4pC0 Q 1I (CO+ (2p-1)Qp1 4(Co) = 0. This satisfies 5.233 (5) with p4 for n 13 -7 P -3 and m= 1 so An choice is correct. For D insert in (3) and use 5.233 (5) A = -2Pl i(Co )Q1 I (Co )/ [(2p-1) (2p+1)Q1 1 (Co)P 11 (Co)} D= -2P 1I (Co) /Q' 1, p-T -7 P -7 13 i "1 "7 The total vector potential is therefore, writing cosh u o for Co , etc. A = -“I
0
Vcosh u, -cos u2] P1 v2(coshun) 1 E (2-61N,L,12(coshuopiL i/2 (coshuo cos nu 2 (2n-1) ( 2n+1)P11_ 2 / 2 (cosh uo) Q1 2/2(coshuo) n=0 N/2
Ans.
Page 100.
Page 326.
Chapter VII.
-41This is a superposition of the fields of coaxial circular loops so the vector potential must be A0, independent of 0, and the nth term must have the periodicity cos Nz (plus harmonics). Thus, from VII- 26 the forms when p < a and p> a must be Ii(Np) cos Nz and K l (Np)cosN z . The potentials must match at p=a and, from 7.02 (5), A must be proportional to
0
when r is small. Hence the form given. -42-
B = (417)-111r2 ds x
ds = i ds sine +j ds cos°
ds x RI = i dscose sin* Ids sine sin* ,/
/1
+k ds cosS(cose sinfzi + sine cos0) cos0 = (p-s cose) /ro , sin0 = s sine/ro ,
-
sin4f= z/R, cos* = ro /R
dsi x R1 =
(i ds • z cos° - 1 ds • z sine + lcds • p sine) /R. laz cos() de pIz sine de _pIp sine de d Bx = --5-- ds, d Bz ---T-s---4ff Tr rids 4m1 *Tr ds' dBy- - 417R3 • _pIz rw r r-a cos() de ds 3 , 31122Slef ma sine dsde 3 B x-417 2.1 0,1-a (z2 +p 2 +s 2-2p s COO) 2) .°Y7417 Co a (za-i-p 2 +0 -2p s cos()) a in ededs ds 1 3 2. where By Dw r 3 B 0 a (z a+ps 2+s2_ 2pscos0) 2 z Y .857.02 ..10 0-1-p 2+s 2 -2pscose)l - r(r-pcose) ra=p2 +z2 I
B z 4rr
(r+p)cot(a/2) z rr-a cosede p.I Ez - 4172p r ( r-fa-a) + 2( tan-1 4r2r4-a r-pcose 1
.r..(t
4r 2p
r
an
___I/c_ ztar c/2)
tan
r+p
(r+p)tan(a/2)
tan-
(r+p)Etan (Ce/2)]].]
,0
Then by Hd.M.F. 4.4.34
( 2P 2 +2 rp)cos (cc/2)sin (a/2) -tan-1 p sina z(r+p) Tr-a sinede sisal ce r-ap .i _ 1 -1 - F2+1+ -atan-1 p Thus B --2 Pn(r-p cos° B Iz x 4rp rTr r-pcose 4ff 2rp z ' Y 411 2rp -a = _az onr+rp co sa 'Az pcosa . At P directions of y and p and of x and 0 coincide tanh pcosa 2r2 rp 4r2rp p cosCX so that Bo =13x=4 Ans. If 0 is the and -EB = B = -21--na tan-1-a-rro - [11-tanh r z 2r2r z azimuth angle measured from the plate edge 0=a. tz z(r+p){1+tan2 (a/2)}
-
-432n+1 sin a I(cos a )P By 10.17(1) dA0 = p.Idzon .(cos 9) for P2n+1 21n+1 g0 (2n+1)(2n+ 2) 1 two ring elements of width dzo at ±zo . The z-dependent terms are a VIV rd a-2n-1 P P1 (cosa). To get A.0, j 2-ri+i (cos a )dz o must be found. a2=c2+4 f 2n-I-1. o dzo 2 rd dzo 3 d Sin a=c/a, cos ci=z0/a, sin3=c/b, cosp=d /b. If n=0: fosin2a7 =c j 0 (c2+4) 2 b cosh. ,db2n+2 a la 1 I o 2n+ i ar -sin B rdsina _ For n> 0 by 10.17 (3) :j --rTrr 1jo (2ny„ 6z0 27TI i 0 -(20:b in P2n (cosh) oa n 2n+1(cosa)dz0 - o zo2n+2 (24; =--[1 2n+1 -P„(cos0) So A0= piobsinBACn PI (cose), Co=c0tP, Ans. Cn - 2n (2n+1) (2n+2) 2n+1 -
(continued)
Chapter VII.
Page 326.
Page 101.
-43- (continued) The solenoid of this problem is the superposition of an infinite solenoid (external field zero) and an identical reverse solenoid with a gap 2d as shown in Fig. 10.17, so the A0 outside the solenoid and inside the sphere of radius d is given by 10.17 (5) and (6). Cnis the same as before but the Cointegral (5) is 1-cosp instead of cosP so Co = -4(cscp-cote). -44Solutions for A0 finite at p=0, P== and zero at z=c that match at p = a are nrpnrz TIM nna nrz nra )-E- if p>a. = nEiCnIi(—E-)Ki( c )cos Ao=jiCnII,( c )K1(--F-)cos-E- if p
6(P(AA8)) By 5.32 (7): Bz-B'z . p bp
nnz 1 c° nrz E C cos---. Multiply by cos-F-dz and integrate P+aainc
from -c/2 to c/2.. Left side is zero except at z=b. By 7.01 (2) fB•ds = pi so that c/2 _p_ CaI cosnr c b. pIcos(nrb/c)=(Cn/a)j cifos 2(nnz/c)dz=i-cCn/a. Thus Cn _ 2 2-pal
nEIT I (nrP/c)K i(nra/c)cos(nrz/c)
Ans.
Flux through loop is 2naA0= N =LI since plea. Energy is ;A lit =W. By 8.13 (3) the force nI fna‘K n tnnbIsin2ncrb is 6W/ob= g1TaI/c)2n-1 I` c ' I` c -45Flux through ring of width dp, dz =0, at p,z is Bz•21Tp dp so ON/6p = 2npBz.
(1) Ans.
Flux through ring of width dz, dp = 0, at p,z is-B0-2nP dz so ON/Oz =-2npBp. 62N O 2N OB, 1 6N oBz . op= 2TrBz+ 2rrpF1; = p + 2Trp-5 (3) = - 2rP rzz
(2) Ans.
-
From 7.01 (3), if i = 0, vxB= 0 so CvxBlo = 0 =
ap a z_
aZ
BP
o2 N 1 ON + 02 N .0. 6P2 P 6P 6z2 From 7.21 (10), normal flux is continuous so N 1 = N2
Add (3) and (4) and use (5):
~
(5)
(III-1)
(6)
Ans.
Ans. (7) From 7.21 (11), tangential flux is continuous or 112 n xBi = 'J i nx Ba (8) Let n= E1 -4-6 =k , o x0 1 =k1712!Axkfr P.a ,koc El = 111x Bz = - 1 Bz -p Bz , n x So by (5) n x B = -EI Bz+lci By. By (1),(2),(8):
6N
112[4
z
(4)
=
By-1)13p •
6N ON 72-1-1,5„, e 21 Ans. p i tr p...),Fr
Page 344. (322)
Chapter VIII.
-12xdx (a+b( 14 -a) Ans. L =,fBdS = a+b-x- (a+b)On(a+b-x)ro 2 0.13(a+b-x) Nr3Trt +2z2n031,.14- b -30d x) _:2-rail-b+[(a+b-x)Pin(a+b-x)-(a+b-30]:3 or L=SA.ds 7.13-ITL 4"
1E2aaiii
-2From 3.132, image loop of strength (p-pv)/(p+pv) gives added self-induct. 2
By 8.06(1),a=b,c=2d:60=131-2ILEY((2-k2)K(k)-2E(k)), k p+pv
k2= a a 2+d 2
Ans.
-3
If b
r ffi+ Anain1P1 (cosa)P(cos 0) by 7.06 (2) s 2nn=1Ln(n+1) ~
outside sphere, if I=1. At r=b, Be=0=o(rA0)/(ror), so n-I -nAn=0 or An=n-2(b/a)n . 2n+1 2 n n+la_ Arls=lpvsina nF111(cos(x)F1(cos0). AL isl2raAola"=17/2 so that /(n a r r=a 2 1 [12n co [1.3.5 2 (2m+1)1 2 tbr 2 r2m+21 AL=Trbpvn.Zi T Ans. 12 ; P111(0) = rbu E - vm0 2.4'6...(2m+2)j L2m+1 {
}
-4 =r/2 From 7.06 (2), M= 2rbA0 a=0 r.b
I=1. Then, since 13111(0)=-(n+1)P114.1(0), n Lv E b ]tit p 1 2 _ rtbp4, r1•3.5—(2m+1) 2m+2r12m+1 = n=1 n(n+1) a m2 n 0/ I '4;2 01. 2.4.6'-(2m)] 2 2m+1La
Ans.
-5The outer shell current density is uniform (neglect Hall effect) so no force T acts on inner wire and self inductance is unchanged by moving it. Thus the concentric value holds for any position. Thus, between wire and shell ra[gvIr 21/rdr pv a LA - - On- . Add 8.09 (2) for inner wire. L= Pv(1+40) pv .-1 13 2rr pv 2r b 177
Ans.
-6-
J1
The same field configuration, crossing the plane of the circuit normally, will satisfy the boundary cond ition before and after the permeable material is in place but a different B value is
needed.
Consider any tube of force be
after inserting the permeable material. The reluctance of half the tube has been changed by the ratio pv/p. The magnetomotance is unchanged (7.18) so the flux ratio is N/N 1 = 2/0_41139= 2p/(P+Pv) for any tube so L/L'= 211/( 1+11v)
Ans.
Page 345 . ( 323)
Chapter VIII.
-7C4/zr,
A=
217 a2 sin2a + C 2 a2 cos 2a
M= LEvacos ar 2i
cosant r2d25.
ccos0 cost 0 = (oos0 -2 + a cost a
-
Page 103.
sin2 Osin2a +(c -a cos0) 2 aacos2 i3 c±sina(c2-aacos2C4)1 B) . A, B= a costa
r2 a cos%
A) (coo
-
and N/A 2 ,B 2 -1= ±(c sin a t%/c2 -a2 cosaa) / (a cos2a) . A+ B = c / (a cos 2 a) e prr r, TT M = iiva cos a (2 11) -1 tjocosok(cosis A) do + Jo oosran (cos 0 B) 4) -
-
-
v= sink ,Ty ,77 (1-A2 )d0 r pr 0 dol _ rrsin2 --- f' cosh dO - j o Ado +J rfAA 27T. - ft an-1( -WI: :2‘ tan-- )1 = -17(A+ 4171 ) J o udv = 00+j o cosh - A cosh - A 2o o 1 M =-Tpvsec a (c+sin,./c 2 a 2 cos2Ct .... -± c sin a _±,../c2 -a 2 cos 2a +c-sinaN/c2-a2cos2a -±c sina T•lo 2- a2 cos 2a u =07z (cos 0 -A), du= -sins6/(cos0 - A), dv= cos0 do,
-
If a >c must choose same sign for ATA 3•7 1 and N 2 -1 to eliminate radical so that 1 s in pvotan -11 . Ans. Chose - for ± since M=0 if a=17/2. M= pv seca(c±c sin a )=Pvc sa -
If c >a take opposite signs for 4/E171 and B2 -1 and - for t to give M = pv secafc-A/c 2 -a2 cos 2a) = pv{c sec a Thic Isec2 a - a2 }
Ans.
-8For uniform field Az0=B(r + CI r-1 ) sine, Azi=BDr sine. At r=b: Azo=Azi so bz+Ci = Db2 , - PI6Azo/ar= 1126Azi /Or so p.i (b2 -CI )=u2 Db2. Thus I'.21/1 / (111+112 ) , = ii 1 jj ( 217r ) •8F14(4) d Fn= i [Li1; C I=C(pi-122 )/(p1+112 )lb 2 . For 11 Be .11 11 2 r p--111 - .1 2 Pa r.217 p r Bs ine]2+ 411? B 2 c os 20B 1 (Pi -11 411 f [ I (pi +p 2) 2 . Fx= No F ncose de =0, F ' = bj F sine de +p2 Y 4nr Jai o n 1.11 —ET 2 r "b21Bsine Ans. F' is on surface of insulation. =2(p2 - P1)j0 Orr (pi +112)sine de= IB [P2Pil Y 11 2+11Li - F”Y = 2 1-1plIB Force on inner wire is IB x 11 14-p Ans. Total force on system is 2 1+1-1 2 = Ft +F" = IB . Ans. FYY Y -9 See VII-17. A00= iB(r +Cr-2 ) sine, A10= -1-BDr sine, C= 2a 3(112 -PO/01 2+211 0 . Flux through sphere is rra 2B1= 2rraA0 at r=a, El=r/ 2.Bi=312213/(12 2+ 2p1). From 8 . 15 AW=ALi/22ppv u-uv 417,33 2 (p-pvi 3pB 2 (p-pv 4ra3 )4rra 3B 2 =____ B B. Ans. DL= 3 2ppv (11+2 ilv) 2111.1v 3 liv (11+211v)
B • Bid v -
- 10SOlve 7.15 (6), (7), (8) and (9) for A2n+i . Substitute in 7.15 (3) for shell Al. (0_01(a4n+2_0n+2) PvI ' 1 rcl4n+2 A•• TT n=02n+11-71 01 _1102 a c61- 1_ ( 31+1102 b 4n-rt • Integrate AI along two wires of unit lengi 4n+2 4n+2 -b 2pv 1. 2_-_07 ) I 1 rs_ 4n+2 Ans. OL = 2A =-- rr L J 4n_ (1111v) 2 b 4n4.2 (g-pv )2 a a+2 n=02n+1 a
Chapter VIII.
Page 345. (323)
Page 104.
- 11CA 10 = +pva 2 (Ci r+r-2)sinO, A20= ÷pva2 (C2 r +
A30 = 4-gva2C4 r-2sinO.
At boundary A10 = A20 112 NrAios )/Or = p1 b(rA 2 0)/br so at r=b: C1b3+ 1 = C2b3+ C3 (1)
11 ( 2C1b3-1)=1-1V ( 2C2b 3 C) (2) . At r=c: C2c3+C3= C4 (3), pv (2C2c3-C3 )= -pC4 (4) (3), (4) C2(p+2pv )c 3+(p-pv)C3 =0 (5) (1), (5) (p-pv )(C2b3+1) = (u-pv)b3- (p-2pv)c9C2 (6) (1), (2) (p.v+2p) C2b3- (p-pv) = 3p.vC 2 (7) . Cia3([3Pv(11-Pv);(11-P-X11 +211nb3+ +21 ) (11v + 2p) c33 11-11v)i.-(2Py+P)vb3+ (2P +P)a3/ =-3pvb3(p -pv ) - (11-Pv )2 c3+ (1-1- P)(11+ 2320C3. C1b3LS: 2 (11-'1 12b3 + ("+2)1v) (171.1v+2P)C3.1 • c3-133) :"2:v' Ans. AL= 2TTaAA1 at r=a so AL=Pv7Tatil-Pj(2A7-" 2b3[(p+2pvXpv+2P)c3- (P-Pv)- b (
2
-12Set t o=g=t, and 11 0=71=71 1 in VII-25. Multiply by 2Tra, setill..-011-1 =a /c 2 , Then
AL=
2n+1 (11-14Ptigo) Ori Cno)1 2PAM c)Pn o) Ans. ITAva 2 I c 1;1n2(n +1) PvCM (71 0 )Pii o -PQn(rl o)Pili (710) -13-
Set t o=
= a i /c. Multiply by 2ra. 11 0=71 i , t= 2 1 1l to in VII-25./117-T1 2n+1 (1 1-1-1v)P1; (ti)Pri(t2) acrIOQA(T10131-11(71o)Pr,(71o) n=1n2 (n+1) Pv4n(Tio)Pn( rlo) -11Qn (rlo)P/11(710)
Ans.
-14Write p=c and z=s in VII-28, multiply by 2rc, and set I=1.
These give
AM = 2pvacro fb(k)K 1 (ka)K 1 (kc)cosks dk
Ans.
-15Write p=a and z=0 in VII-28, multiply by 2ffa and set I=1.
These give
PM
DL = 2pva 2j0cl, (k)(K 1 (ka))2 dk
Ans.
-16dk 2ra 2 --1411(kal)K1(ka2)coskzcoskci rx By VII-29 d z 2 EMF,C, with 11=1 is —1. .11va dz current is Cdz/(2TTa2c) so I 2 =rf4edz/(217a2c7). Total Resistance is 2c R= 2Tra 2c/(2c2). Effective EMI' from 11 =1 alone is IR so that z 24i ia2k M=lva a2 ecrIi(ka)Ki(ka)cos kzsinkcid 411(ka0K(kaa sinkc i sinkc2T , 2 Sk c2 -17-
116---- 2 a
In VII-30 set z=0, p=a so additional self inductance is, taking A at z=0, p=a, - 2k t -2kb -2kt -2kb , F(p2- (1) 1 702e _Itt 2 2iTaA = TTPv a-Jo 01+1.\)2_ 61_42 e - Zkt (ka)) 2dk= Trpva2 131:` {J1(ka)} dk. Expand t-tel -pkt -2kt -4kt 4 -6kt by Dw 9.04. 1-e =1- (1-132 )(e -1g e +p e +•••) .Trpva 2 krocte-2kb_ (1_ p2)n boo2 (n-1) e-2k (b+nt ) 3 ji 12 k . By VII-27, M for coaxial coils -kzr_ 8.06 4a2 of radius a at distance z apart is Tr7a.va2 jo e (ka)3 2dk --=-17[(1-2-)K-E)k2= 1 ka 2 2 (1),(2 4a 2+z2 a _ 1-1 V . So LLD= p[mo (1_0) E R 2(11-1) 1,4n3, Ans P---'n=1" a2+ (b+nt) 2, p+pv 2
_
Page 105.
Page 346-(324)
Chapter VIII.
-18r.co r 2 - 2kt,By VII-30 M 1 2 =2TrbA 3=rrpvab(1-81)j Ji(kb) Ji(ka)e -keLl -8 e idk. -k(c+2nt) =rruvab(1-132) nio2ni e VD) Ji(lca )dk . By VII - 27 , rrpva brJi(ka) Ji(kb)e-lczdk ?ri m k2 4ab k2 (a+b)2±c2 by 8.06(1), (2) . Thus M12= 2us/Z(1-82) -k-;-{(1 -T)K - El ,
where kn = 4ab/( (a+b) 2+(c+2n0 2 ),
An s
13 = ( Pv ) / ( P+P ) •
-19(2 sink) -1) an Current in element du By 7.10 (3), for loop, A0= Py— ITTd 017/2 `r z2+d 2 cos2 0
d
is I '=Nc-idu, in dz it is I=Nc--;dz. Flux linking dz is (Prrd/c)rAfzs (u)du. For unit current dM4Nd 2c If;zfolle2 sin 20 -1)( u2+d 2c os 2ElrOdedu iaNd 2
r11/2
r
0 ( 2sin2e -1)1sinh-C(c -z)/(dcos0)3+sinh-l [z/(dcose) ]) de. M= (N/c)SocdM
so M=pd 2N 2 c -2f 77/2(2sin2 0-3).rsinh-1(z/(dcose))dzde, since arc sine integrals
dz
0 72(2 sin20-1)(csinh-i[c/(d cose)]-..✓ c 2 +d 2cos 20 +d cose}de are same. m= pv2 d2c -2: For arcsinh integral let u=csinh[c/(dcose)] , du= (c2 tanekica+d 2 cos 2 0)de. dv= (2 sin2 0-1)de, v=-sinecose. ludv= luvi 1712 -Svdu=0 c247112 (sin2 eac 2+d 2 cos 2 e)de osin20 =4/7:C c 2+d2 cos29= la7 i.(12 Let k2= d2/(c 2+d2) , ,/ -d2 P so that M= }1(Nd/c)2 ( c 2 +d 2)-1/2(.172[c2 p-isin 2 0 -2(c2 +d 2 )Psin2 e (c2+d 2 )133de+ de} -1)3 - k2 sin2 ecos2 e/P = (2-3sin2 e)P - (1-sin2 0)P 4 so that rrr/ 2p-lsin2 ode. d 2 /c 2 k2/ 2 04A-k2 s in2 Ode = tri2Pd0 - -kr/ 2P- de +to (1-k2 ). Thus
But t (Psinecose) /be = (1-2sin2
(1-k 2 )sin2 0 4P1 2sin2 0+p]do di_ PN 2d f(1- k2)E+ ( 2k2- 1)K- k3 ) 3k(1-k2 ) 3 3P 3P 3 uN 2d k2 If k=sina, (2k2 -1)/(1-k2 )=tan2 a - 1, 77 1 (2 - tan2a M= 3 [csca[(tan2 a-1)E÷K]-tan 2a) Ans.
M=p(Nd/c)2 U rri 2{
-20riD+2 , A0dz, where A is vector potential M = 271,9146, F = -II ' bz= -2TranIV-6-p-j of ring dz at z with unit current I. so F = -2TranII (1141 04 ] - 11-( 2 [ (1 - --i2)K2 - E2 ]) Ans. [ (1 - 2 F = 2grar where k =
4ab (a+b)2 (p+L)2
4ab da
4ab 11=(a+b)2 + p2
4ab
c2
Page106.
Page 346. (324)
Chapter VIII.
-212 sin20-1)de]dudz by 8.03 (2) and 7.10 (3) where p=a=d/2. +d cos a 1 c-a+z c-a+z .. . /1d2nmr/2(2 sin2 0-14r0 t sinh" sinh-1-2— 1 +z-ldz] de, Let uetc. 2 o dcose dcose dcose'. f,a_ rr /dcose . ..1 r(a+b)/dcose -dcose u du sinh-lvdvj= fc sinh-id—C— Lj(cc-a)/dcosesinh Jo cose 4/c 2 +d 2 cos 1 0 jb/dcose
t41/20 pc-a+z 1 lid 2mn j 0 j M = -1 0 .1 b+z
- (c -a) s inh-11=41-- +A/(c -a) 2 +d2COS 203 - (a+b)s inh- a+b +W(a+b)24-d2cos20+bsinh-1-1-dcose dcose dcose r rr / -Wb2 +d2 cos2 O. All integrals have form j (2sin2e-1){Psinh-l— P -A/P2+d 2 cos 2 e}de,„..._ d dcose
The value of this integral as found in VIII-19, if tanft= r=cotal k=sin8=cosC4 is P2 (tan2 13-1)E-K tan 2agcot2 a-1)E-K) d
3d
sin
3cosCt
- 3tseca[(1-tan2a)E-tan2a1G . Thus
M= (Pnmd 3A)nti( -1)nsecag(1- tan2 ari )E(kn ) -tatl aCtnICTI (kr, )
Ans. Where kecos an ,
t ana l = cid, tana2 = (c-a)/d,
a4=
tan a 3 = (a+b)/d,
tan
b/d. Ans.
-22From figure in last problem observe that if one coil is displaced axially 6,13=k,c so that OM/bb = OM /ac. Differentiation of the first expression in last problem rT1/2ra 2sin2 0 - 1 gives, since b and c appear in limits, F = II ' OM/bb = 1 j_ j u 0 c-a+z) +d 2 cos E sinh- a+b snh_ 2sin29-1 dzde= liud2 nmIff:72(2sinae-1)(sinh-idcose c -sinks d cose 'd dcose d cose rrT/2 sin 2 ede _ co_tp [K(k)-E(k)) k=sinp From VIII-19: ri2 (2sin2 0-1)sinh-1---C--de tio c+ • 77= - sing d cose (-1)rcos8 tan13=d/c. Thus force is F =21pd2nmIr sinpn 'nfE(kn)-1((kn)) . If tan cc'= c/d etc. as in VIII-21 above: F =
(-1)n sec2 asinan(E(kn )-K(kn )), kn=cosan -23-
From figure: At= AIcosa = A (c cos0-b)(c2+b2 -2bc cos0 )
_ =-Ai bR/bb.
By VII-26 I= 1, A i=(1.1a/11),10 I1 (ka)Ki(kR) cos kdodk, M= brAtdczs. K i(kR)oR/ob = olCo(kR) /(kbb) so M- V0°I 1(ka)(OS0 Ko (kR)d0/abi cos lad, By 5.35(4): K0(1(11)= K0(kc)I 0 (kb) +2mE1Km (kc)Im(kb)cos A 0-integration, 0 to 2r, removes the summation so that 217j:1 i (ka ) (kb)K0 (kc)cos kddk . Ans. From Watson b2m-2s+l a 2s+lt2m+2 Bes. Fctn. p148: I I(t 11 ( tb)= =m-0 s- 04m+I (m+l-s):(m-s): (s +2): s: • By 5.40 (3): $0t2p cos ztKo (ct)dt = i(-1)1311(2p): r -2P-1P (cos()) 2p m+l 2m-2s 2s OD m (-1) a (2m+2)'. P (cos()) b 2rrt+ 2 M 4.4a2 b Ans 2 TTMEOSE0 4m (m+l-s )1. (s+2)'. r2M+3
Ans
Chapter VIII.
Page 347. (325)
Page 108.
-29M12= =
ds •ds By8 .03 (2) . ds i , ds 2 components 0 12 I " are: parallel to c, ds sino and ds' sine (1) .
normal to c in a-plane; ds cos, ds' cos0bosa. (2) So ds•de = ab(cos0costecosa - sin0 sine )do =-ab(cos0cos0' cos + sin0 sinfind0 de. R component parallel to c: Re =c-acos0-bcos0'. R component perpendicular to c and vertical: Bcos a +A-b sine sins = - B cos 8+A-B singe sink. R component perpendicular to c and horizontal: B sina -a sin0+b sinsecos a= B sin (3 -a sings - b sinecos 8. Square R components and add :
R11,= c - 2ca cos0 -2cbcos0' + a 2 cos2 0+b2 cos201 + 2ab cos0 cose R2
B 2 sin 2 B-2Ba sin8 sin0 -2Bbsintkos (3 sine +a sin 20+ba sin2 0cos 2 +2ab sin0 sine cos
RCILL 2 =B2cos 2-2ABcos 13+A2-2Ab sinfitsin0l +2Bb sinficos 13 sine +112 sin20' Si.28 R2 = D 2 -2( a (cos0+B sin8 sin0) + b (cose +A sinfl sine)) +a2 +b 2+2ab(cos0 cos0' + s in0 sinecos 8) Ans.
where R_ =r i 2
-30z =e ja
a .!_rf 2‘ a /a -Trzia - 2sin /a) __pi a zf-2z i cosa+1_ a pelee.rrzia. (z 1 -ej a) ($ 1 -e -ja) pI W= Pe2 i 2r 2rr z f+2z i cosa+1 (rr-")) (11-a ) (z1 2rr (z +e a+2sin(irc /a) z
= _a_e/Izz Ty
11
e
1
/
r =enx/a . If _ a-c z- 2
)
U is Az , the vector potential. Let x=0, y=i(a-c)+E so that -cos(yrtc/a)sin(TrE/a) U_ _g_gn cos (ry/a) - sin(ric/a) _4 -pIgm 2rr • 2rr cos(ry/a) + sin(rc/c) 2sin (-11c /a) For currents without walls U1 = -1142/n . The AL is the Az difference at the wire surface with and without walls Tic pi (17E a)cos (re /a) c I Tic • e = - on(--cot-3= , , I = - Pili 1U2 2r 2r 2a 2r 2a 2sin(ireis)
so that 2a Trc (—tan--) Tic 2a Ans.
,a
Preceding page blank
Page 348.
Chapter VIII.
Page 109.
-31m (-1) a 2m+1F(m, -m-42;b2 a-2 ) 2m+2 k . m=0 4171+ 1m (m+1)
From Watson Bes. Fctns. p 128; Ji (ak)Ji (bk)= b E From Hd. M. F. 15.4.14 and 8.2.5.: b 2 m+1 -1 a2 +b 2 b2 a F(-m, -m-1;2;72) =131. 1--l a 32 Prn.+1 {1717.7
}
rso+13 2 ue (a 2 -b2 )m+1 (m+2): Pm+1Laa-b 23 ba 2131+1
Substitute in VIII-25 m m-Fl co (-1) (a2-132 ) aa +b re° 2m+2 -kd M= 27Tuab E k e Jo (kc) dk P1 (-3j m=0(2m+2)::(2m-1-4):: m+1 a 2 -ba o From Int. Trans. VI (9) p182 in the integral let 2 = 2m+2, V = 0, a =c, E =d, r2=d 2+c 2 -2m-3 and cos a = d/r. Then the integral becomes (2m+2): r P 2m+2 (cos a) so i m(2m+1)::(sa -b2 )m+1 1 r aa+b2 M= 21Vab 3P2m+2 ( c o sa) Ans. (2m+4):: r 2m+3 i m+1‘ a 2 b2 -32(2m+2): By Hd.M.F. 15.1.20: F(-m, -m-1;2;1)- (m+i),. 01+2): J1(ak) of VIII-26 to get {J1(ak)}2= from VIII-31:
M=
2 2m+3(2m+1):: . Insert this in (2111+4)2
co 2m+2 (2m+1)::]ka /fm: (m+1): (2m+4)::). Then m= E0 [2a
m: (m+1): (2m+4)'.:
2m+2
-33Like VIII-31 but with a factor (4AB)-1 and a new k-integral. In VIII-28 write -kr2 L e 2+e -kC2 3-e -1( 4 ) where 01=-A-B+d, 02=-A-B+d, no=A+B+d sink kA sinh kB = 1. (e and Q4=A-B+d. This gives four integrals of the form of VIII-31. Thus we get rco 2m -k t o k Jo (kc)e k s i
4 -2m-1 P 2m(cos as ) sdk = s;1(2m)!rs
4 -2m-1 2 2 where r s =Os +c2, cos as =C2s /r. So VIII-31 integral is 4-(2m): sEirs P 2m(cos as ) • m+1 (_,)sp (cosa) a 2 4 ,2 (-1)m(2m-1)::(aa-b2) q rg - rrmabmn Ans. P' 3 rim2 T 1 m+1 a 2 -b 2 s= 8AB m=0 (2m+2)(2m+4)::
t
-34Use of (J1(ak)3 2 as in VIII-31 gives (2m+2):a2m+3 /{m!(m+1):} for b(a2 -1)2 ) (-1)m (2m-1): (2m+2). a 2m+3 (-1) P(cos a) in VIII-33 so M- namn 8AB m=0 rn: (m+1): (2m+2)(2m+4):: s=1 r 2111+1
1P110.÷i{ a12. Ans.
21
Page 114
Page 363. (435)
Chapter IX. (XII)
1/4
'1
1,
1/4
Lx = lOsin30°sin30°= 2.5 ft. x=212.5 ft. h= l0sin30°cos30°=5sin60°=4.33fi: -2Starts from rest at 0 0 . d2 0/dt 2=-C sine. -1 (dO/dt)2 =C sine e = C(cos0 - coseo) r0 j i,(cos0-coseo ril2 d0 =.../FCT/4. Let 0=20, A/1-cos00 -1+cos0 =Arksin2 0 0 - sin20. Let sin00 do _ singocosu du _ 4/sir1 20, -in20clu If 0=0, 0=0, u=0. If. E*00,0=00, u=17/2 so that Ni1- sin O0 sin u 0.-sina0
sins
,TET/2="517/2 (1-k1sin2u)-Iodu= K(k), k= sin0 0=sini00 . If CI and C3 are two torque constant values, when started at same 00 ,and Tiand T2 are unmagnetized and magnetized periods:
T 2
Ans.
= N/3mg1 + pv4ra 3 /VA/TaiT.
/F271= 3-
dx fz 2 . Get slope dz' ,ci:7— Since u= 1000pv, VII-17C becomes,' setting y=0, (1.+2(a/r) 31x 2= C, r =d2:_3a3rzi:12c _ .512c- .0015=a= 5' Ans. a3 +2d i c = O. (f)3=. 001, x=z=Afi, so -62r° + z lx 2 +(l+ 2 (f.)3 1 2xd-2' dz dz dz dz r r4 -4Replace, as in 9.07 and 9.08, by two spherical current shells. From 9.08 get outside ima 3.43 _ 3 -mq(ba -a3 )._. From 9.08(1), stream function is dipole field Md = 11 = Ma(Pi (cos0)-P0 (cos0)). Inside put n=0 and n=1 in 7.12(4) (P;(3)=0) so that by 7.02(7), Ai=- 3pv Ma(r/a)sin0 represents a uniform field independent of a. Therefore
e
in cavity fields cancel. There is no field. Ans.
I
-5CForce on a from b is Fab .
M al
0 1 2= M 2 COS0 2 /(4TrIlvr 2 ), C2 21 = MI COS0I R4T1pyr2)
I( -cosOrTe - sin01m 6 '1 r- )n ai ) (F,dy =m igg- br on012.1= 4rrilvr4 1112 at e2=2 . (F 21 "Y =-2rbEy )
3M sin201 1 3M M ( 1 =M 6 2rb A i4rf • 2pyr 3 'Or 0 4Trpyr4
T
_m r.
12 = "11. r6032 )411p.vr
-
T 21
-1;171,:r2 3 =-142 (-k21 3= 4
So down force on 1 at r/3 and up force on 2 at 2r/3 are of magnitude 31.1111 2/(41fur3 ) . Ans. -6CWith polar coordinates r, 0, 0; r being the radius vector, 0 the colatitude and 0 the longitude; the megnetomotence of the fixed magnet is clearly independent of 0, i.e. sin0d0 =0. Thus its field lies in the plane PQ and C, so that 2Mc os an M sine' H 3 .Thus tans = tans=-He /Hr' --e . Ans. 411`pyr r rue 4Truvr3
H0 =
on_
-7CFor equilibrium at magnet B there must be no field normal to it. Thus H
rA
=H
rC
sin30° -H
A __ . _
ec
cos30-. H
rA
LcmAcosem _2mAcossoo DiAsln322. br t. -W I. .az —.1 41Tur 3
4ITUr3
,
cos() 2M -b f ljgcos2a l.mr sin30° So 2M =M (1-2)4/372=-■ H =- bm fcci_ c cos30° H ,5M /2. ' A C rC br L Limit. a ' 4rrpr 3 ec rb0L 4-Tur a i 4rpr 3 C ' /4. But M =M so MA :MB :MC J:4:4 or MA /MC Ans B C
Page 364. (436)
Chapter IX.(x11)
Page 111.
-8C-60 M sin Hen=T6os in (0-0) = 3 sin (0-13), r cos(0-0)= d, 4rtpx obi sinesin (0-0)1 . -61-3u 3cos2(0-9) sin(0-0)(1
tin 2Mcos0 Hrn=-F; cos(0-0) - 41.Tpr 3 COS (0-0) N-
mcos 41Tpd 3
)
-Ø " (2 COSOCOS (0-13)
(0-A)
o-e
0=cos2(0-0)(3cos(0-0)sin(0-0)cose
-4-Esin 2(0-03)-cos 2 (93-0)]sine)• cos2(0-0=0, 0-9=1:TT/2 is minimum so -I sin2(0-0)ccite-cos2(0-0)sin19=1 is a maximum.
Thus 3tan2(0-0) = 2tanO Ans. -9C-
Resolve each magnet into components as at right .
Ml sinei
Forces normal to r come from components shown. When
Mi cose i-o
-
-314 COSO
2
1M2 sine2
-3MiM p since i+0
magnets are parallel forces are along r. From IX-5C:
(F12)n-
4rrp.r 4
since
3MiM sine cose 2+ cose sine 314iM p sin (A 1 +9 2 ) (F21)ii= — 4ngr 4npr 4 41lir 4 From 9.12 (2) : OW MiM 2 _ Mit4 (sine2 cose i +2sin0i cos02)=(F2i)ri,T2 =-Oleos; +2 sine2 cos0i) = - (F12 )r 2 1.1 = -601 Lrtiar 3 602-4mir3 (sin sine2 cosei +2 sine, cos02r sine,cos() +2 sine cos() So r i = r2 = r. Add, ri +r2=r 3 sin (01 +02) 3 s in (91 +02) r i sinel cosei +2sine, cos02. tane,+2tane1 Ans. r2 sineicos02 +2sine2 cose1 tane1 +2tane2 -10CLet M= rni M, Z1 =
r = rl r. For force along M need mi-VW. Use vector forms in 1.071.
4rtliF=m1 .Vr-3M.M I - 3r -5(141-rmi.7(M-z)+M-r mr7V.,01-3M- r M'-rmilr-5. But riLA•tr-s= -312.1.1 -Er-4=-3r-4 ose 5 • Vr-5 = -5rni•rjr-e=
-5r"cos0, tn -17 (M-) = (m i-yr) = M•miMj • (mi-vr) =
M•E= Mrcose, 1..r = Mercose',
cosei , M-M'= MMI cos E.
(M er) = tim)
Thus we get
4rrp.F = -3MM'r" cosecos -3r-5(MM'rcos0' +MMIrcos 0 cos E)+3 • 5r-6MI M 2I r 2 cos 20cosel F = 3/-241 (5cos 2 0cose' - cosec - 2cosOcose) /(41Tpr 4 )
Ans.
Differentiate 9.12 (2) . T = -614/4= MN' sine sine' sin*. -mia+r +m'ib+d=0 (1) But m i . di = 0 and 71,1•cli=0 so d i - (1) gives r•cl1 +d=0 or r i •di= -d/r. x mi=di sin€ . Therefore r1 -(mix4)=r 1 - d4sine = - (d /r) sine r))=5.1
(El
r isinesine"sin*= (ri xmi)x (r2xnei)=K2(r
=-EA (d/r) sine. Substitute in T. T= -MM'd sine/ (4Trur4 )
m ixrd Ans.
-11Cns2iM a n2ai sinea sinei =sin(a - 0 2 ) In equilibrium moments on two magnets balance +or(mMi/imsai: 2: 9 = =sing. cose2 -cosasin92= 4/1- (M1 sin91 /ma )2 sina - (M1/M2) sin% cosa, ssin2ecti(1+ (M1/M2)cos a = 4/1- (Mi /M2)2 s in 201 s in a, sin 201(1 + 2 (Mi/M2) cos a sine MI +1.1? + 2M1M2COS a)
Mz
s in 2a,
sine ' M2
sine2 _ M1
N/Mi+Iv11+2M iM icosa
Ans.
Chapter IX. (XII)
Page 364. (436)
Page 112.
-12CPut I' = 0 in 9.12 (3): W = Mrs? /(4rrpr 3) (sinesinee- 2cosecos9'). Couple on second is zero. 4r5C, -bW /be' = 0 = -MCsinecoset+ 2cosOsine')/(41Tur 3 ). Thus 2tan9' = -tan°, cos9'= 2cose/14 OW 1 (cosesin042sine cose') sinel= -sinea1+3cos 2 0. Couple acting on first magnet is 4rrpr be - -MM 3MM' sin20 -3MM'cosesine Substitute for -MM' -cosesine + 4cosOsine Ans. C= 417pr 34/1+3c os 2 0 - Einpr kil+3cos 20 417pr 3 cosel and sine' A/1+3cos 2 9 -13CTotal potential energy of system due to mutual interaction of field
"'re TB
and magnets is, from 9.12 (3), if C = M 2 / (4npr 3 ),
R 6W 0 OW W=C(sinesine1 +2cosecose')+MH(1-sine) + MH(1-sine') be = be' when 0=9'= 2 since: tow bw = ' C(sinecosel +2cosesin01 ) -MHcose' . 60 be 6 2WC(cosesine'+2cos0' sine)-MHcosO and — 6 2W oe2 = C(-sinesinEr+2cosecosep+MHsine and w,-2 = C(-sinesine'+2cosecose')+MHsine' •
b2 w ye-67 - C(cosecose' ow [ - 2 sine sine')= -2C if 0=9'=n/2. 2w ]2 02w 02w. oo m From Wood p117 or Osgood p178 for a minimum be_ and el be2 be' 24( 0 .' rr 02w 02w Li 10 2WI When 0= Er= - tie2 = ag 2 • Thus 2 rb0b2eil taLi/l 1-1tEb3riJl0;4 < Or la 1602 1 < 2) 2w/68 2=6 2w/be'
When 0=et-17/2
-C+MH.
. 5,
,
3Mr3 Ans. So 2C-MH+C < 0 or H > 3C - 4rrp -14C-6 licos9 Mcose z ..j-Fv2 The.yz component is r 'cos0= y., sine ar 417pr 2 - 2' 771-3 ir ' y Myz z Mz2 4Mcose ,Vc77zT _ Mcose NA-77 7-71 Aff2.___7 5. 5,E =pE so Ery- rtpx3 • -, r . rz rz rpr 4npra ' r +z rpr 2) ricosEll Msine -2Msine M(3 247 2 ) E Msin0 2,22se Myz 'sine= µ rbe 417pr 2 - 4rpr3 ' E0z-417 r 3 2nUr5 ' ey 2nur3 4572.77-2npr 5 M(2z -a -y ) M(2yz+yz) = 3 Angle with z axis Thus E z2rur 5 'E 2rrpr 5 2 5• 3 y2z Thus 0= tan-- 2z2 Ans. Position is stable with magnet in direction - a -y2
r
-15CFor 1 and 3, 9=
2.
For 2 and 4, e=0,17•
1 r2
2 =
3
-b rcosel rbe 4frpr f
MsineM H 417µr -4npr? -
[Mcosel Mcose M roe 4rrprazi 2rrpx s - 2n 7.732
r2 = 1.26 r1
H
31
Ans.
44,
-Page 365. (437)
Chapter IX.(XII)
Page 113.
-16 CTake components parallel and perpendicular to AD =4.b secO. F' = 411pb 3F = 2Mcos30 ° • 8cos3 30°=16Mcos 4 30°=9M F' =Msin30°• 8cos30°=-VSM AL F ' = 2Mcos30° • 8cos3 30° sin30°=-9-2 M F1 =714cos30°- 8cos30° cos30°=V3-M Br Br_L =-Msin30°• 8cos3 30° sin30°4/3 M F130 = -Msin30° • 8cos 330°cos30 °=-41 FB0.1 =-2Mcos30°•8cos 3 30°cos30°= Fc~~ = 2Mcos30°• 8cos 3 30°sin30° 4 F'
FA
FCO =Msin30° • 8cos s 30°cos30° 41 Fc' al.=Ms in30°• 8cos 3 30°sin30°413-M F =
it 41703
(9
e ___,_,_ 9 e el
4.7-M-71
18M, 4Trilb -Y F-1- = 514 4‘ --1"t..3 '— : ;r--:1
-
-
41b3
141324 + 27 #551M 62 0 , -14011 F sine sine. , For pendujum 4rrgb 3 41"rp.b 3 : IFtr - -M 4770 3453i 00 b 20 4 rri.ab 3 d mg/ sine or -oT.sine. = E . Same period for same w so Ans.
F= jr7FF- F1 2 II ---d 20 ML 2 dt 2
-17C-
m2 S
_/42
n(150 ° 0)cos (8 -30°)+2cos(150°-0)sin(0-
TAC 41.4. 1r 1r Isin(6-3°)33s(15°°- 150o, C 30°)). TAB-OTT-4+2cos(0- 30°)sin(150°-13)). Add , equate to 0; 0= - c os(0 -3/3)sin(150°-0)+sin(0-33°)nos(150°-at,-$
So sin (20-30°-150°) =0, sin(20-1)=0, 0 =0, 1/2, TT, 317/2. Ans. Positions e=77/2, 3r/ 2 are stable for each magnet lies along lines of force of others. Positions 0 =
are unstable for each points against are
field of others.
-18C-
Write potential energy W from 9.12 (2) .
a
41tua 3W = M 2(2_ 2[-cowl cos02+2sine1 sin03 -cos02 cos04 -1-2sin02 sin04 ]
da A 7-02)cos(7 - +03) -• sin i n (f0i)sin( Z4-02)- 2c os (7-0)cos (Z+0) -sin (E-03)sin (;c03)-2cos (7rr rr rt r Tf rr rr - s in (.- 03 )sin (4+04 )-2cos (4-03)cos (4+00-sin (T-04)sin (4+0)- 2cos (4-04)cos (4+00 3 '
Pick normal modes.
. Let 0 = 01 = -02= 03=
g‘i
r -04 .4rrlia 3W= M2(2 7(3sin2 0-1) -69.n 0440)cos (E-0)-6sin(;;O)cos (44))
= 3M2 (24-(sin2 0-4) -2s ir; • 4mk 26 20/ot2 = -OW/a0=-3M 2 2 Tsin20/ (41%1a Ans.
so T = 2111M-ksligrumk 2 a 3 2172/3 II. =
a
3M 2132-ri (4fiaa.3 )
3
Let 02=04 =0 , 0 = 01= -03 . 41nla =11.2( -2i(2+ sin 20) 2sin
(
-0)+sin (111-0)+2col;-0)+2ccs
3
3
/4,12{..2 7 (2+s in 2 0) -6 c os0}. 2mk2b20/0t2=-6W/b-M2(-2 7sin20-1-6sine)/(41Va3)PJ1428(2 7-3)/(21Va 3)
so T = 211M-1N/4111imk 2a 3 / (3-2-3/2)
Ans.
III. Let 0= 01=02=03=04 . 4Trpa 3W= M 2( 2 7(3sin20-1)-4[ s (E-0)sin (;-0)+2cos ( -0)cos(+0)] = m2{ 4(3 3 in 2e_r) 6cos20}. 4mk 2b2 0/Ot2= -6W/oE*-3M2 (2 3sin20+4sin20)/(4Trila 3 ) -3M2 0 (2 7+2)/(Inia 3 )
T = 2114-1,•/41Tumk 2 a 3 / [ 3 (2+2' 3/2 )]
Ans.
Chapter IX.(XII)
Page365.(437)
Page 114.
-19CIn passing around the path there must be at least four points a,b,c,d at which the needle lies tangent to the path and has irrevocably completed a quarter turn. Since lines of force can cross only at a charge or neutral point, those lines adjacent to a must enter on one side and leave on the other. Similarly for b,c,d. In passing from a to b we come to the boundary: line between those looping a and those looping b. We meet the same line again between b and c, between c and f and between d and a. Thus this same line of force has one or more branch points, each of which is a neutral point. -20CFrom 9.07 a uniformly magnetized body's potential is reproducable by a shell coincident with its surface around which current flows in planes normal to x, its density being uniform in the x-direction.Each current filament pi is replacable by a uniform magnetic shell fitting the surface and bounded by this filament. Thus the whole potential is reproducable by a superposition of such shells giving a variable strength shell Mx. -21CFrom 9.08 (2) and 9.12 (5), steel sphere gives same field as a needle of moment 4
-srpvMa°, where M is sphere magnetization. If needle moment is M, then from 9.12 (2) 4fipvMa3 m W= (sine, sines - 2cosei cos02 ) since \P is zero if they are parallel. The couple is 4Trpyr Mma 3 = - 3 (COSei S itle2 2sinel cosed T= Ans. ui r d(m/R) {toss 6 r > a 4rpd0o= dx a br
-22Csine 6 a: raini+i pn(cos 0)= m -e _n i+2 -7 j (-(ni-F)pPn(p)+(1 ar a
(p)
from 5.154 (5). Refer to axis 01 ,01 by 5.24 (5). Write
= ma-3n iZ0-
m co A.,n+1 0 n for n1+1. 4rrpvdn0=---i a nrinCri (2 6°)-(1 .1111 1 Pm(cose1)411 (cose)cosm(0-01) (m inn ): = -85nEint-gin+ 5n (0140. But m = Si (000 a2 sinei del d01 = -a aSi (el,951)dpidgii . Therefore m(
cose) E0 (2-8°)-(inral! EI rcosmo1 cosm(01 -0)41 5.231 (3) m 2 1+ 1 (im): b o 4rri a 1+1 2i 0 2rr P (cose)mk Amcosszi = 1 21.+1 \r/ 2i+lO si (" ) ' n°-(2i+1) ;Iv r S i (e4) Ar " M 1 m SP r 1 r < a 47rpvd0o= 2.; cnicosOlh(p)+sina OC(p))=-1rz P (p). a ni 0 a a n0 a nrl Write n+1 for n1 and refer to axis el , 01 by 5.24 (5), n m cct rn 4TTpvdc)0 a gn= =— E0 (n+1)(rn ) E (2- 6 °)-(1 P (cosed Pn (cose)cosm(0-03). This is same as m= 0 m (n-Fm): n ry ,\ r for -i.C.ai i+1 before with (i+1)(--) so 0 _ Ans. a °- (2i+l)pv a Si‘"P"
4rivao=itfi
P
Chapter IX.(XII)
Page 366.(438)
Page 115.
-23CFrom 9.08 a uniformly magnetized sphere has a dipole field. Hr ( r -2cos0) /Or -2cos0 - 2 cotO = 2tano. tan8=-r Ho rO(r-2cosO)/be - sing
Ans.
-24CLet 0 = E(An rn + Bnr-n-l`c•
(1). Horizontal component along meridian is Hm=a-1 60/o9
so Hm = E(Anari-l-Bn a-n-2)OSn /O0
(2). Since An and Bnoccur in same combination, they
cannot be found independently. However, if magnetic field is due to internal causes, An=0 and Bncan be determined from the simultaneous equations (2). -25Cds = 0. If no magnetomotance is present, this equation may be satisfied • on a closed curve through the unknown point and a known point by assuming a uniform field covering both. Values at the unknown point estimated in this way for several known points, weighted inversely as their distance,are: (110p=(r1+r2+r3)(r 1
and (Hy)p =(rl+r241.3)(r11
. Ans. If, on a
circle through the unknown points, the line integral i ti:weig hted x according to the length bx 1...y ‘ AC bx by , BC + m )- -r‘ty 4K-- +Hy v-)-of arc it controls: 0=0-1-1R-+H., rfor the line 3 os3 2rr .eutsi - 71 bs1)21Tr 4-(Hxib;2 Y2 os2 2 rr XVS3 integral around the circle. This gives one relation between NI, Hyl, Hx2, Hy 2, Hx3, Fly3 . Ans. -
-26CMsin (0/2) ,2Mcost(17-F93)/23 sin(V2)( 1 .14 H Hb= 4rua 2 4rub 3 41711 a3 b 3 ' Mcos(0/2) ( 2 1 ). 0 tan 3(95/2)-2 Hb b 33 -2a tana- Ha.E777stan2 2tan 41111 a 3-175 3 (S/ 2)-1 6 =i1T+22-0-a, 0= 11 x,8=111 ik a. tan (6+4Xtan( -a)=(tana+1)/(tanCX-1). let T= tanCI\ 2 X -1+8T+6T 43T4+1-8T-6T 2-3T 4 -16T X 7 Ans. Scot (8+-2)= cotr6tarr2-3tan3X tan( -6+7X/ =- 1+8T+6 T 2+3T 4- 1+8T+6T 2+3T 4 --2+12T 2+6T 4 -1 2-
-
-
-
-27CThere are 2n+1 rational harmonics Sn(0,95) of order n, so if 0= is1+: 11_ S2+1. - 2S F13 ( al Ii+f -arfS2 there are 16 terms. H, V and 6 values give a Wbr, , b0/60 and an/b0 at one station, so 4 stations give 12 relations and 8 at 4 more give 4 more relations. 28Cr
,..,
2
ri= 1f1
„6 Si+t= r3 S2+1.-a r iS3+tfi S4 . From 27C this requires 3 observations from each of 8 station r
„,
4
e
-
29CI'Mr-2Pi(co se) + Nr -4P3(cos e). (Mr-2- -}Nr-4)c ose + a 1 Nr-t os . 60/rbe= -1Mr4+20-4--Ncos 203r-5sine =- (Mr-3- 1Nr-5+ 3Nr-5) sine +3Nr-5sin 30. But sine = cos), so A=-(Mr.' 11-6Nr-5 , B= 2.Nr-5; N = 2a 5B/ 15, -A=Ma -3+ 4B/5, M= - a 3 (4B+A). 60/br = - 2Mr-3131 -4Nr-5P3= (-2Mr-3+6F5IT )cos0-10Nr-Aos31 4 or V= 2(-Ma-3 +3Nalsinx - 10Na-5sin 3\ = 2(A + 4B)sinx - -B sin 3 X Ans.
3
Page 366 .(438)
Chapter IX. (XII)
Page116.
- 30CFrom 9.09:
-3/42 2 rrp2x) (
11+11v
a M2(11-111) w= -F dx - 32,rtpv a3 (p.f pv) - 31-
Force and torque same as for image real magnet Mi= M(p-uv)/(pi-pv) sin% = -sine" cosOecos0 . From 9.12 (2), for fixed magnets
Ans. k--------al I( a , A( _ / e1
W = M 2 (p.-pv)(sine1 sine2 -2cose1 cos%) /(41iir 2(.1+12.7)) From 9.12 (3): F= 32(-singe, - 2cos 201) (11-AO 4rtuva 1 (u+p.v) T = - bW
a Al
(co sei sine, +2sinO, cos02) (p-pv) 4Truva I (il+iiv)
641p.a 4 (u+ uv) -M2 (u-pv) sin20 3217pva3(p+pv)
Ans.
Chapter X. (XI)
Page 408.
Page 117.
-1C- (p415) 1.1 6B From 10.00 (9) 7r- FE2 =7 2Bz where B is a function of x only. As in 10.01 jak jcut 6e -'1JPW'rx +DeN iPaVrx . When x=-±h, Bz=B0 e a solution is Bz =77/ , C=C0 e , Th+Boetr iu / ii= Do ejCpt soBo=A0 e-nv -1" - (I) 7 h and. A0 =B0 . Solving for A0 gives 3/ 1 ec o sh ((dLl ./.;1 ,3 h)) eja it 1/7h) so that B - B0 Ar71A0 =B0 (e Nij'll tr h+eN/WrhI 1= B sech(ju
X
(1)
3ITTrr = (1+j) -if. Multiply (1) by conjugate complex and get amplitude by Dw 655.2.
But N/
cosh2mx+cos2mx cos(pt+8) where p=co and ma=4pip/T. Ans. cosh2mh+cos2mh Find p by multiplying numerator and denominator of (1) by coshC(1-j)3c A.1177rhj so that B=13
°
latter is real. By Dw 651.08 numerator is ifcosh[4.Wr(h+x-jh+jx)]+cosI[i7iTr(h-x-jh-jx)]) Ta Expansion by cosh(a-jb) = cosha cosb -j sinha sinb gives sinhrm(h+x)Jsinrm(h-x)]- sinh[m(h-x)] sin[m(h+x)] imaginary tan real - cosh[m(h+x)]cos[m(h-x)]+cosh[m(h-x)]cos[m(h+x)] A.
mh and mx small:
B. mh and mx large:
B = B0 coscut since
= 0.
B—P Bo e —InhA/COSh2MXCOS2MX
Ans.
Ans.
cos(wt+p) and when x=0 this gives
the field at strip's center falling off exponentially with thickness. Near the surface -m(h-x) h and B-0-B0 e cos(cot-8) so that field falls off exponentially with distance from surface. Also tan8=m(h-x) so phase changes linearly with distance from surface.
-2- (p416) Outside the sphere the potentialof loop and eddy currents is, by 7.13 (2), n -n-1 1pni(oe jtut . Ae6=ECAn(r/c) +Brr Inside by 10.04 (10) the potential is = E(CTI NT3In _4(iliir)Pnl /on + B) n l= so An(a (Cn -
-
eia)t . At r=a : A = A In
and lib (rArjs) /or = pvb(rgis) /Or .
(4/jp a) (1) • and p( (n+1) (a /c) liAn - na-II B n}
= pv(21 1/5/-5- +Nrjr: 7•11 1 11=Pv (Cni:/-5)(i-I -(n+i)I + 2 n+-F n+7 n-I-2 n-1-2 An( -nlIvI...
+17151.1v ain
uF 2
p (n+1) I 11 (a /C)n= Bn (1111vI
i) n- 2
-Ar5pv aI
so that -
la -n-1 .
n-7 2n+Ii+T
4137311vI_-(nliv+n(n+1)1I+ a
Write I for I i and I for I so B + n-T n+T n (jam-p)rd+ r- (p.„-g)nI+-Nrip ThvaI -b\r— aI_-fnuv+Ii(n+1))I+ Cn =4/a I+C(1.31-pv )nI+ jppvaI.)
cr
(2n+l)p.11(a/crAn (p-pv )rd+ +
Putting An value in from 7.13 (2) gives
2n+1 I+(4/75r)P 1 cos a n=1 n(n+1) (p-pv)nI+( jpa)+
A;z1= ppvIVa7 27sina
Am
n ejczt
Ans.
Chapter X.(XI)
Page 408.(416)
Page 118.
-3At t=0 surface eddy currents keep zero field inside. Field at infinity unperturbed by sphere. Solutions 10.08 (4) and (5) give uniform field inside plus dipole field outside maintained by surface eddy currents. At t=o the inside field
-F
A B just equals the static field inside due to a uni- form field of induction B. Superimpose
A and B for C. A is 10.08 (4) and (5) at t=0. B is 10.06 (10) and (11), a static field C is the suddenly established field at t=0. Thus the fields are: at 3pBr rA Br (P-Pv)a 3 B -qst Jo(ksr)e ‘a sine + sine, A 2 E ie A0i-2(p+2pv) t 4 00
Ans.
From 10.08 (4) and (1): A00=0 at r=a when t=0, and from 10.08 (5) and (1) A0i=0 for all r values when t=0. At t== get static values of 10.06 (10) and (11).
Ans.
-4From 10.00 (9), (p/T)N5jot= V2Z or (p/T)dBz /dt= 6 2Bz / op 2 +bBz /00. This is identical with 10.02 (1) so 10.04 (4) and (5), with B for i is a solution. From 10.00 (5) Pt= V41, and, since Bz is a function of p only, 1 bBz. Nr _12 — It _WITp)..ja) berWTp)+beii(N/FP) B0 cos(wt4a) -p Io N3Ta)."- wr. beri(V-p-p)+beig(A5p) i0 = -g op tang=-
ber0 (45a)(beri (N/Fp)+bei l (ViSp)1+bei0(V15a)(ber,(mo0)-beil (VTp)) bero (Vpa)tberi (Vrp)-bei l (Vrp)j+bei o (Vpa)tber i Wg0+bei l (450).1
Ans.
Ans.
5 ( 1-e -R(T+t)/Li c_tiv Isinatin Use 10.16 (6). f(t)= P'(cosa), Ans=Pil,(cose . ' n 2n(n+l)an n 6 -C(T+OrTe (C-N)T 1 Let N= (2n+l)o- R dT Ilvb 1 L = C LARis_it=0 =CnAnsTe 'Jo CCnAns r -CT -NT, ric° .E-CCnAns e-CT{ e (C-N)T..1 } . r Le -e J. Momentum is M=JFdt, so C-N C-N o TA:le -CT (Prnsina+BA,cosa)(N-E2C)(N-C) M= 2rasinaICear(1-e -C e-NT) dT=2raICdha i C- 0 _1 2CN(C+N)(C-N) n+1 -n-2 ._b n+ sinaLr_ 2,P1/11 a Cnsinatcsc^6(sinePC21 Brnaina= b u--5 ---- p=a ---7 171 Opt .' 4 )41 1 n+1 -n-2 x r r 1, =b a Cn sina n(n+1)Pn (cosa.) by 5.15 (1). Bencosa=-Cnbn+1 cosaPj(cosa)1.11.7 1rag h 2n+1 OPn(P) =11-0Ticom/sira bp • B (3 PA(cosa)sin2a{n(n+1)Pn+npPO O=la an np 2an(n+1) a 1 1 2n+1 = i pva_ I(n+lr'sina(b/a) P,1(cosa)P' n+1(cosa) -N-2C f(2n+1 t1.+2pvbR3Pvb gv b{(2n+1)TL+2pvb0 2N(N+C) 2(2n+1)q(2n+1)fL+pvbR3 2(2n+1)q(2n+1)IL+pvbR}
r
M=
sin2 a 0312n+1(2n+l)a+2pvbR E 2S n=1 (n+1) (2n+1) (2n+l)fL+ pvbR PA(cosa)P4i(cosa)
Ans.
Page 119.
Page 409. (416)
Chapter X.(XI)
-6From 7.10 (1): A= pvMsinecosca/(4rfr 2) and, in 10.10 (4)
ff(t-T),P,(2+2f11-1710) sincgt-T)dr or V pv>>.cur / =-etr-1 pvMp{p 2 +(c+z+2filv-I T)21 1cosw(t-T) • A0 - 417 °14 : 1D ( p 2+ (C+Z+2r ).1. 71 T) 2 ) 2 • F or I 4 f2 u,T2r+f pyl(c+z) 3 u„copM sin= I dT 1.103 ,pMsincut sinco(t-T)Zsinwt so Adz 4rr 4rr roc{ p 2+ (c+z+2sliV .02} 2 4 4r 2 11/47 2 p2A/p 2+Kc+z +2sPVT- 32 c+z 1 41.114sinta I - cose 2 s (c+01 p kiwpM sinwt [ sin=[4f2 _A‘DOM l sine 8171r 2f p2 . 417 477•4f 2 p 2 2f ,442+(c+02 2+(c+z) aj 8rrBr = qcoM(fr)isinciftsine(r sine)-1 =Nuti(fr2)- inuft, Be=0. Total fields are Br+14. = p,Msine ).6Mcosfio EvMcose sin= Ans 4 rtr Be rd Be = rs- sinciatf sin= Ans. --T8r fr 21 -7 'typos incot 2 P..& C2 1 -IhicoM sinat r From 10.09 (3) i0=pv - Ibz z=o 1 4rIoNfp2-1-c 2 4-1' 4- 1.17 c2J -4r,co a+c2)3/2. p4IP.4 2(13 2 2r14 20,2 [-1 1.„ .. Eir 2c (0 2.2rpdp Ans. 32r rr 2 J 0 (32+c2)3 -- 32rTr2 12c a +217C7 641Tic2 -
,
01
to
r
-8pvm(T-t+T)p
dT • b ( PA0) From 10.10 (5) A0 P . 4rfrcp 2+ (C+Z+2f p;1T) 2 / 2) pap t 2p.„.m t F T)d'r IlvMdT r PSM T) BP = S n ot 4iff(c+z+2Vp&'11.)3Ip=0 Jo 2T1T(c++2fT)3 EriTfc+z+2f u • 3 }
-
- rsI 2 P= J-7 dA.
- -
o. a a t p'
)+ 3 02rih(cM+tzt)li{11(vc(+c+ zz )+ft2f t) 2
t
When t>T, field dies out as if z increased at constant rate z =z0+2Sp.71 (t-T) from value at t=T, sb 130=22.rr -1 )4M{pv-(c+z 0)+f(2t-,r)ilpv(c+2,:)+2f(t-T))-2(pv (c+z0)+2T3T2 Ans. -9M Pv114 9 , Fx-4TTre 3131/MIM x . By 1.071 (6) for electric dipoles: F=-1--f. For magnetic, by 7.00 F-341.rr By 4rrer If x=0 when t=0, x--vt so by 10.10(5): Fx-
311,142 Orr
V (t T )d T prd 2 )3/21 . tJ 0 Cv 2 ( t - T) a+ (2c+2
for integrand: 6/6T = -6/tit +(f/p.v )6/6c, SO
Note that
_r_ p
pvLT& 4(0+4 f 21.1).1T 2+8 c f piT/T+4c 235/2] By Dw 380.019 xdx -3116M2v p4 so F and 380.013 Jo (a2x2+aix+a)6/2- 24c3(gyNra-2-+21)x 211(2c)4 2 +2f32 Ans.
r
10 Split dipole into components M sin0 in direction of motion and Mcos0
/c
normal to it. Msing5 exerts no force on Mcos0. Calculate separate forces and add. From -9- above: 1F4< - 3/1"M2v(t-1") sin 20= sin 20dF0 4r1R 5 x..Livt-12 c22g. c 62 COSO pvM2COS2 0 x 311,1,42c 0 s 20 co se (3-5cos20) 411,R4 Orr Ox2 R3.1 -
6x2 R2
- 4rr
J__
MsingS Mcos0 1.1
_J`c
:C
‘---12rT --
i2sT
_ v(t-T)
where cose= v(t-r)/R, sine= 2(c+r11v -1T)/R, R 2 = v2 (t- T) 2 +4(c+rlivir)2.
5cos3OdT 5v 3 (t-r)3 ciT 26 r (t-T)diNote that: — to R4 - bti:Iva(t-T)2+4(c-l-fp /-71T)137/ 2 v 6v OtJofv2 (t -T)44(c +W-09 5/2] so that Fx= KLX+FIIX= CF 0sin ays+C3F0 +1.7 26 (F0 v-1)/Ov]cos 201 +
2cos 20 [1
n 2v 2
0 = [1 (v4v9-41‘+2f)1/44v2-1-45 D.F
LIS COS 20
a 4. 4s.2] F0
Ans.
Chapter X. (XI)
Page410. (417)
Page 120.
-11C(cose). If )144,pliso From From 7.13 (2) A'= 3E1 n=ln(n+l)g111 (coset)Pril 22 By 10.1 _ pIansing a _ uuMIan r e°(2rH-1) an MI Pn(cosu)cose cos (cot -En ) c 4f" fn & 10.14' n 4rf-A-2. w=411af2 n1.20 -12In steady state, fields and eddy currents rotate uniformly. Replace rotating field by two alternating fields at right angles and 90 °out of phase. Torque is constant so find it when By is zero by using interaction of Bx with eddy currents produced by By. From 7.02 (7), in 10.16 (4): n=1 and C0=+B. Set cot=41T. In 10.16 (15), i0=k1 B sine mg sine = -gico arB sine/ (9r 2+ptra 2o3 2) . Torque on ring element is BITR2iad9 so that g ,,17/2 B.Tragsin-u.7koaSBsine 9 T?B2 CUa4 s in:0 de 67B2Ccoa4 . T dT= ade = 9f2+4a2m2 a 2 co ) 2 ( 9 Br2
= 2f dT -
Ans.
-13- 21TS B2 2 aa2 W2 317fB 2(02a 4 By By 10.16 (16) : n=1, C1=-2-, cos ei-49.0+ P4 v af-D sza)7 . PL---r 6 • 4 • a 952+4,92032 912+34s2a)2 Pv -14sine 3B r B 11.16 (13): Aio= 2 By sin sin (cot- e) . B= B sine sin(cot- e)=49/4 —T-TTff iva co sin (cot- e)
Ans.
Ans.
-15In 10.20 AI=Bp sine,coscot sinE=-2r/Af 2+(o2 p4a2 .
-S2 i sine sine sin(Lut -e), tane= n=1. Inside field is A'+A pB uvom Ans. so Bi= -2 $13 sin (wt-e) /A/452+(o2utoa
162A 2oB 2Ssint3 2A • /iyalacos(ent-e). Set cot=3 as in -12- above. From 10.20 (2), at p=a i==---'--• uva uva ,v4S•+w-pi•ja. 2r 2 T 8tura 3B 2
452 4tow a2rosin20
41703ra 3 B 2 dB- 43.2+ coap4s2 Ans.
-17-
2n+l [12] cos (2n+1) 8 jat From 7. 09 (1), Alz= Lvegz--1-'2 . From 10.20, =-Ve z . From 4.02, Al IT r 1 2n+1 n=0 P 2v/ E (b/P)2n+2 2 (2n+1)5 si __ -2(2n+l)r sin encos(2n+1)0 sin (cot - en). tanen Ao+A'=- 17 n=0 2n+1 (2n+l)hp+utrcoa' n_ •4/4 ' " /iva)a Ao+A,. SI E 1b1 2n+1 cos (2n+1)0 cos (cot-en ) Ans. 11 n=0,,,(4 (2n+1)4 V +Ti fito‘
Chapter X. (XI)
Page 410. (417)
Page 121.
-18By 10.16 for ring, p=ho, e=-1= -±§c i{Trrh 3(A401-4
2rdh Ns
h )
-1-2---2 bPi(g) .g 4.c2 -oc ,= g=h 2 g . From VII-21, if 10 =C1,./-(
d g so -±d (2+A°)—
srir8Ci 1Pi(jC0)
(1)
C)Pi( g) if Co< C
LP .J...(1) LITER 241._ipt C-4C0) _ 2 c1Ao ( a t ‘ =C 0) crii2+0 ilic1Ai1+t§V-40PRiCo)Q1 (iGo) hsPi (go) (iCo) c1Vg 2+C 2 At surface of shell, :When t> 0, d(A)/dt = 0 from (1). 1•2 151 h, 2c, go dt =A0e2gotifjp(l+C8)Pla Co)Q1aCo) 2g1, dt I A0 I C=o jP(1+C 2)11 (jC0)Q1(jC) rID 46*
Imposed field is A;25 = 2ric1 BP1(jC)P1(g) so that at t=0 AO, A0 where
Ao 4jc i BP1(3)PICA*10C0)}-1Q1(jC) • =
BPI(g)PI(i
By 10.00 (4), i0 = -(dAoidt)/Kti lg o ) at C = Co.
2g0 ti( jp(l+C DPI (iCo) Qi(iCo))
10 h2g o 2.thi-( 1+CPV.i D go)4(.40) e
1.Co) e pct Arrcm2+c8Q10
where
3
-2cati(a-Fcg)Taco)) Bisql+GN/57Ie -Ngo t Ans. N-F Iig 2+C 2
2
3
C2)3/2QP CO
-19From 7.02 (6) potential of x-directed field is A=Bp sings eicpt so that outside cylinder,
AZ +Az= B (p+ air') singie jult (1) . From 10.03 (3) inside Az=ZI1 (..4115p) sin0 di:14 (2) At p=a, AcrAi so
CI 1 (Nrjr3a)= g(a+r)a-1)=2e.r.5a-6(10 -12) (3). Also pvoAi /6p=p6A0 /Op so that
1.1011:FII(Nri-P-a)=P1.3- ( 1-3a -2 11v4.1iPa(I0+1 2) (4). (4)+(p/A) • (3) eliminates D so that e .4pB[VIT(01+4v)10 - (11-11v)10]-1 Ans. D=a2((p-liv)Io-(P+Pv)I23/[(Pillv)I0-(1-1-11v)I2) Ans. -20By 10.00 (4), i=T o t-
cu2 CC 1 ci i(VIT)p) sin0e jept. By10.04,Pav- 2T
Lk 12-4Thera ber 2 ()=01+1102here+"1(p+pv)2bei11-2 (p2- p4)beiobei2+(p-pv)2beil =((n+pv)ber0+(p-pv) be r2) 24(p+pobeio+03-1.0bei
aE-k=16p 2B 2p -IR1.1±}1v) 2101*o -(P 2-i4) (101*2+I*0 I 2) + (p-pv) 2 ber
sin20 pdpdta=7,F CC j:I iI iPdP•
' 1.
81pacuB2(ber ,(/a)beilifip-a)-berikr5a)bei(.,,rp-a) From 5.323 Or 10.04 (11) Pav Nig[(p+pv)te ro-F(p-pv)ber 02-i-t(p+tly)bel +(p-p.v) be i 2]23 4rpa 2cuB 2 (bern (Vga)bei2 (A/Ta)-beic(ViDa)ber 2 (Vip-a)) ° If p is small, by Dw 828.3,828.41 820.3 [(p+pv)bero+(p-pv)ber2] 2 + ap+Ildbe io+41- p.v) be i2 I and 820.4, ber 2 O , beyr.(x/2) 4)=-Pa 2 /8, ber0=1, bei0=Pa 2 /4 so that Trisza 4B 2 noco2a 4 B 2 2B 2.2.21 4r _ Ans. 11' 13a P av 20141) 2T 8 (p+Pv) 2 2 01+Pv) 2 T -21A rotating field is replacible by two alternating fields at right angles 90°out of phase 15P)e iC25 eitpt (real part). Ai=idIi(4.75p)cosoe ja1t-f-ni(.15p) sinfzfe j (c11t+4")I (real part) = ..d.I2(4/3IT= 2 * , =-7-dai /dt=- T I i (V3ISP )e j°e jcpt at any instant. Pr A° v t=f1ZerS pdpd0=Thr . —C J pdp
o
ro This is twice the average power in the last problem so that, since T = PAvt /c131 81 ►pa 2B 2 (berkitia) bei2(Aia) -be io ( T5a) ber, (447 a)) TAns. ((p +gv)bero+(p-pv )ber 2 1 2+((p+pv)bei0 +(p-pv )bei 2} 2
Page 122.
Page 411. (418)
Chapter X. (XI)
-22-
m nrz T° 3s in.--c-, A 20= InfoAnii (q)K i rEll sinn7z planes : A1 0= In E0 Anl i fIV)K i t72cL give Bz=0=P-1 6(PA0)/OP at z=0,c. At p= a, nrz _ 1% _An a sin__c_.inio _ A nr A Lim. 23K0Erlic... 71 n0 I. .L.n`0"11c This is zero except at loop. Multiply by sin (prz/c)dz and integrate from 0 to c.
E
B 1z_ B3z
"
. nTtz
C.-
i(Biz-132z) z.ds in (pTTz /d)d z = s in (prd /c)pI = -(I/a)Apc/2 so Ap = -(2pa/c)sin(prd/c) otiris so A10 =-2pac-IIriioI l(nra /c); (nr p/c)sin (Wird /c) sin (nrz/c) Planes alone. To make Bp =0-7i at p=b, add A40 =2Pac-inE0 Ii(nra/c)Ki(nribic)Ii(nr p/c){I i(nrb/c)}-lsin(nrd /c) sin (nrz/c) Only Ir(nr0/c) are permitted since KinVP /c)=co if p=0. For self inductance
d
decrease compute 2fra (Allimages+40) at p=a, z=d. Positive images at distances -±2nc, -m
the nth positive and nth negative image give at p= a and ki=a2La 2 +(nc-d)23-1 . E 4 C - co ni
2dfi
E
n=1. Los c
a \laK erb)(11(n rrb
c'd )rsina n"
Ans.
2ct.
-23b[p(AciA1)) Power loss in the curved wall can be found from last problem. BzPOP
At p=b
Bz=211CainfolicilIit4u} g(o(nic7b)Ii(141b )+Io(j1 r CL 13 )Ki(n 713))
2d
-fr-l p i sinrj-L ‘rc z . From 5.32 C ÷ ril b ))'sinr
and 5.33 (4) the nrb/c terms are c (nrrbIl c . By 10.02 (8), -15= (2u2 y8)-'SsBedS, dS=2rrbl 21113 4p2a 21 2w, Ii(n Trak ) cs. ord _ 2ra 21 2 rIi(nra/c) sin (nrd /c)] 2 2p2y6 2c 2 n=ili(nrrb/C) 2 1n c p . For the flat ends bcy6 n=1L ii(nrb/c)
i
a solution orthogonal in p is needed which suggests ordinary Bessel functions. Below loop: A r 0=Inci1AnJr(knp) sinhticn(c - d) sinhkn z, Above: A295 = InekAn Ji(knp)sinhfkn(c -z) sinhknd
.
A10 =A2 0 at z=d, and Bp2 -Bp 1 'InEiltriAnsil (knOtainh[kde -CCOShknd +COSh[kn(C••CSinhlind) =InElicriAnJi (knOsinhkn c . Multiply by Ji(kn p)pdp and integrate 0 to b. By 5.296 (7) left side is zero except at loop so J1 (kna) may be taken out of integral, giving rkni° Ic.n pJi(krbo) d(knp) sinhk.n c = 2IAnknb 2( Jo (knb)) 2s inh knc . aJr(kna)iBpd s=paI Jr(lc.na) =I(Anika An= 2p.I a Ji(kna)/(kn[bJ0(1cnb)] 2sinh
. On top at z=c, Bp=-OA0/6 z=-IrikknAnJ i(knp) s inh krfl.
On bottom at z=0, Bp=InriknAnJz(knp) sink kn(c- d). Square B p ,multip ly by 2rtp d p, integrate B 2d S= 2rI 2 nii-liAtiktib 2 Do(knb)3 2sinh2 knd , Sot *S=2171211E14Arikrib 2[Jo(knb))asinh2kn (c d) . 1 4p 212a 2 = kg J_0(knb) J (k,-,a)12 r - 2nb 2 sinh 2kn(c- d)+s inh 2 knd E r 2112y .• By 10.02 (8) . P- 2 b n - lk/itJo(knb)34sinh2knc 0 217a 21 2cci n ..11(kna) (sinh2 k,(c- d) +sinh2 k,d) Ans. (knb) 2sinhaknc y6b 2 n=f1 -
op
_
{
}2
Chapter X. (XI)
Page 411. (419)
Page 123.
-24From 7.02 (8) for primary field, A0=4Bpoand N=2,7p0A0 so that
C = -dN/dt = - jumBP02 eiwt .
Primary i0=C/R = jcorBpe(2Ttpl 'eiwt dpo . Vector potential due to ring element of primary current, 7.10 (4), is dAl;=11(217k)-1jaSoPoriPT::/7{(1-2k2)K-E}eicadpo at z=0. k2 =4 PPo(P+Po)-2 Integration (checked by differentiation*) from p0 =0 to
La i
po =a
gives
-1 = -1-jcouB0 (Rs) s rie 3a)t [(po-p)(2p2+0)K+ (po+p)(2p2- p8)Ero W2p o 1 jwtga-p)(2p2 +a2)K +(a +p)(2pa_ a
=
411 2
Ai0=-( jatio /(21/pc))• 2 ►p so that 2)E) where k2 = 4ap/(a+p) 2 . Total 10 is 10 +kilo .
373e
14. 44.0(B/f)(psincut + Ecop/(617,P)][(s- P) (2 P2+a2)K + (a+ p) (2pa -aa)Eb scut }
Ans.
*Check: bk 2p (p -pa) P)(2P 2+14) K+ (P+Po) (2P 2 -ps) E] -- 2P2+3P1-2PP_O K + (Po TPX2P 2 +P g) 3p 3p 3p bPo k(P+Po)3 t'Po 3P P -Pn [(Po+P )2,E_ + 2 P 2-3PA -2PPo E (P+Po3)p(!P2p2-03P+ Ap) •o(P ) Po) (E-K) Use Dw 789.2 and 789.1 2 po(P+Po) (Po -P) 3p 2 po(Po+P)(2P 2+3p02 -2p0P)- (P-Pc) 2 P Po(2 P -Po) K2Po(Po+p)(2p 2 -3pt-2ppo) - (p+po)2ppo(2p+po ) E 6 pPo(P+Po) 6PPo (P+Po) + 2 2'212 " ) j K_ 3 0o (Pt +1:2)K+-3PO -3PPo E _ Po (P 2+13A)K P (Po+P)E p 3p P 3 P (P+Po) P(P+P0) PPo (P+Po) P+Po = po jp- [s{-1-k1K - 2rcE}1=
((1-12-1c2.-)K - Ei Q. E. D. Note: the bracket in Aio is simpler
in terms of the modulus m=p/a. []=a 3[2(1-m 2 )K(m)+2 (2m 2 -1)E(m)} 25From 7.02 (9) for a uniform field 4j= - -IpPi(t)Plo(). Superimpose for eddy currents an A0 that vanishes at r=2:.. From 5.271 (10) this gives Ao= -3jaB(P1(jC) +C Ql(jC)3P1(g) = -1-2-Bp[1-jC(cot - 1C- l± c2)) from 5.23(11) 4aBV1-4 2.,ff..-11(1-jC[cot -1 C -C(1+C 2)-9). B must be tangent to disk so at C=0, Bc=0=-b (h3 A0) /hi h2 og=- O(07:T2 A0)/al5ERbg so jC17/2=1. Bb 1 A0=4-Bp(1-117[cot" C- C (1+C 2 ) -1]) . At C=0 Brb(h3A0)/h2h3bC= g 2+C 2 2oC[(1+C2){1727(c°t-IC-1---2)}]
17/ n Bo r 1+0 c=02vir=iir 1+ +)] 2,/a2-021.0+v) 21—
pi
from 10.02 (7) i0=210 for both faces. Ans.
-26No external magnetic field sources. Let A0 arise from eddy currents. Apply dN d (2rr d aA) 2Tra (1) ----arsiodz t Ohm' s Law to ring of width dz at z. E= Use Ampere's Law pv iodz = 'BoI zdz - IBi lzdz = p-idzfb(pA9Vbp-o(pAs)i/bp) Eliminate i0 from (1) and (2) so that at p=a : -r-(11-13-1° - (PA6) = dt p=a (3) POP POP
(2)
But A00 = So 8(k)K1(k p )Ii(ka) coskzclk,
(4)
r
A01. =5 e(k)Ki(ka)I1(ko) cos kz clic
By 5.32 (7) left side of (3) is Ze(k)C-kKo (ka)I i(ka)+kKi (ka)Io(ka)icos kz dk = .3-1/0 e(k)cos kz dk (5). Put (5) and (4) into (3). Result holds for all z-values and (ka)Ii(ka) de/dt. Integrate from e(k)=80(k) f [PvaK1(ka)4 a)3 8(10= 90(k)e-feri An at t=0 to e(k)=e(k) at t=T1 then bn--€2 L2- ©o(k) iivaK l (ka)I (ka)
hence for integrands. Thus ©(k)= pvar
Page 124.
Page 412. (419)
Chapter X. (XI)
-27-
so
CO
From 1G.14 (6) to (8), if the inducing field is
Ckl a,t)cos kzdk, the present affect
of the eddy currents formed at a timeT before the time t (present) during an interval f6Eit(k, a, t-T) -rT/ (gvaI l(ka)Ki(ka)) coskzdkdT The total eddy current A0 e dT is dA0 - jo bt /(pvall(ka)K2(ka)) ia , t f"r e cos kzdkdT. at time t due to all past changes is A0 = -4,10 bC4,(kbt 4P-7172 -28For loop alone, by VII-26, A.1716 = pr ilIcoscPtSeaTi (kb)K i(kp)cosk2 dk, p> b. From last -T P problem, due to eddy currents, at p=a, A0- r Ii(kb)1(2(ka)coskz:sinw(t-T)e dTdk. co r.co /a FF 2={sin(cut-*)} / T- in t egr a I : sincbtfo cosar e TPdT-COSCOtj sinane-TPdT= (P sin= -cocoscut ) #1.77 r.cto tar4=4.0/P. Thus at p=a: A =(colzbI/IT),1 (sin(iDt-*)I4b)K 1(k.icoskz/4/7.-7 P -F2)dk so when p
OD
A0
* sin
Pb I
0
0
(cot -T) Ii(kb)(Ii(k a)}"' K Ikani (kP)cos kz dk. Instantaneous induced C in loop
is 211bIdA dt at p=b, z=0. Resistance is in phase term (coscot) and reactance is out of co phase term ( s incut ) . 0R= 2pb2 ctio sin * cos *ai(kb)3 2 KIM/ Ii (ka)) dk, where tan*=cDpaI1(ka)1(.2(ka)/t co Ans. 211b2,10 sin2* [I 1(16)3 2[K1(ka)/I1(lca)3 dk Ans. -29-
u2='4
On the line u2 =0 in the surface cosh u1 = cosh u o = c/b the vector
u 2=C
potential of VII-40 becomes, for unit current I =1, A0p= p{A.-7 b/f2b}Pli/2(c/b)(Q11/2(c/b))-11-(2-8191 )(2n-1-1)-l(2n-1)-'
(c/b)
L = 2Tr(b-Fc)A0p . For a more rapidly convergent series take A0 on the circle u2 = zn so that odd terms are eliminated. Then L = 217pA0pt. From 4.13 (2): bb
at u2 =V=rt/2 and cosh uo = c/b, p= a sinhuo /coshu0 . From 4.13 (3) a coshu9 /sinhu0=c so p=a 2 /c=(c2 -b2)/c=c cos 2a • When I=1 at P1 ,
_
L a -u2 1 ;rr/ 2 (2-6:)QA-v2(c/b) ( 1 2 ;:.,°(-1TQl2n-1/2(t) isj "V=11A/2b Q1 (c/b) n=0 (4n-1)(4n+1) ' . 2b "1/2 nt.1(4n-1)(4n1-1.)QL1/2(4 , , u1 IA v 1 -Wip—)4. ej By VII-39 the factor before the bracket is the A0 of a loop of wire of A
L.I/2(c/b)
E
pn=u„fipi
hc—C
■ I'
radius a= ,,/c2-b 2 given by 7.10 (4) A;6 = -ri-V-a-Ti3-(1-12-k 2 )K - E) where k 2= 4aP/C (a+p) 2+z 2 P = a COSa)2 = asina so 2cosa/(1+cosCz)=10= cosa/cos2 (a/2). lik 2=1/(l+cosa) A;jr. V.+CO S a 2c s a
y
k2)K _ E} 2) t a i - u cooaa ik2)K-E1 . 2Tracosaitif = 217c2cos 2aA0 so a co sa{a. n 2 1 L = 2p.c cos a co4a (1- lik2 )K- El[1 - 2nE_ (-1) Q2n" 1 /2(csca) Ans. sina = b/c. 1(4n-1)(4n+1)Q12/2(csca) i ( -
{
Chapter X. (XI)
Page 412.
Page 125.
-30- (p419) Put p=0 in VII-28 and add ring potential VII-26 to obtain A = pall% I N kblrfkb)Ii(kb)Ki(ka) b (PAO) b (PKi0c4) --kpK0(kp) lif;15 Tr ol-N"' k" l_ kbK0(kb) I 16)1(1 (kp) icos kz dk. Bz - pbp / bp palpcK,Oca)10(kP)-kt•Ki(ka)I0(c.p)Ko(kb)Ii(kb)+k2bIo(kb)Ii(kb)K,(ca)1C0(4) O(pI,(kp) op -kplo(kp). Bzcos kz dk 1 - kbK0(kb)Ii(kb) pal rkK, (ka) I 0 (kb) kz cos dk At p=b second and third terms cancel so iBz rp=b - rf Jo1-kbK0(k b)I 1 (kt) From 10.02 (7) IBz I p.b=pic/3 and by 5.32 (7) 1-kbK 0 (kb)I 1 (kb) = kbK i(kb)10 (kb) so that ece i0 = al/ (br)3j0 (ka)/K 1 (kb)3coskz dk Ans. -31- (p419) Let Ac be second term in 30 above which is due to eddy currents. Then (kb)I (kb)kl (k a)) 2 i 0(b) K i(ka)) 2 dkAns. From AL = -21Fra4 = 2pa2rec'kbI° dk- 2pa2r 1 , 0 co K 2(kb) o kbK i (kb)I0 (kb) (
10.02 (6)
the energy loss is 2TTb(y6)-11:(i20 ) edz =w= Trb(y6)-1f- co ildz. W=11:) r(k)coskz dkrdz Yu-00 ,co = xl- liT, l yf j_1(x) piosx;1(y)cosyz dydzi dx. By Fourier ' s integral theorem we have
-r
',1'°° (f (x)3 2 dx. 2r4'(x)= rcos(ux),f i( t) cos utd t du so that W- 2.1---
CO
07
Y 8 - co
Thus from -30- above
-2Tr 2 b a2 1 2ra'VK,(ko)In(kb)12,, 2a2 1 2 r 12 I (ka ) 2 IT— y6 . 172 y8b Jo K1(kb) dk. TF=Ri! , q=2 so that LI 0 biqk 1 io(kb) OK ]
R = 4a 2 (y6b)-1.1:(Ki (ka)) 2[K1 (k.b)) -2dk
Ans.
-32 By HdMF 9.6.25 (Irl, x=X, z=P ) Z(P 2+u2)-312 cosXu du= (X/p)Ki(X p) jo (p2+u2 r3/2 cosXu du so by 12.2.3 1:(p2+u2)-3/2 sinu du From 6.1.17, r(+).-zift from 12.2.4 (v=0), (p2-i-cu2),3/2sinXu du . By Dw 200.03 and 201.01 as X-c--0 0 = - (xmdcvf (A.p) - zeo) + p ) 2 z.,10)1212 S(.12.1 1-202. i2)3 pvpi coc:34i7 112) 32 z stc02_1T -2± Xc 2 + -?;7 . Thus 10.12 (5) is
L_1=L1+ 4
c.7vc._2 so
Tivii[(XI( (XP ) -
2 1 sin* -
(xp)-L,(xp)]
wpxt+c P 2 }cos41
Ans.
-33co (a2-b2 11141Th, [a 2+b2] [1 2m+ 2 2M+ 2 E From VIII-31 : J1(oz) Ji(bz) = mieo on+i)! (m+2) rm+1 a 2 _b2 m= 0 Cm z -kc dk. Then by IT vI (11) p135 : From 10.13 (2) : A0+A = - 4pvami Cmjo k2m+ 3 ( jcPk-l+krle 0 (
= jcP/X, p
= c. J:=
From HTF vII (4) p 145: Thus A0+A'0 =
) 2m+3 feiPEi(-jp) +2331 (s-1): (-jp) -s ),
CDC
P =— X.
Ei(-jp) = Ci(p) - j si(p), see 10.11 (5).
I r(1232m+3(e jp (ci p j si ,‘_21+3 (s s=1 (-jp)s 2 m=0 s'ul c
Ans.
Chapter X.(XI)
Page 126.
Page 413.
-342n+1 °' ja3t P n4.1 (cos0)e (real part) From VII-43 the imposed field is A°.= E C -P n=0 n v isinft Pcos8) L( where Cn pb2no2n(2n+1)(2n+2) if n> 0 and Co =-Ellviocost3. From 10.05 (10) inside P _Fi (cos0)eiwt .V"'•--7 IT P=47315 'ia 2n+ 32(Pr) 2n 2n+1+isnr-2n-2, _ +1 (cos0)e3mt . By 10.06 (4) at boundary Outside sphere: AnI2n4(Pa)=(cna2.1-1+Dn2n-Z),5 (1) and pv6(rA)/br =pb(rA8)/or so r=a, 4 = 43 so sphere the solution: 46= EAnr-1'21 . r (Car
AvAn t-[(2n+1)/W5]I 2o+2 + P4/iI 2o4)
p(Cn (2n+2)
a2n+l_ _Y Dn (2n+1)a-2D-2 3
(2)
(1)- (2n+l)p/ii + (2) gives An (p-pv )(2n4-1)I 2o+ri + Pia-I 2o4 = pen (4n+3)a2n+1 (3) 2n-1-1 r Ans. /t(K-1)(2n+1)I, 3 (Pa) + PaI2n4(Pa)) Ln+7 By 5.231 (3) From 10.06 (7), 10.07: a= 17T(p/p) 244r2 sin0 dr de . r rp fp 2(2n+1)(2n+2) Then from 5.322 (1), J-11 2+1‘ 4n+3 pa so that Al 0 r(19...,,..2.(1;r)I,_,A(Pr))dr 1 2n+-12.3 4 (a/P)(131 2n-1-4I2n-1-1 +131 2n-1 ‘-"' 2 so An = CnK(4n+3)a
„,dp_
7= Trpap-2 ni°0 (2n+1) (2n+2) (4n+3)-1An'An ( - 131 2n- 41. 2n-4 -35-
2n f-
Ans.
-
Two equal alternating fields like 10.17 (5), Aox along the sphere's x-axis and Aoy along its y-axis 90 °degrees out of phase with Aox produce a field rotating about the sphere's z or 9 axis. By 5.24 (5) both can be referred to the z-axis with factors jm(fil+cot) ejm(0+juk-r/2). e No r-terms appear in 544 (5) so A is the real part of and 2n+1 2m+1 ' j (2m+1)0+jcpt 2m+1 (0 (2m +1) csc 0 P 2o+1(cos 0)- s in0 P 2,1+1(cos0) )e A =nEomrioCmnr by 10.17 (5) and 10.19 (4),(5). A is finite at r=0. Eddy current field must vanish at 2n-2 for 6mnr2n+1. r=co so it is A with BmnrBy 10.05 (10), interior field is A with 5r)Arr for 6mnr 2n+1 . Both limn and iimn are complex. At r=a2Ai = io and AmnI2n+3 ("12n-2.y n o (rAi )/Or = KO(rio) /ar . Thus C a2n+1+15 Arn I /Nei and K(2m+2,13 2n÷lCinn -K(2n+l)a-2n1 2n+22r =Amn t-iI 2o+.3_+abI 2n+3/oa)/45 =An {-(2n+1)I 2n4+a,,FE3I 2o.F.Oba by 5.321 (1). Eliminate iitm, 2 2n1-22 Amn=(4n+3)Ka 2Cmn /R2n+1)(K-1)I 2n+2+4/5aI 20÷0 Ans. The power used is constant so
dP=2d13=Ti•Idv by 10.00 (10) where by 10.00 (4) Ti=-jcii. -15=trA'i •iiir 2 sin 20drded0. Put 2n+1 and 2m+1 in place of n and m in 10.18 (4) for the angular integral and use 5.323 p+1 217 for the r-integral so j 1,rn •A sin 0d0d0= 217(2n+1) (2n+2) (2n+2m+2):/{(4n+3)(2n-2m):3 . Sari
3(kpr)I 2
3(k r)dr= 2n+7
.„2+ ,15ai
L.n 2 4,.nT-2
312n-FTil• Power used is 2n-17
1113 2a co n (2n+1)(2n+2)(2n+2m+2): p jOrrX0Anin •'mn (4n+3) (2n-2m): L a4/715i 2n412n-14 +417)A 2n-l-ii 2n-143 By 10.19 (9) the torque is T= Pico. Ans.
Ans.
Chapter XI.(XIII)
Page 442.(464)
Page 127.
Magnetic vector is normal to incident plane. Radiation in r-direction is
Ye
h
proportional to r- in0. Antenna intercepts amplitude proportional to sing.-1
By 11.)8 (14) reflected amplitude is reduced by factor tan(.9-e")cot(0+0") so by Dw 401.08,9 a_ e 24 in20 t an (0+0" )_sin 20- s in20" s inOcsc0" - sec 6%/1,- s in20" 2e"h-sres/(eLe 2+4h2 tan(0-0") sin20+sin20" sin0csc0"+sec9sT10"-n/c"/E+1/elta-ersin‘ti ,Veucose 26"hA.4 12e"h-,164/(e"-E)d2+4h2 d2 I cl° d2 +4h2e- 24,1;241112 IccE 2 so Ircc [ • 4h2+d 2A/4h 2+d 22 c"h+V"=Id-(4h2 +d2)3 (6-ev )d2+4h2e +2€ ev" Oth tAi( e" - e ) dr4-Th
Are5JOI9
d12 p
-2 By 11.08 (19) amplitudes transmitted are [Ei•2v"/(v+v")]1211/(v+v”)]=4vv"E/(v+v")2 . If n2=e/E 8 16n 2 16v 2v" 2 coskirsin2 0-n2 Ans. Phase difference from 11.10 (6) is tanAns. I t(v+v ) 2 (l+n) sin2 0 -3I.Time lag between routes a l b and acb is 2ac-itan*sin0-2asec*•c/v =2ac-1 sinesec*(sin*-cs0;0=2aelsin0cot*=tl. Phase lag is cue= 6 so 6=t'2re Nac= (4ran/70cos* where n=sinO/sin*. From 11.08, for (1)
B in incident plane and (2) E in incident plane, we have Entering: (1) Ei=E2sin4/cos0csc(*+0), (2) E i=E2sin4icosOcsc(*+0)sec(04). After n reflections (1) E0=Ei t -sin (0 -*) csc(0+0in, (2) E0=EJ-tan(0-40cot(04)f Emerging amplitude is (1) En=2sinecoscsc(04)E0, ( 2) En=2sinOcos*csc(04)sec(0-0E0, which is, in terms of E: (1) En
(04)
_ -4sin0sin4fcosecos* sin11 sin 20+0 • sinn(04) '
-4sin0sin*cosOcos* tann(04) (2) E n sin2 (0+0=2(04) tann(0+0
Then by Dw401. 405.08, 405.0
(1) En=(1-b2 )bn, (2) En=(1-bi)14 where by 11.08 (4) and (14) b and b l are fractions refl. 3e2j6+b3e3j6 +.31 e jcutE=b 1_ (l_b2)ej8 0 (b2e jo)niejti3t E 1 E without enterkrig . Add : Er-1g- (1-b2)[be3ki-b k2 2(1-cos6) p 2 2 i; 0 411 2 3in(8/2) lE jwt_b .1-e Ans. =13(1-(1-b2)3 ro 1-b2 JO' 1-2b2 cos8+b4''° -(1-132) 2+ 4b2 sin2(8/2) e 1-b2 e36 II.Boundary value solution. r= -xcos0+zsin0, EI OE2, 121=112 , X1 =nX.2 . Ey only. r. o 2rj (scos0-zsin0)4e -2rj ( scos0+zsin0 ) , eXi .5e 2rj Oicos0-zs la) _ e s' where %x= x etc.' 2rj (nxcos0" sin0) -2rj (nxcos0"4-2s ine) . +Ce n sine" = sinU . Ei= Be at x=0 i1 =i2 so E044=i+C (1).By 11.01 H ilt =Hot or(VXE)t=(VXTA so Nalcos0(44)=Ailicos0"(i-6) +60.15e qcosOsecrin( ButiVITsin0= asin0". so sine"cos0a4,0)= sinecos0"(ii-6) (2) At x=-a, Be q qcosOsecErin (4) where crib/ 2 (See I above) . H2t=H3t gives sin0" cos0Cie-q-6e9= sinecose"De" Eliminate 5: isin(0-0")eq+'C'sin(0+0")eq=0 so 6.be 2 cli (5) From (1) i o+I=i(l+be 2q) and from (2) sin0Pcos0(50-A)=sinOcos0"i(1-be 2(1). Eliminate B. h-b(1-e 2q)E0/ ( 1-b2 e 2q ) as in I.
In initial II equations write H forE. At x=0 141 =H2 so ;04141+61 (1)%47;(nx110i=4-e-TODXFOit 51eqcosOsec0" ( so sin20(ii0-4=s'in20"(ii1 -C1 ) (2)'. At x=,a 3)' and eqcos6sece" e -q)=sin20"51 (4)' By Dw 401.08 and 401.09 61 =b1 e 2c i1 The Al solution is the same as in I.
(5)1
Chapter XI.(XIII)
Page 443.(465)
Page 128.
-4Proceed as in 11.07 using v" and v instead of p"E" and pe and get 11.07 (3) as before in the form sine/sine"=v/v". But v" is now a function of 0" so the ratio of the sines is nolonger constant although the normal to both incident and refracted waves is in the incident plane, since from 11.03 n= C1(DXOTT=C3 (EA) and since D and E are not, in general in the same direction, the refracted Lay may not lie in the incident plane. -5Let r =
jmi +kn') v' be general ray vector to wave front and let ro=(1.201-kno)v 2
be wave vector normal to wave front. r•ro =v1. Terminals of r and ro lie in plane '% normal to r 0 if r• (r-ro )= 0 or if (i'Ll-nIn)vv-v 2=0, which is v2 /v'=1'.e+n'n=costi• This is 11.05 (3). Q.E.D. From symmetry (see Fig.11.05) a line tangent to the surface at two points must be normal to the xz plane. In 11.05 (4), let r = vt, a= vi t, b= v2t and (1) c = v3t and multiply through by r2 which gives x2a 2 (r2 -a2 )-1 +y2 b 2 (r 2-b2)-1 ±z2 c2(r2 _c 2 ) = Q Putting over a common denominator and dividing out r2 gives r2 ( x 2 a 4y2 b 4z 2c 2)..a2(b2.1.c1) x2 _132(a21..c2)y2_c2(a2+132)za_l_a2bac2.0 (2). To get coordinates of surface points at which surface is normal to xz-plane, set bS/Oy=0 or 2ybS/0(y 2) =O. This gives x2a4y2b4z2c4b11.2_132(a4c2 )=0 or 2b2r2+(a 2_13 2) x 2-(b2_ c 2) z 2=b2( a 4c 2)
(3)
This is the equation of an ellipsoid. To get section normal to optic axis, -4/7:7x1 A/j--c2 , rotate z' to coincide with optic axis. From 11.04 (8):x=xicose-b z=z'sin0=fa1-7b2x'Aia27a. Substitute in (2) and get circle r2=1(a2+x2 ) Q.E.D. -6The angle between the optic (z) axis of a uniaxial crystal and a normal
A n to its face is a. Since v1 =v2 ¢v3 , we take n in the xz plane which makes an angle 0 with the plane of incidence. Incident beam direction cosines are /,,m,n. AB 2=a 2 tantt+a2 tan2 0-2a2 tanatanecos0=a 2(sec 2a+sec 20+2seca secO•) so n = cosacos0-sinasinecos0, (1) similarly n"=cosacose"-sinasinercosczi ( By 11.04 (7), (v 2 -14)(v 2-v2i )(i,"2 -i-m"2)+ (v2 -v23 )2 n"2=0. But £"2+m" 2+n"2=1.
v sine E. sing s in e . If v=vi ,sinO"=c .IE so (va- vi)ctv2-v3+n112( vs _0_vt)=0 (3) Arts. Also s in e" -lvo e VV1 E e sin10"-e l E sill@ 1 EV 1 EV For other sol.put n" from (2) in (3), use vi=e—c2 ,v1=7- c 2 so n"=-•f---i-- 1 1 3 VI -v3 ev ( e 3-gi)sinie =coe2acos2 014- 2sinacosasinegcosOgcos0+sinaasina egcos20 or cos2a-I-E1 (E3 - 041<sin2acos20 - cos 2Ct - e l e3rev(e3 -E1 )]"-bsc 2 0)sin2 OZ=2sin.Ctosasinagcosqcos0. or M+Nsin2 0g=2Psinegcoseg or finally (N2+4P2)sin4q+2 (MN - 2P2)sin2 q-FM2 = 0 (5) so sin2 eg=(2P2-1411t2P0/1271•131(N;+•4P15. -If 0
0>i
Where 14=coe za+e i je 3 -0-1, N=sin2 acos 207cos2a-Fe 2 e3 ( Ev (e3 -Osin2 0-1, P=sinacosacos0. Ans. When optic axis is normal to face cp0, M=E 1 /(e 3-0 , N=-1,,-E i c 2 (ev (E 3 -€ 3)sin 20)-1, P= 0. E,e,isin‘y y(e3 - Ei)sin20 Ans. = -= --1-sing e" g3+(6 3 _E1) Ev sin ae e N E3 - E 1 ElE3-FE4E 3- E i ) sin 2e -
Page 443.(465)
Chapter XI . (XIII)
Page 129.
-7-one171:74 X 4- j (La -81 z) 724tt).= pie io 'Ai /60= quo 14.41f3M3, 2
j (wt- Wz) x])e 71A 2=112 E 2O 2 40/60=c 2f-iiplIcos[N/14=Tx]+15 ,44737 81sin[ frz). At , B2y.Capicos41-13'2x ef(mtx=a eiAix=e2 A ax, B157=C f3fe 4131.1- x
°It -131z)
Ci e i e-' 137"Tq a = cac a cos-' ...A 2 a (1 ) AlZ =A2z) Cr/P12- Pie
5117pa.ce41718T dri
0/31-71a] (2)
p 2Biy=p i B iy also gives (1). (2)/(1) £243'‘ -8?=E1 .411-1ratan[431-(312a] (3)
Solve for P.
From 11.09 (4) v =a)/8' Ans. To find p. graphically let 0=8' 2 -8?, v2=81-812•
EAU= tanva
Then u2 +0=81-8!=wa(p2e 2 -ii1 e i)is constant. Plot (3) u=(€1/e 2 )v tan va, ,/-3 7.1 =1.1489. 44 3a3=1.681 811=81+4 = pa -vi. If (3' 2 -1= 1.32,1
=1.296.
4.59.56z 1.296tan1.296=1.296.3.546=4.598. To fall to 0.1 of its value at -1.1489x/a -1.1489 -3.451 -1.1489 3.451 x=a, must have 0.1=e /e /e . x-11489 3.005a -8oaA, 4 =,, v241,-,i m-poe i-iElb t a ) V" '-'2=-14 (1 '-'2‘(-142 r.; )sin(N4TRIxle j°1t-ft) p, 2 siriL I-,,/Ti-172a) ( 1) P2e 2 i At x=0 A 2y=0, At x=co A iy=0. At x=a A iy=A ay so Cie 48 12-11 a=u ia -pi a paBiz =p1B2z so -p2CrIFITA-....ep =Ca llioi rfT2 cosfa) (2). Bix =B2x also gives (1). 2-
A /T-I31 f x+j (cot -pto
'
(2) /(1) "-IP -112"43TLiff =ilisiifl i icot{-41-7Fie)=-ui1T- 1312cot (II -.11T-71 a) (3). Thus VT-Taa>17/2. Example: Let p1 =u 2=pv, c2 =4t1 , 4/812 a2 -1 = (4-81 aa2 ) cot(Tr) If I3'a=1.03,1 .%/7.= 0.247
.,17 4. 3"Tai= 1.713, cot(7i-1.713)= 0.1427, 1.713.0.1427 =0.244 rz 0.247.
pi
: Pa= 1 : 1.03 : 2. 131= aqv, v=c0/1.03. To fall to 1/10 of the value
at x=a must have 0.1= e
-0.247x/a
/e
-0.247
=e
-2.55
/e -0.247 0.07808 0.7811
x =(2.55/0.247)a = 10.33a. Graphical solution for 81 : Let u= 8' 2 -8
v=
.1-v cot(r-va) u2 -1-v 2=131-132, = co(pa e 2 -pico is constant. Plot ul112 Coordinates of intersection give answer. -9Charge leaks off line at ends A l and B. Each leakage edge has line wave velocity v =±(14) 1,as shown before and afte passing center, and drops line potential by V so that i= (V-0)/R =AV/Zk and AV= ZkVo (R+Zk). Make Fourier expansion of line potential V+LV. V+4,V = V0 (1+ Zk(R+Zk)"iraioAnisin[(2m+1)17z/k) . Multiply by sin[(2p+1)17z/L], int egrate from vt to co .ilm E02m+1 cos(2 m+ rt s in L-vt, solve for Amand obtain V+LV=V0 [1-1.4Zk[r(R+Zk)] m= L When vt = ,Q process repeats with Vi=1.70 -24V0 =(R-Zk)V0 /(R+Zk), etc. so that the line potential is Vs -
va r _zkis iR4L, R+Zk R+Zk
L
(
.4)ss . (2.+1), (2m+orrt] when sgpEL < t < ( s+1)4/Tit in cos ,-„ m= 02m+1 vuEL
Ans.
Page 444. (466)
Chapter XI .(XIII)
Page 130.
-10R
Potentials: Incident wave Ei reflected wave Cr , across R after reflection CR.TE, ,
From 11.16 eimoves along line putting charge on it. When R is reached leakage begins as in last problem, causing a reflected wave er=ei(R-Zk)/(R+Zk). Leakage edge moves back toward battery with velocity v=(.31) T. This charge decrease means reverse current. Thus = (R-Zk) / (R+Zk) and = 2t 0/ (R+Zk) . ./M+2,0=ii-ir. In table below: ®= CR=Ci+ery iR=CR/R= 2; n=0
n=0 n=1 n=2 Potentials: nth incident= Currents:
C
-sec, eTo nth
()Co -08E0 c1)82C 0 zk=11 S 2E0 R+Zk sn) Ek+R
DiR =
Battery end always Co so
refl...00
Total=
13E0
n=1
n=2
_e2 c0
eto
el,(1 - 8)
0(1-8+82 )
when (2n+1),ITIT < t <2n+3)414
Ans.
-11VRis potential across Co and i0 is initial E0 current. R v E --11R+zk . Across line V.rio zk- C9k 4k 1213 zk VR=C 0-i 0 R, jo'iR= R+Z k • Consider V and -V waves leaving 0 and2/ colliding at z =L.
0
7/
i
+it
+V
; -V
i
i
-
From diagram, first reflected wave has zero potential and current +2io . After i1 leaves battery at z=0, 2i0 adds nothing to line potential and effective EMF of battery is CI =Co -2i0 R
0(Zk-R)/(Zk+R),
i1=e1/(R+2,0=Co(Zk- R)/(Zk+R) 2 . When is starts,
to line potential. Effective battery EMF is Ca=e1-2i1 it
2i1 adds nothing
a -2€ 0(4-R)R/(Zk+R) 2=C0 (Zk-R)/(Zk+R)1,
i2=C/(k+R)=E0(Zk-102(k+R)-3. Total current between nth and (n+l)th reflection is In =
+ • • •+in = eosgo(2-8 s° )
(zk-R)s (zk+R) -8- 1
Ans.
-12Z z satisfies 11.14 (2). Z= V(x,y)F(z-ct). c=(pg) T. If f(x)= aF(x)/tx, then by 11.01 (I = 7 -416z= V (x, y) (z-ct). A =it Intl /bt = LcA/FEV(x,y)f (z-ct). By 11.01(4(3) is:=A-c'a:(527•Adt)d bt,i/bz= AlTreV(x,y) f' (z-ct). Z(Z•A)=A/Tre,Z2 V(x,y)f i (z-ct)+0,41oz2 . 624vo z2 = pe owbo so r.f6 2Alot 2dt2= p€A. Thus VrCEIc.x'Z'aV(x,y)f(z-ct) =15X22V(x,y)f 2 (z-ct)
QED
-13By 4.13 (5) and 11.14 (11), L= !IOC= -I (p/r)cosh((c2 -a2 -b2 )/(2ab)) Ans. See Fig. 4.13. By 11.18 (5), (6) Ri=111(2`rfarl[coshLal+cosh2f 2 ). By 4.13 (3) a= a sinh2771 = b sinh242 1 , 217U c= afcoth211+ coth— 2),- b a- = )11
2TrU 2TrU 2r711 = a cosh-1 1/ sinh -2 c + b cosh— pi2 • 111 pi 2TrU 2111.1 j a 2 2rU c2 -a2+b2 l = b 2 -a 2+a b cosh c-a cosh-1= b 1 +—sinh 2 pal cosh 1pi 1= b2 pI pi 2ac pI 2TTU
a p
2ru
c2-a2+b2
_ 2rrul
-2c 2 (b2 -a2)+02_0 )2 _4a2c2 _Pie2c2(a2+b2)+(b2-a2), . a = ajcosh2 22-21 1 = 2ac pI 2c T(a+b)(c2 -(a-b) 2] T c 2 (a+b)-(a+b)(a-b)2 Ans. R1= 8-1 so Ri=cainabe4s/c 4 -2c 2 (a2+132)+(ba_a2)2 2ra6 2abc cosh---2 pI
2bc
Page 131.
Page 444.
Chapter xi. (XIII)
-14- (p466) Capacitance per unit length is C =[U] /IV2 -1/1 1 4.11 (4). From 4.22 (4)
Fig. 4.22
U 70
VI = cosh-li, V2=cosh-11-, [1.1]= 217. By 11.14 (11) L=am =-1.{cosh-1 11 - cosh-1-g} = -1-(1-1/11)&((b+r r-cT)/(a+AM-c 2 )). Ans. By 11.18(5) RfIa= Riff i iids-421ids) From 4.22 (3) W= 4(pI/77) sin-1 i, z =1 c sinf271W/(iiI)}, dz/dW = (217c/(pI))cos 27
ILIVII El cos 277(U-1-jV) 217(U- jV)] -i_ I a21TU + sinha 2r1 7 f pI cos r Pl2 " ..vi 2ds = y Id z I 277c [ pI s bs I pa - 2rfc cos pI PI
I‘c-b--->"
dU 2TtU. 4I pe 0 < x
kdx T Iair filds=- 1T _,...--........- so R = -2. }-1[I Ks-3 +-11({§3] 9 R' ' T1 1T 2 c 04 1-1(2 sinax b a a i 772
Ans.
-15-(p466) Assume uniform and neglect edge effects since a ,G< b. From 2.05 (1) C = eb/a, L = pe/C = pa/b. Current is uniform over two faces so that Ri=c01.i= 2R'/b =2T/(b6). Substitute in 11.18 (1) for L, Liand Ri to get Zk and in 11.18(3) to get
r.
I. 4-----b-0 Ans.
-16- (p466) By 4.13 (5) and 11.14 (11): L=pe/C =1(p/r)cosh1(824-ba-c 2)/(2ab)}. By 11.18(5)(6) 1j1 =b sinh R=R1(2ra)-l(cosh[2TrUi/(pI)]+cosh[277U 2/0-115])• By 4.13 (3) a = a sinh 2-
2,
27U 2TrU a 277U 211U 27U 217U p 1+ bcosh- a . via) - , I-,= sinh-iiya csch--ip 1, c= -a cosh-f c = a(- cot1-7 1-14-coth-fpI 277U 2t7U aa 277U bcosh-2=acosh-1 +c = bil+ _ s inh2___i ..,ih 2 _aa +aacosha 1 . Square, cancel cosha term. 13 2 pI pI pI pI RI 2171.1 C 2 _b2 +a 2 2t7U c2 -aa +1)2 -cosh 1 . a = aVcosha {27*U1 /(PI)} -1 =4/c 2 4 -2c2 (824-b2 )+(ba-aa)s /c . ' cosh pI 2= 2bc pI 2ac T (sa-b)(c2-(a+b)23 T1 T c2(a-b)+(b+a)(ba-a2) Ans. - 217ab64/c4_2002+h2)+02_a2)2 RI = 8 so Ri= coLi = 21 a6 2abc 17
For a lossless line y=0 in 11.15 (2), (15) so =j Thus the characteristic impedance is, from and 71:=#117i. }
4--2 a--4 4-213-).
-› co
11.15 (15), Zk= jcppi:/(1C) =Tle/C where C is the capacitance per unit length. For strip in slit as shown, from 4.29 (10), Zk ='1e/C= r)K(^/1-(b/a) 2 )/(4K()) Ans. Notice that if walls are lossless but medium is lossy this still holds. (a 0 0 in 11.15 (15)). -18From 4.29 (9) From 4.29 (10)
Zk = Tle /C = r1K(b/a)/K(1-0 ,1-7777) Zk
-1971KW1-(b/a)2
C
8K(b/a)
2r 4-2r or
Chapter XI.
Page 445.
Page 132.
-20From 4.30 (2). A=a, B=a-b. Zk = rig /C= +711C(7)/K(1 ,1-711/a)2)
‘-b-)
-21In 4.30 A=a, B=b, C in 4.30 (2) is doubled, r 2 = ab, A.F3TX = b/r. Zk= Tie /c = 4r1K(
(I:./r)2 )/ K(b/r) - 22-
U=0
From 4.30 (3) k="iF iga(w+gXh-w), C is from 4.30 (5)
g
U L. —
Zk= TIE/C= +71K(k)/K( 1:1E2) - 23 From 4.30 (6) k= R(c i+c 2)/(R 21-c i c2 ), C from 4.30 (7)
Zk = ne/c =+rnMaff. 72 )/K(k) -
- 242b
C from 4.31 (4), k= cosq(1+X)nt/b3 from 4.31 (2). X is given implicitly by 4.28 (7): esc{i(l+X)11c/b}= cothC-1(1+X)Trc/(bX)). Zk= TIE/C4TIK(k)/K(Nr1-72)( -252b
Zk= rie/C=ITIK(e -rralb)/KW1-e -2 Tratb
C from IV Prob.61.
CO
- 26Zk = Tie/c = T1t(d/c ii2 .i h2)/K(h/#412 1 h2 )
C from IV Prob.62.
-
CO
124'h
- -
-27-
4-2d --->" V= 0 4-2c-4
Zk = rle/c= 1TIKCoostrrc/(2d)3) /Ksin[rc/(2d)]} 0,1
C from IV Prob.63.
- 28C from Prob.64-IV. k=(X27Fb2 -h)/b.
E—L=1- -. VT:7 0 142-,I2h
Zk= ne/c= rac(k)/K(45:17 .1 ) - 29-
C from Prob.65-IV.
M=coty+cotp, N=coty+cota, V is potential.
Zk =r1E/C= 411C( 7:(7/N))/K(g5) A/i4,
U= 0
- 30C from 4.29 (9), b a from IV Prob.66. Zk= TIE/C=r11((b/a)/KPr-73/a).) - 31-
d U=
C from IV Prob.79, k= cos(rc 2 (1+X)(213 2)-1 ), X from 4.28 (6) with b2 ,c2 for b,c. ..4/7-72 ) Z k =71E/C=4;r1K(k)/K(] - 32k= cos(-1.2-17(1+X)), C from IV Prob.80. X from 4.28 (6) with
for c and -liN/F for b.
d
Zk = Tle/C= iTIK(k)/K(1 .17:172)
U =0
U0 CO
/71, 2,1
Page 446.
Chapter XI.(XIII)
Page 133.
-332 )= bK(k) From IV Prob. 82, (5), (6) and (7) (a-c)K(NT-7 determines k and ki=45:-72 . Let P = 2akI K(k')/(1713). Then
-4P 21
B _ 1[, -
14(1+Pa ) 2 kI 2 le P 2
ne • `k
rliC(45TA C 4K(411-(B/A) Ans. -34- (17,p466)
To make an E loop (open circuit) at far absorber boundary need reflector (2n+1)X/4 meters beyond it. From 11.15 (13) to match at front face of absorber, Zk = qcothfcl. If d is very small this becomes Zk = Z4/(lid). From 11.15 (15)
Zk =
4= jad/(a1 + jut),
= at+ jpi
and (1) becomes
= icull'ACe+13')2(1 3= {(yt + yze)(1}-1. This is satisfied 111. Ans. only if led >> Jule'. In the latter case y'd
7
V=0
r Uoc+.
U= 0
Chapter XII. (XIV)
Page 134.
Page 482. -1-(1,p501)
wt -13r) . Then by III Prob. 2, jpr bcobes203 811EVx:i zz= 9(2) zze jca te [(13- jr-1-j 13rJ sin0cos0-(pr -2 - jr-5e-Nlj (a3t-DY) =0 e = VX(V)d) -3jr-3 +3pr-2-Fj B2r-13 sine/cos°. From 11.01 (9)
From 12.01 (13) 8rrei z=q 22(13r -i-jr-2)(r cos2 0 - sinOcos0)e j
(s in 2 0 cos0) /be
ry
Er =q(2)2(3cos 2 0 - 1)( 3pr cos (cut-pr) + (3-132 r2) sin (cut-pr))/ (817er4 )
Ans.
8TrZEr = 42)z ej (13t -Pr) -3jr-3 +313r-2 +jp2 r-1(r sine)
81EE - -q(2) eiwt-l-in20bC(-3jr -2+ 3pr -1+ j ps)ii 13r 3/ (rbr) zz 2 Ee= q z sin2e( (6 pr -p3 r3) cos (cut -pr) + (6-302 r 2) sin (cut -pr) )/ (1617Er4 ) Ans. e i (cot -Pr) 3jr-3+ 3pr-2+ j par -1)-isin20 8TreB0-- j Flee) zz B0= citz21;1psin20[(3- p 2 r 2 )cos (cut-pr) - 3prs in (cot-13/)3 / (161Tr 3) Ans. -2-(3,p501) From 12.01 (16) 16rEr 2 Ziy = cgy (pr-j)(i sin0 + j cos0)sin0e j (wt-Pr) . From 12.01 (11), (12) 1617432cy = qTy (f3r -i-jr-2)(r I sin20sin20 + sinOcos0 sin20 +0sinecos295)e j(wt-Pr) . From III Prob . 2, 161re (vxzicy)r .cgti(p) (sin2 A) isin860]cos20 - cos06(sin20)/o0)e3 (cot-Pr) 0. ( r component) = 42) xy'(13r"2 jr -3 )(2cosOcos20 - 2cosecos295)e3(cut-Pr) = 13r -sinecos20bE (13-jr-)e jPrDrorie jwt . 16re (VxZ32(y) e=qTcy ((pr -2 -jr-3 sine (sin20) /O0 -
=4y sinOcos20(3Pr-2 - 3jr.-3 + jp2r-1)e j ((a-Pr) 16rre(VxZ,2cy)0=q1x2)y[sinOcos0 sin20 b[(13- jr -1) e ifir]/ror - (13r-1- jr-2 )b(sin2 0)/roesin20e-jPr le jult j (cot -pr) -cos0 cos20 • 16rre (VXV ) ) e . =q;c2)y sinecosesin20 (-3(3r-2 +3jr-3- j132r1)e From 11.01 (11) B =
(v)a) /6t = jculie(WZ)=jprIe(VxZ) from 12.01 (9).
02) pri j pr -2+ 31-3- 13 2r -1)sinecos20 ej (ca "13°' e -- xy
1617B
130 = -Be cosO tan20
Be = cicdyprif (3-13 2 r 2 os (cut-pr) -313r sin (cut-pr) )sin0cos20/ (1617r 3 )
Ans.
Ans. Br = 0. Ans.
From 11.01 (9) 'E. =Vx(Vx'i ) and from III Prob. 2, Er =[o(sin0A0)/o0-bAe/b0] /(r sine) . 1617CEr= cicg (3pr-3 - 3jr-4 + jp ar-2)(- cscO[a(sin 2 Ocos0) /be] s in20 - ( cos 20 ) /60 e (wt-¢r) =q 2 (313r -3_ 3jr-4+i fizr -2 )sin20 (1-3cos2 0 + 2)e j (a3t-13r) = 413,(3f3r-3-3jr-4 + j (32r-2) 3sin20sin29sei (c13t " xy Ans. Er= q()2c)y 3sin20 sin20[3prcos (cut-pr) + (3: P 2r 2 )s in (c13-13r ) ) / (16rer 4 ) jcut From III Prob. 2 and 11.01 (9), E0=b(r (VXZxy )0)/rbr=4)ysinOcos20?)((3(3r-l- 3jr-al-j132)e jr3r)/rbr ierti (wt -Pr)/(16Tre) E0- cf2) x ysinecos20( -613r-3 +133 r-2 + j(6r-4- 3f32 r-2)}e j Ans. = qxy 2 sinOcos20( (P3 r3-613r)cos (cut-13r) + (3(32 r2 -6)sin(cDt-Pr))/(1617Er4) Ee= 46 cos() tan20 Ans.
Chapter XII.(XIV)
Page 483.
Page 135.
-3From XII Prob. 1 need sin 20 in Ee. bCP2 (cos0)3/60 = Pi(cos0) = z sin20. By symmetry need even function TM waves independent of 0. From 12.11 (5) for expanding wave: Wtm =ECa P2n (cose)k2n (j8r). Then from 12.14 (1): Ee = jcuAe = -ja)b 2 (rW tm )/rbrbe = -j(DE-Cn(b[rk ar (jfir)]/r6r}Pin (cos0) by 5.23 (6). If jiia<<1, then near origin use last term in 5.37 (12) summation at r=a. Ee= jrnECnP 2n (cos0){(4n)1./(2n):}(-1)nt:{(213.a)-nypaoa=-ca-2ECTIPin(cose)[(4n):/(2n-1)1.3(-1)n(213arzn) since catil= c by 12.08 (9). Ee = 0 except at 0=a. Multiply by Pln(p)dp and integrate -4. 2_13_2..221) 1 to 1 by 5.231 (3): -2sinaPi2n(cosa)rEed0= 2sinaPin(cosa)Vo/a- a (4n+15(2iy. 8a Eliminate the n=2 term, next in size to the n=1 quadrupole, by setting. C2=0 so that Pi(cosa)= (5/4)sin2a(7cos aa-3)= 0 by 5.23 (10). Thus cosa=.577 and Cl --1 a3v° sinasin2a. 6 re-J[3F 3 . iCa jCbt Ans. At great distance use first term in 5.37 (12). Eee a sin20 e jcuCiYoi (wt j(mt-Br) 3 = jCOCI C-e -sin 20 of which the real part is E =- 16133 aa (Vo/r)sinasin2asin2esin(cot-pr) -4From 11.01 (3) and 12.13 (5) as r
E0= -6A0/at= -cBe.=4-cop8a2R-11sin0(cos[cut-0(R-2-cos0
-coslint-8(R42-cos 0)31 = -icupPaarlIsin(cut-BR)sin{(Bb/2)cos0} by Dw 401.07. (a) For zero at 60°: lifibcos60°= 7}- pb = rt.
b = 4r/13 = 417•X/(217) = 2T.
Ans.
(b) For maximum at 60° : o(sinesin(22-0bcos0)) /60= 0 = cosesin(lobcos0) -+Bbsin2 0 cos(+13bcos0). sin60°= 4/5/2, cos60°=1/2.
a . z pb cos4-f3b =isin+pb or
tan4-I3b= ffib. tan 1.3241743.9716, 3.1.341=3.9722. 8b=5.296 Ans. b=P rx=0.8431X Ans. -5-(5,p502) Find field in xz plane at time t. It rotates uniformly so in any other plane it differs in phase by 0. Dipole components are: Mx=Mcoscut, My.--Mcos(cot-7) Bo is entirely the OyB-component from My . In 12.01 (8) put cot-3 for cot. Be=-cupM(41Tr 2)"1 CrBsin(cut-Br)-cos(u3t-80). B0 is entirely from Mx . It is the negative of the Ox -component. Write 417-0 for a in 12.01 (8). Then B0= cup140-Trr2ricose{rScos (cot -13r)+sin(cot-Br)) In any other plane the phase lags by 0 so Be = -conM(417r2)-1(rB sin(cot-pr-0) - cos (wt-Br-0)) Ans. Similarly
Bo =u312.4(4.ffr 2)-1 frpcos (tot-Br-0)+sin (cot -(3r-0)) cos()
Ans.
Br= O.
Ans.
Ee and Er are entirely from Mx . Write 2 -0 for 0 in 12.01 (6) and (7). Then M sin0 Er =-iii 2T sC cos (cut-Pr-0)-
E0 is entirely from My of which it is the 05, component. Write in 12.01 (7).
O
McosO sin(wt -13r-0)) , Ee= ----5f(1-132 r2)cos(cot-13r-0)-Prsin(ait-pr4TrEr
for 0 and call for tut
E0=M(4Trer 3)-1 {(1-132 r 2)sin(cut-fir-0)+ prcos (cot -Pr -0))
Note: M = i o dz o and from 12.01 (9) 13=a)./r ie = cop/7.
)}
Ans.
Ans.
Page 483. (502)
Chapter XII. (XIV)
---.
-62rtp 2n+1 dTr=m-ri -[E zXBo]dz, el=717, rarz L-z, rbr4 1+Z. 2n+1 loops= (n -14)X. Use 12.03 (17), (19). cospL = O.
Page 136. -4
-
4.____ z - iff -
...----,
<
22 , sin(cot- pr a)= s in ilk -(n+ -1)11 +13z)= (-1)ncos(cut+ ez), s ingot- prb)=-singot- (n++)rr-flz)= elicos (cot- PO c440(4)n[col(ot+iz) ' ez)1 c I , [Lcosotco 3z zsinwtsinezi cos 12.03 (17) Ez 2+tzT -p'n i" Z _ n cp2 Ie •. 2TTOdz 1 icos2 f3z LcpIdidz a (2n+1)Trz 411(itz,cos 12.03 (19) B0=1.11.0 (2TY p) '(-1) coscutcosftz. d11 - 4 ,172p •i• 12-za 22 The end loop radiates about twice the power of the center one. -7From 12.08 the number of oscillators in the x-direction is mx= 2, and the phase angle 4=3=
g.
r • a = hsinecos93. pi= (n +i)17 so from equation
above 12.08 (2), F= - pio cosc(n+12-)rcose)/ (21Tr sine), 13=217/x.. From 12.084 i (5):irr- _ p2iicos2((n+l rcosejsin2(417(sinecos0-1)) if a = X/4 411zrzsinze.2psiesinzar(sinecos0-1)) xi = pcIecos2 fi-(2n+1)!Tcose)cos 2 (4-17(sinecos0-1)) /(2172 r 2 sin2 0) )
\
‘ 1 . . .41
If X # X/4, r•a = a sinecos0. Substitute a for X/4 before sine cos0
=
pcItcosz((n+1)71cose)cos2(10[(4a/X)sinecos0-1])/(2112 r 2 sin2 0) -8-
p2r
15= iri217fi r2sineded0. For 0 integration with a=X/4, j cos 2 t (1-sinecos0)17/41d0 o Orsin e)2n+if 2n+1 1,241 ° r+i(-1)" 0d0=11-H). ° 2-TrsinecosOld0= cos =J -11+cosVE17(1-sinecos0)J)d0=17+SrsinC o 2 (2n+1): o 0 ar pc3 rrr pc I rtii+c os [(2n+1)17c o s 0] sinede P =2T -Tc2 17r 2 jo cos t(n + i)17cosejcscede=- j 411 Jo 1-cos 20 This is exactly twice 12.04 (3) with 2n+1 for n, so that
7= +17 -313.cIi[C+k[2(2n+1)Ty - Ci[2(2n+1)77]). Maximum is at 0=11/2,0=0. "max 2172 r 2
m•
) °-16772 r2
G. Irn (Do
1617 2r 2 . ;xi/ pcIg 277 2r 2 0
This is twice directivity of single antenna. (See below 12.08 (6)) -9First 12.08 (5) factor same as problem 7. F =,41--° cscecost(ri+1)77cose). Number of X2 oscillators in y-direction is my=p, all in phase so T is zero. /4 Spacing is b=X/2=77/13. r t •lb= -ixsinesin0 =rre-isinesin0, so last 12.08 (5) factor is sin 2 (ipr sine sin0) sinesin0) . From last problem r217 sin2( 1prsine sin0) i 0-factor integral is t o fl+sin(2?sinecos0)} s. a, k -ETT sin0sin0).c10 2n+1 . r2i7siL/ 12 •yrsine sings)do+ e sin0 0) _22)7_,_ "s in 2(jp rrsine _pnlansh cos 2n+10d0. In last integral; 2sin 2 (-ersinesin0) n=0` (2n+1). 2 1 integrand at 0 is the negative of that at TT-0 so that integral 0 to 2
Ef
cancels integral from
7
to r and result is zero. With p single in phase
antennas along the x-axis, the first integral would appear in P without the 1/2. The F factor in 12.08 (5) is the same so the directivity of the pairs is twice that of the single antenna array.
Chapter XII. (XIV)
Page 484. (502) -10-
For F-factor (above 12.08 (2)), 284./2, (3/ = 2TTA.= Tr/2,
F= uo Icos(ircose) /(2rrsinO) . Number of z-directed oscillators is mz=p. r•k= cos0, c= X/2, all in phase so C=0. Substitute .p.3aL.c211/ 2(-in 12.08 (5): Ans. 4r 2 r 2 2pNif.ii sin ae sins (rcos0) -11nIo zi. cosnoto sin= Current density is i= To sinn0coscut. Charge density, 0.4-c— p211 Rzr-acosy=r-acos(0-00)sin0. 411A0 = pa j [i]10cos(0-9$0)d00 . At great o distance neglect 4,and Ar . Let u= 7-0-F001du=d0o , let z = Pa sine. TT
4TrAo=paI0 0 s in n(u+0 --12os (cot -f3r+flas in u sine) s in u)(r-a sin u sine)-'du r2r TT p2r paIo V-1 ( C [cos n(0 J 0sin• nu cos(z s in u) s in u du +sib n(0- 2 )j0cos nu cos( z sin u) sin udu r -S [cos n (0-T) 0sin nu s in(z sin u) sin u du+ sin n(0-7) cos nusin(z sinu)s in u du ] ) r. 6 r2rr where C = cos (cot-13r) and S = sin (cot-Or). j =6zjo sin nu sin (z sinu)du, then by 12.17(5), n=2m+1 2rr (z) = 0 if n 2m+1. .1 and =0 by symmetry. = z PJ2m+ Z 0 sinnusin (2m+l)u du = 2TrY 3 2n1+1 p 6 r2r 6 cc j 4=-Ti0 cosnucc6(z sinu)du=-6-z- f(2-g)J2m (z)J 0 cos nucos2mudu= - 2Tqn(z)= 0 if nO2m. 12.17 (4).
g_
cosC(2m+1)0-i(2m+1)r)=(-1)msin(2nF1)0; sin (2m0 -mr7) = -sin (mr-2m0) = (-1)m sin 2m0. So if n is Laic m odd: A0= 2 (4) f2m+1(13asin 0) sin(2m+1)fzi cos (cot -f3r) Ans. If n is even Ao
(-1)m Y2m (13a sinp a 0)iosin,(2n)0) sin (cot-f3r ) Ans. so that f or all n ' s we have A0 Zr Jn(Pa sin0) sin n0 sinquil-kot-13r) -12-
Ans.
From next problem, (13); when a<< X, n=1 and if sphere is perfectly conducting, pi=0 and E.= . Ai = 3j133 a 3 , Bi = 2)4 so angular factor in 12.15 (20), since P =-cote by 5.233 (6), is
f ). 3 13 3a 3[9.( - c os 0+2 ) - o ( 1- 2c os 0) s . Thus E0 is zero when cos° = s or
=17/3 and E0 is zero at 0=0 or 0=11 so E has only 0 component when 0 = 7/3 in unpolarized beam scattered from perfectly conducting sphere. -13118a/3, J_312 (13a)--,12I 8a = 2rra/X << 1. From 5.31 J ota (f3a)—)N513a7/7 3;717(Pa)-2, .114(13a)—>,./21i7FT, 3T/T7/ (Pa) . By 5.31 (10): j1(l3a) -313a/3, jo(fia)-> 1, ni (fia) --> -(13a)--2, no(13a) -(13a)-1 . J.-1/2 (13 a)--)N5F From 5.31, as n increases, jn (8a) decreases if faa is small and nn (fia) increases. Thus 12.15 (16) increases with n so An decreases and we keep only A1 . From 12.16 (17) : Al =
[ . -13ae, 81a/ (38a) + 8'az/(f3a)2 + -e)f3'a/(382a2)r_11.,313'a(-132 a2z 2j(el-s)f33 01 j pati l3'a/3 - 13'ael3a/3 -(61 -e)/(P'Pa2 / 9) IM133 a3(3e -3e-e1 +e)fia*0 3(2E-131)
For B1 write pi
L J
for e i , E so
-p)8 3 a 3/ (2p+pi ) To identify Al and B1 observe
that, if a << X, phase differences vanish over sphere so static solutions hold. (continued)
Page 484.(503)
Chapter XII.(XIV)
Page 138.
-13- (continued) For dielectric sphere V0 = E(r+Ar-2) cose, Vi= ECrcose. At r=a, V0 = Vi so aa+A=Caaand
g o bV0 /6r = tioNli /eir so eo(a 3 - 2A)=Ea3C and A= (Ev-e).23/(2ev+e). From 1.07 (1) for electric dipole 417;r2 V= Mcose. Thus Me= 4rte0 A= -4Treo (El -Er)) Ea3 /(2e0+e1 )= 61748-3k . For uniform C' E sin0=-1KIErsin9 so from 7.06 (3) A00=-2-0-(r+172 )sini3, induction B= E/c by 7.02 (8) A0=
(1) By 7.22 (11 Aoi= 1 1(E/c)Drsine. From 7.22 (10) at r=a Bro=Bri=Vx&i• Thus a3 +0=Da 3 po (Vx&ii) e-=pi (7x 60 )13 or pob(rA0 )/br = piti(rA00)/br so 21.10Da3=pi(20-C' ) (2) 2310 (1) -(2) gives 2p0a3 +2p0C1- 2pia 3-1-giC1 =0. C'= 2(p.i-pn )a2 /(2110+110.
From 7.10 (1)
Ao =TIII(a/r)2 sin° = pMsine/(411r2) (loop M definition). So scattered part of A00 is 2E(pi-pn)a 3 sine By comparison with A0 this may be written Mmsine where Mm is r2 2c(21.10-Fui)r 2 • 417a2 E(pi-un ) 6r ; Mm=pc (21.10+pi) --TICT "1= 17r2I where nr 2 I is magnetic moment of loop of radius r and current I.
j(cot px) = Ee -
-14j(wt ppcos0) -
=c132wte• By 12.17 (6) we write mn(2 _6pjn(pp) cosn0. By 5.293(10) c_ew _ icut P-directed diffracted wave is E te-e n=0 At p=a, E-= 0 °=jabWte/bP from 12.16(2), c P 2W te=EejwtE(-9T1(2- 6A)B11(Jn(130 - iYn(PP)}cosnie. Plane Wave: Ey= cB z
=Ee
so Bn = -Jn(fia)(41 (f3a)-jYn(fta)r. E jcot cc' w=w E cos n0 ( -_nn C2-t(JII(PP)J111 a) -iin(OP)Yti(Pa)-a-kBa)Jna30+i-Vga)Yn (130) te+w te=cp2 e n=0 _ Jri (Pa) -PiA(Pa) . LiEe jca °:, (-g(2-4)0n(PP)Y11(f3a)-J,;(8a)Yo(Sa)i cosn0.From 12.16 (6):E0---ptDa ool ,E0-jcogTd p. c 13 2 ng0 Jri(Pa) - Piii(Pa) 0 cz r,,,n E jt Ep=pp e pay (pp)J/ [A(Ba) - jY;1 (13a )Ds in n95. (2-6n)n Ejn(13014.100(
3
-
E0=EeiwtEt(-j)1/ (2-61: )[4(SP)Y1;(13a)-JI;(8a);;((3p)3/ [4(8a)-j;11 (13a)])cosn0. By 12.16(5) Bz=821.4 Bz=-jcEeiwnt o f(-1)(2-4)[Jn (Sp)Yri(fia)-JA(13a)Yn (8P)]/[Ji; (Pa)-jYt'i (fia)]) cos np
CHECK: Bz satisfies scalar wave equation so Liz = B0ei Dt-13r) . Use 12.16 (5) Bz=B oei ma tno i (-j)(2-6g)(Jn (Bp)+C[Jn (14)-jYn (Bp)])cosn0. At p=a, DDo/bt=jwEE0=(VXH6=-6Hz /apf Thus 4,(0s)-1-C[4(Pa)-jYA(Pa)]= 0 so Bz=-jBoeicIXE(_J)11(2-61D jn(13°J. IrIVIEla)-j(Pa)Yn(PP)cosnO Ois) -iY;(13s) 15are identical in form with W'teand W te of last Incident and scattered waves W' and W" te teo problem. Wave inside cylinder has form c132 Wtei' E(- 1) n (2-6 1%)Cn•In (P t P) cos noSeica.At p=a: sAp 0=c 1 Api, A00=A0i, Bz0=1.1Bzi, so eWte0=E I Wtei, Niteo/OP = bwte ibP 1.11 2w teo=1113' W tei• First and last are identical since fil=w2 jaC, 1313 r--032 11'e i . Thus EDn (13a)+Bn[Jn (13s)-jYn (13.3)3) 13[4(13a)+Bri[J1;(13a)-jYA034)=-01 Cri 4(3 1 a). Multiplication of the first by Cn-in(Va), 8' .1111 (81a)and subtraction of the second times e' (Pia) eliminates ; and gives J' a J' J. a -nreTJ B Ans. -iYn(Pa) I. V-TET-Tn(Pla)[ 1Jni(13 a) -Pft; (Pa)] -Nri'Jil(Pja)Dn(Pa) From 12.16 (4) E0=-jual/pb0,
grejogibp,
From 12.16 (5) iiz=f1 2*
Ans.
Chapter XII . (XIV)
Page 485. (503)
Page 139.
-16E z = -By/
(cot
- ppcos0) . From 12.16 (4) Az=-13 2 Wtm=jEz fin 12.17(6)
= Eei (wt-Px).= Ee - j gip coo._ i 12 jEe -cl)tna.;0 (-1)11 (2-61?) Jn (f3p) cos n95 . Scattered wave is in
all32cm = - jEeiwte
p-direction so by 5.294 (10), 5.296 (8) and 12.16 (2) it has the form 42 wtine j EeicptnkC n Ejn (PP)-iYn(PP)) cos no. At
p = a, Ez= jcuf32W = 0.
so -(71)n(2 - q)Jn (13a) = Cn(Jn (fia)-jYn (13a)) whith determines Cn. W=Wtmom so (013214 = -jEeiwtnE30 (-if (2-6,1)(-Jn (PP) Jn ( Pa)-Fi (PP)Yn(P -1-Jn(PP)Jn(Pa)-j Jn(Pa)Yn(1340))cosn0/(-Jn (Pa)+Yn (Pa)) 1 - (032 )" Eeiwtn 0(- i)1 (2-69 )Dn(PP)Yn(Pa) -Jn( 3a)Yn( 3 P)icosn0/Pn(Pa) -iin( 3a)) From 12.16 (4): Ez = jwfl2 Wtm . From 12.16 (5) 130 =-82 6Wtm/p695, Bo=filoWtm/bP Ans.
Ans.
CHECK: E z satisfies scalar wave equation. Ez=EeiwtE(-j)n (24)[..Tn(PP)+CnDn (PD)- iYn(Pacosr95 . At p=a Ez=0 so Cn=-Jn (f3a)/(Jn(f3a)-jYn (f3a)) and total Ez is, canceling Jn (f1P)Jn (13a) out, -jEe ja3tE(-j)n (2-6,9,)prooyn (i3a)-J03.)yn o303(in (13a)-jYn (i3a)ricos nO
Ezt
Ans.
-17 As in 16 E 2 = -By Nife= Ee3
(cut-f3x)=Ee 3 (wt-f3pcos0),
w LI= -co-113-2Ee ici3tE (-j)n(2-6Pn (13p)cosn0, w tmo
Az =
-13 2w tm= jEzAi)
jw-I p-2EeiwtEBn (Jn (f3p)-jYn (13p)1cos n95 ,
W tmi= jco-'13-2 EeiwtECnJn (f3b)cosnO. At p=a Bp0=Bpi , VB00=p130i, Ez0=E zi , or f32tWtm/aO93=13',twtni/a6, 1'13 2 oWtm/bp=p1312 bW tmi /bP, I32Wtm=13' 2 W -. First and third relations are identical, namely, 132H-j)n(2-6,?,),In (f3a)+Bn[Jn(13a) - jY(f3a)JH312CnJn(f3la).Second is 1314(j)11(2-611)..t(ia)+Bn[SnQ3a)-jY'(flaa =f312 0Cn4(13'a). Multiply first by 11131JE;(13'4 and second by
sn(m) and subtract to obtain
-6t4)(,17 1 (Pa).1,;(f3b)str iiJi; (W.-In (Ph))
n
Jn (Pa)(4(Pa) -ilin (Pa))-“trer4(81 2) (Jn(Pa) -j Yn(f3a)3
r;= jo43 awtm, Bp =13 2 6wtm /Pd0, B0 =
2awtramp
.
Ans.
Where ■ replace Ir ie and.■ ./77 ) 13/w and pi/co before Jn and J
Ans.
CHECK: Ez0=Eeiwt EHT(2-6g)(Jn(00+ cn[Jn(pp)- jYri(pp)]) cos no, Ezi= EdDtE(-?(2-611°)Iii Jn(f3p)cos n95. At P=a: Ezo=E ziand p'1300 =pB0i or
E zo /bP=POE z i/6P so Jr(Pa)tCntsin(Pa)-iYn (Pa)}=DnJn(13'a) (1)
and f3p'f4(8a)+Cn[4(13a)-it(04=P IPIDnNI (P la) (2) From (1) and (2), replacing before, Cn-
} .NriirJn(Pa)..T tli (A)- ,./77 1 4(Pa) ,In(Pb) J;i( f3a) 42(Pa) -“riTijn(13a )Hn(22 (P a)
p/03 and f37w as
Ans. Where He(fia)=Jn (f3a)-jYn (f3a) 5.293 (10)
-18From symmetry B has only 0-component giving TM wave. Wtm=nfoAmk,_ 1013r) P2n4400e4M6 1 by 12.12 (5) . From 12.11(2),7 .06 (4): A0- rorbe 7.nE _oAnPin+a(1 -51.(rkan+I(jer))e ja3t '
From 5.37 (16)-Egrk 2n+10130)= k 2n+I(jfir)- j13rIc2n+2013r)+(2n+1)k2n+i(jf3r) At r=a, E0=-julA0=0 when 0< 0
IOW
E0 = Vcsc0/(aPintaniY)= OURabe). Mult Ao'by P 111+1(p)dp=s ined ( 1) 2n+i (P)) , integrate -1 to 1. -
-
rY „ d(P zn+1 (11))==n a [(2n+2)k2n+1(1f3a) jpak2n+2(J13a)V11CP1n +1GO) 2 du 1 a Pint an (Tr/ 2). 11/2 (continued) 2V 0
-
Page 140.
Page 485. (504)
Chapter XII. (XIV)
-18- (continued) V0(4n+3 )P2n+I(c0sY)
(2/n tan[Y/2])(2n+1)(2n+2) [(2n+2)kan+1 ( jp,a)-jpak 2n+2 (j ea))
V° Cn
E
Er= jcD (2n+1)(2n +2)r-2W ttn by. 12.15 (6) (cose) Ans . Eo = Vo r-leja't n (O[rk an+i(jPr)]/6r)13 rg C Then by 12.11 (3), ro(2n+1) (2n+2) Cnk 211+1( j f3r) P 2n+ (cose) Ans. Er = -Vo r"'eia n =-Vojeku-leiwtnciocnk 211+1 (jf3r)Pin+i(cos 0) Ans -1713 (copi)-I Zk.= pilz taniy
B = -VX13 2rW trn so B= 132 (rWrm) irbe =132OWtm/60, By 12.06 (8) ;(10) Zk= -041 (1317)-IP/ntaniY
so -19
Gap edge current is, from 18, Ia=2Trap -'sinY
2TTaVo jB2 siny i t
e-u) n ocnkan+1 (jfta)Pin+a (cosY)
-
m 1 I 2raje 2 sinycc) „ , 2pcu nid0cnK an+i (jf3a)P 12n+ i (cosy). From 11.15(12) 1"4k=jcDak=jo3 IL n`g° 41- 2V + (Dp2intan(Y/2) 1 +ZkZf tan(3a cos8a +j(Zk/ZT,)sinf3a Ans. By 11.15 (13): Zi-Zk zkZL 1+ j tanf3a (Zk/ZL)cosea +j sinea 1317 1
where
=i 13
Zk 2Ttajf3asinY i0..2E1 i° (4n+3)1c,n+ 1 cif3 a)Pin-i-i (cosY)P,T1+1(coeN) zi, 2pci) L 1713.1n=0(2n+1) (2n+2)f.(2n+2)k2n+i (jea)-jea k 2n+2 0130 3 ifia(4n+3)k,n+i sin ipa)Pin+I cosy)P2n+,(cosy) = Ans. n=0 (2n+1)(2n+2){(2n+2)1t 2n+ I (jea)-jeak 2n+2 (jean
,E
(
(
-20-
(2) Wi=CAe3(uYt-km z)J0 (unp/a) (1) W0=Ge 3(wt-icnz) Di (vnP/a)Yi(inb/a)-Yi (vnP/a)J i(vnb/a)) This gives A125="bwte/613=0 at p=b from 12.16 (4). At p=a, E00=E0i or No /bp-410p Thus unCnJi(un)= vnC"(Ji(vn)Yi (vnb/a)- Yi(vn) Ji(vnb/a)) (3) 111 Bzo = 413zi so that pqq.,J0(un)=111 qVgJo(vn )Y 1 (vn b/a)-Yo(vn)J1 (vnb/a)} (4)
Take the ratio (3)1(4).
11' J i(un)( unJo(un)ri =p[Ji (vri);(vnbia) Yi(vn)..Ti (vnb/a)]/ (vn EJ o (vn) Yi(vnb / a) - Yo(vn)Ji (vn b/a)]) -
From propagation equation or 12.16 (1), since kl, > 0, k,20.2=a2(020e- 14=a:tope - vs. Thus a 2cD2p' e> ut2i , a2 e.D2 pe >v121 . Velocity is v 2=0)2/ kri=a32a 2 /(w2aapie -uti)and vii=uti+a 2cD2 (pe-p'e 1 t -kZ) ..• In iii: Eg5=jcD6Wi /bp=-jcap/(jkn)=jcpuna-2Cnej (CD J o (unp/a). By 12.16 (5) Bz=(cD 2pc-k.g)W 2 -2c1n (a)t-lin z) Jo (un P/a). Let Cn=jcDa"qunCA, then -knE0=coBp=kri Cnej (wt-knz)Ji(unP /a) and = una
d
jcuaBz = unCnei (wt-kn z)J0 (unpia) . -21Yi(vnb/a) and J1 (vnb/a) approach zero as b-'m so wave vanishes there. However all the equations hold with Jo(vn-P/a)-iYo(vn0/ a) in place of 0 in right side of (2) above, and since Jo (x)-jY0(x) .7-4•31 (x)- jYi(x) e j(x-r") there is now a radial component Ji(vn Jo(vn)+Yo (vn)Yavtl) ,iLKYJaL -11a6m) ) (11 ) of velocity unP/a at p=03, giving P unto (un ) VnP0(vn) -jYo(Vn)) vnt J8(vn)+Y8(vn)) 2 • .“17n)Yi(vn) o -Ji(vn)Yo(vn) . Thus J o (vn )Yi (vn) --Ti (vn)Yo(vn)=0=frvi, vn=0, J i (vn)=0. vn(J8(vn)+Y8(vn)) At p=a bw/Op=0 or A0=0 at p=a, which gives metallic boundary at p=a. Also if kn is finite kria 2=a 2co 2p.e-vA requires that either p or s be infinite in outer medium.
Chapter XII. (XIV) Wi = C e
j(mt-krIz)
Page 486.(504)
Page 141.
- 22j(cpt-knz) c7-•• =, K0(X)+Ze-X so J0(un13/0, Wo= Ce Ko(Vn Pia), Ki(X),,
wave dies out as p-)co. From 12.16 (4) at p=a, E930=E0ior bWo/bp=bWi/bp, or from 12.16 (5) Bp i=Bpo . Both give C l unVo (un)=Cvniq (vn) (1). From 12.16 (5) 111Bzo=nBzi=n1C(.02ne-kri)K0(3.0 =p.C(a)2).fel -k4)-70 (un) (2). From propagation equation or 12.16 (1): kti = co 2 pe+40-2 (3). Put (3) in (2) to obtain -1.1, (4), multiply by pp' to get P j13(1/11) - 11'.-7 1(un)
2a-2
cvilico(vo=poupow (4). Take ratio (1) to _ _w(e; (vn)_ PK 1 (Vn) (5). This may be written
unJ0(un) VDKo (vn) vn K ) unJo(un) awo . ii'voJ 1 (un )K0 (vn)+puoJo(un)K1(vn)= 0. By 12.16 (4), E0= jcuoW / p . By 12. 16(5) Bp=-6-p = -jkobr
n1(.0[5_l e i (a3 t "Icn z
[nip] ej (cot -knz a1
_ _u_ma *o [vnpl j(cot-kn z).21.3p a- e so Eo - a ClCi -kr) From 12.16 (5) Bz=(w 2ne-k.12.1 )W= -vi2i a-2 C1C0(vnP/a)e i(a3t-knz) . Let C' = joAh/a. Then j (alt-knz) knE0 =-cnBp= kriC' e i(tA) t -1(n z)K i (vnP/e) -icDaBz =vriC e Ko (v01 /8) 6W
V c;
-23In 20 let e=1.990=1.99ev . p=p1=1.1/47, wa=10 8then u2..0,1.02.32 (Pi el -P4)=1018(1-1.99)Avev:2--' - 11. 0.4076 JIG) Y1(3.46) u 2=1, v 2=12, u=1, v3.46. Check ratios: 0.573, „Tom - 0.578. 3.46Y0(3.46) 3.46.2.054 w2a2 20 2 1018 0989'10 18 , v = 3.145•10 8m/s Ans. V -1,t121 =w2a2iii 1 (10 1-8/9).10-18-1 - (L 240.72.1018 8. = 10 ° . The u 2-v 2=10 18 (1.72-1)nvey9-10 18 ' 1(2.65) _ K o(1) 1.436 Let u 2=7, v 2=1, 11142.65, v=1. Ratio. "41.431. 2.6151jj ;v11K 10 (1) 711-1 o (2.65) w2 w2a2 1" - k ?,-032a2pe+1 - (10 18/9)010-16 +1 Z 8.26.10 18 , v = 2.88- 108m/s Ans. 4
In 22 let E' =1.72e=1.72ev . P=11'91v,
_j(a)t-k.n z)., funfl L a "3. "
_ „ ralo
Also €1 Epi=CEpo, Ei Ni/bp=e6Wo/Opby 12.16 (4), pB0i=111 B00,
26Wi/bP=p1 826140/60 or
- 25(„ , WO=Ce' PtiCnZ)1 (2) 0 L a .1 "L a J lot_ a jJoL a Ji " At p=0 Ezi = E zo, t1/42114 i=v121W0 From 12.16 (4), u#,C1.70(un)= vriC(-To(vn)Yo(vrt bia)-Yo(vn)Jo(vnbie) (3)
etwitv=eic
by 12.16 (5). Thus t'un CiJi(un )=E-vn C(Ji(vn)Yo (vnb/a)-Y i(vn)J0(vnb/a)) (4) Take ratio (3)1(4):
unJo(un)_ vilDn (vn)Y0(vnb/a) -Yn(vn)J11(vnbia)3 By 12.16(5), if r < a B0=13.2 g Ans. tn1 Ei Ji(un) e(Ji (vn)Yo (vnbia) -Yi (Vn ) (vnbia)) Ep=jw• - j kn• (un ia) Cej (13 t-knz)4(unP/a) so Ep.okritlne -1C'e j (13t-knz)f -Ji(un Pia)) =(.6kn /812 )B0 Thus Ep= jCoei(cut-icri z) Ji (unP/a)= kn (cuple)-1 B0 Ans.
From 12.16(5) Ez= yo(8 2 -4)Wtra
From 12.16 (1) Eej(.04a-2 C e j (a) t kn z)Jo(un P/a)=un (kna)'ICnei(mt-icilz).30(unp/a) Ans. w2a2pe > V4. V2=032/14, (152a2/0132a2py.. a 2 kn 2 = a2c112We- tifi= a 2cD 2lie -vA so w2 a2ile> - 26-
Analysis goes through with no change with J0 (vnp/a)±jY0(vnp/a) in right side of (2) in last problem. This now gives radial as well as z propagation. Either C or C' in (3) and (4) must now have an imaginary part to match phase as well a amplitude at r =a.
Chapter XII. (XIV)
Page 486 . (505)
Page 142.
-27e j (cut -knz
e—x so wave Ko )Jo (un P/a) , Wo = Cei(cpt-krIz)K o(17nP/a) (1). dies out at p=co . At p=a, Ezi=Ezoso C' (3' 2 -kt)J0(112-kiai)K0(vii )C by 12.16 (4) . Then from 12.16 (1) kt21 = 13'2- (un /a) 2 = 82 +(vn /a) 2 so C' utiJ0 (un ) = -CvAK0(vn)
(2) By 12.16(4) etii=et1
if e'Epi =eEpo . Also pBoi=11'13930 so by 12.16 (5) (1312/111 )tikk/bP =0 2/0 Owo/Op or et 1.1i/bP= eoWdbp Thus e'unCJ;(un)= - el unni (un)= Cevnlq(vn )=-CevnK i (vn)
(3)
Take ratio of (2) to (3)
eliminating C and CI to obtain eunJ0 (un )K i (vn) + e'vnJi(un) K0(vn )= 0
(4)
By 12.16 (4):
E p=-ja)Ap=ja3o 2Wtmiopbz=ja.)(vn /a) (- jkii )Ce-1(a)t-kn z)lq(vnp/a)=Cne j (mt-lcil z)Ki(vnp/a)=kn (coper1B9i Ez= ja3(13 2 -kri)Wtm=-3.0(vn /a)2Wtta = - jo3(vn ia) 2 Cei(x-rt)K0(vii where Cn=-coknvnC/a. B0=[3 26W /op =jvn(kno-icn e i (a)t-knz) K0(viiP/a) (6) v=a)/kii=coa/Wcza aapz+vt?
iwczx aa nf El _1161
Ans.
-28In 25 let e=2i= Ey) P=111= }111) OM= 0.85-10 8 , thi-A=co2a2 (if ei-ge)=0.72•1018(1-2)(9.10 18)-1 = -8, (Oa 2 2. 3.1/0(3) 1.742 1...J0 (1) 1.739 .„ 1.742 Un= 1, Vn-3, 2e , y1(3) co2a 2piel- urt e l - e' ' v e' ' e l ..7 1 (1) [0.72.10 18 1 1 -1 0.72.1018 v2=0.72•1018 - 10.3.10 18 . Ans. v = 3.21'10 8 m/sec 9.10 18 '.1 8- 1 -290.98.1012(1.81In 27 let E l =1.81E = 1.81ev , u= p'=iiv . um = 0.99.109, 14,1 414 = w2 a 2 (ii, E , _ 116 )- 2 29.1018 0.698 1 • K0(1) 0.700 03 a 2.8.10(2.8) . v 2- coa a ile+1. = 98.0.09 = 8.82, un=2.8 , vii= 1. e e ' - eK i (1) 1.81eJ1(2.8) - 0. 98.1018 = Ans. 8.24.10 18 . v = 2.87.10 8 m/sec. = v 10.89 - 1 -30W 1= Ci 81-1 e -
/F17 - z
H62) (13113)e j(Dt , W 2= Cs13"1COS (f N/-q---1 12Z)HP2) (iii13)e ir-Dt TM 12.16 (4)
(AW-72446° &i= ciCer11727— PillOv-IL(P?+13' 13?) 13i11-162ne 41312-Pfzei wt. 1k...2= ,41:7 1 3,2z)Hriziejcpt From 12.16 (5) 0B1 = Cl2 -14(131-Pi +13'2) fr1cos( 1 1..e143ta- gzH621, ejwt, I p 7pt PI a ji-T Bo 2= C2f31 cos ( ffi-1372z)H6 2/ 1 ei(l)t i 2C2s in (AAIT7875a) At z=a : Api =A02 , 'FY -,187717ia 1 = 81 C2 e 2 cos(4:1711 a) (2). P2 Brih=PIP0 2 also gives (2) . El Az1 = e2Az 2.9 Wel e (1) / (2) gives e24/1312- 0i=e1ilri ( 3) 1312=coaillE1, 14= 01126 2 . Equation (3) f -117 tan ( is identical with the equation used to determine
0'=
cco/v in problem 7 . of XI.
-31 TE 12.16 (4) . W2=( -C2inein( iFT2z)1162)(P1 P)a ja)t WI= ('- c1/131 ) e -#1721zHic,2)(13' P)eicpt _fiTiT f Api i= Cie l z11(2)1(8 1 p)e ja)t , AO 2 = C2s in (Vir-Trez)HP(131p)eic". By12.16 (5) Cifp.pi
418431
2-131 z eiwt. 13,2=C2(-p i,43T-Ticos(Viir-02z) HiP(13'p)Hh2)(13'p) sin (41:171 f z) eiw at ial HAPID)) At z=a A0 1=A0 2 , Cie -IP-7273? a =C;in (,./filr it" Cle-'1131 2 li a) (1 ) 1-13Bip =111 B2 13) )12•13I• 1 -1
= -Ty/T-7i C2cos(Ta) (2) get: 112
B 1Z = B2z also gives (1) . Take the ratio of (2) to (1) to
= -1.1/21 N T-13T ot(.■ a) (3) which is identical with 8' =cav eq. in prob.8 r3T:-812 i
See next page for graphical solution method for 30 (3) and 31 (3).
Page 487. (506)
Chapter XII. (XIV)
Page 143.
-30- and -31To solve graphically for velocity let u2 = 1312-13!, 172=^*pi 1:0 21so 122-1-v 2 =1311 2 =032 (112e 2-341). For 30 plot u= (s2 /E1) v tan va , for 31 plot u= -
1
.r2 _ .
04v 2.131-01 u= ell, tanv, uv
f
/112 ) v cot va. Let intersection value be 110 v0. Then p'=NqrI-uB
Or. 13'1 ,51174. Velocity is 03/13'. Ans.
ea
uo +vo (30; fuo +vo (31) v u=03.2/1i2)vcotv;
-32j (cot -(3)
j (cut -P'z)
. Then by 12.16 (4): W2=C2P;'130(P2P)e (1)t-Piz) A A2 j Ki(P113)+1,5911(0(PIO ej !, i= 1 = ‘-'21:-r E,1JA PI 14 201- 21k Ol.P21), e 03t-131z) = where RI (p20= J1(Pap)Yo(P2a) - Y1(P213) jo(P2a) is zero at p=a. At Pb; Azi =Aza 'Ci Pillco (1310e
so C1p1Ko(pib)=-02p 2 Ro(p 2b) and e lApi= e 2 Ap 2 so e 1 K i(p ib) = e 2 C2Ri(p2b) • Take ratio to eliminate C1 and C2 and obtain -cop1 K 0(p03)R 1 (pab)= e lp2K /(pib)Ro(p2b). Numerical Example.
111 E
Let a=1, b=2, p 2=pi =µv, E 2 =4E i =4Ev and pib= 1.076=201 , Bf=0.2891=0/c 2. Then 14=1.1564, pi+p1=0.8673, 131
.7187, (32=1.075, pi =1.156-0.516=0.640, p2=0.80, pi2=0.227, 131=0.477.
2fr1 Ko (20= 211-1 K0(0.954)=0.2964. Yo(2p2 )=0.4204, Yo(p 2 )=-0.0868, J0(2p 2 )= 0.4554 Ri(2p 2 )= -0.0495+0.293=+0.240, so pi Ko (PIO/ K1 (NW= 0.328. 277-1Ki (2p0= 2Tr1K 1(0.954)= 0.4358 . Yi(2p 2) -0.3476, Jo (p2)=0.8463, Ji(2p2)= 0.5699 Rv (2p2 )=-0.0395-0.3550=-0.3945, so p aRo (1:121))/4R1 (p2b) =-0.329. -33From 12.20 (2) V= Vok(p /b) /k(a/b), r R-p sine cos0, E = -Vo /(0/71(a/b)) I -pcos0)+IRcos0 - Izsinfz$
(9(E) = (isin0 - 1cos0 ) -110/[p2/72(a/b]), r= 1(Rs
(t2X E,)x rLi = (i.cospt cos() - lsinecos0+1zsb in0 cos e) (- Vo /[0 eal( a /b) ) . ]
r
13V0 j (cc t -1312) f'277 _ C(: ej 13p sinecos0pd °do From A Jo pR 12.18(1) x 2fran(a/b) e f32Vo cosesine (, 2 _a 2)ej(Cllt -13r ) P Vocos° j(cut - BR)re j Jcos0 (1+j psin ecos0 ) dp d917-1 4cur ( a /b ) ""21icayn(a/b) e j13 2 V0 sine 02_ a2)e j(cut -Pr) j(cDt -Pr ) -1[3 2sin20 ose- Ay sineA Vo ( b 2 -a 2)e Y-4coR k(a/b) 4coRpin (a/b) A8A0=Axsine+Aycos0 =0, Az = 0 Ans. -34n)(E is parallel to and t XE)XR is normal to y-axis. Et.Ee
j(cp t-pyisina) Ax=AzFricsce'=Acosecsce'.
Az=-AxR-lcsce' =-AsinesinOcscEr. Ar=AxcospSsine+Azcose = 0. ArAcos0cose-Azs in0=Acos0 sine'. A0=-Axs in0=-Acosesingfc see'. -13E c 0 s0 j(cD t - p R) cc' j (3(xicos Osine + (sinOsine-siral Put into 12.18(1) : Aej je e dxfly 2rmR since, from 12.19(1) rkfR-xi cososine-yi sint cos°. The integral is identical with that in 12.19(5) so comparing coefficient in front gives -cosh cosesin0 A_ A = Ai A0 — cosO e cosasino n '
COO
cos a
b
Ans.
Page 144.
Page 487. ( 506)
Chapter xII. (XIV)
- 35 Let xi= pcossOiand x= p singsi in 12.19 (5) integral so exponent is ji3p(cosoSsinecos01 +(sins:6 sine - sin e)sin0i) and dxidyi= P d PO I= dS . Then, if Qa= sinae+sin'a - 2s inesinecos0, j(a) t - (3R) r r bej f3Qpcos (Ora') sa dpds6 Expand by 12.19(6) pEc2os r weRsin 0 e the integral becomes Ae44 pb pp pb p217,. and integrate: jj=j j 1. ..10(3Qp)+2 c4 jn Jn(8QP)cosn(s6 i -e))PdPdf6i= 211,I a PJ0(PQP) dp. By 5.302(2) a Ecos a sink A : A0 Ans . ( aJi(eQa) -bJi(f3Qb)) ei(cDt -eR) - -cos.1° 1 coRQ -Er - 36Integral same as 35 but front factor is that of 34 instead of 12:19 (5). Thus . - cot A . Ecosg. 0 (DRQ (b.,11 (8Qb)-aJi (8Qa)) e i (wt..") cos() -37At great distance only phase difference is important. R?=R2+c2 -2cRcosY=R 2 +c2 -2cy,R1
c
R-V . 11,1=R2+c2+2cRcosy, R2 zR4.
Incident phase differs by an amount that shortens R 2 ,1engthensR1. Alti=c sine, AR2=-c sine so phase factor is approximately ej(cot-13R)te-j pc(i-sine+e jfic t-sine))
ej(cDt-f3ERi+csin a) +e j(cot- R 2- csina])
= 2e j(cl)t- PR)cos{pc(-s in a)} Ans. If = sine, phase diff.is 0 (OK). - 38If holes lie at x=±c, incident phase is same for both openings so cx 3(cDt- pRke -jficx/R+e jPcx/R3. 2e j(cut-8R) cx c0s(pcx/R) I .R22111+ T. e RizR-T,
1f--c
C14-
Ans.
-39Msine Incident by 12.01(7) is Ee = 4rrea3 ( (1-82 a2 )cos(a)t-fir)-8a sin(cut-f3r)) = Q sine = (4rEa 3 )-1 MO-n2-2 p a fj8a)e i(czt -Pa)sine. Wtm=eiwtiAnPn (cos8)kn(j8r) by jat c4;0_111')6a-lsi f2) .r]An 12.11 (5). Diffracted wave is Er- ja)Ace=jcoo 2(rWtm)/rtabr=pe ad ---=F 1 --I be Measure time from dipole. Equate E0 at r=a, Multiply Qsin0 by Pp(u)du and integrate from +1 to -1. QsinO =jcoeiCpti -AnPill(COSO)6(akri(jPa))/aba . By 5.154 (5) we have 12PA(P)d11= C41112(1-P2)1V1(11)d11=Qr1(n+11)5 QSAi312,11--7i2n+1 pi11 Pn- 1 (11) -Pn+1 (OPP
and by 5.231 (3) this is 'tr-le =-jcoAnCO[alcn(jf3a/aba)e-a) j 1P/11 (p))2dp=2jalAnCO[akn (j8a)]/aba)e3Wtn(n+1)/(2n+1) by 5.231 (3) so An=Qejcpt( - (n-1 )-1(11Pn-1 -13m)-(n+2)- 1 (10/1+1-Pn3 11:/( 2 jubEakn(241/abal by 5.154 (5). p2 2n+1 111Pn- i (11a) - lit Pr- 1 (111) 112Pn+i 012) 'Pi Pn+ 1 (Pi ) 111 -(n-1) (n+1) (Pn (42 )-Pn (111)) n-1 n+2
o
An- M(1-82a 2 + jiita) e-jpa ((a1+1)Pn(11) - (n+2 )PPr -1(11) -(n-1)11Pr.4.10)3 1331,21 2 jcoa-Xn+2 )(n-1)6Lakn(it3 a)Voa 4176a 3 M(1-8 2a 2 +10a) [ (2n+1)Pn(1.1)-(n+2)11Pn- ID1) - (n -1)11Pn+1(11)) /2 -ji1a N e 81 jwEa 2 (n+2)(n-1)OLakri (iPa)Pba
Ans.
• ▪ Chapter XII (XIV)
Page 488. (506)
Page 145.
-40Incident by 12.13(5) is Ari-pI(b/r)2 sine[cos (cut-I3r) - f3r sin(cot-pr)). Thus at r=a incident field is E0=-+juvI(b/a)2sin0(1+jf3a)ei (wt.-Pr) (real par Diffracted: ilite=pIb 2 (1+jf3a)nkAn Pn(cosO)kn (jfitr) 12.13 (1) . E0=jab(rW)/r60. Multiply by P'(cose)d(cose) and - jf3a 1 2 j (1-p )Pn (p)dp to 1 by 5.154 (5) or Prob. 39. Ta-2 e integrate n(n+l)e- j a r 2 j (17,11 ( )1) -P +1 (p)) dp gives -kri (j13a)Xn1.:(13n1 (p)) 3dp= (2 +1)4 -I (n-1-2 )PPn- (P)+(n-1)Pn+i (11) - (2n+1)Pn e -rilf3a (n-1) (n+2) 8a2 kn (.9a) L
SO
r
Oa)-423n.1 V1 . Ans
-41V0=-jaAszi=f (P), = E0 (-1coses ino
cos0,
1 E0=1 E0 cosst-lEassingS, ( t2X&)XF,A
cosOcosfe+Icsinesin0) . .1;and k terms go out on 0 to 217
keep p in-phase term only. rivR-psinecos0 integration. As R A _ L-L. j (wt -PR) (-8cose)Sa pf (p)S2rre- jfitp sinOcos0cos0d0d p . Replace "0-2TrRe arr r.2TT exponential in in te gr and by 12.17(6). =f (4,(3psin0)+2?(4,7n(130sine)cosn2f)cos0 j (cut-13R). For loop =- j 2rui (PP sinO) . Thus A0= 1311-1cos Oro p f (p)11(13psine)dpe current with f3b small, from12.13(5), tV-TpI(b/r1 )2sinEr ei (at - 130 which becomes
so
at z=0 ipIb2 p (p 2 1-c2)-3/2e) (at 13.4Pa÷C2> . Neglect 11,/p 2+c 2 . p is small so let J1(8psin0)=2i8p sine. ra P 3 dp P2pIb2 A 2 in213 ej (cut -8R) [a2+2c2 2 cos() sinAjo (0+0)3/2— p a pib 77117T fc - c Ans . `19:5- 8R 16R - 42 2)x ri , r1= p cos*-k sin*, ki=i cosgf+j sin0 d* cos* +k f 2(yo)cos*cos0÷ i f2(y0 ) sin* r
E= I f l(Y0)-k± 2 (yo),n=i, Ci)(Q..fi+kf2)2xri= (k
(rricE)xT.,1=,8 1f 1 (Ye)
N
TT
to 7. component vanishes on '-integration from jcut Substitute A_e raf 4 -ii3rcos*dzdy0 cos0 f2(YEC•11-2r co..Ld-t; z(Yo) in 12.18(1) ra e r cos*= itt = ,JR2+y,22-2Ryosino, r=g/cos*, dS = dye d; ds cos0=rd*,dz cos2* r.17/ 2 -jr3Rtcsc*, . Infinite integral is R'"'j j (cos*-13R1 )e av Let cos*=sechu, f• • -S:j (1+ j f3h Riacoshu) e-j8R1coshucki d*= csc* cschu sechau du= seChudu so Re-l = jrrirl jille Hiv(ORD by 5.331 (3) . A= -We e iwttitef i(y0)+k cosief 2 (yo))11c0(1310dy0„ a A.
Chapter XII. (XIV)
Page 489. (507)
Page 146.
-43For standing wave magnetic field at perfectly conducting surface is doubled so that From 4.23 (2) W = -Bo (z±A/Z2-a2) where U is scalar magnetic potential. At y= 0: 1By1=16V/bx1=1OW/bzii.p. = -Bo[lta/f2-a 2 ]i.p. = xBoa72772= (-j/w)oEz/bx. From 12.18 (5): Ez= ja)B0,17:71. In 42, f 2 (y0)= jcpBorai:71 If p >>a, Hio(R)z No(p)• Ez = -jakz so .
Ez = IfoBo cos0' eiwtHP(Bp)la.sig:"AdY0 y 130:2 '‘/IThO y so Ez+P2a2E0 cosfe HP)(13p)eiwt . Ans. 1 If le =N/R2 +yg -2Ryo cos (ir -0)#p, then HP)goes under integral. By figure, le cos0!= Rcos0. By HTF I (15) p177, Cacos(417-0))=secgisinfn(Ift-0)), so from HTF II (28) p101 cosfeHlakiliN) = 2nsintn(i1,-0)3y;I JEI(By0)Hga(R). But x-lJn (x)=xn" sEApx 2P so integration -a to a removes even n-values. Write 2m+1 for no then cos (2ur+1)0=sin[(2m+1)(41T-0)}. Thus integral is Posfellf2)(pRIWT-Tridy0 =2E (2m+1) cos (2m+1)0 Iran 2ig. 2 (R) where Ita= 2,1ao y0-1.17;TJ 2ak+2 (fly0)dy0 -a ,1 •xJ 2÷i(paAN)dx, where yo=a4fcco dyo = adx/(2vg). From 5.293 (3) x -1 ,1 2m +2(BaN/27)dx =ajo x 07---(-1)r(13a)2m+ar+1 p I
- E r:rci m+r +2) 2 2111+2r+1,1.\7 2 -7 2 xm+r-'dx. By HTF I (11) p 10, t (1-t)v-idt =B(u,v) 2M+ 2r+ I B (M+r±1 e•Dr ( 13a / 2) 21- a ,= aEtho so that =B(m+r+D-12.). Thus I= rE0 r: (2ur+r+1)'. CID Cl
Ez=4.BawB o eiCptin
k (2rn+l)cos(2nr3-1)0 H2)1(R) Em. Ans. If we take r=m=0, B(-12.1 4)=77.,
then Ez=4-13 2 a 2curB oeia3tHi2)(R) which, with B0 =E0/a) gives original approximate result -44From 4.23 (2), if V is the potential function, W=4E(z+4zr), c2 . In problem 42 set f i(y0)=4y0E0 /.177q. Ex=-[aW/bz]i.p= ixE0 /,./-7 re dYo • HP)(1310 HP7(13 (R- Y0 sin0 )) aYo HIV (PR')/N/P-T Aci=.+NIICE0eici5t Hi2j(BR)-Byo singi[H12) (13R)}'+ • • • Neglecting yi compared with R 2 , only second term survi d integration from -a to a so A4;64BsinizINtraEoeiwtHP(BR)r a _y8=43\rire Sill0E0e ja3tHfal(13R)Trasid E
=-2-j1713 2a2E o singS R b3R)elwt Ans. 00 -45b (R'dA,A) -— 4 =Ce-' CHP)(13R1)/N/77Thdy° . dBZ- RIbRI ,-From 4.22 (3) W= sin-1 (z/a), bW/bz=(a 2 - z 2) 2 A l 0 a R-1 o[pRi }qv( pR)] /b (13R)=13110(2)(fiRD . At slit center Rt =yo. Let yo =asine, dyo = a cos° de 2 pa HU) 77: d 7-4.72- = dO so Bo=BC j a a2-y2 dy0 =0C.I174-42)(I3a sine) d9=BCSJo(Basine)de- jfiCfY0(Pa sine) de
=14
r/A_ //n(- ceBa sinEr)dEr where fla is very small so 132a2 terms =13c jr fr/2de- 2.1TT2c j wi (a Pa sinO) dO - -1•1IT1-2117/ 0
are neglected, El' =-0, em(-1)=j1,, ka= -0.11593. Use Dw 868.1 so Bo = 13C(17+21170 / d0-V,Vr(af3a)d0
jliS TT orksine de) = pc( 21,- 2 jk(apa)+ 2 .j0n2)= 213E {1,-jk(aPa/2)) so C=B0 /(213[TT-j242(-1a0a)]) and d
jcot jraHiM (WR2"02-2Ryo sine) dy0 A --P• TrCe jCi3tHi (13p) since 1 7.7 =Tr . 0 -- Ce o✓a -yo2 R=P -SP°2 a a -y , -a B = LLELL-qa) _ _____", Rc1 Te j(cot f3R 41r) Ans. If neglect (Ba)2 get B z Re) z P.Ce r1BR 1 1 ___110 {212j(cut-BR-rt) If Pa is neglected: B --0. z psco Um (pa) BR e (2)
(Opp)} )1 (pp)
-21
-
-
Page 147.
Page 489.(507)
Chapter
-46-
_44_
-45_
By 44: Bz=-1j17132a 2 B0sin as ing$ Hid(PR) e ja3t Bcara
Be Els ina-M-)
jpt By 45:B -1occsa lcf2)(8R) e zzr j2n(YPa/2))
11T
I L11- - - -
Add amplitudes. Take PR >> 1. Let 13a be so small that 2(17 - jOn(y P a/2)).cz -2 jk (pa). Then ,
Bz= 447130C-13i a2 71's in a s ingIeir" + jcOsaf2A(13.3))127(rTPR)ei (a't -PR-TIT) since, as R becomes infinite rPR) e j0R+jr/21 Hiv oR)..0je3r/2HotajcoR) so that Hp) ( f3R) e TT/ 2,/-7— 2/(1 (wtBz= ijBo(-12-132a 2 sinasin0 + cosCak(Pa)}✓217/(PR)ej PR-Trr) . TT varies with B2 so sina IT= 1-7-( [cos a /k (Pa)] - 4132a 2 sin a s in0)2 no 2I3R -47By 43, Ez= T17 1 P2a 2E 0cos 0 11/2)(13p)eialt . B30 sina is zero on screen. As p-'›ce ,
Ri8N 4_ 3 .,,,,
IT
by 5.295 (8), Hi(2)(f3P) = NriT(T--V3Pe j PP+ T irr• Ez-4132 a2E0 posps 2rei4-"" . T3-, E0=Eicosa. 17is proportional to E2 so ft= 4-17P 3 a4 R-1 cos2 0cos2a fro Ans -48,Ec E = -1 by' =-Ec1 =_LACC ._LE .-1 _ _ -EP r, )+-9-. By By 5.272(6): V=Ez(1 ---m 17C t hiogic=o 'Rh 1 G=0 TT/c 2 -p 2 At great distance (nXEt)xr i = Et(Oi cos0 +ft)Js in0cos 0) . By symmetry A0=0. j (wt-pr) rr E pe p cos0ei"cos0 sine pdp cif/3 . Neglect p compared with R rzR-0 oos0 sine 2rvi) JJ (R-p cos0 sine) 4/C2 Subs. in 12.18(1) Ao= 1T in amplitude factors (denominator) and observe that real part of exponential for integral range -Tr to 11 is zero so Aia= [1 j cos0 sin ( pp cos0 sine) Of] dp ei (43"13R) . By 5.302(:. u corn( ca p2 o r11COSt PSinO)dp ej (Lot -(3R) ICP 2 _.1 J ( 1.........___ J i (z)= j —sin(z cost) dt so AA. =--. Let p=csint, dPa,;277FT=dt so u o 4c .•17 2 TT , PEc 2 rri 2 . 2 J3Ep a A0 =- ayftR j sin tJi(pc sine sint) dt J312 ( Pc sin() ) , by IT II (19) p361. 03R.I2Trfip sine o But by 5.31 (3)T.77(S)J 3 /2 (x) =3C-1 (x -4 sin x - cos x) = x -2 ( sin x - x cosx) . Thus sine) - Pc sine cosaic s ine)) e j (wt-I3R) ____? 13 2c 3E singej (wt-pR) A - pEc 2 ( S in (I3crwRf3 aurR 2c 2 s in 2 0 3c -00
Ans.
-49See 5.271. Static field form is
n= Bciopi(jc) + csQl(jC)) Pi(g) poso.
_-----
.-E.C2 , Qi(jC)=TrViTT 5.23(11). -------....„,..--• At C=00, Pl(jC)=jN/7-C i- 2 , Q 1 (jC)=0. At C=-40, Pl(j6)= j./i nc.. -7r-Bp cos0 = Bx. nc= Bx(-1+17C) = 0. so C=r-1. n= Bei[jpia 0+ 40 a 0)pl(g)coso IT Bc 1x. 3.i.-r(-7 + tan-1 C +1—142) C-- AFT-C2,ff-Ticos0= which becomes by 5.23 (11): n = -- (TT -cot-1C 1 C+11 +1+412 2Bx 2Bx 1.42 2Bx 2c1 =_ Bx r 2 V1. FC 2 an B In hole at C=0 Bt=Bx=-, _4..2 ..., C +. 2 circ imic 2-p2 irrfr ca-y2 . 2 Bn=co/V+C 26C C-sOcirL1+0 1 r 4/E -
Where B, the standing wave value is twice the incident Bi. From 12.18 (3) bEy /bx=yoBz . Ey=-2ja3BT7-17-72 =- 2 j pE orriNc777. (2yE)xr1 = (Ai sin0 +0 cos0 cose)E. r ^4: R-p cos0o sin°. (continued)
Chapter XII.(XIV)
Page 490.(508)
Page 148.
-49- (continued) By 12.18 (1)A 0 if r >> a
21132 E0 sin0e j(cot-13r) Scr 217Alr— laeiPPcc69"in°0dod 21r2co R - p cospisin0 oo
By 5.302 (3) jrcros (zcost )dt = ITJ0 (z), frs in (zcost)dt = O. Thus Ao=(2 j 0 2 E0s in0 /(TUR))ei ((zit- r3R) rcAlj:7 0/ 4(13P sinO)pd0. Let p=csint, do=c cost dt. The integral is then Hd.M.F. 11.4.10, ,
0.
j=ric3.3.0 ( 3csin0 sint)cosat sint dt = TT/2{13c s in0) 3 / 2 J312(Pc sin 0)G3. Use 5 o A0 _ 2 j0 2 E 0 c3 s in0 j(cot e - f3R) sin (13csin0)1c sine cos (f3c sine) 0 3 es ins@ 11(.0R (rraA))eifmt-13r). E0sin0 / 3 2c P AO = cot0cos0 Ao !See cnX9s,,r,i sine) - pc sinOcos(fic sine) ___-0 j132c 3Ex os0 cose j (cot-0R) 2 jE, cos0 cosO j tot-13R) sin e . A0= e f3c +0 MLA singe TTE3u3R -50If the incident wave electric intensity is ED, then from 49, writing 2E0 for E0 , E0=jcoA0 =213 2 a 3 cosa cos0coseE0/ (11TR)„ E0=213 2a3cosasin0E 0/(17R)
E6+? _ 404a6cos 2a E0
r R2 2 2
(sin,fs+cos'20 cos291To = !i142421172R2 22 (1-cos 20 s
V2
rFrds=sro S o (1- sin 29cos 2 0Rsn ?
.
Bcos a
0-521-
2r 2r 1604 a e cos 2aCR4 (1-§cos0d0=CRrr) 2so P3r T6 52-
Take entire B tangential and normal to incident plane, let 0 be measured from this plane. It is then Ift-0' where 0' is the 0 of problem 49. Write 2p2 a 3 E0 cos0 j(cot -13R) 2f32a3E0cos0 sin0 j (cot- DR) 2E0 for E0 . From 49: Eo= e , E0= e TTR rR There is also Esina normal to sheet so by 48 E0
TT= ((q +E4)/Eg)lfo=C paa3/(rR)) a s in20cos 2 0
f32a3E0 s in0 sinsj (cut - f3R)
rit
e
2COS0 + sin a sin0)9fT0
Ans.
-53
2r,. = CR2.10 t-sin20 .sitds cs2r2(4s +4cos20- Tfc os0s in izs cos2 0 + (-2cos0 + sine sina) )R2sin0d0 3 o + Is in 2 a) do = cR2 (jr-4- 4r + irsin 2a)= i1TcR2 (4+sin2a) . Ans. 15 = +04a6r-1(4+ s in 2 a)-Tro -54Assume slot fields to be: E= t-al'sinp(2-z1) cost z1>0, E=I?sinf3(L+z1)coscot z1<0. n xE = j.,YE = -kE, r = i(x-xi)+iy +k (z-z1), R2 =x 2+yai-(z -z 1) 2 (nXE)Xrx= (ty-1(x-x1)) Thus Pi a/2 1+i Br) ti. y- ' (x-x ) Br 12.18 (1) gives AI = e- j ,12-1 2,2e dxd sin $- z dz -2Tracz 0 a/2 (x-x1)2 +y2+ ' If x1 is very small, integrand is constant over dx so that If(r)dx-ff(R)Sdxfaf(R) 1C ynt fri ° 1+j j0R j f3Rs in AF Zi)dZi+S 3 e sini3(L+zi)d z then 2 Ty co R i? 0 R3 , _ y- j x) r r(1-±1).E/2 (1+j f3R) +z) e j (cot-(3R) dz Ans. 2r 3--J (1 +1)//2 83 . 6A Take curl of 12.03 (3) where z1 =1-z1, z1 =0, di= -dzi . B=1.bAz- z-c cz -
(..y -pc)e
This is half antenna. r2 = xa -1-y2 +(z -zi ) 2 (continued)
An
Chapter XII . (XIV)
Page 490. (509)
Page 149.
-54- (contijued)
r/sinp(/-zOcos (wt-PI° dz1 +Scsps in0(1- z) s in (wt-PR)dz 1 .
B =_E,I.o(i ‘,_ . sA )12 411 v'w '
R3
2''' 4 0
Real part of E for
ena y -Ix) fsi.in0 -zOcos (wt-pRia.rAspino:::::)sin (wt-pR) d; . . half antenna is :.....EV2= '•... Jo Jo 2r R2 Rs 2E 4r C Thus 13 ;12 and E 1,2 have same form Ea lot= - 71; Tri Bant = - —r Substitute in 12.03 (19) E E0 = -gifT) (2cos p/ sin (wt -pr ) -s in (wt -pra ) -s in (wt- Prb) I
e
Be Eft; _„*.,5
"Us lot--71:1-
cospLsin (at -Pr) sincut-pro + sin (wt - Prb) 2 r ra rb -55. 4,2 1„.. _ 4sca TT 4 . ,,,,. " P 11/8 Tpant pii (Bant frOm 54 and 1Tant from 12.04 (2))
=
B
-55-
z
2nrc
(thos).
4p/e = 360110. Thus Zant = 2IFI 2f7ant dS, Z:lot =Cai.f 2TrslotdS • Zann1ITrandS/ To find radiation resistance let Pt = 417, r a=1-z, rb=l+z in 12.03 (17), (18)
1 z B0 = iiiril.0(cos= cos Pz ) / P • Ez= -lc prrlIo(ic oswt cos pz - z siniwt sin2 Pz) / a a ..z 2) , P = ill E. B cost - s inwt coswt,,[1, z sinilizxPzdz . First term is Rant 2rpdz =V.0 c os hal i "L 2p pc r/ z sinpz cosPz ---12-L SillEE d z dz =-P4f 1 si-1 i 2_ g "Xane -FT 4.2 4 o /-z L+z pc rrh inrn,, _i_ r 2sinrud,:i] = _ si(277) Let /-z = u.t, dz= - /du, .e+z = vi, dz = Ldv.. 20/ = Tr• X ant=4TTLJo u u 2r uu -r-J1 by 10.11 (5) . so Xant = 4rrC 1 r+si (217)3= Orr 1.418 = 30r.1.418 = 42.5 Ohms. R slot= 4-(p/OR 2 ) = 36001F2 -73.13/(73.13 2 +42.52)Z 2600000/7160 = 363 Ohms. . ant/ (R2ant +Xant Agrees with measurement(Putnam, J.I.E.E. v95,1948,PtIII, pp290-294) Xs lot is negative. evaluated in 12.04 (5) .
.
-56-
. 727.! 2T 6 r. , r 3 .__ 0 r121:_r k_ V 2u = peb 2u/ot 2 12.01 (3). V= r-1 e J (wt -pr) j pr)e i (ca -Pr)] {re ar = c-ifir-x+ipr-2 _132r-1 ) ei (wt - Pr ) =_ 02 V By 3.06 (4) 0=4 (-0320 -0024)dv = 407/0-07‘11)•n dSi
+f% 070-93v*).n dS0 . where Si is inner and So is outer surface, and 0 = r-1 ejPr . e..•
...
70 = ... (r-2.fifir-1)e- iPrr
-1 . As sphere becomes very small * becomes constant and so does V*.
Thus, since dS diminishes as r-2, second term vanishes. Small sphere integral is 2r r ri -12 e-iftrasinOded0 = -2*pr 2r(1+j8z.) e--i Pr03 --i• -411% . 413 fo So ip + ri r+0 o ) ) • r SO 41p= "41tr-T fsa:e iPrVir - 07(r e i P n dS Ans.
-57Let U originate from some point outside q so that e jux.4teja3t . 7*. _ k,r_1 2e jpv _, -jpri , U = r - lei (wt - Ord . r_ 1 e j pr 2 e ,r10 1 1e ti 1 1 + jpie- jpri Component normal to surface S ... is n.r so if pr is much greater than one, ir3. -1-4i r1 ,I, . - 11 r 1e-jp(r4T0 f2,E1 52'rrJl dS Ans. TP 4r 0S ri r L ri
Page 150.
Page 491. (509)
Chapter XII• (XIV)
-58(Yi -Da+ zt =R?-2(xixt+yiy')+ (x' 2+y' 2), r2= -x')2+ (y-y)2+z 2 2x'x1111 2-257STIRI2+xl2R12 +y'2R12 =R2- 2 (xx' +yyD+ (x' 2+y' a), r1=R =R1 —AIR-11 —y' yilt11+()Ci 24-V2)+R11. r=R-xxR-1-yly R 1+ (x' 2 +5712)1R.1. y 01 2 02 ri +r = 113-y11+11)+2.1- +TR, = R1+R2+F(xt,y; R, R1) . r=
Substitute in 57. 8=211A.. Keep F (x', 12LIN e- 21ij (R+RI VX /
j (04. P 2XRRI‘ R
4,
r
e
)only in phase term.
-2frjF(x',y' l ri 1 1171 )/Xds,
Ans.
-59„ 1 1 -xi R—p'cos0 since Parallel beam so R1 = co , R>> a . xs = pose. At y=0, F(x , y ,T-1,Ti. )= -,e2Trjp'cosOsina/X= n • Rift= cosaln • RI / RI =-1 cos(2rpbose sina) - j sin(2rOccis esina). •• e-21T3RIA.1 /RI The sine term in exponent goes out in 0 to 2!r integrativi Let -2TrjR/X r -10cos9sin 211 CA be So , the incident amplitude so 41=1 ! jOle d3d0 -11-1)-(1+cosa)e 2XR 0o -2rjR/X r_,2Vpsina Integrate by 5.302 (3) to get li= j__ 2trrj 0pulA x ) dpi . Integrate by 5 2941 2xR (l+cosa)e et0
to get 4'-
OS/
ti
(l+c,os a )2{:ilfraxsina)] ia(l+cosa) e-2f7jR/X , L.nct)4,0 2T . Io -ito*i. I-Yv 2Io. ulc----)7 2R since —*-4:22 10 2 I 2rra since)] Ans. or I = --eacot2g- J X 1‘ -60-
1
11 Put x= x„ = y = yi = 0 in 58. Then F (x', yl, iiqz.i ) = °I2a-{± +1 or Forarge 1 ri. - 11 ---1 R Ro. R and R1 , Lt-and .41 - 1. In 58 gyp =--1-- e -2nj (R+Ri)/Xys e - jffp'2(R-1+ R-1VX dS = 2rrp' dp' 1 "" dS. ao 1' URI a Ri ,ry r e-i CP' 22p,dp, = ri e-jCxdx... _ jrr re-j C as- 1 11 ) so that 'J C` o x2 .1 , * rf 2 2(1 - cosCa2 ) 1 , ,..,..1)2 (1 cos Ca 2 +j sin Ca2 ) (1 0:21 o:C -jesi_n2C rajR 2 ), rk Y pIbpM X. R RI rr2 Or -t- K (R+RI ) a /1 + 1 )1 4 4 sin 2(Ca a/ 2) If s in 21:1-al then 2X (17t Ri /j . Ri (R+R') 2 — (R +R') 2 1= (4R_R*2 s in Ans. 2 [h— : (R +R
Page 545.
Chapter XIII. (XV)
Page 151.
-1- (6,p548) In 13.00 (7) write 132 = a)2 p(e jyar2) for wage. r= i.A.Dip(c-iyw-1)-13,2an=j#1132-1432in- ja)uy 2 n- +jum.ry. Use Dy. 58.1 where r =A413 2-131m)2 4a)21.12 y2 . Take real part positive. = a' + ji31=4/[3m p' = ,1(3 1312rin )2+0 p2 y2+132. pnin fiaf,i032 fitino2 +a) 20 ya 132+ flan 37/45 2. These ■
formulas are valid above cutoff where (32> 131 and below where 132 < finin • - 2-(7 ,p548)
ce =4(32_
fittn) / 2
LA 4. [asap; y 2/0. 2E2(a)2_ (l n )]) + - 2 inn]
written: a te=
---1
_
"to.
(w2-wriin)(1)2P2Y2/[2412ea ((IA- 40'3
]. v2= --. From 13.00 (11 ), this may be
An s .
Milti
[ C132
2 a)
•
Pg
• CO
6k• vm = 13, v/c1 2.. 03 21.12y2
+ f 0Zp2y Ep e 2 (032 -w6)]3 +TT w2p2y2 v ITY77,71 n [1 8122 E 2 on 2_40 21
mn
- 1/2 from 13.00 (12), so v' -----4" 1[ + rn-451 Tfn 4g2e 2 0)2- ahtr, )2] y 2,411 1 r VM al 2 • V n 1- ( vnir -T))t 8(0 = vmb. 8(EMI) 21 a) - 3- (23,p551) . mftx W = Cmn sin-a-cos b e- j fiLnz, A = 72( W= j bW/oz - k bW/bY=C
nrry
cos b
ev
WIT .
-ginz
sin ce`
A is normal to y=0, y=b; vanishes at x=0, x=a, and has components parallel to the loop onl•
-
B = 7je (17X VAT)
.
2W b 2W
bW, _
b2fAT
b2W
b2W
b 2W
b 2W
-
-75-4 +3 E37 +1:50x( 3 bz -t:b7 "11 + 573 +1,67877+1Sb7Ebz) - L032W+--i 3+16, y xa z
= (.0 2- nA a r-r2)W - 12a-11rCmn (jcsin.12b+k ji3IncosT3cosm÷ Tx e"-"Ilanz -4- (27 ,p551) nr A . nn, 8 _ n1137 + 2-) sin-b-m--2)-2cos b For z-sides of loop y0 factor is: sin 7(y° 1
nro
(1)
nro-cos 2 when 8 is small by 13.03 (3). For y-sides of loop the z0 4---T factor is: ei 13'61(z°
ei 131;in(z°+2-6) =-2je3 PInz0 sinpmin 2 =-313m'n8e 3P•rinzo (2)
- j13minz0=/±(2-8pI6 . , . mffx0 nry0_ pIdS 0 By 13.03 cr MTN° !Irv., cos b 313mn8 m)•2 sin-a- cos--Ed-a (4), (5) ,InnYe 13 ab 2-8 j1316132ab•2 . nrry0 By 13.03 t; • s innb ry0 .)„.1.1IdS 2nr1 sinmrx0 sin ,, a (13), (14)'mnz' (3 abi3mn Id 2ab j pihnb p I2a db S zE(2..64) mrrxo 13.03 (6), (14) A= Xy+Az=-Esin— a cos
1-4 timnrr
, mnrr
2=20
SUbstitute in
jrrr3mno , b sin-i-Co ab Pmna rerrl - j13M (z-z0) (1f3f4t1 )2 • n TT) mrx +j p 2 - 1 1 -2 + )Sinnla COST C + k (ALIT j • s in— sin-E- e n a b fithn b Prim ni sinuln) (2 _ 60‘ , VITO . mrrx - j13' ( z-z0) mn cos b Ans. SO = olds E E C OS •7 7 -D .). _n_11 a ab m=ln=0`
A
r
u
Chapter XIII (XV)
Page 546. (552)
Page 152.
-5m"a6 47(xo -i) =2cosn4rx°sin By 13.03(5), xo factor on ty sides is: sinrc(x0+4)- sinni =(nytTo /a)cos (mTrxo /a) . For Ox interchange x and y in 13.03(5) and permute tam . nippy raft . _ m2r2)sin-cos s, ipIdxo loi 2-6° nryo I., j, k thus: Ace 11[- i.....(13 4- — 1311,10 sin b cos---a a a 2 b J .-132ab m" ln0 j131n (z-z0). mr MITX Ea] mn114 r s,M . MITX b sin;-- it,j13Ea ' n— a sin—sin +' For yo factor on e ab c° 4 a 8 b nil' n Ox sides: -sinTir-T (y0 +1)+ sin-b-(y0 -1)=---r8 b-cosnfTY b° . Add t., and A. 8 2= dS. Thus na mrx 33:41° 21 mrxo nryo [.nrrml m ip_I■ :11 ai ci 2=8.fillin Tra 2 mar2 cos a cos b 3..,--1) t-a-i- + 13 -0--)sin b cos a — p2ab m=On0
A=
-j13' (z-z ila :..) bln (Ma:mrerf a . MITX 23- 4 _ tIrtx a 2_En2172 n 2rr01.1.,--i-cos al T( m a _p —}sin—cosh cos ] e mn ° ) m b b2 b -Ls. p ab a nt rsy j131 (z-zo) mlifx i n!drilla mtrx . mr raft A - pIdS °' 4° 2-8° -8° cos a- 3—sin—cos 3e mn E E ----n-mcos ---°cosn-'1°. Ans. ..,.. a ..,te a b ' •..• b ab m=0n=0 13mn b a -6Legs, being normal to y-axis, have no component along the TE 10 electric field which parallels the y-axis. Thus all TE 10 excitation is by side of
x
length c and its image in end surface. Thus, from 13.03 (5) we have . 11X -131oz j pito zfc c2 d yo_ 2pIc s. Trxo . , Ka rx ,. . , ; jpI0a . 1 -1.1310ab in a sinpio zosinie a-(z3 sinBio z0)e LA to - .1., 131'013 2ab sin a sin. 2 , tIA 2rxo 4..- EXB 211=2 12 . - xd y . 7 .— P =Joas0bTrd a- sin 1310zo sin2 frX =1 2 2 sin Ey= ja)Ay, Bx=-6zY= j13;0 AY' II = 1 • p• pio a b 21 1 1T ., . ET C 212 .2 ilbco sin2 ,_,oz R! o Ans. eV= ii2- p to= 4 92 x72 - :: -- , 13;0= —X. [L-1 P 215--sin 2_T P ioz0 sin21-°. R=T1 4a2 1i oab sin
-7Only component dyo of element cd0 excites TE 10 mode. dy0=ccos9d13 zo= coosO. Excitation is independent of Yo for this mode. Thus
II3c ■
by 13.03(6), (5) A io-i feiPl°c"secose a —3, 03 2ab sin 217fa9sinne-j3;°z
de.
z
e / i ,,, "211 1 J o cos (q0cose)cos0d9=0, 217j..11(qoc) 5. 302(3) - c j50s in ( flitocos0 ) cosede= rr prr since from 5.302 (1) : Ji(v) = - dJo(v) iclv= -Trl d[j; cos(v cose)dei /dv = 11-1,1 0 sin(vcose)cos8d8 Ey. _ jrzAy= -jo3.21711.1c
Pio ab it. EXB I 2 .11 I r.p.
__,I_ 2rra' cpI .fr.d. sin a Ji(1310c)sinliab sinillni(Bloc) sin—e- j 13;°z x e-i l311°z . Bx=-tg-Y-
2TT2 pIkoc2 Plioa 2 b2
2 rx , ]2 rb r 2 12cuca rd s in — a [Ji (13iloc)] sin2-7. 7= j j Trdxdy= P i (:110 c) sin 2.1-r-Y o 0 ND ab
2? >1_ 2a 72c 2 [_ , i 2. .2I'd , ,.. .., . . the guide . wave length, is . R=T-2- = _, J i moc) sin 7 Ans. Pio = LTI.A.g where Ag, Noab often much greater than c so that, if d= -lia, as ph, nears zero R-)12a)112c4 pia /(2ab) so, if A is area of loop, R.couA 213 ito /(2ab). For half loop (semicircle)of area A', A= 2A' so 1 2w2 A'2 2cDpAi 2(3,In for small 131'0 4 where c = (pc) a . cab ab
Chapter XIII. (xv)
Page 546.
Page 153.
-8-(31,p552) Component of Ai of image loop tangent to driving loop is At = -A0cosa =-A,0 (nbcos0 - c)/./n 2 b 2 +c2 -2nbc cosszl = Anc . From problem 44, real part of 1412)(PP) is propagated, so At = -j1TcuI0 a-1(bRioc)J1(PR)J1(I31c) sine . = jrcP.Io(Pia)- 1D J0010 Mc] Ji(131c)sin2 (Trd/c) . From 5.298 (3) : J0(13111)=J0(13F)J0(13pb) re IT rrr + 2 raf iJm (Pic),Tm (13111b)cosm93• Rn= en/I = cjojcuAtdro. Since jo cosm0 dO is zero, total image R is Rim=1112cupc2 a-1(Ji(Pic)sin(17d/a)} 2 2nE1J0(P1nb) -. By problem 45, due to loop itself, (a=m), R0=112cupc2 a-1(Ji(P i c)sin(17d/a)) 2 . Total R is then R = Rim+ R0 = 2174.D pc' a-1Di (Pic) sin (1d /a)) a
i+ necTo(tinb)). .
By Watson
Bessel Functions p 633 or Magnus and 0. p44 we have ji+ milJ0 (mp)= 1/p so
y=0
R= 217 2cupc(piabriJ1(pic)sin(rd/a)) which agrees with last problem. See problem 45 for reactan, of loop without walls at y=0, y---b. For reactive effectof images take imaginary part of HP for propagated mode and all terms for other modes. From problem 44 we have 17d 2 C° A9:5 a t -Y1(1301n)4131c)sin2T+ rpE2Ki (PpRri)11( 3pc) sin"-T-)eja)t Need tangential component integrated. For first term see last problem. im 1 1117,7 iom „ = -17J i(Pic) Y0(13inb) ctio 1- cAPI Lv = 13; bctio ( 40310 Yo(flinb)+2:i1Jm (010;1 ( pinb) cosn0
pr i x=x0+coojoAt do=x0 4rwcP. (pic)sinV ] 2 +pi2K 0 (ppnb)[Ii(ppc)sinai-P-] a ) a n12° = (p nb)Pi -9- (36,p553) Superimpose vector potentials of image wires on reactance of single wire given in problem 33. Since r <.< X , pa in 33 is small so, from wire itself, by 5.293 ( 5 ), Xo-2,rrivey Or) J A/TibrIPr20.in(aPr) cup.bk pr) wire 217 1(pi)- 2-217rATE1T Neglect J1 (Pa) compared with Y1 (Pa) in the denominator of problem 32.
+7 +
+
-10-(37,p553)
++
t'+ 2d -
For wire at distance p: jXn=jcuAzb, Xn - b r.2.1y pi Add effects of 21 rap ° images as _shown: X= (22/7 4&r- + 2nE ly0(2n pa) -:.V0[213(na- a)]-42-6r3X4(2ftsin2a2+ez) t Yo(2f3A/(na-d)2+c2 )) Ans. Note: By 5.293(5), C-0712=2/n cc= -0.11593. 5.32(5).
A
t
•a , -
z
•
*±
•• i4c>tclo,
/ 2d
The wire and its close image is a short section of the bifilar line of 2X 0 2 ../ c 46 +.1c 2 -14-1,_,••• 221127: ,„:"2 11.18 (4) so its reactance X0 =coL is given by TI Onfir+Y0 (2pc) 7=ff utr wpb-T1 r 1,-.11 c which agrees with 9. By HMF 9.1.79 V= ol a =3: o ( 4 32n aa2 +432c2)=11-4 t ( 23n a)J2m( 2 pc ) . youpna)J3 y2 221 ZY0 (20na)-Y0 (2pna)-4-P4-" --- 2y2(2pnoW..—1aYI -2 (2pna)=Y0(23na)-Yi ( 2 Pn5a-(-1)mY (2pna)by 5.294(6) neglecting (13c)4 terms. Combine first and third summations in last problem, omitting .n=0 term already included in X0 . In same way:Y0(44320e-d)2+4132c2)---yof2Kna-03 p2c2Y1(213(na-d)) na-d Yif2p(na-d))] Thus X= 11211EC 0 11.-1 TTf3c 2 (213na)+11c2 Ans. n=1 'na 2r n=-02 2 (na - d)
EY
If d= 4a:
X= -IPa 12 cosh-1 e 2 r 213 ra 211 1 (-1)InY m" (nigia)
Ans.
Chapter XIII. (XV)
Page 547. (554)
Page 154.
-11- (38) a x, 13.02 (3) t= 13.02 (1) Ute=ECrancosuirr
--T-ce i Ninz 4 c1 11"Cain sinm
x3e-ii3,Inz. min. m2
13.02 (4): B= E c.mfii"mknsinm14.c+kg,„cosargr— x2 a te m 1 n Set z=c, multiply through by isin(pyTx/a)dx, integrate x=0 to x=a, --=-1.1. 73/2111 C40 . sine Bxdx =-4sinErgTd 3-731 V:sina 11 -21 rxdxC.I;in=::•11--
-+1. i",„E 31Tsinq-Td .
3 ~mc = 2 sinh jj3thc. C =LlP1 e3l3mc sin ad . For closed end add image -I at z=-c. m m10 (3 j2a4mElsin(m7d/a) :L arx-e-J56z If c >z, interchange c arAariggng ign When z >c gte= sinhjgle sinni When z> c ilte= -(2pI/A)miisin(m1Td/a) sinh(jf3Lc)(-i sin (mrx/a)+k am17/131a)cos(mrx/a)) 0
!te="
(2pI/a) E1 sin(mrd/a)(icosh(j8k)sin(mrx/a)+kam1743na)sinh(jl3Lz)cos(mrtx/a)3eiNle m= -12-(39)
By 11.17 (3) = (40'1 E B (real part).
=.1 cE- 2'1)2121
2
÷ xdx 0 -1L ard-sin2 (3LeZa sin2m
No transmission if 133'11 is imaginary and such terms vanish when z is large. Transmission i•r R-cDp 2b g sin 21 . md RI2cupbI 2 n sine 2glesin2mrd if (n+l)v>2av>nv is 2 = a a m=1 a m=1 NI
pm
-l3-(40) r= 0.5mm=0.0005m, 2+.= 15cm=0.15m, from 13.02 (2) 810 =11f/a=10r. 0.1 =27.6 f = 13io/ (Pe )=317' 106, X 0 =21178rg=0.2m. p .2ri (0.15)=41.9, p0= IRT370 cii3v=1'.257•10 10. From 12: 100ohm= (2.514.10 10 4.257.101/(5.27.6) sine 27.6c\
.05
=229 sin2 27.6c. sin27.6c= 0.661, 27.6c=0.7222, c =0.02615m= 2.62cm. By (5)p197 ka=-0.115 2ka/1'f = -0.074. In -9- kOr=k0.021=-3.863. (20730/17=-2.460, (20nafir)/TT=-2.534. Five nearest images: 1 neg. 0.05248 = 2.19, -Y0 (2.19)=-0.521. 2pos.-* 2Y0 (4.74)=-0.518. 2neg. -> -2Y0 (4.19)=+0.180.Sum= -2.534 - 0.521 - 0.518+ 0.180= -3.393. Here 4.19 = pa and 4.74 = 2134/0 . 25a 2+0 .026 2 2 . Reactance is 051,403pb • 3. 39=0 .02.0.85 .1. 257- 10" • ce =266 ohms. L=2.12.10" henries. Will get same results with wire half a guide wave length farther from end. c'= (1/27.6)+0.026=0.114+0.026=0.140M=14cm. Phase velocity vp=a)//3;=f3c/81=1.52c. Signal velocity vs=131c/8= 0.66c. -14- (41) Potential across 2 due to 1 is jcoM12 I1=El b (1). Current 12 in 2 gives equal and opposite potential to this. Potential due to I2 at 2 is 2pb . mrdi . mrd2 . , a mE1131;1 sin—a- sin—a- sinclinc 2e -E1 b=12 (R 2+ jcp1,2) = I2Z 2 . From (1) and -11-: 1412 =— where ea < c1 . If c1 < c2 , interchange el and cal but field and source points are also interchanged so that M21 = M12 . Additional potential at 1 is jcuM12 I2 =Ve 1 . Original el ca2 mf9 4-Zi Z a co 2 M 2 ZII 1 . Ziwas so total is C1={ Z2 'a
Chapter XIII.(XV)
Page 548.
Page 155.
-15-(42,p554) In last problem d1 = d2= ia. To be real (use numbers from -13-) M12 Psinftici cosfeica =(2.1.257.0.02-0.661.10 9/(2.7.6.0.1)) cos131c2 . Let plc2=1". c2=17/27.6=0.11390 14=-1.204'10"e , 12m2= (1.257.120.4)2 = 22940. Since 13114:2=r) R2=0 by -13- so Z2 = jX 2 . From -14-, lei = R1, 0142 X; =X2 From -13- Xl= 261 ohms. Therefore if X1=0, X2=w2M2 /X=22940/261=88.0 ohms. 2 1 From -9-, -(2/17)0/m=6.13738 and 13a= 41.9•Q.1= 4.19 so that X2= luoub (-(2/17)(in41.9+ka) (
+2Y0(4.19)+Y0(2 -0.1139.4.19) -2Y0/14. 19 1+4132c 2• • • 3.
2Y0 (4.19) = 2•-0.0901= -0.180.
Ye (9.55) = 0.1609, -2y0(^/4.192 +4132 cl )=-2Y0 (10. 42) = -0.0972.
-On(41.9r) =11(0.0425
+(4.1.114)/(1.257•10 1 °• 1.257 • 10 8.0.02)) = 1.0561T/2 =1.660, 41.9r = 1/5.260. The 0.0425 is from 0.0738 -0.1800-0.0972+0.1609 =0.0425.
r=0.00454m=4.54mm. Ans.
-16Free wave length X=0.005. From 13.02 (2) cut off Xmn= 4/4/4;1 1712. This gives the following modes up to cut off: m n
Xmn
0 7 0.57
0 8 0.50
1 7 0.55
1 8 0.485
2 6 0.55
2 7 0.495
3 5 0.51
3 6 0.47
4 0 0.50
4 1 0.496 cm.
mn values for transmissable modes are: 01, 02, 03, 04, 05, 06, 07, 08 ; 10, 11, 12, 13, 14, 151 '16, 17; 20, 21, 22, 23, 24, 25, 26; 30, 31, 32, 33, 34, 35; 40 . TM waves. [13.02 (7),(8)]. Excited by legs only because bar and its images cannot excite Ex and therefore not By. But to close loops in transverse plane must have both Bx and Bye YA, From odd symmetry of magnetic field of legs Bx is even and By odd in y-direction about y=1. Thus odd n's are out. E2 is zero at legs for n=0,4,8sothese are out.1
R
E z is also zero at legs for even m-values so m= 0,2,4 are out. Thus the TM pass modes are TM12 , TM32and TM12 .
TE waves. [13.02 (3),(4)]. Excited by crossbar where Ez is zero. From even symmetry of gradient on bar only modes with loops at center can occur. Thus n must be even so that cos(nr/2) # O.' Only modes for which Ey has no nodes on wire can be excited. Thus sin(mr/2) 0 0 and m is odd. For even n there must not be the same number
e
of negative and positive loops on the wire which throws out n=4. Thus the TE pass modes are TE 10 , TE12 , TE 12 , TE30, TE32. -17o2w 62w] to 2W k b2W rry,r oatT = 4 2 Ex = ° if ki3 2— W 2X.1-.)(ZW tmx oy 2 + bz2 - Zbxby -•••• oxbz ntmx 4y oz - by x=0,a y=0,b z. kf-c -) WI= Ai cospix sinEn b W 2=A2 cosp2(a-x) sinTe7t3im n r2 nry nr rm e- jOlnz • A1 = i (-Pt sinpix cos -kjr sinpixsin-I 2 )A1 cos pix sin-r +j pn b2 pnp1 A1 n2r2 any nr A2= i(-13 1 ----)Acosp2 .(a-x)sin b -jpal-A2 sinp2(a-x)cos b +Up' 2 pnp2 A 2sinp2 (a-x)sin, e-j13;nz . n b
r
(continued)
Chapter XIII.
Page 548
Page 156.
-17- (continued)
zBi=-j 3f j3pnAlcospix
j
n _ kf 2 A2 7rcospix cos
e pn
, By* 0 at x=0, y=0,b.
B2 same with subscript 2 for 1 and x-a for x. At x=c: Ez 1=Ez 2 and Er=Ey2 give
E PIA' sinpic =-p2A2 sinpa (a-c) (1) elAl cospi c =e 2 A 2 cosp2(a-c) (2) comes from ElExi = E 2-X1 • Since B1=111 113, .132= 1102 Of =ni cul i t , and 1322 =u2coae 2 , equating tangential Li on the two sides of the boundary also gives (2). (2) /(1) gives p2 E1cotp1 = -plE2 cotpa (a-c) (3).
Equate
(5' -p2 (4) so pi velocities so IV = pf-(n/13)2 r t2- fig=o32)12e 2 -n2r2b-2-13 r - (n/b)2 -pi Insert (5) in (3) to eliminate piand p2 and solve (3) for Pin . For cutoff fii)n is zero so pf=colp1 e1 -n 21?2 b-2, pi=uti2E 2 -na1?2b-2 (6). Put (6) in (3) and solve for cutoff frequency coc . 18
b214. orw o2w 2bW 2bW -7)cix vwtmy= o375-y- (-672+6—zr)"1.5 V e/P2W traf "9") E 7+14 b2w
z TM W2=Aisin(mrx/a)cosqiye 3 14612 , W2 =A2 sin(mrx/a)cosq 2(b-y)e-j1311cl 1 -.113 1 z . mrx . ATI2r2 rork mr mrx MCI e jgignciA i sin-r icira—Aiccs—a—sinchy -j a + Pula)A isin--- cosqiy F ITIff 2 d mrx mem 111RX a A2cos--a--sinqb-y)-1m +fim2c0A2sin-Tcosq 2(b-y)-kjqdraciAl sin y sing 2(b y)] e- ji3mgz 112— A2= jamq z . B2* 0 at x=0, a, y=0. B2 same but subscript 2 to n 1 Aisinosq y+kpfAi -fcosglyle lit 13 2 13 mq
[
-
1 and y-a for y. At y=c: Ex1 =Ex2 and Ezi =E z2
-
-
singic=g2A 2 sing2(b-c) (1), tiEy1=e 2Ey2
and magnetic relations give E1Al cosq2c = E s A2 cosq2(b-c) (2). From (2)/(1):E02cotg1c=-E 2qicotg2 (b-c) Equate velocities : p'm2q=p?- (mr/a)2-gi2=131-(m Tr/ a)2 -
(4), so q;=11-(mr/a)2-
ql=f31 -0117/0-1362q
Insert (5) in (3) to eliminate q1 and g2 and solve (3) for ►3 ng . For cutoff femq=0 so E - (mr/ a) 2, qi =41.12E 2 (mfr / a) 2 (6) Insert (6) in (3), solve for cutoff frequency. ow ,o2w 62w / 62W jy6-7c2 + . Ex=0 at y=0, Ez=0 atc=0,i TE 7X.:324t ey aye = 4z ±k1 tey" -
MITX1 -jpi
ncx- e-j141q z . +1s.A2-a-sing 2(b-y)sin—a-- e — r a xe-i1311 z W i=Ai singlycosm w2=A2sinq 2 (b-y)cosmrx mrx mr mrx A I= [L.j13m1ciA l singly cos-i-- +k A l -Fs in---1 e j131qf A., 2.= [1, jiVaciA 2sing2(b-y)cos-amV(12 mrx] -j pm'qz mrx mrx mr .13,1 =Ltgi--a-Alcosqlysin-r- +j [11414-F--) A1 singly cosg- -.Is jit/incigi cosqly cos--a- e m 2 mr inrx +j03// ON] - jprniq z B2=[ -iq "21A2 singa(b-y)cosm --7 rx 3-+14 jrijnqq 2 cosq 2(b-y) cos -a- e a ..,.. —A 2 a 2cosq2 (a-y)sin--... 1 012 +--7--
At y=c: Ezi =Ez2) Exi= Ex2 and By1=By2 give Ai singIc =A a sing2(b-c) (1) u2 B.x1 =ui Bx2 and (2) . (1)gives -112% cotg ic=piq 2 cotg 2(b-c) (3) 11 2Bzi' 111 BZ 2 give -1.12 %AI cosq2c=p0 2 cosq2( )-c) (2) Equate velocities: (314 =13? - (mr/a)2 -q12=31- (mr/a)2 -qf (4) so qi=032 pi ci - (m1T/a)2 -(31;14 and Oa = cn2p2C2 -(mr/a)2-13ni'll (5) Insert (5) in (3) to eliminate q1 and q2 and solve (3) for I3mIci For cutoff pgiq is zero so q?=culpi e 1 -(narr/a) 2 , q22 = col u2e 2 -(mr/a)2
(6). Insert (6) in (3).
Solve for cutoff frequency. Multiply (3) by -tan qi c tang 2 (b-c) and obtain ilig 2tan qic= -112% tan q 2(b -c)
Ans.
Chapter XIII.
Page 548.
Page 157.
-19Split Fig.13.08b into halves by the dotted line shown shown so that the
1r,I Itb 4,1
shunt susceptance is B/2 in each half circuit. The iris structure does not
z >
involve x and there is no incident Ex so higher order iris waves also have no Ex . By 11.01 (13) an x-directed magnetic Hertz vector with the incident wave x-dependence is rx -1171; z rx n (2) z e Anz (1) From 11.01(13) g =- Erm0 sin--cos b e n -m ir on sinlreos 3.7 =a zn a -2 !Tx .n172 Trx . Eta jpin z b i-j-Ecoslrsin b e Jen 40322n+ b2 (3). The variational formula
n
13.07 (4) is the same forevery thin slice normal to the x-axis and looks like a slice for a parallel plane guide for which (2) and (3) become bEy=-E(2-68)Vncos(nry/b) and bHx=E(2-6DIn cos(nry/b) (4).
By 13.00 (11), (12) and 13.02 (2), with .a infinite,
I/V= 1/Zk=N/T e gzjb/xg)/NA12-(zbAg)2 (5). Ey=0 on the axis and Hx=0 e /A/1- (3 .g/Xin)2 =4/7— in the window so multiplication of (4) by cos(n11y/b) and integration from 0 to TT gives b pc Vn=j0E y (y)cos(nry/b)dy, In =jc Hx (y)cos(nry/b)dy (6) From 7.01 (2) and 1.06 (5) the current In charges the iris to potential Vn . In=-YnVn and Vn=-ZnIn (7) where Yn and Zn are the iris admittance and impedance for the nth mode. If
0.4:y4ze,
Hx in (4) is
zero so shift the n=0 term to the left side in (4), and substitute for In from (7) and
rc cos b cos b aa' „ (8) E (y )cly = lYn.-1 Compare 13.07 (2). Multiply by E(y) and integrate over the aperture so that for Vn from (6) to obtain
-12-I0V0 =-nkYnaE(y)
s°= -n=1YnVncos 27Y b
cosni -3-7cosn1 -2 -7E(yt )dydyi (9) Divide by Ve from (6), use (7) and (5)
rfic
12
0 B 8b cc 1,10E(y)cos(nrep)dYi for 14-l and 11.15 (15) forte. B°= = Yk %g n-1 4/112-(2b/Xg)2[4E(y)dy)2
An E/y value exact at long wave lengths appears in 4.22 now carry primes aperture is at V' = 0,
(10)
(I).
The
y 1 =0 in the right half of Fig. 4.•22b. Insert V! and y' in 4.22 (6)
and take oU l /bx' with a'=c. Write Try/b=0, TTc/b= 0 and dy = bdOPIT. Then 4.72Cos(40) E ,- cos(4ry/b) (11) 4/sin2 (41%/b)-sin 2 (41/y/b) 4/C00-cos° The (10) numerator integral, expanded by Dw 401.06, gives two Mehler Legendre function integrals, HMF 22.5.36 and 22.10.11, so that ZE(y)cos(nry/b) dy is
-DO
4fib ro cos (n+4)0 dO 4. 4,5b recos (n 2r.44/C-00-cose ' 211'io4/cos0-cog0
Tb[Pn[cos ( rrc/b)]+Pn _i[cos(Trc/b) ]3= ibIn (rc/b)
The (10) denominator integral is bP0[cos(trc/b)] so that B°= Note that, when
Xg = 1:0,
s..;i1 k1L1(.7 ))2
2-
(12)
(2b/Xg) 2 (13)
(13) must equal 13.06 (7), which gives
407qcsc(.i1Fc/b)3=ni1{Ln (ffc/b)} 2 /n
(14)
Add the left side of (14) to (13) and subtract the right side to get 2b rc (15) B°=-(4,0ncsc]F+ E [(n 2 - 2,2„13511.- n-131 L'n X D n=1 2 g
In all cases tested four series terms or less gave four correct digits in B° . The static
B°of 13.06 (7) is a lower limit.
Chapter XIII.
Page 549.
Page 158. ya
Y3 Ui0
-201 U=-17/2 U=11/2 3§ Shift right, z2 =z3+7(d 2 -c2 ). a2 =a3=4(d +c 2 1 V=0 z3=a3 sinW 4.22(3). z2 = a 2 sinW+-f(d 3 -c2). ftz i Bend x2 axis up at ±b2 by 4.22 p90. za =b2 sinff;. Let b2=1. = a2-17(da-c 2) sin (12-rzi)= a 2 sinW++ (d2-c2 ). When zi= 14= FIT so sindrdi
E—b1
When z1 =-"c1 , W=-717 so -sin(i'ffci/b1)=-a2+ i(d1c2). Add and subtract to get s in( .-12-17di/b1)-sin(iTtcl/b1)=d 2 - c 2 and sinerdi/q+ sin(rc i /b1)= 2a2. Thus sinW=
U -Tr/2
U 11/2
2sin07zdb,)-(d2- c)= 2 s in(;Trzi/b,)- s in(01d ,/b)+ sin(+17c,/b,) 2a1 tin(ird1/b1)+ sin(71c1 b)
V=0
zi= z-b1 shifts origin to iris bottom. When zi=bi ,z=2bi=b. z=0 at z1=-b1 . c so 2b1=b. c+id=1)1+di or di=c+i(d-b); c-id=.b1 -c1 or c1 =4(cH-b)-c.
db -1=-cosf4. sin(-1-11di /b0=sinS(c+ 2 )]
sin(12-17zi/b1)= sinElTr(zi-bi)/bj=sin
VO 4-d
.sin[11 ,(c+ 1d)-irr].-cos[ - 77(2c+d)/b]. sin(ircA)= sin[I-T(V2-c)]=cos[211(d-2c)/b] - s in(12-Tidi/ + s in(ici/b1)= cosCi11(2c+d)/b1+cos[4rr(d-2c)/b]=2cos (17c /b) cos (-in:lib) by Dw 401.06 s in (-irdi/b3) + s in(iricA)= rcos[417(2c+d)/13]+ co s[1211 (d-2c)/b] = 2s in Mc b) sin (ird/b) by Dw 401.07 Thus : sinW= (-cos (Trz/b)+ cos (re /b)cos (% 1-11d/13)//f sin (ffc /b) sin(Ind /b)
Ans.
Insert W=jV, z=jy. Let V-► co and y-go, then only first term in numerator survives a nd_ sinW -> jeV=-erY/b csc(17c/b)csc(i1Td/b). Take logarithm, neglect Efftj and k(-1), then Trc rd, B0.,4b Bel a V= b -ktsinTsitr-53- 1 Ans. Potential difference is U2 -1.11 =17 so _x:72(sinihin35 :3) Ans. 13.06( -213- cos (11c7a) dz' _ C cos(rre/a)- I- 2 c os (rre/a) where 1) 1 1 +cos(rY ca) da l Nizt-1 Nii7.51 ' z1l+cos(Tre/a) Check: If z1 =0, z1=-b; if z'=c', z1 =-1; if zi =a, z1 =+1.
NiZ771 =
3g
!cos i/14'= 2Cbos;zi n/siniC a: - s in2 rx1 . 2 2 1-rc_ Trx b Omit primes. -c< x
W=
cost 4 7 + ( 1 c o sTi-V-)cos714I-
[
17c / Trx 6 c< x
3§.
= 4Ccos4 12%St1211-:t2 dt = iCcos l 41 Substitute in13.06 (11), (3). i 211' a ° cos4(Trcia) "c -a )= p.g tan 2/1:-/(1+csc21-) rrc /(1+sec2-2 -jacoeu xg Z°= - --• 2a ' 2 N. r sin 2tirraa)C1-cos2(+11'cia)) -
41
a3 (l+csc2r21) As with 13.06 (15) and (16) B° =--:s3 cot2 3-
Ans.
Page 550.
Chapter XIII.
Page 159.
-22CiBz (x)cos(mrtx/a)dx) 2ASBz (x)cos(11x/a)dx)2
From 13.08 (10) Bk= -aXgn:g24/1From -21-: Bz (x)=- c
--11. f-d 4
o s(111x/a) Wsin2(41Tx/a - sin2(irc/a) = Clsin(irfx/a)asinai-Thaa)-sin2(ilt/a)
rr Ec4 sin (4-1x/a)cos pniTx/a) , „ ;sr sin-Ocosm0 , d0 „ r sin(m-4)0 -sin(m-1,-)0 0 UX L4 L.4.1 4/cose-costo 0 cWsin 2(ir x/a) - sin2(irc/a) L.A... 2 (2232 rm (cose)-Pm- (cose) a = CBCPm (cose) -Pm_ i (cose) . In BEI denominator m=1 so B°=-fig cose -1 As in 13.08 (12), this is the static value if ? is infinite. Write in terms of d instead Need )
-
of c cose= cos(Trc/a) = cos(17(a-d)/a)=cosITcos(Trd/a) +sinrisin(Trd/a) = -cos(rd/a) so that Pm (cose) -Pm- (cose)= (- 1)D1Pm(c os (77d /a))- (-1)131-1Pm- i (cosOld/O) Pm(c04)+Fm-i(cost))(-1321. Upper limit is B°=-(Xg/a)mE24/1-4a2/(mT.)2(Pm[cos(fki/a)+Pm_ [cos (M /a)])r 2cos 2(4.rd/a)j2 . Assume, as proves for the symmetric inductive window by 13.06 (14), that when X=co this espression is the static value of Bo so for rapid convergence we have B0 =-X —a-g Lc° t 2(irrdia) [lA-CSC2 (l-rd/a)]-4sec4(4rd/42W1-4a2/(m%)2 -1) (Pm[cos (11d /a)]± Pm_ 1 Ecos(17d/0
rrd
Let *=---: Bo= (T.g/a) [cot 2 4f ( 1-csc 241) -4sec4 4/ nE2W1-4a2/(4)2-1)(Pm(cos 24)+Pm_ (cos24/)) 21 Ans. r12-B2 U: Bx(x)sinBix dx] 2 ) For lower limit by 13.08(6): Icol= +1311 :42(1:Bx(x)sinprnxdx)2/((311 B x(x) is like Bz(x) but with terms in radical reversed so that we need 1, , resin(iN/a) cos (m1Tx/a)dx c ic cos (m-T)v -cos km-I-Iso d0 = C2 (Pm(COSe) (C ose) . Same 0 Vcos0-cos0 /s in 2Ore. /a) - s in2(inxia) -1 _r co Pn-, (cos()) -Pm- (cose) 2 where 6.17c/a. a assumptions as before give: (Bo ) =— 2sin2 (0/2) N.g bm'-(271/702 r Assume this gives static value when X=3,and let y=417c/a, so that 1 -1cot2 y + csely P (cos2y)-Pm- t(cos2y))2] l+sec 2y 4m=2 (2a PO' m
E
mjj
-23dW Like Fig. 13.06 but with two openings. T i = 4/4 -13!4/4 -1 • xi -di dW oV aV = 2=bf(1-100, dzi ayl-Pbx1 j1 41)7774) 4/7— -1 . Let xi= bf-(1-bDt b b kat dt 1 k2 =1If V=0 on walls; i t dV=0 so net flux dx 1 = 41k4 t 2 rbl (xi-dt25_12:1 r f b through window is zero. "%ii7712 E2'131 l :7 1 4. 3" 1 -40 15 1 1dt d2 1 or b So 0. Thus di2=bfE/K dt -k2 t2 97 u Niti 1 dz' rrz' ow Ow bz z 2 -d2 2d 4 sin 7 2 3,. • z' rr 5,11 -1 T z ,z1 = zA - b4 bz' = bzi bz' 71;T:117 dz
__
,
OW I tz'/d1 )-E/K1 bit sin2( r F z., =0 if x' =±oi so bi sinIZ:=1 , k2= cos25 . -Vbf sin2 ( -frzUdI - 1 icL .c sin2(41Tx 1/d1 )- (E/K) sine (2;rtzI/d1)- (E/K) dW y Bz = 0 d z 4/sin 2(irrzl/d') -sin2(-1rTrcl/dI ) • -x Aisin2(irtcYcr )- sin 2 (irtICl/d) (E/K) sin 21
=x - d
sin(12-rtx1/d9 ) = sin (-iirx/d 1 - .irr) = -cos(22.rx/d1 )= -cos (rx/a) (continued)
Ans.
Page 160.
Page 550.
Chapter XIII.
-23- (continued) cost (Trx/a) - (E/K) cos2 CTTx/a) - (E/K) 11c It( Let u sinia- = cos -i-, -C C Bx Nisin 2 (4T1c/a)-cosa (44Tx/a) aicosa(Prx/a)-sin2 (-1-Tic/a) ' (-a/r)sin(irc/a)du= sin(Trx/addx, when x = (a-c)/2 u=1; x=a/2, u= 0. /2 u 2aCc rain . 2 77c TTE) _ aCia k21 rI E) 2a Csisin -rIc /a G.I 2sin2 IfIc/a)- / Trx =--2 rr- 0 ;csin-a-dx 1(a-c)/2 TT T-- 2a -2K - 2 '' - ' -2K sin(rc/a) 1-u
01 ...2cr_.,ji
Let ucos(-inc/a) = sin(1Tx/a) 4(a/1T )cos(-irfc/a)=coi(1rx/a)dx., x= (a c)/2, u=1, x=0, u=0. loasi-Tfc /a)fl- (E/K)-u2cos2(7t/a)lu 2aC E Tr cos2rre 2j Bzcos— a dx = 2aes 0 2 TT ((1-K) 2-4 TT cos (4i rrc/a)A/PFff o from 13.06 116) E4( 1-ka )K 1e3Y.026 Ba. Byr.(3E 11 2k2)K 3;oa• (2-1c2)K- 2E Z =,jCUL o'=" i so )K col. a E- (1-43(2 where k=cos(rc/a) (11),(15) s 2TT (1-1(2)K- 2E )‘. E-1(1-k2)K -
-
-
-24From 13.08 (10): Ba= -(Xg /a) i3,11-4a2/(1IX)2 LiBz (x)cospoix dx/(1Bz (x)cospix dx)3 2, then by -23Bx+jBz= C(cos2(1Tx/a)-(E/MMina(irc/a)-cos2 (Trx/a)) . Write 2n+1 for m since B is odd function. (cost (fic/a)-q/K)3dx [i. f c 0 s 2ft/ a)-(Etiolcos{en+1)1Tx/a}dx/r 4 a2 B°=-:?Sg , /1(2n+1)2)NF Jap (cos 2ME/a) - s in 2( Trc /a)) 1/ 2 Pap j-cos 2MX/a ) - s i n2( irda )11/2] a n-IA/
ra lc, dO4rdx. I 2a3/4/cos(Ticia)+cos(277x/a) dx. Let 0=2..i a +i-b cosEfn+1)Sap= C 2I(oa-c)/2 (4c° s2LTx 0=17(a-c)/a=0=17d/a,cos0=cos(17-7)=-cos-7. 1 Sap= C3je0cospicos (n495 (cos0-cos0) 1do + (1-2E/K)recosgn+o
f x= a-c
'
3(coss6- cos0) 2do
C Pecos(n+--)0d04..10 cas(n-)1046 + 2(1_ 2E)scos(n-44)95d0 - CEs (Pn+ I (COSO) + Pn _ i (cose) K 0ilcosref-cos0 0 4coscif-cos0
= '4 J OVCOSO-COSEJ .
+2(1-2rap= C5 (cosO+ 3 - 4E/K). Thus if k = cos (+17c/a) and 0 = frd/a, K )P n (cos0)). If n= LA_ Bo ? ■ . g [E-1,-(17k 2)K + 4a2 1 113,4_1 (cos0)+1",.. i (cos0)+2(1-2E/K)Pn (cos0)] UPPER 2n+1)2)4.2 -j cgs0 + 3- 4E/K a E-(1---2-k2)K n 1 LIMIT -1 rf cc' .rBx(x)sink2n+1)Trx/aidx/.1Bx(x)sin(Trx/a)dx) From 13.08 (16): where 13-6-=KgnE i
f
'I[(2n +1)Tri a ] 2 -(21T/x)2 r.cpa/ 2 (cos(2rfx/a) + 1-(2E/IQ3si.rn+l)rx/a)dx Let 0=21Tx/a; x=- (a-c)/2, 0=-11-17(c/a)
J
,(a-c2 -cos (1Tc/a)=cos(11-11c/a)=cose, 0=1Td/a. //cos7fc/a) + cos(2,Tx/a) {cos0+1- (2E/K)}sin(n+ 16-)0 46 rrsin(n+4)0+sin(n-1)0+2{1- 2E/K))sin(n+1 95 do rrr =C z o Vcos0-cos0 cose-cosill Jo =C4 (13n.f 1 (cos0)+Pn _ 1 (cos0+ 2[1-(2E/K)]Pn (cose)) = C4 f (9) 1 1 a {E-:il.:+k2)K °' LOWER Ans. =- E•40.-k2 )K + nr17(2n+1) 2 -(2a/4.) 2 2n +1] f (In} LIMIT
Page 161.
Page 551.
Chapter XIII . (XV)
-25-
(44 , p554)
In 13.02 (1) take TE 01 -TE 10 . Ute=c(cos-lial - cos-7), vc= --v/a from 13.02 (2). Tangential E is zero when y=a-x since cos (IT-43)=-cos0 and sin(1,-6)=sine so - j 1310 z si n a + sin-r . sin le C{i. sin on s Ex-Ey=0 (13.02 (3)) and lEx l = lEy l. rr Alsina a 3+..s. k Ir a[cos= a -cos73e-il3;°-9. Use 13.01 (6), /7 131e-a* 13 te= Ute 2+bUte a -71s te.A ds = dx so a--, r 2 On diagonal Ute=-2Ccosrrx
2rr2c 2 ,..- pa rrx rac 2 112 ds- a2 OTJ ITC TTx 42. On one side — =— sin— , J2 sin2 —dx =
irrx a Tr2C 2 a) [--1 ds = . U=C(1-cos— x 2a a Uads = 4aC 2 . Include two sides. On diagonal U= -2Ccosi-T-, fU 2ds =4"i- fa cos2 7-dx =212a .
jsys*
S
a
a
—
as
e
r
a
Over area U2 = C2 (cos? - cos-7)2 . Take one half integral over square. SULdS is rx a ra a Trx ulds= 1c2S (cos a - 2cos a cos-if- + cos2-a--)dxdy = C222-02- - 0+ cl= 1-C2a2 so from 13.01 (6) 2 0 j0 2 2 v2 rr2 0 . V 1(2-k5XV10/V)2 al r r _L_ a= 1- (Vio/v)2 aa2C 3....r Ans. (1+,511- V2 -0J+ 3 aC2 V2 II 2.1TC")/(4/ v2 + 2pv6 (- 172 a iiv6a[1- (via/V) 2 3 1/2
f
-26- (45,p544) rig Mc a . Itx . 2m In 13.02 (7).• (8) take TM . U=C(sinsin ) by 13.02(1). al andTM 12 a +sin—sina sinflay +c oaTraxsin211V, . . . 2Trx 2a -1 E=CCual34(2cos2 a ' )+).(sin a cos+2sinrr -2-c a c os a )j +1,z-V(sin214x-sin7 + sincsin-rrine-i °1z . When y= a- x, Ex-Ey=0,Ez=0=U. a On sides Et=0, U=0. On diagonal (oU/on)2 = (Ou ) 2 + (b uThy)a= 4(1Icia) 22s ine(fi/a) 2c2 2 27.17 2c2 r, bil, 2 , _ 81117 2 C 2 ra , or/d _wili___, 6U 2TIC . 3Trx rsin Todsai2dx, j ( On ) as - --F--jo sln a xa 00 1,, D'Vz 2a _r Ox- -a a Trx rx "rf6u, 2dx 4 rrC 2 r TrC , 21tx ,, rx,_ 217C On one 5ide, y=0 :OU/on.--7c ksin-7—f-Lsin-a- —a--sin-5-(1+cos-w-)..p_b-n-1 - --a--j (sin2u +sin2ucosu . Trx . 2M7 4rrc2 1 1 3 5 ff 2 C 2 radS= 2rx . La ,..+4sinirLX Q. rara, . 2 -+sin2ucos2u) du =-7sin 3 -- (-2-r+-2-11+717)=7---. JU 22 J o joksin a-sin a sinisinTsin-r . Trx 2ry 1 aa a a a2 C 2 +sin2 -i-sin2 )dxdy = TC2 ( -i•-2- +2.0 + 2 2)---- (Half integral over square). Substitute in
r
13.01 ( 7 ) : 1321 =r5-a.- , Vc=
,v/32va .
a 41... 2-1.1,./..12 1 a=6
Ans.
kVe VJ
-27- (32,p552) By symmetry Bp is an even function of 0, so for a TE wave, from 13.09 (6), tangential
e'ECnni Prm( -ei 3mnJn mnP) cc sm0 +01pJm (3mn P)sinm0)e iPm'nz . Multiply through by (-pipmn4(13mn P)cos m0+61Jm (f3mnP) sinm01pdpdo, integrate 0 to a,-1T to V. (
3
r
From 5.296 (11) right side is i Pm'nCmn t-i(Prtn a a- n2 ) Jit(f3mna)ie ii3:nn z° . 210 since JM(I3mna)=0. On left side sinmo goes out from -e to E leaving -SmnOo d0Jrh(Smn po).4. Thus for TE wave : Cmn=e i /3mtnz°(2-5g) j 13mnPo JEt2(13mn00)p1d0/(21713TIm (13ina2 -m2) Jr1(13mn a)}
Ans.
For TM wave, by 13.09 (8), Bp is an even function of p so that l ( pranp ) sin mo 3e inn = EDmnib ( - (3 2 i-)3m ( Pmn P)cos mc6 + fzir3213ranJni
Multiply through by
(43m (13mnP) cosm0 +0 flmnJM(f3mnP)sinm03pdpd0. Integrate p=0 to a, 0=-Tr to TY. From 5.296 (11 right side is, since Jm(13mn a) is 0, 132 Dmg-iffimn aJr'n (Pmna)) 2 (211/(2-6r0 °))e-j13mn z° and left side is, since sinm0 goes out, -drom Jm (13mnP)-lip.I. Thus we have Dmn= -uId0m Jm(Smn po) e iPmn zo i c r[ vimn4m(omna)]2)
Ans.
Chapter XIII.(XV)
Page 551. -28-(33,p553)
Must generate waves so E0 is odd about 0=00 . Use -27- for arc elements. dpn,, r TM WAVES. For arc elements Jrn rOmn(Do+-TJPmn (Po d72-1-1 =Pron4 (Pm nr3o) dPo • From -27- Dr;m=-pIOnindpo d0m4 (PmnPo)eiPlulnz°/[ff[OPmnaJM(Pmna)]2). Radial dpo is negative if 0>0 so in 13.11 (3): -sin m(0-480)+sinm(0+80)=2cosm0 sin180 Pe•rn80cosr0 so coefficient of cosm0, -Dmn in13.09 (9)0 comes from Cmn of 13.11(3) with eqg omitted. The result is -Dmin which cancels the arc terms above. Thus NO TM wave. 8c orprin cJ4,(filmne 6c erc rta)] -8mn (c---2-)4[Pinn(c-TE WAVES. For arc elements finin (c+T)4[11mn (c+2 2-)= d({3miic) Q (Pronc) =C8mncJm "(fininc)+Jm t n (pinnc))8mn dc. Substitute in -27-, letdS = c dc do Cm'n= juIdS (2-41)(134nc24(8mnc)+pninc4(13Tanc)) ei 131:11nz°/ (217c2 gin (81na2-m2)4(13mna) . For dpo sides when 0›.0, current is negative so in 13.11 (2) and 13.09 (9) for B 0-component, -Dmn(-sinm(0-450)+sinm(0+-100=-2Dmncosm0 sinm = -mDmn 80cosm0. So coefficient of cosm0 in 13.09 (9) comes from Dmn of 13.11 (2) multiplied by m80. Write 2-8% for 2. C;l'in=-jm2 p.I(2-54)dS Jm (filmnc)eif34Inz421fc 2 13m1n(Brgn =-8rfinc2 Jm (8mnc) by 5.291(3) so Cran=Crnn+Cm "n-
-m 2) [-Tm (8mn a)]2) Add 13m 'm2Jm 2 nc2 4-F8Enn c.Jm
- j pIdS (2-8A) flignJra(f3„,,c)
217131n(f3An aa -m 2) pm (puma)] 2 e
j Om'n zo
Ans.
-29-(34,p553)
, a, TE WAVES come from dpo only. By 13.09(6)0 13.10(3): ei/34m`zo-1:2)-illtnn(204) ,.e iPmn zo -muIdSJni8„,nc) i prrin 6e j (3Mnzo . Put into 13.10(3) j8/ zo n Ans. Dmn" rc(81na2-m2 ja ( mne)32 e Let dS= adpo TM WAVES. As above, for radial 8 write j8thn8e 3/31nnz° for e jl3r6z0 in 13.10 (4). so ;it n = ndS (2- 8/) (I2mnc) e nz°/(2178mn[8a4(13mna)] 2 ). Outer radial 8 is negative so -Jm[f3mn(c+1)] Jm[Poin(c4).]= - Pmn6 J42 (13mnc). Substitute into 13.10 (7). Let dS=6dzo . qin= -juI8mndS(2-8/)Jmi (8mnc)e il3ron z° /(21,BL[13a4(13inna)] 2 ).
Prim +
Pmn 0dal
_Orini+0An _ 2 Pmn0riln 0mnPln
so r
-ipids(2-6,00,141(13mnc) u.r1, -Cln' ''rnn 2r0mn1341n[aJM(Pmna)32 -30-(46,p555)
Ans.
From 13.09 (1),(2) Ute= CJI/2(8no)cos12-0, 42 031 0)=0. E0=jus, Ep= 16 a((.0)• From 5.31 (2) J1/2 (v.,r277(-; -s1-1:71"-T +cosvlivV= 00 so tanv= 2v. J;i2 (v)-).. s/2inv
n-
CJI/2 (13 1P)3 2 -4. 811/(217p). Thus the integral for ate is logarithmically infinite at the lower limit so the attenuation is infinite. When tanitu a=2Pna, 80211.1656 Ans. -31- (46,p555)
sin 1 pZ From 13.09 (1),(2) Utm= CJ//2 (80)sin10, Ju2 (pn a)=0. From 5.31 (2) Joi (13np)=JI 01713 2 p so 811a=1'7=21TV11 a/v by 13.00 (8). Cuts off at VII = iv/a. By 13.01(7),Wirdp at 0=0 2 C 2 sin2 811 0 occurs in atrn . But -0so aimlogarithmically. 21703 0 P 50 0=0
3 p-0 -P 2rp
Chapter XIII . (XV)
Page 552. (549)
Page 163.
- 32-(8) By 12.16 (2), 5.295 (8) an expanding wave depending only on p is Wtm= Ce 3cut{Jo (0)- jY0 (13p)3= Ce jCptH6,2)(130), so B0=133C e imt(4 (PO- jY,')(13p))=-Rs cePt H(12t130) by 12.16 (5) . At p=a, 271aBo=pieica from 7.01 (2) so -217a8 311M8a)=0I, C=
27703HIV(8a)). From 12.16 (4) Az=-132 Wtm=1I4,23(8p)/(217a8Hitat8a)}
J1(0.036217)=J10.1137 0.0569, Y1(0.1137)=5.69, so 1J1(pa)J<. 0.011Y 1(Pa)1. For Hanke 1 Function 42)(v) see 5.293 (10). -33-(9) Pi( J J +YoYA-1-1[-J IY 0 +J03/131 jcot =IZi,Ziam
(I3a)) Va=-Ei=jcP(Az)a./.- jcPuI O 2tTaurlifetJi(Pa)-jYaia)) e PLO, Yn -JOY') p-t(2/(170,a )) (J (13a) J (Pa) +Y0 (13a)Y (pa) Xi° 2WITaMa • '121,lifeMa cm/pc aM - 34-( 10)
Pi by 5.293 Trkijal3a 2M
In 12.16 (2), (4) : Az=flawtm=CJ0(PP)+DY0(13P)• At p=b, A z=0 so D= -J0(8b)C/Y0(8b) At 0=a by 12.16 (5)
,
2TraB0=pIc °scut= 217a fq Yo(13b) JO(f3a)-J0(13b)YA3a))coscot /Y0(13b) •
so C = 1IY0(13b)/(2 Tra1No(1b)JO(13a)- J0(13b)V0(134), V = "EL = jw lAz Ia./ = IZi so that Jo(PP)Ye(f3b)- Jo(Pb)Yo( PP) A z-C -Ira FM( Pb) - (131DYin (00)_ Y0 0313) 211a13( 4(8a ) Yo(pb) -4(1313)4(8a))
Ans. X -=
_ .
- 35- (11) E ju.YpAz • (.13 jwp CJ0(80)+ DYn(PP) From 12.16 (5) Tiz (poi_ Dylo o - 30 bwtmibp = p
Az
I3P) +MY° (13P) YIN ji(poi _mylop) (
(1)
At P=a Ez/H0 = Z a• IA= -fillvJo(13a)+Za Ji(13a))/ (jpvY0(13a)+Z aYi(1a)). Substitute this in (1)
, _ jPV Z `b
03b)Y103a ) (13a) Yo(P))+ P 2V 2 D0(13b)Y003a)- Jo (Pa) Yo (I3b)) jp.vpi(13b) Yo(f3a) - Jo(13a)Yi(f3b)1 +Z a ( (8b)Yi(13a)- J1 (80111 01)))
_ Pypq (Pb)Y0(13a)-Jil (13a)Y0 M) )+iZaCJI (Pa)Y0 (13b) -J0 (131D)Y1(13a)} -EvZa tJi(13a)Y1(13b)-J1(13b)Y1 (13a))-j(J0 (8a)Y1 (8b)-J1(813)110 (Ha)} -36- (12) (
Short circuit at p=a means Za =0. Thus first term in denominator and last term in numerator in Zb are zero in -35-. To make Zb infinite remaining denominator term must be zero so J0 (13a) Yi(13b)=.11(13b)Y0(8a). From11.09 (4) 13= 211,/TIE NE for the frequency Vc , so Jo(21151fiEVc a)Y1{2tr., "Vcb} = Ji (217/Tavcb) Yo{217/r ieVea}
Ans.
Section
-37-(13) Refer to -35-. At outer edge H0=0 so Za is infinite. Other terms are small so that Zb = j(Ji(3a);(8b)-J0(13b)Y1(13a)) If ev[Ji(13a);(f3b)-Ji (13b)Y1(13a) . To pe Vz requires that make Zb=O at B = 2111s.r— ieve b)=J0 {21 Ji{2114Tivc a}Y0[2r.s.(iT
vcb)Y3 (2776/Trevc a)
Ans.
1-tJ,
Chapter XIII. (XV)
Page 553 .(549)
Page 164,
- 38- (14) Potential across outer flange edge in last problem. From -34- at p=b: )30 = - 13(CJ I(13b)+DY1 (pb)) . At p=a: s=0=C.11(13a)+DY1 (Pa) . Thus C= 1-(P01D= PI" Pa) (P a) 21713 JA3a) Yi(j3b).11(13b)) pIJI(Pa) Djcakz,t.-julf, . Across outer edge, from -34-, V=(CJ0(f 3a>4D74103a)). 2Trbp(J i (pb)Yi(pa) -171(Pb) Ji(Pa)} p=carTi=cz/v. From 5.294 (9), Jo (13 a)Y1 (3a) - Yo ( 3 8)-11 (3a) = -2/(170a). L = t. V° =
(fia)Ji(Pb))=It/( 117 2coeab[J1 (Pa)Y1 (Pb)-Y1(13a)Ji(3b)]) V = 2Trbrii31 a /Pl(Pa)Y1(PE)-Yi - 39- (15)
Ans.
AZ = ECmR14 (13c)Jm(f3p)sinMO, Xz =ECmR14(13p)Jm(pc)sinMo . These give Az = AZ when
p=c and
by choosing M=mrf/a1 there is no tangential E at 0= 0 and 0 =a' • To give a diverging wave in the 0-direction, RM has the form Ftm ((3p) = Jm(Pp)-jYm(13p) so B0- )3'0 = oA/op - bAz /Op = iettPm( 3c)41( 30 ) -i74.1(3c)JA( 3 0) -Jk( 3 C).4.1(13 c)±jYk(pp)*(pc)}13 sinM0. At p=c by 5.294 (9), B0 = jPE C m(YV -Y'J)sinM0 = (2j 0/ (TrP c))E Cmsin (mTTO/Ce )= 2j /(Trc)ECrnsin m170/Ce. Multiply by sin(p110/a t )c d0, integrate 0 to a'. pisin(mrra/a')e ici3t= rel 2.c •c•4a1 cm. so Cm=(-jr-p.I/a t )e ia't sin(mlia/a1 ). Thus the vector potentials are: p
M
(13c)
For radiation resistance use real part: R = jczAzb =(co1rubict i) mE lpm(13c)3 2 sio 2ma For reactance use imaginary part: jX = (jcuAzB/I)=-j (corlib/a1 )mti Jm(Pc)Ym(13c)sin2 Ma Reactance expression probably diverges. Let angle subtended by wire be 2a/e and try mean value of Az at a=c, 0=a+1 and ps=a-g- so x= - (corrlibia' )rnf rim(po)Ym(po) sin/4a cosMa/c. -40- (16) By last problem: Ez=-YzAz = -(jurtp.I/a 1 )ejwtmEiJm(Pc)Rm(pp) sinMa sinM0, where M=mr/a 1 • B0 = -bAz /op =
)e juAinci rTm(i3c)IiA(PP)siomasinmo . it=1(fgzi50 /4)1 = cross products
+ (2J41721112/a' 2 )triiiPm(13032gm(130)K1m (pp) sin 2Ma sin 2 M0 • From 5.295(8) iii,21 (pp)fig.23 '(pp)= jrpp so TT=(2corriiI 2/(2& 2p))mEipm(Pc)3 2sin2 Masin2 M0 + cross products. 7 = +RV = broa ffp do . Since cross product integrals are zero: 13= (curnibi 2/a"),A(Jm (pc)) 2sin2m sinaMO (JO R = (wial'Th/Cx' )mcgipm (Pc)] 2 sin2 Ma Ans. -41- (17) By 8.08(1.1) L11I 2=Si•Adv. By 5.30 and -33- Yo(13a)41i(Prn 422 +C) as Pa-40. By 5.293 (5) Y1(13 a)- -2/ (17(3a). Ri=‘,,FTEb i. /0 213 a2)) • (fffia/21}= 2-4.01113 Xi= ..17MbY (i. 0(Pa) Yi(13a))/ 2 rta [111 ( Pa)] 23=
-ivri/e(kifta+C) / (217)=-(DpbOirli-f3a+CY(2rT)
For image wires Z = V/I = -Eb/I = jalAb/I. ±image distances Are pn+=2c since
o n _=2o sin( (s+i)(rr/o)-al, where 0 t's -.tn-1. From problem 32 at distance d, H .L.)Ab 'c jccliIbH 0(2) (pd ) jwplb rff3a, , mph (23 I I2Traf3H1 21(Pa) 2rral3I 2j HP030 = 4 Ho (I3d) so that Z of images is 1 04.1.19 r n in sfr 4 tzt1}1,5(213c sin--)- E 1.02t2Pc sin (5+1) j r-na)). Total input impedance is then n s=0 ° z sr R+____ n-1 2 , will)(1_2740A2a+C) si042)(2 0c , (s+1)17-na sE0H 0 L2Pc sin Ans. n 4 ])
Chapter XIII. (XV)
Page 165.
Page 553. (550) -42- (18)
By5.295 (8), From 12.16 (2): W= e iwt [Alinf l)(PranP)+ B42)(P mn P))(CsinT +Dco Pinn mustbe real for radial propagation. But PrAn =P2 -(nr/b)2 from 12.16 so p > nrrib. Az Thus cp > nr/(b) or v> n/(2b) Ans. -
43 (19) -
z, B2= b(pA0 )/pbp and from 12.16: p>c Wt e = e j6t3tECnIC0 (Pn p)I i (pnc)sinn: jwt nrrz rCnKi(pnc)I0(11np) p>c = -6W/OP pc, A0 = (2cp.I/a)ni/Ki (pnP)I2(Pnc)sinn4Td-sinn
fq= (nr/a)2 -wape = (nr/a) 2- (277/X)2 . pn= (2rr/x)A/4-(nrria)2 -1
Ans.
-44-(20) From 12.16
= f32 _ crt/ a )2>o,
= (nr/a)2 -02 > 0 if n>1. If X.=a, P?=(2r/a)2 -(r/a)2 =3(Tria) 2 .
If X=2a, 14=0. if X=a, pi=0, so piand 132 are just on the verge of propagation. H, (P io)Ji (plc) sinr; tic2cnKoor,
inonosin -n-;E)e jwt
A00= -6W/bp so that 1-.+z)ejwt _ A 0 --(cill p lit 2) 03 PV itc)sinill n12 PnCriK1 (PnP) Ii(3nc) sin z-, e jwt . p < c, These give A0=A i at P=c ci Hp.)03,0J,(13,0sin7-+ 2 f3nc„K ,03 ,G3no P>t ' W te = ( Ci
Cn is evaluated in problem -43-. Use first term: IBzil - 1Bzo l p f-c- (CA Do(1310 ..710:310- jlio(fif)4 1310 - J0(131C) J1010+141310 Yi(1310Sit114e jcpt which by 5.293 (9) is e jwt 2jf3iCisin7-/ (re) . Multiply by sini1;-, integrate, sinr4§Bds = pi° eiwt sinV -Vsinn--1:?) "-1 K (13np)Ii ((3nc)sinn = -11 C ejwt . So X =-j-rliI0 (c/a)(42)(P i P) J1(P ic)sin-ac n2 1 TTc I 0 rd TT 'rrcp.I0 pi=7,:14-(X/a)2 , Pn =-7.4/(nX/a) 2 -4 Ans. Here CI=
Ans.
-45 -( 2 1)
0, If a <>■ .<2a only lowest mode propagates giving Rr .IR=Vr.p.=-CxEc=jwaclAotfp
By -44- : R = ( jwac/I)-(-jrcp1/a)(...Ti(P ic)sin 1413 2 =(rwa c 2p/a) Ji(Pic) sin17,41 ) Ans. d rrd c. 2 nrrd nrrz 1703a C 211 Ji (P ic)Yi(Pic)sin2-a-X= fwacA0/ I + E K1(1311 I ( finc a n=217 , ,.4_ ' , nr A0 is infinite at z=d where series diverges. Evaluate 4-74 : . a From - 44 -, pnrr at d-z=r which is surface of wire of radius r. 1 nrrd nrz 1 nr(d.2,) _ 1 nr(d+z) 1 , a By 5.322(1), (2) K i(pnc)Ii(Pnc) -i-g n. r---/ - 2pn c -2nrc. sin--g-sin a -icos a Tcos a 2WNIC 42PrIC So after a few terms, by Dw 603.2, the series takes the form 2r a r E 1 nrr(d- z) .t::1-. on ,,,...r(d+z)_ 2s,Trd ._Trzl a sin(rd/a) sinrzi,_ a rk 2sin E 'cosT-2.1-1-2)-1 bin a I a -1 2TTc Fr14rrc Ln= gicos a a .1- 2r2c I 2a L' a`u 2a n=ln rrd rrn pm sin(rr(d+z)/(2a) i r Take this sum plus the. difference _a 3 sin— a sin-a-J. At z+r: d+zgv2d, d-z=r. .. . t he ser ies to obtain rwac2 nrrd Trd 2a rrd 2 c° a X= a PI-EJ1( pc)Y1( pc) + 7.acjain 2 7+fin;2[K i(f3nc)i1(13nc)- -a-- ]s in2 a-- +22c fr [Par s i n-a- +Pm— rrr ]3 Ans. 2nrc
Chapter XIII. (XV) 2r X=1.5a, c=a, cc=r/10, d=a/2: Pn=1.5
Page 166.
Page 554. -45-<21,p550 4rr 9 n2.1.a.5a) 2 1= 16 3a
9n
finc=;-PrT-T.6if 9T ntl.
11314=4/irr=2.77, 1/33c1= +A/F,r=8.44. n=1, J1 (B 1 c)=0.4195, Y1 (j3ic)=0.2531, JI Y1 =0.1062,[]=0.326. c. =0.1592. - 0.0531, D=0.0058. * n=3, I1 (13 3c)=606.2, K1(B 3 c)=0.00006188; I1K1 =0.0589, - .7alliTc= a/(4rc)=0.0796, 0=-0.326+0.0531+0.0796k-22-PA3•2=4.02. r=Le-4.02_0.0180a 1.571 Trr/ 217c 1 rr a 2 3.10 8 •4r 7 (0.4195)2=274.5 ohms. Ans. 10 "=171.5a a;• 10 . -46-(22,p550)
0.01145a Ans.
Simplest TE mode is TE ica -TE an . B is given by ,13.20 (7) where we write for 101: m=m=1; x,z for x,y; 1.,1c for I., j; a,d for a,b. For 011: m=n=1; y,z for MT-1 . riz 1 rx rCr , . Tfx .. rtv, Trz x,y; 1,1 for L.,1; a, d for a,b. B=i-tataiz.sirra-+2sinvcos-d--td[cosi--cos a j sin d 3 a rz d2 B2dydz ld co a-a-+ 7i 0.-cosTrsin2 On side x=0: B2=13 2y+B2,..orrocoafain27 z . oo r2 C2 a d ,1T2C 21- ra . 2 =d + d2 ral d rr2C 2(a2+ 3d 2) osin a y ---ij0 (1-cos a • On diagonal y=a-x and 4 ad 2d {E 2 2d a +2a 2a} 1 rrC 7sii.12 Ttz ) M( i nz ..(1 rl so B1-7(-d-)2{4sin2 -a—cos TI- 4cos 2 BE _{?3'}} 2+ 2 + Bi, sinr(r) = cos114:2(1r2 Ca(a 2 +2d 2). d +z41 2c2 {12._ar On both ends Soajoa (Bi+Bpdxdy (z=0) dS=65dxdy, SslydS=. tad r2 C2 a2+3d 2 _j_J(a2 +2d 2 ) a2 = (rrC/d)2,07(sin2 + sina -jir Y-)dxdy=(rCa/d) 2 . Tot al SBtadS+— 2a 2a ' d2 712C 2 (2a3+ (1-1711)a2d+(3+2)03 . From 13.20 ( 1) c=2Ttftv.,/a2 +d 2 /(ad). For volume integral 2ad 2 let u=17x/a, v=rry/a, w=rz/d, dxdydz=(aad/TP) d udvdw, so that d 27 3( d22+_; 2 fir rrrn sink; +s inavcoszw+cos2u+cos 2 17 20TraC2(a2 +d2) IB2 dV- a 2dC "sin2 w)dudvdw---2 7c d2 4d 2r Jo "J o do a2
7). la
cos a.
—a--
2iarr2C2(a 2 +d 2) •2ad3
pad(a 2+d 2 ) 13.18(8) Q-4dpIerr 2C2[2a3+(l+,12-)a 2d+.(3+21f)da) - p16(2a3+(14•12)a2d+(34.122)da)
From
Ans.
-47- (54,p556) TM 210 +TM 120 . Add the fields m=2, n=1 and m=1, n=2 of 13.20 (7) 3, -7c{i.. (2sinr cos n + si n 2x cosv -j(cosrsin am a + 2cos 2r sina a)3 . On side x=0, a a 5raadC2 0 +4, 2) fB2dS = rr2C 24:(sio-22a7Y + 2s innY)2 . On diagonal face, a dy = fr2C 2dq+ 2 q= Bx +By, dS= Idx•d, y=a-x, cos? = -cos7c, cos2— = cos m, sin7 =sin7, 4.-- a ---> gm ra . 2rrx r 2rx rx 8rtx sin a = -si.n--a-. j B2t dS =ATiraC 2dr2(2sin-5rrxcos-a-2rx + sin---5-cos7-)2dx= 2A5r2C ad j 4 sin -dx o =5r2 C 2ad/A5. Losses on two ends same as on one complete square. Bx and By give equal contributions so that, since cross terms are out, 2.172c 2rar.a 577 2a aca rx IB2t dS = 1 j (4sina— a cos2 ; +sin2 21T—X aC 0 S 2 7 )dxdy= 2 r2c202+ -41. a2)_ wo o 2 Total SB dS = rr 2c 2 (5 ad + 5adA5+1-aa )= 2ar2 C2((2 +A.ii)d+ a). IB2dv=d1B 2dS (on one end)= •1r 2 a2C2d t 2p• 5rr 2C2aa d • 2 pad From 13.18 (8): QAns. )264 • arr2C25{(2+,12)d+ a} - pl6{(2 +A/2)d + a}
Chapter XIII. (XV)
Page 554. (556)
Page 167.
-49-(56) Ax=0 at z=y and y=0 if Ax=C[sinp(z-y)sinq(z+y)- sinp(z+y)sinq(z-y)) (1)
E
2b
where p and q are positive. It is independent of x so e' in 13.00 N 4r2 b2A (16) is zero. By 13.00 (3) --53-04-+ -1— Ei-T x2=0=-2(p2+ c12)AX+TrAx ( 2) so this is a possible oscillation if p2 +q2 = 2(rA.)2 (3). To meet the 2b---0
2b
boundary conditions at,y=b, where Ax=O for all z-values: sill(P+9)z sin(p-q)b=sir(p-q)zsin(p+q This is true only if p-q=vand nandp+q=-7 1 so that p- (n+131 2b
and q- (m n)" .To meet conditions at z=y+2b: Ax=C(sin(2pb)sin(2qb+2qy)-sin(2qb)Sin(2pb+2py)=0. This is met by p and q values
already found. But p 0 q because Ax would then vanish entirely. Therefore m=n,n=0 are out. 3r 17(Z-y) TITO 31z+b) . -Y) . If n=1, m=2 or n t h enAx=Cvlin3 then Cr 2b sin b ). sin b sinr(z -=2b, 2b Mode shown at right is impossible because fB.ds around point on dotted line is not in general zero which requires a current normal to the paper on this line. Lowest possible mode with only Ax is therefore -50- (57) Use results of -48-. Four side walls yield 4.
(5rr 2bdC2 ), Two diagonal walls yield
2.-11(5,firr 2 bdC2), Eight end faces give 4 •i-(511 2b 2 C22)bs2 o2 total losses are25b rd2 bC2((2+.1f)d + 2b) .
SB 2 dv = 44(raC)2 d so by 13.18 (8): 0--lee) • 4 .115;217 1, 2c2t(2+,72) d+2b3-p'6((2+1:/12)d+2b)
Ans.
-51- (58) Vr,(cos (cut - Or )+ cos (cut+Or))
By 12.06(9)for standing waves: E=rb0- *tanla/
(rsin0) For shortcircuit at r=d, cos(2rd/X)=cos8d=0. fantan(a/2)) (rsin0) • (2p+ 1)Trr coact, So cos( (2p+l)r/2)=0, X=4d/(2p+1). Ans. Ee 2d -rr sine (2p+ C sinczt . C 2 C1. Over two cones 1134. 2rr sinci dr sin B0---2-2 d1) rr r sine I du dr 2 u du (2:3+1)rr . dr _ . r . Let u=7 t sin 77 = 41:rCIs1— Shn2 1 (2p+l)rr, --7=7,4rcscaC1(02p+1)r 2d sin 2a 0 = 2flcscaC 21((1- cosu)/u)du= 2rcscaC2 Cy +k[(2p+1)17]-Ci[(2p+I)r]) Hd.M.F. 5.2.2 y=0.5772. 2C 2 fl/2 2rd 2sin0 2 si ao de - 4rqpnit an (0/2% On sphere: j BcidS= 7-2,6 17/2 = -4rC 2em tan(a/2) 2Vacos Or coscut
2d du, so integral is - ., dr =1,32d v1,12Trr2sinedrd0=2MC sin2 (2P+ 2d1 ”Tr drIcre/22csc0d0. Let u= 9ri u , pr g 2Tad sin a Pri(co t(af) 8d c -.j sio2(2p+1)ud u4ncotc)-2I--- 2nd cpeicot3. Q— Ans. p'6(y+fin[(2p+1)TTJ -Ci [(2p+l)r]+2 sin al2fizc o t (a/2))
-52- (60) IdZI From 4.22 (3) : zi = a 1sinW, im 1 =a llcosi*a 1 4/cos (U+jV) cos (U-jV)=a0/cosh2V- sinaU by Dw 401.14. [U]=2r, M 2-1-m2= al eV2 ) Mi-Ern 1 = al eV1 by 4.22(4) so V2-V1-242 ,42---I : Fn ,urrmi. du 4K rril2 K M i =a i coshlfi, M2=a i coshV2 . § II dW ili Jo .11-(ai /M 1)2sinaU - 711 where the K1 , Wm2-m2 d (M,K,+MtK2) modulus is -70-= . From 13.22 (6) Q=----/ (1 + MI 2).1'6 1■ 12 2rMiM2Ca(M2+M2)-11241011-R41)3)
M -41 )
!(-mr>:
Chapter XIII . (XV)
Page 555. (557)
Page168.
-53- (61) [1.1]= 277, 4.13 (5) Uo- 1.y-cosh-I
a2 +b2 -c2
2ab
W oz , 4.13 (1) z=a,cotTs, -6,77
a
W
I dz dW
sidIrldV=2rrcoshU. From 4.13 (3) 1 a, • 1 dWI ' jaisinT(V+JU) 2sin T(VcoshU-cosV (c +bcoshU2)2 c = a, (cothUo -cothU,), al= a sinhU, =b sinhU o so c = a coshUi - b coshUo 4.138 a 2-b2_ c2 Fi g• =a2 cosh2U, =a2+ a2 sinh2 1.11=a 2+b2 sinh2 U2 . Thus c2+2bc coshU2 -1-13 2=a 2 , coshU2 = 2bc a2- b2+c2 1 c 4 _ 4a2c2 ,_._21_,(aa..b2 )2-2c2(a2-1432) +c4 = T- . a1= wicosh2 U, -1 -1-0102-b2)2+2c.2(a2 coshU,-
lac
•
B 13.22(6): 0 -= 2116/
1 lid /11+ d • 2r(coshUi+coshU,) d (a-b)[(a4b)2-c2] 4abRcosh-12a2+b2 -c2)/(2ab)]j 4.2rr • cosh-11(al+b2-c2)/(2ab)JJ 211'6 /
Ans.
-54-(62) Resonance in wave guide sections occur only with propagated waves. For resonance the capacitor sees the two Z' impedances in parallel. These are
z'
Z'
GTE
shortcircuited at the ends so ZL in 13.06 (2) and (3) is zero and Z' = j sin (B;o d)sec (13;0d)=j tan (Nod). From 13.00 (12) and 13.02 (2): (310 =[7-111=‘,/132-(rr/a)2 =11s./(2/X) 2 - (1/a)2 = 2r/Xg . For resonance ijtanI330d= j/(cue) or cue tent:31'0=2. Insert C° from 13.06 (7) to get 1 = (4b/Xg)tan(217d/Xdkcsc(irc/b) Ans. 4(41%02 or If d
16+X 4--
'y
c:
au /ax=
94.773
r 1/1
13.2988
13.3). x.=3.692 g
Ans.
-56- (64) The static field in a guide with iris is shown :
between x=0 and x=b in Fig.4.22(b). For potential
1
Vo across the gap, the gap fielq is, from 4.22 (6), COS(7/110)) So in the notation of this o3-c [ . _I 'I''i' TrXib1 _012-Vo (2Vo/r)6sill -sin rra/bj - b A/cos (rx/b)-cos(rra/b) problem at z=d we have T
Eycs c:i atx=ii(Vo /b)coiTIA/cos(rry/b)-cos(rtc/b), 0
As in 13.00
and -3-, g standing wave solution with E = 0 at x = 0 is W=CinsirtLra cosnTjYcospiri z cosczt . 13 1n =15-ri n =/12 -(n 7T/b)2- (r/a)2 so a solution. with a single standing wave, (310 real, is co W = s in (rrx/a) ( C io cosSilo z +nEiCincos Binycosh pgriz cosan ) . 47- zx 1.-W-= Sti1(7X/8) LI [- 13110 Cio sin 1134z1
+ E Iliqn1Cin cos piny sinhINn a - ).5 nii Pinc Insinprly coshliii biz). 1-1
Ey=cosin7(-13:0 C10 s in 10;0 zI
+nE ilp,IniCin cos piny sinh Ifqb zr) (2). Equate to assumed value (1) at z=d, multiply by dh sbc nry rd"rff—rre rc cosSin ydy. Integrate 0 to b. Left side: b cosRcos -r-vcosi3- -cos F dy, let *=1, 0=-b1 , e 1 Nfiv b Pe i so left side is 73 a r jScosiAlicosni4cos*- coseplisig --'6— -17-11 r cos (n Ili+ cos(n-1)y ... 0 Use Hd.M.F.
rc
Tic
-_ io (Pn(cos -5-)+Pn_ 1 (cos-F) ) . n=0,
2r Jo L2170 reco s-el ild* 1 b J
o„/cos* toss -
(continued)
cos -cos rc o Vo Po (cosT)
-
22.10.11,22.5.
Page 555.
Chapter XIII.
Page 169.
-56- (continued) 1 17 c s c god . Right side, at x=Ta . n=0 -coND Cio sin plod Jo dy= -wpi'oCiob so -a3p'oCio = 2- Nh4P0(co s f-) cos27 3dy= a311313n1Cn sinhil;nd•-. Thus col(3;n1Cn = le csch Ble d. Then n> 0, a3113;n1C nsinh B'indfob
E
rrc Vnsin (T1x/a) [sin 113inzi + 2cos PlnY sinhii3 i!rizIr cosmic)) )3costat] 1.Pn(casT +Pn-i(aas-Esin ifi;odf n=1 sinhirqnd I 2b rc,1 r + Pn _ i (cos-b-L, Vo sin(rrx/a) 2Bin cosh IfirinzifPn(cosT") si cosa3t. L ,-. d . h!pit Ez = b n=1 131n nPinYsin E Y
Note that Ey=0 on iris is rigorously satisfied in both static and wave guide solutions but Ey in window is only approximate. -57In 4.22 (6) dW/dz at y=0 is valid for 0 <x
As in problem 54:
TT rr X = 2aXg iAi raTr14.2 t g (2) Ans. This approximation is best for large d. By 13.16 (3) Bit0=-;=. a g -59Try X = 3.7840 in (1) above. fhlo d=4rAg=190.275°. 190.275-180=10.275. Cot 10.275°= 5.5163. t an 2(4rrcia )=t an2 11/ 8=tan 2 22.5°= (0.4142)2 =0.17152. 2acot 13;od tan2(22-rrc /a)=415.5163 ' 0.17152 =3.7846 X8 so (1) is satisfied. From (2) 7). =47s.g /A/767— F7N.g = 15.1384/4/T1.35- = 2.7491
Ans.
Page 556.
Chapter XIII.
Page 170.
-60From 13.06 (8): PI= 2r/kg , din ----A/cu 2us-(m1T/a)2 I Pin= mr/a, 8 = 2r/X. Write Cm for -+mcilCmcosh 811:az sech140-s1.4- --- lc °scut Amfem in 13.06 (9) , (10) : Bx=picosplic sec rid sinliax
NCrn sinifi'mzsechgldcosm--7-x3costot . a.„'Cisin * OlcseciTidcosi7x --a--mE3Fm where m= 2n+1 is odd. Bz =f-PPi i Bx is even and Bz odd about 7a. Note that Bz is zero at z=0. By 13.06 (14) the static fiel sings df sin0 -sin(2rx/a) of Fig 13.06d is Bx + jB = z cos rh. c'-77-g=—'--2" .snl A.Tiin2(/2)-sina(0/2) •✓cos0 - cosh where 2Trx/a = rf +0, rrc /a = 0. if x=a, 0=17. If x=(a+c)/2, 0=0. dx= 4ad0/Tr. Assume static value of Bx on iris where i(a+c) < x 0 but if n=0: C1 = (1+cos7) = 4cos2 (i-rc/a) (2)
Let Bx=B X /C I, B Z=B Z/C1 then .8, sin81 z cosrx+ ci pmCmsinh f:qz cosmrrx cospiz_si rx+ E Cmcosh Orlizsinmrx Bx= a cos= Ans, a coscut Bz-- filicosfq z cosfq d a m=3Cicosh 8;c1 a m=3 f3; Cicosh 86d -61Write Cm for Ampm in 13.06 (9), (10) where m= 2n+1. Thus
E
Bx= ((pi /131)C1cos 13'iz c sc 81d sinria-x + (13/h/l3m)Cmcoshp,hz csch 13'ind sing!) coscut ... m3 mrx, rrx Bz = (Coin f) z csc8;d cos - mE 3cmsinh 13/In z csch8m'd cos7-3cos cut . Assume Bz=0 on the iris and has the static value in the window, where, in the notation of the last problem, Bz =Nrisingsacos0-cos0 if a/2< x <(c+a)/2, Bz = 0 if (c+a) /2 < x< a. Multiply through by n mrx 1 x = (-1) sin(n+ 7)0 and integrate from 0 to eo using Hd.M.F. 22.10.11 and 22.5.36, ma 3 1 1 1111 r0,1 sin0 sin (n+7)0 d0 _ (-1)na [re cos (n+-P0 d recos(n- '7)0 ](-1)na (cos 0)- Pn _ i(cos0)) -2r Jo .,/cose-cos0 Anrr Jo A,/costf-coss ° •JcPicosU-cosoid° - 2 (Pn+1 ner TTx. Right side: Cml „cos2a2n+u—a--jdx =cmit sin2(2n+l)u du = Cm a/4. So as in -60-: a/2 17/2 n+1 2n + 1 rrc rc (-1) sin17c a -Pln+I (cos-a-), C1 = 2(1+cos-a-)= 4 cos 22rc/a, as before. Thus C211+1=2 n(n+1) mrx sin8; z cosrx r Crnsinh13;nz r cosI3,z sinrx z sinmrrx coscut B Bx tos--5-I cos= cisinh Nad sinhf3m ' d a ' z s in pi d 3 litmC2 sink a m =3 d a Pi sin fri +m
cos
Page 557.
Chapter XIII. (XV)
Page 171.
-6 2Like 13.28 solution but with second form instead of first in 13.28 (3). By 13.28(7),(8): 2 p2 z2 (a2+b2)(82-P2)- z2 a2 k E .. 6Ez or E =-°-- k zpa E kfl---1 k / let S - a2 (a 2 +b2)' (7XE);6=D----P ° a2+b2- a2(a2+1W z- .- a2 a2-1-b2 J = a2(a2 +b2) oz
5
(7 XE )0=S (pa2 + 2p (a2 +b2 )}= S (3a2 +2b2 )P. Let zo=(b/a)Ara2 -71/. i(Vx E)4 dv = 2rrjoaf oz°d zP 2•0 d p.(3a2+291) = 21-r (bia X3e 2+b2) Sr3 3(a2- p 2 )d p = 4rba4 (3i+b2) 2/15. f(Eg +El) dv = 2Trif aSzcta4z2 p 2 + (a2+b2)2 (a2 - p 2) 0 .5 hp 5 1 _2024.b2)(a 2_ p2).92 z2.L.,3 a 4z41Jpdpd z=-21750Crablp 302_0 z)5/2 41202.fb2)14692.. p04/2 .. 2bi(a ai-b2)P (a2- P 251-125*-0,3"d -573[1.5(a2 +b2 )2-10 ( .92 +b2 ) b2 +3b43 p (a2 -P2)7} d P . .-2rrjaC--ab3P3(P2pq+11 [ --- 15 a°+20a2 1:p2 +8b 4 0 3 o _24(5a2+4b2)(3a2+ 2132) fE 2dv = 2r(-- 3171,-(a2- P2)5/ 2- 4.0 (a 2 ..1) 2)3/ 2 3 + Tfa (15a4+20a2b 2+8b4)T( a2.. p2)V2 13 a ]
By 13.28 (3) (32= S (Z E)0dV/rE2 dV= -0
a,,___,.. 2.788 ija"13-
2ba4 :4)1. 382+2112 a ° 5a2+4b2. - 63- (65 ,p557)
Ans.
<------- 2b ---).
In 13.29 (11) the crosshatched area has been included in the coaxial line of length c so the capacitance (radial) of this
4E-- 2a
1 c
region must be subtracted from the fringing capacitance given
in Problem 60 of Chapter IV. Thus LC per unit circumference is h 2 +k2 k _1 h h _3 k h , k=b-a=0.005, a=0.01, b=0.015, d=h=0.001, c=0.02. 1 LC1 = 2EN02/71- +. tan it +i t tan h IthIc -171 )--gl . 2E 0.26 AC1 = rT + 5 t an-1 + Igtan-15}-4 = Et-PO.262+0.987+0.275)-0.23= (3.048-0.2) Etrr = 2.420e/rt. AC= 2rrAC1 a =0.0484e . Co= e{(rta 2/h)+2'rfaCi }= (0.31416+0.0484)E = 0.3626E. By 13.29(11) fitan0.02S = 2rrE/{CO2A}=6.2832/ (0.3626Ek1.5)=42.8. 13 = 40.6 . Circumference is 2rra. Area is rra 2 . rT a_2 4.okh 4 2+ hk k2+4h ak ..1 +4k ahton_1 k 27(ah ) Total capacitence is C = s{ t an Ans. - 64- (66,p557) From 13.00 (16), 13.20 (6) Az =1Ez=1-1-00 sinr4-sin7. To get Co in terms of C1 : square Az, ab 4d integrate over v, equate to abdCi -(1CPSoarsin 2 7-sin2 dxdy =- 0321 = abdCf or C° = --24 ) 1. CO
Thus 4=-2j sin1:4)-(isinTrbY-,
qi
(1311
2jd sinlIsin-Z31 as x and y approach a and b. Substitute
E-1 = iav2, (..p 2=4 and Q from 13.20 (10) into '13.30 (7) 4v 2cuud 2 s in2(rra , /a) sin2(13,/b) vpyorrd (82 +b 2) 3/ 2 Z= R + 2(ab(a2+b2)+2d(a2 +b2)} abdw 2 2y6 1.1 2 v2d2 (82+" 2171 sin 2 T-1 R'= 2=R+ ab(a2 +b2)+ 2d (a3+b3) sin
From 13;20 (1) at resonance cuab = Tri.P42 4-b 'I so that Ans.
-65- (67,p557) d<< a, d<
c a -• +
c'a +6
neglect J1(f3c). Then Az from image a distance p from wire is Az= 4-11'0 J0030+Yo(3P)}.For wire Az=1(uI/Tr)Pinc • jZ
cod /I)rAz= R+jX. .
Include all images but wire itself. Thus exclude m=n=0. Then X is Eat EL' 12.„ ° 64)Y(2134/n 2 a24m2 b2 )-Yo(24/132(na-a1)24(13mb) 2-) (1-om 1.p. 217 '" 4
z
-Yo( 2,103n02+132(rab-b1)2)
+Yo (2f3•1(n a - a 1)2+(mb-b1)2)} Summed over m. Ans.
b _g
Page 558.(558)
Chapter XIII . (XV)
Page 172.
-66- (68) From 13.20 (1) V =212.!
+0 04 - 7.5.2.234 . 108 =1.676 . 10 9 . c13.352.17•10 9 , 13=11.17r, 0101 10"4=1.611•10", 0.05 3 /2= 0.01118. Thus by 13.20 (10):
6=0.0660/
Q = { 3.1.257.5.8* 1.611•rr• 2.1.118•10"8+7 "8 " 2" 2}/{2(2.5+4•0.9)•10"2-2}= 9100 and by -64-, bi a isinq1 2+16 -4- 2 )/(2'5+4•0.9} sin2 rl Rr=(2.5.8.1.611. 1.2572 .9-05.10 2+2+7 ' =390,400sin2
sin2-7bT1-3' . Also R={ 4 • 9.1.257 • 20/ (3.352r• 2) )9100:-391,000 which checks.
M=2.828.1.257•2•10" 8 "2" 2+2 /(1.117 2r 2 .2)=2.890•10"sin7l sinr43-1. Cavity resistance is R11 =4•10" 2- 6+12+2-3/(3,352r• 8854 • 2• 9.08=2.366 • 10-3ohms . Re(3.352r.2.89)2 103/2.366=391,000 check 0.0016_ sinrai _ for one wall. 71 0.016, al Tr -0.000512o s in2 "al= 100 0•016Y €=,.17075=10volts 1 a a 390000' sin(rb1 /b)=0.016, b1 =0.001024m. If al /a =b1/b, then sinT41=4./17.1.6=0.4, -7-1 =0.4118, a1=0.0131m, b1=0.0262m. Potential multiplication is csc7-1= 62.5.
V=10•62.5=625volts.
-67 (69) By 13.20 (2), 13.30 (5) B1nIpnp=1VOy-bz .Integrate over loop area-see below. rrnJ nrtb rr iLd +1r2-x 2 bA 2 a 2) ',nig . rerrb 1 V. pr(d-14j 1 sin M = Siginpdxd z=—sin b2 o c4r2 - x2 bzM d z dx - 13mnb Sin-. b o vi ch rr L,4___ 2-x2 7 etc ix prr(d, -4/r 2-X 2) mrrx r- 11n sin (nrbi/b) jtrzoos .., - j sin cos a dx= -,,/8 sin cos d o d o Omni) /-2rnr . nrrbi p cirtd1 rlsinpTIrcose cosmrr sine where x=rsin0 = -8i3nr nbs i nb cos cos ed 0 a d Jo the--rcose dO
={rr 3•Jr3pnr/(13mnpmpbd)} sin"' cosc-' Ji(Stapr). Ans. Integrations follow. Met/ Method god I. Let M= mrfria, P = prrr/d y4.4.1 2+P 2 =13mpr, tenet= M/P, 01=0-a, 03=04a. e , r rrr 1 rii = ij cos (msine) sin (Pcose) cosed8= -j (sin (Pcos0-1-Msin0)+sin(Pcose-MsinCco43d0 Jo 1 rrr+ a rr ): d1 = Tir , -Cel sing3mprcosplacos(01+a) doi+-4-.6 singlinprcoszs 2 ) cos (02-a) d02 1 ' TT - a rr+a rra cosa cosar sin _ __4 ___ sinajr_ds in oc impr cos 0 )sinsly300.....27_ sin p cows 2)co4620 I3mpr cos0)cosOi d01 =-4---j_ a ( +s ins TT-KX s in , ■ 13mpr cos0 2)sings2 d02 . Second and fourth integrals cancel, other s +, periodic in ¶. 21 -a Tr-a co s a [r • (pmprcos01)cos Oid01 +Sr_asin (Omprcos03 ) coo 3dq Let 03= 2TT-02 so ..1 - .--7 4 -Lasin
. cz
r
07 ___a_ T2 Tito sa b rl ii" 1 =-1.2•cosovj osin (13mprcoso)coso dth i b( r ) LlTi o cos (13mpr cost) dt] = - ircosa J1 (Pmpr) -2a crT 1 (Pmpr) • vmp 2m+1 rt r ¶T/2 m+1 P , . 2m+2 I, co 0(10. Then Method II. Let Rs =- -19 and expand: j =40(-1) cos (Mcosgosin (2m+1) : '-' 0 2m+1 ap (4) m+1p2m+1 r co (1)111P r(2m+1):: by 5.302 ( ) jarfl (m) - 2m-40 2"`m:M m+1 jrn+1(m) m;) (2m+1): Ze+1 s5VIM 2+P a lap r-—rfP • M X c'' 1-X2 mJr/1+1 (M) 1-X2 P 2 Mor X = Let 7P E 1.— PlmJrn+i1(4) M M 2M• prm=0 2 m. 2 m. 2M 2Mm=0 L 2M j
f=
-TTP
TI T? 2
pdJ1(13mpr) then by Hd .M.F 9.1.74 =,4-Ji (13mpr) - frinff
Page 173.
Page 558.
Chapter XIII.(XV)
-68- (70,p558)
ri3 • Bdv rA.Adv To get AL from given A use 13.30 (1) . From 13.18 (8): SA•Adv=jt-21-6--J 02 . Then
Z(N a s in2 Mxcos2 Ny-IM2 cos2 Mxsin2 Ny)cos2 Pinzni:dxdydz
from 13.20 (5): SA•Adv = 1302inpCl2j0ai403
Let x= au/TT, y= bvirr, z= dw/Tr so dxdydz = abdirr3dudvdw. M=mTT/a, N=nrrib, P=prr/c. 2 sin • 2 3 JITSTT mucos2 nv +M2 cos2 musin2 nv)cos2 pw dudv dw 0Sr 0(N SA•Ad v= f3m2npCl2(v/rr ) in/8 .SA0 ., o = f4inpcta v(N2 +1,1 2)/8 = 13n2Inpe 2v g A dvCtiorv. So to get AL divide 13.20 by
prunp f3 (L)/8. Thus if (1---03ranp) AL= I fj/ (13mnp l3mn CiMCOSMXSiTINy + INSir1MXCOSNy)PSit1PZ h.6, r-nrarr 2 sin(nTrb,/b ) rr [rd+N/r2- X2---° -kqmsinMxsinNycosPz] . M°=SSB° dxdz = NI 8 51 cIz dx -A172 .7-x2 do Jd PmnpPmnad mnpy The integral is, from-67-:
mpr nrrb lord M °0 = . NrilTr4 ,D 0 A 2 sin-g--I cos a (i 1Ji(f3mpr) PmnpPmnPmpau -69-( 71, p558)
Ans.
From 13.18 (8): cz 2ptj'A'Adv =1B•Bdv, In the TM01 mode B= Beo2pEC'sineji(1301 r) fl and at resonance p = poi so that SA•Adv=f32C 2S0sin 30d01021101r 2 ji(pr)dr. By 5.31 4rr rr 2 (10) and 5.296 (7) = 02c' 23-Soa rr r2J061/2(pr)dr = p2c'2-- •T3-3 Jor13.3 uJ3/2(u)du ! b a-z/212.038)] Loot = 3p 17C 2 r4 24-21J32/2(8a)+Jd2(8a)j- -}f3 aJ3,2(Pa)J1/203 s)). For TM01. By 13.24(5) l ba Area 2f3a =0=J1/2 (13a)-1(13a)-1J3/2 (13a). Then by 5.31 (2) Ji/2 (pa)=/4 7 sinBa, J3/2(f3a)= 3 Ji/ 2(fla)=-23,4T3aTrr sinfig T (4p2 a2 -9)sin2f3 a C'2= 1.tra 3C1.210r . Thus C' 2 = 18a2„2, Thus SA •Adv = 23;2 V C...Or /[(4132a2 -9) sine pa) CI
6aCnor 4/2(413 2 a2 -9)sinPa
B°
68 2 a sine j (Pr) ../2(4p2a2-9)sinPa
M° = BS ° -
__.22 3A/713-2L a17.-9
since B° links a small area S at 0 As 17/2, rr..-a where j1 (tir)--P.4sinpa. From 13.30 at resonance where 13 = Poi , Z= R+ (M°)2 Q/ (gym) so writing 4a3y6 for (p18) -1 in 13.24 by 13.18 (8 (mo)20 4,32 p2 y6 0132 a 2+9)sa 2 n 2+9) so by 13.30 (7) Z-R-I-'wEv.‘ - R + Y1--la6 O Ans. , roi02 (48 -tm-- ul80 12Tra2(4p2a2-9) R, _ f32 a2(4132 a +9) p ..811 f32 a2 =7.530, 2 =141,900, RI = 5.252.1044 By 2. 13.24 (3). and 5.31 C 12(4p2 a22- 9) E ya4 • a ,,a _PP01a sin 3 u cot ÷ cousu 4 sinu du) po, a Dw 431.] V =2J cEdr =441.4hia u-1 Ji (u)du=4(1.)Cj u ldu = t AB o o u 0 u 0 u 441.1. =2wC'(- j1(1300) +Si(Po1s)3• Across wire e=--dN/dt=c0130 S so V/C = ( 2 /( 311 S))qSi(Pois) iii(Poia)J -13 • 01..38824 758 = At resonance 6 ( aJ3/2 (001 4 iba is 0, 1301a= 2.744, Si (Pois)/ii(Poia) 4.710 and V/C = 7.42/ (13021S)Psa 2/S. Ans. - 70p2 E2=p1E1 so
".
= 22 .Displacement is continuous so E1 E 1=g 2 E2 . Tangential
P1
H is also always continuous so H1 =H2 . In TE 101 there is only an Ey and a tangential H so the instantaneous magnetic energy ratio Min/Warn) is 4/ (p2 Hi) = /122 . The instantaneous electric energy ratio Wie /W2e is E 3 EME2 El) = Ei /E2 =E2/CI =Pi /112 so total ratio is W1/W2 = pi /p2 Ans. (W1 =Wie
= (111 /p2)(W2e 4W 2m) = (Pi /P2)1423
Chapter XIII.
Page 558.
Page 174.
-71-
4—s -.
Use 13.20 (6) and (7), m=n=1, p=0, a=b.Superimpose TEoli , TE011 and TE0 o modes to give Ec directed along one of the three cube diagonals. On the z=0 plane: E no= -k Eo sinr4sin7, 13= 1: ax -±1)) . The pressure is 1 °-(i sin-7-(cos7±1 )-±pin7(cosia tua rara B2dxdy= 2r2E2n a pa 172 n 3a2 f dx dy since cross terms are out. P = pw 2a 2 4 tido 2p 2w.D 2a 2 ro o 6 a Anaoa .ffx = 322E2 2pea2 sinsinviz-dxdy E2c = E2x+ E2y+ E2z . Tension is sara-g-2-dxdy.4p . 4TT2 8 o o 2 00 2 l 3-1e82 e a2 _ eE a 8 . Net pressure is Ans. rms 24 12 4 8 °a
2
2
-72 Lowest mode is Ey = Eo sin--sin-/7-1, Bx =
fa) PzY = Trjc0E:sin-gTrxc o s7
rrx rz -1 OE rrE r BadS B = — Y= --acos—sin-a-. By page 371: -i = j 2 -Z jw b x 2p-y8 jcza 13 . 1 r4 Arar,2_,si 2 ,, 2,z E 8 ,,..„ 2,x 4sasarr2 d a cos -a- dxdz 2p21,6 t. 4J0cuz a 2 n a z+ o ow aa 2-'-n
a a4
a/2 /4
3,r2 E 4p2 y6w rrz Wire EMF from lowest mode is C = SEyds/.1=rEydz= Edna 8 sin— a dz = a (E0 /11)(1-cosi). P =12-CI0 = 4aEo I o 1T -1(1-cos8) (2). From (1) and (2) E0 = 3w2P 2ay6I0 17-3(1- cosii). Then from 13.20 (1): w 2 = 2r 2 /(p.ea 2 ). Substitute in (2) to get 1 =-Ipy6(err2 )-1(1-cosir-)I20 Ans. -73From 13.20 for TE io mode in cavity a =1, b= 0.5, d= 2. From 13.18 (8) Q is = 172C(-1 (ad)-Isin7cos7+.1s a-kosr-Igxsinril 2lisiB2 dv 1 I. :is . By 13.13 (8) =.-=1- cuoy. By 13.20, co = Trv117-11-IrrvIC Q = p, 8j17.27 4 B te
0
Q=1TvpdyjSBadv/(21B2 dS). Actual cavity has threedvolumes, six y=0 walls, four x=0 walls and four z=0 walls. For y = 0: loaio ((ad)-2sin2 /La axcos2 T+a-4cos2--sin2rixd a pa 5 = IT4 C 2 t(ad)-2 +a-43 1-ad =114 C 27*-1= ir4 C 2 . For x = 0:jo 4dx•b = Tr4 C 214 (ad)-asinagdx 4174C 2( ad )-2 a = /T 4 o/16. For z= 0: brod Bidz= rr4 C 2-kfod a- in2 7dz =4174 .id C 2= 2r4 C 2 . dv=bdxdy
z
16 rtypthyj 15 Trvp6y5 3 /2 Q2 8'60+4+32 32
Ans.
Chapter XIV.(XVI)
Page 175.
Page 587.(581)
-1 d 2z dz dp_pIqdp mdz_pIg 2fizo A 0 when p =a, mdta =Bqdt From 14.05 - 2rp dt' -dt - 2r a' (8) -c 1/2-=
=b
entire velocity is z-directed.and entirely due to fall through potential V. Is 2 Thus .221-[*efii-t Ans. a = qV. V 1824cr12412rd -2]
' 4---b-4 IL* :OB 1
-
The motion is similar in all planes through the x-axis. Choose the xz plane. Transform to axes moving in the x-direction with velocity v. Use 14.16 (4) and (5). Fields of moving observer are: Ey =x(Ey-vBz)„ q=lc(Ez+vBy), BY =(0 y +4z) / 31 =1 (3z -4Y) where 441-0 /02=1,0=1/"If Eo>cB0) then in the xy plane Ey >vBzand we can choose 13/c =Bz/Eyso 131 vanishes. Thus there is always an accelerating field for the charge in the O'system so it continues to leave the x-axis and never returns. -3Similar to last problem except that cB0>Eco. Consider xy-plane then from 14.16 (4) and (5 Ey ' =x(Ey-vBz), Elz =x(Ezi-vBy), 14= 24(By+ -Ez), Bzr =X(Bz
u
.E)7). Ey -
Jai
Ygn(bia) 1 jjz= 2ITY • To eliminate E' NrTBz =1 ...r.:7 321.3.1/(21715)• Y take v=Ey/Bz=21TV/[ pIrm(b/a)] so B1 -K(1-132)Bz=1cravY.1„.._ q,v7zr_P_L ] dx i . These are the equations of motion for S% d(lcm‘qc) Jr.:niEL 1.).'L 32___ dt -(1' " 2Try dt ' dt " 2ny dt Integrate, putting 1/x= N17-1307=4/c2-vp-v 12 2/c, where vi and NTI2are the initial v; and v;„ velocities for 5. Then 2Trmo(v-v) =c/1-n24(T.:17.712IPni. If vyis zero when y =b, then
vX = nc is entire velocity at p =b, so that 2rm o(n-v1)=AIT/1-
which gives
2 a , where, from 14.02 (3) and 14.02 (4): 217m0(.j1 2+v12 2 -v1)= laq4c 2-v1 2-v 12 2.117132/n-12 1.7= (v 1-cP)/(1-1.71 13/c) and v'2 =v2,17137/(1-v1 13/c). Here vl and v3are the initial velocitie in the x and p directions and 13= cEy/Bz=cEp/F0 • -4-
d(mr2 e) -Bobri vc1;17 (1) From 14.04 m=m0/1-(v/c)2(2) Integrate, dt assuming that
o=o
at r=ro . r26=-Bqb(.4F117-49i71,7)/(mox) (3)
From energy conservation, 14.07 (2), (x1 -1)moc2= qV
r /
(4)
Velocity is constant so at outer cylinder, where r=riand v is all in A direction. v= r1 0= Dil-(1/k)2from (2). Substitute in (3): ric,A17-7 1=(Bqb/m0)(17i-b2-4,/ /17 0) so Xi = 1 "IL 1-b2 4 1 /T1-b2Put )2 (5) (5) into (4) to obtain: mor ic] (N/717
/m ac2 B2 b2 V = M 0 C 2 (X1 - 1 ) = C 2 r2 q
m Dc2 b 212 --
Ans.
Chapter XIV.(XVI)
Page 176.
Page 588.(581) -5-
I
2a Beam moves in x-direction with velocity v. Vq =(x-1)ma c2 by (1 T Va 1+1 1,7 / (2m 0 c2) 14.07 (2) so x =1 +--'-2-, p2 = (x2 -1)/x2 =2Vq . Let S' move with the beam. 'floc K 'floc
) 3 -
He then sees only an electric field cr'/(2rrevp) where cr' is the linear density in his systei His observed acceleration is 1:0=
2 (t-- . The density seen by stationary S is xcr l = a 2r evmaa 1 q I = I/v = I/(c0) so that, from 14.02 (6), since ux' =0, ior x2.2fremaa • xc0 • __AI_ [1+ vci-2 ina.slif;+_is vi_i +-1711 2}..1 Ans. moc J 2Vq 2Tryn a ac 2m 0c2 L mc2.1 ano c 2j - 4TfEva m a V
J
-6-
dr dds e (2) r if S . Central force so d (mr 2 e)ldt = O. mr 2e = po = p. (1) --7 :rib= pr= r2de — 2 Wk 1 (3) where Wk is kinetic energy. Thus if Wt -7 3i Then from 14.07 (2): x 2 = --I +1 -1 [m ac 1 is total energy: W k= Wt - qQ/(4Trevr)=Wt - cps/ (4rrev) (4). But 132 -c20e2+i. 2) m 21.9 2/ r 2 ± m 2i. 2 3 /r )2 +O. 1 S 2 p 2 i-pr2 (13/0 2 + Pr* 02 (6) - ( m 2 cZ / 1.,p2 1 (5) m2c2 m8c2 1-02 - mic2 .
a
Note that m=mj/(1-02 ). Substitute in left side of (6) from (3) and (4), in right side
qQ/ (4rre) 1 2
P del l • Only variables are s,0. - [s2 + — = 1 + 11 de nio l---moc 2 12 rds1 2_ ir qQ 1 -1 ) 52 - 211e mclQp[1 11t-1 12-115-2-[ 1 +---2] . Differentiate with respect to e a +IsieJ _orrEvpc i ma c rly 21 S 4- p 2
from (2). Then: [1+ Wt -
and dixiue out
ds d2s
dO" d0 2 =
inacIQ [1 +-iLt 1 ] -v2 (s-A). Thus s=A+Bcos(s0+8)
1'Tg E2 vpc) 1 s 1- 4revp 2
ma c 2 - '
If s is maximum at 0=0, 8=0, and s =A + BcosyO Ans.
where y 2= 1- [qQ/(4rrevpc)]2 Ans.
-7Each particle travels on spiral on surface of cylinder. In worst i
i
-M2p ir--
case, to get through second hole, particle must make a complete number of loops. Time needed is t = 2nrrp/(vsina)=2nrrn/(Bq) sincot 4:/a my sina/P=Bq. a = t •vcosa= 2ffmnvcosa/(Bq). But cosakrl, sinaAla, Ak.
Ans.
Independent of a if a#0: Maximum beam 1E--- a 4mva aq2aa ; diameter is Ib+4p DAlb +4mva/ (Bq) = -q--2rminv+b- nff +B so B 2.12rmnv/(aq)
.0"
= b +2a sina/(nrr) if a3 terms are neglected. -8-
A
a
Minimum B gives one complete loop between A and B. Write a+ 2d= (aa+b)/a for a in last problem. B 211mva/[q(aa+b)] Ans. Particles going on a helix of radius p will get through if the arc chord for arc length pe is
2 less than b/2. 0 _ 217b e arc 2 0 2Tra + 2d' - a a +b' F- a2 chord 2 4sin2÷0 Ans.
4
Chapter XIV. (XVI)
Page 588. (582)
Page 177-.
-9Let observer move along x-axis. From 14.13 (4) and (5) Eqc =Bx = 0, BYt =x(By+-PEz), Bz=x (Bz-fEy), E)Ic=Ex=0, 4=x(Ey-vBz), El=x(Ez+vBy). But E = A6 [---q--6t1 -11 B =-F6 [Ellej-l so Ey=- 11Z. E =4-- [---2-k-r-1-1 Y -..,y 2rre r 2 ' z -.37 2if r 2 2 I pe Z Z 21TE r 2.,' 02 / _I -6 [ E4n I ll so E -----B Let v=-pei so Ex I =Ey 1 =0 and IC = 1BYThz 2Tr- r2 ' ° z -peI Y' [ utI21 ]
Since v
Ans.
-10, First find velocity parallel to wires for S . d(xmovx) dt' =c1(BIST-B"JM d _[MA, _ bAL 1._ A:.. qd i r. Integrate: xm o (v;c - y;c0 ) =.-q(4 -tqc0 ) =c10:71 by " TZ -"J _A )=._ RA I A-Ilal =x[1---21-- B Bye xx x o 2rricw"r 2 r 10' since from last problem B'' y,z (PeI)2
=B Bye /x so
4 2, .
But a particle with zero vx for S has initial velocity -vx for S'. Thus from last problem v i = _2_292 ._-1___IsLon r_rjo + I - Px+-ri/ [1+vxc y r11.2 0 -Pict 14.02 02 (4) so 2.... b 14 2_ 072_ xo pE i 1 . vx 2rrx2 m0 r r10
as!
2' 10
.142 1riL EY1 . =- Vx [1 1 2T7P 2c1Cio •'2/4 Write ■
0
vx. [ ..ipn
for for 1-gand divide through by -ig2 czroo- to get
o] ra r io /
[Qa l rmo ] r 2 ra p Ans. 10 +2 -11-
From 14.05 (8) d(xmo vz)/dt = q(ByX-Bx}r)= q[ (bAz iox)dx (aAz ioy)dyjidt = -
-
-
qdAz /dt.
Integrate: Imo (vz2 -vz1)= q(A 2 -A1 ) so vz2 = vz1 - mogc N/c 2 -v2 (A 2-A1) Ans. -12For two orthogonal fields By /Bz =-Ez /Ey or Ey /Bz = -E z /By (1) Then by 14.16 (4) and (5) 4 =K(Ey vBz ), El =x(Ez+vBy) -
By= x (By+ -E z ), )3 = x(Bz - fEy)
(2)
(3)
If (El< c 1131 , set v= (Ey/Bz)-(Ez /By). Then from (1) and (2) both Ey=0 and q=0. Since E3ic IBT set fi/c = -By /Ez = Bz /Ey . Then from (1) and (3) both By= 0 and BZ = 0. -13s = acoswt, v = —= dt -wasincot. Substitute in 14.11 (7). Neglect a compared dt r, [B] with r so that [r] 0 etc. A z =- ? ii vsin[w(t 4)] (1) where M= qa. Substitute in 14.11 (7), keeping first term in a since is largest f(t) term. a -1 Er] [ r] = r - a co sOc os (wt -E) . 4.= (4TrEv r[l - ycosecoscu(t - )][1+Vsinw(t - -c-)cose)
-4
_o 4
0(1COSW(t- r
cos
Esino(t -E))
(2)
Variable terms in (1) and (2) are identical with those in 12.01 (6), (7), (8).
Page 589.(582)
Chapter XIV.(XVI)
Page 178.
-14ds = cue sin(licut), substitution in 14.11 (7) gives s = 2a cos(2cut) 2 d —t u2 (sin 2 II(t-I) ) COSe ) 1 sin1(t-r )cosej-1 -(1-4sinl(t-r )cose)-1 A z =- le sin-(t-r ) [f 1 (4 2u)T2 ai 2a 2 dr o OA g.u 2qa 2 co E ) since — = F' — = sine. neglecting ---retc. Bo =--R--z= . (2- • 2cosci(t4)sinl(t-E) •2 r c up dt r 21Trc r -1 r CD CD T um 1103 3Q SO Bo = arrro ---q— „Ter E '1.-±41cosecosi(t-E)H1T-e-sin-7(t-E)cosq sin2esincu(t-E) Ans. 0=4 enn 2
f r ‘ Qa) 2rrecr 2 C°S2e sincl4t - a.' ' I. . _d_Ae . neglecting r3 etc. r3 etc. Ee _ terms in 1 _dAe dt - rbe ...' - dt E r 4'n'Er 2 r Cl) r Q cu3 dAz cu r cu ( t-E) E0-- sine-- = 1.*(1)2 sinecoset7, 2 si1r2(t-E)cos2(t- ) -=8Trerc3 sin2esina dt 2nrc
2(t ^, ---c-I--(1-' 1= cos 2 0 sini(t rc 2rrer
)cos-
-;') 3
-
q
= 2rrer
Le _ 1 E ec2 Ans.
-15d (mp 2 )
dt
r
A
- Pq(Bovo -Bzvo ) = pqt-
1...02A0)
z .dt - -p-
Op
_LI dt}
- qcj--(1 r -421/. Integrate from dt
p1, z1, cu i to p2, z 2 , CD 2 ; MO Pi - IIIPICO2 = -q(P 1 A0 (P i , zi) p 2 A0 (p2 ,z2 )). 16 -
Ans.
From 13.02 (4) and (3), when n = 1, m= 0, y= B/2: Ex= jEo e i Niz. By = jco-lpz i Eo e j13011 z =c0-1 13 1 Ex . Then from 14.16 (4) and (5) E x' =R(Ex-vBy), 13,= x (By-vc-2 Ex) =14 (B,, i /0))- (v/c) 2 3Ex . If u)-1 8c: i = c-2-v, observer S' with z-velocity v= caN i cu'l sees only Ex' =IcE0 (1-cunVo1v) = RE° (1-v2 c -2 ) . By 13.00 (2) and (12), 8021 = p2 - pie, = (a.)/02 (1-(v/02 3 so 1/1( = eh i /co. From 14.01 (7), wt = x (cut' - cuvz7c2 ), 80 iz =x1 (u)vzI/c 2 )- (aw2 t I /c 2) 3
so wt-pl,, z= lc( 1- (v/c)2 lue = cott hc = c1301 t' • Eix = Eo chi 03-1sin(cfilli t'). The S' equation of motion is: d(mv')/dt= cl30 0)-1qE 0 sin(c1301t')
Ans.
Integrate from mv10 to mvlx
and assume ‘4(<< c so that initial S' velocity dominates throughout. By 14.04 (1), (4) m 7,-, Km° .
xmo (vx' - v0).,:r., cu(c80 ) im° (q-v0 ) = -qE 0w71 (cos c80it' - cos c80A) •
clEr (cos cp 01 t' - cos chi t') }dt'. Sdx' = Stvl0 + cf3" m ocu x 1 = x =volt'- tio) - clE° —tsinc8oit' - sinc80
, = ys _
1,70
-
mow2 cf3,,vo
. Thus X = Vgt t -
-
-cpc,
tDcos coo -el •
aEo.rt si.n cut - sincut0 -co(t-t0 )coscut0 )
mco,'
=
=(-1)ttio = coo,. Ans.
401
ERRATA STATIC AND DYNAMIC ELECTRICITY - 3rd Edition, lst - printing
Page 179
Most corrections are made in later printings. Page
Corrections. (Put symbols in italic type)
xxii Line 13. "capacitivity" for "capacivity". 3 101
Line 15. "9.1" for "9.0" Eq.(2) "sn" for "sin". Line 1 below (2) "sines" for "signs" Fig. 4.29(a) Interchange "a" and "b" twice.
102
Line 4 from bottom. "Erdelyi" for "Erdeyli" twice.
105
Line below (4). "c" for "C". Page 115 Prob.64,1ine3. "x-axis" for "y-axis".
116
Line 1 Prob.68. "tube" for "plate".
118
Prob.79. "sn k()" for "sn ()" in z equation. Line below "1/2" for "1/4". Next line. "2c" for "c", "b2" for "2b2" and "2b" for "b". Prob.80. "8" for "4" in z equation.
119 Line 2. A./21,,57) for vc.05. Prob.81. In k equation,
,45 for
NT813.
Prob.82 "[ti 11] for "N/1-(B/A):' 7/ 17/a" in numerator only. Prob.83. In C equation:"(b/a) a" for "(b/a)2 120 Ref. "Erdelyi" for "Erdeyli" Page 159 Eq.(12). "(2n):" for "(2n!)". 180 5.292,1ine 2. "jkl " for "jk". Page 211 5.40,1ine 9. "Zeit" for "Zett.". 215 Line 5. "HTF" for "ITT". Last lineV o "Ves"' "-ex" for 216 Eq.(5). "C; n", "qN "C in","Cm" on right side only. "Con", for "Con"' Eq.(6). "C' sn" for "Csn" and add " sfn Csn =C' sn". Page 217 "C3)"for"(2)" at t4 221 Line 16 "251 or Magnus" for "261 or HMF". Line 18. Add " for field point insert (-1)non right side of (2) and (3),". Page 230 Prob.62. "47E" for "47". 240
Prob.128.Insert"Qzn(j")"'in right side of V equation.
242 Prob. 131. "(-1)n" for "(-1)" twice. 241 Prob.129."concentricsphere"for "sphere' Prob. 132. Last line "5.41 (5)" for "5.41" 243 End of line 9 from bottom, "Art. 5.41" for"last problem". 245 "Erdelyi" for "Erdeyli" twice. 259 Last term in (5).u_ a2“ for "82" in numerator. "r 12" for "r" in denominator. "r2s+1"for r 2n+1"," !„"I'(2n+2s+2)fi for"_ , 271 Eq.(3). (2s+2):" for"(2s+1) I (2n+2s+2). 2n+2s+3 In Csn and C of Eq.(4) write"(2n+2):"for "(2n+1):': nn 275 "7" for "I" twice in answer to 9. 276 "0" for "6°in answer to 18. 279 Prob.33. "6°" for "8 ","8°" for "8 ", "1263" for "1268". "(U -4-2n+2):" for os n on 284 Eq. (1) Boldface "A" and "A'". "(242n+2)" .
Page 180
0
4. A
eN
4.
Page 181
Corrections. 291 Last term of Eq.. (4). "4" for "2r" and "64" for "128". 313 Figure 7.29 should be 315 Fig. 7.31 to be replaced by one on next sheet. Line 3 from bottom: "2n'I'" for "2nI". 316 Line 8 should be "Thus,taking zero order terms from 7.05 (7)," Eq.(1) "Bo cl/p"for "Bo c". Eq.(3)"B oc' 2W for "Bo c". 317 Eq.(11). "-Bm" for "Bm". Eq.(12). "-Cm" for "Cm". 324 Prob. 31. Eq. for Cn:"..1 1" for "Jo" in numerator only. Prob.32. Lines 2 and 3: "pI" for "p". Prob.34. Eqs. for xi andyi, "crilmy" for "cnsmx". 326 Prob.42.
1B0 = Bz"
for
"B p =p3z".
327 Prob.44. Eq. for A0: "2p" for "p". Force Eq. "p" for "i". for."8". Prob.27. " 2-" for "8". AB AB 1 Prob.28. "c" for "v" "-" for "4". ' AB 348 Prob.32. Insert "(2m+2):" in numerator. 347 Prob.26.
Prob.33. " 17 " for "2.e, "p3= d-A+B" for "p3= d+A-B" 8A B for "2r". Prob.34. Insert "(2m+2):" in numerator. IL8iTB" 381 Line 2, second word is "the". 366 "MAGNETISM" for "ELECTROMAGNETIC INDUCTION' 394 Eq.(5). "Ct" for "5" in exponent 395 Eq.(8). " =+" for " =-". Eq.(10) "=-" for last "=". 397 Below Eq. (2). "s >1" for "n >1" 401 "10.16" for "10.14" eight times on page. Eq.(6), "=+" for "=-". 405 Eq.(19). "P 2" for uP2u. 412 Last line: "ka" for "kc". 413 Prob. 32 Line 2, "and -" for "and
" Prob.34: "10.06" for "10.16".
414 Top line: "5.231 (3)" for "5.23 (1)" Prob. 35: "[(2m+1)0+wt]" for "m(0+at)" three times. 415 Eq.(1): "v" should be bold face. 416
Eq.(4) "dt/e" for "dt".
417 Next to last line "(2) and (8)" for "(4) and (9)". 424 Line 2 above Eq (5): "wave" for "waves". 425 "11.08 Intensity of" for "11.08 Intensity or", at top and bottom. 427 Heading "INTENSITY OF" for "INTENSITY OR". 442 Prob.l. "radio" for "ratio",in line 2. 443 Prob.9. Add bracket at end of formula.
Page 182 Corrections. 1 445 Prob. 20. "K(b7a 7)(K[(1-ba-1)73) 1" for "K[(1-ba-1)2j[Ko a a 2) 3-i I, Probs. 22 and 23. "4.30" for "4.31". Prob. 24. "0-1 " for "0". 446 Prob.26 Interchange "h" and "d", in lines 1 and 2. Prob,27. "63" for "62" and "a=lic7/d" for "tana= c/d". 7j" for "K(k)(1(1(1-k 2)TW". Prob.29. "[K(k)]-1K[(1-k2) Prob. 31. "±xy" for "xy". 447 "der Optik" for " der Optic". 448 Between line 2 and line 3 from bottom insert "solutions give little information on the input reactance. The radiation from orifices is often found by assuming initial field values which are then" 450 Eq.(15)."Z" should be bold face. 451 Eq. (27). "473" for "73" twice. 4r/ 2r2 459 Line 3. "403.4"for "403.2". Eq.(3) "-Ci " for"A-Ci ". X 465 Figure 12.08
>
5C
4411r01-
110 -11110
468 Line before Art. 12.12 "positive" for "postive". 469 Over Eq. (1). "0-component" for "0 =component". 482 Prob. 1. "167" for "4r" twice. 483 Prob. 2. "-E0" for "E0".
Prob.3. Line 2. "20e1 for "2a".
490 Prob. 50. "4" for "16" and delete "9". Prob. 51. "16" for "64" and "3Tr" for "277". Prob. 52. "Wk" for "404" and "r 2" for "972". Prob 53. "4" for "16" and "317" for "27r". 492 Last line. "XI" for "XL". 495 Eq.(1) and line above it. "P" for "P". 497 Top line. "13.01 (6)" for "15.01 (6)": 499 Eq. (5). "4" for "on". 501 Eq.(13). Delete "j". "(2712, )" 504 Eq.(5). for "272" 511 Line 9. "11.01 (13)" for "11.00 (14)" Line 3 below (3). "(0-1" for "p". 513 Eq.(13) "Ao" for "AO" 523 Eq.(4) "72T" for "V 2T". Eq.(9). 1171P for "7U" twice. 524 Eq.(8). ",2U" for "VU" 525 Line below Eq.(14). "VU" for "VU". Eq.(17). 'WI for "7U". 528 Line above Eq.(5). "10.02" for "11.02"
Page 183 Corrections. 529 Eq. (6) . "S(kX17U)2" for "(itxpu)2“. 541 Eq.(1). All subscripts are "i". Line 3 below Eq.(1)."11.01" for "11.18". Eq.(2)."wl" for"0". Line over Eq.(3).Bold face "1". Eq.(5).Bold face "A". 544 Eq.(5). "L" for "1".
542 Eq.(3)."Bp" for "Bo ".
546 Line 3. "j0:In" for "jp;m". Prob.6, line2. "x ozo" for "x0y01t. 547 Prob.12, on summation. "n" for "op". 556 Prob.57. "+" for "-" in three places. 557 Prob.62. "B" for "b" in Ez. "b2)" 'for "b2" twice and "4" for RI". 558 "I" andl" bold face. In
he
"cos" for "sinflinProb.67.
Prob.68 line 3. "Ny" for "My". Last line "cos" for last "sin". Prob.69 In Ri. "Logi" for"8201" and
566
upti n
for
"010".
Eq.(1)."q for "Fr".
572 Eq.(3). "[r]2" for "[r]3" 574 Over Eq.(4). "+" for central "-". "Jr," for "rr" 577 Last line "[P]"for "P". 583 Line 18. "nonmagnetic" for "nomagnetic". 589 Prob.15. Last line "pl " for "pi ". 599 Under"Antenna" transfer "in cavity,543-545" to just above "directivity" 604 After"Current, density of". "247" for "248". 614 "Phasor, 369" for "Phasor, 391".
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