Permanent Magnet and Electromechanical Devices: Materials, Analysis, and Applications

Permanent Magnet and Electromechanical Devices Academic Press Series in Electromagnetism Edited by ISAAK MAYERGOYZ, U...

Author: Furlani E.P.

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Permanent Magnet and Electromechanical Devices

Academic Press Series in Electromagnetism Edited by ISAAK MAYERGOYZ, UNIVERSITY OF MARYLAND, COLLEGE PARK, MARYLAND Electromagnetism is a classical area of physics and engineering that still plays a very important role in the development of new technology. Electromagnetism often serves as a link between electrical engineers, material scientists, and applied physicists. This series presents volumes on those aspects of applied and theoretical electromagnetism that are becoming increasingly important in modern and rapidly developing technology. Its objective is to meet the needs of researchers, students, and practicing engineers.

Books Published in the Series Giorgio Bertotti, Hysteresis in Magnetism: For Physicists, Material Scientists, and Engineers Scipione Bobbio, Electrodynamics of Materials: Forces, Stresses, and Energies in Solids and Fluids Alain Bossavit, Computational Electromagnetism: Variational Formulations, Complementarity, Edge Elements M. V. K. Chari and S. J. Salon, Numerical Methods in Electromagnetism Göran Engdahl, Handbook of Giant Magnetostrictive Materials Vadim Kuperman, Magnetic Resonance Imaging: Physical Principles and Applications John C. Mallinson, Magneto-Resistive Heads: Fundamentals and Applications Isaak Mayergoyz, Nonlinear Diffusion of Electromagnetic Fields Giovanni Miano and Antonio Maffucci, Transmission Lines and Lumped Circuits Shan X. Wang and Alexander M. Taratorin, Magnetic Information Storage Technology

Related Books John C. Mallinson, The Foundations of Magnetic Recording, Second Edition Reinaldo Perez, Handbook of Electromagnetic Compatibility

Permanent Magnet and Electromechanical Devices Materials, Analysis, and Applications

Edward P. Furlani Research Laboratories Eastman Kodak Company Rochester, New York

San Diego San Francisco New York Boston London Sydney Tokyo

This book is printed on acid-free paper. Copyright 2001 by Academic Press All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt, Inc., 6277 Sea Harbor Drive, Orlando, Florida 32887-6777. Explicit permission from Academic Press is not required to reproduce a maximum of two ﬁgures or tables from an Academic Press chapter in another scientiﬁc or research publication provided that the material has not been credited to another source and that full credit to the Academic Press chapter is given. Academic Press A Harcourt Science and Technology Company 525 B Street, Suite 1900, San Diego, CA 92101-4495, USA http://www.academicpress.com Academic Press Harcourt Place, 32 Jamestown Road, London NW1 7BY, UK http://www.academicpress.com Library of Congress Catalog Card Number: 2001089410 International Standard Book Number: 0-12-269951-3 Printed in the United States of America 01 02 03 04 05 06 EB 9 8 7 6 5 4 3 2 1

This book is dedicated to my parents Edward and Helen for their love and sacriﬁce, my wife Karen for her patience and encouragement, and my two children Edward and Amanda for the joy they give me.

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Contents

Foreword Preface 1. Materials 1.1 1.2 1.3 1.4

1.5 1.6 1.7 1.8 1.9

1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20

Introduction Units Classiﬁcation of Materials Atomic Magnetic Moments 1.4.1 Single Electron Atoms 1.4.2 Multielectron Atoms Paramagnetism Ferromagnetism Magnetostatic Energy Demagnetization Field Anisotropy 1.9.1 Magnetocrystalline Anisotropy 1.9.2 Shape Anisotropy Domains Hysteresis Soft Magnetic Materials Hard Magnetic Materials Ferrites Alnico Samarium-Cobalt Neodymium-Iron-Boron Bonded Magnets Magnetization Stability

2. Review of Maxwell’s Equations 2.1 Introduction

xi xiii 1 1 2 6 7 9 14 17 19 23 23 27 27 31 33 36 39 44 46 48 50 51 53 55 66

73 73 vii

viii 2.2 Maxwell’s Equations 2.2.1 Constitutive Relations 2.2.2 Integral Equations 2.2.3 Boundary Conditions 2.2.4 Force and Torque 2.3 Potentials 2.4 Quasi-static Theory 2.5 Static Theory 2.5.1 Magnetostatic Theory 2.5.2 Electrostatic Theory 2.6 Summary

3. Field Analysis

Contents 74 75 77 79 82 82 85 87 87 89 91

97

3.1 Introduction 3.2 Magnetostatic Analysis 3.2.1 Vector Potential 3.2.2 Force and Torque 3.2.3 Maxwell Stress Tensor 3.2.4 Energy 3.2.5 Inductance 3.3 The Current Model 3.4 The Charge Model 3.4.1 Force 3.4.2 Torque 3.5 Magnetic Circuit Analysis 3.5.1 Current Sources 3.5.2 Magnet Sources 3.6 Boundary-Value Problems 3.6.1 Cartesian Coordinates 3.6.2 Cylindrical Coordinates 3.6.3 Spherical Coordinates 3.7 Method of Images 3.8 Finite Element Analysis 3.9 Finite Difference Method

97 97 102 110 112 116 118 126 131 135 141 144 144 153 161 162 175 180 185 190 200

4. Permanent Magnet Applications

207

4.1 Introduction 4.2 Magnet Structures 4.2.1 Rectangular Structures 4.2.2 Cylindrical Structures 4.3 High Field Structures 4.4 Magnetic Latching 4.5 Magnetic Suspension

207 208 208 218 265 268 273

Contents

ix

4.6 4.7 4.8 4.9 4.10 4.11

282 286 306 312 318 325

Magnetic Gears Magnetic Couplings Magnetic Resonance Imaging Electrophotography Magneto-Optical Recording Free-Electron Lasers

5. Electromechanical Devices 5.1 Introduction 5.2 Device Basics 5.3 Quasi-static Field Theory 5.3.1 Stationary Reference Frames 5.3.2 Moving Reference Frames 5.4 Electrical Equations 5.4.1 Stationary Circuits 5.4.2 Moving Coils 5.5 Mechanical Equations 5.6 Electromechanical Equations 5.6.1 Stationary Circuits 5.6.2 Moving Coils 5.7 Energy Analysis 5.8 Magnetic Circuit Actuators 5.9 Axial-Field Actuators 5.10 Resonant Actuators 5.11 Magneto-Optical Bias Field Actuator 5.12 Linear Actuators 5.13 Axial-Field Motors 5.14 Stepper Motors 5.15 Hybrid Analytical-FEM Analysis 5.16 Magnetic MEMS

A. Vector Analysis A.1 A.2 A.3 A.4 A.5 A.6

Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates Integrals of Vector Functions Theorems and Identities Coordinate Transformations

335 335 335 337 338 341 354 354 358 361 363 363 366 376 383 393 402 406 413 421 437 446 455

469 469 473 476 479 485 491

B. Green’s Functions

495

C. Systems of Equations

497

C.1 Euler’s Method

498

x

Contents C.2 Improved Euler Method C.3 Runge-Kutta Methods

502 503

D. Units

509

Index

513

Foreword

This volume in the Academic Press Series in Electromagnetism presents an in-depth, self-contained, and up-to-date treatment of permanent magnets and related electromechanical devices. The emergence of modern rare-earth permanent magnets has led to the tremendous proliferation of devices that utilize these magnets. Currently, there does not exist any book that presents comprehensive treatment of permanent magnet technology and covers material aspects, ﬁeld analysis, and wide-ranging device applications of permanent magnets. This book, written by Dr. Edward P. Furlani, represents the ﬁrst and successful attempt to give the exposition of all these issues within one volume. The book reﬂects the author’s extensive experience as well as his intimate and ﬁrsthand knowledge of the area of permanent magnet technology. The unique feature of this book is its broad scope and the proper and delicate balance between theoretical and applied aspects of permanent magnet devices. The book is intended for readers without extensive experience or knowledge in the area of permanent magnets. It reviews the physics of magnetic materials and the basics of electromagnetic ﬁeld theory. Then, it presents in detail the analysis and design of the wide range of permanent magnet devices such as magnetic gears, couplings, bearings, magnetic resonance imaging devices, free-electron laser magnets, magnetooptical recording systems, permanent magnet motors and stepper motors, and, ﬁnally, magnetic microactuators (MEMS). The book contains a large number of practical design examples that are completely worked out. This feature will help readers to follow all the computational and design steps involved in the development of new devices and will be very beneﬁcial for graduate students and practicing engineers. I believe that this book will be a valuable reference for both experts and beginners in the ﬁeld. Electrical and mechanical enginers, applied physicists, material scientists, and graduate students will ﬁnd this book very informative. Isaak Mayergoyz Series Editor xi

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Preface

Permanent magnet and electromechanical devices can be found in a wide variety of products that have an impact on our daily lives. Examples of these include audio and video players, telecommunications equipment, personal computers, printers, copiers, automobiles, appliances, power tools, data storage devices, and biomedical apparatus. Furthermore, the use of modern rare-earth magnets is proliferating at a remarkable rate, which is giving rise to new products and markets and stimulating research into a variety of new devices. However, despite the substantial theoretical and practical interest in this technology, there is no single comprehensive text that adequately treats the subject matter required for the research and development of such devices. Instead, one needs to extract crucial information from a variety of texts, each with its own notation and style and emphasis of presentation. It is hoped that this book will eliminate the need for this cumbersome approach. This book is a comprehensive and self-contained exposition of the theory and methods used in the analysis and design of permanent magnet and electromechanical devices. It is intended to be used as both a text and a reference for researchers, professors, students, and engineers who are engaged in the research and development of new and conventional permanent magnet and electromechanical devices. It is written to provide the reader with a ﬁrst-principles understanding of the underlying theory of such devices, and to enable him or her to analyze, design, and simulate practical devices. The salient features of the book are as follows:

· Extensive discussion of magnetic materials including the physical

·

principles of magnetism, domains, hysteresis, soft and hard magnetic materials and their properties, and the magnetization and stability of permanent magnets. Comprehensive presentation of analytical and numerical methods for the analysis of steady currents, permanent magnets, and xiii

xiv

Preface

· · · · · · · ·

magnetic circuits. Topics include magnetostatic analysis, the Maxwell stress tensor, the current model, the charge model, magnetic circuit analysis, boundary-value theory, ﬁnite element analysis, and ﬁnite difference analysis. Over 60 examples of practical permanent magnet and electromechanical applications with detailed solutions. Analytical analysis and design formulas for the ﬁeld distributions of the most common rare-earth permanent magnet structures. Analytical analysis and design formulas for the performance of rare-earth magnetic couplings, gears, and bearings. Comprehensive presentation of the theory of electromechanical devices with numerous solved examples. Analytical analysis and design formulas for linear and rotational actuators. Analytical analysis and design formulas for permanent magnet brushless dc motors and stepper motors. Presentation of a hybrid analytical FEM approach for the analysis of electromechanical devices. Introduction to magnetic microactuators.

This book consists of ﬁve chapters and four appendices. The organization of the material is illustrated in the following diagram. Materials Review of Magnetic Field Analysis Theory $ Maxwell’s $ Analytical and Properties Equations Numerical Methods Appendices Permanent Vector Analysis Electromechanical $ Magnet $ $ Green’s Functions Devices Devices ODE Systems Units Book Organization

The ﬁrst three chapters provide a theoretical background, while the ﬁnal two chapters emphasize the analysis of practical devices. The ﬁrst chapter covers magnetic materials. It begins with a brief review of the different units of magnetism, followed by a summary of the major classiﬁcations of materials. This is followed by a discussion of the basic mechanisms of magnetism starting at the atomic level with an analysis of an isolated single electron atom. The atomic magnetic mo-

Preface

xv

ment is principally due to the orbital and spin motion of the electron. The discussion is extended to multielectron atoms where the spin and orbital moments of the constituent electrons couple to give a net atomic moment. The theory is then generalized to the case of an ensemble of atoms. If the atomic moments are noninteracting, the ensemble exhibits a bulk paramagnetic behavior. If the atomic moments are interacting, there is a cooperative alignment of the moments (ferromagnetism) below the Curie temperature. This ferromagnetic behavior is combined with the concept of magnetic domains to explain the B-H (hysteresis) curves of bulk magnetization. A survey is given of commercially available soft and hard magnet materials and their properties. The chapter concludes with a discussion of the magnetization and stability of permanent magnets. The second chapter contains a brief but thorough review of Maxwell’s equations. The ﬁeld equations are presented in both differential and integral form along with the constitutive relations and boundary conditions. The scalar and vector potentials are introduced and shown to provide an alternate and often more tractable formulation of ﬁeld theory. Quasi-static ﬁeld theory is discussed. Here, there is a partial uncoupling of the ﬁeld equations that simpliﬁes their solution. Static ﬁeld theory is also discussed, and it is shown that the magnetic and electric ﬁelds uncouple into separate magnetostatic and electrostatic ﬁeld equations. The third chapter builds upon the second. It consists of a comprehensive and systematic treatment of the various analytical and numerical methods used for the analysis of steady currents, permanent magnets, and magnetic circuits. Magnetostatic ﬁeld theory is presented in some detail, including a discussion of force, torque, energy, and inductance. The current and charge models for magnetic materials are introduced. These are used to reduce a permanent magnet to an equivalent source term that can be analyzed using the ﬁeld equations. Magnetic circuits are discussed, and the concept of reluctance is introduced and used to transform a physical circuit into an equivalent lumped-parameter circuit. Various analysis methods are also discussed, including boundary-value theory, the method of images, ﬁnite element analysis (FEA), and the ﬁnite difference method. These methods are demonstrated via solved examples. The ﬁnal two chapters cover permanent magnet applications and electromechanical devices, respectively. These chapters follow a problem solving approach in which numerous practical examples are worked out in sufﬁcient detail to be of immediate use to researchers and practitioners in the ﬁeld. Many of the examples contain design formulas that cannot be found in any other text.

xvi

Preface

The fourth chapter deals with permanent magnet applications. These include bias magnet structures, high ﬁeld structures, latching magnets, magnetic suspensions, and magnetic gears and couplings. Analytical methods are used to derive design formulas for many of these applications. Speciﬁcally, analytical formulas are derived for the most common rare-earth permanent magnet structures, and the performance of rareearth gears, couplings, and bearings In addition, various miscellaneous applications that utilize magnets are discussed, including magnetic resonance imaging (MRI), electrophotography, magneto-optical recording, and free-electron lasers. The ﬁfth chapter covers electromechanical devices, with an emphasis on magnetically linear, singly excited devices with a single degree of mechanical freedom. Quasi-static ﬁeld theory is used to derive Kirchhoff’s circuit laws for actuators with stationary circuits, and circuits with coils that move through an external magnetic ﬁeld. The circuit laws are combined with the laws of rigid body dynamics to obtain the coupled electromechanical equations that govern linear and rotational actuators. The theory is applied to numerous practical devices, including magnetic circuit actuators, linear and rotary actuators, permanent magnet direct current (dc) motors, and stepper motors. The chapter concludes with a brief introduction to magnetic microactuators. In addition to the ﬁve chapters, there are four appendices. Appendix A contains a review of vector analysis. The topics covered include vector calculus in the Cartesian, cylindrical, and spherical coordinate systems, line and surface integrals, Stokes’ theorem, the Divergence theorem, various vector identities, and coordinate transformations. In Appendix B the free-space Green’s functions for Poisson’s equation are introduced. These are used to obtain the ﬁeld distributions of free-standing permanent magnet structures. Appendix C contains a discussion of various numerical methods that are used for the solution of ﬁrst-order initialvalue problems. Speciﬁcally, the Euler, modiﬁed Euler, and Runge-Kutta methods are discussed. These are used for the analysis and design of electromechanical devices. Last, Appendix D contains a summary of the different systems of units and conversion factors that are used in the analysis of permanent magnet and electromechanical devices.

Acknowledgments Finally, I would like to acknowledge and thank the individuals who assisted me in the preparation of this book. First, I would like to thank

Preface

xvii

Svetlana Reznik for many insightful discussions on soft and hard magnetic materials. I am also grateful for the patient and diligent work of Christopher Devries in the preparation of numerous drawings and ﬁgures. Special thanks go to Dr. Isaak Mayergoyz for his encouragement and support during this project. I would also like to thank Marsha Filion, Julie Bolduc, and the rest of the editorial and production staff at Academic Press for their help and guidance with the preparation of the manuscript. Last, I would like to thank the research managers at the Eastman Kodak Company for their support. Edward P. Furlani

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CHAPTER

1

Materials

1.1 INTRODUCTION In this chapter we study the principles and properties of magnetic materials. Our study begins with a brief review of the basic units of magnetism. This is followed by a summary of the major classiﬁcations of magnetic materials. Next, we consider the basic mechanisms of magnetism. We start at the atomic level with an analysis of an isolated single electron atom. Our analysis shows that the electron’s magnetic moment is due to both its orbital motion and spin. The same analysis is then applied to an isolated multielectron atom where the spin and orbital moments of the constituent electrons couple to give a net atomic moment. After discussing single atoms we consider a collection of atoms. Speciﬁcally, we apply classical statistical mechanics to determine the bulk magnetization of an ensemble on noninteracting atomic moments. The ensemble exhibits a bulk paramagnetic behavior. A similar analysis is applied to an ensemble of interacting atomic moments. In this case, there is a cooperative alignment of the moments (ferromagnetism) below the Curie temperature. Following this, we introduce the concepts of magnetostatic energy, magnetic anisotropy and domains, and then use these to explain the B-H (hysteresis) curves of bulk magnetization. After describing bulk magnetization, we discuss soft and hard magnetic materials. For soft materials, we give a brief description of the most common materials and list their key physical properties. This is followed by a survey of commercially available hard materials (permanent magnets). The chapter concludes with a discussion of the magnetization and stability of permanent magnets. The topics covered in this chapter are summarized in Fig. 1.1.

1

2

CHAPTER 1 Materials

FIGURE 1.1

Organization of chapter topics.

1.2 UNITS There are three systems of units that are commonly used in the magnetism. These are the CGS or Gaussian system, and two MKS or SI systems that are referred to as the Kennelly and Sommerfeld conventions, respectively. Throughout this book we use SI units in the Sommerfeld convention [1, 2]. In the Sommerfeld convention the magnetic ﬂux is in webers (Wb), the ﬂux density B is in teslas (T) or Wb/m, and both the ﬁeld strength H and magnetization M are in A/m. The units for the SI (Sommerfeld) and CGS systems are as follows: Symbol

Description

SI

CGS

H

Magnetic Field Strength

A/m

Oe

B

Flux Density

Tesla

Gauss

M

Magnetization

A/m

emu/cm

Flux

Webers

Maxwells

(1.1)

3

1.2 UNITS

The conversion factors for these systems are 1 Oe : 1000/4 A/m 1 Gauss : 10\ T 1 emu/cm : 1000 A/m

(1.2)

1 Maxwell : 10\ Webers Additional conversion factors are discussed in Appendix D. We can obtain an intuitive feeling for the magnitude of the ﬁelds (units) described in the foregoing text by considering the following physical examples: 1. Field strength H: Consider a straight, inﬁnitely long wire carrying a current i : 2A. The conductor generates a tangential ﬁeld strength H : 1 A/m at a radial distance r : 1 m from its center. As another example, consider a long solenoid that has n turns per meter and carries a current of 1/n A. It generates a ﬁeld strength H : 1 A/m along its axis. 2. Flux density B: Consider an inﬁnitely long conductor that carries a current i : 1 A perpendicular to an external B-ﬁeld. A force of 1 newton will be imparted to each meter of the conductor when B : 1 T. 3. Flux : Consider a single turn coil with 1 Wb of magnetic ﬂux passing through it. One volt will be induced in the coil when the ﬂux is uniformly reduced to zero in one second. The fundamental element in magnetism is the magnetic dipole. This can be thought of as a pair of closely spaced magnetic poles (Section 3.4.1), or equivalently as a small current loop (Section 1.4) (Fig. 1.2). A magnetic dipole has a magnetic dipole moment m. In the Sommerfeld convention m is measured in A · m. In the Kennelly and CGS systems it is measured in Wb · m and emu, respectively (1 emu : 4 ; 10\ Wb · m). Magnetization M is a measure of the net magnetic dipole moment per unit volume. Speciﬁcally, it is given by m M : lim G G , V 4 where m is a vector sum of the dipole moments contained in the G G elemental volume V.

4

CHAPTER 1 Materials

FIGURE 1.2 Magnetic dipole: (a) magnetic charge model and H-ﬁeld; and (b) current loop and B-ﬁeld.

If a magnetic dipole is subjected to an external B-ﬁeld, it acquires an energy E : 9m · B (1.3) and experiences a torque T : m ; B.

(1.4)

In both the Kennelly and CGS systems these are given by E : 9m · H and T : m ; H. In the Sommerfeld convention the ﬁelds are related by the constitutive relation B : (H ; M)

(Sommerfeld convention),

(1.5)

5

1.2 UNITS

where : 4 ; 10\ T m/A is the permeability of free space. In the Kennelly convention, the constitutive relation is B: H;J (Kennelly convention), where J is called the magnetic polarization and is measured in tesla. Notice that J : M. The constitutive relation (1.5) simpliﬁes for linear, homogeneous and isotropic media. For such materials, both B and M are proportional to H. Speciﬁcally, B : H,

(1.6)

and M : H, (1.7) K where and are the permeability and susceptibility of the material, K respectively. These coefﬁcients are related to one another. From Eqs. (1.5), (1.6), and (1.7) we ﬁnd that or

: ( ; 1), K

(1.8)

: 9 1. (1.9) K The constitutive relations (1.6) and (1.7) need to be modiﬁed for nonlinear, inhomogeneous or anisotropic materials. A material is magnetically nonlinear if depends on H, otherwise it is linear. For nonlinear materials, Eqs. (1.6) and (1.7) become B : (H)H, and M : (H)H, K respectively. A material is inhomogeneous if is a function of position, otherwise it is homogeneous. In inhomogeneous materials the permeability and susceptibility are functions of the coordinate variables, : (x, y, z), : (x, y, z). Last, a material is said to be anisotropic if K K depends on direction, otherwise it is isotropic. In anisotropic materials Eq. (1.6) generalizes to B : H ; H ; H , V V W X B : H ; H ; H , W V W X B : H ; H ; H . X V W X A similar set of equations hold for Eq. (1.7).

6

CHAPTER 1 Materials

1.3 CLASSIFICATION OF MATERIALS Magnetic materials fall into one of the following categories: diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, and ferrimagnetic (Fig. 1.3). Diamagnetic materials have no net atomic or molecular magnetic moment. When these materials are subjected to an applied ﬁeld, atomic currents are generated that give rise to a bulk magnetization that opposes the ﬁeld. Bismuth (Bi) is an example of a diamagnetic material. Paramagnetic materials have a net magnetic moment at the atomic level, but the coupling between neighboring moments is weak. These moments tend to align with an applied ﬁeld, but the degree of alignment decreases at higher temperatures due to the randomizing effects of thermal agitation. Ferromagnetic materials have a net magnetic moment at the atomic level, but unlike paramagnetic materials there is a strong coupling between neighboring moments. This coupling gives rise to a spontaneous alignment of the moments over macroscopic regions called domains. The domains undergo further alignment when the material is subjected to an applied ﬁeld. Finally, antiferromagnetic and ferrimagnetic materials have oriented atomic moments with neighboring moments antiparallel to one another.

FIGURE 1.3

Classiﬁcations of magnetic materials [16].

1.4 ATOMIC MAGNETIC MOMENTS

7

In antiferromagnetic materials the neighboring moments are equal, and there is no net magnetic moment. In ferrimagnetic materials the neighboring moments are unequal, and there is a net magnetic moment. Rigorous treatments of these materials require a full quantum mechanical analysis that is beyond the scope of this text. We are primarily interested in ferromagnetic materials, and for our purposes it is sufﬁcient to review some of the key principles and results. The interested reader can ﬁnd more detailed presentations in numerous texts [1—7].

1.4 ATOMIC MAGNETIC MOMENTS In this section we study the magnetic moment of an isolated single electron atom. We start with a classical analysis. Consider an isolated atom with an electron of mass m and charge e moving in a circular orbit C of radius r with a linear velocity u (angular velocity : u/r) as shown in Fig. 1.4. The electron completes an orbit in : 2/ s and has an orbital angular momentum L given by L:m r;u C : m r. (1.10) C The orientation of L is given by a right-hand rule in which the ﬁngers of the right hand curl in the direction of the particle motion, and the thumb points in the direction of L (Fig. 1.4). The orbiting electron deﬁnes a closed loop of current i where e i: .

FIGURE 1.4

Atomic system.

8

CHAPTER 1 Materials

The current i deﬁnes the ﬂow of positive charge and therefore circulates in a direction opposite to the motion of the electron (Fig. 1.4). The current loop gives rise to a magnetic dipole moment (Fig. 1.2b). In general, if a current i circulates around an enclosed area ds, it gives rise to a magnetic dipole moment m, m : ids

(magnetic moment).

(1.11)

The vector ds deﬁnes the orientation of m relative to the circulation of i. This is given by the right-hand rule, which states that if the ﬁngers of the right hand follow the direction of current, then the thumb points in the direction of ds (Fig. A7). The magnitude of the moment is m : ids,

(1.12)

and is measured in units of A · m. Let m denote the magnetic moment of the electron due to its orbital * motion. The magnitude of this moment is given by Eq. (1.12) m : ir * e : r e : r. 2 This can be expressed in terms of L. Speciﬁcally, from Eq. (1.10), and the rules for determining the orientations of m and L, we ﬁnd that * e m :9 L. (1.13) * 2m C Notice that m is antiparallel to L as shown in Fig. 1.4. Amazingly, Eq. * (1.13) holds even at the atomic level when L is replaced by its quantized expression. If the orbiting electron is subjected to an external B-ﬁeld, its moment m acquires an energy * E : 9m · B * e : L · B, 2m C and experiences a torque T:m ;B * e :9 L ; B. 2m C

9

1.4 ATOMIC MAGNETIC MOMENTS

FIGURE 1.5

Precession of a magnetic moment about an applied ﬁeld.

As shown in Fig. 1.5, the torque causes L to precess around B. In classical mechanics, L can take on a continuum of orientations relative to B. However, when quantum effects are taken into account, the orientation of L is restricted to a discrete set of values as we shall see.

1.4.1 Single electron atoms Atomic systems are governed by quantum theory with electronic states speciﬁed in terms of wavefunctions [8, 9]. The wavefunctions are obtained by solving Schrödinger’s equation

9

; V : E , 2m

where m is the mass of the orbiting particle, V is the potential energy, and E is the total energy. The wavefunctions are indexed by a discrete set of quantum numbers (n, l, m , m ) that distinguish the allowable J Q energy levels, orbital conﬁgurations, etc. The principle quantum number is n and this speciﬁes the energy of a given orbit or shell. Observables such as energy and angular momentum are represented by operators that act on . The observed value L of an observable L for a given state

10

CHAPTER 1 Materials

(n, l, m , m ) is obtained by computing the expectation value of its J Q operator L , L

:

*(n, l, m , m )L (n, l, m , m ) dv, J Q J Q

(1.14) where * is the complex conjugate of . It follows from Eq. (1.14) that observables are also quantized and indexed. We now discuss the angular momentum of an atomic electron. Consider a hydrogenlike single electron atom. There are two contributions to the angular momentum: one due to orbital motion and the other due to spin. From quantum mechanics we know that for a given value of n the electron’s orbital angular momentum L has a magnitude

L : [l(l ; 1)] .

(1.15)

In Eq. (1.15), l is the orbital angular momentum quantum number with allowable values l : 0, 1, 2, . . . , (n 9 1), and : h/2 where h is Planck’s constant (h : 6.6260755 ; 10\ J s) [3]. Electrons with l : 0, 1, 2, 3, . . . are referred to as s, p, d, f, . . . electrons, respectively. For example, the n : 1 shell has an s electron, the n : 2 shell has s and p electrons, etc. Notice that the magnitude of L is restricted to a discrete set of values. This is in contrast to classical theory, which allows for a continuum of values. The orientation of L is also restricted. Speciﬁcally, its projection onto a given axis (say, the z-axis) is given by L : m , (1.16) X J where m is called the magnetic quantum number, and is restricted to the J values m : l, (l 9 1), . . . , 0, . . . , 9 (l 9 1), 9l. Thus, for a given value of J l there are 2l ; 1 possible orientations of L relative to the axis (Fig. 1.6a). An electron also has a spin angular momentum. The concept of spin was originally proposed to explain the multiplet structure of atomic spectra. Initially, it was thought that spin could be explained in terms of the electron spinning about an internal axis. However, predictions based on this model do not yield the correct magnetic moment. Therefore, spin is viewed as a purely quantum mechanical phenomenon. The spin angular momentum S is quantized and has a magnitude S : [s(s ; 1)] , where s : 1/2. The projection of S along a given axis is also quantized with S : m , (1.17) X Q where m : <1/2 (Fig. 1.6b). Q

1.4 ATOMIC MAGNETIC MOMENTS

11

Spatial quantization of angular momentum: (a) orbital angular momentum (l : 2); and (b) spin angular momentum [4].

FIGURE 1.6

The total angular momentum of the electron, which is denoted by J, is the sum of its orbital and spin angular momenta J : L ; S. It has a magnitude

J : [ j( j ; 1)] , where j : l < 1/2. The total angular momentum depends on the magnitude and relative orientations of the orbital and spin angular momenta. We now consider the electron’s magnetic moment. There are two contributions to the moment: one due to the orbital motion and the other due to spin. Recall from the classical result of Eq. (1.13) that the orbital moment m is proportional to L. This same relation holds at the atomic * level, e m :9 L, * 2m C but now the magnitude of m is quantized, * e m :9 [l(l ; 1)]

* 2m C : 9 [l(l ; 1)],

(1.18)

where : e /2m is the Bohr magneton. Notice that m is restricted by C *

12

CHAPTER 1 Materials

the allowable values of l. If the electron is subjected to an external H-ﬁeld, its orbital magnetic moment couples to H, and the electron experiences a torque that causes m to precess around H. However, the * orientation of m relative to H is restricted to a discrete set of values as * described in the preceding. Speciﬁcally, the projection of m along H * (denoted m ) is *& m : 9 m , *& J where m is as speciﬁed in Eq. (1.16). Thus, for a given value of l there J are 2l ; 1 possible orientations of m relative to H (Fig. 1.6a). * The magnetic moment m due to spin is given by 1 e m : 9 S. 1 m C It has a magnitude m : 92 [s(s ; 1)]. (1.19) 1 The spin also couples to an external ﬁeld, and has a projection : 92 m . 1& Q The electron’s total magnetic dipole moment m is the sum of its orbital and spin moments (Fig. 1.7): m

m :m ;m * 1 e :9 (L ; 2S) 2m C As J : L ; S, this can also be written as

Notice that m

e m :9 (J ; S). 2m C is not antiparallel to J.

FIGURE 1.7

Addition of angular momenta.

(1.20)

(1.21)

13

1.4 ATOMIC MAGNETIC MOMENTS

From Eq. (1.20) we see that m depends on both the magnitude and relative orientations of L and S. However, L and S are neither independent of one another nor constant in time. Rather, they are coupled via a spin-orbit interaction. This interaction can be understood by considering a reference frame attached to the electron. In this frame, the nucleus appears to revolve around the electron. Therefore, the electron is subjected to a magnetic ﬁeld B due to the orbiting charged nucleus. L Moreover, B couples to the spin magnetic moment m with an energy L 1 E . m · B . As B is parallel to L and m is antiparallel to S, we have 1 L L 1 E : L · S, where is the spin orbit parameter. Notice that E depends on the angle between S and L. Consequently, a torque is imparted perpendicular to both L and S causing them to precess. We may visualize this as a precession of L and S about the total angular momentum J, which is constant (absent any external inﬂuences). It follows from Eq. (1.20) that m also precesses around J. Therefore, it is time dependent and cannot be speciﬁed as a ﬁxed vector. However, we can deﬁne a constant average-value vector m . Speciﬁcally, as J is constant, we can specify m by taking the time-averaged value of m in the direction of J (Fig. 1.8). This is given by m

J J : m · . J J

(1.22)

Substituting Eq. (1.21) into Eq. (1.22), and making use of the fact that J : L ; S $ S · J : 1/2(J · J ; S · S 9 L · L), we get

(1.23)

j( j ; 1) ; s(s ; 1) 9 l(l ; 1) . 2j( j ; 1)

(1.24)

m

:9

e gJ 2m C

where g:1;

FIGURE 1.8

Spin-orbit coupling and the orientation of m

[9].

14

CHAPTER 1 Materials

This is called the Landé g factor. From Eq. (1.23) we see that the magnetic moment of an electron can be determined given its angular momentum. This generalizes to the case of multielectron atoms, which we study next.

1.4.2 Multielectron atoms In this section we consider the magnetic moment of an isolated multielectron atom. When multiple electrons are present, each must occupy a different orbit as deﬁned by the four quantum numbers n, l, m , and m . This is the Pauli exclusion principle which states that no two J Q electrons may have the same set of quantum numbers. The magnetic moment of a multielectron atom can be determined from the total angular momentum J using a relation similar to Eq. (1.23). Since the atom is isolated, J is a constant with a magnitude

J : [J(J ; 1)] . The projection of J onto a given axis is J : M , X ( where M : J, J 9 1, J 9 2, . . . , 9(J 9 2), 9(J 9 1), 9J. To determine J ( or J we need to know the allowed values of J. These depend on the X orbital and spin conﬁgurations of the constituent electrons, and the way in which their respective moments couple. There are two different modes of coupling. These are known as Russell-Saunders and spin-orbit coupling,

L ;S (Russell-Saunders) J : or (1.25) (L ; S ) (spin-orbit). G G G The rule for evaluating Eq. (1.25) is that vectors with the strongest coupling are summed ﬁrst. Russell-Saunders coupling applies when the orbital-orbital (L -L ) and spin-spin (S -S ) couplings between different G H G H electrons are stronger than the spin-orbit (S -L ) coupling of each separG G ate electron. This applies to all but the heaviest atoms. On the other hand, spin-orbit coupling is used when the spin-orbit coupling of each separate electron dominates the couplings between different electrons. Russell-Saunders coupling applies to most magnetic atoms. In this coupling the total orbital angular momentum is the vector sum of orbital momenta of the individual electrons L :L . G G

1.4 ATOMIC MAGNETIC MOMENTS

15

It has a magnitude

L : [L (L ; 1)] , and its projection along a given direction (say, the z-axis) is L : M , X * where M : L, (L 9 1), . . . , 0, . . . , 9 (L 9 1), 9L, and M : m . * * G JG Similarly, the total spin angular momentum is given by

This has a magnitude and a projection

S :S . G G

S : [S(S ; 1)] ,

S : M , X 1 where M : S, (S 9 1), . . . , 0, . . . , 9(S 9 1), 9S, and M : m . The net 1 1 G QG angular momentum of the electrons in a closed shell is zero,

and

L : 0, G

S : 0. G Therefore, only the partially ﬁlled shells contribute to J . An example of Russell-Saunders coupling is shown in Fig. 1.9. Recall that the total angular momentum is J : L ; S . This has a magnitude

J : [J(J ; 1)] . In Russell-Saunders coupling J is restricted to the following values: J : L ; S, L ; S 9 1, . . . , L 9 S . To evaluate J we need to know the ground state orbital and spin conﬁgurations of the electrons. These are determined using Hund’s rules, which state that: (i) the spins S combine to give the maximum G value of S consistent with the Pauli exclusion principle; (ii) the orbital vectors L combine to give the maximum value of L compatible with the G Pauli principle and condition (i); and (iii) the resultant L and S combine to form J , with J : L 9 S if the shell is less than half ﬁlled, and J : L ; S if it is more than half ﬁlled. When the shell is exactly half full, L : 0 and J : S.

16

CHAPTER 1 Materials

FIGURE 1.9

Russell-Saunders coupling [9].

Finally, for an isolated atom J is constant. We determine an average magnetic moment of the atom m by taking the projection of its magnetic moment m along the direction of J . Following an analysis similar to that for Eq. (1.22) we obtain

m : g (J(J ; 1), where g:1;

J(J ; 1) ; S(S ; 1) 9 L(L ; 1) . 2J(J ; 1)

Notice that if the net orbital angular momentum is zero (L : 0), then J : S and g : 2. On the other hand, if the net spin angular momentum is zero (S : 0), then J : L and g : 1. The projection of m along the direction of an applied H-ﬁeld is denoted by m &

m : g M , (1.26) & ( where M : J, J 9 1, J 9 2, . . . , 9(J 9 2), 9(J 9 1), 9J. Thus, m is ( & constrained to a discrete set of orientations relative to H. This completes our study of isolated atoms. In summary, we have found that the magnetic moment of the atom is due principally to the magnetic moments of its constituent electrons. These, in turn, are due to both orbital motion and spin. The magnitude of the atomic moment can

17

1.5 PARAMAGNETISM

be determined from the total angular momentum, and this is obtained by summing the contributions of the individual electrons using RussellSaunders coupling, subject to Hund’s rules.

1.5 PARAMAGNETISM Recall that paramagnetic materials consist of atoms that have a net magnetic moment, but negligible coupling between neighboring moments. In this section we study the response of such materials to an applied ﬁeld. This serves as a prerequisite to our subsequent discussion of` ferromagnetism. Consider a paramagnetic specimen with N noninteracting atomic magnetic moments per unit volume. We analyze its response to an applied ﬁeld H . Initially, we treat the moments classically assuming ? that they can take on a continuum of orientations. Later, we impose quantum constraints. Let m denote the magnetic moment of a single atom. We seek an expression for the magnetization M, which is the total magnetic moment per unit volume in the direction of H . Each moment m couples to H ? ? and acquires a magnetostatic energy E : 9 m · H ? : 9 mH cos(), (1.27) ? where is the angle between m and H . We model the material as an ? ensemble of noninteracting atomic moments [10]. Speciﬁcally, we use Boltzmann statistics to describe the probability p(E) of a moment having an energy E at a temperature T p(E) : exp : exp

9E k T

mH cos() ? , k T

(1.28)

where k is Boltzmann’s constant. Physically, the alignment of m is opposed by thermal agitation and Eq. (1.28) gives the probability of m obtaining an angular orientation relative to H at temperature T. The ? magnetization is given by M:

L

m cos()n() d,

(1.29)

18

CHAPTER 1 Materials

FIGURE 1.10

Langevin function.

where m cos() is the projection of m (with orientation relative to H ) ? onto the ﬁeld direction and n() d is the number of atomic moments per unit volume with angular orientations between and ; d. Speciﬁcally, n() d : 2C exp

mH cos() ? sin() d, k T

(1.30)

where C is a normalization constant determined from L n() d : N. An evaluation of Eq. (1.29) yields M : NmL

mH ? , k T

(1.31)

where L() is the Langevin function 1 L() : coth() 9 , (see Fig. 1.10). Notice that 91 L() 1 and that lim L() : 1. (1.32) A From Eqs. (1.31) and (1.32) we ﬁnd that for low T and high H , M ? approaches its maximum value Nm, which means that the atomic moments tend towards perfect alignment in this limit.

19

1.6 FERROMAGNETISM

In most cases 1, and L() can be represented as an inﬁnite power series of the form, L() : 9 ; · · · . 3 45 For small we ignore all but the ﬁrst term and obtain NmH ?. M: 3k T This gives a parametric susceptibility of the form :

M Nm : . 3k T H ?

This is known as Curie’s law. The expression (1.31) follows from a classical analysis that allows for arbitrary orientations of the moments. However, from Eq. (1.26) we know that m is constrained to a discrete set of orientations relative to H . ? We modify Eq. (1.29) taking this into account and obtain ( m g exp( m g H /k T) ( ( ? M : N K(\( ( exp( m g H /k T) ( ? K(\( This can be written as M:M B (

gJ H ? , k T

(1.33)

where M : NgJ is the maximum possible moment in the direction of H , and ? 2J ; 1 (2J ; 1) 1 B () : coth 9 coth (1.34) ( 2J 2J 2J 2J

is the Brillouin function. In the limiting case when J ; - the allowed values and orientations of the atomic moments approach a continuum, and we recover the classical Langevin result of Eq. (1.31).

1.6 FERROMAGNETISM Ferromagnetic materials consist of atoms with a net magnetic moment, and there is substantial coupling between neighboring moments. These

20

CHAPTER 1 Materials

materials exhibit a magnetic ordering that extends over a substantial volume and entails the cooperative orientation of a multitude of atomic moments. An early attempt to explain this phenomenon was made by Weiss, who postulated the existence of a ‘‘molecular ﬁeld’’ H that is K proportional to the magnetization H : M. (1.35) K In Eq. (1.35) is the mean ﬁeld constant [11]. Consider a material characterized by Eq. (1.35). We seek an expression for M and follow the same analysis as in the previous section, but now H : H . Speciﬁcally, the magnetostatic energy is given by ? K E : 9 m · H K : 9 mM cos(), (1.36) where is the angle between the local moment m and the bulk magnetization M. Substitute H : M into Eq. (1.31) and obtain ? mM M : NmL . (1.37) k T

Notice that Eq. (1.37) does not give M explicitly in terms of T. Instead, graphical or self-consistent methods are used to determine M for a given T. A careful analysis of Eq. (1.37) shows that spontaneous magnetization occurs at low temperatures, and the material becomes saturated with M : M : Nm. However, as T increases, the magnetization gradually decreases until T approaches the Curie temperature, Nm T : A 3k

(Curie temperature).

Near the Curie temperature, the magnetization drops dramatically to zero, and as shown in Fig. 1.11 the material becomes paramagnetic. Below the Curie temperature, the material divides into numerous distinct localized regions called domains. Each domain is uniformly magnetized to its saturation value. However, the bulk magnetization can be zero if the domains are randomly oriented. Domains are discussed in more detail in Section 1.10. If we subject a ferromagnetic material to an applied ﬁeld H , the ? same analysis as in the preceding applies, but now H ; H ; H . The K ? K

21

1.6 FERROMAGNETISM

FIGURE 1.11

Spontaneous magnetization vs temperature.

magnetostatic energy is given by E : 9 m · (H ; H ) ? K : 9 m(H ; M) cos() ? and we obtain M : NmL

m(H ; M) ? . k T

(1.38)

In ferromagnetic materials M H and, therefore, the applied ﬁeld ? does not alter the local magnetization appreciably. That is, H does ? not alter the magnetization within a domain. However, an applied ﬁeld does alter the bulk magnetization of a specimen by altering the orientation and structure of individual domains. This is discussed in Section 1.11. Expression (1.38) is based on classical theory in which the magnetic moments can take on a full range of orientations. If we apply quantum

22

CHAPTER 1 Materials

theory, the orientations of the moments are restricted, and we obtain an expression similar to Eq. (1.33) with H ; H ; M , ? ? Q M :M B Q (

gJ (H ; M ) ? Q , k T

(1.39)

where M : NgJ , and B () is the Brillouin function (1.34). Because ( M H , Eq. (1.39) reduces to ? M :M B Q (

gJ M Q . k T

(1.40)

As J ; - the allowed values and orientations of the atomic moments approach a continuum, and Eq. (1.40) reduces to the classical result seen in Eq. (1.37). Although the Weiss mean ﬁeld theory explained some key features of ferromagnetism such as spontaneous magnetization below the Curie temperature, it nevertheless lacked a rigorous foundation. This would come later with the discovery of quantum mechanics [12]. Speciﬁcally, Heisenberg postulated the existence of an exchange interaction between spin angular momenta. This is characterized by a potential energy (exchange energy) of the form, w : 92J S · S GH G H : 92J S S cos() G H

(exchange energy),

(1.41)

where J is called the exchange integral, S and S are the coupled spin G H angular momenta, and is the angle between S and S . Notice G H that if J 0, the exchange energy is minimum when the spins are parallel ( : 0). However, if J 0, the exchange energy is a minimum when the spins are antiparallel ( : ). Therefore J 0 gives rise to ferromagnetic behavior, and J 0 gives rise to antiferromagnetic be havior. The exchange force falls off rapidly and is rarely effective beyond the nearest neighbors. The forementioned analysis describes the local ordering (spontaneous alignment) of atomic moments and the temperature dependence of this ordering. However, it does not describe the hysteretic behavior of bulk materials. For this, we need to understand magnetic domains and their behavior in an external ﬁeld. In the next few sections we introduce the concepts of magnetostatic energy, demagnetizing ﬁeld, and magnetic anisotropy. These are key to understanding domains. Following this, we study domains speciﬁcally and then use them to describe hysteresis.

23

1.8 DEMAGNETIZATION FIELD

1.7 MAGNETOSTATIC ENERGY In this section we discuss the magnetostatic self-energy of a magnetized specimen, and the energy it acquires in an external ﬁeld. Consider a specimen of volume V with a ﬁxed magnetization M. It possesses a magnetostatic self-energy that is given by W :9 Q 2

4

M · H dv +

(self-energy),

(1.42)

where H is the ﬁeld in the specimen due to M (i.e., due to the + continuum of dipole moments of which the specimen is composed [3]). The self-energy equation (1.42) can be derived by considering the energy required to assemble a continuum of dipole moments in the absence of an applied ﬁeld [6]. Notice that the energy density associated with W is Q w :9 M·H . (1.43) Q + 2 The self-energy equation (1.42) plays an important role in the formation of magnetic domains. We discuss this more in Section 1.10. If a magnetized specimen is subjected to an applied ﬁeld H , it ? acquires a potential energy W : 9 ?

4

M · H dv. ?

(1.44)

This can be viewed as the work required to move the specimen from an environment with zero ﬁeld to a region permeated by H . The express? ion (1.44) can be used to determine the force or torque imparted to a magnetized body by an external ﬁeld. From Eq. (1.44) we see that the energy density due to the coupling of M to H is given by ? w : 9 M · H ? ? : 9 MH cos(), (1.45) ? where is the angle between M and H . ?

1.8 DEMAGNETIZATION FIELD When a specimen is magnetized, a self-ﬁeld develops within it that opposes the magnetizing ﬁeld. This is called the demagnetization ﬁeld.

24

CHAPTER 1 Materials

We study this in some detail because it plays an important role in the magnetization process. Consider a uniformly magnetized specimen with a volume V and a surface S. Its magnetization M gives rise to surface poles (Section 3.4). These, in turn, give rise to a demagnetization ﬁeld H within the B specimen. The ﬁeld H is proportional to M, but in the opposite B direction. For example, along the x-axis we have H : 9N M , (1.46) BV V V where H and M are the x-components of H and M, respectively, and B V B N is theV demagnetizing factor along the axis. If the specimen is V subjected to an applied ﬁeld H , then H inside the specimen (denoted ? H ) is the vector sum of H and H , ? B H :H ;H . (1.47) ? B For example, consider the bar magnet shown in Fig. 1.12. Outside the magnet H and B are in the same direction (Fig. 1.12a,b). However, inside the magnet H : H , which opposes both M and B (Fig. 1.12c). B Demagnetizing factors depend on the permeability and shape of a specimen and are difﬁcult to obtain in closed-form. However, analytical expressions have been derived for ellipsoidal shapes. Ellipsoids have

Field distribution of a bar magnet: (a) H-ﬁeld; (b) B-ﬁeld; and (c) B, H, and M inside the magnet.

FIGURE 1.12

25

1.8 DEMAGNETIZATION FIELD

FIGURE 1.13

Ellipsoidal shapes: (a) general ellipsoid; and (b) prolate ellipsoid.

three demagnetizing factors N , N , and N corresponding to the three ? @ A orthogonal axes with lengths a, b, and c, respectively (Fig. 1.13a). These factors satisfy the following relation, N ;N ;N :1 (ellipsoid). (1.48) ? @ A For the special case of a sphere where a : b : c we have N : N : ? @ N : 1/3 and A H : 9M (sphere). (1.49) B Another shape of interest is the prolate spheroid with a semimajor axis c and semiminor axis a (i.e., a : b c) (Fig. 1.13b). The demagnetization factor along the c-axis is

1 k N : ln(k ; (k 9 1) 9 1 , A k 9 1 (k 9 1 where k : c/a [2]. For a long cylinder that is magnetized along its axis, the demagnetizing factors can be approximated by N : 0, and A N : N : 1/2, where the c-direction is along the axis of the cylinder. ? @ Finally, consider a magnetized specimen with H : 9NM. Its selfB energy is given by Eq. (1.42) with H : H . Speciﬁcally, we have + B W : 9 M · H dv Q B 2 4 N : M dv. (1.50) 2 4 If M is constant throughout V, then Eq. (1.50) reduces to

NMV W : . Q 2 We demonstrate some of these principles in the following example.

26

CHAPTER 1 Materials

FIGURE 1.14

Sphere with linear permeability in an external ﬁeld.

EXAMPLE 1.8.1 Determine the magnetization and dipole moment of a sphere of linear isotropic material in free space that is subjected in an external ﬁeld H (Fig. 1.14). SOLUTION 1.8.1 Let and denote the permeability and susceptibility of the K sphere, and let R denote its radius. The ﬁeld inside the sphere is H :H ;H B : H 9 M. Notice that we have used Eq. (1.49). Recall from Eq. (1.7) that M: H . K

(1.51)

Therefore,

As : / 9 1 we have K

H H : . 1 ; /3 K

3H . H : 2 ; / The magnetization inside the sphere is obtained from Eq. (1.51), M:

3( 9 ) H . ( ; 2 )

The dipole moment m is m : Volume ; M : RM.

(1.52)

(1.53)

27

1.9 ANISOTROPY

Therefore, we ﬁnd that m:

4R( 9 ) H . ( ; 2 )

(1.54) )

1.9 ANISOTROPY A material with magnetic anisotropy is one in which the magnetic properties are different in different directions. The principal classiﬁcations of anisotropy are as follows: 1. 2. 3. 4.

magnetocrystalline anisotropy; shape anisotropy; stress anisotropy; and exchange anisotropy.

A thorough discussion of these can be found in Cullity [5]. In this section we review magnetocrystalline and shape anisotropy. These play an important role in the magnetization process.

1.9.1 Magnetocrystalline anisotropy Magnetocrystalline anisotropy is an intrinsic property of a material in which the magnetization favors preferred directions (easy axes). For example, iron (Fe) has a body-centered cubic lattice structure, and there are six easy, equally preferred directions of magnetization: [0,0,1]; [0,1,0]; [1,0,0]; [0,0,91]; [0,91,0]; and [91,0,0] (Miller indices) (Fig. 1.15). The preferred alignment within a material constitutes a lower energy state and is primarily due to spin-orbit coupling. Speciﬁcally, atomic electrons couple to the crystal lattice in such a way that their orbital moments tend to align along the crystallographic axes, and because of spin-orbit coupling the spin moments tend to align as well. The most elementary form of magnetocrystalline anisotropy is uniaxial anisotropy. If a specimen with uniaxial anisotropy is polarized at an angle relative to its easy axis, the anisotropy energy density is w K sin() (J/m). (1.55)

Notice that energy minima occur when : 0° and 180°. Consider the response of a single domain specimen with uniaxial anisotropy when it is subjected to an applied ﬁeld H . Assume that H ? ?

28

FIGURE 1.15

CHAPTER 1 Materials

Crystal structure of iron and magnetization curves along various

axes.

is directed at an angle with respect to the easy axis (Fig. 1.16). We analyze the behavior of this specimen following the presentation of Stoner and Wohlfarth [13]. Assume that the specimen is magnetized to saturation with M at an angle with respect to the easy axis. Further assume that demagnetization ﬁeld H is negligible. Therefore, inside the B specimen H : H in accordance with Eq. (1.47). As the specimen ? constitutes a single domain, its energy density w() consists of two parts: the magnetostatic energy density due to the coupling of M to H as given ? by Eq. (1.45) and the magnetocrystalline anisotropy energy density w

as given by Eq. (1.55). The total energy density is w() : K sin() 9 M H cos( 9 ) Q ? : K sin() 9 M [H cos() ; H cos()], (1.56) Q V W where H : H cos() and H : H sin(). We identify the values of V ? W ?

29

1.9 ANISOTROPY

FIGURE 1.16

Specimen with uniaxial anisotropy.

that represent energy minima using the second derivative test where w()/ : 0 identiﬁes the critical points, and w()/ 0 conﬁrms a minimum. We consider the case where the ﬁeld is applied along the easy axis (H : 0) and obtain W w() : 2K sin() cos() ; M H sin() : 0, (1.57) Q V and w() : 2K cos(2) ; M H cos() 0. Q V

(1.58)

Notice that minima occur when : 0 or . When : 0 a minimum occurs when 2K H , (1.59) V M Q and we have M : M cos() : M . When : a minimum occurs when V Q Q 2K H (1.60) V M Q and we have M : 9M . The conditions (1.59) and (1.60) identify the V Q intrinsic coercivity H (or switching ﬁeld) of the specimen. Speciﬁcally, H is given by 2K H Y . M Q The intrinsic coercivity is an important property of permanent magnet materials. It gives a measure of the ﬁeld required to magnetize and demagnetize a specimen (Section 1.19). 9

30

FIGURE 1.17

CHAPTER 1 Materials

Response of a specimen with uniaxial anisotropy: (a) M vs H; and

(b) B vs H.

We apply these results to the behavior of a single domain specimen with uniaxial anisotropy. Assume that the specimen is initially magnetized along its easy axis with : 0 (M : M x ) (Fig. 1.17a). We apply a Q reversal ﬁeld in the direction : (H negative). As we gradually V increase the ﬁeld (in a negative sense), the specimen remains polarized along its easy axis (at an energy minimum) as long as Eq. (1.59) is satisﬁed (i.e., as long as 9H H 0). When the ﬁeld increases to the V

1.9 ANISOTROPY

31

point where H : 9H then Eq. (1.59) is violated and M ﬂips orientation V to : (M : 9M x ). When this happens Eq. (1.60) is satisﬁed and the Q specimen is once again in a state of minimum energy. The specimen stays in this state as H becomes more negative. V Next, we gradually decrease H (making it less negative). The V specimen stays with M : 9M x for all negative ﬁeld values. Eventually, Q H reaches zero. We then gradually increase H from zero until we reach V V H : H . At this point, Eq. (1.60) is violated and M ﬂips orientation to V : 0 (M : M x ) in order to satisfy Eq. (1.59). Once this occurs the Q specimen is again at an energy minimum. The specimen remains in this state for any further increase H . V This entire sequence of events can be summarized in graphical form by plotting M vs H (Fig. 1.17a). Notice that this plot has the form of a square loop. In particular, the specimen starts out with M : M . When Q H is reversed and reaches H : 9H , the magnetization abruptly re verses direction to M : 9M . When H is reversed again and increases Q to H : H , the magnetization abruptly reverses back to its original value M : M . Thus, except for the two points H :
1.9.2 Shape anisotropy Shape anisotropy is not an intrinsic property of a material. Rather, it is due to a geometrically induced directional dependency of the demagnet-

32

CHAPTER 1 Materials

FIGURE 1.18

Specimen with shape anisotropy.

ization ﬁeld H within the specimen. For example, consider a specimen in B the shape of` a prolate spheroid with a semimajor axis c and semiminor axis a (Fig. 1.18). Assume that the specimen is subjected to an applied ﬁeld H ? at an angle with respect to the c-axis. Further assume that the specimen is uniformly polarized to a level M with M at an angle with respect to the Q c-axis. The ﬁeld inside the specimen is H : H ; H . The energy density ? B w() consists of the self-energy density w as given by Eq. (1.43) with Q H : H , and the magnetostatic energy density due to the coupling of M + B to H as given by Eq. (1.45). Thus, the total energy density is ? 1 w() : 9 M · H 9 M · H B ? 2 : M[N sin() ; N cos()] 9 M H cos( 9 ) A Q ? 2 Q ? : MN ; K sin() 9 M [H cos() ; H cos()], (1.62) Q Q V W 2 Q A where M K : Q (N 9 N ). Q ? A 2

(1.63)

Notice that except for the leading constant term, Eq. (1.62) is formally similar to Eq. (1.56) of Section 1.9.1. We follow the same energy minimization analysis presented there except that here the c-axis plays the role of the easy axis. We ﬁnd that the intrinsic coercivity (switching ﬁeld) is given by 2K Q H Y M Q : M (N 9 N ). (1.64) Q ? A

33

1.10 DOMAINS

FIGURE 1.19

The B-H loop for a specimen with shape anisotropy.

From Eq. (1.48), and the fact that N : N for a prolate spheroid, we have ? @ 2N ; N : 1. Therefore, ? A

H M (shape anisotropy). (1.65) Q From the discussion following Eq. (1.61) we know that in this case, H : H , and the demagnetization curve is linear over only a portion of` A the second quadrant as shown in Fig. 1.19.

1.10 DOMAINS We have seen that the coupling of atomic moments in ferromagnetic materials gives rise to a spontaneous alignment of the moments below the Curie temperature (Section 1.6). Thus, one would expect ferromagnetic materials to be magnetically saturated at room temperature. However, we observe that these materials are often unmagnetized at a macroscopic level. This can be explained by the concept of magnetic domains [1, 11]. Domains typically contain 10—10 atoms. The atomic moments within a domain are subject to a magnetocrystalline anisotropy that favors spontaneous alignment along certain preferred crystallographic axes (Section 1.9). Thus, the moments within a domain align parallel to

34

CHAPTER 1 Materials

one another and the domain represents a local region of magnetic saturation. A macroscopic or bulk sample of material consists of a multitude of domains that vary in size, shape, and orientation. The magnetization of the sample is deﬁned by the collective structure and orientation of all its domains. For example, it is unmagnetized if the domains are randomly oriented, and saturated if the domains are uniformly aligned. This explains the apparent contradiction between spontaneous ferromagnetic alignment of atomic moments and the existence of macroscopically unmagnetized ferromagnetic materials. In homogeneous materials, domains form in such a way as to minimize the total energy of the specimen. For example, consider the block of material shown in Fig. 1.20. The magnetostatic energy of this block is highest in conﬁguration a, and lowest in conﬁguration c. In inhomogeneous materials the domain structure is affected by inhomogeneities such as inclusions, dislocations, crystal boundaries, internal stress, etc. Inclusions are regions with different magnetic properties than the surrounding material, and dislocations are line imperfections in the crystal lattice. Neighboring domains are separated by a transition layer called a domain wall (Fig. 1.21a). These are also called Bloch walls since Bloch

Magnetized material: (a) high energy state; (b) intermediate energy state; and (c) low energy state.

FIGURE 1.20

1.10 DOMAINS

35

Domain structure: (a) domain wall; and (b) domain wall movement in an applied ﬁeld.

FIGURE 1.21

was the ﬁrst to study their properties [14]. A domain wall has a ﬁnite thickness and an energy associated with it. The wall thickness and energy follow from an analysis of two competing effects: magnetocrystalline anisotropy that favors spin alignment along a preferred axis (thinner wall) and exchange interactions that favor parallel alignment of neighboring spins (thicker wall). The orientation of the spins within a wall varies gradually across its width so that the spins at either side of the wall are aligned with the respective adjacent domains. For example, if two neighboring domains are polarized in opposite directions, the spins in the wall between them rotate 180° across the wall from one side to the other (Fig. 1.21a). Domain walls can move in response to an applied ﬁeld. For example, if an external ﬁeld is applied parallel to the magnetization in a domain, the spins within the wall experience a torque that tends to rotate them toward the ﬁeld direction. As the spins rotate, the wall effectively moves with the aligned domain expanding and antialigned domain contracting (Fig. 1.21b).

36

CHAPTER 1 Materials

The wall motion can be impeded by the forementioned listed inhomogeneities. For example, an inclusion lowers the energy of a domain wall when they intersect. Thus, a domain wall will tend to ‘‘stick’’ to an inclusion when it contacts it. Similarly, localized regions of inhomogeneous strain (which can be caused by defects and dislocations) give rise to local energy barriers that impede the motion of a domain wall. The energy barriers are caused by the interaction of the local magnetostriction with the magnetic moments of the wall. The various impediments to the movement of a domain wall tend to make its motion jerky and discontinuous. This effect is named after Barkhausen who was the ﬁrst to observe it. Domain wall motion plays an important role in the magnetization process. We discuss this next.

1.11 HYSTERESIS For the analysis and design of permanent magnet applications we need the macroscopic or bulk magnetic properties of a material. These are expressed in graphical form by plotting the magnetic induction B or the magnetization M as a function of H. Either of these plots can be obtained from the other using the constitutive relation B : (H ; M). A typical B vs H plot is shown in Fig. 1.22. This plot is nonlinear and multivalued, reﬂecting the fact that the response of a material depends

FIGURE 1.22

Hysteresis loop.

1.11 HYSTERESIS

37

Distribution of domain orientations at various points along a hysteresis loop [4].

FIGURE 1.23

on its prior state of magnetization. The B-H plot is called a hysteresis loop. The word hysteresis is derived from the Greek word meaning ‘‘to lag.’’ The energy expended in traversing a hysteresis loop is equal to the area of the loop. Detailed discussions of this phenomenon have been given by Jiles [1] and Mayergoyz [15]. For our purposes, it sufﬁces to review the magnetization cycle plotted in Fig. 1.22. Consider a specimen with an isotropic distribution of preferred axes orientations (magnetocrystalline anisotropy). Initially, it is unmagnetized with its domains isotropically oriented as shown at point O in Fig. 1.23. We apply an H-ﬁeld, and as H increases from zero in the plus direction (weak ﬁeld), the domains aligned with H grow, and the antialigned domains shrink due to domain wall motion. As H increases further (moderate ﬁeld), the magnetic moments in the remaining unaligned domains rotate (or ﬂip) into alignment along preferred axes that are in the general direction of H. For sufﬁciently high H, the magnetic moments along the preferred axes in domains that are not strictly aligned with H rotate away from these axes and align with H (point S) (Figs. 1.22 and 1.23). At this point the material is saturated, and has achieved its saturation induction B . The segment OS is called the initial magnetizQ ation curve. As H decreases from its saturation value, the moments along preferred axes strictly aligned with H retain their orientation. However, the moments in domains with easy axes offset from H, rotate back towards the nearest easy axis, away from H. This results in a distribution of domain alignments about the plus direction. Even when H is reduced to

38

CHAPTER 1 Materials

zero, there is still a net alignment in the plus direction, and the specimen exhibits a remanent (residual) induction B (Fig. 1.23). P Now, as H is reversed to the negative direction, the moments in domains oriented along the plus direction are ﬁrst to rotate (or ﬂip) into alignment along the negative direction. As H increases further (in the negative direction), a sufﬁcient number of domains obtain a predominant negative orientation, thereby offsetting the magnetization of the remaining positively oriented domains so that the net induction in the specimen is zero, B : 0. The value H at which this occurs is called the coercivity and is denoted H , H : 9H . Notice that although B : 0 at A A this point, the distribution of domain orientations is not isotropic as it was in the initial demagnetized state (Fig. 1.23). As H increases beyond 9H (in the negative direction), a progressA ively larger fraction of the domains obtain a predominant negative orientation until saturation is reached (point S). From this point, H is decreased to zero and a fraction of the domains reverse their orientation, resulting in a remanent induction 9B . As H increases from zero in a P positive sense, a sufﬁcient number of domains obtain a predominant positive orientation, thereby offsetting the magnetization of the remaining negatively oriented domains so that the net induction in the specimen is zero. This occurs when H : H . As H increases further, the A domains once again align themselves with H until saturation is achieved. Figure 1.22 is called the major hysteresis loop. There are also minor hysteresis loops as shown in Fig. 1.24. These are formed by cycling H within the range 9H H H . A A

FIGURE 1.24

Minor hysteresis loops.

1.12 SOFT MAGNETIC MATERIALS

39

This section completes our study of the principles of magnetism. In the remaining sections we discuss various magnetic materials.

1.12 SOFT MAGNETIC MATERIALS Ferromagnetic materials are classiﬁed as either soft or hard depending on their coercivity H . Soft materials are characterized by a high permeaA bility and a low coercivity (H 1000 A/m), which makes them easy to A magnetize and demagnetize. Hard materials have a relatively low permeability and a high coercivity (H 10,000 A/m), which makes A them more difﬁcult to magnetize and demagnetize. The difference between these two materials is best illustrated by comparing their hysteresis loops. These are shown in Fig. 1.25. In this section we give a brief survey of common soft magnetic materials. More detailed treatments can be found in the texts by Jiles and Bozorth [1, 7]. Soft magnetic materials are used as ﬂux conduits to conﬁne and direct ﬂux, as ﬂux ampliﬁers to amplify the ﬂux density across a region, and as magnetic shields to shield a region from an external ﬁeld. The most commonly used soft magnet materials are soft iron, alloys of iron-silicon, nickel-iron, and soft ferrites. They can be found in numerous devices including transformers, relays, motors, inductors, and electromagnets. When selecting a soft material, the properties that are weighed are its permeability, saturation magnetization, resistance, and coercivity. A higher permeability and saturation magnetization are desired for ﬂux conﬁnement and focusing. The resistance and coercivity are important for high-frequency applications. A higher resistance

FIGURE 1.25

The B-H loops for soft and hard magnetic materials.

40

CHAPTER 1 Materials

reduces eddy currents, whereas a lower coercivity reduces hysteresis loss. Soft materials are magnetically linear over limited portions of their B-H curve where the permeability is constant (independent of H). However, they are nonlinear over other portions where is a function of H, : (H). The B-H curves for various soft materials are shown in Figs. 1.26 and 1.27.

The B-H curves for various soft magnetic materials: (a) H 400 A/m; and (b) H 400 A/m ([3] of Appendix A).

FIGURE 1.26

41

1.12 SOFT MAGNETIC MATERIALS

The B-H curves of various soft magnetic materials.

FIGURE 1.27

Various measures of permeability are used. Speciﬁcally, there is the absolute permeability :

B H

(absolute permeability),

that changes along the B-H curve. There is also the relative permeability (relative permeability), : P that is normalized to the free space value : 4 ; 10\ Wb/A · m. A plot of vs H for silicon steel is shown in Fig. 1.28. This plot shows the P nonlinear behavior of the material. There is also the differential permeability

:

dB dH

(differential permeability),

which is the slope of the tangent line to the B-H curve. The initial permeability is evaluated at the origin of the initial magnetization G curve (segment OS in Fig. 1.22), dB : G dH

. & Other measures of permeability include the initial and maximum relative permeability and , respectively. GP

P

42

CHAPTER 1 Materials

FIGURE 1.28

Relative permeability for silicon steel vs H. P

Similar deﬁnitions apply to the susceptibility of soft materials. Speciﬁcally, there are the absolute, differential and initial susceptibilities that are given by M : K H dM : K dH

(susceptibility), (differential susceptibility),

and dM : KG dH

, +&

respectively. Soft iron was the original material of choice for electromagnetic applications, and is still widely used as a core material for dc electromagnets. However, for ac applications it has been supplanted by materials with a higher resistivity because these have lower eddy current losses. Commercially available soft iron typically contains low levels of impurities such as carbon (C) (0.02%), manganese (Mn) (0.035%), sulfur (S) (0.015%), phosphor (P) (0.002%), and silicon (Si). Iron with such impurities has a coercivity on the order of 80 A/m (1 Oe), a saturation magnetization of 1.7 ; 10 A/m, and a maximum relative permeability

1.12 SOFT MAGNETIC MATERIALS

43

of 10,000. However, these properties can be improved by annealing the iron in hydrogen, which removes the impurities. Such treatment can lower the coercivity to 4 A/m (0.05 Oe), and increase the maximum relative permeability to 100,000. The properties of soft iron also degrade when it is subjected to mechanical stress, such as when it is ﬂexed or bent. However, these properties can also be restored by annealing, which relieves the internal stress. Iron-silicon alloys typically contain 2—4% silicon. The addition of silicon has the beneﬁcial effects of increasing the resistivity and reducing both magnetostriction and magnetic anisotropy. For example, an alloy with 3% silicon has a resistivity 14 cm, which is approximately four times the resistivity of pure iron. Consequently, these alloys are superior to soft iron for low-frequency applications such as transformer cores. Their increased resistivity reduces eddy current loss. Iron-silicon alloys have an additional advantage in that their reduced magnetostriction reduces acoustic noise (transformer hum). Furthermore, their reduced anisotropy gives rise to increased permeability in nonoriented alloys. Alloys with 1—3% Si are used in rotating machines, whereas alloys with 3—4% Si are used in transformers. Iron-silicon alloys are inferior to soft iron in two respects. Speciﬁcally, they have a reduced saturation induction B , and they are more brittle. This latter property limits the level of Q silicon to 4% for most applications. Iron-silicon alloys have coercivities in the range of 5—120 A/m, and a maximum relative permeability in the range of 4000 to 20,000, depending on the percentage content of the silicon [1]. Nickel-iron alloys are used for a variety of applications such as inductors, magnetic ampliﬁers, and audio frequency transformer cores. Commercially available Ni-Fe alloys contain 50—80% Ni, and are characterized by a very high permeability. These include Permalloy (78% Ni, 22% Fe) with : 8000 and : 100,000; Supermalloy (79% Ni, 16% GP

P Fe, 5% Mo) with : 100,000 and : 1,000,000; and Mumetal (77% GP

P Ni, 16% Fe, 5% Cu, 2% Cr) with : 20,000 and : 100,000 ( and GP

P GP are the initial and maximum relative permeability, respectively).

P The resistivity of Ni-Fe alloys decreases from a peak value of 72 cm for 30% Ni to 15 cm for 80% Ni [1]. Permalloy has a coercivity H : A 4.0 A/m (0.05 Oe), and a saturation magnetization M :0.86;10 A/m. Q The corresponding values for Supermalloy are H : 0.16 A/m and A M : 0.63;10 A/m. Q Finally, there are the soft ferrites. These have a cubic crystalline structure with a chemical formula of the form MO · Fe O where M is a transition metal such as iron, nickel (Ni), magnesium (Mg), manganese, or zinc (Zn). Most soft ferrites have a relatively high resistivity (up to

44

CHAPTER 1 Materials

10 m), a relatively low permeability (initial relative permeability in the range of 10 to 10,000), and a relatively low saturation magnetization (M : 4 ; 10 A/m) as compared to the soft magnetic alloys. Their low Q permeability and saturation magnetization render these materials inferior to soft magnetic alloys for dc and low-frequency applications. However, their high resistivity makes them the material of choice for high-frequency applications. Soft ferrites are sometimes categorized as either nonmicrowave ferrites (for use with audio frequencies to 500 MHz), or microwave ferrites (for use from 100 MHz to 500 GHz). Many soft ferrites have a permeability that is relatively constant up to frequencies of between 10—100 MHz. This makes them useful for broadband telecommunications applications such as specialized transformers and inductors [1, 4].

1.13 HARD MAGNETIC MATERIALS Hard magnetic materials are characterized by a low permeability and high coercivity, typically 10,000 A/m. This latter property makes them difﬁcult to magnetize and demagnetize. Such materials are referred to as permanent magnets because once magnetized they tend to remain magnetized. Permanent magnets are used as ﬁeld source components in a wide range of products including consumer electronic equipment, computers, data storage devices, electromechanical devices, telecommunications equipment, and biomedical apparatus [2, 16—18]. The properties of primary importance in the selection of a magnet are those that deﬁne the magnitude and stability of the ﬁeld that it can provide. These include the coercivity H , saturation magnetization M , A Q and remanence B , as well as the behavior of the hysteresis loop in the P second quadrant (Fig. 1.29a). This portion of the hysteresis loop is called the demagnetization curve. The points (B, H) on the demagnetization curve deﬁne an energy product BH that obtains a maximum (BH) for some point in the

interval 9H H 0 as shown in Fig. 1.29b. When a magnet is used as A a ﬁeld source it becomes biased at an operating point (B ,H ) on its K K demagnetization curve. The operating point depends on the circuit in which it is used. It can be determined from the load line of the circuit. This intersects the demagnetization curve at the operating point (B , H ) K K as shown in Fig. 1.30. The load line is discussed in Section 3.5. We show there that it is desirable to bias a magnet at the point of maximum energy (BH) . This

minimizes the volume of the magnet and reduces its cost. The energy

1.13 HARD MAGNETIC MATERIALS

FIGURE 1.29

45

Second quadrant B-H curve: (a) demagnetization curve; and (b)

B · H vs H.

FIGURE 1.30

Demagnetization curve and load line.

46

CHAPTER 1 Materials

FIGURE 1.31

Second quadrant constant energy contours.

product for a given material and circuit can be determined from second quadrant constant energy contours (Fig. 1.31). These are usually plotted on the demagnetization curves supplied by permanent magnet vendors. In the following sections we give a brief survey of various commercially available hard materials. These include ferrites, alnico, samariumcobalt (Sm-Co), neodymium-iron-boron (Nd-Fe-B) and bonded magnets. A more extensive treatment of hard materials is given by McCaig and Clegg, Parker, and Campbell [2, 16, 17].

1.14 FERRITES Hard ferrites are the least expensive and most commonly used permanent magnet material. The development of ferrite magnets dates back to the 1950s [19]. These magnets are made from ﬁne particle powders with compositions of the form XO · 6(Fe O ) where X is either barium (Ba), strontium (Sr), or lead (Pb). They are fabricated using powder metallurgical methods and are commonly referred to as ceramic magnets. The fabrication cycle starts by either wet or dry mixing the correct proportions of iron oxide (Fe O ) with the carbonate of either Ba, Sr, or Pb. The mixture is then calcined at a temperature of between 1000 and 1350 °C. The calcined material is then crushed and milled into a ﬁne powder. Isotropic magnets are formed by drying and pressing the powder into a desired shape, with subsequent sintering at a temperature in the range of 1100—1300 °C.

47

1.14 FERRITES TABLE 1.1

Properties of ceramic magnets

Ceramic $ B P

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P Density (kg/m) Curie temp (°C)

1

5

7

8

10

0.23 2300

0.38 3800

0.34 3400

0.385 3850

0.41 4100

147 1850

191 2400

259 3250

235 2950

223 2800

259 3250

203 2550

318 4000

243 3050

231 2900

8.36 1.05 1.1 4982 450

27.0 3.40 1.1 4706 450

21.9 2.75 1.1 4706 450

27.8 3.50 1.1 4706 450

31.8 4.00 1.1 4706 450

Anisotropic ferrite magnets are fabricated using a ﬁne grain powder with the grain size on the order of a single domain (:1 m). The powder is mixed with water to form a slurry, and the slurry is pressed and then sintered. An alignment ﬁeld is applied during the compression of the powder, and the water acts as a lubricant to facilitate the alignment process. A shrinkage of 15% is common during sintering. Finished magnets are produced by grinding down the sintered material. Ferrite magnets are characterized by a high coercivity and a nearly linear second quadrant B vs H curve. This is due to the dominant magnetocrystalline anisotropy of the constituent particles. Ceramic magnets are produced with a wide range of properties. These depend on various process variables such as grain size, milling sequence, and heating schedule in both the calcination and sintering phases of production. Demagnetization curves for different grades of ceramic magnets are given in Section 1.20. Various magnetic properties are listed in Table 1.1. Iron oxides are also used for various applications. The two most commonly used oxides are ferric oxide (Fe O ), also known as magnetite or lodestone, and gamma ferric oxide (Fe O ). Ferric oxide has the following properties: H : 7.16 ; 10 A/m (90 Oe); M : 4.8 ; 10 A/m; A Q density : 5000 kg/m; and T : 585°C. Some of the corresponding A properties for Fe O are H : 21 ; 10 9 32 ; 10 A/m (270—400 Oe), A : 4700 kg/m, and T : 590 °C. A

48

CHAPTER 1 Materials

1.15 ALNICO Alnico alloys date back to the early 1930s. Their principal components are iron, cobalt, nickel, and aluminum (Al) along with lesser amounts of other metals such as copper (Cu). Alnicos are hard as well as too brittle for cold working. Production methods are limited to casting the liquid alloy, or pressing and sintering the metal powders [2]. Once formed, the alloy is subjected to controlled heat treatment so as to precipitate a dispersion of ﬁne magnetic particles throughout the Al-Ni-Fe-Co matrix. These particles are elongated and rod shaped with a dominant shape anisotropy that is manifest in the relatively high coercivity of the ﬁnished magnet. Alnico magnets can be either isotropic or anisotropic, depending on whether or not the magnetic particles are oriented during their formation. These magnets exhibit a wide range of properties depending on the nature of the heat treatment, and whether the original alloy was prepared by casting or sintering. Demagnetization curves for various grades of Alnico are shown in Fig. 1.32a,b. The properties of Alnico magnets are listed in Tables 1.2 and 1.3.

TABLE 1.2

Properties of sintered Alnico magnets

Alnico (sintered) $ B P

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P Density (kg/m) Curie temp. (°C)

2

5

6

8

0.71 7100

1.05 10,500

0.94 9400

0.76 7600

43.8 550

47.8 600

62.9 790

119 1500

45.3 570

49.3 620

65.2 820

134 1690

11.9 1.5 6.4 6837 810

24 3.0 4.0 7000 900

23 2.9 4.5 6892 860

36 4.5 2.1 6975 860

1.15 ALNICO

49

Demagnetization curves for Alnico magnets: (a) Alnico 1, 2, 3, and 4; (b) Alnico 6, 7, and 8 [1].

FIGURE 1.32

50 TABLE 1.3

CHAPTER 1 Materials Properties of cast Alnico magnets

Alnico (cast) $ B P

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P Density (kg/m) Curie temp. (°C)

2

5

6

8

0.75 7500

1.24 12,400

1.5 10,500

0.82 8200

44.6 560

50.9 640

62.1 780

131.3 1650

46.1 580

54.1 680

63.7 800

148 1860

13.5 1.7 6.4 7086 810

43.8 5.5 4.3 7308 900

31.0 3.9 5.3 7418 860

42.2 5.3 2.0 7252 860

From these tables we see that Alnico magnets have impressive magnetic properties. Unfortunately, they also have poor physical properties. As already noted, they are extremely hard and brittle. Thus, production of ﬁnished magnets with tight tolerances requires tedious and costly machining.

1.16 SAMARIUM-COBALT The development of samarium-cobalt (SmCo) magnets began in the 1960s [20]. Samarium is one of the rare-earth elements that form a transition group with atomic numbers ranging from 58 (Ce) to 71 (Lu). The development of SmCo resulted from research that was directed toward the formation of alloys of the rare-earth elements with the transition series ferromagnets: iron, cobalt, and nickel. The two most common SmCo materials are SmCo and Sm Co . The ﬁrst samarium-cobalt magnets were made by bonding SmCo powder in a resin. Subsequent production entailed the formation of the base alloy using either a reduction/melt or reduction/diffusion process. In the reduction/melt process Sm and Co are mixed and induction melted to form the alloy. The cast alloy is brittle, and readily ground into a ﬁne grain powder.

1.17 NEODYMIUM-IRON-BORON

51

In the reduction/diffusion process samarium oxide (Sm O ) and cobalt powder are reacted with calcium (Ca) at approximately 1150 °C to form a compound: 10Co ; Sm O ; 3Ca ; 2SmCo ; 3CaO. The CaO is separated from the compound via a sequence of steps starting with a reaction with water, followed by a gravimetric separation of the hydroxide, followed by an acid rinse, and then drying. Once in powder form, powder metallurgical methods are used to form the desired magnet shape. Samarium-cobalt magnets are characterized by high coercivities and nearly linear second quadrant demagnetization curves. This suggests that the dominant magnetic mechanism for this material is magnetocrystalline anisotropy (Section 1.9). However, the grains in SmCo powders are typically 5—10 m in diameter, which is approximately an order of magnitude larger than single domain size. This relatively large grain size makes it energetically favorable for the formation of domain walls within each grain. Thus, each grain may comprise multiple domains. In SmCo , the domain walls move with relative ease within each grain. The high coercivity of these magnets can be explained by the limited nucleation sites for domain walls, and the pinning of domain walls at the grain boundaries that hinders the growth of domains from grain to grain. In Sm Co , the high coercivity is attributable to the pinning of domain walls. The magnetic properties of SmCo and Sm Co are given in Table 1.4. Demagnetization curves for these materials can be found in Section 1.20.

1.17 NEODYMIUM-IRON-BORON The development of neodymium-iron-boron (NdFeB) followed that of SmCo and dates back to the early 1980s [21, 22]. The initial motivation for its development was a desire for a cost effective alternative to SmCo. This was based on concern over the cost and availability of cobalt. Also, neodymium is much more abundant than samarium. The NdFeB powders are produced using various methods including a reduction/diffusion process, and a rapid quenching (melt spinning) method. The ﬁnished magnets can be either isotropic or anisotropic [17]. The reduction/diffusion process is similar to that of SmCo (Section 1.16). The ﬁnished magnets are produced from the powder via the steps of pressing and sintering, followed by heat treatment, and then grinding

52

CHAPTER 1 Materials TABLE 1.4

B P

Properties of samarium-cobalt magnets

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P Density (kg/m) Curie temp. (°C)

SmCo5

Sm2Co17

0.83 8300

1.0 10,000

600 7500

480 6000

1440 18,100

558 7000

128 16 1.05—1.1 8200 700

192 24 1.05 8100 750

into the desired shape. The powder can be aligned with a bias ﬁeld as it is pressed. The ﬁnished magnets have a high coercivity that is attributed to domain wall pinning at the grain boundaries. The rapid quenching or melt spinning method is commonly referred to as Magnequench [23, 24]. In this process the NdFeB alloy is melted and forced under argon (Ar) pressure through a small oriﬁce onto the surface of a water-cooled revolving metal wheel. This yields a rapidly quenched thin ribbon of the alloy 35 m thick and 1—3 mm wide. This material has a very ﬁne microcrystalline structure consisting of randomly oriented single domain NdFeB grains with a diameter on the order of 0.030 m. The ribbon is milled into thin platelets 200 m across and 35 m thick. There are three grades of Magnequench known as MQ I, MQ II, and MQ III. In the MQ I process the milled platelets are annealed and then blended with an epoxy resin. The compound is pressed into a desired shape and then oven cured. The ﬁnished magnets are isotropic due to the random orientation of the NdFeB grains within the platelets. In the MQ II process the magnets are formed by hot-pressing the platelets at 700 °C. This results in a nearly fully densiﬁed magnet that is also isotropic. In the MQ III process magnets are formed in a two-step sequence. First, the milled platelets are hot-pressed as in the MQ II process. Next,

53

1.18 BONDED MAGNETS TABLE 1.5

B P

Properties of Magnequench

(T) (G)

H A (kAm\) (kOe) H (kAm\) (kOe) (BH)

(kJm\) (MG Oe) Density (kg/m) Curie temp. (°C) Maximum operat. temp. (°C) P

MQ I

MQ II

MQ III

0.61 6100

0.80 8000

1.18 11,800

424 5.3

520 6.5

840 10.5

1200 15.0

1280 16.0

1040 13.0

64 8 6000 312 125 1.15

104 13 7500 312 150 1.15

256 32 7500 312 150 1.05

the hot-pressed material is subjected to a second hot-pressing in a die with a larger cross section. This process compresses the height by close to 50%, and is known as die-upsetting [25]. As the material thins, internal shear stresses develop at the granular level, and these tend to align the preferred magnetic axes parallel to the pressing direction. Thus, the ﬁnished magnet is anisotropic. The properties of the Magnequench materials are given in Table 1.5.

1.18 BONDED MAGNETS Permanent magnets made from fully dense materials tend to be hard and brittle. Consequently, substantial machining is required to form these materials into ﬁnished magnets with precession tolerances. A cost-effective solution to this problem is the use of bonded magnets. These magnets are fabricated by blending a magnetic powder with a binder material to form a compound, and then molding or extruding the compound into a desired shape. Typical binder materials include rubber, resins, and plastics. Bonded magnets are formed using compressionmolding, injection-molding, and extrusion. Compression-molded magnets are made using a thermosetting binder such as epoxy resin, whereas

54

CHAPTER 1 Materials

injection-molded magnets are formed using a thermoplastic binder such as nylon. Extruded magnets are produced using an elastomer such as rubber. Bonded magnets can be made rigid or ﬂexible depending on the volume fraction of powder used. Typical volume fractions are 80—85% for compression-molding, 60—65% for injection-molding, and 55—60% for extrusion [17]. The most common bonded magnets are ferrite based. These are usually fabricated using a ﬁne powder (1-m particle diameter) made from sintered BaFe O . The particles in this powder form as thin platelets with their planes perpendicular to the magnetic easy axis. Isotropic grades of these magnets can be produced by injection-molding the powder/binder compound. Typical energy products for these magnets are on the order of (BH) : 4 kJ/m (0.5 MG Oe).

Anisotropic magnets can be produced using an injection-molding or extrusion process with an alignment ﬁeld applied during the formation of the magnet. In the injection molding process, the alignment ﬁeld is applied across the mold cavity while the powder/binder compound is still ﬂuid. The alignment ﬁeld can be provided by coils embedded in the mold itself. The energy product of anisotropic injection-molded ferrite magnets is typically (BH) : 14 kJ/m (1.76 MG Oe).

A lesser degree of alignment can be achieved by applying mechanical pressure to the powder/binder compound during the formation of the magnet. The pressure tends to orient the platelets so that they face the pressing direction. Such mechanical alignment is sometimes implemented in the extrusion process. Magnetic properties of various grades of bonded ferrite magnets are listed in Table 1.6 [16]. Bonded magnets are also made using samarium-cobalt materials SmCo and Sm Co . Such magnets have reduced performance relative to their sintered counterparts, but they are far less brittle and more easily machined. SmCo powder lends itself to the injection-molding and extrusion processes where gate sizes limit the particle sizes to 5—10 m. Energy products on the order of (BH) : 60 kJ/m (7.5 MG Oe) are

achieved with these magnets. The Sm Co particles suffer a reduction in magnetic properties if their diameter is reduced much below 40 m [26]. This material is best suited for compression molding where energy products on the order of (BH) : 130 kJ/m (16 MG Oe) are achieved.

Additional properties of bonded SmCo magnets are given in Table 1.7 [2]. Bonded magnets can also be made using NdFeB powder. The Magnequench MQ I material is one example of this (Section 1.17). This section completes our survey of permanent magnet materials. In the remaining sections we discuss the magnetization and stability of magnets.

55

1.19 MAGNETIZATION

Properties of bonded ferrite magnets (isotropic (i), anisotropic (a))

TABLE 1.6

Bonded ferrite $ B P

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P

TABLE 1.7

Flexible (i)

Flexible (a)

Rigid (i)

Rigid (a)

0.17 1700

0.25 2500

0.14 1400

0.30 3000

127 1600

175 2200

84 1050

191 2400

239 3000

239 3000

— —

223 2800

5.57 0.7 1.05

11.9 1.5 1.05

3.18 0.4 1.05

15.9 2.0 1.05

Properties of bonded samarium-cobalt magnets

Bonded SmCo $ B P

(T) (G)

H A (kAm\) (Oe) H (kAm\) (Oe) (BH)

(kJm\) (MG Oe) P

SmCo5

Sm2Co17

0.65 6500

0.86 8600

460 5750

497 6250

620 7790

800 10,050

80 10 1.1

130 16.2 1.1

1.19 MAGNETIZATION Permanent magnet materials are of little use until they are magnetized. In this section we discuss the magnetization process. It is widely held that the ﬁeld H required to saturate a magnet is 3—5 times its intrinsic Q

56

CHAPTER 1 Materials Magnetizing ﬁelds of various permanent magnet materials

TABLE 1.8

Hs Magnetizing ﬁeld $ Ceramic ferrite Alnico (2,5,6) Alnico (8,9) SmCo Sm Co NdFeB MQ 1—MQ II MQ III

kA/m

Oe

796 199—318 636 1193 1989

10,000 2500—4000 8000 15,000 25,000

2784 1989

35,000 25,000

coercivity H . Here, H is the ﬁeld inside the material. The applied ﬁeld Q H must be greater than H to compensate for the internal demagnetiz? Q ation ﬁeld H (Section 1.8). Speciﬁcally, B H :H ;H , Q ? B or H :H 9H . ? Q B Recall that H : 9NM where N is the (geometry dependent) demagnetB ization factor and M is the magnetization. For a saturated magnet H : 9NM . Therefore, the (external) ﬁeld required to saturate the B Q magnet is H : H ; NM . ? Q Q The values of H for various materials are listed in Table 1.8. Q Conventional materials such as Alnico and hard ferrites have lower values of H and can be magnetized with ﬁelds H 8.0 ; 10 A/m. Q Such ﬁelds can be obtained by passing direct currents through conventional solenoids (Fig. 1.33). Magnetizing ﬁxtures are also used to magnetize conventional materials. A typical ﬁxture consists of a coil wrapped around a soft magnetic core that has a gap in which the magnet is placed (Fig. 1.34). To polarize the magnet, a direct current is applied to the coil and this generates a ﬁeld through the core. The core guides and focuses the ﬁeld across the magnet to produce the desired polarization pattern. High-grade core materials are available that are capable of carrying ﬂux densities of up to 2.0 T, which is sufﬁcient for the

57

1.19 MAGNETIZATION

FIGURE 1.33

Magnetization of a bar magnet using a solenoid.

magnetization of low coercivity materials (i.e., B : 2 T ; H : B/ : 16 ; 10 A/m H ). Q The design of such ﬁxtures entails the following steps: (a) estimate the demagnetization ﬁeld H in the magnet assuming it is saturated; (b) B compute the applied ﬁeld H : H 9 H required to saturate the magnet; ? Q B and (c) determine the current required to generate H assuming the ? magnet is an air gap. We demonstrate this procedure using the ﬁxture shown in Fig. 1.34. This circuit (without the coil) is analyzed in Example 3.5.3. If the circuit is ideal ( -), and there is no leakage, then the demagnetization ﬁeld in the magnet is essentially zero, H : 0. ThereB fore, to saturate the magnet the applied ﬁeld must equal the saturation ﬁeld H : H . To determine H we assume that the permeability of the ? Q ? magnet is . Then, the magnet itself represents an air gap, and the ﬁeld through it is given by ni B: , l K

or

H:

FIGURE 1.34

ni , l K

Conventional magnetizing ﬁxture.

(1.66)

58

CHAPTER 1 Materials

FIGURE 1.35

Components of an impulse magnetizing system.

where n is the number of turns, i is the current, and l is the length of K the magnet (height of the gap) (Example 3.5.1). Thus, to saturate the magnet we choose ni : l H so that H : H . K Q Q While conventional magnetization methods are appropriate for low energy magnets, they are impractical for rare-earth materials. These require magnetizing ﬁelds on the order of H : 2 ; 10 A/m (25,000 Oe), Q and the current required to generate such ﬁelds cannot be sustained without overheating and damaging the conductors. The nominal limit of current density for continual operation is generally taken to be 1.5 to 2 A/mm Fortunately, the magnetizing ﬁeld needs to be applied only momentarily. Therefore, rare-earth magnets can be magnetized using impulse magnetizers that provide a high current pulse of short duration. Such magnetizers are relatively inexpensive and can operate from standard power lines. A typical impulse magnetizing system consists of a charging circuit, a capacitor bank for energy storage, a thyristor or ignitron tube for switching the current, a step-down transformer, and a magnetizing ﬁxture (Fig. 1.35). An equivalent electrical circuit for a basic magnetizing system (without the transformer) is shown in Fig. 1.36. The charging circuit consists of a voltage source V and a resistor R . The ignitron tube BA is represented by a toggle switch, and the capacitor bank is represented by capacitance C. The magnetizing ﬁxture is represented as an RL load.

FIGURE 1.36

Equivalent circuit of an impulse magnetizing system.

59

1.19 MAGNETIZATION

The magnetization process entails two distinct steps. In the ﬁrst step, the capacitor bank is slowly charged to a predeﬁned voltage. This corresponds to the switch being set to position (1) in Fig. 1.36. The switch is held in this position until the capacitor bank C is charged to the voltage V . The duration of the charging phase is determined by the resistance R . In the second step, the ignitron tube is activated, which in effect toggles the switch to position (2). The current from the capacitor is discharged into the ﬁxture in the form of a single high-current pulse. This produces a peak ﬁeld of short duration across the magnet that is sufﬁcient to magnetize it. Impulse magnetizing ﬁxtures typically consist of a coil wound on a nonmagnetic support structure that houses the magnet. There is usually little beneﬁt in using a soft magnetic core because it would saturate at a level below the required magnetizing ﬁeld. Moreover, such cores are conductive and can therefore generate eddy currents that oppose the magnetizing ﬁeld. Without such a core, a ﬁxture is magnetically linear and the magnetizing ﬁeld that it generates scales linearly with the applied current. It is instructive to study the electrical response of an impulse magnetizing system. Consider the circuit of Fig. 1.36. After the capacitor is charged to the voltage V , the switch is moved to position (2), which sends current to the ﬁxture. The current is governed by the equation di R di i ; ; : 0, dt L dt LC

(1.67)

with initial conditions i(0) : 0 and di/dt : V /L. The solution to Eq. R (1.67) is given by V i(t) : 9 [e\RO 9 e\RO] 2L

(1.68)

1 : , R/2L ;

(1.69)

1 : , R/2L 9

(1.70)

where

and : ((R/2L) 9 1/LC. Notice that . Initially, the current surges through the ﬁxture and rises with a time constant . After reaching a peak value, the current decays with a time constant . The system is critically damped when : 0, and overdamped when 0.

60

CHAPTER 1 Materials

Thus, the system is overdamped when R2

L C

and the current through the ﬁxture will be unidirectional without oscillations. This is desired for magnetization. A ﬁxture needs to be designed with appropriate values of R and L so that sufﬁcient energy with be transferred from the capacitor bank to the magnetizing ﬁeld. To saturate a magnet, the ﬁxture must provide a ﬁeld H throughout the magnet. Let E denote the magnetostatic Q

energy required to saturate the magnet. We estimate E using Eq.

(3.66), E

: H ; (volume of the magnet), ?

where H 3H , and is the permeability of the unmagnetized material ? ( for rare-earth materials). The energy stored in an impulse mag netizer is E : 1/2CV when its capacitor bank is fully charged. The energy dissipated through the resistor R is E : 1/2R i dt, where i is given by Eq. (1.68). Thus V , C, R, and L are chosen so that E E

; E . Moreover, the ﬁxture needs to be designed to render the desired magnetization pattern throughout the magnet. The design of an impulse magnetizing ﬁxture is straightforward [27]. First, a coil structure is chosen that will render the desired magnetization pattern within the magnet. Next, an initial wire gauge is selected for the coil. Third, an analysis is performed to determine the resistance R and inductance L of the coil as well as the average B-ﬁeld that it generates across the magnet per ampere of applied current. This latter value is referred to as the tesla per amp ratio, or TPA. This ratio is used in a subsequent analysis to determine the level of current required to saturate the magnet, B : H . Q The coil resistance R is easy to determine given the length and gauge of the wire. However, the inductance and TPA calculations are more complicated. The topology of the coil often precludes a simple analytical evaluation of these values. However, they can be determined using numerical methods such as ﬁnite-element analysis (FEA) (Section 3.8). The FEA for ﬁxture design is essentially a steady-state magnetostatic analysis. Brieﬂy, the ﬁnite-element method amounts to dividing the ﬁxture geometry and surrounding region into small elements (collectively known as a mesh) and assigning to each element a polynomial that will approximate the behavior of the magnetic potential in that element.

61

1.19 MAGNETIZATION

The coefﬁcients of these polynomials are arranged in a matrix, and their values are obtained via the solution of a matrix equation that contains all the information of the problem, including the location of each element relative to its neighbors, the underlying ﬁeld equations, and the boundary conditions. This is discussed in detail in Section 3.8. Once R, L, and TPA are known, a circuit analysis is performed to determine the current through the ﬁxture and the ﬁeld that develops across the magnet. A thermal analysis is also performed to determine the temperature rise in the ﬁxture. The temperature T is predicted using the formula T(t) :

1 R J dt, c

where , c , and are the density, constant-volume speciﬁc heat, and conductivity of the conductor, and J is the current density through it. For copper at room temperature, : 8950 kg/m, c : 383 J/(kg °C), and : 5.8 ; 10 mho/m. The calculations are performed for a series of wire gauges and an optimum wire gauge is selected based on achieving the appropriate ﬁeld strength, the sharpest pole transitions, and a tolerable ﬁxture temperature (220 °C). We demonstrate the design procedure via a simple example. Consider the bipolar disk magnet shown in Fig. 1.37. Such magnets are typically used for clamping or holding applications. Assume that the magnet is made from NdFeB MQ I material that requires a magnetizing ﬁeld H : 2785 k Am\. This corresponds to a saturation ﬂux density Q B : 3.5 T (Table 1.8). The design starts with a wire conﬁguration design such as the one shown in Fig. 1.38. The wires are arranged so as to induce the desired bipolar magnetization across the magnet. They are connected at their terminals to form a continuous circuit. Several magnets are to be magnetized at once and therefore the length of the ﬁxture is much greater than its width. We perform a two-dimensional (2D) FEA of the cross-sectional geometry of the wires. This cross section is shown

FIGURE 1.37

Bipolar disk magnet.

62

CHAPTER 1 Materials

FIGURE 1.38

Wire conﬁguration for magnetizing a bipolar disk magnet.

in Fig. 1.39a. The ﬁrst step is the generation of the mesh (Fig. 1.39b). Once the geometry is meshed, a nominal value of current is deﬁned for the wire regions and the appropriate boundary conditions are imposed. In this case the ﬁeld is set to zero far from the ﬁxture (Dirichlet condition). The ﬁeld problem is then solved, and plots of the ﬂux lines and the B-ﬁeld are obtained (Fig. 1.40a,b). A ﬂux line is a line that indicates the direction of the B-ﬁeld at any point along its length (Section 3.2.1). The inductance L and TPA are obtained from this analysis. The TPA is computed by dividing the ﬁeld generated at the center of a pole by the number of amps of applied current. At this point V , R , R, L, and TPA are known. A circuit analysis can now be performed to determine the optimum wire gauge. A sample output of an analysis is shown in Fig. 1.41. The vertical axis has been divided into 10 parts, and the top of the scale corresponds to the values B : 5.0 T, I : 50,000 A, and TEMP : 200 °C. The negative swing

in current is prevented in practice and therefore only the positive half of the cycle is shown. The analysis shows that a peak ﬁeld of 4.5 T is achieved with a maximum current of 25,000 A. The operating temperature of the coil reaches 140 °C, which is acceptable since the maximum working temperature for common grades of insulation is typically 200 °C.

1.19 MAGNETIZATION

FIGURE 1.39

FEA mesh.

63

Fixture analysis: (a) cross section of the wire conﬁguration; and (b)

64

CHAPTER 1 Materials

FIGURE 1.40

Fixture analysis: (a) ﬂux lines; and (b) B-ﬁeld.

65

1.19 MAGNETIZATION

FIGURE 1.41

Analysis of the magnetization process.

As a further example, consider the multipole disk magnet shown in Fig. 1.42. Eight north-south alternating magnetic poles are arranged in pie-shaped zones with vertical magnetization over each zone. This type of magnet is used in axial-ﬁeld permanent magnet motors (Section 5.13). A ﬁxture for this magnet is shown in Fig. 1.43. The coil is supported in a nonmagnetic structure and is wound to provide the desired magnetization. The resistance of the coil is easily calculated once an initial wire gauge is speciﬁed. The inductance and TPA are calculated using FEA. Once these have been determined, parametric circuit calculations are performed to determine the optimum wire gauge.

FIGURE 1.42

Multipole disk magnet.

66

FIGURE 1.43

CHAPTER 1 Materials

Exploded view of a magnetizing ﬁxture for an eight-pole disk

magnet.

1.20 STABILITY The operating point of a magnet depends on its load line as well as its demagnetization curve. These, in turn, depend on the operating environment. A good design must account for variations in the environment so that the performance of the magnet remains within an acceptable range throughout its working life. This is the issue of stability. A magnet’s load line is a function of the magnetic circuit used. Thus, load line stability needs to be considered on a case-by-case basis. However, the stability of the demagnetization curve depends on factors that are common to all applications. In this section we brieﬂy review and summarize these factors. A thorough discussion of this subject is given by McCaig and Clegg [2].

67

1.20 STABILITY

The magnetic properties and hence demagnetization curve of a magnet depend on many variables including temperature, pressure, and the applied ﬁeld. The magnetic properties change due to variations in these variables, and such changes can be classiﬁed as reversible, irreversible, or structural. Reversible changes occur when the magnetic properties change in response to a variation in a variable, but return to their original state when the variable returns to its initial value. For example, a magnet’s hysteresis loop changes with temperature, and for a limited range of temperatures these changes are reversible and essentially linear. Such changes can be described by reversible temperature coefﬁcients for the remanence B , coercivity H , and intrinsic coercivity H . The coefﬁcients P A are expressed as a percentage change per °C. All magnets suffer a reduction in B as temperature increases. However, the change in H can P A be positive or negative depending on the dominant anisotropic mechanism of the material. Most magnets exhibit a decrease in H with A temperature (a negative coefﬁcient for H ). However, ceramic magnets A have a positive coefﬁcient of 0.02%/°C. The designer must account for these changes to ensure that a magnet’s operating point remains within an acceptable tolerance over the range of operating temperatures. The demagnetization curves for ceramic and samarium cobalt materials at various temperatures are shown in Figs. 1.44 and 1.45, respectively. The temperature coefﬁcients for various materials are listed in Table 1.9 [16]. In an irreversible change in magnetization, the magnetic properties change in response to a variation in a variable, and remain changed even when the variable returns to its initial value. To restore the original properties the magnet must be remagnetized. For example, a ferromagnetic specimen undergoes an irreversible change when its temperature is elevated above the Curie temperature (Section 1.6). Once above the

TABLE 1.9

Temperature coefﬁcients of various permanent magnet materials

Maximum service temp. (°C) Curie temp. (°C) B P Reversible coefﬁcient (%/°C) H Reversible coefﬁcient (%/°C)

Alnico 5

Ferrite

SmCo5

NdFeB

520

400

250

150

720

450

725

310

90.02

90.20

90.04

90.12

90.03

;0.40

90.30

90.60

68

CHAPTER 1 Materials

Demagnetization curves of ceramic magnets (dashed lines : M): (a) ceramic 5; (b) ceramic 7; and (c) ceramic 8 [16].

FIGURE 1.44

1.20 STABILITY

69

Demagnetization curves of sintered SmCo materials (dashed lines : M): (a) SmCo ; and (b) Sm Co [17].

FIGURE 1.45

70

CHAPTER 1 Materials

Curie temperature, the specimen becomes paramagnetic. Its domains retain a random orientation as it cools, which renders it unmagnetized. The specimen can be remagnetized to its original state provided that its metallurgical structure has not been altered. Elevated temperatures can also cause changes in domain structure. In fact, immediately after magnetization a specimen undergoes a relatively slow process in which its unstable domains relax to a lower energy state via a variety of mechanisms including domain boundary movement. Such irreversible effects are collectively known as magnetic viscosity. This accelerates with temperature. Another example of an irreversible change is when an external demagnetizing ﬁeld H is applied that exceeds the intrinsic coercivity H of the specimen. The magnetization irreversibly ﬂips to the opposite direction. Again, the specimen can be remagnetized to its original state. Structural changes entail a permanent change in the metallurgical state of a magnet. Such changes cannot be undone by remagnetization. Corrosion and oxidation are examples of structural changes. Samarium cobalt, ceramic and Alnico magnets are resistant to oxidation, but rare-earth magnets are not. Oxidation is especially severe in sintered Nd Fe B. Oxygen diffuses into this material at elevated temperatures, causing an oxidized layer to form. The layer thickness d grows according to the relation d : f(T)(t, where f(T) represents a nonlinear temperature dependence. The oxidized layer has a lower intrinsic coercivity than the body of the magnet, and is more easily demagnetized by the magnet’s internal ﬁeld. This layer degrades performance because: (a) it reduces the working volume of the magnet; and (b) it shunts the ﬁeld from the magnet’s interior. Oxidation can be controlled by varying the composition of the material, and/or coating the ﬁnished magnet. Common coating materials include nickel, zinc (Zn), and aluminum (Al) with thicknesses of 10—20 m, or epoxy with a thickness of 20—30 m. References 1. Jiles, D. (1991). Introduction to Magnetism and Magnetic Materials, London: Chapman and Hall. 2. McCaig, M. and Clegg, A. G. (1987). Permanent Magnets in Theory and Practice, 2nd ed., New York: John Wiley and Sons. 3. Craik, D. (1995). Magnetism: Principles and Applications, New York: John Wiley and Sons.

1.20 STABILITY

71

4. Chikazumi, S. and Charap, S. (1964). Physics of Magnetism, New York: John Wiley and Sons. 5. Cullity, B. D. (1972). Introduction to Magnetic Materials, Reading, MA: Addison-Wesley. 6. Morrish, A. H. (1983). The Physical Principles of Magnetism, Malabar, FL: R. Krieger Publishing Co. 7. Bozorth, R. M. (1993). Ferromagnetism, New York: IEEE Press. 8. Messiah, A. (1961). Quantum Mechanics, Amsterdam: North Holland. 9. Alonzo, M. and Finn, E. J. (1968). Fundamentals of University Physics, vol. III: Quantum and Statistical Physics, Reading, MA: Addison-Wesley. 10. Langevin, P. (1905). Annales de Chem. et Phys. 5: 70. 11. Weiss, P. (1907). L’hypothèse du champ moléculaire et la propriété ferromagnétique, J. de Phys. 6: 661. 12. Heisenberg, W. (1928). Zur theorie des ferromagnetismus, Z. Phys. 49: 619. 13. Stoner, E. C. and Wohlfarth, E. P. (1948). A mechanism of magnetic hysteresis in heterogeneous alloys, Phil. Trans. Roy. Soc. A240: 599. 14. Bloch, F. (1932). Zur theorie des austauschproblems und der remanenzerscheinung der ferromagnetika, Z. Phys. 74: 295. 15. Mayergoyz, I. D. (1991). Mathematical Models of Hysteresis, New York: Springer-Verlag. 16. Parker, R. (1990). Advances in Permanent Magnetism, 2nd ed., New York: John Wiley and Sons. 17. Campbell, P. (1994). Permanent Magnet Materials and their Application, Cambridge: Cambridge Univ. Press. 18. Watson, J. K. (1980). Applications of Magnetism, New York: John Wiley and Sons. 19. Went, J. J., Rathenau, G. W., Gorter, E. W., and Van Oosterhout, G. W. (1952). Ferroxdure, a class of new permanent magnet materials, Philips Tech. Rev. 13: 194. 20. Strnat, K., Hoffer, G., Olson, J., Ostertag, W., and Becker, J. J. (1967). A family of new cobalt based permanent magnetic materials, J. Appl. Phys. 38: 1001. 21. Sagawa, M., Fujimura, S., Togawa, N., Yamamoto, H., and Matsuura, Y. (1984). New material for permanent magnets on a base of Nd and Fe, J. Appl. Phys. 55: 2083. 22. Sagawa, M., Horosawa, S., Yamamoto, H., Matsuura, Y., Fujimura, S., Tokuhara, H., and Hiraga, K. (1986). Magnetic properties of the bcc phase at grain boundaries in the Nd-Fe-B permanent magnet, IEEE Trans. Magn. 22: 910. 23. Croat, J., Herbst, J. F., Lee, R. W., and Pinkerton, F. E. (1984). Pr-Fe and Nd-Fe-based materials: a new class of high-performance permanent magnets, J. Appl. Phys. 55: 2078.

72

CHAPTER 1 Materials

24. Lee, R. W. (1985). Hot-pressed neodymium-iron-boron magnets, Appl. Phys. Lett. 46: 790. 25. Lee, R. W., Brewer, E. G., and Schaffel, N. A. (1985). Processing of neodymium-iron-boron melt spun ribbons to fully dense magnets, IEEE Trans. Magn. 21: 1958. 26. Satoh, K., Oka, K., Ishii, J., and Satoh, T. (1985). Thermoplastic resin-bonded Sm-Co magnet, IEEE Trans. Magn. 21: 1979. 27. Lee, J. K. and Furlani, E. P. (1990). The optimization of multipole magnetizing ﬁxtures for high-energy magnets, J. Appl. Phys. 67(3).

CHAPTER

2

Review of Maxwell’s Equations

2.1 INTRODUCTION Maxwell’s equations govern the interaction of charged matter, and the behavior of electromagnetic ﬁelds. They provide a fundamental understanding of a wide range of phenomena including the magnetic interactions that are of interest to us. In this chapter, we review electromagnetic ﬁeld theory and Maxwell’s equations. This review is not intended to provide a working knowledge of the theory because it is assumed that the reader already possesses such knowledge. Instead, it is written to provide an overview of the theory, with an emphasis on the reduction and simpliﬁcation of the ﬁeld equations for the applications that we intend to study. We start with Maxwell’s equations in their most general form. These are presented in both differential and integral form along with constitutive relations and boundary conditions. We then introduce the scalar and vector potentials, and show that they provide an alternate, and often more tractable formulation of electromagnetic ﬁeld theory. Next, we specialize to the case of quasi-static ﬁeld theory. Here, there is a partial uncoupling of the ﬁeld equations that simpliﬁes their solution. Finally, we study static ﬁeld theory in which all time dependence is negligible. We ﬁnd that the magnetic and electric ﬁelds uncouple into separate magnetostatic and electrostatic ﬁeld equations. Permanent magnet and electromechanical devices are analyzed using magnetostatic and quasistatic ﬁeld theory, respectively. We study the methods and techniques for analyzing such devices in Chapters 3 and 5, respectively. Throughout this text we use SI units in the Sommerfeld convention. These and other units are discussed in Appendix D.

73

74

CHAPTER 2 Review of Maxwell’s Equations

2.2 MAXWELL’S EQUATIONS Electromagnetic ﬁelds arise from sources of charge and current, and are governed by Maxwell’s equations. We begin with the ﬁeld equations in differential form, Maxwell’s Equations ;H:J;

D t

·B:0 ;E:9

(2.1) (2.2)

B t

· D : .

(2.3) (2.4)

In these equations J (A/m) and (C/m) are source terms, and H, B, E, and D are the ﬁelds. These are deﬁned as follows: Field E D H B

Description Electric field intensity (V/m) Electric flux density (C/m) Magnetic field intensity (A/m) Magnetic flux density (T)

(2.5)

In deﬁnition (2.5), V/m stands for volt/meter, C/m stands for Coulomb/meter, A/m stands for ampere/meter, and T stands for tesla. The ﬁelds (2.5) are vector-valued functions of space and time, for example, B : B (x, y, z, t)x ; B (x, y, z, t)y ; B (x, y, z, t)z . V W X As each of the four ﬁelds has three components, the ﬁeld equations represent a system of equations for 12 unknown ﬁeld components. However, while these equations are consistent, they are not all independent. Therefore, they do not provide the 12 independent scalar equations required to determine the 12 unknowns [1, 2]. Speciﬁcally, the two divergence equations can be obtained from the two curl equations if one imposes the continuity of charge equation (2.6). Thus, for a complete

75

2.2 MAXWELL’S EQUATIONS

theory the ﬁeld equations must be augmented by additional independent equations that take the form of constitutive relations. These are discussed in more detail in Section 2.2.1. The source terms in the ﬁeld equations are the free current density J (A/m), and the free charge density (C/m) [2]. Free charge is charge that has been separated from its atomic or molecular source and is capable of moving through a material. The free current density J is due to the motion of free charge. It accounts for such phenomena as the motion of conduction electrons in conductors (conduction current), the motion of conduction electrons and/or holes in semiconductors (conduction current), the migration of positive or negative ions through liquids (electrolytic current), and the motion of electrons or ions in a vacuum (convection current) [3]. The ﬁeld equations embody the conservation of free charge as expressed by the continuity equation, ·J;

: 0. t

(2.6)

This is obtained by combining the divergence of Eq. (2.1) with the time derivative of Eq. (2.4). On the other hand, if Eq. (2.6) is viewed as an independent equation, we ﬁnd that the two divergence equations (2.2) and (2.4) can be obtained from the two curl equations (2.1) and (2.3) as noted in the preceeding text. In addition to free charge there is also bound charge. Bound charge is charge that is displaced from its atomic or molecular source, but does not separate from it. Bound charges arise during the polarization of dielectric materials. The term D/t in Eq. (2.1) is called the displacement current density and accounts for the motion of bound charge.

2.2.1 Constitutive relations As already noted, Maxwell’s equations by themselves do not provide a complete set of equations for the ﬁelds. To complete the theory, the ﬁeld equations must be augmented by additional independent equations that are speciﬁed by the following constitutive relations: Constitutive Relations B : (H ; M) D:+ E;P

(2.7) (2.8)

76

CHAPTER 2 Review of Maxwell’s Equations

where M (A/m) is the magnetization, P (C/m) is the polarization, and : 4 ; 10\ T m/A and + : 8.854 ; 10\ F/m are the permeability and permittivity of free space, respectively. The magnetization M was discussed in detail in Chapter 1. It represents the net magnetic dipole moment per unit volume of a material. Speciﬁcally, M is given by m M : lim G G , V 4 where m is a vector sum of the magnetic dipole moments contained G G in the elemental volume V. Similarly, polarization P represents the net electric dipole moment per unit volume. It is given by p P : lim G G , V 4 where p is a vector sum of the electric dipole moments contained in G G the elemental volume V. From Eq. (2.8) we ﬁnd that the displacement current density consists of two components D E P :+ ; . t t t The addition of D/t to the ﬁeld equations is due to James Clerk Maxwell and is perhaps his most notable scientiﬁc contribution. This term, by virtue of its ﬁrst component + (E/t), provides for the propa gation of electromagnetic ﬁelds in free space. As already noted here, its second component P/t is due to the motion of bound charge in polarized materials. In stationary, linear, homogeneous and isotropic media the constitutive relations reduce to B : H

(2.9)

D : +E,

(2.10)

and where and + are the permeability and permittivity of the media, respectively. There is an additional constitutive relation that relates J to E, J : E, where is the conductivity with units A/V · m.

(2.11)

77

2.2 MAXWELL’S EQUATIONS

The constitutive relations (2.9)—(2.11) need to be modiﬁed for nonlinear, inhomogeneous or anisotropic materials. A material is nonlinear if , +, or depend on the ﬁelds, otherwise it is linear. For nonlinear materials Eqs. (2.9)—(2.11) are written as B : (H)H, D : +(E)E, or J : (E)E, depending on the nonlinearity. A material is inhomogeneous if , +, or depend on position, otherwise it is homogeneous. In inhomogeneous materials, , +, or are functions of the coordinate variables, for example, : (x, y, z), + : +(x, y, z), or : (x, y, z). Finally, a material is said to be anisotropic if , +, or depend on direction, otherwise it is isotropic. In anisotropic materials Eq. (2.9) generalizes to three equations: B : H ; H ; H , V V W X B : H ; H ; H , W V W X B : H ; H ; H . X V W X Similar relations hold for Eqs. (2.10) and (2.11).

2.2.2 Integral equations So far we have studied the differential form of Maxwell’s equations. However, the ﬁeld equations can also be written in integral form. The integral equations are usually more useful for applications that manifest a high degree of geometric symmetry. The integral form of ﬁeld equations is obtained as follows: Take the surface integral of the curl equations (2.1) and (2.3) over an open surface S bounded by a contour C, and then apply Stokes theorem (A.42) to the left-hand side and obtain

H · dl :

!

J;

1

D · ds t

(2.12)

and

!

respectively.

E · dl : 9

1

B · ds, t

(2.13)

78

CHAPTER 2 Review of Maxwell’s Equations

Similarly, integrate the divergence equations (2.2) and (2.4) over a volume V with a closed surface S, and then apply the Divergence theorem (A.45) to the left-hand side. This gives

B · ds : 0

(2.14)

1

and

D · ds : dv, (2.15) 1 4 respectively. Equations (2.12)—(2.15) constitute the integral form of the ﬁeld equations. The ﬁrst equation (2.12) is a generalization of Ampere’s circuital law, which states that the circulation of the magnetic ﬁeld intensity around any closed path is equal to the free current ﬂowing through the surface bounded by the path. The second equation (2.13) embodies Faraday’s law of electromagnetic induction, which states that the electromotive force induced in a stationary closed circuit equals the negative rate of increase of the magnetic ﬂux linking the circuit. In simpler terms, this law states that an E-ﬁeld is generated by a time-varying magnetic ﬂux. The negative sign in Eq. (2.13) implies that the induced E-ﬁeld will act to oppose the change in magnetic ﬂux. This is known as Lenz’s law. Faraday’s law provides the basis for the behavior of many important devices including transformers, generators and electromechanical devices. The third equation (2.14) states that the total outward ﬂux of the B-ﬁeld over a closed surface equals zero. Recall that the ﬂux of a vector ﬁeld B through a surface S is deﬁned by, :

B · ds.

1

Flux of B through S

Equation (2.14) implies that there are no isolated sources or sinks of magnetic ﬁeld, that is, no magnetic monopoles. The fourth equation (2.15) is Gauss’ law. It states that the total outward ﬂux of the D-ﬁeld over a closed surface equals the total free charge enclosed by the surface. The differential and integral forms of Maxwell’s equations are summarized in Table 2.1. In the next section, we use the integral form of the ﬁeld equations to determine the boundary conditions for the ﬁelds at material interfaces.

79

2.2 MAXWELL’S EQUATIONS TABLE 2.1

Differential and integral form of Maxwell’s equations Maxwell’s Equations (with Eqs. 2.16)

Differential Form ;H:J;

D t

Integral Form

H · dl : !

;E:9

B t

·D:

1

·B:0

J;

B · ds : 0 1

E · dl : 9

!

1

D · ds : 1

D · ds t

(2.16)

B · ds t dv

4

2.2.3 Boundary conditions We often need to solve Maxwell’s equations in a region consisting of various different materials. However, to obtain a ﬁeld solution we need to know the behavior of the ﬁelds at the material interfaces. The boundary conditions are determined by evaluating the integral form of the ﬁeld equations over control volumes at the material interfaces. Speciﬁcally, Eqs. (2.12) and (2.13) are evaluated around an inﬁnitesimal Stokesian path, and Eqs. (2.14) and (2.15) are evaluated over the surface of an inﬁnitesimal Gaussian pillbox (Fig. 2.1). We determine the boundary conditions for Eqs. (2.12) and (2.13) ﬁrst. Apply Eq. (2.12) around the Stokesian path of Fig. 2.1b and ignore the contributions from the inﬁnitesimal end portions. This yields D H l 9 H l : J lh ; L lh, R R L t where H (i : 1, 2) are the components of H tangential to the interface. GR Take the limit as h ; 0 and obtain H 9H :J , (2.17) R R where J : lim J h is the free surface current density at the inter F L face (into the page as shown in Fig. 2.1b). If both regions have ﬁnite conductivity (which is true for all applications we study), then J : 0 and the tangential component of H is continuous H :H . R R

(2.18)

80

CHAPTER 2 Review of Maxwell’s Equations

Boundary analysis: (a) Gaussian pillbox and Stokes path at the material interface; and (b) ﬁeld components.

FIGURE 2.1

Exceptions to this occur at an interface of a perfect conductor or superconductor. We apply a similar analysis to Eq. (2.13) and obtain E :E , (2.19) R R which shows that the tangential component of E is continuous at an interface. Next, we determine the boundary conditions for Eqs. (2.14) and (2.15). Apply Eq. (2.14) over the surface of the Gaussian pillbox of Fig. 2.1b and obtain This implies that

B A 9 B A : 0. L L B :B . L L

(2.20)

81

2.2 MAXWELL’S EQUATIONS

Thus, the normal component of B is continuous at an interface. A similar analysis applies to Eq. (2.15) and yields D 9D : , (2.21) L L Q where : lim h is the free surface charge density at the interface. F Q This shows that the normal component of D is discontinuous when a free surface charge exists at the interface. If both regions are perfect dielectrics (zero conductivity), then : 0 and the normal component of Q D is continuous, D : D . If, on the other hand, region 1 is a perfect L L conductor (inﬁnite conductivity), and region 2 is a perfect dielectric, then D : 0 (E : 0 in a perfect conductor), and D : , where resides on L L Q Q the surface of the perfect conductor. The boundary conditions are summarized as follows: 1. The tangential component of H is discontinuous at an interface by an amount equal to the free surface current density at the interface (H 9 H : J ). R R 2. The tangential component of E is continuous across an interface (E : E ). R R 3. The normal component of B is continuous across an interface (B : B ). L L 4. The normal component of D is discontinuous at an interface by an amount equal to the free surface charge density at the interface (D 9 D : ). L L Q The derivations of the boundary conditions are summarized in the following diagrams:

H · dl :

!

J;

1

1

1

t

B

E · dl : 9 !

D

t

· ds

· ds

B · ds : 0 1

D · ds :

4

$ dv

$

Integrate around Stokesian path across interface

Integrate over surface of Gaussian pillbox at interface

$

$

H 9H : J R R E :E R R

B :B L L D 9D : L L Q

82

CHAPTER 2 Review of Maxwell’s Equations

2.2.4 Force and torque While the ﬁelds themselves are of interest, their interaction with charged matter is equally important. In particular, for electromechanical analysis we need to know the electromagnetic force and torque imparted to steady currents. To determine this, we study the behavior of a charge q moving with velocity u through a region permeated by E- and B-ﬁelds. It is well known that the moving charge experiences a Lorentz force given by F : q(E ; u ; B)

(Lorentz force).

(2.22)

In the continuum limit, Eq. (2.22) generalizes to f : E ; J ; B

(N/m),

(2.23)

where is the free charge density and J is the free current density. Notice that f is a force density. We are especially interested in the force and torque imparted to steady currents by an external B-ﬁeld. From Eq. (2.23) we ﬁnd that the Lorentz force on a current distribution J is F:

J ; B dv,

(2.24)

4 where V is the volume occupied by J. It follows that the torque is given by T:

r ; (J ; B) dv, (2.25) 4 where r is the vector to the point about which the torque is computed.

2.3 POTENTIALS Maxwell’s equations can be solved directly for the ﬁelds. However, it is often more convenient to obtain the ﬁelds using potential functions. Speciﬁcally, the four coupled ﬁrst-order ﬁeld equations can be rewritten in terms of a pair of uncoupled second-order equations by making a change of variables to the vector and scalar potentials A and [1]. In this section we discuss the potentials, their governing equations, and their relation to the ﬁelds. For our purposes, it sufﬁces to consider Maxwell’s equations in an unbounded, stationary, homogeneous, isotropic and linear medium with

83

2.3 POTENTIALS

constitutive relations B : H and D : + E. The derivation of the equations for the potentials is as follows: Apply Proposition A.5.1 to Eq. (2.2) and introduce the vector potential A, · B : 0 $ B : ; A.

(2.26)

Next, substitute Eq. (2.26) into Eq. (2.3) and obtain

; E;

A A :0$E; : 9 t t

or E:9

A 9 . t

(2.27)

Notice that we have used Proposition A.5.2 to introduce the scalar potential . Substitute Eqs. (2.26) and (2.27) into the remaining Eqs. (2.1) and (2.4) and obtain

(2.28)

A 9 · A ; + : 9J, t t

(2.29)

A 9 ( · A) : 9J ; +

A ; , t t D term t

or A 9 +

and ;

( · A) :9 . t +

(2.30)

Notice that we have applied the identity ;;A : ( · A)9A to obtain Eq. (2.28). Now, recall from Theorems A.5.1 and A.5.2 that A is uniquely speciﬁed if both its curl and divergence are speciﬁed (Appendix A). The former is speciﬁed by Eq. (2.26). We specify the latter by imposing the Lorentz gauge condition · A : 9+

t

(Lorentz gauge).

(2.31)

84

CHAPTER 2 Review of Maxwell’s Equations

Finally, substitute Eq. (2.31) into Eqs. (2.29) and (2.30) and obtain two second-order equations for the potentials A 9 +

A : 9J, t

(2.32)

9 +

:9 . + t

(2.33)

It is well known that Eqs. (2.32) and (2.33) have solutions 4

1 4+

A(x, t) :

J(x, t 9 x 9 x /u) dv

x 9 x

(2.34)

(x, t 9 x 9 x /u) dv,

x 9 x

(2.35)

4

and (x, t) :

4

where A(x, t) and (x, t) are time-retarded potentials, and u : 1/(+ is the speed of light in the medium. In free space, + : + , : , and u : c : 1/( + : 3 ; 10 m/s. It can be shown that A(x, t) and (x, t) satisfy the Lorentz condition (2.31) as long as J and satisfy the continuity condition (2.6). Thus, given J and we can solve Maxwell’s equations by ﬁrst solving Eqs. (2.34) and (2.35), and then using Eqs. (2.26) and (2.27) to obtain B and E. The ﬁelds D and H are obtained from the constitutive relations D : + E and H : B/. For certain time-harmonic problems it is possible to reconstruct the ﬁelds from A(x, t) alone. The boundary conditions for and A must be compatible with the boundary conditions for the ﬁelds (2.17)—(2.21). Speciﬁcally, and A must be continuous at an interface. If there are discontinuities in or + at an interface, then the normal derivatives of and A must be discontinuous in order to ensure that the normal component of D : 9+ ( ; A/t), and the tangential component of H : 1/( ; A) are continuous across the interface [5]. Maxwell’s equations govern a broad range of phenomena that span the entire electromagnetic spectrum. It is useful to categorize such applications as high-frequency, quasi-static, or static. For high-frequency applications, the ﬁeld equations (2.1)—(2.4) need to be solved in their entirety. However, for quasi-static and static applications, certain terms are negligible and as we shall see the analysis simpliﬁes.

85

2.4 QUASI-STATIC THEORY

2.4 QUASI-STATIC THEORY Quasi-static ﬁeld theory applies at low frequencies (long wavelengths) when the dimensions of the region of interest are small relative to the wavelength of the electromagnetic ﬁeld that permeates it. When this is the case, we ignore the ﬁeld’s ﬁnite speed of propagation and assume that any change in the ﬁeld is felt instantaneously across the region. For example, consider a system that resides in a cubic region of free space that measures (3 m on each side. Assume that the system generates a time-harmonic electromagnetic ﬁeld with period and wavelength . The ﬁeld propagates across this region with a velocity c : 3 ; 10 m/s. Therefore, any change in the ﬁeld is felt throughout the region within 10 ns. The system is said to be quasi-static as long as the ﬁeld changes much more slowly than the time required for it to propagate across the region, that is, as long as 10 ns, or f : 1/ 100 MHz. Alternatively, because f : c the system is quasi-static as long as 3 m, which relates back to our original deﬁnition. Quasi-static theory governs an important range of applications including electrical circuit analysis, electromechanical devices, and eddy current phenomena [6]. We study these in some detail in Chapter 5. Here, we brieﬂy review the quasi-static ﬁeld equations. From a mathematical perspective, the quasi-static approximation amounts to ignoring the displacement current term D/t in the ﬁeld equations. When this is done, the ﬁeld equations simplify as follows:

;H:J

·B:0 D :0$ B t ;E:9 t · D : .

(Quasi-static)

(2.36)

Notice that in the ﬁrst and second equations in (2.36) there are no time derivatives. Thus, even when the current source J is time dependent, the ﬁelds H and B are determined as if the system were static. Also notice that the ﬁrst equation implies that · J : 0. The third equation in (2.36) is known as Faraday’s law. It plays an important role in the analysis of electromechanical devices (Chapter 5). The ﬁelds can be obtained by direct solution of (2.36) as long as the constitutive relations (2.7) and (2.8) are taken into account. However, the ﬁelds can also be obtained using the potentials introduced in Section 2.3.

86

CHAPTER 2 Review of Maxwell’s Equations

To see this, follow the same analysis presented there, except that D/t : 0 in Eq. (2.28). Then, Eqs. (2.32) and (2.33) reduce to A 9 ( · A) : 9J(t),

(2.37)

( · A) (t) :9 , t +

(2.38)

and ;

respectively. The potential A needs to be uniquely speciﬁed, and we do so by imposing the Coulomb gauge condition, ·A:0

(Coulomb gauge).

(2.39)

Substituting Eq. (2.39) into Eqs. (2.37) and (2.38) gives A : 9J : 9/+ .

(2.40)

In free space (no material boundaries), the solutions to Eq. (2.40) can be written as A(x, t) : 4

4

J(x, t) dv,

x 9 x

(2.41)

and

1 (x, t) dv. (2.42)

x 9 x 4+ 4 Notice that there is no time retardation in Eqs. (2.41) and (2.42) and, therefore, any change in a source is felt instantaneously. The procedure for obtaining the ﬁelds is as follows: Given the source terms J(x, t) and (x, t), evaluate Eqs. (2.41) and (2.42). This gives A(x, t) and (x, t). Once the potentials are known, E and B are obtained using (x, t) :

E(x, t) : 9

A(x, t) 9 (x, t), t

and B(x, t) : ; A(x, t). The remaining ﬁelds are obtained from the constitutive relations D : + E and H : B/.

87

2.5 STATIC THEORY

2.5 STATIC THEORY In static ﬁeld theory there is no time variation. Amazingly, this restrictive theory applies to a wide range of important phenomena involving steady currents (magnetostatic theory), and stationary charge distributions (electrostatic theory). We make extensive use of magnetostatic theory in our study of permanent magnet structures and direct current devices. In static ﬁeld theory the time-dependent terms in Maxwell’s equations are negligible: D B : : 0. t t

(2.43)

When this is the case, the ﬁeld equations uncouple into magnetostatic and electrostatic equations as shown in the following diagram: D :0 t B :0 t

Maxwell’s Equations $ Uncouple

;H:J (Magnetostatic) ·B:0 ;E:0 (Electrostatic) ·D:

The uncoupled magnetostatic and electrostatic equations along with their constitutive relations can be solved independently for the magnetic and electric ﬁelds.

2.5.1 Magnetostatic theory The equations for magnetostatic analysis can be summarized as follows: Magnetostatic Analysis Differential Form ;H:J

Integral Form

!

·B:0

H · dl :

J · ds 1

B · ds : 0 1 Constitutive Relation B : (H ; M)

88

CHAPTER 2 Review of Maxwell’s Equations

The ﬁrst integral equation is Ampere’s circuital law, which states that the circulation of the magnetic ﬁeld intensity around any closed path is equal to the free current ﬂowing through the surface bounded by the path. The second integral equation states that the total outward ﬂux of the B-ﬁeld over a closed surface equals zero. This implies that there are no isolated sources or sinks of magnetic ﬁeld. The magnetostatic equations can be solved directly for the ﬁelds. However, it is often more convenient to obtain the ﬁelds using the vector potential A introduced in Section 2.3. The equation for A follows from the analysis presented there, with the additional constraint that D/t : B/t : 0. In this case, Eq. (2.32) reduces to A 9 ( · A) : 9J.

(2.44)

Here A is uniquely speciﬁed by imposing the Coulomb gauge condition (2.39), which gives A : 9J.

(2.45)

The solution to Eq. (2.45) can be expressed in the integral form using the free space Green’s function G(x, x):91/4(1/ x9x ) for (Appendix B). Speciﬁcally, we have A(x) :

4

J(x) dv.

x 9 x

(2.46)

4 Once A is known, the B-ﬁeld follows from B(x) : ; A(x) : :

4 4

; 4

J(x) dv

x 9 x

J(x) ; (x 9 x) dv.

x 9 x

(2.47)

4 The H-ﬁeld is obtained from the constitutive relation H : B/. A magnetostatic ﬁeld contains energy. In the absence of dissipation mechanisms, the energy stored in a ﬁeld equals the energy required to create it [7, 8]. It can be shown that the energy supplied to a system of currents and magnetizable materials in creating a ﬁeld B : ; A is given by W : K

4

J · dA dv

(J)

(2.48)

89

2.5 STATIC THEORY

or W : K

(J). (2.49) H · dB dv 4 If the system is linear in the sense that A is proportional to J, then Eqs. (2.48) and (2.49) reduce to 1 W : K 2 and 1 W : K 2

A · J dv

(J)

(2.50)

4

B · H dv (J). (2.51) 4 Notice that in Eq. (2.50) the integration extends over the region occupied by the source, which is usually ﬁnite, whereas in Eq. (2.51) the integration typically extends over all space. From Eq. (2.51) we ﬁnd that the magnetostatic energy density for a linear system can be written as 1 w : B·H K 2

(J/m).

(2.52)

This completes our review of magnetostatic theory. A more comprehensive presentation is given in Chapter 3.

2.5.2 Electrostatic theory The equations for electrostatic analysis are as follows: Electrostatic Analysis Differential Form

Integral Form

;E:0 · D :

E · dl : 0

!

D · ds :

1

dv 4

Constitutive Relation D:+ E;P The ﬁrst integral equation provides the basis for Kirchhoff’s voltage law, which states that around any closed path the algebraic sum of the

90

CHAPTER 2 Review of Maxwell’s Equations

voltage drops equals the algebraic sum of the voltage rises. The second integral equation is Gauss’ law, which states that the total outward ﬂux of the D-ﬁeld over a closed surface equals the total free charge enclosed by the surface. The ﬁelds can be obtained by direct solution of the electrostatic equations. However, they can also be obtained using the scalar potential introduced in Section 2.3. The equation for follows from the analysis presented there if we impose the additional constraint D/t : B/t : 0. In this case, Eq. (2.33) reduces to : 9 . +

(2.53)

The solution to Eq. (2.53) can be expressed in integral form using the free space Green’s function for (Appendix B), that is, (x) :

1 4+

4 Once is known, E is obtained from

(x) dv.

x 9 x

(2.54)

E(x) : 9(x) :9

1 4+

1 : 4+

(x) dv

x 9 x

4 (x)(x 9 x) dv.

x 9 x

(2.55) 4 Here D is obtained from D : +E. Notice that E : 9 implies that the E-ﬁeld points in the direction of decreasing potential. Consider the line integral of E between two points a and b. As E : 9, we have

@

@

· dl ? : 9 , (2.56) ? @ where and are the potentials at a and b, respectively. The electrical ? @ potential energy E of a charge q at points a and b is E : q and ? ? E : q , respectively. Thus, the potential difference 9 is the @ @ @ ? energy expended in moving a unit charge from point a to point b. If no magnetic ﬁeld is present, the potential difference is independent of the path taken. The electrostatic ﬁeld contains energy. Speciﬁcally, it can be shown that the energy supplied to a system of charges, isolated conductors, and ?

E · dl : 9

91

2.6 SUMMARY

dielectric materials in creating a ﬁeld E : 9 is given by 1 W : C 2

M

(J), (2.57) d dv 4 where is the free charge density [7]. When the system is linear in the sense that is linearly proportional to , then Eq. (2.57) reduces to 1 W : C 2

dv

(J).

(2.58)

4 We substitute · D : , make use of the identity · (D) : · D ; D · , and obtain 1 W : C 2

D · E dv

(J),

(2.59)

4 which gives the electrostatic energy in terms of the ﬁelds. Notice that the integration in Eq. (2.58) extends over a region containing the free charge, which is usually ﬁnite, whereas the integration in Eq. (2.59) typically extends over all space. In addition, Eq. (2.59) implies that the electrostatic ﬁeld has an energy density 1 w : D·E C 2

(J/m).

(2.60)

2.6 SUMMARY We have presented an overview of classical electromagnetic ﬁeld theory. We summarize the key results for future reference. Electromagnetic ﬁelds are produced by sources of charge and current and are governed by Maxwell’s equations: Differential Form ;H : J ;

D t

Integral Form

H · dl :

!

·B:0 ;E : 9

B t

·D:

J;

1

B · ds : 0 1

E · dl : 9 !

1

D · ds : 1

D · ds t

B · ds t dv

4

Ampere’s circuital law Conservation of flux Faraday’s law Gauss’ law

92

CHAPTER 2 Review of Maxwell’s Equations

The ﬁeld equations are consistent, but not all independent. For a complete theory they must be augmented by the constitutive relations B : (H ; M), and D : + E ; P. (2.61) In stationary, linear, homogeneous and isotropic media these reduce to B : H and D : + E. There is an additional constitutive relation that relates the E-ﬁeld to the current density J : E. Maxwell’s equations imply the conservation of charge, ·J;

: 0. t

(2.62)

Moreover, if a charge q moves with velocity u through a region permeated by E- and B-ﬁelds, it experiences a Lorentz force F : q(E ; u ; B).

(2.63)

In the continuum limit, Eq. (2.63) generalizes to f : E ; J ; B

(N/m),

(2.64)

where is the free charge density and J is the free current density. When solving Maxwell’s equations in regions consisting of different materials, we need to impose boundary conditions at the material interfaces. The boundary conditions for the tangential components of H and E are H 9H :J , R R (2.65) E :E . R R For most practical applications J : 0 and H : H . The boundary R R conditions for the normal components of B and D are B :B , L L (2.66) D 9D : . Q L L The ﬁelds can be determined by direct solution of Maxwell’s equations. However, it is often more convenient to make a change of variables to the scalar and vector potentials and A where B:;A

(2.67)

and E:9

A 9 . t

(2.68)

93

2.6 SUMMARY

FIGURE 2.2

Simpliﬁcation of Maxwell’s equations in terms of the ﬁelds.

94

FIGURE 2.3

CHAPTER 2 Review of Maxwell’s Equations

Simpliﬁcation of Maxwell’s equations in terms of the potentials.

In stationary, linear, homogeneous and isotropic media we impose the Lorentz condition · A : 9+

, t

and ﬁnd that the potentials satisfy the inhomogeneous wave equations, A 9 +

A : 9J, t

9 + :9 . t +

(2.69)

95

2.6 SUMMARY

These equations have solutions 4

1 4+

A(x, t) :

4

J(x, t 9 x 9 x /u) dv,

x 9 x

(2.70)

and (x, t) :

(x, t 9 x 9 x /u) dv.

x 9 x

(2.71) 4 The B- and E-ﬁelds can be obtained from Eqs. (2.70) and (2.71) using Eqs. (2.67) and (2.68). Once B and E are known, H and D are obtained from the constitutive relations. Finally, electromagnetic applications can be broadly categorized as either high-frequency, quasi-static, or static. In Figs. 2.2 and 2.3 we illustrate the reduction and simpliﬁcation of the ﬁeld equations for these categories in terms of the ﬁelds and the potentials, respectively. Note that these equations are written for a stationary, linear, homogeneous and isotropic medium. References 1. Jackson, J. D. (1962). Classical Electrodynamics, 2nd ed., New York: John Wiley and Sons. 2. Cheng, D. K. (1989). Field and Wave Electromagnetics, 2nd ed., New York: Addison-Wesley. 3. Lorrain, P., Corson, D. P., and Lorrain, F. (1987). Electromagnetic Field and Waves, New York: W. H. Freeman and Co. 4. Reitz, J. R., Milford, F., and Christy, R. (1979). Foundations of Electromagnetic Theory, 3rd ed., New York: Addison-Wesley. 5. Silvester, P. P. and Ferrari, R. L. (1990). Finite Elements for Electrical Engineers, 2nd ed., New York: Cambridge Univ. Press. 6. Mayergoyz, I. (1998). Nonlinear Diffusion of Electromagnetic Fields with Applications to Eddy Currents and Superconductivity, New York: Academic Press. 7. Johnk, C. T. A. (1988). Engineering Electromagnetic Fields and Waves, 2nd ed., New York: John Wiley and Sons. 8. Morrish, A. H. (1983). The Physical Principles of Magnetism, Malabar, FL: R. Krieger Publishing Co.

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CHAPTER

3

Field Analysis

3.1 INTRODUCTION In this chapter we study the theory and methods used in the analysis of steady currents, permanent magnets, and magnetic circuits. We start with a review of magnetostatic ﬁeld theory, which includes a discussion of force, torque, energy, and inductance. Next, we introduce the current and charge models for magnetic materials. These are used to reduce a permanent magnet to an equivalent source term. Following this, we study magnetic circuits, introduce the concept of reluctance, and show how to transform a physical circuit into an equivalent lumped-parameter circuit. Finally, we discuss various analysis methods including boundary-value theory, the method of images, ﬁnite element analysis (FEA), and the ﬁnite difference method.

3.2 MAGNETOSTATIC ANALYSIS Although the magnetostatic ﬁeld equations were presented in both differential and integral form in Section 2.5, we repeat them here for convenience. The ﬁeld equations in differential form are and

;H : J,

(3.1)

· B : 0,

(3.2)

where J (A/m) is the free current density, and Field

Description (units)

H

Magnetic field intensity (A/m)

B

Magnetic flux density (T)

(3.3)

97

98

CHAPTER 3 Field Analysis

The ﬁeld equations in integral form are

H · dl : !

J · ds,

(3.4)

1

and

B · ds : 0.

(3.5)

1

The magnetic ﬂux through a surface S is given by :

B · ds;

(3.6)

1

ﬂux is measured in webers (Wb). The magnetostatic ﬁelds and their units were discussed in Section 1.2. We repeat the following physical descriptions for convenience. 1. Field strength H: Consider a straight, inﬁnitely long wire carrying a current i : 2 A. The conductor generates a tangential ﬁeld strength H : 1 A/m at a radial distance r : 1 m from its center. As another example, consider a long solenoid that has n turns per meter and carries a current of 1/n A. It generates a ﬁeld strength H : 1 A/m along its axis. 2. Flux density B: Consider an inﬁnitely long conductor that carries a current i : 1 A perpendicular to an external B-ﬁeld. A force of 1 newton will be imparted to each meter of the conductor when B : 1 T. 3. Flux : Consider a single turn coil with 1 Wb of magnetic ﬂux passing through it. A voltage of 1 V will be induced in the coil when the ﬂux is uniformly reduced to zero in one second. Equation (3.4) is Ampere’s circuital law. It states that the line integral of H around any closed path C is equal to the free current ﬂowing through the surface bounded by C. Equation (3.5) states that the total outward ﬂux of B over a closed surface equals zero. This implies that there are no isolated sources or sinks of magnetic ﬁeld. The integral equations are easily derived from the differential equations and vice versa (Section 2.2.2). However, neither set of equations alone is sufﬁcient to determine the ﬁelds. This is because there are six unknown ﬁeld components (three for both H and B) and at most only four independent equations (three for Eq. (3.1) and one for Eq. (3.2)). Therefore, the ﬁeld equations must be augmented by an additional

99

3.2 MAGNETOSTATIC ANALYSIS

independent equation that takes the form of the constitutive relation B : (H ; M). (3.7) For linear, homogeneous and isotropic materials, B and M are proportional to H. Speciﬁcally, B : H,

(3.8)

and M : H, (3.9) K where and are the permeability and susceptibility of the material. K Notice that and are related to one another. Speciﬁcally, from Eqs. K (3.7), (3.8) and (3.9) we ﬁnd that : ( ; 1), K

(3.10)

and : 9 1. (3.11) K When a ﬁeld solution is sought in a region consisting of different materials, the appropriate boundary conditions must be applied at the material interfaces. In Section 2.2.3 we found that the normal component of B is continuous across an interface B :B . (3.12) L L We also found that the tangential component of H is discontinuous by an amount equal to the free surface current density J : LQ HR 9 H : J . (3.13) R LQ In the applications that we study, J : 0, and the tangential component LQ of H is continuous: H :H . (3.14) R R For many applications, we can determine the ﬁelds directly from the ﬁeld equations. This is especially true when a high degree of symmetry is present. For example, we can use Eq. (3.4) to determine H when there is sufﬁcient symmetry to evaluate H · dl around a closed path bounding the source J. The following examples illustrate this approach. Example 3.2.1 Determine the B-ﬁeld outside an inﬁnity long wire of radius R with current density J (Fig. 3.1).

100

CHAPTER 3 Field Analysis

FIGURE 3.1

Inﬁnite wire.

Solution 3.2.1 Use cylindrical coordinates with the axis of the wire along the z-axis, J : J z . Apply Eq. (3.4) to a closed circular path of radius r R. From X symmetry we know that H is in the direction and that it has a constant magnitude H (r) at a ﬁxed radius r. Thus, we obtain

L

H r d :

L 0

J r dr d, X

which reduces to or

H 2r : J R, X H (r) :

I , 2r

(3.15)

where I : J R is the current in the wire. In free space, B : H and therefore, X I B : . (3.16) 2r ) Example 3.2.2 Consider a long tightly wound cylindrical coil (solenoid) with N turns and with its axis along the z-axis (Fig. 3.2). Let I and L denote the

101

3.2 MAGNETOSTATIC ANALYSIS

FIGURE 3.2

Solenoid.

current through and length of the coil, respectively. Determine the B-ﬁeld along the axis of the coil. Solution 3.2.2 Use cylindrical coordinates. Apply Eq. (3.4) to a rectangular path with one side of length S along the axis of the coil and the other parallel side just outside the coil. This gives

H · dl : N

S I, L

! where N(S/L)I is the total current passing through the rectangle. The ﬁeld components along the radial edges of the rectangle and just outside the coil are negligible relative to H on the axis. Therefore, we ignore contributions along X these segments and obtain S H S:N I X L or NI H : . X L Because B : H, we have

N B : I, X L

where N/L is the number of turns per unit length.

(3.17) )

Example 3.2.3 Determine the H-ﬁeld inside a toroid of material of height h and inner and outer radii R and R , respectively, which is wrapped by a tightly

102

CHAPTER 3 Field Analysis

Toroid with coil [9].

FIGURE 3.3

wound N turn coil carrying a current I. Assume that the material is linear with a permeability (Fig. 3.3). Solution 3.2.3 Use cylindrical coordinates with the toroid centered with respect to the z-axis. From symmetry we know that H : H . Apply Eq. (3.4) to a circular path inside the toroid and obtain

Thus,

H · dl : H 2r : NI.

H :

NI 2r

(R r R ),

(3.18)

NI . 2r

(3.19)

and B :

)

3.2.1 Vector potential In the previous examples we determined the ﬁelds by direct solution of the ﬁeld equations. However, it is often more convenient to use the vector potential A as was discussed in Section 2.3. We repeat the derivation of A for convenience. Consider a stationary, homogeneous and isotropic material with a linear constitutive relation B : H. Apply Proposition A.5.1 to Eq. (3.2) and introduce the vector potential A · B : 0 $ B : ;A.

(3.20)

Substitute this into Eq. (3.1) and obtain A 9 ( · A) :9J.

(3.21)

103

3.2 MAGNETOSTATIC ANALYSIS

Next, uniquely specify A by imposing the Coulomb gauge condition · A : 0. Substitute this into Eq. (3.21), which gives A :9J.

(3.22)

Thus, we ﬁnd that the magnetostatic ﬁeld equations (3.1) and (3.2) and the constitutive relation (3.8) collectively reduce to a single second-order equation (3.22). The derivation of Eq. (3.22) is summarized in the following diagram: A9( · A):9J Coulomb Gauge $ A :9J. · A:0

;H:J B:;A $ $ · B:0 B:H

The solution to Eq. (3.22) can be expressed in integral form using the free-space Green’s function G(x, x) :9(1/4)(1/ x 9 x ) for the operator (Appendix B). Speciﬁcally, we obtain A(x) :

4

J(x) dv.

x 9 x

(3.23)

The integration in Eq. (3.23) is over the source region (region of nonvanishing J). Once A is determined, we can compute B using B : ;A, which gives B(x) :

4

J(x);(x 9 x) dv.

x 9 x

(3.24)

It is important to note that Eq. (3.24) applies to problems in which the source term J is speciﬁed in a homogeneous space characterized by a uniform permeability . It does not apply when material interfaces are present. The derivation of Eq. (3.24) is summarized in Eq. (3.25): A :9J $ A(x) :

4

4

J(x) dv

x 9 x

B : ;A

B(x) :

4

J(x);(x 9 x) dv.

x 9 x

(3.25)

104

CHAPTER 3 Field Analysis

In free space : , and Eqs. (3.23) and (3.24) reduce to J(x) A(x) : dv, 4

x 9 x 4 and

B(x) : 4

(3.26)

J(x);(x 9 x) dv.

x 9 x

(3.27) 4 For thin current ﬁlaments or wires, J dv ; Idl, and Eqs. (3.26) and (3.27) become I A(x) : 4

!

dl ,

x 9 x

(3.28)

and I B(x) : 4

dl;(x 9 x) ,

x 9 x

(3.29) ! where C is the circuit path. This is known as the Biot-Savart law. For surface currents, Jdv ; Kds, where K (A/m) is the surface current density. If K ﬂows along a surface S, then Eqs. (3.26) and (3.27) reduce to A(x) : 4

K(x) ds ,

x 9 x

(3.30)

K(x);(x 9 x) ds .

x 9 x

(3.31)

1

and B(x) : 4

1 The following examples demonstrate the use of these equations. Example 3.2.4 Consider a current I ﬂowing in a wire of length 2L along the z-axis as shown in Fig. 3.4. Determine the vector potential A(x, y, 0) at any point in the x-y plane. Use A to determine B : ;A. Solution 3.2.4 We apply Eq. (3.28) using cylindrical coordinates. In this case dl : dzz , and

x 9 x : (r ; z.

105

3.2 MAGNETOSTATIC ANALYSIS

Current along the z-axis.

FIGURE 3.4

We obtain I A(r) : 4

*

dz

\* (r ; z

z

I : [ln z ; (z ; r!] * z \* 4

I (L ; r ; L : ln z . 4 (L ; r 9 L

(3.32)

Next, we compute B, B : ;A : ;A z X A 1 A X r 9 X . : r r

(3.33)

From symmetry we know that A / : 0. Combining Eqs. (3.32) and (3.33) X we have B:9

I (L ; r ; L ln r 4 (L ; r 9 L

I L : . 2r (L ; r

(3.34)

106

CHAPTER 3 Field Analysis

Inﬁnite current sheet with height 2h; (a) perspective of sheet; and (b) cross-sectional view. FIGURE 3.5

For a long wire r L, and Eqs. (3.32) and (3.34) reduce to

and

I A(r) : 9 ln(r) ; C z , 2

(3.35)

I B : , 2r

(3.36)

where C is a constant. These are the usual results for an inﬁnite wire.

)

107

3.2 MAGNETOSTATIC ANALYSIS

Example 3.2.5 Consider an inﬁnitely long current sheet of height 2h with a surface current density K : K z (Fig. 3.5); K has the units of A/m. Determine the vector potential A at any point in the x—y plane. Use A to determine B : ;A. Solution 3.2.5 We use the results of the previous example. Speciﬁcally, we treat the sheet as a collection of inﬁnite line currents. The current for an inﬁnitesimal section of height dy is given by I : K dy. From Eq. (3.35) we know that the contribution from this line current at y is I dA : 9 ln [(x ; (y 9 y)] X 2 K dy :9 ln [x ; (y 9 y)]. 4 therefore, K A (x, y) : 9 X 4

F

ln [x ; (y 9 y)] dy

\F K : 9 (y 9 y) ln [x ; (y 9 y)] 9 2(y 9 y) 4 ; 2x tan\

(y 9 y) F . x \F

This reduces to

K A (x, y) : 9 (h 9 y) ln [x ; (y 9 h)] ; (h ; y) ln [x ; (y ; h)] X 4 9 4h ; 2x tan\

(3.37)

(3.38)

2h x x ; y 9 h

.

Next, we evaluate B, B : ;A :

A A X x 9 X y y x

K x ; (y 9 h) 2h x : ln x ; 2 tan\ y . 4 x ; (y ; h) x ; y 9 h

)

108

CHAPTER 3 Field Analysis

Current loop.

FIGURE 3.6

Example 3.2.6 Determine the B-ﬁeld along the axis of a circular current loop of radius R and current I (Fig. 3.6). Solution 3.2.6 We apply the Biot-Savart law (3.29) with dl : R d , x 9 x : zz 9 Rr , and x 9 x : (R ; z. We obtain IR B(z) : 4

L

d z , (R ; z)

which reduces to B(z) :

IR z . 2(R ; z)

B(z) :

m , 2(R ; z)

(3.39)

This can also be written as

where m : IRz is the magnetic dipole moment of the current loop.

)

The forementioned examples demonstrate the use of the vector potential in obtaining the B-ﬁeld. However, it can also be used to determine the ﬂux . Speciﬁcally, the ﬂux through a surface S can be expressed in terms of the line integral of A around a path C bounding S.

109

3.2 MAGNETOSTATIC ANALYSIS

Speciﬁcally, :

B · ds

1

:

(;A) · ds

1

:

A · dl, (3.40) ! where in the ﬁnal step we have used Stokes’ theorem. In two-dimensional (2D) problems the vector potential can be used to visualize ﬂux lines. A ﬂux line is a line that indicates the direction of the B-ﬁeld at any point along its length. In addition, in 2D problems ﬂux lines coincide with equipotential contours of A. To see this, consider a 2D problem in Cartesian coordinates. The vector potential is in the z-direction A : A (x, y)z . Consider the change in A along an inﬁnitesiX mal segment dl : dxx ; dyy that lies along a ﬂux line: dA : A · dl. (3.41) X Because dl is along a ﬂux line, it is parallel and proportional to B. Therefore, dl : B : B x ; B y , V W where is a constant of proportionality. However,

(3.42)

B : ;A A (x, y) A (x, y) X x 9 X y . y x Therefore, from Eq. (3.42) we have :

dl :

A (x, y) A (x, y) X X x 9 y . y x

(3.43)

Substitute Eq. (3.43) into Eq. (3.41) and obtain dA :

A (x, y) A (x, y) A (x, y) A (x, y) X X X X 9 x y y x

: 0. This shows that for two-dimensional problems, A is constant along ﬁeld lines, or equivalently that equipotential contours of A coincide with ﬂux lines.

110

CHAPTER 3 Field Analysis

3.2.2 Force and torque So far, we have studied methods for determining the magnetic ﬁeld for a given current source. In this section, we study methods for determining the force and torque on a current source given an external magnetic ﬁeld. When a particle of charge q moves through an external ﬁeld B it experiences a Lorentz force F : q(u;B ).

(3.44)

We generalize this to the case of currents. Consider charges per unit S volume moving with velocity u. This gives rise to a volume current density J : u. The force on each charge is given by Eq. (3.44), and S therefore the force per unit volume is f : J;B (force density),

(3.45)

where f is a force density (N/m). The total force on a conductor with a current density J is obtained by integrating f over the volume of the conductor: F:

f dv.

(3.46)

4

For thin current ﬁlaments or wires, J dv ; I dl, and Eq. (3.46) reduces to

F:I

dl;B .

(3.47)

This line integral is evaluated in the direction of the current ﬂow over the length of the wire as described in Section A.4. The force on an inﬁnitesimal length of the wire is dF : I dl;B .

(3.48)

In the special case when a wire of length l is perpendicular to B , the Lorentz force reduces to F : IlB .

(3.49)

111

3.2 MAGNETOSTATIC ANALYSIS

If we know the force density f, we can determine the torque using T: :

r;f dv

4

r;(J;B ) dv, (3.50) 4 where r is the vector from the point about which the torque is computed. We apply Eq. (3.50) to thin wire carrying a current I. The torque on a segment of the wire of length dl is dT : I[r;(dl;B )], and the total torque is obtained via integration, T:I

r;(dl;B ).

(3.51)

(3.52)

It is important to note than when evaluating Eqs. (3.51) and (3.52) the force term (dl;B ) must be evaluated before the cross product with r is taken. Also, Eq. (3.52) is evaluated in the direction of the current ﬂow over the length of the wire as described in Section A.4. We demonstrate the use of Eq. (3.47) in the following example. Example 3.2.7 Determine the force per unit length between two parallel wires separated a distance d and carrying currents I and I , respectively (Fig. 3.7). Solution 3.2.7 We use Eq. (3.48) to compute the force. Choose a coordinate system with I along the z-axis. This current gives rise to a ﬁeld B across I I B : 9 x . 2d The force is given by F : I z ;B I I : 9 y 2d

(3.53)

Notice that the force is attractive, it pulls the wires together. If the currents are in opposite directions the wires repel one another. ) Other common conductor conﬁgurations are shown in Fig. 3.8. Figure 3.8a shows two parallel current sheets carrying currents I and I ,

112

CHAPTER 3 Field Analysis

FIGURE 3.7

Force between two parallel wires.

respectively [1]. The force between the conductors is

h d h I I 9 ln 1 ; F : tan\ h d 2h d

(parallel current sheets).

Figure 3.8b shows two coplanar current sheets. The force between the conductors is

I I Q>U w ;x F: ln dx (coplanar current sheets). 2 w w x Q

3.2.3 Maxwell stress tensor In the previous section we used the Lorentz force to determine the force and torque on steady currents. An alternate and more general approach entails the use of the Maxwell stress tensor [1, 2]. In this approach, the force density equation (3.45) is given by f:

1 · T.

(3.54)

113

3.2 MAGNETOSTATIC ANALYSIS

Force between conductors: (a) parallel current sheets; and (b) coplanar current sheets [1].

FIGURE 3.8

Here, T is the Maxwell stress tensor T : T x x ; T x y ; T x z ; T y x ; T y y VV VW VX WV WW ; T y z ; T z x ; T z y ; T z z , WX XV XW XX

(3.55)

where x , y , and z are the Cartesian unit vectors. Such an expression is called a dyadic, and it is but one way of expressing the components of the second-order tensor T. From Eq. (3.54) we have f:

1

T T T T T T VV ; VW ; VX x ; WV ; WW ; WX y x y z x y z ;

T T T XV ; XW ; XX z . x y z

If we substitute J : ;H into Eq. (3.45), and consider linear media with a constitutive relation B : H, the components of T are as

114

CHAPTER 3 Field Analysis

follows [2, 3]: T

VV [T] : T WV T XV

T

VW T WW T XW

T (B 9 B ) B B B B V VX V W V X T : B B (B 9 B ) B B . W WX W V W X T B B B B (B 9 B ) X XX X V X W (3.56)

The force on a body is computed using Eq. (3.46), that is, F:

1

· T dv.

1

T · n ds,

4 This volume integral can be written as a surface integral using the Divergence theorem. Speciﬁcally, we ﬁnd that F:

(3.57)

1 where is the permeability of the medium where the integration takes place, n is the outward unit normal to the bounding surface, and the integration is performed over a surface S immediately surrounding the body. In principle, S should be the surface of the body itself. However, in practice (especially in FEA), S usually encompasses the body but is slightly offset from its surface. To illustrate the use of Eq. (3.57), consider a surface with a unit normal n : n x ; n y ; n z . According to Eq. (3.57), the magnitude of V W X the surface force in the x direction is 1 F : V

(T n ; T n ; T n ) ds. VV V VW W VX X

(3.58)

1 Equation (3.57) is particularly useful for evaluating the force on soft magnetic materials that have a high permeability ( ) in free space. Because of the high permeability, the ﬁeld components tangential to the surface are negligible. We choose a surface immediately surrounding the body so that the integration takes place in free space. Then Eq. (3.57) reduces to F:

1 2

1

B n ds, L

(3.59)

where B is the component of B normal to the surface. Notice that the L direction of F is along the outward normal to the surface [1]. The following example demonstrates the use of the Maxwell stress tensor approach.

3.2 MAGNETOSTATIC ANALYSIS

FIGURE 3.9

115

Magnetic circuit actuator: (a) perspective; and (b) side view.

Example 3.2.8 Consider the actuator shown in Fig. 3.9a. Determine the force on the bar. Assume that all materials are linear with permeability , and that there is no leakage ﬂux. Solution 3.2.8 The magnetic circuit of Fig. 3.9a is analyzed in Example 5.8.1. The ﬁeld in the gap is given by Ni B : , E 2g

(3.60)

where N is the number of turns, i is the current, and g is the gap length. This ﬁeld is constant across the gap, and perpendicular to the surface of the bar. Thus, B : B in the gap and zero elsewhere as there is no leakage ﬂux. Choose L E a rectangular surface of integration immediately outside the bar as indicated by

116

CHAPTER 3 Field Analysis

the dotted line in Fig. 3.9b. Substitute Eq. (3.60) into Eq. (3.59) and obtain

1 B n ds L 2 1 1 [2B A ]n : E E 2 NiA E n : 4g

F(g) :

(3.61)

where A is the cross-sectional area of the gap. Notice that the direction of the E force is the same as n (the outward normal to the surface of the bar in the gap region). This unit vector points toward the actuator, and hence the force tends to move the bar towards the actuator, thus reducing the gap. )

3.2.4 Energy A magnetostatic ﬁeld contains energy. In the absence of dissipation mechanisms the energy stored in the ﬁeld equals the energy required to create it. The energy supplied to a system of currents and magnetizable materials in creating a ﬁeld B : ;A is given by [4] W : K

J · dA dv (J). 4 This can also be written in terms of the ﬁelds W : K

(3.62)

H · dB dv (J). (3.63) 4 If the system is linear in the sense that A is proportional to J, then Eqs. (3.62) and (3.63) reduce to 1 W : K 2

1 W : K 2

A · J dv (J),

(3.64)

B · H dv (J),

(3.65)

4

and

4

respectively. In linear, homogeneous and isotropic materials, B : H

117

3.2 MAGNETOSTATIC ANALYSIS

and Eq. (3.65) becomes 1 W : K 2

H dv (J),

1 W : K 2

B dv (J).

or

(3.66)

4

(3.67) 4 Therefore, the energy density for a linear system can be written as 1 w : B · H (J/m). K 2

(3.68)

Recall from Section 1.7 that a magnetized specimen possesses selfenergy. Speciﬁcally, if a specimen of volume V has a ﬁxed magnetization M, then it possesses a self-energy W :9 Q 2

4

M · H dv (self-energy), +

(3.69)

where H is the ﬁeld in the specimen due to M (i.e., due to the + continuum of dipole moments comprising the specimen). The selfenergy equation (3.69) can be considered to be the energy required to assemble a continuum of dipole moments in absence on an applied ﬁeld [5, 6]. Notice that the energy density associated with W is Q (3.70) w :9 M · H . + Q 2 In addition to its self-energy, a magnetized specimen acquires a potential energy when it is subjected to an applied ﬁeld H . This is ? given by W : 9 ?

4

M · H dv. ?

(3.71)

This can be viewed as the work required to move the specimen from an environment with zero ﬁeld to a region permeated by H . From Eq. ? (3.71) we see that the energy density due to the coupling of M to H is ? w : 9 M · H ? ? : 9 MH cos(), (3.72) ? where is the angle between M and H . ?

118

CHAPTER 3 Field Analysis

3.2.5 Inductance In this section we introduce the concept of inductance. Consider a linear magnetic system in the form of an electrical circuit that consists of a voltage source V, a switch, resistor R and an ideal (no resistance) N turn coil (Fig. 3.10). When the switch is open there is no current and the system contains no energy. When the switch is closed, a current I ﬂows, and an energy W is supplied by the source. A portion of this energy is Q dissipated in the form of heat W : IR, and the remainder is stored in the magnetic ﬁeld W : K W :W ;W , Q K where 1 W : K 2

A · J dv (J)

(3.73)

4 (Section 3.2.4). For a linear system, both A and J are proportional to I and Eq. (3.73) can be written as 1 W : LI, K 2

(3.74)

where L is called the self-inductance of the coil. The unit of inductance is the Henry (H), which equals one volt second/ampere (1 H : 1 Vs/A). It is important to note that the expression (3.74) only applies to a magnetically linear system. There is no simple deﬁnition of inductance for nonlinear systems [2].

FIGURE 3.10

Series RL circuit.

119

3.2 MAGNETOSTATIC ANALYSIS

We can express the inductance in terms of other ﬁeld variables by substituting either Eq. (3.65) or (3.73) for W in Eq. (3.74). Speciﬁcally, K we ﬁnd that

L:

1 I 1 I

A · J dv 4

(linear system)

(3.75)

B · H dv. 4

The ﬁrst integral in Eq. (3.75) is over the volume of the conductor (the region of nonvanishing J), whereas the second integral is over all space (the region of nonvanishing B and H). The second integral in Eq. (3.75) can be decomposed into two separate integrals, that is, one for the region V inside the conductor, and another for the region V outside the conductor. In this case, we have L:L ;L

where 1 L : I

B · H dv

(internal inductance),

(3.76)

4int

inside conductor

and 1 L : I

4ext

B · H dv

(external inductance);

(3.77)

outside conductor

L and L are called the internal and external self-inductance, respec tively. It can be shown that

1 L : I

B · ds

1

ext

outside conductor

:

" , I

where S is the surface area external to and bounded by the conductor, and " is the ﬂux linkage "Y

1ext

B · ds

(ﬂux linkage),

120

CHAPTER 3 Field Analysis

FIGURE 3.11

Magnetically coupled circuits.

" is the total ﬂux linking the conductor. For many applications L L and we use the approximation LL

" : . I

(3.78)

So far, we have considered the inductance of a single circuit in isolation (self-inductance). However, two or more circuits can be magnetically coupled to one another giving rise to mutual inductance effects. For example, consider the two circuits shown in Fig. 3.11. The current I through circuit 1 gives rise to a ﬂux " through circuit 2 and vice versa. If the circuits are magnetically linear, the coupling can be expressed in terms of mutual inductance coefﬁcients M and M as follows: " M : , (3.79) I and " M : . I These coefﬁcients can be determined using

1 M : A · J dv I I 4 1 : B · H dv, I I 4

(3.80)

121

3.2 MAGNETOSTATIC ANALYSIS

and

1 M : A · J dv I I 4 1 B · H dv. : I I 4 Moreover, it can be shown that M :M . The total magnetic energy of the circuits can be written as 1 1 W : L I ; L I ; MI I , K 2 2 where M : M : M . We demonstrate some of these concepts in the following examples. Example 3.2.9 Determine the internal self-inductance per unit length of an inﬁnitely long straight wire of radius R. Assume that the wire has a permeability . Solution 3.2.9 Let I denote the total current through the wire. The current within a radius r R (inside the wire) is i(r) : I

r . R

Consider a length h of the wire. From Example 3.2.1 we know that H and B at a radius r inside the wire are H (r) :

i(r) , 2r

(3.81)

and i(r) B (r) : . 2r

(3.82)

Substitute Eqs. (3.81) and (3.82) into Eq. (3.76), and integrate over the length of wire h. This gives L : I

F

L

h : (H). 8

0 i(r) r dr d dz 2r (3.83)

122

CHAPTER 3 Field Analysis

Parallel wire transmission line: (a) reference frame; and (b) cross section ([2] of Chapter 2).

FIGURE 3.12

Thus, the internal self-inductance per unit length is L : (H/m). 8

(3.84)

Example 3.2.10 Determine the internal and external self-inductance per unit length of a transmission line consisting of two inﬁnite parallel wires (Fig. 3.12). Solution 3.2.10 Consider a length h of the transmission line. We ﬁrst compute the internal inductance. This is just the sum of the internal inductances of the individual wires. From Eq. (3.84) we have L : . 4

(3.85)

Next, we compute the external inductance. We need to determine the ﬂux linkage between the two wires. This depends on the ﬁeld component perpendicu-

123

3.2 MAGNETOSTATIC ANALYSIS

lar to the area between the wires. Therefore, we need B which is the sum of the W ﬁelds of the individual wires B : B ; B, W W W where I B : , W 2x and I . B : W 2(s 9 x) The total ﬂux between the wires is given by

Q\?

(B ; B dx W W ? hI s9a : ln . a

":h

The external inductance per unit length is " L : hI

s9a : ln . a Thus, the total inductance per unit length is L:L ;L s9a : ; ln . 4 a

(3.86)

If the radii of the wires are small relative to the separation distance (a s), then Eq. (3.86) reduces to

s L ln a

(H/m).

(3.87) )

Example 3.2.11 Determine the self-inductance of a long, tightly wound solenoid with N turns, length L and radius a (Fig. 3.2).

124

CHAPTER 3 Field Analysis

Solution 3.2.11 From Example 3.2.2 we know that the ﬁeld on the axis of the solenoid is N B: I . L The ﬂux through the coil is Ba. Because each turn of the coil intercepts this ﬂux, the ﬂux linkage is " : NBa N : I a. L From Eq. (3.78) we ﬁnd that the inductance of the coil is N L: a (H). L The inductance per unit length is L:

N a (H/m). L

)

Example 3.2.12 Determine the self-inductance of a tightly wound N turn coil wrapped around a toroid of material with a height h and inner and outer radii R and R , respectively. Assume that the material is linear with a permeability (Fig. 3.3). Solution 3.2.12 From Example 3.2.3 we know that the ﬁeld inside the toroid is H :

NI 2r

(R r R ).

The ﬂux linkage is

0 H dr 0 hNI 0 1 : dr 2 r 0 hNI R . : ln 2 R

" : h

125

3.2 MAGNETOSTATIC ANALYSIS

FIGURE 3.13

Coaxial solenoids ([3] of Chapter 2).

The inductance is estimated using Eq. (3.78), which gives L:

hN R ln 2 R

(H).

(3.88)

Example 3.2.13 Determine the mutual inductance beween the coaxial solenoids shown in Fig. 3.13. Solution 3.2.13 Consider the inner solenoid. It has a current I , length l , and N turns. The ﬁeld along its axis is given by Eq. (3.17) (Example 3.2.2), N I . B : l Thus, the ﬂux through the outer solenoid due to I is B A, where A : R (R is the radius of the inner solenoid). This ﬂux passes through each of the N turns of the outer solenoid. Therefore, the ﬂux linkage is " : B N A, N N I R. : l The mutual inductance is given by Eq. (3.79), " M : I N N (H). l Now consider the outer solenoid. It has a current I , length l and N turns. : R

126

CHAPTER 3 Field Analysis

The ﬁeld along its interior due to I is N I . B : l This ﬁeld passes through the interior of the inner solenoid, which has an area A : R. It creates a ﬂux linkage with N l /l turns of the inner solenoid. Therefore, the ﬂux linkage " is l " :B N A l N N I R. : l The mutual inductance M is " M : I N N : R (H). l Notice that M : M as expected. )

3.3 THE CURRENT MODEL The current model is used in the analysis of permanent magnets. In this model, the magnet is reduced to a distribution of equivalent current. This is then input into the magnetostatic ﬁeld equations as a source term, and the ﬁeld is obtained using standard methods for steady currents. The derivation of the current model starts with the magnetostatic ﬁeld equations (3.1) and (3.2). Apply Proposition A.5.1 to the solenoidal ﬁeld B and introduce the vector potential B : ;A.

(3.89)

Substitute Eq. (3.89) into Eq. (3.1), taking into account the constitutive relation (3.7). This yields A 9 ( · A) : 9 (J ; ;M). Next, impose the Coulomb gauge condition · A : 0 and obtain A : 9 (J ; ;M).

(3.90)

127

3.3 THE CURRENT MODEL

The form of Eq. (3.90) suggests the deﬁnition of an equivalent magnetic volume current density J Y ;M. Notice that because Eq. (3.90) is a K linear equation, the potential A (and hence B) can be obtained as a superposition of the solutions for J and J separately. K The derivation of the current model is summarized in the following diagram: A 9 ( · A):9 (J ; J ) K ;H : J B : ;A (J Y ;M) K $ $ ·B:0 B : (H ; M) Coulomb Gauge ·A:0

A : 9 (J ; J ). K If there is no free current (J : 0), and if we assume an inﬁnite homogeneous material (no boundaries), then the solution to Eq. (3.90) can be written in integral form using the free-space Green’s function G(x, x) : 9(1/4)(1/ x 9 x ) for the operator (Appendix B). Speciﬁcally, we ﬁnd that A(x) : 4

J (x) K dv.

x 9 x

(3.91)

Notice that in Eq. (3.91) the integration is over the region containing the source J . We compute B : ;A and obtain K (x 9 x) dv. (3.92) B(x) : J (x); K

x 9 x 4

To obtain Eq. (3.92) we have used J (x) 1 ; K : 9J (x); , K

x 9 x

x 9 x and

1 (x 9 x) :9 ,

x 9 x

x 9 x

where indicates differentiation with respect to the unprimed variables. If the magnetization M is conﬁned to a volume V (of permeability ), and falls abruptly to zero outside of V, then Eqs. (3.91) and (3.92)

128

CHAPTER 3 Field Analysis

reduce to A(x) : 4

4

J (x) K dv ;

x 9 x 4

1

j (x) K ds,

x 9 x

(3.93)

and B(x) : 4

4

(x 9 x) dv ; J (x); K 4

x 9 x

1

(x 9 x) ds, j (x); K

x 9 x

(3.94)

respectively. In these expressions S is the surface of the magnet, and J K and j are equivalent volume and surface current densities. These are K deﬁned in the following: Equivalent Currents J : ;M (A/m) (volume current density) K j : M;n (A/m) (surface current density). K

(3.95)

We demonstrate the current model in the following examples. Example 3.3.1 Determine the ﬂux density along the axis of a cylindrical magnet that is polarized along its axis with a uniform magnetization M : M z . Q The magnet has a radius R and length L (Fig. 3.14a). Solution 3.3.1 We use cylindrical coordinates. The equivalent current densities are computed using Eq. (3.95). We readily determine that J : ;M : 0. (3.96) K To evaluate the surface term, we ﬁrst evaluate the unit surface normals n : and then compute

z z : 0 r

r:R

9z z : 9L

j : M;n : M (r : R). (3.97) K Q The only nonvanishing source term is a surface current that circulates around the body of the cylinder (Fig. 3.14b). We substitute Eqs. (3.96) and (3.97) into

129

3.3 THE CURRENT MODEL

FIGURE 3.14

Cylindrical magnet: (a) physical magnet; and (b) equivalent surface

current.

Eq. (3.94) with x : zz , x : Rr ; ; zz and x 9 x given by Eq. (A.15) with p (0, 0, z) and p (R, , z). This yields L (zz 9 (Rr ; ; zz ))R d dz B(z) : M ; . Q 4 [R ; (z 9 z)] \* We are interested in the z-component of B, which is given by

B (z) : X 4

L

\* This is readily integrated yielding

M R d dz Q . [R ; (z 9 z)]

M z z;L B (z) : Q 9 . X 2 ((z ; L) ; R (z ; R

(3.98)

We demonstrate Eq. (3.98) with some sample calculations. Calculations: We apply Eq. (3.98) to a solid cylindrical magnet. First, we calculate B along the line 2 z 12 mm with R : 5 mm, L : 20 mm, and X M : 8;10 A/m (B : 10,000 G) (Fig. 3.15). Notice that there is a rapid Q P decrease in the ﬁeld near the surface of the magnet. Next, we compute B at a X ﬁxed height z : 5 mm and vary the magnet radius 2 R 30 mm (Fig. 3.16). Here, we set L : 20 mm and M : 8;10 A/m as before. Notice that B Q X

130

CHAPTER 3 Field Analysis

FIGURE 3.15

Component B vs z for a solid cylinder. X

initially increases with R, but then peaks (at R < 17 mm), and gradually decreases. The reason for this is as follows: As R increases, the projection of the ﬁeld along the z-axis increases, but the ﬁeld source (equivalent surface current) moves farther away from the observation point. For small values of R the ﬁrst effect dominates and the ﬁeld increases. After a point, the second effect dominates and the ﬁeld decreases. )

FIGURE 3.16

(z : 5 mm).

Component B vs magnet radius R for a cylindrical magnet X

131

3.4 THE CHARGE MODEL

The current model is useful for computing the force and torque on permanent magnets. The force and torque can be determined using the basic relation for the force on a distribution of currents in an external ﬁeld as described in Section 3.2.2. The procedure is as follows. 1. Use the current model to reduce the magnet to a distribution of equivalent volume and surface current densities J and j as K K speciﬁed in Eq. (3.95). 2. Determine the force using F:

4

J ;B dv ; K

1

j ;B ds. K

(3.99)

where V and S are the volume and surface of the magnet, respectively. 3. Determine the torque using T:

4

r;(J ;B ) dv ; K

1

r;(j ;B ) ds, K

(3.100)

where r is the vector from the point about which the torque is computed. It is important to note that when evaluating Eq. (3.100), the force terms (J ;B ) and ( j ;B ) must be evaluated before the cross product with K K r is taken. We will use Eqs. (3.99) and (3.100) in our study of electromechanical devices in Chapter 5.

3.4 THE CHARGE MODEL The charge model is another useful method for analyzing permanent magnets. In this model, a magnet is reduced to a distribution of equivalent ‘‘magnetic charge.’’ The charge distribution is used as a source term in the magnetostatic ﬁeld equations, and the ﬁelds are obtained using standard methods. The derivation of the charge model is as follows: Start with the magnetostatic ﬁeld equations for current-free regions ;H : 0 and · B : 0. Next, apply Proposition A.5.2 to the irrotational ﬁeld H and introduce a scalar potential , K H : 9 . K

(3.101)

132

CHAPTER 3 Field Analysis

Finally, substitute Eq. (3.101) and the constitutive relation B : (H ; M) into · B : 0 and obtain : · M. (3.102) K In the absence of boundary surfaces we can represent the solution to Eq. (3.102) in integral form using the free space Green’s function G(x, x) for (Appendix B). In particular, we ﬁnd that (x) : K

G(x, x) · M(x) dv

:9

1 4

· M(x) dv,

x 9 x

(3.103)

where x is the observation point, x is the source point, operates on the primed coordinates, and the integration is over the volume for which the magnetization exists. If M is conﬁned to a volume V (of permeability ), and falls abruptly to zero outside of this volume, then Eq. (3.103) becomes 1 (x) : 9 K 4

1 · M(x) dv ; 4

x 9 x

M(x) · n ds,

x 9 x

(3.104)

4 1 where S is the surface that bounds V, and n is the outward unit normal to S. The form of Eq. (3.104) suggests the deﬁnitions of volume and surface charge densities given in Eq. (3.105): Charge Model : 9 · M (A/m) K : M · n (A/m) K

(volume charge density),

(3.105)

(surface charge density).

The derivation of the charge model is summarized in Eq. (3.06): ;H : 0 H : 9 : 9 K K K $ $ ·B:0 B : (H ; M) ( : 9 · M). K

(3.106)

If the magnet is in free space, B : H and from Eqs. (3.101) and (3.104) we have B(x) : 4

4

(x)(x 9 x) K dv ;

x 9 x 4

1

(x)(x 9 x) K ds.

x 9 x

(3.107)

133

3.4 THE CHARGE MODEL

The B-ﬁeld due to an isolated permanent magnet with a known magnetization M in free space is obtained from Eq. (3.107) by ﬁrst evaluating and , and then performing the indicated integrations. K K From the preceding analysis, we ﬁnd that the ﬁeld at a point x due to a ‘‘point charge’’ Q (x) at x is + Q (x)(x 9 x) B(x) : K . 4

x 9 x

(3.108)

where Q (x) : (x)V K K

(volume charge),

(3.109)

or Q (x) : (x)A (surface charge), (3.110) K K and V and A are the volume and area elements containing (x) K and (x), respectively. In one dimension, Eq. (3.108) reduces to K Q (x) K B(x) : . 4 (x 9 x)

(3.111)

We demonstrate the use of the charge model in the following example. Example 3.4.1 Compute the ﬁeld above a rectangular bar magnet of width 2a, depth 2b, and height L. The magnet is polarized along its axis with uniform magnetization as shown in Fig. 3.17a. Assume that the magnetization is M : M z . (3.112) Q Solution 3.4.1 We ﬁrst determine the charge densities. From Eq. (3.105) we ﬁnd that : 9 · M : 0. K To evaluate we ﬁrst compute the unit surface normals K z z : 0 n:

9z

z : 9L

<x

x :

From Eq. (3.112) and the fact that : M · n we have : M for the top K K Q surface (z : 0), and : 9M for the bottom surface (z : L). We evaluate B K Q X

134

FIGURE 3.17

CHAPTER 3 Field Analysis

Rectangular bar magnet, (a) physical magnet; and (b) equivalent

charge.

along the z-axis due to the top surface of the magnet (z : 0) using Eq. (3.107) with x : zz and x : xx ; yy , B (z) : X 4

?

@

\? \@

Mz Q dx dy. [x ; y ; z]

(3.113)

Notice that the integrand in Eq. (3.113) is an even function of x and y, and therefore B (z) : X

?

@

Mz Q dx dy [x ; y ; z]

Mz ? b : Q dy (y ; z)(b ; y ; z M by ? : Q tan\ z(b ; y ; z

M z(a ; b ; z : Q 9 tan\ 2 ab

.

In the last step we have used tan\(x) ; tan\(1/x) : /2. This is the ﬁeld due to the top surface. A similar analysis applies to the bottom surface with z

135

3.4 THE CHARGE MODEL

replaced by z ; L. The total ﬁeld is given by

M (z ; L)(a ; b ; (z ; L) B (z) : Q tan\ X ab 9 tan\

z(a ; b ; z ab

.

(3.114) )

3.4.1 Force The charge model is also useful for determining the force and torque on a magnet in an external ﬁeld. An expression for the force in terms of magnetic charge can be derived from Eq. (3.99). Speciﬁcally, start with F: :

4

J ;B dv ; K

1

j ;B ds K

(;M);B dv ; (M;n );B ds, 4 1 and then apply the vector identity

(;U);V dv 9

4

(3.115)

(n ;U);V ds

1 :

[U;(;V) 9 U( · V)] dv ; (U · )V dv 4 4 to Eq. (3.115) with U : M and V : B . This gives F:

4

(M · )B dv,

(3.116)

where

(M · )B : M ;M ;M B x V V x W y X z

; M ;M ;M B y W V x W y X z ; M ;M ;M B z . X V x W y X z

To obtain Eq. (3.116) we have used the fact that · B : 0, which is

136

CHAPTER 3 Field Analysis

always true, and the fact that ;B : 0, which holds because the source of B does not overlap the region occupied by M. From Eq. (3.116) we ﬁnd that f : (M · )B

(force density),

(3.117)

is a force density. Apply the identity

(U · )V dv : 9 4

( · U)V dv ;

4

(n · U)V ds

1

to Eq. (3.116) with U : M and V : B and obtain

F:9

4

( · M)B dv ;

1

(M · n)B ds.

From Eq. (3.105) this can be rewritten as

F:

4

B dv ; K

1

B ds. K

(3.118)

We can use Eq. (3.118) to compute the force on a magnet. The procedure is as follows: 1. given a magnet with a magnetization M, determine the equivalent charge densities : 9 · M and : M · n ; and K K 2. substitute and into Eq. (3.118) and determine the force F. K K We demonstrate this procedure in the following example. Example 3.4.2 Determine the activation force of the solenoidal actuator shown in Fig. 3.18a. Assume that the magnet has a ﬁxed and uniform magnetization M : 9M z . Q Solution 3.4.2 We compute the activation force using Eq. (3.118). Consider a cylindrical magnet of radius R and length h (Fig. 3.18b). The equivalent charge densities are obtained from Eq. (3.105). We ﬁnd that : 9 · M : 0. K

137

3.4 THE CHARGE MODEL

Solenoid with permanent magnet plunger: (a) device with circuitry; and (b) magnet in isolation.

FIGURE 3.18

To evaluate , we ﬁrst determine the unit surface normals K

9z z : 0

n :

z z : h r

r : R.

Next, we evaluate : M · n and obtain K

M z:0 Q : K 9M z : h Q The ﬁeld B

due to a solenoid was derived in Example 3.2.2. Speciﬁcally, B

N : i , L

where i is the current, N is the total number of turns, and L is the length of the solenoid. Only the north pole of the magnet is in the ﬁeld of the solenoid.

138

CHAPTER 3 Field Analysis

The force is therefore F:

1

B ds K

N : i M L Q

0

L

r dr d

N : i M R. L Q

)

Often, it is not possible to evaluate Eq. (3.118) analytically. When this is the case, the force is obtained numerically by discretizing the volume V and surface S of the magnet F : (x )B (x )V ; (x )B (x )A , (3.119) K L L L K N N N L N where V and A are indexed volume and area elements, respectively. L N Combining this expression with Eq. (3.108) gives the force between two separate magnets

(x ) (x )(x 9 x ) H V V F: K L K H L (3.120) H L 4

x 9 x L H L H (x ) (x )(x 9 x ) I A V (3.121) ; K L K I L I L

x 9 x 4 L I L I (x ) (x )(x 9 x ) H V A ; K N K H N (3.122) H N 4

x 9 x N H N H (x ) (x )(x 9 x ) I A A , (3.123) ; K N K I N I N

x 9 x 4 N I N I where the j and k indices label the discretization of one magnet, and n and p label that of the other. Once two magnets have been reduced to discretized charge distributions, we use Eqs. (3.120) to (3.123) to compute the force between them. From this analysis, it follows that the force between two discrete ‘‘point charges’’ Q (x ) at x and Q (x ) at x is K K Q (x )Q (x )(x 9 x ) F : K K 4

x 9 x

(3.124)

139

3.4 THE CHARGE MODEL

where the charge Q is computed using either Eq. (3.109) or (3.110) K depending on whether it is volume charge or surface charge. In one dimension Eq. (3.124) reduces to Q (x )Q (x ) F : K K . 4 (x 9 x )

(3.125)

We demonstrate the use of Eq. (3.125) in the following example. Example 3.4.3 Consider two thin cylindrical magnets oriented along the x-axis as shown in Fig. 3.19a. Both magnets have a radius R and length L and are polarized to a level M . Assume R L and determine the force between the two Q magnets as a function of their separation d. Solution 3.4.3 We compute the force on the magnet to the right. Because the magnets are thin (R L), we can treat the magnets as dipoles to ﬁrst order (Fig. 3.19b). The charge at each pole face is given by Eq. (3.110), which reduces to Q : <M R, K Q where the < sign refers to the north and south poles, respectively. The force is obtained by summing the forces between the charges as given by Eq. (3.125). We obtain

Q 1 2 1 F(d) : K 9 ; 9 . 4 d (L ; d) (2L ; d )

(3.126)

Force between two thin bipolar cylindrical magnets: (a) physical system; and (b) equivalent charge model.

FIGURE 3.19

140

CHAPTER 3 Field Analysis

FIGURE 3.20

Magnetic dipole.

The force on the magnet to the left is of the same magnitude but opposite direction as shown. ) Finally, consider a short thin cylindrical magnet of radius a, length d, and magnetization M as shown in (Fig. 3.20). From the analysis in Q Example 3.4.3 we know that there are surface charges of 9Q and ;Q K K at the left and right ends of the magnet, where Q : M a. These two K Q charges form a magnetic dipole with a dipole moment given by m:Q d (dipole moment). K The ﬁeld of the dipole can be obtained from Eq. (3.108), which gives

r r Q B: K 9 ; , (3.127) 4 r r where r : (1/2)d ; r and r : 9(1/2)d ; r. When d r, Eq. (3.127) can be approximated by [7]

3(m · r)r m B: 9 . 4 r r

(3.128)

Consider two dipoles m and m separated by a distance r d (Fig. 3.21). The force imparted to m by m follows from Eq. (3.117). Speciﬁ cally, we have F : (m · )B . We substitute Eq. (3.128) into Eq. (3.129) and obtain [1]

(3.129)

5(m · r)(m · r)r 3 F : (m · m )r ; (m · r)m ; (m · r)m 9 . 4 r r (3.130)

141

3.4 THE CHARGE MODEL

FIGURE 3.21

Two separated dipoles.

3.4.2 Torque The charge model is also useful for determining the torque on a magnet in an external ﬁeld. We derive a formula for the torque as follows: Start with the torque formula (3.100) from the current model T: :

4 4

r;(J ;B ) dv ; K

1

r;(j ;B ) ds K

r;[(;M);B ] dv ;

1

r;[(M;n );B ] ds.

(3.131)

Next, apply the identity

r;[(;U);V] dv 9

4 :

r;[(n ;U);V] ds

1

r;[U · V ; U;(;V) 9 U( · V)] dv ;

4

U;V dv, 4

with U : M and V : B . This gives T:

r;(M · B ) dv ;

4

4

M;B dv.

(3.132)

Here, we have used · B : 0 and ;B : 0. From Eq. (3.132) we see that if B is uniform (B : 0), then the torque reduces to a couple T:

4

M;B dv

(uniform B ).

(3.133)

142

CHAPTER 3 Field Analysis

Finally, apply the identity

r;(U · V) dv; U;V dv:9 ( · U)(r;V) dv; (n · U)(r;V) ds 4 4 4 1 to Eq. (3.132) with U : M and V : B . This yields T:

(9 · M)(r;B ) dv ; (n · M)(r;B ) ds. 4 1 From Eq. (3.105) we have T:

4

(r;B ) dv ; K

1

(r;B ) ds. K

(3.134)

We can use Eq. (3.134) to compute the torque on a magnet. The procedure is as follows: 1. given a magnet with a magnetization M, determine the equivalent charge densities : 9 · M and : M · n ; and K K 2. substitute and into Eq. (3.134) and determine the torque T. K K We demonstrate this procedure in the following example. Example 3.4.4 Determine the torque on a bipolar cylinder in a uniform ﬁeld (Fig. 3.22a). Assume that the cylinder has a magnetization M at an angle Q with respect to the x-axis (Fig. 3.22b). Solution 3.4.4 We use Eq. (3.134) to determine the torque. First, determine the equivalent charge densities Eq. (3.105). Let R and h denote the radius and length of the cylinder, respectively. The magnetization is given by M : M [cos()x ; sin()y ], Q where is the angular rotation of M relative to the x-axis (Fig. 3.22b). It follows that : 9 · M : 0. The surface charge is given by K : M · n K : M [cos() cos() ; sin() sin()], (3.135) Q

where we have used n : r : cos()x ; sin()y . The external ﬁeld is B : B x : B [cos()r 9 sin() ].

(3.136)

3.4 THE CHARGE MODEL

143

Bipolar cylinder in an external ﬁeld: (a) perspective of the cylinder; and (b) cross-sectional view.

FIGURE 3.22

Substitute Eqs. (3.135) and (3.136) into Eq. (3.134) and obtain

T:

F

L

M [cos() cos() ; sin() sin()] Q

;(Rr ;B [cos()r 9 sin() ])R d dz L : 9M hRB [cos() cos() sin() ; sin() sin()] dz Q : 9M hRB sin()z . (3.137) Q

This is the torque on the magnet as a function of its rotation angle .

)

144

CHAPTER 3 Field Analysis

3.5 MAGNETIC CIRCUIT ANALYSIS Magnetic circuits consist of ﬁeld sources such as magnets and coils, and ﬂux conduits made from soft magnetic materials that guide and direct the ﬂux. In this section we develop equivalent lumped-parameter models for analyzing such circuits. These models follow from the integral form of the magnetostatic ﬁeld equations

H · dl :

!

J · ds

1 :I ,

(3.138)

and

B · ds : 0. (3.139) 1 In Eq. (3.138), I is the total current passing through the area S, which is bounded by the contour C. It is important to note that in order to evaluate the left-hand side of Eq. (3.138), we need to know H · dl. That is, we need to know the orientation and magnitude of H relative to the integration path dl. For most practical applications, we choose dl based on the circuit geometry and assume that H is parallel to dl, which reduces H · dl to the scalar product H dl. This is one of the key assumptions of circuit analysis. Another is that the ﬂux generated by the ﬁeld sources is conﬁned to the circuit with negligible leakage. Recall that the ﬂux through a surface S is given by :

B · ds (magnetic ﬂux). (3.140) 1 These assumptions are difﬁcult to verify a priori, and need to be carefully weighed when analyzing a circuit. However, even when they are not strictly valid, the performance of a magnetic circuit can usually still be estimated with reasonable accuracy.

3.5.1 Current sources Consider a magnetic circuit consisting of a discrete set of segments, each with a different permeability, and a ﬁeld source in the form of a coil. A circuit with four segments is shown in Fig. 3.23. We apply Eq. (3.138) to

145

3.5 MAGNETIC CIRCUIT ANALYSIS

FIGURE 3.23

Magnetic circuit with four segments.

a circuit with m such segments and obtain

K H · dl : H l : I . (3.141) GG ! G To simplify the analysis, assume that each element has a linear response B : H , and constant cross-sectional area A . Then, from Eq. (3.140) we G G G G have B : /A , and Eq. (3.141) becomes G G G K l G :I . (3.142) A G G G G or K R :I , G G G

where

l R: G , (3.143) G A G G is the reluctance of the i th element. If the cross-sectional area and permeability of an element vary over its length, then Eq. (3.143) generalizes to an integration RY

dl A

(reluctance),

where and A are the permeability and the cross-sectional area along the path. The reluctance of different rectangular sections are shown in Fig. 3.24. If there is no ﬂux leakage, : and Eq. (3.142) reduces to G K R :I . (3.144) G G

146

FIGURE 3.24

CHAPTER 3 Field Analysis

Reluctance of rectangular sections: (a) solid element; and (b) air

gap.

In addition, if the current source is an n-turn coil carrying a current i, then I : ni, and Eq. (3.144) becomes :

ni . R ;R ;···;R K

(3.145)

We can use Eq. (3.145) to compute the ﬂux density in the ith element. Speciﬁcally, B : /A , which gives G G G 1 ni B: , G A (R ; R ; · · · ; R ) G K

(3.146)

(recall that : ). G In most applications Eq. (3.146) simpliﬁes because the reluctance of a few elements dominates the denominator. In particular, soft magnetic circuit elements have a high relative permeability when they operate P below saturation (Fig. 3.25). Therefore, these elements have a much lower reluctance than gap regions, and can usually be ignored when gaps are present. Magnetic circuits have many similarities to electrical circuits. Recall that electrical circuits are governed by Kirchhoff’s voltage and current laws R I : V , G G I G I

(3.147)

I : 0. G G

(3.148)

and

147

3.5 MAGNETIC CIRCUIT ANALYSIS

FIGURE 3.25

The B-H curve of soft magnetic material.

Kirchhoff’s voltage law (3.147) states that around any closed path the algebraic sum of the voltage drops equals the algebraic sum of voltage rises (applied voltage). The current law (3.148) states that the algebraic sum of the currents ﬂowing into a circuit node is zero. The analogous laws for magnetic circuits are

and

R: n i , G G II G I

(3.149)

: 0. G G

(3.150)

The ﬁrst law states that around any closed path the algebraic sum of the products of the reluctance and ﬂux equals the algebraic sum of ampereturns (A · t). The second law states that the algebraic sum of the magnetic ﬂux ﬂowing into a circuit node is zero. An example of Eq. (3.150) is illustrated in Fig. 3.26. The magnetic circuit laws (3.149) and (3.150) are summarized in the following:

!

H · dl : I

B · ds : 0

$ $

R: n i , G G II G I : 0. G G

(3.151)

148

CHAPTER 3 Field Analysis

FIGURE 3.26

Algebraic sum of ﬂux into a node is zero.

In keeping with the electrical analogy, we introduce the notation V K for the source term in a magnetic circuit; V is known as the magK netomotive force (mmf) and is given by V : ni K

(3.152)

for an n-turn coil. The unit of V is the ampere (A). However, it is K often speciﬁed in terms of ampere-turns (A · t) because of the form of Eq. (3.152). If a circuit contains one source, then the ﬁrst equation in Eq. (3.151) reduces to R :V . G G K G

(3.153)

If there are N sources, then , R : V I , K G G G I

(3.154)

where V I is the magnetomotive force (mmf ) of the kth source. K Recall that resistors in parallel give rise to an equivalent resistance 1 1 1 : ; ;···. R R R If we work with conductance G : 1/R, then the equivalent conductance is G :G ;G ;···.

149

3.5 MAGNETIC CIRCUIT ANALYSIS

FIGURE 3.27

Permeance of gap regions: (a) angular gap; and (b) radial gap.

Similarly, when parallel reluctances are encountered, it is convenient to work with permeance P, P:

1 R

(permeance).

The permeance of a rectangular element of length l, cross-sectional area A, and permeability is P:

A . l

Other geometries of interest are shown in Fig. 3.27. Speciﬁcally, the permeance of an angular gap between radially oriented inclined surfaces (Fig. 3.27a) is given by w P : ln(r /r ). Similarly, the permeance of a radial gap between cylindrical surfaces at radii r and r with axial length w that subtend an angle is (Fig. 3.27b), P:

w . ln(r /r )

Other geometries are discussed in Reference 8.

150

CHAPTER 3 Field Analysis

FIGURE 3.28

Magnetic circuit with current source.

Example 3.5.1 Consider the magnetic circuit shown in Fig. 3.28. Determine the ﬂux density in the gap, and the inductance of the coil. Assume that the core has a permeability and that there is no leakage ﬂux. Solution 3.5.1 We determine the ﬁeld in the gap ﬁrst. Choose an integration path along the axis of the circuit as indicated by the dotted line. Apply Eq. (3.138) to the path assuming that H is parallel to it. This gives

H · dl : ni,

!

or H l ; H l : ni, (3.155) AA EE where l and l are the lengths of the path in the core and gap, respectively, and A E H and H are the ﬁelds in these sections. Notice that I : ni. Next, apply Eq. A E (3.139) at the core-gap interface. As there is no leakage ﬂux, all the ﬂux that passes through the core also passes through the gap : , A E B A :B A (no fringing ﬂux at gap), (3.156) A A E E where A and A are the cross-sectional areas of the core and gap. Substitute Eq. A E (3.156) into Eq. (3.155), making use of B : H and B : H . This gives A A E E ni B : E (A /A )(l /) ; (l / ) E A A E ni : . (3.157) (A /A )( /)l ; l E A A E

3.5 MAGNETIC CIRCUIT ANALYSIS

151

If we assume that A /A 1 and / 1, then Eq. (3.157) reduces to E A B : ni/l . It is instructive to rederive Eq. (3.157) using Eq. (3.144). E E Speciﬁcally, ni , (3.158) R ;R A E where R : l /A and R : l / A . If we substitute R , R , and : B A A A A E E E A E E E into Eq. (3.158) we obtain Eq. (3.157). When R R , we obtain E niA E : ( ). (3.159) l E Inductance: Next, we determine the inductance. From Eqs. (3.156) and (3.157) we have :

ni . B : A ( /)l ; (A /A )l A A E E Thus, the ﬂux through the core is

(3.160)

:B A A A A

niA A . (3.161) ( /)l ; (A /A )l A A E E Because this ﬂux passes through each turn of the coil, the ﬂux linkage is :

niA A . ( /)l ; (A /A )l A A E E Finally, the inductance of the coil follows from Eq. (3.78), ":

L:

(3.162)

" i

nA A . (3.163) ( /)l ; (A /A )l A A E E If there is no gap, l : 0 and we obtain E nA A L: (no gap). (3.164) l A On the other hand, if there is a gap, and the permeability of the core is high ( ), then L can be approximated by nA E. L: (3.165) l E :

152

CHAPTER 3 Field Analysis

Calculations: We demonstrate the theory with some sample calculations. Consider a magnetic circuit with the following parameters: n : 110, i : 1 A, l : 10 cm, l : 1 mm, : 1000 , and A : A : 1 cm. The ﬁelds and ﬂux A E A E in the core and the gap are as follows: Core

Gap

100

100,000

H (A/m) B (T)

4;10\ 4;10\

(Wb)

4;10\ 4;10\

If the circuit has no gap (l : 0), then we ﬁnd that E Core H (A/m)

1100

B (T)

1.38

(Wb)

1.38;10\

)

Example 3.5.2 Determine the inductance of the actuator shown in Fig. 3.29. Assume that the core has a permeability , and that there is no leakage ﬂux. Solution 3.5.2 We follow the same analysis as in the previous example. First, apply Eq. (3.138) to a path around the actuator circuit (dotted line) assuming that H is parallel to the path. This gives

H · dl : ni

!

FIGURE 3.29

Magnetic circuit actuator with a current source.

3.5 MAGNETIC CIRCUIT ANALYSIS

153

or H l ; 2H x : ni. (3.166) AA E Next, apply Eq. (3.139) at the core-gap interface assuming that : , A E B A :B A (no fringing ﬂux at gap). (3.167) A A E E Substitute Eq. (3.167) into Eq. (3.166) with B : H and B : H , and A A E E obtain ni B : E (A /A )(l /) ; (2x/ ) E A A ni : . (A /A )( /)l ; 2x E A A The ﬂux through the circuit is : B A . Therefore, E E niA E : . (A /A )( /)l ; 2x E A A We are interested in the inductance of the coil. This can be determined from the ﬂux linkage ". Because the ﬂux passes through each turn of the coil, the ﬂux linkage is " : n. The inductance follows from Eq. (3.78), and is given by L:

" i

(3.168)

nA E . (3.169) (A /A )( /)l ; 2x E A A Notice that the inductance in a function of the gap spacing x, L : L(x). Because / 1, we can estimate the inductance using nA E. L(x) ) 2x :

3.5.2 Magnet sources Permanent magnets are frequently used as source elements in magnetic circuits. The analysis of a circuit with a magnet source is similar to that of a current source. However, magnets are different than current sources in two respects. First, the mmf for a current source (coil) is known and is given by ni, but the mmf for a magnet is not known a priori. Instead, it depends on the magnet’s operating point (B , H ), which is a function K K

154

CHAPTER 3 Field Analysis

of the circuit itself. Second, the magnet itself is physically part of the circuit, and therefore represents a circuit reluctance. The procedure for the analysis of permanent magnet circuits is demonstrated in the following example. EXAMPLE 3.5.3 Consider the circuit of Fig. 3.30. Derive an expression for the operating point of the magnet (B , H ). Determine the ﬂux density B in the K K E gap. Derive an expression for the magnetostatic energy in the gap. Assume that the core has an inﬁnite permeability -, and that there is no leakage ﬂux. Furthermore, assume that the magnet has a linear second quadrant demagnetization curve of the form B :B ; H , (3.170) K P K K where : B /H , and B and H and the residual magnetization and K P A P A coercivity, respectively. SOLUTION 3.5.3 Apply Eq. (3.138) to a path around the circuit and obtain

!

H · dl : H l ; H l ; H l : 0, A KK EE

(3.171)

or H l ; H l : 9H l , (3.172) A EE KK where H , H , and H are the ﬁelds in the magnet, core, and gap, respectively, K E and l , l , and l are the lengths of the paths in the respective sections. Here, K A E I : 0 and all of the ﬂux is supplied by the magnet. In fact, the magnet can be replaced by an equivalent source equal to 9H l (because the magnet operates KK in the second quadrant, H is negative and 9H l is positive). We compare K KK Eq. (3.172) with Eq. (3.155) and ﬁnd that 9H l plays the role of ni. However, KK

FIGURE 3.30 Magnetic circuit with magnet source.

155

3.5 MAGNETIC CIRCUIT ANALYSIS

H is not known a priori. Instead, it depends on the operating point of the K magnet. Operating point: The operating point (H , B ) is determined as follows. First, K K as the core has a high permeability, its ﬁeld H is negligible compared to H K and H . Thus, the second term on the right-hand side of Eq. (3.171) can be E ignored, which gives H l : 9H l . EE KK

(3.173)

l B : 9 H K , E Kl E

(3.174)

It follows that

where we have used B : H . Because there is no ﬂux leakage, : or E E K E B A :B A . K K E E

(3.175)

Combining Eqs. (3.174) and (3.175) we obtain A l B : 9 E K H K A l K K E

(load line).

(3.176)

slope of load line

Equation (3.176) deﬁnes the magnet’s load line. This line has a slope equal to 9 (A /A )(l /l ) as shown in Fig. 3.31. The operating point is the intersec E K K E

FIGURE 3.31

Demagnetization curve with load line.

156

CHAPTER 3 Field Analysis

tion of the load line with the second quadrant demagnetization curve (3.170) A l B ; H : 9 E K H . P K K A l K K E second quadrant demagnetization curve

(3.177)

load line

This gives 9B P H : . (3.178) K (B /H ; (A /A )(l /l )) P A E K K E Notice that as the gap goes to zero, the H-ﬁeld in the magnet goes to zero, that is, lim H : 0. As the gap goes to inﬁnity, H approaches the coercivity JE K K lim H : 9H . From Eq. (3.178) we determine the equivalent source term JE K A Bl PK 9H l : (equivalent source). K K (B /H ; (A /A )(l /l )) P A E K K E Gap ﬁeld: To determine the ﬁeld in the gap, substitute Eq. (3.178) into Eq. (3.174) which gives l B K, P B : E (B /H ; (A /A )(l /l )) l P A E K K E E or B P B : . (3.179) E ((l /l )(B /H ) ; (A /A )) E K P A E K Gap energy: The magnetostatic energy density in the gap is given by Eq. (3.68), : B · H . K E E Because B and H are colinear and constant throughout the gap, the total E E energy is (3.180) W : B H l A , K E EE E where l A is the volume of the gap. It is instructive to represent W in terms E E K of the magnet’s ﬁeld. To this end, substitute Eqs. (3.173) and (3.175) into Eq. (3.180) and obtain W : 9B H l A K KK K K : 9B H ; (volume of magnet). K K

(3.181)

3.5 MAGNETIC CIRCUIT ANALYSIS

157

Notice that W 0 because H 0. Thus, for a ﬁxed volume of magnet K K (volume : l A ), the gap energy is maximized when the magnet’s energy K K product B H is a maximum (B H ) ) K K K K 3.5.2.1 Energy product. From Example (3.5.3) we see that: (a) the operating point (B , H ) of the magnet depends on the circuit (gap dimensions, K K and so on); and (b) for a given volume of material the energy supplied to the gap is maximum when the energy product is a maximum. Thus, efﬁcient circuit designs are realized when the load line of the circuit intersects the second quadrant demagnetization curve near (B H ) . K K 3.5.2.2 Equivalent circuit. When a magnet has a linear second quadrant demagnetization curve B :B ; H , (3.182) K P K K it can be replaced by an equivalent circuit element consisting of a source and reluctance term. To see this, consider the circuit of Example 3.5.3 (Fig. 3.30). From Eq. (3.182) and : B /H we have K P A B H : K9H . (3.183) K A K Substitute this into Eq. (3.173) and obtain B H l ; K l :H l . (3.184) EE K AK K If there is no ﬂux leakage : B A : B A , and Eq. (3.184) reduces to E E K K (R ; R ) : H l . (3.185) E K AK Here, R : l / A is the reluctance of the gap, and E E E l (3.186) R : K K A K K is the equivalent reluctance of the magnet ( : B /H ). Comparing Eq. K P A (3.185) with Eq. (3.153) we see that the magnet can be thought of as an equivalent source with an internal reluctance:

V :H l (mmf), K AK Equivalent Circuit (3.187) $ l for Magnet R : K (reluctance). K A K K It must be emphasized that Eq. (3.187) only rigorously applies when the second quadrant demagnetization curve is of the form (3.182).

158

CHAPTER 3 Field Analysis

It is instructive to compute B using the equivalent circuit model E (3.187). From Eqs. (3.185) and (3.186) we obtain

or

Hl AK B A : E E (l / A ; l / A ) E E K K K A H E K A : , ((l /l ) ; (A /A )) E K K E K

B P B : , E ((l /l )(B /H ) ; (A /A )) E K P A E K which is the same result as Eq. (3.179) in Example 3.5.3. The equivalent circuits for both current and magnet sources are summarized in Figs. 3.32 and 3.33. A comparison of magnetic and

Magnetic circuits and ﬁeld relations: (a) coil and core with no gap; (b) coil and core with a gap; and (c) magnet and core with a gap.

FIGURE 3.32

3.5 MAGNETIC CIRCUIT ANALYSIS

159

Magnetic circuits and equivalent lumped parameter models: (a) coil and core with no gap; (b) coil and core with a gap; and (c) magnet and core with a gap.

FIGURE 3.33

electrical circuits is summarized in Table 3.1. A magnetic circuit with both current and magnet sources is reduced to an equivalent lumpedparameter circuit in Fig. 3.34. Finally, for many applications, magnetic circuit analysis gives a reasonable estimate of circuit performance. Obviously, more accurate predictions require a more rigorous analysis that takes into account factors such as saturation, ﬂux leakage, and so on. Frequently, an optimum design can be determined from intuition based on prior experience. For example, consider the magnetic circuits shown in Fig. 3.35. Which of these provides the strongest ﬁeld in the gap region? The

160

CHAPTER 3 Field Analysis Comparison of magnetic and electrical circuits

TABLE 3.1

Magnetic circuit

Electric circuit

H B : H

E J : E

NI V : K Hl AK :

R:

B · ds dl A

V I:

R:

J · ds dl A

Magnetic circuit with a current and magnet source: (a) physical circuit; and (b) equivalent lumped-parameter circuit.

FIGURE 3.34

3.6 BOUNDARY-VALUE PROBLEMS

161

Magnetic circuits: (a) least efﬁcient; (b) somewhat efﬁcient; and (c) most efﬁcient.

FIGURE 3.35

least efﬁcient circuit is (a), which has the most leakage ﬂux, the second most efﬁcient circuit is (b), and the most efﬁcient circuit is (c). In general, the closer the magnets are to the gap, the more efﬁcient will be the device.

3.6 BOUNDARY-VALUE PROBLEMS Boundary-value problems arise naturally and frequently in the study of magnetic ﬁeld theory. In problems of this type we seek the solution to a

162

CHAPTER 3 Field Analysis

partial differential equation in a given region, with constraints on the solution speciﬁed on the boundary of the region. In particular, the value or normal derivative of the solution might be speciﬁed on the boundary. Boundary-value problems can often be solved analytically when the boundary coincides with constant coordinate surfaces in an orthogonal curvilinear coordinate system. In this case, the solution can be separated into a product of single variable functions, and the partial differential equation can be reduced to a system of ordinary differential equations. This is called the method of separation of variables. For our purposes, boundary-value problems are classiﬁed as either Dirichlet, Neumann or mixed. In the Dirichlet problem, the value of the solution is speciﬁed on the boundary, whereas in the Neumann problem the normal derivative is speciﬁed. In mixed problems, the value of the solution is speciﬁed on some parts of the boundary, and the normal derivative is speciﬁed on other parts. In this section, we study boundary-value problems for Laplace’s equation : 0, in Cartesian, cylindrical and spherical coordinates.

3.6.1 Cartesian Coordinates We start with Laplace’s equation in Cartesian coordinates, ; ; : 0. x y z

(3.188)

Assume that (x, y, z) separates into a product of single variable functions (x, y, z) : X(x)Y(y)Z(z).

(3.189)

Substitute Eq. (3.189) into Eq. (3.188) and obtain Y(y)Z(z)

dX(x) dY(y) dZ(z) ; X(x)Z(z) ; X(x)Y(y) : 0. (3.190) dx dy dz

Next, divide Eq. (3.190) by X(x)Y(y)Z(z), which gives 1 dX(x) 1 dY(y) 1 dZ(z) ; ; : 0. X(x) dx Y(y) dy Z(z) dz

(3.191)

As each term in Eq. (3.191) is a function of a single variable, they are separately equal to a constant. If this were not the case, a variation in x

163

3.6 BOUNDARY-VALUE PROBLEMS

in the ﬁrst term in Eq. (3.191) would require a change in the remaining terms to preserve the equality. However, this is not possible because the remaining terms are independent of x. Thus, Eq. (3.188) reduces to the following three ordinary differential equations: 1 dX(x) : 9k ; V X(x) dx

(3.192)

1 dY(y) : 9k ; W Y(y) dy

(3.193)

1 dZ(z) : 9k. X Z(z) dz

(3.194)

and

These equations need to be solved subject to the prescribed boundary conditions, and the constraint k ; k ; k : 0. The constants k, k, and V W X V W k can be positive, zero or negative, and therefore k , k , and k can be X V W X real or imaginary. The solutions to Eq. (3.192), (3.193), and (3.194) can be written as follows:

A x;A A sin(k x) ; A cos(k x) V V or X(x) : A eHIVV ; A e\HIVV A sinh(kx) ; A cosh(kx) or A eIV ; A e\IV B y;B B sin(k y) ; B cos(k y) W W or Y(y) : B eHIWW ; B e\HIWW B sinh(ky) ; B cosh(ky) or B eIW ; B e\IW

(k : 0) V (k 0) V , (3.195) (k 0 with k : ( k ) V V

(k : 0) W (k 0) W , (3.196) (k 0 with k : ( k ) W W

164

CHAPTER 3 Field Analysis

and

C z;C C sin(k z) ; C cos(k z) X X or Z(z) : C eHIXX ; C e\HIXX C sinh(kz) ; C cosh(kz) or C eIX ; C e\IX

(k : 0) X (k 0) X , (3.197) (k 0 with k : ( k ) X X

where j : (91, sinh(kz) : (eIX 9 e\IX)/2 and cosh(kz) : (eIX ; e\IX)/2. The constants A , B , and C (i : 1, . . . , 4) are determined from the G G G boundary conditions. A particular solution to Eq. (3.188) can be written as (x, y, z) : X (x)Y (y)Z (z). IVIWIX IV IW IX A general solution can be expressed as a superposition of all such , that is, IVIWIX

(x, y, z) : C (x, y, z). III III IVIWIX V W X V W X The constants C are determined from the boundary conditions. IVIWIX Often, the geometry of a system is such that one of its dimensions is much greater than the other two, and the ﬁeld is essentially invariant along the greater dimension. When this is the case, the ﬁeld problem reduces to a 2D problem: ; : 0. x y

(3.198)

The general solution to Eq. (3.198) can be written as (x, y) : (a x ; a )(b y ; b ) k : k :0 V W

; (c sin(k x) ; c cos(k x))(d eILW ; d e\ILW), (3.199) L L L L L L L k :k 0 L V

where k ; k : 0, and k 0. We demonstrate the theory in the followV W V ing examples.

165

3.6 BOUNDARY-VALUE PROBLEMS

FIGURE 3.36

Cross-sectional view of a rectangular magnet between ﬂux plates.

EXAMPLE 3.6.1 Figure 3.36 shows a cross-sectional view of a magnetic circuit. The circuit consists of a magnet resting on (and attached to) a soft magnetic (ﬂux) plate, and a second ﬂux plate positioned a distance g above the magnet. Assume that the plates have inﬁnite permeability and that they are magnetically connected with zero reluctance far away from the observation point (at inﬁnity). Further assume that the magnet is ideal with a linear second quadrant demagnetization relation of the form B : (H ; M).

(3.200)

Determine B in the gap region above the magnet. W SOLUTION 3.6.1 From symmetry we know that there is no variation in the ﬁeld in the x or z directions. Therefore, the problem reduces to a 1D Dirichlet boundary-value problem. Because there is no applied current, we use the magnetic scalar potential formulation (Section 3.4). Speciﬁcally, we use Eqs. (3.101) and (3.102), which reduce to d K, dy

(3.201)

dM d W. K: dy dy

(3.202)

H:9 and

There are two regions to consider, the magnet (region 1), and the gap (region 2). In these regions we have M:

M y Q 0

(region 1) (region 2).

(3.203)

166

CHAPTER 3 Field Analysis

We ﬁnd that the volume charge density is zero in both regions dM W : 0. :9 K dy Therefore, Eq. (3.202) reduces to K : 0. y The general solutions to this equation is given by Eq. (3.196) with k : 0, that W is, (y) : a y ; b K

(region 1),

(3.204)

and (y) : a y ; b (region 2). K The coefﬁcients are determined from the boundary conditions.

(3.205)

Magnet-plate interface (y : 0): We assume that the ﬂux plates have inﬁnite permeability ( -). Therefore, the H-ﬁeld in them is negligible, and the potential is constant. As the plates are connected (at inﬁnity), they are both at the same potential. Without loss of generality, we set this potential to zero, that is, : 0. As the potential is continuous at the magnet-plate interface, we have (0) : K : 0. From this we ﬁnd that b : 0. Magnet-gap interface (y : lm): At the magnet-gap interface the potential and normal component of B are continuous (B : (H ; M )). These conditions L W Q give (l ) : (l ) K K K K

(continuity of ), K

and

d 9 K ;M Q dy From these we ﬁnd that

WJK

d K : 9 dy

WJK

(continuity of B ). L

a l :a l ;b , K K or b : (a 9 a )l K

(3.206)

167

3.6 BOUNDARY-VALUE PROBLEMS

and or

9a ; M : 9a , Q

M :a 9a . Q Combining Eqs. (3.206) and (3.207) we obtain

(3.207)

b :M l . (3.208) QK Gap-plate interface (y :g ; lm): At the gap-plate interface the potential is continuous, that is, (l ; g) : K K : 0, which gives or

a (l ; g) ; b : 0, K

b . a :9 (l ; g) K Finally, substitute Eq. (3.208) into Eq. (3.209) and obtain Ml QK . a :9 (l ; g) K Therefore, the ﬁeld in the gap is

(3.209)

(3.210)

d H :9 K E dy

or

l K , :M Q (l ; g) K

l K (gap). (3.211) B : M E Q (l ; g) K We can also determine the ﬁeld in the magnet. Combining Eqs. (3.207) and (3.210) we have

l K a :M 19 Q (l ; g) K g :M . Q (l ; g) K

168

CHAPTER 3 Field Analysis

Therefore, d H :9 K

dy g : 9M (magnet). (3.212) Q (l ; g) K Notice that H is in the opposite direction to the magnetization. This is the

demagnetization ﬁeld. Notice that this ﬁeld goes to zero when the gap is reduced to zero, that is, lim H : 0. ) E EXAMPLE 3.6.2 Figure 3.37a shows a partial cross-sectional view of a magnetic circuit. The circuit consists of rectangular multipole magnet, the top of which is attached to a soft magnetic ﬂux plate. A second ﬂux plate is positioned a distance g below the magnet. Assume that the ﬂux plates have inﬁnite permeability, and that the magnet is ideal with a linear second quadrant

Magnetic circuit: (a) magnetization pattern with reference frame; and (b) reduced geometry.

FIGURE 3.37

169

3.6 BOUNDARY-VALUE PROBLEMS

demagnetization curve given by B : (H ; M). (3.213) Determine B in the gap region. W SOLUTION 3.6.2 This problem constitutes a 2D boundary-value problem. As there is no applied current, we use the magnetic scalar potential formulation H : 9 , K

(3.214)

and : · M, (3.215) K (Section 3.4). The 2D geometry can be reduced by exploiting the symmetry that results from the repeating magnetic structure. Speciﬁcally, the ﬁeld is symmetric about the vertical center lines of each pole. We consider a reduced rectangular region extending from the center of one pole to the center of the neighboring pole as shown in Fig. 3.37b. The reduced geometry consists of three regions with

M y Q M : 9M y Q 0

(region 1) (region 2)

(3.216)

(region 3).

We ﬁnd that : 9 · M : 0 in all three regions, and Eq. (3.215) reduces to K K; K : 0. x y The general solution to this equation is given by Eq. (3.199). We rule out the k : k : 0 solution because of the periodicity of the geometry. Thus, we obtain V W (x, y) : (S eIW ; S e\IW) sin(k x) K ; (C eIW ; C e\IW) cos(k x) (region 1), (3.217) (x, y) : (S eIW ; S e\IW) sin(k x) K ; (C eIW ; C e\IW) cos(k x)

(region 2),

(3.218)

(region 3),

(3.219)

and (x, y) : (S eIW ; S e\IW) sin(k x) K ; (C eIW ; C e\IW) cos(k x)

respectively. The coefﬁcients in Eqs. (3.217) to (3.219) are determined by imposing the boundary conditions. Notice that the three regions in Fig. 3.37b are collectively bounded by four lines that form a rectangular area. Two sides

170

CHAPTER 3 Field Analysis

of the rectangle, that is, the y-axis and the line x : l, represent lines of symmetry along which the normal component H if zero. The boundaries that L form the top and bottom of the rectangle are bordered by a ferromagnetic material with inﬁnite permeability ( : -). Consequently, the tangential component H vanishes along these boundaries. We evaluate the boundary R conditions region by region. Region 1: One of the boundaries that border region 1 is the y-axis along which the normal component of the ﬁeld vanishes: H(0, y) : 9 L

(x, y) K x

:0

(g y h).

V

This condition yields S eIW : 9S e\IW. (3.220) Region 1 is also bounded by a ﬂux plate ( : -) at y : h. The tangential component of the ﬁeld is zero along this boundary, that is, H(x, h) : 9 R

(x, y) K x

:0

(0 x l/2).

WF

This condition is satisﬁed when C eIF : 9C e\IF. Applying Eqs. (3.220) and (3.221) to Eq. (3.217) gives

(3.221)

(x, y) : C sinh(k (y 9 h)) cos(k x), (3.222) K where C is a constant. Region 2: The conditions for region 2 are similar to those of region 1. Speciﬁcally, the normal component of the ﬁeld is zero along the symmetry line deﬁned by x : l, and the tangential ﬁeld is zero along the upper boundary deﬁned by y : h: H(l, y) : 9 L

(x, y) K x

:0

(g y h),

(3.223)

(l/2 x l).

(3.224)

VJ

and H(x, h) : 9 R

(x, y) K x

:0 WF Applying the ﬁrst condition to Eq. (3.218) gives S eIW : 9S e\IW,

(3.225)

171

3.6 BOUNDARY-VALUE PROBLEMS

and n k : l

(n : 1, 2, 3, . . .).

(3.226)

Then, the second condition is satisﬁed when C eIF : 9C e\IF. Relations (3.225) to (3.227) reduce Eq. (3.218) to (x, y) : C sinh K

(3.227)

n n (y 9 h) cos x , l l

(3.228)

where C is a constant. Magnet-magnet interface: At the interface between the two magnets the normal component of B and the tangential component of H must be continuous, that is, B(l/2, y) : B(l/2, y) L L

(g y h),

(3.229)

H(l/2, y) : H(l/2, y) R R

(g y h),

(3.230)

and

where BG(l/2, y) : 9 L

G(x, y) K x

(i : 1, 2),

VJ

and HG(l/2, y) : 9 R

G(x, y) K y

(i : 1, 2). VJ

From these relations we obtain C :C ,

(3.231)

and k :k . (3.232) Therefore, the solutions to both regions are identical. We form a general solution for these regions using superposition, that is,

n n (x, y) : C sinh (y 9 h) cos x . (3.233)

L

l l L

172

CHAPTER 3 Field Analysis

Region 3: For region 3 there are two boundaries about which the ﬁeld is symmetric and on which the normal component of H is zero. These boundaries are vertical lines deﬁned by x : 0 and x : l. Therefore, H(0, y) : 0 L

(0 y g),

(3.234)

H(l, y) : 0 (0 y g). L When these conditions are imposed, we ﬁnd that

(3.235)

and

S eIW : 9S e\IW,

(3.236)

and n k : l

(n : 1, 2, 3, . . .),

(3.237)

respectively. Also, as the bottom of the region borders a ﬂux plate, the tangential component of the ﬁeld vanishes there: H(x, 0) : 0 R

(0 x l).

(3.238)

This implies that C : 9C . (3.239) We call this region the gap, and form its general solution via superposition,

n n (y 9 h) cos x . (3.240) (x, y) : C sinh L l l L The coefﬁcients C are determined at the magnet-gap interface. L Magnet-gap interface: At the interface between the magnet and the gap the tangential component of H must be continuous, that is, (x, g) : H (x, g) (0 x l). (3.241)

R R This condition, along with the fact that the functions sin(nx/l) are orthogonal on (0, l), implies that H

C

L

sinh(ng/l) :C . L sinh(n(g 9 h)/l)

(3.242)

In addition, the normal component of B must also be continuous, that is, B

L

(x, g) : B

L

(x, g)

(0 x l).

(3.243)

173

3.6 BOUNDARY-VALUE PROBLEMS

Because B

L

:

(H ;M ) L Q (H 9M ) L Q

(0x l/2) (l/2 x l)

(3.244)

Eq. (3.243) reduces to

nx M n Q C K(n, h, g, l) cos : l 9M l L Q L where K(n, h, g, l) : cosh

(0 x l/2) (l/2 x l)

ng ng n 9 sinh coth (g 9 h) . l l l

(3.245)

(3.246)

Again, exploiting the orthogonality of cos(nx/l) on (0, l) we ﬁnd that

J

nx J nx dx 9 cos dx . (3.247) l l J After integrating Eq. (3.247) we obtain 2M Q C : L nK(n, h, g, l)

cos

4lM (91)L\ Q C :9 L (n)K(n, h, g, l)

(n : 1, 3, 5, . . .),

(3.248)

which fully deﬁnes the solution for the gap region. Speciﬁcally, substitute Eq. (3.248) into Eq. (3.240) and obtain (x, y) : 94lM Q

(91)L\ ny nx sinh cos . l l (n)K(n, h, g, l) L

(3.249)

From Eq. (3.249) we ﬁnd that the vertical component of the ﬁeld in this region is B

W

(x, y) : 94 M Q

(91)L\ ny nx cosh cos , (n)K(n, h, g, l) l l L

(3.250)

where g : the gap, h : g ; t , t : height of the magnet, and K(n, h, g, l) is K K as speciﬁed in Eq. (3.246). Calculations: We demonstrate the use of Eq. (3.250) with some sample calculations. Consider a structure with l:4.8 cm, t :1.28 cm, and M :3000 K Q (A/m). We compute the gap ﬁeld B (x, y) at y : 0.375 mm and 0 x l, W for three different gaps g : 6, 8, and 10 mm (Fig. 3.38). A similar calculation is performed with y:2.5, 5 and 7.5 mm, for a ﬁxed gap g:10 mm (Fig. 3.39). )

174

CHAPTER 3 Field Analysis

FIGURE 3.38

Component B in the gap vs gap height g. W

FIGURE 3.39

Component B in the gap vs y. W

175

3.6 BOUNDARY-VALUE PROBLEMS

3.6.2 Cylindrical coordinates We now consider Laplace’s equation in cylindrical coordinates, that is,

1 1 r ; ; : 0. r z r r r

(3.251)

Again, assume that separates into a product of single variable functions (r, , z) : R(r)()Z(z).

(3.252)

Substitute Eq. (3.252) into Eq. (3.251), divide by Eq. (3.252) and obtain

1 d dR 1 d 1 dZ r ; ; : 0. dr r d Z dz Rr dr

(3.253)

Notice that the last term in Eq. (3.253) is a function of z only. Therefore, from the same argument that follows Eq. (3.191) we have dZ 9 kZ : 0, dz

(3.254)

and

r d dR 1 d r ; kr ; : 0. R dr dr d

(3.255)

The last term in Eq. (3.255) is a function of only and we set it equal to a constant, that is, 9n. Thus, d ; n : 0, d

(3.256)

and Eq. (3.255) reduces to r

d dR r ; (kr 9 n)R : 0. dr dr

If there is no restriction on , we must have () : ( ; 2), which restricts n to integer values. The solutions for () and Z(z) are of the form () :

A ;A A sin(n) ; A cos(n)

(n : 0) (n " 0)

(3.257)

176 and

CHAPTER 3 Field Analysis

B z;B B sinh(kz) ; B cosh(kz) or Z(z) : B eIX ; B e\IX B sin(kz) ; B cos(kz) or B eHIX ; B e\HIX The radial solutions are of the form

(k : 0) (k 0) (3.258) (k 0 with k : ( k

C ln(r) ; C (k : 0, n : 0) C rL ; C r\L (k : 0, n " 0) R(r) : C J (kr) ; C J (kr) (k 0, n " 0, n " integer) (3.259) L \L C J (kr) ; C Y (kr) (k 0, n : 0 or integer) L L C I (#r) ; C K (#r) (k 0 ; k : j#) L L where J (kr) are Bessel functions of the ﬁrst kind of order
!L

(91)K(kr/2)K!L , (kr) : m!$(m < n ; 1) K

(3.260)

cos(n)J (kr) 9 J (kr) L \L , sin(n)

(3.261)

and Y (kr) : L where $(p) :

xNe\V dx

(p 0),

is the gamma function $(p ; 1) : p! when p is an integer. The remaining functions I (#r) : j\LJ (j#r), L L and K (#r) : jL>[J (j#r) ; jY (j#r)] L L L 2

177

3.6 BOUNDARY-VALUE PROBLEMS

are the modiﬁed Bessel functions of the ﬁrst and second kind, respectively. The functions Y (kr) and K (#r) diverge when their respective L L arguments are zero. Therefore, they are discarded when the solution region includes r : 0. When k is real (k 0) the two independent radial solutions are J (kr) L and Y (kr), or J (kr) and J (kr) depending on whether or not n is an L L \L integer. As noted in the preceding, n must be an integer when takes on its full range of values. However, if the solution region spans a restricted range of (e.g., wedge-shaped), then n need not be an integer. In either case, the asymptotic behavior of these functions is oscillatory. Thus, the solution (r, , z) is exponential in the z-direction and oscillatory in the r-direction. When k is imaginary (k 0) we deﬁne k : j# and the two independent radial solutions are I (#r) and K (#r). These functions exhibit L L exponential asymptotic behavior. Thus, the solution (r, , z) is oscillatory in the z-direction and exponential in the r-direction. For applications in which there is invariance along the z-axis, Eq. (3.251) reduces to a 2D equation:

1 1 r ; : 0. r r r r The foregoing analysis applies with k : 0, and a particular solution is of the form (r, ) : (a ln(r) ; a )(b ; b ) n:0

; (c rL ; c r\L)(d sin(n) ; d cos(n)).

(3.262)

n"0

If the region of interest includes the full angular range 0 2, then a general 2D solution is of the form (r, ) : a ;b ln(r); (a rL;b r\L)(c sin(n);d cos(n)). L L L L L

(3.263)

We demonstrate the two-dimensional theory in the following example. EXAMPLE 3.6.3 Determine the magnetic ﬁeld of a bipolar cylindrical magnet that is polarized perpendicular to its axis (Fig. 3.40). Assume that the length of the cylinder is much greater than its diameter. This justiﬁes a 2D analysis.

178

CHAPTER 3 Field Analysis

Bipolar cylinder: (a) perspective of cylinder; and (b) cross-sectional view with reference frame.

FIGURE 3.40

SOLUTION 3.6.3 We solve a 2D boundary-value problem. Choose a coordinate system with the magnetization along the x-axis, which gives M : M x Q : M [cos()r 9 sin() ]. Q

(3.264)

Because there are no current sources we can represent H in terms of a magnetic scalar potential H : 9 , K

(3.265)

where : 0, (3.266) K (Section 3.4). Let (r, ) and (r, ) denote the solutions to Eq. (3.266) K K inside and outside the magnet, respectively. From symmetry, these solutions must be even functions of , G(r, ) : G(r, 9) K K

(i : 1, 2).

(3.267)

They must also be well behaved at r : 0 and r : -, respectively, (0, ) -, K (-, ) -. K

(3.268) (3.269)

179

3.6 BOUNDARY-VALUE PROBLEMS

The general forms for (r, ) and (r, ) compatible with these conditions K K are (r, ) : a rL cos(n) K L L

(r a),

(3.270)

and (r, ) : b r\L cos(n) (a r), (3.271) K L L which follow from Eq. (3.263). The coefﬁcients a and b are determined from L L the boundary conditions imposed at the surface of the magnet (r : a). Speciﬁcally, the potential (r, ) and the normal component of B must be continuous K at this surface, that is, (a, ) : (a, ), K K

(3.272)

and [9(a, ) ; M] · r : 9(a, ) · r , (3.273) K K respectively. These conditions yield the following expression for (r, ): K M a (r, ) : Q cos() (a r). K 2 r Outside the magnet B : H, and therefore M a Q cos() B (r, ) : P 2 r

(a r),

(3.274)

(a r).

(3.275)

(a r).

(3.276)

and B (r, ) :

M a Q sin() 2 r

Also, because B : ; A we have M a A(r, ) : Q sin()z 2 r

Calculations: We demonstrate the use of Eqs. (3.274) and (3.275) with some sample calculations. Consider a cylinder with a radius a : 2.5 mm, and a magnetization M : 4.3;10 (A/m) (B : 5400 G). The ﬁelds B (r, ) and Q P P B (r, ) are computed for r : 3.8 mm and 0 (Fig. 3.41). Notice that B (r, ) peaks at : 0. This shows that the radial ﬁeld component peaks at the P center of the north pole as expected. )

180

CHAPTER 3 Field Analysis

FIGURE 3.41

magnet.

Components B and B vs for an inﬁnite bipolar cylindrical P

3.6.3 Spherical coordinates Finally, we study Laplace’s equation in spherical coordinates:

1 1 1 r ; sin() ; : 0. (3.277) r r sin() r sin() r r We assume a solution of the form (r, , ) : R(r)()%().

(3.278)

Substitute Eq. (3.278) into Eq. (3.277) and obtain %

dR R d d% R% d ; sin() ; : 0. (3.279) dr r sin() d d r sin() d

Next, multiply Eq. (3.279) by r sin()/R(r)()%() and obtain

1 d dR 1 d d% 1 d r ; sin() ; : 0. R dr dr % sin() d d d

(3.280)

181

3.6 BOUNDARY-VALUE PROBLEMS

We reduce Eq. (3.280) to a system of equations by introducing separation constants m and n(n ; 1), d ; m : 0, d

(3.281)

d dR r 9 n(n ; 1)R : 0, dr dr

(3.282)

1 d d% m sin() ; n(n ; 1) 9 % : 0. d sin() sin() d

(3.283)

If the full range of is allowed (0 2), then m must be an integer to ensure that () is singled valued, that is, () : ( ; 2). Equation (3.283) is known as Legendre’s equation, and the solutions %() are ﬁnite at : 0 and only if n is an integer. When m and n are integers, the solutions are of the form () : A sin(m) ; A cos(m),

(3.284)

R(r) : B rL ; B r\L>,

(3.285)

and %() : C PK(cos()) ; C QK(cos()), L L where PK(cos()) and QK(cos()) are the associated Legendre functions of L L the ﬁrst and second kind, respectively. The functions QK(cos()) are L singular at : 0 and , and therefore are excluded for problems that include these angles. The general solution to Eq. (3.277) for regions that include : 0 and encompass the full azimuthal range 0 2 is of the form L (r, , ) : [A sin(m);B cos(m)](C rL;D r\L>)PK(cos()). L K K L L L K (3.286) The sum over m ends at m : n because PK(cos()) : 0 for m n. L For problems with azimuthal symmetry, there is no dependence and m : 0. When this is the case, Eq. (3.277) reduces to

1 1 r ; sin() : 0. r r r r sin()

(3.287)

182

CHAPTER 3 Field Analysis

If the solution region includes : 0 or , the general solution to Eq. (3.287) is of the form (r, ) : (A rL ; C r\L>)P (cos()), (3.288) L L L L where P (cos()) : P (cos()) are called the Legendre polynomials. The L L ﬁrst few Legendre polynomials are as follows: n

P (cos()) L

0

1

1

cos() (3 cos() 9 1) (5 cos() 9 3 cos())

2 3

EXAMPLE 3.6.4 Determine the H-ﬁeld due to a uniformly magnetized bipolar sphere of radius a in free space (Fig. 3.42a) [7]. Assume that the sphere is polarized along its diameter and is made from a permanent magnet material with a second quadrant demagnetization curve of the form B : (H ; M). (3.289) SOLUTION 3.6.4 Use spherical coordinates with the sphere oriented with its magnetization along the z-axis as shown in Fig. 3.42a, M : M z . Q The H-ﬁeld can be expressed in terms of a scaar potential H : 9 , (3.290) K Because : 9 · M : 0, we have K : 0 (3.291) K (Section 3.4). Given the orientation of the sphere, will have no dependence. K Let (r, ) and (r, ) denote the solutions to Eq. (3.291) inside and outside K K the magnet, respectively. These functions must be well behaved at r : 0, and r : -, respectively, (0, ) -, K

(3.292)

(-, ) -. K

(3.293)

and

183

3.6 BOUNDARY-VALUE PROBLEMS

Bipolar sphere: (a) magnetization and reference frame; (b) B-ﬁeld; (c) H-ﬁeld; and (d) internal ﬁelds.

FIGURE 3.42

The general forms for (r, ) and (r, ) compatible with these conditions K K are (r, ) : A rLP (cos()) K L L L

(r a),

and (r, ) : C r\L>P (cos()) K L L L We apply the boundary conditions

(r & a),

H(a, ) : H(a, ) F F

(H : H), R R

B(a, ) : B(a, ) P P

(B : B), L L

and

184

CHAPTER 3 Field Analysis

at the surface of the sphere. This gives (A aL 9 C a\L>) d L L P (cos()) : 0, a d L L

(3.294)

and (9 ; M z ) · r : 9 · r , K Q K respectively. The second condition (3.295) reduces to

(3.295)

0 : M cos() 9 C a\ Q 9 (A naL\ ; C (n ; 1)a\L>)P (cos()). (3.296) L L L L The Legendre polynomials P (cos ()) are linearly independent relative to one L another. Therefore, each of the terms containing P (cos()) or d/(d)P (cos()) L L in Eqs. (3.294) and (3.296) must separately be zero for each value of n. In particular, when n : 0 we have d P (cos()) : 0, d and C a\ : 0. Therefore, C : 0. The coefﬁcient A is undetermined from these conditions, but it represents an additive constant that may be set to zero without affecting the ﬁelds. For n : 1 we have A 9 C a\ : 0 and M 9 A 9 2C a\ : 0, Q which we solve simultaneously and obtain C : M a Q

(3.297)

and A : M . (3.298) Q Continuing in this fashion we ﬁnd that the remaining terms A : C : 0 for L L n & 2. Thus, : M r cos(), K Q

185

3.7 METHOD OF IMAGES

and

a : M cos(). K Q r The corresponding ﬁelds are H : 9M z Q

(r a),

(3.299)

and

a H : M [2 cos()r ; sin() ]. Q r

(3.300)

Notice that inside the sphere, H is in opposite direction to the magnetization. This is the demagnetization ﬁeld (Section 1.8). Finally, we determine B from Eqs. (3.289) and (3.299): B : (H ; M z ) Q : M z . Q The B- and H-ﬁelds inside and outside the sphere are shown in Fig. 3.42b and c, respectively. The orientations of B, H, and M inside the sphere are shown in Fig. 3.42d. )

3.7 METHOD OF IMAGES The method of images is used to solve ﬁeld problems when the source is near a material with a high permeability, and the solution region manifests a high degree of symmetry. In this method, the material is replaced by a number of carefully chosen image sources that render the appropriate boundary conditions at the interface. Both the position and magnitude of the image sources need to be determined. Once the image sources are known, a ﬁeld solution is obtained by taking the superposition of the solutions for the individual source terms (real and image) as if they were in free space. The solution is only valid in the region outside the material (where the real source is located). The advantage of this method is that the original boundary-value problem (with material interfaces) is reduced to a free-space problem (without material interfaces). The validity of the method is based on the uniqueness of the ﬁeld solution. Speciﬁcally, because the image sources are chosen to render the prescribed boundary conditions, and because the ﬁeld solution satisﬁes these conditions and the ﬁeld equation in the solution region, it is unique.

186

CHAPTER 3 Field Analysis

Wire above a ferromagnetic plate: (a) physical system; and (b) equivalent system.

FIGURE 3.43

We demonstrate the method of images using a simple example. Consider an inﬁnite wire carrying a current i a distance h above of an inﬁnite nonconducting plate that has inﬁnite permeability ( -) (Fig. 3.43a). In the plate, H 0 but B is ﬁnite, and as the media is nonconducting there is no surface current. From Eqs. (3.12) and (3.14) we know that at the surface of the plate the normal component of B must be continuous and the tangential component H must be zero, that is, B : B, L L (3.301) H : 0, R where B and B are the normal components of B inside and outside L L the plate, respectively. These same boundary conditions are obtained if the plate is replaced by an image current directly beneath the source, a distance h into the plate as shown in Fig. 3.43b. The magnitude and direction of the image current needs to be the same as the source so that the boundary conditions (3.301) are satisﬁed at the surface of the plate. Therefore, the original problem is transformed into one in which there are two wires in free space separated by a distance 2h. The ﬁeld solution to this equivalent problem is the superposition of the solutions of the individual wires, and is valid above the plate (in the region containing the real source). A similar analysis applies to the case of a current ﬁlament near wedge-shaped regions [1]. The source and image currents

3.7 METHOD OF IMAGES

187

Source and image currents for wedge-shaped regions: (a) rectangular corner section; and (b) 60° wedge [1].

FIGURE 3.44

for a rectangular corner section, and for a 60° wedge are shown in Fig. 3.44. The method of images can be summarized as follows: 1. Determine the magnitude and position of image sources that render the boundary conditions (3.301) at the material interface; 2. replace the material by the equivalent image sources; and 3. obtain the ﬁeld solution as a superposition of the ﬁelds due to all the sources both real and image as if they were in free space. This method is demonstrated in the following examples.

188

CHAPTER 3 Field Analysis

EXAMPLE 3.7.1 Compute the force on a wire carrying a current i above a ferromagnetic plate as shown in Fig. 3.43a. Assume that the plate has an inﬁnite permeability -. SOLUTION 3.7.1 We replace the plate by an image current as shown in Fig. 3.43b. The problem reduces to ﬁnding the force between two parallel wires carrying a current i and separated by a distance 2h. This was solved in Example 3.2.7 and the force is given by Eq. (3.53) with d : 2h i F : 9 z . 4h

(3.302) )

EXAMPLE 3.7.2 Compute the ﬁeld due to a cylindrical magnet of radius R and length L that is polarized along its axis with uniform magnetization M : M z , Q The magnet is resting on an inﬁnite plate. Assume that the plate has an inﬁnite permeability - (Fig. 3.45a). SOLUTION 3.7.2 We replace the plate by an image magnet (Fig. 3.45b). The image magnet is identical to the source magnet, and positioned as shown. From symmetry, we know that the ﬁeld due to the two magnets will satisfy the boundary conditions (3.301) at the interface. The resulting magnet structure is equivalent to a single magnet of length 2L (Fig. 3.45c). Thus, the ﬁeld is given by Eq. (3.98) with L replaced by 2L:

z ; (2L) z M B (z) : Q 9 . X 2 ((z ; 2L) ; R (z ; R

(3.303) )

EXAMPLE 3.7.3 Determine the force on a cylindrical bar magnet a distance z above an inﬁnite ferromagnetic plate that has an inﬁnite permeability (Fig. 3.46a). The magnet has a radius R and length L and is magnetized to a level M along its axis. Assume R L. Q SOLUTION 3.7.3 We replace the plate by an image magnet as shown in Fig. 3.46b. The image magnet is identical to the source magnet, and it is easy to check that the ﬁeld due to the two magnets will satisfy the boundary conditions (3.301). The force between the source and image magnet was determined in Example 3.4.3. It is given by Eq. (3.126) with d replaced by 2z,

Q 1 2 1 ; 9 , F(z) : K 9 4 (2z) (L ; 2z) 4(L ; z) where Q : M R. K Q

(3.304) )

3.7 METHOD OF IMAGES

189

Cylindrical magnet on a ferromagnetic plate: (a) physical system; (b) magnet and image; and (c) equivalent system.

FIGURE 3.45

EXAMPLE 3.7.4 Compute the ﬁeld above a rectangular bar magnet of width 2a, depth 2b and length L, which is polarized along its axis with uniform magnetization M : M z . Q The magnet is resting on an inﬁnite plate that has an inﬁnite permeability - (Fig. 3.47a). SOLUTION 3.7.4 We replace the plate by an image magnet as shown in Fig. 3.47b. The image magnet is identical to the source magnet, and from symmetry we know that the ﬁeld due to the two magnets will satisfy the boundary

190

CHAPTER 3 Field Analysis

Cylindrical magnet above a ferromagnetic plate: (a) physical system; and (b) magnet and image.

FIGURE 3.46

conditions (3.301) at the interface. The resulting magnet structure is equivalent to a single magnet of length 2L as shown in Fig. 3.47c. The ﬁeld for this structure is given by Eq. (3.114) with L replaced by 2L,

(z ; 2L)(a ; b ; (z ; 2L) M B (z) : Q tan\ X ab 9tan\

z(a ; b ; z ab

.

(3.305) )

3.8 FINITE ELEMENT ANALYSIS Finite element analysis (FEA) is currently the most popular numerical method for the solution of electromagnetic ﬁeld problems. There are two reasons for this. First, FEA is a powerful method that is readily adapted to a wide variety of applications. Second, there are numerous commercially available FEA software packages that enable a user to solve complex problems without having to develop his or her own application

3.8 FINITE ELEMENT ANALYSIS

191

Rectangular magnet on a ferromagnetic plate: (a) physical system; (b) magnet and image; and (c) equivalent system.

FIGURE 3.47

speciﬁc algorithms. In this section, we brieﬂy review the FEA method, and present simple examples to demonstrate its use. More detailed presentations of electromagnetic FEA can be found in numerous excellent texts [10—15]. The ﬁnite element method entails the following steps: 1. Divide the solution region into a ﬁnite number of subregions called elements, and label the nodes that deﬁne each element; 2. select an approximating function for the ﬁeld solution in each element (usually a polynomial);

192

CHAPTER 3 Field Analysis

3. express the solution in each element as a function of the nodal values and the spatial variables in that element; 4. deﬁne an energy functional for the ﬁeld equation and evaluate this functional over each element. This gives an expression for the energy in each element in terms of its nodal values; 5. assemble a global energy expression as a sum of the energy of the individual elements. Reduce this expression to a function of the unknown nodal values eliminating redundant and prescribed nodal values; 6. minimize the global energy expression with respect to the unknown nodal values. This results in a system of algebraic equations for the nodal values; 7. solve the system of equations of Step 6 to determine the nodal values; and 8. reconstruct the desired solution from the nodal values using the expressions of Step 3. We demonstrate this procedure by solving a 2D magnetostatic ﬁeld problem consisting of a rectangular conductor in free space (Fig. 3.48a). We solve the problem using a vector potential formulation (Section 3.2.1). The governing ﬁeld equation is Eq. (3.22), which in two dimensions reduces to A (x, y) : 9 J (x, y). X X

(3.306)

Because A (x, y) and J (x, y) are scalars, we drop the subscript z for the X X remainder of this section, that is, A (x, y) ; A(x, y) and J (x, y) ; J(x, y). X X Step 1: Divide the solution region into triangular elements and label the vertex nodes of each element (Figs. 3.48b and 3.49). Step 2: Select an approximating function for the desired solution in each element. We assume that the potential is a linear function of x and y. Speciﬁcally, let AG(x, y) : a ; bx ; cy

(3.307)

represent the potential in the ith element. Step 3: Express AG(x, y) in terms of the spatial variables in the ith element and the corresponding nodal values AG (k : 1, 2, 3). Let (x , y ), I (x , y ), and (x , y ) denote the coordinates of the nodes for the ith

3.8 FINITE ELEMENT ANALYSIS

FIGURE 3.48

193

Conductor in free space: (a) rectangular conductor; and (b) FEA

mesh.

element. The potentials at these nodes can be written in the following matrix form: AG 1 x y a AG : 1 x y b . (3.308) AG 1 x y c An expression for the coefﬁcients a, b, and c can be obtained by inverting

194

CHAPTER 3 Field Analysis

FIGURE 3.49

Triangular element with labeled nodes.

Eq. (3.308) as follows: 1 x y \ AG b : 1 x y AG . (3.309) c 1 x y AG We express AG(x, y) inside the ith element in terms of its nodal values by substituting Eq. (3.309) into Eq. (3.307). This gives a

1 x y \ AG AG(x, y) : [1 x y] 1 x y AG , 1 x y AG or AG(x, y) : AGG(x, y), I I I where 1 [(x y 9 x y ) ; (y 9 y )x ; (x 9 x )y], G(x, y) : 2G 1 G(x, y) : [(x y 9 x y ) ; (y 9 y )x ; (x 9 x )y], 2G and 1 G(x, y) : [(x y 9 x y ) ; (y 9 y )x ; (x 9 x )y], 2G

(3.310)

195

3.8 FINITE ELEMENT ANALYSIS

Here, G is the area of the ith element

1 x y 1 G : 1 x y . 2 1 x y The G are called element shape functions. It is easy to verify that I 0 k"j G(x , y ) : I H H 1 k:j

and that G(x, y) : 1. I I We represent the source J(x, y) in a similar fashion, that is, JG(x, y) : JGG(x, y), I I I where JG (k : 1, 2, 3) are known values of the source at the nodal points. I Step 4: Construct an energy functional compatible with Eq. (3.306) and evaluate it for a single element. In this case we use the energyrelated functional W(A) :

1 2

A ds 9

AJ ds.

(3.311)

We evaluate Eq. (3.311) over the ith element and obtain WG(AG) :

1 2

AG ds 9

AGJG ds.

(3.312)

Integration over ith

This can be written as

1 AG G · G ds AG I I J J 2 I J 9 AG G · G ds JG. I I J J I J We rewrite Eq. (3.313) in the following matrix form: WG(AG) :

WG(AG) : AG2KGAG 9 AG2PGJG,

(3.313)

196

CHAPTER 3 Field Analysis

where AG AG : AG AG AG2 : [AG AG AG] and KG : IJ

· ds I J

Integration over ith element

and PG : IJ

G · G ds. I J

Integration over ith element

Step 5: Next, assemble the energy of the elements, and form a global energy expression: , W(A) : WG(AG, AG, AG), (3.314) G where N is the total number of elements. The sum needs to be modiﬁed because some of the nodal values are redundant or prespeciﬁed. Speciﬁcally, nodal values that are common to more than one element are redundant, and nodal values on the boundary are prespeciﬁed. Let N denote the number of unknown nodal values and rewrite Eq. (3.314) in terms of these unknowns: W : W(A , A , . . . , A ). (3.315) , Step 6: Minimize Eq. (3.315) with respect to the unknown nodal values A : I W(A , A , . . . , A ) , :0 (n : 1, . . . , N). A L This leads to a system of equations for the unknown A : G , C A : 0, (3.316) GI I I

3.8 FINITE ELEMENT ANALYSIS

FIGURE 3.50

197

Flux plot for the rectangular conductor.

where the C are known. The N unknown nodal values are determined GI by solving Eq. (3.316). Step 7: Once the nodal values have been determined the solution is obtained using Eq. (3.310). A typical FEA output is shown in Fig. 3.50. This is a ﬂux plot for the rectangular conductor problem. The lines represent equipotential contours of A (x, y). These are also known as ﬂux X lines (Section 3.2.1). We now present a FEA of a permanent magnet structure. Consider the cylindrical (axisymmetric) structure shown in Fig. 3.51. This is an axially polarized permanent magnet shell with an attached ﬂux plate. Such magnets are used for various applications including the focusing of charged particle beams [16], and have a ﬁeld distribution in their interior similar to that of a solenoid.

FIGURE 3.51

Axially polarized cylindrical shell with attached ﬂux plate.

198

CHAPTER 3 Field Analysis

Cross section of magnet structure: (a) right-hand side; and (b) ﬁnite element mesh.

FIGURE 3.52

The shell has inner and outer radii R and R , respectively, and is centered about the z-axis. We assume that the ﬂux plate has a linear constitutive relation B : H, where . Because the structure is axisymmetric, it sufﬁces to study the cross-sectional element shown in Fig. 3.52a. The full geometry is obtained by rotating this element 360° around the z-axis. The mesh for the element is shown in Fig. 3.52b. Notice that the mesh is reﬁned in the interior of the magnet in order to render a more accurate ﬁeld solution in this region. In general, a denser mesh gives a higher degree of accuracy. Figure 3.53 is a ﬂux plot for the magnet. The full 3D ﬂux distribution is obtained by rotating the crosssectional plot 360° around the z-axis. Finally, we evaluate the axial ﬁeld

199

3.8 FINITE ELEMENT ANALYSIS

FIGURE 3.53

Flux plot.

component B along the interior of the magnet at two different radii r : 0 X and r : R /2 (Fig. 3.54). Notice that B is highest in the center of the X magnet as anticipated. At this point, we list several commercially available software packages for electromagnetic analysis. Some of these utilize the ﬁnite element method, while others utilize the boundary element method.

FIGURE 3.54

Component B vs z for two different radial values. X

200

CHAPTER 3 Field Analysis

1. Maxwell — Ansoft Corporation, Pittsburgh, Pennsylvania, USA. 2. OPERA — Vector Fields Inc., Aurora, Illinois, USA, or Vector Fields Ltd., Oxford, UK. 3. MagNet — Infolytica Corporation of Canada, or Infolytica Ltd. in the UK. 4. AMPERES — Integrated Engineering Software Corporation, Winnipeg, Manitoba, Canada. 5. Flux2D/3D — MAGSOFT Corporation, Troy, NY, USA. 6. ANSYS — Swanson Analysis Systems Inc., Houston, PA, USA. 7. COSMOS/M — Structural Research and Analysis Corporation, Los Angeles, CA, USA.

3.9 FINITE DIFFERENCE METHOD The ﬁnite difference method is one of the oldest and most developed numerical methods for the solution of partial differential equations. In this method, the solution region is discretized into a ﬁnite number of nodes, and the differential equation is approximated by a ﬁnite difference equation that relates the value of the solution at a given node to its values at the neighboring nodes. This discretization transforms the original boundary-value problem into a system of algebraic equations that can be solved using standard methods. In this section we brieﬂy review the ﬁnite difference method. A comprehensive treatment of this method can be found in numerous texts [15, 17, 18]. The ﬁnite difference method entails the following steps: 1. Assign a grid of nodes (mesh) to the solution region and its boundary; 2. approximate the differential equation by a ﬁnite difference equation that relates the value of the solution at a given node to its values at neighboring nodes; 3. construct an algebraic system of equations for the nodal values taking into account the prescribed boundary values and/or initial conditions; and 4. solve the system of equations of Step 3, and obtain the nodal values of the solution. We demonstrate the ﬁnite difference method using Poisson’s equation in two dimensions: (x, y) : (x, y).

(3.317)

201

3.9 FINITE DIFFERENCE METHOD

FIGURE 3.55

Finite difference grid for a rectangular region.

This equation occurs frequently in magnetostatic ﬁeld theory. In particular, it is the governing equation for the vector potential for 2D problems: A (x, y) : J (x, y), X X (Section 3.2.1). It is also the governing equation for the scalar potential in the magnetic charge model: (x, y) : 9 (x, y), K K (Section 3.4). Consider Eq. (3.317) in a rectangular solution region R in the x-y plane deﬁned by (0 x L) and (0 y H). As a ﬁrst step, discretize R. Choose discretization parameters x and y, and then assign a grid of nodal points for R and its boundary S. In this case, S consists of the four line segments (x:0, 0y H), (x:L, 0yH), (0xL, y:0), and (0 x L, y : H). Label the nodal points with indices (m, n) as shown in Fig. 3.55, where x : mx

(m : 0, 1, . . . , M),

y : ny

(n : 0, 1, . . . , N).

and

Notice that (x < x, y < y) : (m < 1)x, (n < 1)y), and so on. Thus,

202

CHAPTER 3 Field Analysis

coordinate points are identiﬁed with indices (x, y) ; (m, n) and (x < x, y < y) ; (m < 1, n < 1), and so on. It follows that functional values are also indexed (x, y); , KL and (x < x, y < y) ; , and so on. K!L! Next, we develop a ﬁnite difference equation for Eq. (3.317). To this end, consider the Taylor series expansion in x for the bivariate function (x, y): (x < x, y) : (x, y) < <

(x, y) 1 (x, y) x ; x x 2! x

1 (x, y) 1 (x, y) x ; x ; · · · . (3.318) 3! x 4! x

We can obtain three different approximations for (x, y)/x from Eq. (3.318). These follow from the ;x, 9x and the difference of the ;x and 9x expansions, respectively, that is, (x, y) (x ; x, y) 9 (x, y) : ; O(x) x x

(forward difference),

(x, y) (x 9 x, y) 9 (x, y) : ; O(x) x x

(backward difference),

and (x, y) (x ; x, y) 9 (x 9 x, y) : ; O(x) x 2x

(central difference),

where O(x) means that the error of the approximation is of the order of x, and so on. Clearly, the central difference approximation is the most accurate. We obtain similar expressions for (x, y)/y: (x, y) (x, y ; y) 9 (x, y 9 y) : ; O(y) x 2y

(central difference),

Once we have assigned a nodal grid to R, we can express the derivatives in terms of the nodal values, for example: (x, y) 9 K\L : K>L x 2x

(central difference).

Second-order derivatives are obtained in a similar way. For example, an expression for (x, y)/x is obtained from Eq. (3.318) by adding the

3.9 FINITE DIFFERENCE METHOD

203

;x and 9x expansions. This gives the central difference approximation: (x, y) (x ; x, y) 9 2(x, y) ; (x 9 x, y) . x x Using the second-order expressions we discretize Eq. (3.317) and obtain 9 2 ; 9 2 ; K>L KL K\L ; KL> KL KL\ : . (3.319) KL x y This can be further simpliﬁed by choosing x : y : h, which yields

K>L

; ; ; 9 4 : h . K\L KL> KL\ KL KL

(3.320)

Equation (3.320) represents a system of linear equations. To solve this system, we need to represent the unknown nodal values in the form of a column matrix (vector). To this end, we introduce a new index p Y m(N ; 1) ; n for m : 0, 1, . . . , M, and n : 0, 1, . . . , N. Using this index, Eq. (3.320) becomes ; ; ; 9 4 : h . N>,> N\,> N> N\ N N

(3.321)

This expression is only valid at the interior points m : 1, 2, . . . , M 9 1, and n : 1, 2, . . . , N 9 1, with ; and ; . It is K,>>L KL K,>>L KL instructive to examine the vector : ' , ,> ,> [] : ' ,>, ,> ' ' +,>,

(3.322)

204

CHAPTER 3 Field Analysis

In Eq. (3.322), the boundary values are identiﬁed by m : 0 $ ( , , . . . , ) (x : 0), , m : M $ ( , , . . . , ) (x : L), +,>> +,>> +,>>, n : 0 $ ( , , , . . . , ) (y : 0), +,> ,> ,> and n : N $ ( , , , . . . , ) (y : H). +,>>, , ,>>, ,>>, As a ﬁnal step, assemble Eq. (3.321) into matrix form: (3.323) [A] · [ ] : [(], where [A] is a sparse matrix (many zero elements), [ ] is the vector of unknown nodal values at the interior points, and [(] is a vector containing the nodal source values and the known boundary condiN tions. There are many approaches to solving Eq. (3.323), and the most popular ones can be broadly classiﬁed as iterative methods and direct matrix methods. The iterative methods involve making an initial guess [ ], and then solving iteratively for successively more accurate solutions [ ]I. The direct matrix methods entail the direct solution of Eq. (3.323). For example, if [A] can be inverted, the solution is given by the following: [ ] : [A]\[(]. A detailed description of the various solution methods is beyond the scope of this book. However, they are described in detail in several texts [17, 18]. References 1. Moon, F. C. (1984). Magneto-solid Mechanics, New York: Wiley. 2. Lowther, D. A. and Silvester, P. P. (1986). Computer-aided Design in Magnetics, New York: Springer-Verlag. 3. Woodson, H. H. and Melcher, J. R. (1985). Electromechanical Dynamics Part II: Fields, Forces and Motion, Malabar, FL: John E. Krieger Publishing Co. 4. Johnk, C. T. A. (1988). Engineering Electromagnetic Fields and Waves, 2nd ed., New York: John Wiley and Sons. 5. Morrish, A. H. (1983). The Physical Principles of Magnetism, Malabar, FL: R. E. Krieger Publishing Co. 6. Craik, D. (1995). Magnetism: Principles and Applications, New York: John Wiley and Sons. 7. Reitz, J. R., Milford, F. J., and Christy, R. W. (1980). Foundations of Electromagnetic Theory, 3rd ed., Reading, MA: Addison-Wesley.

3.9 FINITE DIFFERENCE METHOD

205

8. McCaig, M. and Clegg, A. G. (1987). Permanent Magnets in Theory and Practice, 2nd ed., New York: John Wiley and Sons. 9. Zahn, M. (1987). Electromagnetic Field Theory: A Problem Solving Approach, Malabar, FL: R. E. Krieger Publishing Co. 10. Silvester, P. P. and Ferrari, R. L. (1990). Finite Elements for Electrical Engineers, 2nd ed., New York: Cambridge Univ. Press. 11. Sabonnadiere, J. C. and Coulomb, J. L. (1987). Finite Element Methods in CAD: Electrical and Magnetic Fields, New York: Springer-Verlag. 12. Hoole, S. R. H. (1989). Computer-aided Analysis and Design of Electromagnetic Devices, New York: Elsevier Science Publishing Co. 13. Salon, S. J. (1995). Finite Element Analysis of Electrical Machines, Boston: Kluwer Academic Publishers. 14. Sykulski, J. K. (ed.) (1995). Computational Magnetics, New York: Chapman & Hall. 15. Sadiku, M. N. O. (1992). Numerical Techniques in Electromagnetics, Boca Raton, FL: CRC Press. 16. Furlani, E. P. (1994). Analysis of a sintered NdFeB magnetic lens, J. Magn. Magn. Mat. 134: 117. 17. Smith, G. D. (1985). Numerical Solution of Partial Differential Equations: Finite Difference Methods, 3rd ed., Oxford Applied Mathematics and Computing Science Series, New York: Oxford University Press. 18. Ames, W. F. (1992). Numerical Methods for Partial Differential Equations, 3rd ed. New York: Academic Press.

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CHAPTER

4

Permanent Magnet Applications

4.1 INTRODUCTION Permanent magnets in one form or another can be found in a wide variety of equipment ranging from consumer products, to industrial machinery, to research apparatus. In particular, they are used in audio/ video equipment, personal computers, printers, copiers, automobiles, household appliances, power tools, industrial motors and generators, and various biomedical apparatus. Permanent magnets have several advantages over conventional (current-driven) electromagnets. The fundamental advantage is that they can provide a relatively strong magnetic ﬁeld over an extended spatial region for an indeﬁnite period of time with no expenditure of energy. Of course, the ﬁeld they provide is ﬁxed whereas the ﬁeld of an electromagnet can be changed by adjusting the current. Another advantage of permanent magnets is that they can be fabricated with a wide range of structural properties, geometric shapes, and magnetization patterns. They are also relatively inexpensive on a per unit basis depending on the material used. Permanent magnets have an additional advantage over electromagnets in that their performance scales well with size. Speciﬁcally, if we change all the linear dimensions L of an electromagnet, while keeping the ﬁeld strength at all the rescaled observation points ﬁxed, the current density must be adjusted by a factor of 1/L (Section 5.16). Therefore, as the size of an electromagnet is reduced, a point is reached where overheating precludes viability. Moreover, if the dimensions of the conductors are increased to compensate for this, the ﬁeld strength at the

207

208

CHAPTER 4 Permanent Magnet Applications

rescaled observation points will decrease. By comparison, if we change all the linear dimensions of a permanent magnet, the ﬁeld strength at all the rescaled observation points remains constant (assuming the magnetization is constant). Thus, there will be a dimension below which a permanent magnet will be the only viable ﬁeld source. In this chapter we discuss various permanent magnet applications. We consider applications in which a magnet is used alone as a passive ﬁeld source or in conjunction with other magnets as part of a functional device. Speciﬁcally, we discuss bias magnet structures, high ﬁeld structures, latching magnets, magnetic suspensions, and magnetic gears and couplings. We also discuss various miscellaneous applications that utilize magnets including MRI, magnetooptical recording, electrophotography and free-electron lasers. Throughout the chapter, we present solved examples of various applications. Many of these contain formulas that are useful for design and optimization.

4.2 MAGNET STRUCTURES Bias magnets are used to provide a prescribed ﬁeld distribution over a given region. Such magnets are usually either rectangular or cylindrical in shape and have bipolar or multipole magnetization patterns. In this section we study the most common bias magnet structures. We derive ﬁeld solutions for these structures and demonstrate the solutions with practical calculations.

4.2.1 Rectangular structures The most common permanent magnet structure is the rectangular bar magnet. We developed a limited ﬁeld solution for this structure in Section 3.4, where we used the charge model to obtain the B-ﬁeld above the magnet along a line of symmetry (Example 3.4.1). In this section, we derive general ﬁeld solutions for two different rectangular structures, an inﬁnitely long magnet and a ﬁnite bar magnet. The ﬁeld solutions are derived in the following examples. EXAMPLE 4.2.1 Determine the B-ﬁeld outside a long rectangular magnet (Fig. 4.1). Assume that the magnet has a magnetization M : M y . Q

209

4.2 MAGNET STRUCTURES

Rectangular magnet.

FIGURE 4.1

SOLUTION 4.2.1 Assume that the magnet is inﬁnitely long. In this case the analysis reduces to a 2D ﬁeld problem. Choose a reference frame as shown in Fig. 4.2a. We use the current model to determine the ﬁeld (Section 3.3). First, we reduce the magnet to an equivalent current distribution. We apply Eq. (3.95) and ﬁnd that the volume current density is zero, J : ;M : 0. To determine K the surface current density j we need to identify the unit surface normals. K These are n : and n :

y (y : h) 9y (y : 9h)

x (x : w) 9x (x : 9w).

As j : M ; n and M : M y , we ﬁnd that K Q 9M z (x : w) Q j : K M z (x : 9w). Q Thus, the magnet reduces to two parallel inﬁnitely long current sheets of height 2h as shown in Fig. 4.2b. The ﬁeld solution for a current sheet was derived in Example 3.2.5. We repeat it here for convenience,

K x ; (y 9 h) 2hx B (x, y) : ln x ; 2 tan\ y . 4 x ; (y ; h) x ; y 9 h (4.1) Here, K is the surface current density (A/m). The solution (4.1) applies to a current sheet at x : 0. However, for the magnet there are two current sheets,

210

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.2 Inﬁnite rectangular magnet: (a) cross-sectional view with reference frame; and (b) equivalent surface currents.

one at x : 9w, and another at x : w. We obtain the ﬁeld of the magnet by taking a superposition, B(x, y) : B (x ; w, y) 9 B (x 9 w, y).

(4.2)

Notice that the minus sign in the second term reﬂects the fact that the current at x : w is ﬂowing in the negative z-direction (into the page as shown in

211

4.2 MAGNET STRUCTURES

Fig. 4.2b). We evaluate Eq. (4.2) and obtain the ﬁeld components,

M (x ; w) ; (y 9 h) (x 9 w) ; (y 9 h) B (x, y) : Q ln 9 ln V 4 (x ; w) ; (y ; h) (x 9 w) ; (y ; h)

(4.3)

and

M 2h(x;w) 2h(x9w) B (x, y) : Q tan\ 9 tan\ W 2 (x;w);y9h (x9w);y9h

(4.4) where we have substituted K : M . Equations (4.3) and (4.4) give the B-ﬁeld Q at any point outside the magnet. Calculations: We apply Eqs. (4.3) and (4.4) to a magnet with w : 10 mm, h : 5 mm, and M : 8.0; 10 A/m. This value of magnetization corresponds Q to sintered NdFeB with a remanence of B : 10,054 G. We evaluate B and B P V W along two different horizontal lines that span 92w x 2w. These are located 0.5 and 2.0 mm above the magnet, respectively (Fig. 4.3). Notice that closer to the magnet (0.5 mm above the surface), the vertical component B is higher near W the ends of the magnet than at its center (x : 0). However, farther away the opposite is true. The reason for this is as follows: Near the magnet, the ﬁeld is predominantly vertical across the entire surface and has a greater magnitude closer to the current sheets, which are at the ends of the magnet. However, as you move away from the magnet, the vertical projection of the ﬁeld decreases dramatically near the ends of the magnet and this results in a peak value in the center. Finally, notice that the horizontal component B peaks above the current V sheets (x : <w) as expected. ) EXAMPLE 4.2.2 Determine the B-ﬁeld outside a rectangular bar magnet (Fig. 4.4a). Assume that the magnetization is M : M z . (4.5) Q SOLUTION 4.2.2 Choose a coordinate system centered with respect to the magnet (Figs. 4.4b and 4.5). Let (x , x ), (y , y ) and (z , z ) denote the positions of the edges of the magnet with respect to the x-, y-, and z-axes. We use the charge model to determine the ﬁeld (Section 3.4). First, reduce the magnet to an equivalent charge distribution. From Eqs. (3.105) and (4.5) we ﬁnd that the volume charge density is zero, : 9 · M : 0. However, there K is a surface charge density

M (z : z ) Q : K 9M (z : z ). Q

(4.6)

212

CHAPTER 4 Permanent Magnet Applications

The components B and B vs x: (a) y : h ; 0.5 mm; and (b) V W y : h ; 2 mm.

FIGURE 4.3

The B-ﬁeld follows from Eq. (3.107), M B(x, y, z) : Q (91)I 4 I W V [(x 9 x)x ; (y 9 y)y ; (z 9 z )z ] dx dy I ; . [(x 9 x) ; (y 9 y) ; (z 9 z )] W V I

(4.7)

213

4.2 MAGNET STRUCTURES

FIGURE 4.4

Rectangular magnet: (a) geometry and polarization; and (b) refer-

ence frame.

The x-component: B follows from Eq. (4.7), V M W V (x 9 x) dx dy B (x, y, z) : Q (91)I . V 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I Integration with respect to x gives

M W dy B (x, y, z) : Q (91)I>K . V 4 [(x9x );(y9y);(z9z )] I K W K I

214

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.5 Cross-sectional views of a bar magnet with reference frame: (a) x-y plane; and (b) x-z plane.

The remaining y integration can be evaluated by making a change of variable to : y 9 y. The resulting ﬁeld expression is M B (x, y, z) : Q (91)I>K ln[F(x, y, z, x , y , y , z )], V K I 4 I K

(4.8)

where (y 9 y ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I F(x, y, z, x , y , y , z ) : . K I (y 9 y ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I

215

4.2 MAGNET STRUCTURES

The y-component: B also follows from Eq. (4.7), W M W V (y 9 y) dx dy B (x, y, z) : Q (91)I . W 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I Integration with respect to y yields

M V dx B (x, y, z) : Q (91)I>K W 4 [(x 9 x) ; (y 9 y ) ; (z 9 z )] K I I K V . The remaining x integration is evaluated using a change of variable : x 9 x. The resulting ﬁeld expression is M B (x, y, z) : Q (91)I>K ln[H(x, y, z, x , x , y , z )]. K I W 4 I K

(4.9)

where (x 9 x ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I . H(x, y, z, x , x , y , z ) : K I (x 9 x ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I The z-component: B is given by X M W V (z 9 z ) dx dy I B (x, y, z) : Q (91)I . X 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I The x integration is performed using a change of variable : x 9 x,

M W V\V z9z I B (x, y, z) : 9 Q d dy X [ ; (y 9 y) ; (z 9 z )] 4 W V\V I M : Q (91)I>L>(z 9 z )(x 9 x ) I L 4 I L W dy ; , (4.10) W [(y 9 y) ; (z 9 zI)]((y 9 y) ; b

where b : (x 9 x ) ; (z 9 z ). The remaining y integration is performed L I using a change of variable : y 9 y. This gives M B (x, y, z) : Q (91)I>L>K X 4 I L K (x 9 x )(y 9 y ) L K g(x, y, z; x , y , z ) , (4.11) ;tan\ L K I (z 9 z ) I

216

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.6

Bar magnet.

where 1 g(x, y, z; x , y , z ) : . L K I [(x 9 x ) ; (y 9 y ) ; (z 9 z )] L K I The formulas (4.8), (4.9) and (4.11) give the ﬁeld at any point outside the magnet. Calculations: We apply the theory to a magnet with a : 10 mm, b : 20 mm, and c : 10 mm. We set x : 0, x : a, y : 0, y : b, z : 9c, z : 0 mm, and M : 8.0;10 A/m (Fig. 4.6). We evaluate B across two different surfaces Q X above the magnet, one at z : 0.5 and the other at z : 2.0 mm. The computed data are shown in Fig. 4.7. Notice that closer to the magnet, B dips near the X center of the magnet, but farther away it peaks there. This can be understood from the current model. Speciﬁcally, the magnet can be reduced to equivalent current sheets along its edges. Near the magnet the ﬁeld is predominantly vertical and has a greater magnitude closer to the current sheets, which are at the edges. However, as you move away from the magnet, the vertical projection of the ﬁeld decreases dramatically near the ends of the magnet and this results in a peak value at the center. ) The ﬁeld solution derived in Example 4.2.2 can be used to determine the ﬁeld of multipole rectangular structures such as the quadrupole structure shown in Fig. 4.8. This structure consists of four block magnets and the ﬁeld is given by B(x, y, z) : B (x, y, z), G G where B (x, y, z) is the ﬁeld due to the ith block. G

4.2 MAGNET STRUCTURES

FIGURE 4.7

at z : 2 mm.

217

Magnetic ﬁeld above a bar magnet: (a) B at z : 0.5 mm; and (b) B X X

218

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.8

Quadrupole magnet structure.

4.2.2 Cylindrical structures Cylindrical magnets are the second most common magnet structure. They are used in devices such as encoders, actuators, motors, electrophotographic copiers, and magnetic couplings and gears. Cylindrical magnets can have various magnetization patterns including bipolar with magnetization either through the cross section or along the axis, and multipole with radial or axial magnetization (Fig. 4.9). In Sections 3.3 and 3.6.2. we developed partial ﬁeld solutions for the bipolar cylindrical magnets (Examples 3.3.1 and 3.6.3). In this section we consider the full range of magnetization patterns and derive general ﬁeld solutions that

Cylindrical magnets: (a) bipolar magnetization; (b) multipole radial magnetization; and (c) multipole axial magnetization. FIGURE 4.9

219

4.2 MAGNET STRUCTURES

FIGURE 4.10

Bipolar cylinder: (a) geometry and magnetization; and (b) refer-

ence frame.

are valid at any point outside the magnet. The ﬁeld solutions are derived in the following examples. EXAMPLE 4.2.3 Derive an expression for the B-ﬁeld outside a bipolar cylindrical magnet of ﬁnite length (Fig. 4.10) [1]. Assume that the axis of the magnet is along the z-axis. Let R denote it radius, and let z and z denote the positions of its bottom and top edge, respectively. Furthermore, assume that the magnet has a magnetization M : M x . Q

220

CHAPTER 4 Permanent Magnet Applications

SOLUTION 4.2.3 We use the charge model of Section 3.4. First, we reduce the magnet to an equivalent charge distribution and then determine the ﬁeld by treating the charge distribution as a ﬁeld source. The equivalent charge distribution follows from Eq. (3.105). We ﬁnd that the volume charge is density zero, : 9 · M : 9 · M x : 0, and that the surface charge density is K Q : M · n K : M x · r Q : M cos(). (4.12) Q The B-ﬁeld is computed using

X L M cos() Q B(x) : 9 R d dz, 4

x 9 x X where acts on the unprimed variables. Radial component: The radial component is given by

M R X L B (r, , z) : Q cos()[r 9 R cos( 9 )] P 4 X ; g(r, , z; R, , z) d dz,

(4.13)

where g(r, , z; r, , z) :

1 (r ; r 9 2rr cos( 9 ) ; (z 9 z)

.

We evaluate the integration numerically using Simpson’s method and the z integration in closed form. We obtain M R , B (r, , z) : Q (91)I>S (n) cos((n)) P 2N L I ; [r 9 R cos( 9 (n))]I(r, , z; R, (n), z ), I where N is the Simpson’s method mesh number (even), (n) :

n 2 N

(n : 0, 1, 2, . . . , N),

(4.14)

221

4.2 MAGNET STRUCTURES

and

S (n) :

(n : 0) (n : 1, 3, 5, . . .) (n : 2, 4, 6 . . .) (n : N).

Function I in Eq. (4.14) is deﬁned by

X

g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z ). I X I Evaluation of Eq. (4.15) gives

IY

where

(4.15)

F(r, , z; r, , z ) I if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z ) I if

(4.16)

r : r, cos( 9 ) : 1, z " z I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 ) Notice that the functional form of I depends on the relationship between the coordinates of the ﬁeld point (r, , z) and the source point (r, , z). Speciﬁcally, I is given by F as long as r ; r 9 2rr cos( 9 ) " 0. If this condition is violated, then the lower ratio is used. Azimuthal component: The azimuthal ﬁeld component is given by

M R X L B (r, , z) : Q cos() 4 X ; sin( 9 )g(r, , z; R, , z) d dz.

(4.17)

As in the radial case, the integration is evaluated using Simpson’s method, and the integration in z is expressed in terms of Eq. (4.16). The resulting

222

CHAPTER 4 Permanent Magnet Applications

expression for B is M R , B (r, , z) : Q (91)I>S (n) cos((n)) 2N L I ; sin( 9 (n))I(r, , z; R, (n), z ). (4.18) I Axial component: The derivation for the axial ﬁeld component is similar to that for the previous two cases. Speciﬁcally,

M R X L cos()(z 9 z) B (r, , z) : Q X 4 X ; g(r, , z; R, , z) d dz.

(4.19)

Once again, the integration can be expressed as a discrete sum, M R , B (r, , z) : Q S (n) cos((n)) X 2N L X ; (z 9 z)g(r, , z; R, (n), z) dz. X The remaining integral can be evaluated analytically. This gives

(4.20)

M R , B (r, , z) : Q S (n) X 2N L ; (91)I cos((n))g(r, , z; R, (n), z ). (4.21) I I Calculations: We demonstrate the ﬁeld solutions (4.14), (4.18) and (4.21) with some sample calculations. Consider a solid cylindrical magnet with R : 2.54 mm, L:50 mm (z :925 mm, z :25 mm), and M :4.3;10 A/m Q (B : 5400 G). We evaluate B and B at r : 3.8 mm for a series of angular P P values : 0, 10, 20, . . . , 180°. These data are shown in Fig. 4.11. Next, we analyze a cylindrical shell geometry with inner and outer radii R and R , respectively (Fig. 4.12). The hollow interior of the shell is taken into account using the principle of superposition. Speciﬁcally, we take the ﬁeld due to a solid cylinder B (x), and subtract the ﬁeld due to an oppositely magnetized inner 0 core B (x), 0 B(x) : B (x) 9 B (x). (4.22) 0 0 The cylindrical shell is symmetrically positioned with respect to the x-y plane

223

4.2 MAGNET STRUCTURES

FIGURE 4.11

The components B and B vs for a solid cylinder. P

(center at z : 0) with its top at z : 3 cm. The parameters used in the analysis are as follows: M : 4.3 ; 10 A/m Q R : 2.0 cm

(inner radius)

R : 4.0 cm

(outer radius)

L : 6.0 cm

(height).

The components B and B are computed at z:0 and 2.7 cm at a radius r:5 cm P for a series of angular values : 0, 10, 20, . . . , 180°. These data are shown in Fig. 4.13. Notice that the ﬁeld decreases in magnitude as the observation point moves away from the center (z : 0 cm) towards the end of the cylinder. Finally, the ﬁeld components B and B are evaluated at r : 5 cm and : 0° for a series P X of points z : 0, 0.25, 0.5, . . . , 5 cm. These data are shown in Fig. 4.14. Notice the behavior of the ﬁelds as the observation point passes over the end of the cylinder (z : 3 cm). )

224

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.12

Cylindrical shell geometry and reference frame.

EXAMPLE 4.2.4 Derive an expression for the ﬁeld outside an inﬁnitely long cylindrical magnet with a linear second quadrant demagnetization relation of the form B : H ; M (r, ). (4.23) Q The magnetization M (r, ) is well behaved but otherwise arbitrary [2]. Let R Q and R denote the inner and outer radii of the cylinder, respectively (Fig. 4.15). SOLUTION 4.2.4 Since the magnet is inﬁnitely long the problem reduces to a 2D boundary value problem (BVP) in cylindrical coordinates. Boundary value problem: The magnetic ﬁelds H inside the magnet, and H and H in the regions V and V outside the magnet can be represented

4.2 MAGNET STRUCTURES

FIGURE 4.13

The components B and B vs for a cylindrical shell. P

FIGURE 4.14

The components B and B vs z for a cylindrical shell. P X

225

226

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.15

Cross section of an inﬁnite cylinder with arbitrary magnetization.

in terms of scalar potentials , and where K K K H : 9 HI : 9I (k : 1, 2). (4.24) K K The BVP for and I follows from: (a) the constitutive relation B : K K H ; M (r, ); (b) the ﬁeld equations Q · B : 0 · BI : 0 (k : 1, 2), (4.25) (c) the boundary conditions for B and H at the media interfaces; and (d) the constitutive relations BI : HI (k : 1, 2), (4.26) which apply outside the magnet. Speciﬁcally, inside the magnet we solve : 9 , K K where is an equivalent volume charge density, K : 9 · M (r, ). K Q

(4.27)

(4.28)

Outside the magnet we solve I : 0 K The boundary conditions are : I K K

at

(k : 1, 2). r:R I

(4.29)

(k : 1, 2),

(4.30)

and where

I K 9 K : I n n

at

r:R I

(k : 1, 2),

(4.31)

227

4.2 MAGNET STRUCTURES

I : nI · M (r, ). (4.32) K Q Here, nI : (91)Ir are the outward surface normals at r : R . We also I require that (0) -, K

(4.33)

and (-) -, (4.34) K which restrict the form of the solutions in regions V and V , respectively. Separation of variables: We employ the method of separation of variables to solve the BVP (4.27)—(4.34). Accordingly, we represent the solutions to Eqs. (4.27) and (4.29) as follows: (r, ) : [V (r) sin(i) ; U (r) cos(i)], K G G G (r, ) : rG[V sin(i) ; U cos(i)], K G G G

(4.35) (4.36)

and (r, ) : r\G[V sin(i) ; U cos(i)]. (4.37) K G G G The volume charge density and surface term can also be expressed as Fourier series, (r, ) : [C (r) sin(i) ; D (r) cos(i)], K G G G

(4.38)

and I() : [AI sin(i) ; BI cos(i)], (4.39) K G G G where C (r), D (r), AI and BI are determined once the magnetization M (r, ) G G G G Q is speciﬁed. Substituting Eqs. (4.35) and (4.38) into Eq. (4.27) yields the following ordinary differential equations for the unknown coefﬁcients V (r) and G U (r), G dV 1 dV i G ; G 9 V : 9C (r), (4.40) G dr r dr r G

228

CHAPTER 4 Permanent Magnet Applications

and dU 1 dU i G ; G 9 U : 9D (r). G dr r dr r G

(4.41)

These equations are of the form df(r) 1 df(r) i ; 9 f(r) : g(r), r dr r dr with homogeneous solutions f (r) : rG and f (r) : r\G. Solution: We use the method of variation of parameters and obtain general solutions for Eqs. (4.40) and (4.41) of the form

f ())g()) P f ())g()) d) ; f (r) d), W(f , f )()) W(f , f )()) 0 0 where c and c are constants and W(f , f ) : f f 9 f f is the Wronskian. Application of this method to Eqs. (4.40) and (4.41) yields f(r) : c f (r) ; c f (r) 9 f (r)

P

1 P ) V (r) : E rG ; F r\G 9 G G G 2i 0

r G ) G 9 C ()) d) G ) r

(4.42)

and

1 P r G ) G U (r) : P rG ; Q r\G 9 ) 9 D ()) d). (4.43) G G G G 2i ) r 0 The unknown coefﬁcients VI, UI (k : 1, 2), E , F , P , and Q are determined G G G G G G by imposing the boundary conditions (4.30) and (4.31). For example, the following group of simultaneous equations is obtained for the coefﬁcients UI, G P , and Q : G G i UiR\G> ; P iRG\ 9 Q iR\G> 9 S : B, (4.44) G G G G R G UR\G : P RG ; Q R\G 9 S , (4.45) G G G G and URG : P RG ; Q R\G, G G G

(4.46)

UiRG\ 9 P iRG\ ; Q iR\G> : B, G G G G

(4.47)

and

229

4.2 MAGNET STRUCTURES

where

1 0 S : ) G 2i 0

R G ) G ; D ()) d), G ) R

and

1 0 R G ) G 9 S: ) D ()) d). G 2i G ) R 0 Equations (4.44)—(4.47) can be written in matrix form: 0 1

RG

9

0 1

9RG

91

1 0

91

9R\G R\G

1 0

9

U G U G : P G Q G

S B G; G RG> i R 9RG S G 0

.

(4.48)

B R\G> G i

A similar set of equations is obtained for the coefﬁcients VI, E , and F . The G G G method for solving the BVP can now be summarized: Given M (r, ) inside the Q magnet, evaluate Eqs. (4.28) and (4.32) and determine the coefﬁcients C (r), G D (r), and AI, BI (k : 1, 2) in Eqs. (4.38) and (4.39), respectively. Next, G G G substitute D (r) and BI into Eqs. (4.44)—(4.47), and C (r) and AI into the G G G G corresponding equations for VI, E , and F and obtain two sets of decoupled G G G (four by four) simultaneous equations for the unknown coefﬁcients UI, P , Q , G G G and VI, E , and F . Finally, solve the equations for VI and UI and substitute G G G G G these into Eqs. (4.36) and (4.37) to obtain the desired ﬁeld solution. Calculations: The analysis outlined above is applied to a cylinder with the magnetization as shown in Fig. 4.16. Such magnets are used in the electrophotographic process (Section 4.9) [2]. The magnetization is

9M r Q M r Q M : Q 9M r Q 0

(9 9 ) (9 ) ( ) ( 2 9 ). This can be expressed analytically as follows: M : M()r , Q

(4.49)

(4.50)

230

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.16

Magnetized cylinder and reference frame.

where

and

M() : M cos(i), G G

(4.51)

2M Q (2 sin(i ) 9 sin(i )). M: G i

(4.52)

A solution is sought for the region V outside the cylinder. Therefore, the coefﬁcients of interest are V and U. From the symmetry of the problem, the G G coefﬁcients V : 0 for all i. The coefﬁcients U are determined from Eq. G G (4.48), which gives H(M , R , R , ) G U(M , R , R , ) : G G G G(R , R , ) G

(i : 1, 2, 3, . . .),

(4.53)

where

1 ; 1 K(M , R , R ) H(M , R , R , ) : RG> G G G R M R G ; G 19 9 1; i R M R G> G ;2 ; h* (M , R , R , ) , G G i R

(4.54)

231

4.2 MAGNET STRUCTURES

with

M R G G K(M , R , R ) : R 9 R , G i(i ; 1) R

(4.55)

and with

M 1 R G G 91 h* (M , R , R , ) : G G i R R 1 R G R 9 R 19i R ; R R ln R

and

R G G(R , R , ) : G R

(i 1) (4.56) (i : 1)

92 ;1 9

;2 ;1 . (4.57)

The ﬁeld components follow from B : 9 . Speciﬁcally, K B(r, ) : ir\G>U(M , R , R , ) cos(i), P G G G and B (r, ) : ir\G>U(M , R , R , ) sin(i). G G G The parameters used in the analysis are as follows:

(4.58)

(4.59)

M : 2.0 ; 10 A/m Q R : 8.0 mm R : 12.0 mm : 30° : 60°. (4.60) The value of M corresponds to injection-molded barium ferrite material. We Q compute the ﬁeld components B(r, ) and B (r, ) at r : 16 mm and : 0, P 5, 10, 15, . . . , 180° with : . Figure 4.17 shows the predicted ﬁeld data along with the corresponding data from FEA. We also compute B (r, ) at the P same observation points as in the preceding and with : 1.0 , 1.5 , and

232

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.17

The components B and B vs (; : FEA). P

2.0 . These data are shown in Fig. 4.18. Notice that lower values of give rise to stronger ﬁelds outside the cylinder. ) EXAMPLE 4.2.5 Determine the B-ﬁeld outside an inﬁnitely long cylindrical magnet that has an alternating radial polarization and a linear second quadrant demagnetization curve of the form B : H ; M (). Q

(4.61)

Let R and R denote the inner and outer radii of the cylinder, respectively (Fig. 4.19). SOLUTION 4.2.5 As the magnet is inﬁnitely long, the problem reduces to a 2D boundary value problem (BVP) similar to that of Example 4.2.4. We use the solution method presented there with the magnetization inside the cylinder given by M () : M()r , Q

(4.62)

233

4.2 MAGNET STRUCTURES

FIGURE 4.18

FIGURE 4.19

Component B vs . P

Inﬁnite cylinder with radial magnetization.

234

CHAPTER 4 Permanent Magnet Applications

where M() : M cos(i), G G

(4.63)

and

M , (2k 9 1)i (2k ; 1)i M : Q (91)I sin 9 sin . (4.64) G i N N I In Eq. (4.64), N is the number of poles. The expressions (4.63) and (4.64) give the Fourier decomposition of M(). First, we determine the representations for () and I () by evaluating the coefﬁcients C (r), D (r), AI and BI in K G G K G G Eqs. (4.38) and (4.39). From Eq. (4.28) we obtain () : 9 · M () K Q M() :9 . r From the orthogonality of sin(i) and cos(i) we ﬁnd that C (r) : 0, and that G M D (r) : 9 G . G r Similarly, from Eq. (4.32) we have I() : (91)Ir · M () K Q : (91)IM().

Therefore AI : 0 and G (k : 1, 2). BI : (91)IM G G We seek a solution for the region V outside the cylinder. The coefﬁcients of interest are V and U. From the orthogonality of sin(i) and cos(i) we ﬁnd G G that V : 0 for all i. The coefﬁcients U are given by Eq. (4.53): G G H(M , R , R , ) G U(M , R , R , ) : G (i : 1, 2, 3, . . .), (4.65) G G G(R , R , ) G where H(M , R , R , ) and G(R , R , ) are deﬁned in Eqs. (4.54)—(4.57) G G G with M given by Eq. (4.64). Once the U are known, we compute the ﬁeld G G components B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.66)

4.2 MAGNET STRUCTURES

FIGURE 4.20

235

The components B and B vs (r : 12 mm). P

and B (4.67) (r, ) : ir\G>U (M , R , R , ) sin(i). G G G Calculations: We demonstrate the ﬁeld solution (4.66)—(4.67) via the analysis of an eight-pole cylindrical magnet with R : 1 mm, R : 10 mm, M :4.3 Q ; 10 A/m, and : . We evaluate B and B at r : 12 mm for 0 P 180°. These data are shown in Fig. 4.20. Here, : 0° corresponds to the center of a north pole. Notice that B peaks at the center of the pole, whereas B P peaks at the transition between neighboring poles at : 22.5°. ) EXAMPLE 4.2.6 Determine the radial and azimuthal B-ﬁeld components outside a ﬁnite, radially polarized multipole cylinder (Fig. 4.21) [3]. Assume that the magnet has a magnetization M : <M r , (4.68) Q where the < term takes into account the alternating polarity of adjacent poles.

236

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.21

Radially polarized multipole cylinder.

SOLUTION 4.2.6 The geometry is three-dimensional and, therefore, we need to solve a 3D ﬁeld problem. We use the current model of Section 3.3. From Eqs. (3.95) and (4.68) we ﬁnd that the volume current density is zero J : ; M : 0. K

(4.69)

Therefore, Eq. (3.93) reduces to A(x) : 4

1

j (x) K ds,

x 9 x

(4.70)

where j : M ; n. K

(4.71)

The B-ﬁeld is computed from Eq. (4.70) using B : ; A.

(4.72)

Instead of determining B for the entire magnet, it is simpler to evaluate Eq. (4.70) ﬁrst for a single sector shown in Fig. 4.22, determine B for the sector, and then compute the total ﬁeld as a superposition of the contributions from all the sectors. As a ﬁrst step, we determine j for the various surfaces of the sector. K From Eqs. (4.68) and (4.71) it follows that j : 0 on the back (r : R ) and K front (r : R ) of the sector because the magnetization and surface normals are

237

4.2 MAGNET STRUCTURES

FIGURE 4.22

Sector geometry for a radially polarized cylinder.

either parallel or antiparallel for these surfaces. There are four remaining surfaces to consider:

r (1) r r (2) Q Q top surface : (1) (2) Q Q z : z (2), Q

(4.73)

r (1) r r (2) Q Q bottom surface : (1) (2) Q Q z : z (1), Q

(4.74)

r (1) r r (2) Q Q left side : : (1) Q z (1) z z (2), Q Q

(4.75)

238

CHAPTER 4 Permanent Magnet Applications

and

r (1) r r (2) Q Q right side : : (2) Q z (1) z z (2). Q Q

(4.76)

The unit normals for these surfaces are

n :

z

(top surface)

9z

(bottom surface)

9

(left side)

(4.77)

(right side).

Therefore, assuming the radial magnetization given in Eq. (4.68), the corresponding surface current densities are

9M Q M Q j : K 9M z Q M z Q

(top surface) (bottom surface)

(4.78)

(left side) (right side).

Taking into account Eqs. (4.73)—(4.78), we write Eq. (4.70) as M A (x) : Q (91)H Q 4 H ;

XQ PQ

XQ PQ

z dr dz

x 9 x : ( j) 9

Q

PQ

PQ

Q

Q

x 9 x

r d dr . XYXQH

(4.79)

In this expression, and for the remainder of this example, the subscript s denotes the contributions due to a single sector. Notice that the unit vector in the integrand of Eq. (4.79) is itself a function of , : 9sin()x ; cos()y .

(4.80)

239

4.2 MAGNET STRUCTURES

Therefore, Eq. (4.79) can be written in terms of Cartesian coordinates, M A (x) : Q (91)H Q 4 H XQ PQ z ; dr dz

x 9 x : ( j) Q XQ PQ P Q sin( ) x ; Q r d dr

x 9 x XYXQH PQ Q P Q cos() y 9 Q r d dr . (4.81)

x 9 x XYXQH PQ Q This can be rewritten in terms of cylindrical coordinates by computing the following projections:

A (x) : A (x) · r , PQ Q A (x) : A (x) · , Q Q

(4.82) (4.83)

and A (x) : A (x) · z . (4.84) XQ Q We apply these projections to Eq. (4.81) and obtain the following explicit forms: M A (x) : Q (91)H> PQ 4 H PQ Q sin( 9 ) r d dr, ;

x 9 x PQ Q XYXQH M A (x) : Q (91)H> Q 4 H PQ Q cos( 9 ) ; r d dr,

x 9 x PQ Q XYXQH

and

(4.85)

(4.86)

M A (x) : Q (91)H XQ 4 H XQ P 1 dr dz. (4.87) ;

x 9 x XQ PQ Y QH Equations (4.85), (4.86), and (4.87) give A (x) for a single sector. For the sector, Q B (x) is obtained using Eq. (4.72). As the total ﬁeld is a superposition of the Q

240

CHAPTER 4 Permanent Magnet Applications

ﬁelds due to all the sectors, it can be expressed as , B(x) : (91)Q>B (x), (4.88) Q Q where the (91)Q> term takes into account the alternating polarity of adjacent poles. For the derivation of B and B we use the function P 1 . (4.89) g(r, , z; r, , z) : [r ; r 9 2rr cos( 9 ) ; (z 9 z)] Radial component: The radial ﬁeld component due to a single sector follows from Eq. (4.72): 1 B (x) : (4.90) A (x) 9 A (x). PQ r XQ z Q Substituting Eqs. (4.86) and (4.87) into Eq. (4.90) yields M B (x) : Q (91)H> PQ 4 H XQ PQ r sin( 9 ) dr dz ;

x 9 x XQ PQ Y QH P Q (z 9 z) cos( 9 ) r d dr . (4.91) ; Q

x 9 x XYXQH PQ Q The integrations in r in the ﬁrst term and in the second term can be written in terms of discrete sums using various numerical methods. We use Simpson’s method. Speciﬁcally, let N and N denote the number of mesh points in the r P and variables, respectively, and let r(n) and (m) denote the values at which Q the integrands are evaluated. As r (1)rr (2), and (1) (2), we Q Q Q Q have r (2) 9 r (1) Q r : Q (4.92) Q N P and (2) 9 (1) Q . : Q (4.93) Q N

The integration points are as follows:

and

n r (n) : r (1) ; (r (2) 9 r (1)) Q Q Q N Q P m (m) : (1) ; ( (2) 9 (1)) Q Q Q N Q

(n : 0, 1, 2, . . . , N ), P

(4.94)

(m : 0, 1, 2, . . . , N). (4.95)

241

4.2 MAGNET STRUCTURES

Finally, deﬁne the Simpson integration coefﬁcients S (n) and S (m), P (n : 0) (n : 1, 3, 5 . . .) S (n) : P (n : 2, 4, 6 . . .) (n : N ) P and

(m : 0) (m : 1, 3, 5 . . .) S (m) : (m : 2, 4, 6 . . .) (m : N ). Using this scheme, Eq. (4.91) reduces to

(4.96)

(4.97)

M B (x) : Q (91)H> PQ 4 H , ; r P S (n)r (n) sin( 9 (j)) Q Q P Q L X ; Q g(r, , z; r (n), (j), z) dz Q Q XQ , ; S (m)(z 9 z (j)) cos( 9 (m)) Q Q Q K P ; Q g(r, , z; r, (m), z (j))r dr . Q Q PQ The remaining integrals in z and r can be expressed in closed form. In addition, the contributions from all the sectors need to be taken into account in accordance with Eq. (4.88). The resulting ﬁeld expression is

M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I ,P ; r S (n)r (n) sin( 9 (j))I (r, , z; r (n), (j), z (k)) Q Q Q P Q Q Q L , ; S (m)(z 9 z (j)) cos( 9 (m))I (r, , z; r (k), (m), z (j)) . Q Q Q Q Q Q K (4.98)

The function I in Eq. (4.98) is deﬁned by XQ g(r, , z; r, , z) dz : (91)I>I (r, , z; r, , z (k)). (4.99) Q XQ I

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CHAPTER 4 Permanent Magnet Applications

Evaluation of Eq. (4.99) gives

I Y

where

F(r, , z; r, , z (k)) Q if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z (k)) Q if

(4.100)

r : r, cos( 9 ) : 1, z " z (k) Q

(z 9 z (k))g(r, , z; r, , z (k)) Q Q . F(r, , z; r, , z (k)) : Q r ; r 9 2rr cos( 9 ) The function I in Eq. (4.98) is deﬁned by PQ g(r, , z; r, , z)r dr : (91)I>I (r, , z; r (k), , z). (4.101) Q PQ I Evaluation of Eq. (4.101) gives

K(r, , z; r (k), , z) Q if r sin( 9 ) ; (z 9 z) " 0 or 1 r r (1) (r 9 2r (k)) Q Q p: (91)N 2(r 9 r (k)) 2 r r (2) Q Q if r " r (k), cos( 9 ) : 1, z : z Q or I Y r ; 2r (k) Q 2(r ; r (k)) Q if r " r (k), cos( 9 ) : 91, z : z Q or

1 r (k) Q if r : 0, z : z

(4.102)

243

4.2 MAGNET STRUCTURES

where K(r, , z; r (k), , z) : Q

(r ; (z 9 z)) 9 rr (k) cos( 9 ) Q r sin( 9 ) ; (z 9 z)

;g(r, , z; r (k), , z). Q According to Eq. (4.100), the functional form of I depends on the relationship between the coordinates of the ﬁeld point (unprimed) and the source point (primed). In particular, I is evaluated using F as long as r ; r 9 2rr cos( 9 ) " 0. However, if this condition is violated, then the lower ratio is used. The function I also depends on the relation between the source and ﬁeld points. Speciﬁcally, if r sin( 9 ) ; (z 9 z) " 0 then the ﬁrst term in Eq. (4.102) is used. When this condition is violated, the remaining three cases apply as indicated. Notice that the fourth term in Eq. (4.102) is well deﬁned because r (k) " 0 for Q this case (the ﬁeld and source points cannot coincide). Notice also that this analysis does not include the case in which the ﬁeld and source point are one and the same. The parameters that appear in Eq. (4.98) are summarized here for convenience, r (2) 9 r (1) Q , r : Q Q N P (2) 9 (1) Q , : Q Q N and

(2s 9 3) (1) : Q N (2) : (2s 9 1) Q N s : 1, 2, . . . , N r (1) : R (inner radius) Q r (2) : R (outer radius) Q z (2) : top of sector Q z (1) : bottom of sector Q N : angular mesh number (even) N : radial mesh number (even). P

(4.103)

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CHAPTER 4 Permanent Magnet Applications

Azimuthal component: The azimuthal ﬁeld component also follows from Eq. (4.72), B (x) : A (x) 9 A (x). Q PQ z r XQ

(4.104)

Substitute Eqs. (4.85) and (4.87) into Eq. (4.104) and obtain M B (x) : Q (91)H Q 4 H PQ Q (z 9 z) sin( 9 ) ; r d dr

x 9 x PQ Q XYXQH XQ PQ r 9 r cos( 9 ) ; dr dz . (4.105)

x 9 x Y QH XQ PQ Similar to the radial case, the integrals can be evaluated numerically using Simpson’s method as outlined in Eqs. (4.92)—(4.97). This gives

M B (r, , z) : Q (91)H Q 4 H , ; S (m)(z 9 z (j)) sin( 9 (m)) Q Q Q K P ; Q g(r, , z; r, (m), z (j))r dr Q Q PQ , ; r P S (n)(r 9 r (n) cos( 9 (j))) Q Q P Q L X ; Q g(r, , z; r (n), (j), z) dz . Q Q XQ The remaining integrals in z and r are evaluated using Eqs. (4.100) and (4.102). The total azimuthal ﬁeld component is obtained as a superposition of contributions from all the sectors,

M , B (r, , z) : Q (91)Q>H>I 4 Q H I , ; S (m)(z 9 z (j)) Q Q K ;sin( 9 (m))I (r, , z; r (k), (m), z (j)) Q Q Q Q

245

4.2 MAGNET STRUCTURES

, ; r P S (n)(r 9 r (n) cos( 9 (j))) Q Q P Q L

;I (r, , z; r (n), (j), z (k)) , Q Q Q

(4.106)

where I and I are deﬁned in Eqs. (4.100) and (4.102), respectively. Calculations: We demonstrate the ﬁeld solution with some sample calculations. First, we test Eqs. (4.98) and (4.106) using the 2D ﬁeld solutions (4.66) and (4.67) from Example 4.2.5. Consider an inﬁnitely long, radially polarized cylindrical shell with 10 alternating poles. Assume that the shell has inner and outer radii R : 2 cm and R : 4 cm, and a uniform magnetization M : 4.3 Q ; 10 A/m. The value of M corresponds to a bonded NdFeB material. For 3D Q analysis we need to set the length L of the shell. To obtain 2D results, the length must be much greater than the diameter of the shell (L 2R ). We set L : 80 cm with the bottom and top of the shell at z(1) : 940 cm and z(2) : 40 cm, respectively. The ﬁeld components B and B are computed at a radius r : 5 cm P for a series of angular values : 0, 3, 6, . . . , 36° (i.e., from the center of one pole to the center of the neighboring pole). The data from the two models are compared in Fig. 4.23. Next, we apply the model to a cylindrical shell geometry with an axisymmetric magnetization (Fig. 4.24). We use the values R : 2 cm, R : 4 cm, L : 4 cm, and M : 4.3 ; 10 A/m. The cylinder is symmetrically Q positioned with respect to the x-y plane (z : 0 corresponds to the middle of cylinder). Field values are computed along several different lines. First, B is P computed at z : 0, 5, 10, . . . , 60 mm, with r : 50 mm (Fig. (4.25). Next, B P is computed along the line z : 40 mm at the points r : 0, 5, 10, . . . , 60 mm (Fig. 4.26). ) EXAMPLE 4.2.7 Determine the radial and axial ﬁeld components at any point outside an axially polarized cylindrical shell (Fig. 4.27) [4]. Let R and R denote the inner and outer radii of the shell and z and z denote the axial position of the bottom and top of the shell. Assume that magnet has a second quadrant demagnetization curve B : (H ; M z ). (4.107) Q SOLUTION 4.2.7 We use the current model (Section 3.3). From Eqs. (3.95) and (4.107) we ﬁnd that the volume current density is zero, J : ; M : K ; M z : 0. Therefore, Eq. (3.93) reduces to Q j (x) K A(x) : ds. (4.108) 4 x 9 x 1

246

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.23

FIGURE 4.24

B and B vs (r : 50 mm). P

Cylinder with axisymmetric magnetization.

247

4.2 MAGNET STRUCTURES

FIGURE 4.25

FIGURE 4.26

Component B vs z (r : 50 mm). P

The component B vs r (z : 40 mm). P

248

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.27

Axially polarized cylindrical shell with reference frame.

As for the surface current density, we ﬁnd that j : 0 on the top and bottom K surfaces of the shell because the magnetization and surface normals are either parallel or antiparallel there. There are two remaining surfaces to consider,

(4.109)

(4.110)

r : R inner surface : 0 2 z zz

and r : R outer surface : 0 2 z zz .

The unit normals for these surfaces are n :

9r r

(inner surface) (outer surface).

(4.111)

249

4.2 MAGNET STRUCTURES

It follows that the surface current densities are

9M Q M Q Thus, Eq. (4.108) can be rewritten as j : K

(inner surface) (outer surface).

(4.112)

X L M R (j) d dz, (4.113) A(x) : Q (91)H A

x 9 x 4 PY0AH H X where R (1) : R , and R (2) : R . Recall that is a function of position, A A : 9sin()x ; cos()y . (4.114) Substitute Eq. (4.114) into Eq. (4.113) and obtain

M X L sin()x A(x) : Q (91)H 9 R (j) d dz A 4

x 9 x PY0AH H X X L cos()y ; R (j) d dz . A

x 9 x PY0AH X Note that A(x) has no z-component. It can be written in terms of cylindrical coordinates by computing the projections

A (x) : A(x) · r P

(4.115)

A (x) : A(x) · .

(4.116)

and

This gives

M X L sin( 9 ) A (x) : Q (91)H R (j) d dz P A 4

x 9 x PY0AH H X

(4.117)

and M A (x) : Q (91)H 4 H Radial ﬁeld: The radial ﬁeld

X L cos( 9 ) R (j) d dz. (4.118) A

x 9 x PY0AH X component is given by

B (x) : 9 A (x). P z

(4.119)

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CHAPTER 4 Permanent Magnet Applications

Substitute Eq. (4.118) into Eq. (4.119) and take into account both the differentiation in z and the integration in z. This gives M B (r, , z) : Q (91)H>I P 4 H I ;

L

cos( 9 )g(r, , z; R (j), , z )R (j) d. A I A

(4.120)

where g(r, , z; r, , z) :

1 . (4.121) [r ; r 9 2rr cos( 9 ) ; (z 9 z)]

The remaining integral in can be written in terms of a discrete sum using a simple numerical scheme such as Simpson’s method. Speciﬁcally, let N denote the number of mesh points in the variable and let (m) denote the values at which the integrand is evaluated. As 0 2, we have (m) :

m 2 N

(m : 0, 1, 2, . . . , N).

(4.122)

The integration coefﬁcients S(m) are speciﬁed in Eq. (4.97). Applying this scheme to Eq. (4.120) gives an equation for the total radial ﬁeld component: M B (r, , z) : Q (91)H>IR (j) P A 2N H I , ; S (m) cos( 9 (m))g(r, , z; R (j), (m), z ). (4.123) A I K Axial ﬁeld: The axial ﬁeld component is obtained from

1 B (x) : (rA (x)) 9 A (x) . X P r r

(4.124)

Substituting Eqs. (4.117) and (4.118) into Eq. (4.124) yields

M X L r cos( 9 )9R (j) A B (x) : Q (91)H> R (j) d dz. X A 4

x 9 x PY0AH H X (4.125)

251

4.2 MAGNET STRUCTURES

Equation (4.125) can be simpliﬁed using the same numerical scheme as in the previous two sections. Speciﬁcally, we obtain M B (r, , z) : Q (91)H> X 4 H 2 , ; S(m)(r cos( 9 (m)) 9 R (j))R (j) A A N K X ; g(r, , z; R (j), (m), z) dz . A X The remaining integration in z can be evaluated analytically and will result in

M B (r, , z) : Q (91)H>I X 4 H I 2 , S (m)(r cos( 9 (m)) 9 R (j)) ; A N K

; R (j)I(r, , z; R (j), (m), z ) A A I

.

(4.126)

In Eq. (4.126), the function I(r, , z; r, , z ) is deﬁned by I XI g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z ). (4.127) I XI I Evaluation of Eq. (4.127) gives

IY

where

F(r, , z; r, , z ) I if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z9z ) I if r : r, cos( 9 ) : 1, z " z I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 )

(4.128)

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CHAPTER 4 Permanent Magnet Applications

Calculations: We demonstrate Eqs. (4.123) and (4.126) with some sample calculations. We use the following parameters: M : 4.3 ; 10 A/m Q R : 2.0 cm R : 4.0 cm h : 4.0 cm

(4.129)

where h is the height of the magnet (z : 92.0 cm, z : 2.0 cm). First, B and P B are computed at the points z : 0, 5, 10, . . . , 80 mm, for r : 0.25R , 0.5R , X and 0.75R . The computed data are compared with corresponding FEA data in Figs. 4.28 and 4.29. Next, B is computed along the axis of the magnet where X the radial component is zero (Fig. 4.30). Finally, B and B are computed along P X the line deﬁned by z : 40 mm at the points r : 0, 5, 10, . . . , 80 mm. These data are shown in Fig. 4.31. )

FIGURE 4.28

The component B vs z (; : FEA). P

253

4.2 MAGNET STRUCTURES

FIGURE 4.29

The component B vs z (; : FEA). X

EXAMPLE 4.2.8 Derive an expression for the B-ﬁeld outside an axially polarized multipole cylindrical disk magnet (Fig. 4.32) [5]. Assume that the magnet has a second quadrant demagnetization curve B : (H < M z ). (4.130) Q The < term takes into account the alternating polarity of adjacent poles. SOLUTION 4.2.8 We use the current model of Section 3.3. It follows from Eqs. (3.95) and (4.130) that the volume current density is zero, J : ; M : 0. K Therefore, Eq. (3.93) reduces to A(x) : 4

j (x) K ds.

x 9 x

(4.131)

(4.132)

1 The B-ﬁeld is computed from Eq. (4.132) using B : ; A.

(4.133)

254

FIGURE 4.30

CHAPTER 4 Permanent Magnet Applications

The component B vs z along the axis of the cylinder (; : FEA). X

Instead of determining B for the entire magnet, we ﬁrst evaluate Eq. (4.132) for a single sector shown in Fig. 4.33, determine B for the sector, and then obtain the total B-ﬁeld as a superposition of the contributions from all the sectors. Equivalent current density: As a ﬁrst step, we determine the surface current density j for the various surfaces of the sector shown in Fig. 4.33. The K magnetization for this sector is M : M z . Q

(4.134)

From Eqs. (3.95) and (4.134) it follows that j : 0 on the top and bottom of K the sector because the magnetization and surface normals are either parallel or antiparallel on these surfaces. There are four remaining surfaces to consider:

r : R inner surface : (1) (2) Q Q z (1) z z (2), Q Q

(4.135)

255

4.2 MAGNET STRUCTURES

FIGURE 4.31

The components B and B vs r (; : FEA). P X

r : R outer surface : (1) (2) Q Q z (1) z z (2), Q Q R r R left side : : (1) Q z (1) z z (2), Q Q

FIGURE 4.32

Axially polarized multipole disk magnet.

(4.136)

(4.137)

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CHAPTER 4 Permanent Magnet Applications

FIGURE 4.33

Sector of an axially polarized disk magnet.

and

R r R right side : : (2) Q z (1) z z (2). Q Q The unit normals for these surfaces are

n :

9r

(inner surface)

r

(outer surface)

9

(left side)

(4.138)

(4.139)

(right side).

The corresponding surface current densities are

9M Q M Q j : K M r Q 9M r Q

(inner surface) (outer surface) (left side) (right side).

(4.140)

257

4.2 MAGNET STRUCTURES

Vector potential: We use Eqs. (4.135)—(4.140) and rewrite Eq. (4.132), M A (x) : Q (91)H Q 4 H

XQ

Q

x 9 x

r (j) d dz Q

PYPQH Q X PQ r 9 Q dr dz .

x 9 x Y QH XQ PQ

XQ

(4.141)

In this expression, the subscript s denotes the contribution due to a single sector. The unit vectors r and in Eq. (4.141) are themselves functions of , and therefore need to be considered when performing the integration (Section A.2). From Table A.2 we have r : cos()x ; sin()y

(4.142)

: 9sin()x ; cos()y .

(4.143)

and

Therefore, Eq. (4.141) can be rewritten in terms of Cartesian components, M A (x) : Q (91)H Q 4 H

; 9

XQ

XQ

;

XQ

XQ

9

XQ

XQ

9

XQ

XQ

Q sin()x r (j) d dz Q

x 9 x PYP H Q Q Q cos()y

x 9 x r (j) d dz PYPQH Q Q PQ cos()x dr dz

x 9 x PQ Y QH PQ sin()y dr dz .

x 9 x PQ Y QH

Notice that A (x) has no z-component. It can be decomposed into its cylindrical Q components by computing the projections, A (x) : A (x) · r PQ Q

(4.144)

A (x) : A (x) · . Q Q

(4.145)

and

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CHAPTER 4 Permanent Magnet Applications

The cylindrical components are M A (x) : Q (91)H PQ 4 H XQ Q sin( 9 ) r (j) d dz ; Q

x 9 x XQ Q PYPQH X PQ cos( 9 ) 9 Q dr dz ,

x 9 x XQ PQ Y QH

and

M A (x) : Q (91)H Q 4 H

XQ

(4.146)

Q cos( 9 ) r (j) d dz Q

x 9 x PYPQH XQ Q X PQ sin( 9 ) ; Q dr dz , (4.147)

x 9 x XQ PQ Y QH Equations (4.146) and (4.147) specify the components of the potential for a single sector. The ﬁeld B (x) due to the sector can be obtained using Q B : ; A . Since the total ﬁeld is a superposition of the ﬁelds due to all the Q Q sectors, it can be written as

, B(x) : (91)Q>B (x), (4.148) Q Q where (91)Q> takes into account the alternating polarity of adjacent poles. Radial component: The radial ﬁeld component due to a single sector follows from Eq. (4.133), B (x) : 9 A (x). PQ z Q

(4.149)

Substituting Eq. (4.147) into Eq. (4.149) and taking into account both the differentiation in z and the integration in z yields M B (x) : Q (91)H>I PQ 4 H I Q cos( 9 )g(r, , z; r (j), , z (k))r (j) d ; Q Q Q Q P ; Q sin( 9 )g(r, , z; r, (j), z (k)) dr , Q Q PQ

(4.150)

259

4.2 MAGNET STRUCTURES

where g(r, , z; r, , z) :

1 . [r ; r 9 2rr cos( 9 ) ; (z 9 z)]

The remaining integrals in r and can be written in terms of discrete sums using Simpson’s method. Let N and N denote the (even) number of mesh P points in the r and variables, respectively. Let r(n) and (m) denote the Q values at which the integrands are evaluated. As r (1) r r (2) and Q Q (1) (2), we have Q Q r (2) 9 r (1) Q (4.151) r : Q Q N P and (2) 9 (1) Q . : Q Q N

(4.152)

The integration points are n r (n) : r (1) ; (r (2) 9 r (1)) Q Q Q N Q P

(n : 0, 1, 2, . . . , N ) P

(4.153)

and m ( (2) 9 (1)) (m) : (1) ; Q Q Q N Q The coefﬁcients S (n) and S (m) are P S (n) : P and

(m : 0, 1, 2, . . . , N). (4.154)

(n : 0) (n : 1, 3, 5, . . .) (n : 2, 4, 6, . . .)

(4.155)

(n : N ), P

(m : 0) (m : 1, 3, 5, . . .) S (m) : (4.156) (m : 2, 4, 6, . . .) (m : N). Applying this scheme to Eq. (4.150) and summing the contribution from all the sectors in accordance with Eq. (4.148) gives an equation for the total radial ﬁeld

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CHAPTER 4 Permanent Magnet Applications

component, M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I

, ; S (m) cos(9 (m))g(r, , z; r (j), (m), z (k))r (j) Q Q Q Q Q Q K

, ;r P S (n) sin( 9 (j))g(r, , z; r (n), (j), z (k)) , Q Q Q P Q Q L (4.157) where N : number of poles (1) : (2s 9 3) Q N (2) : (2s 9 1) Q N s : 1, 2, . . . , N r (1) : R (inner radius) Q r (2) : R (outer radius) Q z (2) : top of sector Q z (1) : bottom of sector Q N : angular mesh number (even) N : radial mesh number (even). P

(4.158)

Azimuthal component: The azimuthal component also follows from Eq. (4.133), B (x) : A (x). Q z PQ

(4.159)

Applying Eq. (4.159) to Eq. (4.146) and taking into account the differentiation

261

4.2 MAGNET STRUCTURES

in z and the integration in z yields M B (r, , z) : Q (91)H>I> Q 4 H I

Q sin( 9 )g(r, , z; r (j), , z (k))r (j) d Q Q Q Q P 9 Q cos( 9 (j))g(r, , z; r, (j), z (k)) dr . Q Q Q PQ Again, apply Simpson’s method and sum the contributions from all the sectors. This gives ;

M , B (r, , z) : Q (91)Q>H>I 4 Q H I

, ; S (m) sin(9 (m))g(r, , z; r (j), (m), z (k))r (j) Q Q Q Q Q Q K

, 9r P S (n) cos( 9 (j))g(r, , z; r (n), (j), z (k)) , Q Q Q P Q Q L (4.160) where the variables are deﬁned in Eq. (4.158). Axial components: The axial component of a sector is given by

1 B (x) : (rA (x)) 9 A (x) . XQ Q r r PQ

(4.161)

Substituting Eqs. (4.146) and (4.147) into Eq. (4.161) gives M B (x) : Q (91)H> XQ 4 H

;

XQ

Q r cos( 9 ) 9 rQ (j) r (j) d dz Q

x 9 x XQ Q PYPQH X PQ r sin( 9 ) ; Q dr dz , (4.162)

x 9 x XQ PQ Y QH

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CHAPTER 4 Permanent Magnet Applications

Equation (4.162) can be simpliﬁed using Simpson’s method as above, M B (r, , z) : Q (91)H> XQ 4 H

, ; S (m)[r cos( 9 (m)) 9 r (j)]r (j) Q Q Q Q K ;

XQ g(r, , z; r (j), (m), z) dz Q Q XQ

, ;r P S (n)r sin( 9 (j))) Q P Q L ;

XQ

XQ

g(r, , z; r (n), (j), z) dz Q Q

.

The remaining integration in z can be evaluated analytically. We sum the contributions from all the sectors and obtain B for the entire magnet, X M , B (r, , z) : Q (91)Q>>H>I X 4 Q H I

, ; S (m)[r cos( 9 (m)) 9 r (j)] Q Q Q K ;r (j)I(r, , z; r (j), (m), z (k)) Q Q Q Q

, ;r P S (n)r sin(9 (j)) I(r, , z; r (n), (j), z (k)) . Q Q P Q Q Q L (4.163) In Eq. (4.163), the function I is deﬁned by

XQ g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z (k)). (4.164) Q XQ I

263

4.2 MAGNET STRUCTURES

Evaluation of Eq. (4.164) gives

IY

where

F(r, , z; r, , z (k)) Q if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z9z (k)) Q if

(4.165)

r : r, cos( 9 ) : 1, z " z (k) Q

(z 9 z (k))g(r, , z; r, , z (k)) Q Q F(r, , z; r, , z (k)) : . Q r ; r 9 2rr cos( 9 ) The functional form of I depends on the relationship between the coordinates of the ﬁeld point (unprimed) and the source point (primed). In particular, I is evaluated in terms of the function F as long as r ; r 9 2rr cos( 9 ) " 0. If this condition is violated, then the lower ratio is used. Calculations: We demonstrate the ﬁeld solution via the analysis of a four-pole disk magnet with the following parameters: M : 4.3 ; 10 A/m R : 2.0 cm R : 4.0 cm h : 1.0 cm

(inner radius) (outer radius) (height).

(4.166)

The magnet is oriented with its upper surface in the x-y plane with one of its sectors symmetrically positioned in an angular sense with respect to the x-axis. The upper surface of this sector is taken to be a north pole. The ﬁeld components B and B are evaluated along a radial line that is 1 cm above the surface of the P X magnet and centered with respect to this sector ( : 0). Field values are computed at r : 0, 5, 10, . . . , 45 mm. These data are compared with corresponding data from a 3D FEA as shown in Fig. 4.34. Similarly, the ﬁeld components B and B are evaluated along an arc at the mean radius of the magnet, 1 cm X above its surface. The arc extends from the center of one pole to the center of the neighboring pole. Field values are computed at : 0, 5, 10, . . . , 90°. These data are shown in Fig. 4.35. )

264

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.34

The components B and B vs r (; : FEA). P X

FIGURE 4.35

The components B and B vs (; : FEA). X

4.3 HIGH FIELD STRUCTURES

265

4.3 HIGH FIELD STRUCTURES Conventional permanent magnet structures such as those studied in the preceding typically produce ﬁeld strengths of 1.0 T in proximity to their surface. However, special structures can be fabricated that provide uniform multitesla ﬁelds throughout their interior. For example, ﬁelds of up to 4.0 T have been achieved in disk-shaped cavities 2.5 cm in diameter and 0.5 cm high with structures 15 cm in diameter [6—8]. Such structures can be used for numerous applications such as Hall effect studies. They can also be used to focus charged particle beams if through-ports are provided. The most popular high ﬁeld structures are the ‘‘magic’’ cylinder and sphere [6—9]. These are made from high-strength rare-earth materials such as NdFeB. They are fabricated piecewise by cutting uniformly ` a magnetized block, and then orienting polarized slices of material from and assembling the constituent pieces so as to provide a predeﬁned magnetization pattern throughout the assembled structure. The magnetization pattern is chosen to provide a desired ﬁeld strength and uniformity in the interior of the structure. The ‘‘magic’’ cylinder is shown in Fig. 4.36. It has the form of a cylindrical shell with a magnetization that varies with the polar angle as shown. This structure can be built from a stack of cylindrical sections as shown in Fig. 4.37. These sections can be constructed as follows. First, cut a circular ring from a uniformly magnetized block of material. Next, cut the ring into the desired number of pie-shaped elements. Finally, reorder the pie-shaped elements around the circumference to obtain the desired angular magnetization distribution. The fabrication sequence is illustrated in Fig. 4.37. The ﬂux density in the interior of a magic cylinder is given by B : B ln(r /r ), (4.167) P M G where B is the remanence of the material and r and r are the inner and P G M outer radii of the cylinder, respectively [6]. Notice that Eq. (4.167) implies that there is no theoretical upper limit to the ﬁeld that can be obtained in this structure. However, the slow logarithmic dependency on r places M a practical upper limit of 2.5 T for cavities of more than a few centimeters in diameter. The mass of material required to obtain higher ﬁelds would be prohibitively large. The ‘‘magic’’ sphere is shown in Fig. 4.38. It has the form of a spherical shell in which the direction of magnetization varies with the polar angle as :2. This magnetization pattern is obtained by assembling a collection of melonlike segments. These segments can be constructed by

266

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.36

Magic cylinder: (a) idealized structure; and (b) physical structure.

beveling the ring structures used in the construction of the magic cylinder (Fig. 4.39). The ﬂux density in the cavity of a magic sphere is given by B : B ln(r /r ), P M G

FIGURE 4.37

Fabrication sequence for a magic cylinder [9].

4.3 HIGH FIELD STRUCTURES

FIGURE 4.38

267

Magic sphere: (a) idealized structure; and (b) physical structure.

FIGURE 4.39

Fabrication sequence for a magic sphere [9].

268

CHAPTER 4 Permanent Magnet Applications

where r and r are the inner and outer radii of the shell, respectively [6]. G M Such structures provide uniform ﬁelds of 2.0 to 3.0 T in their interior when the outer radius of the shell is only three to four times the radius of the cavity. For example, a magic sphere with B : 1.2 T and an outer P radius of 3.0 cm produces a uniform ﬁeld of 2.5 T in a cavity with a 1.0 cm radius [7].

4.4 MAGNETIC LATCHING Magnetic latching is one of the oldest and most common applications of permanent magnets. In latching applications, magnets are used to provide a force of attraction between two members. Latching magnets can be found in a variety of products including household appliances, tool equipment, electrical devices, and toys. Such magnets are usually made from ferrite materials because of their low cost. Latching applications fall into one of two categories: contact and noncontact. The distinction between these two is important because the design approach is different for each. In contact latching the magnet is attracted to and in contact with another member (either another magnet or a soft magnetic material such as iron). The latching magnet is often designed to have numerous closely spaced poles to enhance the latching force. In noncontact latching the magnet exerts a force of attraction over a distance. Here, the poles of the latching magnet need to be separated so that the ﬁeld will span the gap between the magnet and the adhering surface in order to provide the latching force. Therefore, the magnetic structures for contact and noncontact latching are quite different. We consider contact latching ﬁrst. Consider a magnet with a remanence B that is in contact with a plate of high-permeability material P such as iron that is unsaturated (Fig. 4.40). The force of attraction between the magnet and the plate can be estimated using the Maxwell stress tensor expression (3.59), which gives BA (4.168) F: P , 2 where A is the area of contact between the magnet and the plate. The force can be rewritten in terms of the ﬂux : B A, P F: . (4.169) 2 A This expression implies that the latching force can be increased by

4.4 MAGNETIC LATCHING

FIGURE 4.40

269

Contact latching: (a) single-pole contact; and (b) multipole contact.

reducing the contact area A and/or increasing the ﬂux . This can be accomplished by tapering the poles of the magnet, or by adding tapered iron pole pieces to the magnet. However, there is a practical limit to this effect because Eq. (4.169) only applies when the material being held is well below saturation. As A decreases the material saturates and the force no longer increases. This is especially true in applications involving thin plates that have a relatively low saturation threshold due to their limited cross-sectional area. One way to achieve a higher latching force in such applications is to use a magnet with numerous closely spaced poles as shown in Fig. 4.40b. If there are N such poles, and each exerts a force F, then the total force is F : NF as long as the material being held is below saturation. We now consider noncontact latching. In noncontact latching the two attracted members are separated by a gap, and the magnet must project its ﬁeld across the gap to exert the force of attraction. Thus, the poles of the magnet must be effectively separated so that the ﬁeld will span the gap. Noncontact latching is often implemented using a magnetic circuit as opposed to a standalone magnet. There are numerous latching circuits, and most of these can be designed and optimized using magnetic circuit theory (Section 3.5). We demonstrate the design method via the analysis of two practical latching circuits.

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CHAPTER 4 Permanent Magnet Applications

Magnetic latching device: (a) magnet with two attached ﬂux plates; and (b) latching circuit.

FIGURE 4.41

EXAMPLE 4.4.1 Consider the magnetic circuit of Fig. 4.41b. A magnet with attached ﬂux plates is attracted to a wall made of soft magnetic material. Determine the force of attraction when a gap exists between the magnetic structure and the wall g 0. Assume that the ﬂux plates and the wall have inﬁnite permeability, and that the magnet has a linear second quadrant demagnetization curve of the form B :B ; H , K P K K

(4.170)

where : B /H . K P A SOLUTION 4.4.1 We use magnetic circuit theory (Section 3.5). As there are no currents we have w H · dl : 0, which gives H l ; 2H g : 0, (4.171) KK E where H is the ﬁeld in the gap. The factor of 2 in Eq. (4.171) is due to the fact E that the path of` integration passes through the gap region twice as shown in Fig. 4.41b. In addition, notice that Eq. (4.171) ignores the path through the ﬂux plates and the wall because H is negligible in these elements by assumption. As there is no fringing, w B · dS : 0, which implies that : . Thus, K E B A :B A (no fringing at gap), (4.172) K K E E

271

4.4 MAGNETIC LATCHING

where A : Lw is the area of the magnet, and A : Lw is the area of gap. K K E E We combine Eqs. (4.170), (4.171), and (4.172) and obtain A B P B (g) : K . E A (1 ; 2( / )(A /A )(g/l )) E K K E K Next, we determine the force using the Maxwell stress tensor (Section 3.2.3). We compute the force on the wall by applying Eq. (3.59) over an area bounding the wall (similar to the area indicated by the dotted line in Fig. 3.9 of Example 3.2.8). The only contributions to Eq. (3.59) come from the area in the gap regions. In these regions, B : B , and therefore Eq. (3.59) reduces to L E 1 F(g) : B(g)n ds L 2 1 1 (2B(g)A )n : E E 2 BA P K : n . (4.173) A (1 ; 2( / )(A /A )(g/l )) E K K E K The outward normal n of the wall points towards the magnet, and, therefore, the force is attractive. )

EXAMPLE 4.4.2 Determine the force of attraction for the latching circuit of Fig. 4.42b. Assume that the ﬂux plate and the wall have inﬁnite permeability and that the magnet has a linear second quadrant demagnetization curve of the form B :B ; H , K P K K

(4.174)

where : B /H . K P A SOLUTION 4.4.2 We use magnetic circuit theory (Section 3.5). First, apply w H · dl : 0 to the closed dotted line path on the right-hand side of the circuit. This gives H l ; 2H g : 0, (4.175) KK E where H is the ﬁeld in the gap. The factor of 2 in Eq. (4.175) is due to the fact E that the path of integration passes through the gap region twice as shown. Notice that we have ignored contributions from the path through the ﬂux plate and the wall as H is negligible in these elements. Next, because there is no fringing, : 2 and we have K E B A : 2B A (no fringing at gap), (4.176) K K E E

272

CHAPTER 4 Permanent Magnet Applications

Magnetic latching circuit: (a) magnet with attached ﬂux plate; (b) latching circuit.

FIGURE 4.42

where A : Lw is the area of the magnet, and A : Lw is the area of the gap. K K E E The factor of 2 in Eq. (4.176) accounts for the fact that the ﬂux through the magnet splits evenly between the two return paths as indicated by the dotted lines in Fig. 4.42b. We combine Eqs. (4.174), (4.175), and (4.176) and ﬁnd that A B P B (g) : K . E A 2(1 ; ( / )(A /A )(g/l )) E K K E K

273

4.5 MAGNETIC SUSPENSION

We use the Maxwell stress tensor to obtain the force (Section 3.2.3). Speciﬁcally, we apply Eq. (3.59) over an area bounding the wall (similar to the area indicated by the dotted line in Fig. 3.9 of Example 3.2.8). The only contributions to Eq. (3.59) come from the area in the gap regions. There are three such regions — a central region immediately beneath the magnet and two end regions. To apply Eq. (3.59) we need to know B (the ﬁeld component normal to the wall). In the L central gap region B : B : 2(A /A )B , whereas in the end regions B : B . L K E K E L E Substituting these into Eq. (3.59) gives

1 B(g)n ds, L 2 1 1 : [2B(g)A ; B A ]n E K K E 2 BA A ; 2A P K K E n . : (4.177) 2 (1 ; ( / )(A /A )(g/l )) 2A K K E K E Because the outward normal at the wall n points towards the magnetic circuit, the force is attractive. ) F(g) :

4.5 MAGNETIC SUSPENSION In magnetic suspension, magnets are used to provide passive levitation and/or positioning of a structure. The magnets can be arranged in myriad conﬁgurations and substantial levitation force can be realized when modern high-strength materials such as NdFeB are used. The two most common suspension conﬁgurations are the radial- and axial bearing conﬁgurations shown in Fig. 4.43 [10]. Magnetic bearings are often used to provide low friction support for rotating mechanical structures as illustrated in Fig. 4.44. They enable substantially higher operating speeds relative to conventional mechanical bearings. It is well known that a passive magnetic ﬁeld alone cannot provide stable support of a structure. Such stability can only be achieved with some form of active control [11, 12]. This is a consequence of Earnshaw’s theorem, which states that ‘‘A body with steady charges, magnetizations, or currents placed in a steady electric or magnetic ﬁeld cannot rest in stable equilibrium under the action of electric and magnetic ﬁelds alone’’ [13]. Earnshaw’s theorem can be understood from the following heuristic argument [14]. Let Q denote the magnetic scalar potential due to a K permanent magnet support structure. Assume that the structure is

274

CHAPTER 4 Permanent Magnet Applications

Cross-sectional view of axisymmetric magnetic bearings: (a) radial bearing conﬁgurations; and (b) axial bearing conﬁgurations.

FIGURE 4.43

designed to suspend a given magnet. Because the suspension point is outside the support structure, we have Q : 0. (4.178) K Now, the suspension force FQ is proportional to the ﬁeld strength HQ : 9Q, K FQ . Q. K If the suspended magnet is to be in stable equilibrium, the net force must be zero, which implies that Q : 0. K

(4.179)

4.5 MAGNETIC SUSPENSION

FIGURE 4.44

275

Magnetic suspension with lateral mechanical constraint.

However, Eq. (4.179) indicates a potential maximum or minimum of Q. K For stable equilibrium, the potential must be a minimum that occurs when Q 0. (4.180) K However, Eq. (4.180) contradicts Eq. (4.178), which implies that it is impossible to provide stable suspension with a passive magnetic ﬁeld. Therefore, to implement magnetic suspension, magnets are usually used along with some form of mechanical structure that constrains the motion of the supported member. An example of this is shown in Fig. 4.44. Here, opposing disk magnets levitate a rotating member, while its lateral motion is constrained by a ﬁxed support with a conventional mechanical bearing that enables rotation. The force between the disk magnets is derived in the following example. EXAMPLE 4.5.1 Derive an expression for the levitation force between two identical, axially polarized cylindrical disks as shown in Fig. 4.45 [15]. The disks have inner and outer radii R and R , thickness t , and are separated by K a distance h (Fig. 4.46). Assume that both magnets have a linear second quadrant demagnetization curve of the form B : (H < M z ). (4.181) Q where the < takes into account the opposite polarizations of the opposing magnets.

276

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.45

FIGURE 4.46

Levitated magnetic disk.

Cross section of disk geometry with reference frame.

277

4.5 MAGNETIC SUSPENSION

SOLUTION 4.5.1 We use the charge model (Section 3.4). Speciﬁcally, we ﬁrst reduce the magnets to equivalent charge distributions and then determine the force using the force formulas (3.120) to (3.123). The volume and surface charge densities and are given by Eq. (3.105). From Eq. (4.181) we have K K : 9 · M K : 9 · M z Q : 0, (4.182) and : M · n K : <M , (4.183) Q where the < sign takes into account the opposite polarity of the charge on opposing pole faces. Now, consider the force expressions (3.120) to (3.123) for a discretized magnet. Because : 0 in both magnets, the volume terms vanish K and the force between the magnets reduces to

(x ) (x )(x 9 x ) I A A . F: K N K I N (4.184) I N 4

x 9 x N I N I Instead of evaluating Eq. (4.184) directly, it is easier to derive the force in the following steps: First, obtain the force between two point sources; next, derive the force between opposing pole faces; and ﬁnally, obtain the total force by summing the contributions from all the surfaces. Point sources: We use cylindrical coordinates. Without loss of generality, one of the charges Q (x ) can be taken to be in the x-y plane at (r , , 0) while the K other Q (x ) is taken to be in a plane deﬁned by z : h at (r , , h). The axial K force between these two point sources follows from Eq. (4.184): Q (r , , 0)Q (r , , h)h K K F : . (4.185) X 4 [r ; r 9 2r r cos( 9 ) ; h] Surfaces: The next step is to evaluate the axial force exerted by one pole face (surface) on an opposing pole face. Again, without loss of generality, one surface can be taken to lie on the x-y plane while the other can be taken to lie on the plane deﬁned by z : h. The idea is to discretize each surface into a mesh of subsections (elements), deﬁne a point charge at the midpoint of each element in accordance with Eq. (3.110), and then compute the force exerted by one surface on the other by summing the contributions from the individual point charges. The area of each surface is (R 9 R). The ﬁrst step is to divide this area into

278

CHAPTER 4 Permanent Magnet Applications

a set of indexed elements. These elements are taken to have equal areas so that they contain an equal amount of surface charge. Because the area of the surface is spanned by the two coordinate variables r and , two indices (i, j) are required. Let N and N be the number of mesh points in these two variables, P respectively. Each element will have the same area A, (R 9 R) . N N P Choose each element so that it spans an angular measure A :

:

2 . N

(4.186)

(4.187)

Let (r , ) denote the midpoint of the element A , and let R(i) and R(i ; 1) G H GH denote its radial boundaries. Then r : R(i ; 1) 9 R(i), G R(i ; 1) ; R(i) r: (i : 1, 2, . . . , N ), G P 2

(4.188) (4.189)

and , : (j 9 1) ; H 2

(4.190)

for j : 1, 2, . . . , N. The end points are computed using the following equation:

(R 9 R) (i : 1, 2, . . . , N ), (4.191) P N P where R(1) : R (inner radius). Equation (4.191) is used to generate all the end point values from which all the midpoint values are obtained. Therefore, Eqs. (4.189) to (4.191) deﬁne a scheme for discretizing the surface into elements and assigning values to the midpoints of each element. Next, consider the magnetic interaction between an element of one surface and an element of an opposing surface. Let the coordinates of the midpoints of the two elements be (r , , 0) G H and (r , , h), respectively. The magnetic charge associated with each element GY HY is R(i ; 1) :

R(i) ;

Q (r , , 0) : (r , , 0)A K G H G H

(4.192)

Q (r , , h) : (r , , h)A, K GY HY GY HY

(4.193)

and

279

4.5 MAGNETIC SUSPENSION

respectively. Substitute Eqs. (4.183) and (4.186) into these expressions and obtain Q : M K Q

(R 9 R) . N N P

(4.194)

The force is obtained by summing the contributions of all the elements in accordance with Eq. (4.185). By exploiting the symmetry of the problem the force reduces to

(R 9 R) F (h) : M X 4 Q N N P , , , h . (4.195) ;N P P [r ; r 9 2r r cos( ) ; h] GY G GY HY G GY HY G Force: There are two magnets to consider, which are labeled upper or lower in accordance with their respective positions. The upper magnet has its bottom surface in the plane deﬁned by z : h and its top surface lies in the plane deﬁned by z : h ; t (Fig. 4.46). The lower magnet has its top surface in the x-y plane K and its bottom surface in the plane deﬁned by z : 9t . The axial force on the K top surface due to both the top and bottom surfaces of the upper magnet is F (h) : F (h) 9 F (h ; t ). X X X K Continuing in this fashion and taking into account the contributions from all the surfaces, the following formula is obtained for the total force between the magnets:

MN (R 9 R) F : Q X 4 N N P , , , (91)K h K K ; P P . [r ; r 9 2r r cos( ) ; h ] GY K K G GY HY G G GY HY where : (j 9 1) ; HY 2 :

2 N

(j : 1, 2, . . . , N)

(4.196)

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CHAPTER 4 Permanent Magnet Applications

and r: G

R(i ; 1) ; R(i) 2

r : GY

R(i ; 1) ; R(i) 2

(i : 1, 2, . . . , N ) P (i : 1, 2, . . . , N ) P

(R 9 R) (i : 1, 2, . . . , N ) P N P (R 9 R) R(i ; 1) : R(i) ; (i : 1, 2, . . . , N ) P N P R(1) : R (inner radius). The terms h and are given by K K h : h ; mt (m : 0, 1, 2) K K h : separation of magnets R(i ; 1) :

R(i) ;

t : thickness of magnets K and

1

(m : 0)

: 2 K 1

(m : 1) (m : 2).

Notice that Eq. (4.196) contains nested summations of elementary functions that are easy to program. However, it is important to note that when programming nested summations, a higher degree of accuracy can be achieved with fewer mesh points (lower values N and N) if the total of a single P summation over one index is evaluated and saved separately (as an intermediate step), and then the totals of the successive summations are added together to obtain the ﬁnal sum. For example, Eq. (4.196) could be evaluated as follows:

MN (R 9 R) F : Q S (m), X 4 N N P K where , S (m) : P S (i, m), G

4.5 MAGNETIC SUSPENSION

FIGURE 4.47

with

and

281

Levitation force vs separation distance (; : FEA).

, S (i, m) : P S (i, i, m), GY

, (91)K h K K S (i, i, m) : . [r ; r 9 2r r cos( ) ; h ] G GY K HY G GY HY Calculations: We demonstrate Eq. (4.196) with some sample calculations. We use the following parameters: M : 4.3 ; 10 A/m Q R : 2.0 cm R : 4.0 cm t : 1.0 cm. K Force values are computed for a series of separation distances with the mesh parameters set to N : 20 and N : 60. The computed data are compared with P corresponding FEA data in Fig. 4.47. )

282

CHAPTER 4 Permanent Magnet Applications

4.6 MAGNETIC GEARS Magnetic gears can be used in place of mechanical gears to reduce undesired vibrations and for applications that require torque coupling between separated members. There are many gear conﬁgurations. One such conﬁguration consists of two separated, radially polarized cylindrical magnets that are constrained to rotate about their respective axes (Fig. 4.48). The magnets are magnetically coupled to one another and when one magnet rotates it imparts a torque to the other, causing it to rotate [16]. The coupling between the magnets is a function of several variables including the number of poles, material properties, dimensions, and separation. Furthermore, substantial torque can be realized if modern, rare-earth materials such as NdFeB are used. We derive a formula for the torque in the following example. EXAMPLE 4.6.1 Derive an expression for the torque between two separated radially-polarized multipole cylinders (Fig. 4.48) [16]. Assume that the magnets have identical geometric dimensions, and that their length is much greater than their diameter. Let N denote the number of poles, and let R and R denote the inner and outer radii of each magnet. Assume that the magnets have different second quadrant demagnetization relations. The demagnetization curve for the magnets in the primed and unprimed coordinate systems are B : H ; M,

FIGURE 4.48

(4.197)

Magnetically coupled cylinders with reference frames.

283

4.6 MAGNETIC GEARS

and B : (H ; M),

(4.198)

respectively, where M : <M r . (4.199) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.6.1 The gear consists of two magnets. We refer to the magnet centered with respect to the primed coordinate system (x, y) as the source magnet. We refer to the magnet centered with respect to the unprimed coordinate system (x, y) as the drive magnet. The analysis proceeds in two steps. First, we use boundary-value theory to obtain a free-space ﬁeld solution for the source magnet. Second, the drive magnet is reduced to an equivalent current distribution and the torque is computed by considering this current to be in the external ﬁeld of the source magnet. Source magnet: As the length of the source magnet is much greater than its diameter, we can determine its ﬁeld using a 2D analysis. The ﬁeld due to the source magnet was derived in Example 4.2.5. Speciﬁcally, the ﬁeld components in the primed coordinates are given by B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.200)

and (4.201) B (r, ) : ir\G>U (M , R , R , ) sin(i). G G G The coefﬁcients U(M , R , R , ) are deﬁned by Eq. (4.53), with M given by G G G Eq. (4.64). The ﬁeld components (4.200) and (4.201) constitute the external ﬁeld in the following analysis. Therefore, in what follows we adopt the notation (r, ). B(r, ) : B(r, ) and B (r, ) : B PY P Y Drive magnet: To determine the torque on the drive magnet we ﬁrst reduce it to a distribution of equivalent volume and surface current densities J and j , K K respectively (Section 3.3). Once this is done, we compute the torque using Eq. (3.100), T:

4

;

r(r, ) ; (J (r, ) ; B (r, ))r dr d dz K

1

r(r, ) ; (j (r, ) ; B (r, )) da, K

(4.202)

284

CHAPTER 4 Permanent Magnet Applications

where V and S denote the volume and surface of the magnet. Notice that B (r, ) is the ﬁeld due to the source magnet represented in the unprimed coordinates. From Eq. (4.199) we have J : ; M : 0 and, therefore, the ﬁrst K term in Eq. (4.202) is zero. As for the second term, there are N sectors, and each sector has two surfaces with current densities that contribute to the torque. These surfaces constitute the radial sides of the sector at angular positions and , respectively. If the magnet is rotated by an angle , then the surface current densities for the pth sector are R rR M z (p) : ; (1 ; 2p) Q N j (p, r, ) : (4.203) K R rR 9M z (p) : 9 (1 ; 2p), Q N where p : 0, 1, 2, . . . , N 9 1. Notice that : 0 occurs when the middle of the zeroth sector coincides with the x axis and its polarization is radially outward as shown in Fig. 4.48. It is assumed that the source magnet is held stationary with one of its sectors in a similar orientation with respect to the primed coordinate system and, therefore, is the relative angular offset of the two magnets. We substitute Eq. (4.203) into Eq. (4.202) and obtain

T() : 2M L Q

, \ 0 (91)N r[cos( (, p))B(r, (, p)) V 0 N ; sin( (, p))B(r, (, p))] dr, W (4.204)

where L is the length of the gear, and (, p) : ; (1 ; 2p). (4.205) N The coefﬁcient 2M takes into account the fact that there are two surfaces at the Q interface between neighboring sectors (one for each sector). The integral in Eq. (4.204) can be evaluated numerically using Simpson’s method. The resulting torque formula is T() :

2M LR* , \ ,P Q (91)NS (q)r(q) P N P N O ;[cos( (, p))B(r(q), (, p)) V ;sin( (, p))B(r(q), (, p))], W

(4.206)

285

4.6 MAGNETIC GEARS

where N is the radial mesh coefﬁcient (even), R* : R 9 R , and S (q) are the P P Simpson coefﬁcient terms

S (q) : P

(q : 0) (q : 1, 3, 5, . . .) (q : 2, 4, 6, . . .)

(4.207)

(q : N ). P

The integration points are as follows: q r(q) : R ; (R 9 R ) N P

(q : 0, 1, 2, . . . , N ). P

(4.208)

To compute the torque we express the external ﬁeld components B(r, ) PY and B(r, ) (Eqs. (4.200) and 4.201)) in terms of the unprimed coordinates, Y B(r, ) Y B(r(r, ), (r, )) cos((r, )) V PY 9 B (r(r, ), (r, )) sin((r, )) Y

(4.209)

and B(r, ) Y B(r(r, ), (r, )) sin((r, )) W PY ; B (r(r, ), (r, )) cos((r, )) Y

(4.210)

In Eqs. (4.209) and (4.210) we use the coordinate transformation relations r(r, ) : (r ; 2rd cos() ; d

(4.211)

and (r, ) : arctan

r sin() . r cos() ; d

(4.212)

The quantities B(r, (, p)) and B(r, (, p)) in Eq. (4.206) are V W evaluated using Eqs. (4.209) and (4.210). Calculations: We demonstrate the torque formula (4.206) via a practical example. To clarify the analysis, we distinguish the parameters for the source and drive magnets using superscripts s and d, respectively. For example, RQ and RQ refer to the inner and outer radii of the source magnet, and RB and RB refer to the inner and outer radii of the drive magnet. We use the following

286

CHAPTER 4 Permanent Magnet Applications

parameters: MQ : 7.2 ; 10 A/m Q MB : 7.2 ; 10 A/m Q d : 80 mm RQ : 10 mm RQ : 20 mm RB : 15 mm RB : 30 mm N : 4 NB : 4. (4.213) The values of MQ and MB are characteristic of sintered NdFeB material. The Q Q torque per unit length is computed with the source magnet ﬁxed and the drive magnet rotated through a series of angular values : 0, 5, 10, . . . , 90°. The analysis is performed with N : 20 and : in Eq. (4.197). The torque data P are compared with corresponding FEA data in Fig. 4.49. Notice that the peak torque occurs at : 45° when the drive magnet is rotated half the angular span of one pole. An analysis is also performed to determine the decrease of peak torque ( : 45°) as a function of separation distance d for a series of values d : 60, 70, . . . , 160 mm. A ﬁrst set of data is computed with : in Eq. (4.197). These data are compared with FEA data in Fig. 4.50. An additional calculation is performed with : 2 . This represents a steeper slope in the demagnetization curve (4.197), which results in a lower external ﬁeld B and a reduced torque as shown. )

4.7 MAGNETIC COUPLINGS Synchronous magnetic couplings are used to transmit torque between two separated members. Couplings are used for various purposes including the activation of devices in sealed containers and the transmission of instrument readings from vacuum or pressure vessels. The two most common coupling conﬁgurations are radial and axial couplings [17]. Radial couplings consist of two coaxial radially polarized cylindrical magnets (Fig. 4.51). Axial couplings consist of two axially polarized multipole disks that face one another (Fig. 4.52). For both conﬁgurations, the transmitted torque is a function of several variables

4.7 MAGNETIC COUPLINGS

FIGURE 4.49

FIGURE 4.50

287

Torque per unit length vs rotation angle (FEA : ;).

Torque per unit length vs separation distance d (FEA : ●).

288

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.51

Synchronous radial coupling.

FIGURE 4.52

Synchronous axial coupling.

289

4.7 MAGNETIC COUPLINGS

including material properties, number of poles, dimensions, separation distance, and the relative angular offset of the magnets [18]. In this section, we derive formulas for the torque of radial and axial couplings. EXAMPLE 4.7.1 Derive an expression for the torque between two coaxial, radially polarized multipole cylinders (Fig. 4.51). Assume that the lengths of the magnets are much greater than their diameters. Further assume that the inner and outer magnets have linear second quadrant constitutive relations of the form B : H ; M,

(4.214)

B : (H ; M),

(4.215)

and

respectively, where M : <M r . (4.216) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.7.1 The coupling consists of an inner and outer magnet. We refer to the inner magnet as the source magnet and the outer magnet as the drive magnet. The analysis proceeds in two steps: First, we use boundary-value theory to obtain a free-space ﬁeld solution for the inner magnet; and second, the outer magnet is reduced to a distribution of equivalent currents, and the torque is determined by considering the equivalent currents to be in the external ﬁeld of the source magnet. Source magnet: Because the length of the source magnet is much greater than its diameter, we can determine its ﬁeld using a 2D analysis. The ﬁeld due to the inner magnet was derived in Example 4.2.5. The ﬁeld components are given by B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.217)

and (4.218) B (r, ) : ir\G>U (M , R , R , ) sin(i). G G G The coefﬁcients U(M , R , R , ) are deﬁned by Eq. (4.53), with M given by G G G Eq. (4.64). In the following analysis the ﬁeld components (4.217) and (4.218) constitute an external ﬁeld. Therefore, we adopt the notation B(r, ) : P B(r, ) and B (r, ) : B (r, ). P Torque: We determine the torque on the outer magnet. Let N , M , and R Q and R denote the number of poles, magnetization, and inner and outer radii of

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CHAPTER 4 Permanent Magnet Applications

this magnet, respectively. As a ﬁrst step, we reduce the magnet to a distribution of equivalent volume and surface current densities J and j , respectively K K (Section 3.3). Once this is done, we compute the torque using Eq. (3.100). Speciﬁcally, T:

r(r, ) ; (J (r, ) ; B(r, ))r dr d dz K

;

4

1

r(r, ) ; (j (r, ) ; B(r, )) da, K

(4.219)

where V and S denote the volume and surface of the magnet. From Eq. (4.216) we have J : ; M : 0. Therefore, the ﬁrst term in Eq. (4.219) is zero. As K for the second term, there are N sectors to consider, and each sector has two surfaces with current densities that contribute to the torque. These surfaces constitute the radial sides of the sector at angular positions and , respectively. If the magnet is rotated by an angle , then the surface current densities for the pth sector are determined using

j (p, r, ) : K

M z Q

9M z Q

R rR (p) : ; (1 ; 2p) N

R rR (p) : 9 (1 ; 2p), N

(4.220)

where p : 0, 1, 2, . . . , N 9 1. Notice that : 0 occurs when the middle of the zeroth sector coincides with the x-axis and its polarization is radially outward as shown in Fig. 4.53. It it is assumed that the source magnet is held stationary with one of its sectors in a similar orientation; therefore, represents the relative angular offset of the two magnets. We substitute Eq. (4.220) into Eq. (4.219) and obtain

, \ 0 (91)N rB(r, (, p)) dr, P N 0 where L is the length of the coupling and T() : 2M L Q

(, p) : ; (1 ; 2p). N

(4.221)

(4.222)

The coefﬁcient 2M takes into account the fact that there are two surfaces at the Q

291

4.7 MAGNETIC COUPLINGS

FIGURE 4.53

Cross section of radial coupling with reference frame.

interface between neighboring sectors (one for each sector). The integral in Eq. (4.221) can be evaluated numerically using Simpson’s method. The resulting torque formula is 2M LR* , \ ,P Q (91)NS (q)r(q)B(r(q), (, p)), (4.223) P P N P N O where N is the radial mesh coefﬁcient (even), R* : R 9 R , and S (q) are P P the Simpson coefﬁcient terms T() :

S (q) : P The integration points are

(q : 0) (q : 1, 3, 5, . . . ) (q : 2, 4, 6, . . . )

(4.224)

(q : N ). P

q r(q) : R ; (R 9 R ) (q : 0, 1, 2, . . . , N ). N P P The ﬁeld component B (r, ) is given by Eq. (4.217). P

(4.225)

Calculations: We demonstrate Eq. (4.223) with some sample calculations. To clarify the analysis, we distinguish the parameters for the inner (source) and outer (drive) magnets using superscripts in and out, respectively. For example, R and R refer to the inner and outer radii of the inner source magnet, and R and R refer to the inner and outer radii of the drive magnet. We consider

292

CHAPTER 4 Permanent Magnet Applications

a coupling with the following parameters: M : 5.0;10 A/m Q M : 5.0;10 A/m Q R : 2.0 cm R : 5.0 cm R : 6.0 cm R : 9.0 cm N : 6.0. (4.226) We compute the torque per unit length for a series of angular values : 0, 5, 10, . . . , 60° (rotation from the center of one pole to the next). The computed data are compared with corresponding FEA data in Fig. 4.54. We also determine the number of poles that render the maximum torque (all other parameters held constant). These data are shown in Fig. 4.55. It is clear that the maximum torque occurs when N : 8. Finally, we compute the peak torque as a function of the aspect ratio L/D of the coupling, where L is the length of the coupling and D : 2R is the diameter of the outer magnet. The same

FIGURE 4.54

Torque per unit length vs rotation angle.

293

4.7 MAGNETIC COUPLINGS

FIGURE 4.55

Torque per unit length vs number of poles.

parameters as in the preceding are used and L is varied from D to 10D. The percent difference between the scaled 2D data and a corresponding 3D analysis is shown in Fig. 4.56. According to this analysis, the scaled 2D value is within 7% of the actual value even when the aspect ratio is 1. ) EXAMPLE 4.7.2 Derive an expression for the torque between two coaxial radially polarized multipole cylinders (Fig. 4.51) [19]. Assume that both magnets have the same number of poles and linear second quadrant constitutive relations of the form B : (H ; M).

(4.227)

where M : <M r . (4.228) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.7.2 The coupling consists of an inner and outer magnet. We refer to the inner magnet as the source magnet and the outer magnet as the drive magnet. The analysis proceeds in two steps: First, we obtain a free-space ﬁeld solution for the inner magnet; second, the outer magnet is reduced to a

294

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.56

Percent error of 2D model vs aspect ratio.

distribution of equivalent currents, and the torque is determined by considering these currents to be in the external ﬁeld of the source magnet. Source Magnet: The ﬁeld for the inner magnet was derived in Example 4.2.6. The only ﬁeld component that contributes to the torque is the radial component, which is determined using M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I , ; r P S (n)r (n) sin( 9 ( j)) Q Q P Q L ;I (r, , z; r (n), ( j), z (k)) Q Q Q , ; S (m)(z 9 z ( j )) cos( 9 (m)) Q Q Q K

;I (r, , z; r (k), (m), z ( j )) . Q Q Q

(4.229)

295

4.7 MAGNETIC COUPLINGS

The various functions and parameters in Eq. (4.229) are deﬁned in Example 4.2.6. In the following analysis this component is an external ﬁeld. Therefore, we adopt the notation B (r, , z) : B (r, , z). P P Torque: We determine the torque on the outer (drive) magnet. Let N , M , Q R , R and z and z denote the number of poles (even), magnetization, inner and outer radii, and end positions of this magnet, respectively. As a ﬁrst step, we reduce the magnet to a distribution of equivalent volume and surface current densities J and j , respectively (Section 3.3). Once this is done, we compute K K the torque using Eq. (3.100), T:

4

r(r, );(J (r, );B(r, ))r dr d dz K

r(r, );(j (r, );B(r, ))da, (4.230) K 1 where V and S denote the volume and surface of the magnet. From Eq. (4.228) we have J : ;M : 0 and, therefore, the ﬁrst term in Eq. (4.230) is zero. As K for the second term, there are N sectors to consider, and each sector has two surfaces with current densities that contribute to the torque. Consider the sector that is centered with respect to the x axis as shown in Fig. 4.53. This sector has surface currents along its radial sides at angular positions : 9/N and : ;/N If the sector is rotated by degrees the contributing surface current densities are given by: ;

R rR M z : ; /N Q z zz j (r, , z) : (4.231) K R rR 9M z : 9 /N Q z zz . It is assumed that the inner magnet is held stationary with one of its sectors centered about the x-axis. Therefore, represents an angular offset between the two magnets We substitute Eq. (4.231) into Eq. (4.230) and solve for the axial component. This gives

X

0

rB r, ; , z dr dz, (4.232) P N X 0 where B (r, , z) is given by Eq. (4.229). The coefﬁcient 2N reﬂects the fact P that each sector contains two surface currents that contribute equally to the T () : 2N M Q X

296

CHAPTER 4 Permanent Magnet Applications

torque, and that the total axial torque is the same for each sector. The integrals in Eq. (4.232) can be evaluated using Simpson’s method. The resulting torque expression is M LR* ,X ,P Q S (p)S (q) X P NN P X N O ;r(q)B r(q), ; , z(p) , (4.233) P N where L : z 9 z , R* : R 9 R , and S (q) and S (p) are the Simpson P X coefﬁcient terms T () : X

2N

and

S (q) : P

(q : 0) (q : 1, 3, 5, . . . ) (q : 2, 4, 6, . . . ) (q : N ) P

(p : 0) (p : 1, 3, 5, . . . ) S (p) : X (p : 2, 4, 6, . . . ) (p : N ) P The integration points are as follows:

and

q r(q) : R ; (R 9 R ) N P

(q : 0, 1, 2, . . . , N ), P

p z(p) : z ; (z 9 z ) (p : 0, 1, 2, . . . , N ). N X X Calculations: We demonstrate Eq. (4.233) via a practical design problem. Consider an application that requires a torque of T : 90 Nm to be transmitted through a 0.45-cm enclosure (i.e., the coaxial magnets must have a 0.45-cm radial gap between them). Assume that the magnets are made from bonded NdFeB with M : 4.3;10, and that they have the following dimensions: Q R : 1.0 cm (inner radius of inner magnet) R : 5.0 cm (inner radius of outer magnet) R : 8.0 cm (outer radius of outer magnet) L : 5.0 cm (length).

297

4.7 MAGNETIC COUPLINGS

FIGURE 4.57

Torque vs number of poles.

We compute the peak torque as a function of the number of poles N for a series of values for the outer radius of the inner magnet, R : 3, 3.5, 4, and 4.5 cm (when R : 4.5 cm there is a 0.5-cm radial gap between the inner and outer magnet). The computed data are shown in Fig. 4.57. The analysis shows that the peak torque (which occurs at an angular offset of : /N ) occurs at 4, 4, 6, and 10 poles, respectively. Notice that only the R : 4.5-cm data has a torque of 90 Nm. Thus, a ten-pole coupling with R : 4.5 cm is optimal for this application. ) EXAMPLE 4.7.3 Derive an expression for the torque between two identical axially polarized multipole cylindrical disks as shown in Fig. 4.58 [20, 21]. The disks have inner and outer radii R and R , respectively, and a thickness t . Let K N denote the number of poles on the surface of each disk, and let h denote the spacing between disks. Assume that both magnets have a linear second quadrant demagnetization curve of the form B : (H < M z ), (4.234) Q where the < sign takes into account the alternating polarity of adjacent poles.

298

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.58

Synchronous axial coupling.

SOLUTION 4.7.3 We use the charge model (Section 3.4). Our approach is as follows: First, reduce the magnets to equivalent charge distributions; next, determine the force imparted by one magnet to the other using Eqs. (3.120) to (3.123); Last, determine the torque. The volume and surface charge densities K and are given by Eq. (3.105). From Eq. (4.234) we ﬁnd that K : 9 · (<M z ) : 0, (4.235) K Q and that : M · n : <M . K Q

(4.236)

299

4.7 MAGNETIC COUPLINGS

Now, consider the force formulas (3.120) to (3.123). Because : 0 in both K magnets, the volume terms vanish and the force between the magnets reduces to

(x ) (x )(x 9 x ) I A A . F: K N K I N (4.237) I N 4

x 9 x N I N I Instead of evaluating Eq. (4.237) directly it is easier to derive the force in the following steps. First, the force between two point sources is obtained. Next, the force between opposing pole faces (sectors) is derived. Third, an equation is derived for the force between a sector of one magnet and an entire surface of the opposing magnet. Last, the total force is obtained by summing the contributions from all of the sectors and surfaces. Because torque is obtained from the force, its derivation is presented as well. Point Sources: We use cylindrical coordinates with one of the charges Q (x ) K in the x-y plane at (r , , 0), and the other Q (x ) in a plane deﬁned by z : h K at (r , , h). The axial force between Q (x ) and Q (x ) follows from Eq. K K (3.124), Q (r , , 0)Q (r , , h)h K K F : . (4.238) X 4 [r ; r 9 2r r cos( 9 ) ; h] The axial component of torque imparted to Q (x ) by Q (x ) is K K T : (r ;F ) · z , (4.239) X where F is the force imparted to Q (x ) by Q (x ), and r is the vector from K K the origin of the coordinate system to the ﬁrst particle. The torque T is given by X Q (r , , 0)Q (r , , h)r r sin( 9 ) K . (4.240) T : K X 4 [r ; r 9 2r r cos( 9 ) ; h] Sectors: Next, we compute the axial force and torque imparted to one pole face (sector) by an opposing pole face. Again, without loss of generality, one sector is taken to lie on the x-y plane symmetrically positioned about the x-axis ( : 0), while the other is taken to lie on the plane deﬁned by z : h, symmetrically positioned about an angular displacement : (Fig. 4.59). Q The idea is to discretize each sector into a mesh of subsections (elements), deﬁne a point charge at the midpoint of each element in accordance with Eq. (4.236), and then compute the force and torque imparted to one sector by the other by summing the contributions from the individual point charges. The area of each sector is area of sector :

(R 9 R) . N

(4.241)

300

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.59

Relative angular offset of two sectors.

In addition, each sector spans an angular measure 2 : . (4.242) N The ﬁrst step is to discretize the area (4.241) into a set of indexed elements. These elements are chosen to have equal areas so that they contain an equal amount of surface charge. Because the area of the sector is spanned by the two coordinate variables r and , two indices (i, j) are required. Let N and N be P the number of mesh divisions in these two variables, respectively. Then each element will have the same area A given by (R 9 R) . (4.243) N N N P Furthermore, choose each element so that it spans an angular measure given by A :

:

. N

(4.244)

Let (r , ) denote the midpoint of the element A , and let R(i) and R(i ; 1) G H GH

301

4.7 MAGNETIC COUPLINGS

denote its radial boundaries. Then r : R(i ; 1) 9 R(i), G

(4.245)

and r: G

R(i ; 1) ; R(i) 2

(i : 1, 2, . . . , N ). P

(4.246)

As the sector is centered about : , Q

: 9 ; (j 9 1) ; H Q N 2

(4.247)

for j : 1, 2, . . . , N . The endpoints are computed using R(i ; 1) :

R(i) ;

(R 9 R) N P

(i : 1, 2, . . . , N ), P

(4.248)

where R(1) : R (the inner radius). Equation (4.248) is used to generate the endpoint values from which the midpoint values are obtained. Therefore, Eqs. (4.246)—(4.248) deﬁne a scheme for discretizing the sector into elements and assigning values to the midpoints of each element. Next, consider the magnetic interaction between an element of one sector and an element of an opposing sector. Let the coordinates of the midpoints of the two elements be (r , , 0) and G H (r , , h), respectively. The magnetic charge associated with each element is GY HY Q (r , , 0) : (r , , 0)A K G H G H

(4.249)

Q (r , , h) : (r , , h)A, K GY HY GY HY

(4.250)

and

respectively. From Eqs. (4.236) and (4.243) these charges reduce to

(R 9 R) . Q : <M K Q N N N P

(4.251)

where, once again, the < term takes into account the polarity of oppositely charged sectors. The force and torque are obtained by summing the contributions of all of the elements in accordance with Eqs. (4.238) and (4.240). The resulting

302

CHAPTER 4 Permanent Magnet Applications

expressions are

M (R 9 R) ,P , F (h, ) : Q X Q 4 N N N P G H , , h ; P , [r ; r 9 2r r cos( 9 ) ; h] GY GY HY G G GY H HY

(4.252)

and T

M (R 9 R) ,P , (h, ) : Q X Q 4 N N N P G H , , r r sin( 9 ) G GY H HY . ; P [r ; r 9 2r r cos( 9 ) ; h] GY G GY H HY GY HY G

(4.253)

Torque: Next, we determine the total axial force and torque taking into account the interaction of all the sectors of both magnets. There are two magnets that are labeled upper and lower in accordance with their relative positions. These magnets are identical and have a thickness t . Each magnet has two surfaces K (labeled top and bottom) and each surface contains N sectors. The upper magnet has its bottom surface in the plane deﬁned by z : h and its top surface lies in the plane deﬁned by z : h ; t . Assume that this magnet is rotated by K an angle as shown in Fig. 4.60. Under this condition, the angular positions

FIGURE 4.60

Rotated magnet surface.

303

4.7 MAGNETIC COUPLINGS

of the centers of its sectors (, n) are Q 2 (, n) : ; (n 9 1) (n : 1, 2, . . . , N ). (4.254) Q N The lower magnet is centered about : 0, and has its top surface in the x-y plane, and its bottom surface in the plane deﬁned by z : 9t . Consider the K sector of this magnet on the top surface centered about : 0. Based on the results of the previous section, the axial force on this sector due to all of the sectors on both the top and bottom surfaces of the upper magnet is , F (h, ) : (91)L[F (h, (, n)) X X Q L 9F (h ; t , (, n))]. X K Q Continuing in this fashion and taking into account the contributions from all the sectors and surfaces, the following formulas are obtained for the total force and torque,

N M (R 9 R) , ,P , F () : Q X N N N 4 P K L G H , , (91)L>K h K K , ; P [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K G GY H HY GY HY G and

(4.255)

N M (R 9 R) , ,P , T () : Q X 4 N N N P K L G H , , (91)L>K r r sin( 9 (, n)) K G GY H HY ; P , [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K GY HY G G GY H HY where

(4.256)

:9 ; (j 9 1) ; , H N 2 (, n) : (, n) 9 ; (j 9 1) ; , HY Q N 2 and (, n) is given by Eq. (4.254). The terms h and are given by Q K K h : h ; mt (m : 0, 1, 2) K K h : separation of magnets

t : thickness of magnets K

304

CHAPTER 4 Permanent Magnet Applications

and

1 (m : 0)

: 2 (m : 1) K 1 (m : 2). The angle is the relative offset of the two magnets. The condition : 0 occurs when poles of one magnet align with the poles of opposite polarity on the other magnet (the torque is zero when : 0). Note that Eqs. (4.255) and (4.256) contain nested summations of elementary functions, which are easy to program. However, it is important to note that when programming nested summations, a higher degree of accuracy can be achieved with fewer mesh points (lower values N and N) if the total of a single summation over one index is evaluated and P saved separately (as an intermediate step), and then the totals of the successive summations are added together to obtain the ﬁnal sum. For example, Eq. (4.256) can be evaluated as follows:

N M (R 9 R) , T () : Q S (, n, m), X N N N 4 P K L where , S (, n, m) : P S (, i, n, m), G with , S (, i, n, m) : S (, i, j, n, m), H and , S (, i, j, n, m) : P S (, i, i, j, n, m), GY and , (91)L>K r r sin( 9 (, n)) K G GY H HY S (, i, i, j, n, m) : . [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K HY G G GY H HY Calculations: We demonstrate Eqs. (4.255) and (4.256) with a practical design. Consider an application that requires a torque of 1.0 Nm to be transmitted through a 10-mm enclosure. To begin the analysis, choose the

305

4.7 MAGNETIC COUPLINGS

FIGURE 4.61

Torque vs number of poles.

following nominal magnet dimensions, R : 1.0 cm R : 4.0 cm t : 1.5 cm K

(inner radius) (outer radius) (thickness).

The coupling is to be made from bonded NdFeB with a 10% concentration of binder by volume which renders a bulk magnetization M : 4.3;10 A/m. The Q peak torque is computed as a function of the number of poles N for a series of separations h : 10, 12, 14, and 16 mm. The analysis is performed with N P and N set to 12. We ﬁnd that the peak torque (which occurs at an angular offset of : /N ) occurs at 6, 4, 4, and 4 poles, respectively (Fig. 4.61). Notice that the magnets can be separated by as much as 14 mm and still render the desired torque. To complete the analysis, the torque and force are computed as a function of for the h : 14 mm separation (Fig. 4.62). The maximum force is approximately 60 N and is attractive (negative) at : 0° and repulsive (positive) at 90°. )

306

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.62

Torque and force vs .

4.8 MAGNETIC RESONANCE IMAGING Magnetic resonance imaging (MRI) is widely used for noninvasive diagnosis of the human body. In this section, we brieﬂy review the principles of MRI, and the role that permanent magnets play in this technology. The MRI process itself is relatively complex and we present a brief and simpliﬁed description of it here. The interested reader can ﬁnd more detailed discussions of MRI in numerous texts [22—24]. In MRI, radio frequency (rf) signals are used to create sectional images of the body. To obtain an image, a patient is ﬁrst placed in a cavity (imaging region) that is permeated by a strong magnetic ﬁeld. The nuclear magnetic moments within the patient’s tissue tend to align either parallel or antiparallel to the ﬁeld, with the former having a lower energy than the latter (Fig. 4.63b). In a given specimen there are a large number of magnetic moments and most of the parallel moments are cancelled by the antiparallel moments leaving a relatively small number of unpaired parallel moments. The interaction between the magnetic

4.8 MAGNETIC RESONANCE IMAGING

307

moments and the ﬁeld causes them to precess about the ﬁeld direction at an angular frequency (Larmor frequency), : B , (4.257) where is the nuclear gyromagnetic ratio and B is the applied ﬂux density. The magnetic moment of the hydrogen nucleus (proton) is used for MRI. Hydrogen is abundant in biological tissues, and the proton has a relatively high gyromagnetic ratio ( : 42.6 MHz/T). The Larmor frequency for hydrogen is 64 MHz when B : 1.5 T. Once the nuclear moments are aligned, the specimen is subjected to a pulsed rf signal that is tuned to the Larmor frequency. The rf signal induces transitions between the lower- and higher-energy proton spin populations (Fig. 4.63c). As the moments transition from the higher- to the lower-energy state they generate an rf signal that is sensed by a coil

Orientation of nuclear magnetic moments: (a) random orientation, (b) alignment in a B-ﬁeld; (c) reorientation due to absorption of rf energy; and (d) reorientation due to reradiation.

FIGURE 4.63

308

CHAPTER 4 Permanent Magnet Applications

and processed to create sectional 3D images (Fig. 4.63d). Spatial imaging can be accomplished by use of additional gradient ﬁeld coils that have ﬁelds that superimpose with the primary ﬁeld to produce a predeﬁned variation in the bias ﬁeld across the specimen. This gives rise to a spatial variation of the Larmor frequency, which in turn gives rise to a variation of the generated rf signal and which can be processed to form an image. There are essentially three different magnetic ﬁeld systems used in MRI apparatus: 1. conventional electromagnets; 2. superconducting coils; and 3. permanent magnet structures. Conventional electromagnets consist of multiple turns of low gauge wire or conducting ribbons usually used in conjunction with soft magnetic cores and pole pieces. Electromagnet coils have a relatively high resistance and therefore they consume large amounts of energy and this causes heating. The heating sets a practical limit on the working current of such coils and, therefore, limits magnetic ﬁeld strength. On the other hand, superconducting coils are capable of producing very stable and uniform ﬁelds of up to 4.0 T. However, they require liquid cryogens and cooling apparatus to suppress their resistance. Such coils are, therefore, expensive and require extensive preparation and maintenance. Permanent magnet systems are competitive with current-driven coils for MRI applications that require ﬁelds in the range of 0.1—0.5 T. They consume no energy and require little maintenance. When designing an MRI magnetic ﬁeld system, the following parameters need to be considered: 1. 2. 3. 4. 5.

magnetic ﬁeld strength; the homogeneity of the magnetic ﬁeld; the temporal stability of the magnetic ﬁeld; fringing ﬁelds; and patient access to the imaging volume.

A wide range of ﬁeld strengths can be used for MRI. For example, whole body MRI has been achieved with ﬁelds ranging from 0.02 T (0.85 MHz) to 4.0 T (170 MHz). Higher ﬁeld strengths are usually desirable because the number of transitions from high- to low-energy states scales approximately linearly with ﬁeld strength. The three magnetic ﬁeld systems can be ranked in terms of their ﬁeld strength as follows: superconducting coil (3 T); conventional electromagnet (0.2—0.4 T); and permanent magnets (0.1—0.5 T).

4.8 MAGNETIC RESONANCE IMAGING

309

A high degree of ﬁeld uniformity is required for MRI. This is deﬁned as the normalized maximum deviation in ﬁeld strength between two points in the imaging volume B B 9 B

. : B B The ﬁeld homogeneity is usually speciﬁed in ‘‘parts per million’’ for a given diameter of spherical volume (dsv). It can also be expressed as a ratio of frequencies using the Larmor relation (4.257). In this case it is given by f 9f f

: . f f Typical requirements for MRI ﬁeld homogeneity are 10 to 100 ppm over a spherical imaging volume with a diameter of 50 cm (50 dsv). Superconducting coil systems have the highest degree of ﬁeld uniformity (10 ppm, 50 dsv), followed by permanent magnet systems (10 ppm, 20 dsv), and then conventional electromagnet systems (30 ppm, 40 dsv). In MRI, data acquisition occurs over time. Therefore, it is important that the magnetic ﬁeld have a high degree of temporal stability. Instability can be caused by variations in the ﬁeld source itself, or by time dependent factors such as the movement of external ﬁeld sources or ferromagnetic materials. Superconducting coils are the most stable ﬁeld source. On the other hand, conventional electromagnets are susceptible to variations in their power source. Permanent magnets also exhibit ﬁeld instability due to ﬂuctuations in the ambient temperature. In this regard, SmCo magnets have an advantage over NdFeB due to their lower temperature coefﬁcient (Section 1.20). The magnetic fringe ﬁeld is another factor that must be considered when evaluating MRI magnet systems. A related issue is magnetic shielding. The fringe ﬁeld can have undesired effects on medical equipment in proximity to the MR imager. Conversely, external ﬁeld sources or the movement of ferromagnetic materials can degrade MR image quality if the system is not properly shielded. These effects can be reduced using active or passive shielding. In active shielding, additional coil sets are arranged inside the imager and activated to counter the fringe ﬁeld. In passive shielding, a soft magnetic enclosure is used to conﬁne the ﬁeld and to prevent stray ﬁelds from penetrating the imaging cavity. The ﬁnal factor we consider is patient access to the imaging region. An MRI magnet system must provide an imaging region that can

310

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.64

Comparison of MRI magnetic ﬁeld systems [25].

accommodate the patient (or body part), a supporting structure (couch), and the rf coils. This space must be sufﬁcient to provide minimal physical and psychological comfort for the patient. For example, in whole body MRI the magnet system cavity is cylindrical with a diameter of 1.0 m. However, the free access diameter for imaging is approximately 0.75 m. In pediatric MRI, the corresponding diameters are 0.5 and 0.39 m, respectively. For research applications, the magnet cavity has a diameter ranging from 0.2 to 0.4 m, and the imaging region has a diameter of 0.135 to 0.31 m. The characteristics of the three MRI magnetic ﬁeld systems are summarized in Fig. 4.64. As already noted, permanent magnet systems can be used for MRI with ﬁeld requirements that range from 0.1 to 0.5 T. The main advantage of these systems is that they consume no power. Moreover, when rareearth magnets are used the size and weight of such structures are greatly reduced relative to conventional materials. The MRI permanent magnet systems are classiﬁed as either yokeless, yoked, or hybrid. Yokeless systems consist of permanent magnets alone. The imaging ﬁeld is conﬁned and focused by the magnetization pattern within the magnets. The ring dipole arrangement shown in Fig. 4.65a is an example of a yokeless system [25]. This is composed of eight permanent magnet segments with magnetizations indicated by the solid

4.8 MAGNETIC RESONANCE IMAGING

FIGURE 4.65

311

Two MRI magnet systems: (a) ring dipole; and (b) H-frame [25].

arrows. This assembly provides a transverse magnetic ﬁeld across the imaging region as indicated by the dotted lines. The ﬁeld generated by these systems is typically 0.3 T. In yoked systems a high-permeability material surrounds the permanent magnet structure. This enhances the ﬁeld in the imaging region and minimizes the fringe ﬁeld. The H-frame structure shown in Fig. 4.65b is an example of a yoked system [25]. In this structure, permanent magnet

312

CHAPTER 4 Permanent Magnet Applications

blocks are supported in a ferromagnetic frame called a yoke. The yoke not only supports the magnets but also serves as a ﬂux return path for the imaging ﬁeld. Soft magnetic pole pieces are also used to enhance the strength and uniformity of the imaging ﬁeld. The H-frame structure is the design of choice for MRI despite its weight and size. Hybrid structures contain yoked and yokeless sections. The design of MRI magnet structures is complicated. In the design of conventional magnet structures, the analyst solves the ‘‘direct’’ problem in which the geometry and magnetization are given, and the ﬁeld distribution is determined. In MRI, the analyst is faced with the more difﬁcult ‘‘inverse’’ problem in which the ﬁeld strength and uniformity are speciﬁed across the imaging region and the geometry and magnetization of the structure need to be determined. There is no unique solution to such problems in that a speciﬁed ﬁeld distribution within a closed region can be obtained using an inﬁnite number of different structures [26]. One design approach is to select one of the many ideal closed structures that give the desired interior ﬁeld distribution and then remove a portion of the material to allow access to the cavity. These openings give rise to undesired ﬁeld nonuniformities that must be compensated for. Remedies for this include the modiﬁcation of the geometry or magnetization, and/or the insertion of additional components such as high-permeability materials.

4.9 ELECTROPHOTOGRAPHY Electrophotography is one of the most widely used methods for image reproduction and printing [27, 28]. The central element of this process is the photoconductor (PC), which consists of a thin sheet (10—50 m) of photosensitive material that is coated onto a ground plane. The PC generates electron-hole pairs as it absorbs incident light. Once generated, the electrons and holes separate and move under the inﬂuence of an internal electric ﬁeld with the electrons moving to ground. This process forms the bases of electrophotographic image reproduction. The electrophotographic process comprises six distinct steps: Charge; Expose; Develop; Transfer; Fuse; and Clean (Figs. 4.66 and 4.67). We brieﬂy review these steps in the order in which they occur: 1. Charge: The ﬁrst step in the electrophotographic process entails charging the PC. The PC (which is initially uncharged) passes through a corona-charging device. This device creates ionized gas

4.9 ELECTROPHOTOGRAPHY

FIGURE 4.66

313

Electrophotographic system [27].

molecules, which are absorbed by the PC and thereby create a uniform charge distribution on its surface. This charge distribution gives rise to an electric ﬁeld within the PC (Fig. 4.67(1)). 2. Expose: To form an image, the charged surface of the PC is exposed to light that is reﬂected from an existing image or produced by a laser writer. A portion of the incident light is absorbed in the PC and electron-hole pairs are created in the exposed regions. These charged carriers separate and drift under the inﬂuence of the PC’s internal ﬁeld. The electrons move to ground and the holes move to the charged surface where they discharge the exposed regions. This produces a charge pattern on the exposed surface of the PC that mirrors the image. This is called the latent image (Fig. 4.67(2)). 3. Develop: During this step, charged pigmented toner particles are transported proximal to the latent image and are electrostatically attracted to regions of the PC in accordance with the local surface charge (Fig. 4.67(3)). This transforms the latent image into a real image.

314

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.67

Electrophotographic process.

4. Transfer: In the transfer step, the toner on the PC is electrostatically transferred to paper. This is accomplished by depositing a uniform charge of opposite polarity to that of the toner on the back side of the paper prior to it contacting the PC (Fig. 4.67(4)).

4.9 ELECTROPHOTOGRAPHY

315

5. Fuse: During this step, the image is permanently ﬁxed to the paper by melting the toner onto the paper surface (Fig. 4.67(5)). 6. Clean: In this ﬁnal step, the PC is discharged and cleaned of any excess toner using coronas, lamps, brushes, and/or scraper blades (Fig. 4.67(6)). Once this is done the PC is ready for the next imaging cycle. We are primarily interested in the development of the latent image since this is the most magnetically intensive step of the process. Most electrophotographic copiers and printers use ‘‘magnetic brush’’ technology to implement this step. While there are various implementations of this technology, they all perform essentially the same function, namely the transportation of developer material from a reservoir to the latent image. The developer material is usually either monocomponent or two component. In monocomponent systems, the toner particles are magnetic and on the order of 5—30 m in diameter. The magnetic transfer ﬁeld couples directly to these particles. In two component materials, there are magnetic ‘‘carrier’’ particles and nonmagnetic toner particles. The carrier particles range from 70 to 250 m in diameter and are much larger than the toner particles. The magnetic transfer ﬁeld couples to the carrier particles and several toner particles are transported by each carrier particle. A typical magnetic brush system is shown in Fig. 4.68. In this system the magnetic toner/carrier material is transported to the latent image by

FIGURE 4.68

Magnetic brush development station [28].

316

CHAPTER 4 Permanent Magnet Applications

riding on the surface of a conductive cylindrical shell. The conductive shell encloses a rotating radially polarized permanent magnet cylinder. The rotating magnet gives rise to a time-varying magnetic ﬁeld at the surface of the shell, which is typically a few tenths of a tesla or lower. This time-varying ﬁeld provides a magnetic transport force that causes the toner particles to move in the opposite direction as shown. In magnetic brush systems the magnetic transport force dominates the centripetal, gravitational and electrostatic forces. When two magnetized toner/carrier particles approach one another they exert a dipole-dipole force on each another. This force is far greater than all the other ‘‘ambient’’ forces, including the magnetic force exerted by the magnetic brush. Speciﬁcally, the dipole-dipole force is typically more than an order of magnitude greater than the force due to the magnetic brush [27]. Consequently, coaligned neighboring toner/carrier particles tend to adhere to one another, forming chains of particles (Fig. 4.69. During transport, these chains behave as a single mechanical entity. They ride on the surface of the conductive shell and cartwheel end over end as they are transported. The dynamics of this are complicated and beyond the scope of this text. We simply derive here the force on an isolated toner particle in the ﬁeld of the magnetic brush. For this, we need the ﬁeld components for an inﬁnite, radially polarized multipole

FIGURE 4.69

Monocomponent toner transport [28].

317

4.9 ELECTROPHOTOGRAPHY

cylinder. These were derived in Example 4.2.5. The radial and azimuthal components are given by Eqs. (4.66) and (4.67), respectively, B (r, ) : ir\G>U (M , R , R , ) cos(i) P G G G

(4.258)

and B (r, ) : ir\G>U (M , R , R , ) sin(i), (4.259) G G G where U (M , R , R , ) is given by Eq. (4.65) with M given by (4.64) (the G G G superscript 2 has been dropped). In the following example we derive the force on a magnetic particle in the ﬁeld of a magnetic brush. EXAMPLE 4.9.1 Determine the force on a spherical toner particle of radius R in the ﬁeld of a cylindrical magnetic brush. Assume that the toner is made from a magnetically linear and isotropic material with permeability and susceptibility . K SOLUTION 4.9.1 The force on the toner particle is computed using Eq. (3.116). We assume that (M · )B is constant over the volume of the toner particle. This gives F:

(M · )B dv 4 (m · )B . where m is the magnetic dipole moment of the toner particle, and B is the ﬁeld of the magnetic brush (multipole cylinder). Speciﬁcally, : B (r, )r ; B (r, ) , P where B (r, ) and B (r, ) are given by Eqs. (4.258) and (4.259), respectively. P From Example 1.8.1 we know that the dipole moment of the toner particle is B

4R( 9 ) H . ( ; 2 )

(4.260)

4 R( 9 ) (B · )B . ( ; 2 )

(4.261)

m: Thus, we ﬁnd that F:

318

CHAPTER 4 Permanent Magnet Applications

We evaluate Eq. (4.261) and obtain F(r, ) :

4R( 9 ) ( ; 2 ) ;

B (r, ) P

; B (r, ) P

B (r, ) 1 B (r, ) P P ; B (r, ) r r r

B (r, ) 1 B (r, ) ; B (r, ) . r r

(4.262)

Finally, we substitute Eqs. (4.258) and (4.259) into Eq. (4.262) and ﬁnd that 4 R( 9 ) ( ; 2 ) ; ii(i ; 1)r\G>GY>U U cos(i) cos(i) G GY G GY ; iiYr\G>GY>U U sin(i) sin(i) r G GY G GY ; ii( i ; 1)r\G>GY>U U cos(i) sin(i) G GY G GY 9 iir\G>GY>U U sin(i) cos(i) , (4.263) G GY G GY where U (M , R , R , ) is given by Eq. (4.65) with M given by Eq. (4.64). G G G )

F(r, ) :9

4.10 Magnetooptical Recording Magnetooptical (M-O) recording is routinely used for digital data storage [29—31]. Just as in magnetic storage, the information is recorded along tracks on data disks. A track is a narrow annulus region a distance r from the center of the disk. Each track is divided into a number of smaller segments or sectors and each segment stores a single block of data of typically either 512 or 1024 bytes. In conventional M-O systems, data retrieval/storage is accomplished using a read/write head that is held stationary while the disk spins (Fig. 4.70). Typical disk rotation speeds are 1200 and 1800 rpm for slower drives, and 3600 rpm for high data-rate systems.

4.10 MAGNETOOPTICAL RECORDING

FIGURE 4.70

319

Magnetooptic recording system.

An M-O storage disk consists of a multilayer stack of thin ﬁlms. The conventional quadrilayer structure is shown in Fig. 4.71. This structure is optically designed to provide near-optimum readback. The functional storage layer in this stack is a thin magnetic ﬁlm (250 Å) that is currently made from a rare-earth transition-metal (RE-TM) material, preferably an alloy based on Tb, Fe, and Co. This material is characterized by a vertical magnetic anisotropy and a high magnetic coercivity on the order of several tenths of a tesla (several kilo-gauss at room temperature). Because of the high coercivity, the ﬁlm is magnetically very stable and

FIGURE 4.71

An M-O recording thin ﬁlm stack [31].

320

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.72

Magnetooptic recording process.

can retain a recorded magnetization pattern indeﬁnitely under normal operation. However, this layer tends to be chemically unstable due to the high afﬁnity of rare-earth elements for oxygen. In practice, it is sandwiched between two chemically stable dielectric layers as shown. Data is recorded in the RE-TM layer using a thermomagnetic process. In conventional systems, recording occurs with the media moving (linear velocity < 10 m/s) between a stationary laser source and a ﬁxed bias ﬁeld source (Fig. 4.72). The ﬁlm is initially polarized in one direction (e.g., the down direction). To record information, the laser is pulsed (duration < 50 ns) and focused so as to heat a local spot of the ﬁlm (diameter < 1m). As the temperature of the spot increases the coercivity within it decreases. Eventually, the spot reaches the Curie temperature (T < 473 K) at which point the local magnetization reverses A and aligns with the bias ﬁeld, which is typically 0.02—0.04 T across the ﬁlm. The temperature at the center of the spot can reach a value as high as 500 K, which, while sufﬁcient to reverse the magnetization, is not high enough to damage the ﬁlm. When the laser is turned off, the temperature decreases back to room temperature, and the magnetization within the spot remains reversed. Thus, the spot represents a recorded bit of information. By repeating this process, regions of the disk are selectively reversed-magnetized and the resulting magnetization pattern represents the recorded information. Data readout in M-O systems is based on the magnetooptic Kerr effect. Speciﬁcally, when linearly polarized light is normally incident on a perpendicular magnetic medium its plane of polarization undergoes a slight rotation upon reﬂection (Fig. 4.73). This rotation is known as the I polar Kerr effect and the direction of rotation depends on the orientation

4.10 MAGNETOOPTICAL RECORDING

321

Kerr effect: (a) reﬂection from a north pole; and (b) reﬂection from a south pole [30].

FIGURE 4.73

of the magnetization within the ﬁlm. To read data, a polarized laser beam is applied to the disk as it rotates and the reﬂected light is collected and processed. The same laser source is usually used for both the read and write process. However, the output power is substantially reduced for readout so as to avoid erasure. To erase information, the bias ﬁeld is reversed and the ﬁlm is heated at a constant power level until an entire sector is erased. The disk makes at least one revolution between the erasure of an old block and the recording of a new one. Therefore, the bias ﬁeld reversal required for the erase and rerecord sequence can be implemented over a relatively long

322

CHAPTER 4 Permanent Magnet Applications

FIGURE 4.74

Cross section of a rotary actuator.

time frame (16 ms at 3600 rpm). It is convenient and common to use permanent magnet actuators for the bias ﬁeld mechanism. In these actuators the magnet that provides the bias ﬁeld is also a functional part of the actuator. A cross-sectional view of a rotary bias ﬁeld actuator is shown in Fig. 4.74. This consists of a rotating bipolar cylindrical bias magnet surrounded by a stationary concentric conductive shell, and a drive coil. When the coil is energized it rotates the magnet to one of two equilibrium positions, thereby producing a bias ﬁeld of desired polarity across the recording media. Rotation of the magnet gives rise to eddy currents in the conductive shell that impart a drag torque and reduce oscillations of the magnet about its equilibrium positions. We present an analysis of this actuator in Example 5.11.1. We work through a design of the bias magnet in the following example [32]. EXAMPLE 4.10.1 Determine the dimensions of a cylindrical bias ﬁeld magnet for use in an M-O system (Fig. 4.10). The magnet must provide a bias ﬁeld of between 0.02 and 0.04 T along a radial length of the disk which is taken to be 100 mm. Assume that the axis of the magnet is positioned 4 mm below the recording layer and aligned with the radial length of the disk. Assume that the magnet has a second quadrant constitutive equation B : (H ; M ). (4.264) Q Let R, z and z denote the radius and respective end points of the cylinder. SOLUTION 4.10.1 For the erase/rerecord sequence the magnet needs to be rotated between two equilibrium positions. We choose these positions to be when the magnetization is at <85° with respect to the x-axis (Fig. 4.74). The magnetization needs to be offset from the vertical in this fashion so that it can couple to the coil’s ﬁeld (which is predominately vertical).

323

4.10 MAGNETOOPTICAL RECORDING

Field Solution: A ﬁeld solution for this magnet was derived in Example 4.2.3. Only the radial component is of interest, and it is repeated here for convenience, M R , B (r, , z) : Q (91)I>S (n) cos((n)) P 2N L I ;[r 9 R cos( 9 (n))]I(r, , z; R, (n), z ), I where N is the Simpson’s method mesh number (even), (n) :

n 2 N

(4.265)

(n : 0, 1, 2, . . . , N),

and

(n : 0) (n : 1, 3, 5, . . . ) S (n) : (n : 2, 4, 6, . . . ) (n : N). The function I(r, , z; R, , z ) is given by I F(r, , z; r, , z ) I if

IY

where

r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z ) I if r : r, cos( 9 ) : 1, z " z , I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 ) We want to evaluate the ﬁeld along the y-axis. For a given observation point (0, y, z) on this axis the ﬁeld is determined by substituting r : y, and : /2 9 in Eq. (4.265), that is, K B (y, , z) : B y, 9 , z , (4.266) W K P K 2

where is the angular position of the north pole of the magnet relative to the K

324

CHAPTER 4 Permanent Magnet Applications

x-axis ( is positive in the counterclockwise direction). We use Eqs. (4.265) K and (4.266) to design the magnet. Calculations: For the design, we assume that the magnet is made from sintered NdFeB material with a uniform magnetization of M : 7.2;10 A/m Q (B : 9000 G). First, we determine a viable magnet radius R. The length of the P magnet is set to 100 mm (z : 950 mm, z : 50 mm) and Eq. (4.265) is evaluated for a range of R values, 0.5 mm R 1.5 mm with : 85°, r : 4 K mm, and z : 0. A plot of B vs R is shown in Fig. 4.75. From this plot it is P obvious that a magnet with R : 1 mm will sufﬁce because its ﬁeld is 0.028 T, which is within the desired range. Having determined a viable radius, we examine the angular dependency of the ﬁeld. Speciﬁcally, we compute the ﬁeld with R : 1 mm, r : 4 mm, and z : 0 for a series of angular offsets relative to the y-axis, 0 90° 9 20°. This corresponds to a magnet rotation K 70° 90°. These data are shown in Fig. 4.76. Note that the ﬁeld is within K the desired operating range over this entire angular range. Thus, there is considerable tolerance in the angular position of the magnet. The next step is the determination of a viable length for the magnet. For this, we set R : 1 mm

FIGURE 4.75

The M-O bias ﬁeld vs magnet radius R (h : 100 mm).

4.11 FREE-ELECTRON LASERS

325

The M-O bias ﬁeld 4 mm above the magnet vs angular offset 90° 9 (z : 0). K

FIGURE 4.76

and vary the length h as follows: h : 100, 104, 108 mm (954 mm z 950 mm, and 50 mm z 54 mm). The bias ﬁeld is computed with : 85° at K a distance r : 4 mm above the magnet along a series of points that span its length. A plot of B for h : 100, 104, 108 mm is shown in Fig. 4.77. Note that P the ﬁeld data is plotted for positive values of z only (the data for negative z are the mirror image). We ﬁnd that a magnet length of h : 108 mm will sufﬁce because the ﬁeld is virtually constant along the speciﬁed length of 100 mm (0 z 50 mm). In summary, the analysis shows that a sintered NdFeB magnet with a radius R : 1 mm and length h : 108 mm rotated an angle of <85° with respect to the x-axis will provide a uniform bias ﬁeld of <0.028 T over a radial length of 100 mm. )

4.11 FREE-ELECTRON LASERS A free-electron laser (FEL) is a device that generates tunable, coherent and high-power electromagnetic radiation from a beam of high velocity

326

FIGURE 4.77

CHAPTER 4 Permanent Magnet Applications

The M-O bias ﬁeld 4 mm above the axis of the magnet vs z.

electrons that moves through a spatially periodic magnetic ﬁeld (Fig. 4.78). The periodic magnetic ﬁeld is provided by a structure that is typically a few meters in length and is referred to as a wiggler or undulator. The beam is created using a variety of apparatus including rf linear accelerators, microtrons, storage rings, and electrostatic accelerators. The basic operation of an FEL is relatively straightforward [33, 34]. The electron beam is injected at one end of the wiggler and travels down its length. Once inside the wiggler, the electrons experience a spatially varying magnetic ﬁeld that causes them to oscillate (wiggle or undulate) back and forth in a plane perpendicular to the magnetic ﬁeld. This oscillatory motion causes the electrons to radiate electromagnetic energy and as the electrons are randomly positioned in the beam, the radiation is incoherent. This corresponds to spontaneous radiation in a conventional laser. Moreover, just as in a conventional laser, the radiation is ampliﬁed and made coherent by placing the wiggler between two parallel mirrors that form a resonant cavity for the emitted radiation` The electromagnetic wave set up by the mirrors interacts with the electron

327

4.11 FREE-ELECTRON LASERS

FIGURE 4.78

Free-electron laser [33].

beam to promote ampliﬁcation and coherence of the radiation. One of the mirrors is partially transparent, and the radiation exits this mirror in the form of a coherent laser beam. The wavelength of a FEL depends on the energy of the electrons, and the period and magnitude of the magnetic ﬁeld. We derive an expression for the wavelength by considering the behavior of a single electron as it moves through the wiggler (without the mirrors). Let L denote the U length of the wiggler, and let denote the spatial period of its magnetic U ﬁeld ( is the distance over which the magnetic ﬁeld undergoes one U complete cycle and returns to its original value). The number of cycles of the ﬁeld along the wiggler is N : L / . U U U The electron moves with a speed v : #c where # is a constant and c is the speed of light. Once it enters the wiggler, the electron radiates energy down the wiggler at the speed of light. At any time t, the wavefront of the radiation is a distance d : (c 9 v)t ahead of the electron. The electron travels the length of the wiggler in a time : L /v. U As it passes through the wiggler it undergoes N oscillations, each of U which generates one cycle of radiation. Therefore, when the electron exits the wiggler the wave packet of radiation that it generates contains N cycles and spans a distance d : (c 9 v) in front of it. The wavelength U

328

CHAPTER 4 Permanent Magnet Applications

of the radiation is therefore *

(c 9 v) N U (c 9 v) : U L U (1 9 #) : . U #

: *

(4.267)

As the electron is moving close to the speed of light, # 1 and Eq. (4.267) can be rewritten as : U, * 2

(4.268)

where :

1 . (1 9 #)

The derivation of Eq. (4.268) ignores the additional path length that the electron travels due to its oscillatory motion. In wigglers with moderate to strong ﬁelds this can be signiﬁcant. The longer path length results in a reduced velocity v, which produces a longer wavelength than predicted by Eq. (4.268). A detailed analysis of this effect yields a wavelength

m c eB M U U : U 1; , (4.269) * 2 E 2m c where E, e, and m are the energy, charge, and rest mass of the electron, respectively, and B : +B, is the rms average magnetic ﬁeld in the U wiggler (the average is taken along the axis of the wiggler). To obtain Eq. (4.269) we have used the relation E : m c. The laser frequency is f : c/ . * * From Eq. (4.269) we see that the wavelength (frequency) of an FEL can be tuned by changing the period and magnitude of the wiggler ﬁeld, or the electron energy. For example, if : 10 cm, and eB /2m c : 1, U U U the FEL emission can range from the infrared (IR) to the ultraviolet (UV) by changing E from 100 to 1000 MeV. In practice, the wiggler ﬁeld is made as strong as possible to maximize the electron’s oscillatory motion and radiation. A typical wiggler ﬁeld is on the order of 0.5 T for : 2 U cm. In practice, the term eB /2m c is on the order of unity. U U

4.11 FREE-ELECTRON LASERS

FIGURE 4.79

329

Cross section of a permanent magnet wiggler.

Wigglers can be made using permanent magnets or electromagnets. Permanent magnet wigglers are made using rare-earth materials such as SmCo or NdFeB. One such wiggler is shown in Fig. 4.79. This conﬁguration of magnets is often referred to as the Halbach conﬁguration [35]. In this wiggler there are two rows of magnets arranged parallel to one another and separated by a distance 2h. The electrons travel along the axis of the wiggler (dotted line), and oscillate in a plane perpendicular to the paper. It is desirable to have the magnetic ﬁeld uniform over the length of oscillation and, therefore, we assume that the magnets are of inﬁnite extent in this direction (perpendicular to the page). The periodic wiggler ﬁeld is obtained by rotating the magnetization of adjacent magnets by 2/N, where N is the number of magnets per wiggler period in one array. In total, 2N magnets are needed to form one period of the wiggler (N magnets in each row). The magnetic ﬁeld for the Halbach wiggler conﬁguration can be obtained using complex variable theory [36, 37]. The ﬁeld solution is B* (x, y) : B (x, y) 9 iB (x, y) U V W sin(-n/N) : 2iB cos(nk (x ; iy)) P U n/N N ;[1 9 exp(9nk L)] exp(9nk h), U U

(4.270)

330

CHAPTER 4 Permanent Magnet Applications

where B is the remanent magnetization of the magnets, k : 2/ , P U S n : 1 ; pN, and i : (91. It should be noted that our notation is slightly different than that found in the literature. Speciﬁcally, the variable N that we use is denoted by M in the literature [33, 35]. The ﬁeld solution (4.270) is dominated by the fundamental term in the series (p : 0), B*(x, y) : 2iB U P

sin(-/N) /N

;[1 9 exp(9k L)] exp(9k h) cos(k (x ; iy)). (4.271) U U U In the limit as L ; -, N ; - and - : 1, B* (x, y) obtains a maximum U on-axis value, B : 2B exp(92h/ ). U P U Another example of a permanent magnet wiggler is shown in Fig. 4.80. This wiggler consists of a sequence of cylindrical ring structures that are formed using pie-shaped magnet segments, each with a different orientation of magnetization. Each magnet segment spans the same angular measure (in this case 2/8 radians). The ring structures are sequentially rotated by this angular measure as shown. This produces a helical ﬁeld variation along the axis of the wiggler where the electron beam ﬂows. There are also hybrid wiggler conﬁgurations that are formed using permanent magnets and soft magnetic materials. One such wiggler is shown in Fig. 4.81. The soft magnetic pole pieces are made from a high-

FIGURE 4.80

Permanent magnet wiggler with a helical ﬁeld [33].

331

4.11 FREE-ELECTRON LASERS

FIGURE 4.81

Hybrid wiggler [33].

grade material such as Vanadium permendur, which has high permeability and ﬂux capacity. The pole pieces are designed to operate at saturation, which tends to smooth out the nonuniformities in the permanent magnets. Hybrid wigglers provide more ﬁeld uniformity and a higher peak ﬁeld than permanent magnet wigglers. Finally, wiggler ﬁelds can also be produced by (current-driven) electromagnets [33]. In fact, the ﬁrst FEL used a superconducting wiggler. Electromagnets have advantages over permanent magnets in that they are adjustable. However, their disadvantages include cost, complexity, maintenance, and the fact that their ﬁelds do not scale well for short wiggler periods. References 1. Furlani, E. P., Reznik, S., and Janson, W. (1994). A three-dimensional ﬁeld solution for bipolar cylinders, IEEE Trans. Mag. 30 (5): 2916. 2. Mayergoyz, I. D., Furlani, E. P., and Reznik, S. (1993). The computation of magnetic ﬁelds of permanent magnet cylinders used in the electrophotographic process, J. Appl. Phys. 73 (10): 5440. 3. Furlani, E. P., Reznik, S., and Kroll, A. (1995). A three-dimensional ﬁeld solution for radially polarized cylinders, IEEE Trans. Mag. 31 (1): 844. 4. Furlani, E. P. (1995). Formulas for computing the ﬁeld distributions of cylindrical shells with axial polarization, Int. J. Appl. Elec. Mech. 6 (2): 103. 5. Furlani, E. P. (1994). A three-dimensional ﬁeld solution for axially-polarized multipole disks, J. Magn. Magn. Mat. 135: 205.

332

CHAPTER 4 Permanent Magnet Applications

6. Leupold, H. A. and Potenziani, II, E. (1991). Augmentation of ﬁeld uniformity and strength in spherical and cylindrical magnetic ﬁeld sources, J. Appl. Phys. 70 (10): 6621. 7. Leupold, H. A., Tilak, A. S., and Potenziani, II, E. (1993). Multi-tesla permanent magnet ﬁeld sources, J. Appl. Phys. 73 (10): 6861. 8. Leupold, H. A., Tilak A. S., and Potenziani, II, E. (1993). Adjustable multitesla permanent magnet ﬁeld sources, IEEE Trans. Mag. 29: 2902. 9. Leupold, H. A., Tilak, A. S., and Potenziani, II, E., (1994). Laminar construction of spheroidal ﬁeld sources with distortion-free access, J. Appl. Phys. 76 (10): 6859. 10. Yonnet, J. P. (1981). Permanent magnet bearings and couplings, IEEE Trans. Mag. 17 (1): 1169. 11. Frazier, R. H., Gilinson, P. J., and Oberbeck, G. A. (1974). Magnetic and Electric Suspensions, Cambridge, MA: MIT Press. 12. Sinha, P. K. (1987). Electromagnetic Suspension: Dynamics & Control, IEE Control Engineering Series 30, London, UK: Peter Peregrinus Ltd. 13. Moon, F. C. (1984). Magneto-solid Mechanics, New York: Wiley. 14. McCaig, M. and Clegg, A. G. (1987). Permanent Magnets in Theory and Practice, 2nd ed., New York: John Wiley and Sons. 15. Furlani, E. P. (1993). A formula for the levitation force between magnetic disks, IEEE Trans. Mag. 29 (6): 4165. 16. Furlani, E. P. (2000). Analytical analysis of magnetically coupled multipole cylinders, J. Phys. D: Appl. Phys. 33: 28. 17. Furlani, E. P. (1996). Analysis and optimization of synchronous magnetic couplings, J. Appl. Phys. 79 (8): 4692. 18. Charpentier, J. F. and Lemarquand, G. (1999). Optimal design of cylindrical air-gap synchronous permanent magnet couplings, IEEE Trans. Mag. 35 (2): 1037. 19. Furlani, E. P. (1995). A three-dimensional model for computing the torque of radial couplings, IEEE Trans. Mag. 31 (5): 2522. 20. Furlani, E. P. (1993). Formulas for the force and torque of axial couplings, IEEE Trans. Mag. 29 (5): 2295. 21. Wang, R. and Furlani, E. P. (1994). Design and analysis of a permanent magnet axial coupling using 3d ﬁnite element ﬁeld computations, IEEE Trans. Mag. 30 (4). 22. Carlton, R. R. and Adler, A. M. (1992). Principles of Radiographic Imaging: An Art and a Science, Albany, New York: Delmar Publishers. 23. Foster, M. A. (1984). Magnetic Resonance in Medicine and Biology, NY: Pergamon Press. 24. Kuperman, V. (2000). Magnetic Resonance Imaging: Physical Principles and Applications, New York: Academic Press. 25. Thomas, S. R. (1993). Magnets and gradient coils: types and characteristics, The Physics of MRI, M. J. Bronskill and P. Sprawls, eds., New York: American Institute of Physics, 56.

4.11 FREE-ELECTRON LASERS

333

26. Abele, M. G. (1993). Structures of Permanent Magnets: Generation of Uniform Fields, New York: John Wiley & Sons. 27. Williams, E. M. (1984). The Physics and Technology of Xerographic Processes, New York: John Wiley & Sons. 28. Schein, L. B. (1988). Electrophotography and Development Physics, 2nd ed., New York: Springer Verlag. 29. Marchant, A. B. (1990). Optical Recording: A Technical Overview, Reading, MA: Addision-Wesley Pub. Co. 30. Mansuripur, M. (1995). The Physical Principles of Magneto-optical Recording, Cambridge, England: Cambridge Univ. Press. 31. McDaniel, T. W. and Victora, R. H. (1997). Handbook of Magneto-Optical Recording; Materials, Subsystems, Techniques, Westwood, New Jersey: Noyes Pub. 32. Furlani, E. P. and Williams, C. (1995). Formulas for optimizing the bias ﬁeld for magneto-optic recording, J. Magn. Magn. Mat. 148: 475. 33. Brau, C. A. (1990). Free-Electron Lasers, New York: Academic Press. 34. Svelto, O. (1989). Principles of Lasers, 3rd ed., New York: Plenum Press. 35. Halbach, K. (1985). Application of permanent magnets in accelerators and electron storage rings, J. Appl. Phys. 57 (1): 3605. 36. Halbach, K. (1980). Design of permanent magnets with oriented rare earth cobalt material, Nucl. Instum. Method 169: 1. 37. Halbach, K. (1981). Physical and optical properties of rare earth cobalt magnets, Nucl. Instum. Method 187: 109.

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CHAPTER

5

Electromechanical Devices

5.1 INTRODUCTION Electromechanical devices convert electrical energy to mechanical energy and vice versa. Common examples of these include generators, actuators, transducers, and motors. Such devices can be found in a wide variety of products including audio and video devices, appliances, computers, automobiles, automated machinery, and power tools. In this chapter, we study the theory of electromechanical devices, as well as various practical applications. Electromechanical devices are governed by a coupled system of electrical and mechanical equations. The electrical equations follow from quasi-static ﬁeld theory, and the mechanical equations follow from Newton’s laws. In the following sections, we derive the equations of motion for electromechanical devices that execute either linear or rotational motion. We apply these equations to the analysis of various practical devices including magnetic circuit actuators, linear, rotational and resonant actuators, axial-ﬁeld motors, and stepper motors. The chapter concludes with a brief discussion of magnetic microactuators. The emphasis of this chapter is on magnetically linear, singly excited (single pair of electrical terminals), electromechanical devices with a single degree of mechanical freedom (single coordinate variable). The interested reader can ﬁnd discussions of multiply excited devices with multiple degrees of freedom in various other texts [1—5].

5.2 DEVICE BASICS We begin with a brief overview of electromechanical devices. These typically consist of three subsystems: the electrical drive circuitry, an electromechanical coupling subsystem, and a mechanical subsystem. The 335

336

CHAPTER 5 Electromechanical Devices

interrelation of these subsystems is depicted in the following diagram: Electrical Electromechanical Mechanical drive circuit coupling subsystem subsystem In many devices, a given component can be considered to be an element of two or more subsystems. For example, in a moving coil actuator the coil can be considered to be an element of all three subsystems. For our purposes, electromechanical devices can be classiﬁed into one of the following two categories: 1. Devices in which all the electrical circuit elements are stationary, and the moving element is either a soft magnetic member or a permanent magnet. These are sometimes referred to as magnetic circuit actuators, or moving magnet actuators, respectively. 2. Devices in which the moving member is a coil that is also part of the electrical circuit. These are sometimes referred to as moving coil actuators. An example of the ﬁrst type of device is shown in Fig. 5.1a. This is a magnetic circuit actuator. In this device the electrical drive circuit consists of a voltage source V , an external resistor R that limits the Q current, and a lossless coil. All of these circuit elements are stationary. The electromechanical coupling subsystem is a magnetic circuit that consists of the lossless coil, a soft magnetic core, and a movable soft magnetic bar that is separated from the core by a gap. The mechanical subsystem consists of the movable bar that is represented by a mass m, and an attached spring. The magnetic circuit is housed in a frictionless frame. The core and spring are ﬁxed to the frame, and the bar is free to move as shown. To activate the device, a voltage is applied to the coil, creating a magnetic ﬁeld through the circuit. This, in turn, gives rise to a force that acts on the bar to pull it toward the core. The motion is resisted by the force exerted by the spring. An example of the second type of device is shown in Fig. 5.1b. In this device all the circuitry except for the coil is stationary (only the coil is shown). The leads to the coil are ﬂexible and allow the coil to rotate relative to the circuit. The coil is positioned in an external ﬁeld that is provided by a wedge-shaped magnet. When the coil is energized, its side elements experience a Lorentz force. This force gives rise to a torque that causes the coil to rotate as shown. Usually, a mechanical component such as a spring is used to provide a restoring torque.

5.3 QUASI-STATIC FIELD THEORY

337

Electromechanical devices: (a) magnetic circuit actuator; and (b) moving coil actuator. FIGURE 5.1

5.3 QUASI-STATIC FIELD THEORY In this section we study quasi-static ﬁeld theory and its application to electromechanical devices. Quasi-static ﬁeld theory governs the behavior of slowly varying ﬁelds, and applies to conventional electromechanical devices that operate at low frequencies and relatively low velocities. This section is divided into two parts. In the ﬁrst part, we consider quasistatic ﬁeld theory in a stationary reference frame and derive Kirchhoff’s circuit laws for RLC circuits. In the second part, we consider quasi-static ﬁeld theory as observed in two different inertial reference frames (in motion relative to one another). We obtain the equations for transforming the ﬁelds from one reference frame to the other. We then use these

338

CHAPTER 5 Electromechanical Devices

to determine the voltage induced in a conductor (coil) as it moves through a stationary magnetic ﬁeld.

5.3.1 Stationary Reference Frames We have already studied quasi-static ﬁeld theory in a stationary reference frame in Section 2.4. We restate the key results here for convenience. In quasi-static ﬁeld theory the displacement current is negligible (D/t : 0), and Maxwell’s equations reduce to: Quasi-Static Field Theory Differential Form

Integral Form

;H : J

H · dl :

!

·B:0 ;E : 9

B t

·D:

J · ds 1

B · ds : 0

1

E·dl:9

!

1

D · ds :

1

B · ds t

(5.1)

dv.

4

For electromagnetic devices with time varying or steady currents the free charge density is negligible and we neglect D altogether. The ﬁrst equation in (5.1) is a generalization of Ampere’s law for slowly varying currents. It implies that · J : 0,

(5.2)

which is Eq. (2.6) for the conservation of charge with the /t term set to zero. Using Gauss’ law, we rewrite Eq. (5.2) in integral form,

J · ds : 0, (5.3) 1 where S is a closed surface. If we apply Eq. (5.3) to an electrical circuit we obtain Kirchhoff’s current law, which states that the algebraic sum of the currents entering a circuit node must be zero. Speciﬁcally, if there are n wires at a node then i :0 L L

(Kirchhoff’s current law),

(5.4)

339

5.3 QUASI-STATIC FIELD THEORY

FIGURE 5.2

The RLC circuit.

where i is the current ﬂowing into the node in the nth wire (i is positive L L into the node and negative out of the node). Next, we derive Kirchhoff’s voltage law for the RLC circuit shown in Fig. 5.2. First, apply Faraday’s law to a closed contour C that passes through the circuit,

E · dl : 9

B · ds. t

(5.5) ! 1 We evaluate w E · dl clockwise around the circuit. In doing so, we ! decompose the integral into integrals over the individual components,

> \

E · dl ; Q

source

E · dl ;

resistor

E · dl ;

inductor

E · dl : 9

1

B · ds. t

(5.6)

capacitor

The line integrals in Eq. (5.6) are evaluated as described in Section A.4. We assume that the source is ideal with no internal resistance. The ﬁeld E in the source can be represented as the gradient of a potential Q E : 9. Therefore, Q > > E · dl : 9 · dl Q \ \

source

where and > \

: 9( 9 ), > \ are the potentials at the positive and negative

340

CHAPTER 5 Electromechanical Devices

terminals of the source, respectively. The source voltage V is deﬁned as Q the potential difference V : 9 and, therefore, Q > \ > V Y9 E · dl. (5.7) Q Q \ Thus, Eq. (5.6) becomes

V : Q

E · dl ;

resistor

E · dl ;

capacitor

E · dl ;

1

B · ds. t

(5.8)

inductor

In the resistor, J : E (Ohm’s law) and, therefore,

E · dl :

resistor

J · dl

resistor

: iR.

(5.9)

The same analysis applies to the inductor,

E · dl :

inductor

J

· dl

inductor

: iR , (5.10) where and R are the conductivity and resistance of the inductor. For the capacitor, we have

E · dl :

1 C

i dt.

(5.11)

capacitor

Substituting Eqs. (5.9), (5.10) and (5.11) into Eq. (5.8) yields 1 V : i(R ; R ) ; Q C

i dt ;

B · ds. t

(5.12)

1 Now, consider the last term in Eq. (5.12). This can be written as d dt

B · ds : 1

d" , dt

(5.13)

341

5.3 QUASI-STATIC FIELD THEORY

where " is called the ﬂux linkage. That is, " is the total ﬂux through (linking) the circuit, "Y

B · ds (ﬂux linkage). 1 The term d"/dt represents the voltage induced in the circuit by a time rate of change of magnetic ﬂux through it. For a simple RLC circuit (with no moving members), " is a function of the current only and, therefore, d" d"(i) di : . dt di dt

(5.14)

We deﬁne the circuit inductance L to be LY

d"(i) di

(inductance),

and then obtain d" di :L . dt dt

(5.15)

Finally, substitute Eq. (5.15) into Eq. (5.12) to obtain 1 V : i(R ; R ) ; Q C

i dt ; L

di . dt

(5.16)

This is Kirchhoff’s voltage law for an RLC circuit. In electromechanical devices, the ﬂux linkage " is no longer a function of the current only. Instead, it is a function of both the current and the position of a moving member. Thus, for the analysis of electromechanical devices the preceding derivation needs to be generalized to include the effects of electromechanical coupling. We discuss this in detail in Section 5.4.1.

5.3.2 Moving Reference Frames In our study of electromechanical devices we will need to determine the voltage induced in a coil as it moves through a stationary magnetic ﬁeld. This requires an understanding of quasi-static ﬁeld theory as observed in two different inertial reference frames (in motion relative to one another). For our purposes, an inertial reference frame is one that moves with constant velocity. In this section, we derive the transformation laws

342

CHAPTER 5 Electromechanical Devices

FIGURE 5.3

Inertial reference frames in relative motion along one axis.

for quasi-static ﬁelds as observed in two different inertial reference frames. Consider two inertial reference frames, a stationary ‘‘laboratory’’ reference frame O with coordinates (x, y, z, t), and a moving reference frame O with coordinates (x, y, z, t). Frame O moves with velocity u with respect to O as shown in Fig. 5.3. Observers in O and O measure different values for the electromagnetic ﬁelds. Rigorous comparisons of these measurements can be made if one knows how the ﬁelds transform from O to O or vice versa. The transformation laws follow from the theory of special relativity. A detailed presentation of this theory is beyond the scope of this text and is also unnecessary for our presentation. For our purposes, it sufﬁces to state the underlying principles of the theory, and then simply write down the coordinate transform relations. The interested reader can ﬁnd detailed presentations of special relativity in numerous excellent texts [6]. The basic postulates of special relativity are as follows: 1. The laws of physics are the same in all inertial reference frames. 2. Observers in inertial reference frames measure the same value for the speed of light. The ﬁrst postulate implies that the quasi-static ﬁeld equations must have the same form in both O and O as shown in Eqs. (5.17):

343

5.3 QUASI-STATIC FIELD THEORY

Quasi-Static Equations Reference Frame O

Reference Frame O

;H : J

;H : J

·B:0 B ;E : 9 t ·D:

· B : 0 B ;E : 9 t · D : .

(5.17)

Here, is the del operator with respect to the primed coordinate, :

x ; y ; z . x y z

Our goal is to determine the transformation laws for the ﬁelds going from O to O and vice versa. We can determine these if we know the transformation relations for the coordinate variables. These follow from the second postulate of special relativity and are given by t : (t 9 #x/c) x : (x 9 #ct) y : y

(5.18)

z : z, where :

1 (1 ; #

,

and u #: . c Here, c is the speed of light in a vacuum. We are interested in low-velocity electromechanical devices for which u c. Therefore, #0

(5.19)

1.

(5.20)

and

We substitute Eqs. (5.19) and Eq. (5.20) into Eq. (5.18), and ﬁnd that for low velocities the Lorentz transformation relations Eq. (5.18) reduce to

344

CHAPTER 5 Electromechanical Devices

FIGURE 5.4

Inertial reference frames in relative motion.

the Galilean transform relations, t : t x : x 9 ut y : y

(5.21)

z : z. This analysis generalizes to the case of an arbitrary velocity as shown in Fig. 5.4. For this case we ﬁnd that t : t and x : x 9 ut, or x : x 9 u t V y : y 9 u t (5.22) W z : z 9 u t. X Now that we know the transform relations for the coordinates, we can determine the corresponding relations for the ﬁeld equations. These will ultimately enable us to compare the ﬁelds in the two reference

345

5.3 QUASI-STATIC FIELD THEORY

frames. We start by considering a scalar-valued function f of the primed coordinates, f (x, y, z, t). By virtue of Eq. (5.22), f can be considered to be a function of the unprimed coordinates, f (x, y, z, t) : f (x(x, t), y(y, t), z(z, t), t(t)). We use the chain rule to determine the derivatives of f with respect to the unprimed coordinates. We consider spatial derivatives ﬁrst and ﬁnd that f f x : x x x :

f . x

It follows that f : f .

(5.23)

Next, we consider the time derivative. From the chain rule we have f f t f x f y f z : ; ; ; y t z t t t t x t :

f f f f 9 u 9 u 9 u . t x V y W z X

This can be rewritten as f f : 9 u · (f), t t or, alternatively, as f f : ; u · (f). t t

(5.24)

The relations (5.23) and (5.24) relate the derivatives in O and O for scalar-valued functions. A similar analysis applies to vector-valued functions. Let A(x, y, z, t) be a vector-valued function of the primed coordinate variables. This can also be considered to be a vector-valued function of the unprimed coordinates A(x, y, z, t) : A(x(x, t), y(y, t), z(z, t), t(t)). It is easy to check that · A : · A,

(5.25)

346

CHAPTER 5 Electromechanical Devices

and that ;A : ;A.

(5.26)

Similarly, by taking the time derivative we ﬁnd that A A : ; (u · )A. t t This can be rewritten as A A : ; u( · A) 9 ;(u;A), t t

(5.27)

where we have used the identity ;(A;B) : (B · )A 9 (A · )B ;A( · B) 9 B( · A). Finally, substitute Eqs. (5.25), (5.26), and (5.27) into the quasi-static equations for O in Eq. (5.17) and obtain ;H : J · B : 0 ;(E 9 u;B) : 9

B t

(5.28)

· J : 0. We have used · B : 0 in the third equation of Eq. (5.28). By comparing Eqs. (5.17) and (5.28) we ﬁnd that the ﬁelds transform as follows: Transformation Relations H : H

(5.29)

B : B

(5.30)

E : E ; u;B

(5.31)

J : J.

(5.32)

In Eqs. (5.29)—(5.32) the variables (H, B, E, u, J) are measured in the stationary frame O, and the variables (H, B, E, J) are measured in the moving frame O [4]. For these relations to hold, the variables must be compared at the same physical point in space. For example, from Eqs. (5.22) and (5.31) we have E(x, y, z, t) : E(x 9 u t, y 9 u t, z 9 u t, t) V W X ;u;B(x 9 u t, y 9 u t, z 9 u t, t). V W X

347

5.3 QUASI-STATIC FIELD THEORY

Notice that we have assumed that u is a constant (independent of position). The relations (5.29)—(5.32) give the transformations for the ﬁelds and the source. However, we still need to know the constitutive relations in O and O. We apply a general principle that says that the constitutive relations for moving media are the same as for stationary media when they are written in terms of the ﬁelds deﬁned in an inertial reference frame moving with the media [4]. For example, for a linear, homogeneous and isotropic media at rest in O we have J : E and B : H. However, in O (which is moving with respect to the media) the relations are modiﬁed by virtue of Eqs. (5.29)—(5.32). Speciﬁcally, we ﬁnd that Constitutive Relations Reference Frame O

Reference Frame O

J : (E ; u;B)

J : E

B : H

B : H.

(5.33)

In our study of electromechanical devices we will need to determine the voltage induced in a coil as it moves through a stationary magnetic ﬁeld. Therefore, we discuss this in some detail. Consider a reference frame O in which there is a static (but not necessarily uniform) B-ﬁeld. Consider an open conductor moving with velocity u relative to O. Assume that the conductor has a conductivity (as measured by observers in both O and O). We want to determine the voltage induced across the terminals of the conductor as measured in O. We apply Faraday’s law to a stationary contour C that passes through the conductor at a given (but arbitrary) instant in time,

E · dl : 9

d dt

B · ds. (5.34) ! 1 We label the terminals a and b, and then decompose the left-hand side of Eq. (5.34) into integrals across the terminals and the conductor itself,

E · dl : !

?

E · dl ;

@

E · dl,

?

terminals

: 9V ;

@

conductor

@

E · dl,

?

conductor

(5.35)

348

CHAPTER 5 Electromechanical Devices

where V is the terminal voltage (as measured in O). As for the remaining integral, from Eqs. (5.31) and (5.32) we have E : E 9 u;B and E : J/ : J/. Therefore, J E : 9 u;B

(moving conductor).

(5.36)

Notice that all the terms in Eq. (5.36) are measured in O (the conductor is moving with respect to O). As the conductor is open, there is no current through it (J : 0) and, therefore, E : 9u;B

(5.37)

Combining Eqs. (5.34), (5.35) and (5.37) gives us

V:9

@

(u;B) · dl ;

d dt

B · ds. (5.38) ? 1 Because B, C, S, and ds are constant in O, the last term in Eq. (5.38) is zero and the terminal voltage reduces to

V:9

@

(u;B) · dl. (5.39) ? It is important to note that if there were a time-varying current in the conductor, the ﬂux through the circuit due to the current would add to the voltage via the last term in Eq. (5.38). We demonstrate this theory in the following examples. EXAMPLE 5.3.1 A conductive bar is in sliding contact with a pair of stationary conducting rails as shown in Fig. 5.5. The bar is moving through a constant uniform B-ﬁeld with a constant velocity u relative to the rails. Determine the voltage across the rails. Assume that the bar and rails are good conductors. SOLUTION 5.3.1 We choose a reference frame O at rest with respect to the rails, and a stationary contour C (dotted line) that runs clockwise through the rails, and passes through the bar at a single instant in time (Fig. 5.5). Apply Faraday’s law (5.5) to C,

E · dl : 9

d dt

B · ds.

(5.40)

! 1 As B, C, S, and ds are constant in O we have d dt

1

B · ds : 0.

(5.41)

349

5.3 QUASI-STATIC FIELD THEORY

FIGURE 5.5

Conductive bar moving along conductive rails.

Substitute this into Eq. (5.40) and decompose the right-hand side into integrals over the various segments and obtain,

>

E · dl ;

\

terminals

E · dl ;

upper rail

E · dl ;

moving bar

E · dl : 0

(5.42)

lower rail

As the upper and lower rails are good conductors ( -) we have E 0 and, therefore,

E · dl :

upper rail

E · dl : 0.

(5.43)

lower rail

In the moving bar we have

E · dl :

(E 9 u;B) · dl,

moving bar

but E : J/ : J/ 0 and, therefore,

E · dl : 9 (u;B) · dl.

(5.44)

moving bar

To evaluate Eq. (5.44), we use u : ux , B : Bz , and dl : dyy (Fig. 5.5). Therefore, 9 (u;B) · dl : 9 9uB dy F : 9uBh,

350

CHAPTER 5 Electromechanical Devices

or

E · dl : 9uBh.

(5.45)

moving bar

Finally, substitute Eqs. (5.43) and (5.45) into Eq. (5.42) and obtain V : 9uBh, where V is the terminal voltage,

V:9

>

E · dl.

)

\

EXAMPLE 5.3.2 Determine the lift and drag force on a wire carrying a current i and moving with velocity u relative to a conductive sheet of thickness as shown in Fig. 5.6a [7]. SOLUTION 5.3.2 We choose a reference frame at rest with respect to the wire (Fig. 5.6b). In this reference frame, the magnetic ﬁeld B of the wire is constant (time invariant), and the sheet moves with velocity u : 9ux . The sheet experiences a time varying magnetic ﬁeld. The current density induced at a point x in the sheet is J (x) : 9uBR(x), W X where BR(x) is the total magnetic ﬁeld at x, that is, W BR(x) : B(x) ; BG(x). W W W Here, B (x) and BG (x) are the ﬁelds due to the current in the wire, and the W W induced current in the sheet, respectively. Speciﬁcally, we have i x B(x) : . W 2 x ; h and BG(x) : W 2

K ()) X d), x9) \

(5.46)

where K (x) : 9uBR(x) (5.47) W X is the current density per unit length ( is the thickness of the sheet). Notice

351

5.3 QUASI-STATIC FIELD THEORY

Wire moving above a conductive plate: (a) wire moving and plate stationary; and (b) wire stationary and plate moving [7].

FIGURE 5.6

that Eq. (5.46) contains a singularity. As K ()) is continuous, we evaluate the X integral using the Cauchy Principal value. Substituting Eq. (5.46) into Eq. (5.47) yields the following integral equation for K (x), X u K ()) 2u X K (x) : 9 d) 9 B(x), (5.48) X w x9) w W \ where w : 2/( ). We solve Eq. (5.48) using the Fourier transforms, eIVK (x) dx, F[K ] : X X \ and

F

K ()) 1 X d) : F[K ]F . X x9) x \

352

CHAPTER 5 Electromechanical Devices

The second transform follows from convolution theory. In addition, notice that F

1 i k 0 : x 9i k 0.

We apply these to Eq. (5.48) and obtain

F[B] u 1 2u W 19 F . F[K ] : 9 X w x w 1 ; (u/w) Next, we apply an inverse transform to Eq. (5.49) and ﬁnd that

(5.49)

2u 1 u B()) W K (x) : 9 B (x) 9 d) . (5.50) X w (1 ; u/w) W w )9x \ The last term in Eq. (5.50) can be rewritten in a more useful form using B : ;A. After some analysis we ﬁnd that

2u 1 u B(x) ; B(x) , K (x) : 9 X w V w (1 ; u/w) W

(5.51)

where i h B(x) : , V 2 x ; h and i x . B(x) : W 2 x ; h Notice that Eq. (5.51) is not limited to a single moving current ﬁlament. Rather, it is valid for a system of moving ﬁeld sources provided they are of limited extent, produce a one-dimensional (1D) current pattern, and travel at the same velocity. The lift and drag forces on the wire are equal and opposite to the forces on the sheet. The force density on the moving sheet is the Lorentz force density (2.23), f : K (x)z ;(B(x)x ; B(x)y ). V W X Therefore, f : 9K (x)B(x) W V X and f : K (x)B(x). V W X

353

5.3 QUASI-STATIC FIELD THEORY

We integrate over the sheet to obtain the total lift and drag force

>

K (x)B(x) dx V X \ ih > K (x) X dx : x ; h 2 \ i u : 4h w ; u

F : *

:F

u w ; u

(Lift force)

and w F : F " u *

(Drag force),

where i F : . 4h A normalized plot of these forces is shown in Fig. 5.7. Notice that the drag force reaches a maximum when u : w. )

FIGURE 5.7

Normalized plot of lift and drag forces vs normalized velocity [7].

354

CHAPTER 5 Electromechanical Devices

5.4 ELECTRICAL EQUATIONS The electrical equations for an electromechanical device follow from circuit theory, which in turn is based on quasi-static ﬁeld theory. In this section, we derive a generalized version of Kirchhoff’s voltage law that takes into account the effects of electromechanical coupling. We study two different types of circuits, stationary and moving coil. In a stationary circuit, all the circuit components are at rest with respect to one another. For these circuits, we apply quasi-static ﬁeld theory in a reference frame ﬁxed with respect to the circuit. The electromechanical coupling is accounted for by a ﬂux linkage term that takes into account the voltage induced in the circuit by a time rate of change of magnetic ﬂux through it. This change in ﬂux is due to a change in the magnetic ﬁeld that can be caused by a change in the current and/or by the movement of a ferromagnetic member that alters the ﬁeld. Mechanical motion is taken into account by considering the ﬂux to be a function of both the circuit current and mechanical displacement. In a moving coil circuit, a coil is moving with respect to an external magnetic ﬁeld. To analyze this kind of circuit, we apply quasi-static ﬁeld theory in a reference frame that is at rest with respect to the magnetic ﬁeld. Mechanical motion is taken into account by the voltage induced in the coil as it moves through the ﬁeld. Finally, it is important to note that the following analysis applies to magnetically linear devices.

5.4.1 Stationary Circuits In this section we study electromechanical devices with stationary electrical circuits. Such a device is shown in Fig. 5.8. It consists of a voltage source V , a resistor R, and a magnetic circuit with a ﬁxed coil Q and a moving soft magnetic bar that alters the ﬂux through the coil when it moves. We analyze the electrical circuit using quasi-static ﬁeld theory. Speciﬁcally, we choose a reference frame at rest with respect to the circuit and a ﬁxed integration contour C along the circuit. At any point along the circuit the electric ﬁeld is given by E:9

A 9 , t

(5.52)

where 9 is due to the accumulation of charge along the circuit, and A/t is due to the time-varying current. We evaluate the line integral

355

5.4 ELECTRICAL EQUATIONS

Electromechanical actuator circuit.

FIGURE 5.8

of Eq. (5.52) clockwise along C and obtain

A · dl 9 t

E · dl : 9

· dl. ! ! ! The ﬁrst term can be decomposed into integrals over the individual components

>

E · dl ;

\ source

E · dl : 9

E · dl ;

resistor

A · dl 9 t

· dl.

(5.53)

coil

Notice that w · dl : 0 because it entails the integration of a conservative vector ﬁeld around a closed path. Also notice that although w A/t · dl is evaluated around the entire path, the major contribution to it is from the inductor. We assume that the source is ideal with no internal resistance and deﬁne the voltage impressed at the terminals in the usual way:

V Y9 Q

> \

E · dl. Q

(5.54)

Thus, Eq. (5.53) becomes V : Q

E · dl ;

resistor

E · dl ; coil

A · dl. t

(5.55)

356

CHAPTER 5 Electromechanical Devices

In the resistor J : E (Ohm’s law). Therefore,

E · dl :

resistor

J · dl,

(5.56)

resistor

: iR. Similarly, in the coil J :

E, and, therefore,

E · dl : iR

.

(5.57)

coil

Substituting Eqs. (5.56) and (5.57) into Eq. (5.55) yields V : i(R ; R ) ; Q

A · dl. t

(5.58)

Now, consider the last term in Eq. (5.58). We can rewrite this as follows:

A · dl : t :

1

1 d : dt :

A ; · ds t

B · ds t B · ds

1

d" , dt

(5.59)

where "Y

B · ds (ﬂux linkage). 1 The symbol " denotes the ﬂux linkage, that is, the total ﬂux through (linking) the circuit. To determine " we need to evaluate B · ds over an 1 area bounded by the circuit path in accordance with Stokes’ theorem. However, this integral can be difﬁcult to evaluate depending on the topology of the circuit. For a simple single-turn current loop, the ﬂux is denoted by and, therefore, " : . For a tightly wound N turn coil the ﬂux linkage is " : N

(N turn coil),

where is the ﬂux through each turn.

357

5.4 ELECTRICAL EQUATIONS

We digress brieﬂy to discuss the second and third step in Eq. (5.59). Notice that we have replaced the partial time derivative inside the integral by a total time derivative outside the integral. We do this because inside the integral, B is a function of both time-independent coordinate variables and other time-dependent variables (including t itself). However, the integration effectively eliminates the time-independent variables, and leaves only the time-dependent variables. Therefore, the total time derivative is appropriate outside the integral. We return to the derivation. Substitute Eq. (5.59) into Eq. (5.58) and obtain V : i(R ; R ) ; Q

d" . dt

(5.60)

mechanical coupling

This is a generalization of Kirchhoff’s voltage law. The term d"/dt in Eq. (5.60) requires some discussion because it is key to the electromechanical coupling. This term represents the voltage induced in the circuit by a time rate of change of magnetic ﬂux through the circuit. The change in ﬂux can be due to a change in circuit current and/or the movement of a ferromagnetic member that alters the ﬁeld. To account for this we write " as a function of both the circuit current and the position of the member. Let x and denote the position of the member when it executes linear and rotational motion, respectively. Then ":

"(i, x) (linear motion) "(i, ) (rotational motion).

Given this functional dependence, the total time derivative of " is

"(i, x) di "(i, x) dx ; (linear motion) i dt x dt

d" : dt "(i, ) di "(i, ) d ; (rotational motion) i dt dt or

d" : dt

L

di "(i, x) ; v dt x

di "(i, ) L ; dt

(linear motion) (5.61) (rotational motion)

358

CHAPTER 5 Electromechanical Devices

where L is the inductance and which is deﬁned by LY

" i

(inductance).

(5.62)

The variables v : dx/dt and : d/dt are the linear and angular velocities, respectively. It is important to note that the deﬁnition of inductance Eq. (5.62) applies only to magnetically linear devices. There is no simple deﬁnition of L for nonlinear devices [8]. Finally, substitute Eq. (5.61) into Eq. (5.60) and obtain

Electrical $ Equations

di(t) "(i, x) V (t) : i(t)(R ; R ) ; L ; v(t) Q dt x (linear motion)

di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt

.

(5.63)

(rotational motion)

These are the electrical equations of motion for stationary circuits with time varying currents and a moving ferromagnetic member.

5.4.2 Moving Coils In this section we derive Kirchhoff’s voltage law for circuits that have a coil moving through an external magnetic ﬁeld B . All the other circuit components are stationary with respect to the ﬁeld. Let O denote the stationary reference frame. The coil moves with velocity u through B , which is constant (but not necessarily uniform) with respect to O. Such a circuit is shown in Fig. 5.9. This ﬁgure shows a linear actuator consisting of a conductive bar of mass m in sliding contact with a pair of stationary conducting rails. The rails are connected to a voltage source V and a resistor R that limits the current. The bar is moving through a Q constant uniform B-ﬁeld with a time-dependent velocity u(t) relative to the rails. This circuit is analyzed in Example 5.6.2. We derive Kirchhoff’s voltage law for a circuit with a moving coil by applying Faraday’s law to a stationary contour C that passes through the circuit and coincides with the coil for a given position of its travel (the dotted line in Fig. 5.9). Thus, C coincides with the entire circuit at a single instant in time. We evaluate

E · dl : 9 !

d dt

1

B · ds

(5.64)

359

5.4 ELECTRICAL EQUATIONS

Moving conductor actuator.

FIGURE 5.9

clockwise around the contour and obtain

>

E · dl ;

\

source

E · dl ;

resistor

E · dl : 9 coil

d dt

B · ds.

(5.65)

The integrals over the ﬁxed components (source and resistor) follow from Eqs. (5.54)—(5.57). We obtain V : iR ; Q

E · dl ;

d dt

B · ds. (5.66) The coil is moving with velocity u through B . The E-ﬁeld in it as measured in O is given by Eq. (5.36), J E : 9 u;B . Therefore,

where R and obtain

E · dl :

J 9 u;B · dl,

: iR 9

(u;B ) · dl, (5.67) is the resistance of the coil. Substitute Eq. (5.67) into Eq. (5.66)

V : i(R ; R ) 9 Q

d (u;B ) · dl ; dt

B · ds.

(5.68)

Now consider the last term in Eq. (5.68). The dominant contribution to this term comes from the coil. Moreover, as the system is magnetically linear we can write B as a superposition of two ﬁelds: the ﬁeld B due G

360

CHAPTER 5 Electromechanical Devices

to the current i and the external ﬁeld B , B:B ;B . G Therefore, the last term in Eq. (5.68) can be written as d dt

B · ds : :

d dt

d B · ds ; G dt

d d " (i) ; dt dt

:L

di d ; dt dt

B · ds

B · ds

B · ds, (5.69) where " (i) : B · ds, and we have used the deﬁnition of inductance G " (i) L : . i Finally, we substitute Eq. (5.69) into Eq. (5.68) and ﬁnd that di V : i(R ; R ) ; L 9 Q dt

d (u;B ) · dl ; dt

B · ds. (5.70) This is a generalization of Kirchhoff’s voltage law that takes into account the voltage induced in a coil as it moves through B . If B is static, d B · ds : 0 and Eq. (5.70) reduces to dt di V : i(R ; R ) ; L 9 (u;B ) · dl. (B static). (5.71) Q dt It is important to note that the line integral in Eq. (5.71) is evaluated as described in Section A.4 with the direction of integration in the direction of conventional current ﬂow. The voltage equation (5.71) applies to a coil in linear motion. However, it can be generalized to the case of a coil that executes rotational motion. Speciﬁcally, if a segment dl of the coil is rotating with an angular velocity at a distance r with respect to a ﬁxed reference frame, then its linear velocity is

u : ;r.

(5.72)

We substitute Eq. (5.72) into Eq. (5.71) and obtain the circuit equation for rotational motion, di V (t) : i(R ; R ) ; L 9 Q dt

[(;r);B ] · dl

(B static). (5.73)

361

5.5 MECHANICAL EQUATIONS

It is important to note that when evaluating Eq. (5.73), the velocity (;r) must be evaluated before the cross product with B is taken. If the rotation is with respect to a speciﬁc axis (which we label the z-axis), then : z , and Eq. (5.73) can be rewritten as di V (t) : i(t)(R ; R ) ; L 9 Q dt

[(;r );B ] · dl ,

(B static), (5.74)

where r is a vector from (and perpendicular to) the z-axis to the , segment dl.

5.5 MECHANICAL EQUATIONS In this section we review the mechanical equations of motion for electromechanical devices that execute linear or rotational motion with a single degree of freedom. Such linear or rotational motion is described by a single coordinate variable that we denote by x or , respectively. The equations for these variables follow from Newton’s second law of motion, m

dx : F(i, x) dt

(linear motion),

(5.75)

(rotational motion),

(5.76)

or d J : T(i, ) K dt

where m is the mass, J is the moment of inertia about the axis of K rotation, and F(i, x) and T(i, ) are the force and torque. The moment of inertia of a point mass m at a ﬁxed distance r about the axis of rotation is J : mr. If the mass is distributed, then J is computed via integraK K tion, J : K

4

r dv,

where and V are the density and volume of the mass, respectively,

and r is the distance of the element dv from the axis of rotation. For

362

CHAPTER 5 Electromechanical Devices

example, if a rigid body of density is rotating about the z-axis, its

moment of inertia is J : K

(x ; y) dv. (5.77)

4 Notice that in Eqs. (5.75) and (5.76) we have chosen the current i and position x or as the dependent variables. This is often a convenient choice, but other variables may be more convenient depending on the application. Regardless of the choice of variables, for lumped-parameter analysis we need to determine analytical expressions for F(i, x) and T(i, ). This is usually accomplished using one of the following methods: 1. Derive an analytical solution for the magnetic ﬁeld and determine the force density using the Lorentz force formula or the Maxwell stress tensor. Then, perform the appropriate volume integration to obtain the total force or torque. This approach is most useful in the analysis of moving coil actuators. It is discussed in what follows. 2. Derive an analytical expression for the ﬁeld energy (or coenergy) and then determine the force or torque via differentiation. This approach is especially useful in the analysis of magnetic circuit or moving magnet devices. It is discussed in Section 5.7. 3. Discretize the independent variables into a discrete set of values, use a numerical method such as FEA to obtain force or torque data for these values, and then ﬁt the data to an analytical formula. This is called a hybrid approach and is discussed in Section 5.15. The ﬁrst method is the most direct approach for determining the force or torque. It is especially useful in the analysis of moving coil actuators. The force and torque imparted to the coil are obtained using the Lorentz force formulas Eqs. (3.47) and (3.52) (Section 3.2.2). Speciﬁcally, if a coil carries a current i and moves with velocity u through an external ﬁeld B , the force and torque imparted to it are F:i

(dl;B ),

(5.78)

and T:i

r;(dl;B ), (5.79) where r is the vector from the origin about which the torque is computed

5.6 ELECTROMECHANICAL EQUATIONS

363

and integration is over the length of the coil. When evaluating Eq. (5.79), the force (dl;B ) must be evaluated before the cross product with r is taken. Equations (5.78) and (5.79) are vector equations. We are interested in both the force in the x direction, and the torque about a given axis, which we choose to be in the z-axis. The magnitudes of force and torque in these directions are F :i V

(dl;B ) · x ,

(5.80)

[r ;(dl;B )] · z , ,

(5.81)

and T :i X

where r is a vector from (and perpendicular to) the z-axis to the , segment dl.

5.6 ELECTROMECHANICAL EQUATIONS In this section we summarize the equations of motion for electromechanical devices. We have seen that the electrical and mechanical equations are separately second-order equations in the independent variables i and x or , respectively. These equations are coupled and therefore represent a second-order system of equations. These coupled equations can sometimes be solved directly. However, for practical analysis it is often easier to reduce them to a ﬁrst-order system of equations. Such systems lend themselves to various numerical timestepping solution procedures such as the Euler or Runge-Kutta methods. These are discussed in Appendix C.

5.6.1 Stationary Circuits In this section we summarize the equations of motion for electromechanical devices with stationary circuits. These include magnetic circuit and moving magnet actuators. The electrical and mechanical equations for these devices were derived in the foregoing text. We repeat

364

CHAPTER 5 Electromechanical Devices

them here for convenience,

Electrical $ Equations

and

di(t) "(i, x) V (t) : i(t)(R ; R ) ; L ; v(t) Q dt x (linear motion)

di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt

.

(5.82)

(rotational motion)

dx(t) : F(i, x) (linear motion) dt Mechanical $ . d(t) Equations J : T(i, ) (rotational motion) K dt m

(5.83)

Since the same variables i and x or appear in both sets of equations, they need to be solved simultaneously. Thus, the analysis of an electromechanical device entails the solution of the following coupled second-order equations: Linear Motion di(t) "(i, x) ; v(t) V (t) : i(t)(R ; R ) ; L Q x dt dx(t) m : F(i, x) dt

(5.84)

or Rotational Motion di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt d(t) J : T(i, ), K dt

(5.85)

For practical analysis is convenient to rewrite the second-order equations in terms of the following ﬁrst-order systems,

365

5.6 ELECTROMECHANICAL EQUATIONS

Linear Motion di(t) 1 : dt L

V (t) 9 i(t)(R ; R ) 9 Q dv(t) 1 : F(i, x) dt m dx(t) : v(t) dt

(5.86)

(5.87)

"(i, x) v(t) x

and Rotational Motion di(t) 1 : dt L

V (t) 9 i(t)(R ; R ) 9 Q d (t) 1 : T(i, ) J dt K d(t) : (t). dt

"(i, ) (t)

Equations (5.86) and (5.87) constitute initial-value problems that need to be solved subject to the following initial conditions, i(t ) : i x(t ) : x v(t ) : v

(linear motion)

(5.88)

and i(t ) : i (t ) : (rotational motion) (5.89) (t ) : , respectively. Occasionally, the initial-value problems can be solved analytically. More often, their solution requires the application of a numerical time-stepping procedure such as the Euler or Runge Kutta method (Appendix C). EXAMPLE 5.6.1 Apply Euler’s method to Eq. (5.87). SOLUTION 5.6.1 We reduce Eq. (5.87) to a set of algebraic equations using Euler’s method as discussed in Appendix C. In Euler’s method, derivatives are

366

CHAPTER 5 Electromechanical Devices

approximated by a simple difference quotient, for example, df f(t ; t) 9 f(t) . t dt

(5.90)

We apply Eq. (5.90) to Eq. (5.87) and obtain i(t ; t) : i(t) ;

t "(i, ) V (t) 9 i(t)(R ; R ) 9 (t) Q L

t T(i, ) J K (t ; t) : (t) ; t (t).

(t ; t) : (t) ;

(5.91)

To solve Eq. (5.91) we need expressions for "(i, ) , L:

"(i, ) , i

(5.92) (5.93)

and T(i, ).

(5.94)

Once these are known, Eq. (5.91) can be solved using the following iterative scheme: i(n ; 1) : i(n) ;

t "(i(n), (n)) V (n) 9 i(n)(R ; R ) 9 (n) Q L t T(i(n), (n)) J K (n ; 1) : (n) ; t (n).

(n ; 1) : (n) ;

(5.95)

The solution method is as follows: Choose a value for t. Then, given the initial conditions [i(0), (0), (0)] use Eq. (5.95) to obtain [i(1), (1), (1)], which gives the solution at t : t. Repeat this process to obtain [i(2), (2), (2)], [i(3), (3), (3)], . . . , [i(n), (n), (n)], [i(n ; 1), (n ; 1), (n ; 1)]. These are the solution values at t : 2t, 3t, . . . , nt, (n ; 1)t, . . . , respectively. )

5.6.2 Moving coils In this section we summarize the equations of motion for electromechanical devices with coils that move through an external ﬁeld B

367

5.6 ELECTROMECHANICAL EQUATIONS

(moving coil actuators). The electrical and mechanical equations were derived in the preceding; we repeat them here for convenience:

Electrical $ Equations

di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt

(u ; B ) · dl

(linear motion) di(t) 9 V (t) : i(t)(R ; R ) ; L Q dt

(rotational motion)

[(;r);B ] · dl

(5.96)

and

Mechanical $ Equations

m

dx(t) :i dt

d(t) J :i K dt

(dl;B ) · x ; F (x)

(linear motion)

[r;(dl;B )] · z ; T ().

(rotational motion) (5.97)

In Eq. (5.97) the term T () represents the mechanical torque imposed

on the system. The electrical and mechanical equations are coupled and can be grouped into the following second-order systems of equations, Linear Motion di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt m

dx(t) : i(t) dt

(u ; B ) · dl

(dl ; B ) · x ; F (x)

(5.98)

368

CHAPTER 5 Electromechanical Devices

and Rotational Motion di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt d(t) J : i(t) K dt

[( ; r) ; B ] · dl

(5.99)

[r ; (dl ; B )] · z ; T ().

The line integrals in Eqs. (5.98) and (5.99) are evaluated as described in Section A.4. The direction of integration follows the current i. In addition, in Eq. (5.99) the velocity ( ; r) and force i(dl ; B ) are evaluated before the remaining cross products are taken. For conventional rotary devices, the analysis is usually performed using cylindrical coordinates with the rotation taken to be about the z-axis, : (t)z . The current ﬂow is often taken to be along the z-axis with dl : dzz . When this is the case, Eq. (5.99) reduces to the following form: Rotational Motion di(t) V (t) : i(t)(R ; R ) ; L ; K (t) Q C dt J

(5.100)

d(t) : i(t)K ; T (),

K dt R

where K is called the electrical constant (or back emf constant) and K is C R called the torque constant. In the MKS system K : K . We derive Eq. C R (5.100) for two different rotational devices in Examples 5.6.3 and 5.11.1. These equations are also derived for a device with a different conﬁguration in Example 5.9.1. For practical analysis, it is convenient to rewrite Eqs. (5.98) and (5.99) in terms of the following ﬁrst-order systems: Linear Motion di(t) 1 : [V (t) 9 i(t)(R ; R ) ; dt L Q du(t) 1 : [i(t) dt m

(u ; B ) · dl]

(dl ; B ) · x ; F (x)]

dx(t) : u(t) dt

(5.101)

369

5.6 ELECTROMECHANICAL EQUATIONS

and Rotational Motion di(t) 1 : [V (t) 9 i(t)(R ; R ) ; L Q dt

[( ; r) ; B ] · dl]

(5.102)

d (t) 1 : [i(t) [r ; (dl ; B )] · z ; T ()]

J dt K d(t) : (t). dt

Equations (5.101) and (5.102) constitute initial-value problems that need to be solved subject to the initial conditions (5.88) and (5.89), respectively. We demonstrate the use of these equations in the following examples. EXAMPLE 5.6.2 Consider a linear actuator consisting of a conductive bar of mass m in sliding contact with a pair of stationary conducting rails as shown in Fig. 5.9. The rails are connected to a voltage source V , and a resistor R that Q limits the current. The bar is moving through a constant uniform B-ﬁeld with a time-dependent velocity u(t) relative to the rails. Determine the equations of motion for the system. Assume that the rails are good conductors with negligible resistance. SOLUTION 5.6.2 This device has a moving conductor and is governed by the linear motion equations (5.101). In this case the coil is the moving bar, and there is no mechanical force (F (x) : 0). Therefore, Eq. (5.101) becomes

di(t) 1 : V (t) 9 i(t)(R ; R ) ; L Q dt du(t) i(t) : dt m

dx(t) : u(t), dt

(u ; B ) · dl

(dl ; B ) · x (5.103)

where L is the inductance of the complete circuit. We need to evaluate the integrals in Eq. (5.103). We use a Cartesian reference frame O at rest with respect to the rails as shown in Fig. 5.9.

370

CHAPTER 5 Electromechanical Devices

Induced voltage: To evaluate the induced voltage, we need to determine (u ; B ) · dl along the bar. From Fig. 5.9 we have B : 9B z , (5.104) u(t) : u(t)x , (5.105) and dl : dyy . (5.106) Therefore (u ; B ) · dl : u(t)B dy. (5.107) We integrate Eq. (5.107) along the bar in the direction of current. This gives

(u ; B ) · dl :

F

(u ; B ) · dl

: B u(t)

dy

F : 9B hu(t). (5.108) Force: Next we evaluate the force. For this we need (dl ; B ) · x along the bar. From Eqs. (5.104) and (5.106) we have (dl ; B ) · x : 9B dy. We integrate this along the bar in the direction of current ﬂow and obtain

(dl ; B ) · x :

F

(dl ; B ) · x

: 9B

dy

F : B h. (5.109) Therefore, the Lorentz force on the bar is F : iB h. Equations of motion: Substitute Eqs. (5.108) and (5.109) into Eq. (5.103), which gives di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 B hu(t)] L Q dt du(t) 1 : i(t)B h dt m dx(t) : u(t). dt

5.6 ELECTROMECHANICAL EQUATIONS

FIGURE 5.10

371

Rotating coil device: (a) perspective of device; (b) end view; and

(c) side view.

These equations need to be solved subject to initial conditions (5.88).

)

EXAMPLE 5.6.3 Consider the rotary device shown in Fig. 5.10a. This represents a basic direct-current (dc) motor. In this device, a magnetic ﬁeld B (assumed constant) is produced between two pole pieces by a ﬁeld current i through a D coil that is wrapped around the pole pieces. A rectangular coil is mounted between the pole pieces and is free to rotate about the z-axis. When a current i passes through the coil it experiences a torque that causes it to rotate as shown.

372

CHAPTER 5 Electromechanical Devices

A split ring with brushes is connected to the terminals of the rotating coil so that the current through it reverses direction every half turn. This ensures that the torque will always be in the same direction, thereby producing continuous rotation. Write the equations of motion for this device. SOLUTION 5.6.3 This device has a moving coil and is governed by the rotational motion equations (5.102). There is no mechanical torque and Eqs. (5.102) reduce to di(t) 1 : [V (t) 9 i(t)(R ; R ) ; dt L Q

[( ; r) ; B ] · dl]

d (t) i(t) : [r ; (dl ; B )] · z J dt K d(t) : (t). dt

(5.110)

where L and R are the inductance and resistance of the coil, and J is the K moment of inertia of the coil about the z-axis. We need to evaluate the induced voltage and torque integrals in Eq. (5.110). We use a Cartesian reference frame at rest with respect to the pole pieces as shown in Fig. 5.10a. We choose a rotation angle , which is measured with respect to the y-axis in a counterclockwise sense about the z-axis as shown in Fig. 5.10b. Induced voltage: We determine the voltage induced at the terminals of the coil by summing the contributions from its various segments. The coil has four segments; relatively long left and right side segments, and two shorter end segments (Fig. 5.10b,c). We evaluate the side segments ﬁrst. To evaluate the induced voltage we need [( ; r) ; B ] · dl along these segments. We know that, (t) : (t)z , and from Fig. 5.10b we ﬁnd that

9b sin()x ; b cos()y b sin()x 9 b cos()y

r:

(left side) (right side).

Therefore, ;r:

9 (t)b cos()x 9 (t)b sin()y (t)b cos()x ; (t)b sin()y

Note that B is in the x-direction, B : B x .

(left side) (right side).

373

5.6 ELECTROMECHANICAL EQUATIONS

Therefore,

(t)bB sin()z ( ; r) ; B : 9 (t)bB sin()z

(left side) (right side).

Because dl : dzz for both segments we have

(t)bB sin() dz [( ; r) ; B ] · dl : 9 (t)bB sin() dz

(left side) (right side).

(5.111)

We evaluate Eq. (5.111) along the side segments with the integration in the direction of the current. This gives

[( ; r) ; B ] · dl :

(t)bB sin() dz

?

left side

: 9 (t)abB sin(),

(5.112)

and

[( ; r) ; B ] · dl : 9

?

(t)bB sin() dz

right side

: 9 (t)abB sin(). Next, consider the end segments. For these we have ( ; r) ; B . z , and dl . r. It follows that [( ; r) ; B ] · dl . z · r : 0

(end segments).

Therefore,

[( ; r) ; B ] · dl : 0. end segments

(5.113)

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CHAPTER 5 Electromechanical Devices

This shows that the end segments make no contribution to the induced voltage. The induced voltage is the sum of Eqs. (5.112) and (5.113). Speciﬁcally,

[( ; r) ; B ] · dl : 92 (t)abB sin()

: 9 (t)A B sin(), (5.114) where A : 2ab is the area of the loop. If there are n loops in the coil, then Eq. (5.114) is multiplied by n. Torque: We now determine the torque. Only the left- and right side segments of the coil contribute to the torque. We make a brief argument to show that the end segments make no contribution. Consider the torque on an element of an end segment, dT : r ; (dl ; B ) · z . X On these segments, dl dr and, therefore, dl ; B . dr ; B . z .

This gives

dT . (r ; z ) · z : 0. X Thus, the end segments generate no torque. Now consider the left- and right side segments. For these we have dl : dzz . Therefore, dl ; B : B dzy

It follows that

(side segments).

9bB sin() dz z (left side) (5.115) bB sin() dz z (right side). We evaluate the integral of Eq. (5.115) along the side segments, integrate in the direction of current ﬂow, and obtain r ; (dl ; B ) :

[r ; (dl ; B )] · z : 9

?

bB sin() dz

left side

: abB sin(),

and

[r ; (dl ; B )] · z :

?

(5.116)

bB sin() dz

right side

: abB

sin().

(5.117)

5.6 ELECTROMECHANICAL EQUATIONS

375

The torque on the coil is the sum of Eqs. (5.116) and (5.117),

[r ; (dl ; B )] · z : 2abB sin() : A B sin().

(5.118)

If there are n loops in the coil then Eq. (5.118) is multiplied by n. Moment of inertia: Next, we determine the moment of inertial. We use Eq. (5.77), which we repeat here for convenience, J : K

(x ; y) dv.

4

The inertia of each side segment is J : K

?

bA dz

: abA : m b, where A and are the area and density of the wire, respectively, and

m : aA is the mass of each side segment. The inertia of each end segment is J : 2 K

@

A d

: bA : m b, where m : 2bA is the mass of each end segment. Summing the contributions from all segments gives the total moment of inertia, J : 2J ; 2J K K K : (2m ; m )b. If there are n loops in the coil then Eq. (5.119) is multiplied by n.

(5.119)

376

CHAPTER 5 Electromechanical Devices

Equations of motion: We are ﬁnally ready to write the equations of motion. Substitute Eqs. (5.114) and (5.118) into Eq. (5.110) and obtain di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 A B sin((t)) (t)] dt L Q d (t) 1 : i(t)A B sin((t)) J dt K d(t) : (t). dt These equations are solved subject to the initial conditions (5.89). Notice that the electrical constant K equals the torque constant K (see Eq. (5.100)). C R Speciﬁcally, K : K : A B sin((t)). C R If the coil had n turns all at the same angle , then K and K would be C R multiplied by n. If the turns were distributed in an angular sense, then K and C K would be obtained by integrating the contributions of the individual turns R over the angular span of the coil. )

5.7 ENERGY ANALYSIS In this section we discuss an energy method for the analysis of electromechanical devices. This method, based on the conservation of energy, is used to determine the force or torque on a moving member. It is especially useful for the analysis of magnetic circuit actuators as we shall see. We start with the basic energy balance relation: electrical energy energy energy mechanical input at : dissipated by ; stored in ; energy device terminals device magnetic field output Apply this to a system that is subjected to an inﬁnitesimal (virtual) displacement and obtain dW C

electrical energy input

: dW

energy dissipated

; dW

DJB

stored field energy

; dW

mechanical energy

(5.120)

377

5.7 ENERGY ANALYSIS

The term dW accounts for energy dissipation due to loss mechan isms such as ohmic heating and mechanical friction. For our purposes, we consider the system to be lossless and treat the loss mechanisms as external to the system. For example, a real inductor can be treated as a lossless coil in series with an external resistance. Thus, without loss of generality, we restrict our attention to lossless systems in which dW : 0. For these systems, energy balance equation (5.120) becomes dW : dW ; dW . C DJB

(5.121)

We consider the various terms in Eq. (5.121). Recall from Eq. (5.60) that mechanical motion couples to the electrical equations via ﬂux linkage ". Moreover, the voltage e induced across the terminals of an electromechanical component such as a coil is e : d"/dt. Therefore, the electrical energy input into the terminals of the component during a time dt is d" dW : i dt C dt : d" i.

(5.122)

Substitute Eq. (5.122) into Eq. (5.121) and obtain dW

DJB

: d" i 9 dW

.

(5.123)

(linear motion) (rotational motion),

(5.124)

The mechanical energy output is given by dW

:

f dx DJB T d DJB

where f and T are the force and torque imparted to the mechanical DJB DJB member by the magnetic ﬁeld. It is important to note that a positive force (or torque) is assumed to be in the same direction as a positive displacement dx (or d). From Eqs. (5.123) and (5.124) we have dW

DJB

:

id" 9 f dx DJB id" 9 T d DJB

(linear motion) (rotational motion).

(5.125)

We desire relations for f and T in terms of the ﬁeld energy W . The DJB DJB DJB form of Eq. (5.125) suggests that we can obtain these if we choose " and

378

CHAPTER 5 Electromechanical Devices

x or as the independent variables. With this choice, W is of the form DJB W (", x) (linear motion) DJB W : DJB W (", ) (rotational motion). DJB The total derivative of W is DJB W (", x) W (", x) DJB DJB d" ; dx (linear motion) " x (5.126) dW : DJB W (", ) W (", ) DJB DJB d" ; d (rotational motion). "

Now, we compare Eq. (5.125) with Eq. (5.126) and ﬁnd that W

(", x) DJB x W (", ) DJB T :9 DJB f :9 DJB

(linear motion) (5.127) (rotational motion),

which are the desired results. Recall from Eq. (3.74) in Section 3.2.5 that for linear systems the energy stored in the magnetic ﬁeld is W : Li (linear systems). DJB Moreover, from Eq. (3.78) we know that L:

" . i

Therefore, W

DJB

:

"i . 2

(5.128)

We use Eq. (5.128) to express W in terms of the independent variables DJB " and x or . This gives

"i(", x) " : 2 2L(x)

W : DJB "i(", ) " : 2 2L()

(linear motion) (5.129) (rotational motion).

379

5.7 ENERGY ANALYSIS

We can use Eq. (5.129) to determine f and T . First determine L(x) or DJB DJB L(), substitute these into Eq. (5.129), and then obtain f and T using DJB DJB Eq. (5.127). Instead of choosing ﬂux linkage " and position (x or ) as the independent variables, it is often more natural and convenient to work with current i and position. When this is the case, the energy has a functional dependency of the form W (i, x) or W (i, ), and Eq. (5.125) DJB DJB becomes

id"(i, x) 9 f dx DJB dW : DJB id"(i, ) 9 T d DJB

(linear motion) (rotational motion),

(5.130)

where d"(i, x) :

"(i, x) "(i, x) di ; dx i x

(linear motion)

d"(i, ) :

"(i, ) "(i, ) di ; d i

(rotational motion). (5.131)

Similarly,

W (i, x) DJB W : DJB W (i, ) DJB

(linear motion) (rotational motion),

which gives

W (i, x) W (i, x) DJB DJB di ; dx i x

(linear motion)

dW : DJB W (i, ) W (i, ) DJB DJB di ; d i

(5.132) (rotational motion).

The functional form of these relations motivate the deﬁnition of the coenergy function W A , DJB W A (i, x) : i"(i, x) 9 W (i, x) DJB DJB (linear motion) (Coenergy)

(5.133)

W A (i, ) : i"(i, ) 9 W (i, ) DJB DJB (rotational motion). We can determine the force and torque directly from the coenergy. To

380

CHAPTER 5 Electromechanical Devices

this end, ﬁrst take the total derivative of W A (i, x). This gives DJB d(i"(i, x)) 9 dW (i, x) (linear motion) DJB dW A : DJB d(i"(i, )) 9 dW (i, ) (rotational motion) DJB or

"(i, x)di ; id"(i, x) 9 dW (i, x) DJB dW A : DJB "(i, )di ; id"(i, ) 9 dW (i, ) DJB

(linear motion) (5.134) (rotational motion).

Substitute Eqs. (5.123) and (5.124) into Eq. (5.134) and obtain

"(i, x)di ; f dx (linear motion) DJB (5.135) dW A : DJB "(i, )di ; T d (rotational motion). DJB Now dW A is formally given by DJB W A (i, x) W A (i, x) DJB DJB di ; dx (linear motion) x i dW A : (5.136) DJB W A (i, ) W A (i, ) DJB DJB di ; d (rotational motion). i

We compare Eqs. (5.135) and (5.136) and ﬁnd that ":

W A (i, x) DJB , i

(5.137)

and that f : DJB

W A (i, x) DJB x

W A (i, ) DJB T : DJB

(linear motion) (5.138) (rotational motion).

The relations (5.138) are the desired expressions for force and torque. Last, we obtain an explicit expression for W A from Eqs. (5.129) and DJB (5.133),

"(i, x)i L(x)i : 2 2

WA : DJB "(i, )i L()i : 2 2

(linear motion) (5.139) (rotational motion).

We can use Eq. (5.139) to determine both force and torque. First

381

5.7 ENERGY ANALYSIS

FIGURE 5.11

Electromechanical system: (a) actuator; and (b) energy conversion

diagram.

determine L(x) or L(), substitute these into Eq. (5.139), and then obtain f and T using Eq. (5.138). DJB DJB EXAMPLE 5.7.1 Set up the equations of motion for the actuator circuit of Fig. 5.11 [1]. SOLUTION 5.7.1 This is a magnetic circuit actuator. Choose a convention in which f (force imparted to the moving member by the ﬁeld) is positive in the DJB direction of increasing displacement x as shown. For the equations of motion use the ﬁrst-order system (5.86), which is written in terms of the independent variables (i, x),

1 "(i, x) di(t) : V (t) 9 i(t)(R ; R ) 9 v(t) L(x) Q x dt dv(t) 1 : F(i, x) dt m dx(t) : v(t). dt

(5.140)

382

CHAPTER 5 Electromechanical Devices

To solve Eq. (5.140) we need expressions for the terms L(x), "(i, x)/x, and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). From Eq. (5.139) we have DJB L(x)i W A (i, x) : . (5.141) DJB 2 The magnetic circuit of Fig. 5.11 was analyzed in Example 3.5.2 from Section 3.5.1. The inductance L(x) was found to be NA E , (5.142) (A /A )( /)l ; 2x E A A where l is the path length in the core, and A and A are the cross-sectional A A E areas of the core and gap, respectively. Substitute Eq. (5.142) into Eq. (5.141) and obtain L(x) :

NA i E W A (i, x) : . (5.143) DJB 2((A /A )( /)l ; 2x) E A A Next, use Eqs. (5.137) and (5.138) to obtain " and f . We ﬁnd that DJB W A (i, x) DJB "(i, x) : i :

NA i E , ((A /A )( /)l ; 2x) E A A

(5.144)

and that f (i, x) : DJB

W A (i, x) DJB x

NA i E . (5.145) ((A /A )( /)l ; 2x) E A A The minus sign in Eq. (5.145) implies that the direction of f is opposite to the DJB direction of increasing x (i.e., toward the actuator and opposite to the direction of increasing air gap). The total force F(i, x) on the moving member is a sum of the forces due to the ﬁeld and spring :9

F(i, x) : f

(i, x) ; f (x), (5.146) DJB Q where f (x) is the force due to the spring. In its initial state, the actuator is Q unenergized with i(0) : 0, and the mass is at rest in an equilibrium position x . In this position the spring force is zero, f (x ) : 0. Thus, when the mass is Q

383

5.8 MAGNETIC CIRCUIT ACTUATORS

at a position x the spring force is f (x) : 9k(x 9 x ). (5.147) Q Notice that when x x (x x ) this force tends to move the mass towards (away from) the magnetic circuit. Substitute Eqs. (5.144), (5.145), (5.146), and (5.147) into Eq. (5.140) and obtain di(t) ((A /A )( /)l ; 2x) E A A : dt NA E

2 NA i E ; V (t)9i(t)(R;R ); v(t) ((A /A )( /)l ;2x) Q E A A dv(t) NA i 1 E :9 ; k(x 9 x ) dt m ((A /A )( /)l ; 2x) E A A dx(t) : v(t). (5.148) dt

This nonlinear ﬁrst-order system has to be solved subject to the initial conditions x(0) : x , v(0) : v and i(0) : i . ) Finally, for linear systems the energy and coenergy are numerically equal, W : W A . This follows from the fact that " : Li for such DJB DJB systems, which gives " Li : . 2L 2 However, for nonlinear systems " is not proportional to i and W and DJB W A are not numerically equal. The difference between W and W A for DJB DJB DJB linear and nonlinear systems is illustrated in Fig. 5.12.

5.8 MAGNETIC CIRCUIT ACTUATORS In this section we study magnetic circuit actuators. In these actuators the moving member is part of the magnetic circuit, and its motion alters the ﬂux through the electrical circuit. The energy method is especially useful for analyzing these actuators. We demonstrate this in the following examples. EXAMPLE 5.8.1 Consider the magnetic actuator circuit shown in Fig. 5.13a. Assume that the magnetic circuit operates in a linear region of its B-H curve

384

CHAPTER 5 Electromechanical Devices

Graphical representation of energy and coenergy: (a) linear system; and (b) nonlinear system.

FIGURE 5.12

with , and that there is no ﬂux leakage. Determine the force on the plunger [2]. SOLUTION 5.8.1 We determine an analytical expression for the force on the plunger using both energy and coenergy approaches. Choose a convention in which a positive f is in the direction of increasing displacement x as shown. DJB Energy: In the energy approach, " and x are the independent variables and the energy is determined using Eq. (5.129), " W : . DJB 2L(x)

(5.149)

5.8 MAGNETIC CIRCUIT ACTUATORS

385

Actuator with moving plunger: (a) actuator circuit; and (b) reference frame for plunger [2].

FIGURE 5.13

We need to determine the inductance L(x). We use the inductance relation (3.75). L:

1 i

B · H dv.

2 i

B · H dv,

(5.150)

4 As , the core and plunger have a high permeability and the H-ﬁeld in these elements is negligible. Therefore, Eq. (5.150) reduces to an integration over the gap regions, L:

4g

(5.151)

386

CHAPTER 5 Electromechanical Devices

where V : wg(d 9 x) is the volume of the gap. Factor 2 in Eq. (5.151) takes E into account the identical integrations over the gap regions above and below the plunger. To evaluate Eq. (5.151) we need to know the ﬁelds B and H in the E E gap. To this end, apply Eq. (3.141) to a path around the circuit through the plunger and obtain 2H g : ni, E which gives ni H : . E 2g Further, B : H . Substitute B and H into Eq. (5.151) and obtain E E E E nw L(x) : (d 9 x). (5.152) 2g Then, substitute this into Eq. (5.149), which gives g" W (", x) : . DJB nw(d 9 x) Finally, obtain the force using Eq. (5.127), f

DJB

:9

W (", x) DJB x

:9

g" . nw(d 9 x)

(5.153)

(5.154)

As 0 x d and " 0, the force is negative, which means that it acts to pull the plunger towards the core (in the direction of decreasing x). Formula (5.154) would be more useful if we had an explicit expression for ". For an n-turn coil we have " : n where is the ﬂux through each turn of the coil. As there is no ﬂux leakage, : : B A where A : w(d 9 x). Therefore E E E E ni ni : w(d 9 x) and " : w(d 9 x). 2g 2 g Coenergy: In the coenergy approach i and x are the independent variables and the coenergy is given by Eq. (5.139), WA : DJB

L(x)i . 2

(5.155)

5.8 MAGNETIC CIRCUIT ACTUATORS

387

Substitute Eq. (5.152) into Eq. (5.155) and obtain in WA : w(d 9 x). DJB 4g

(5.156)

The force follows from Eq. (5.138), inw f :9 . DJB 4g

(5.157)

Notice that the force is independent of x. The two force expressions (5.154) and (5.157) are equivalent. To see this, substitute Eq. (3.78) (i.e., " : L(x)i) into Eq. (5.154), which gives Eq. (5.157). ) EXAMPLE 5.8.2 Consider the cylindrical actuator shown in Fig. 5.14. Assume that the magnetic circuit is linear with , and that there is no ﬂux leakage. Set up the equations of motion for the actuator [2]. SOLUTION 5.8.2 First, choose a convention in which a positive f is in the DJB direction of increasing x as shown. As the motion is linear, we use the ﬁrst-order

FIGURE 5.14

Cross section of a cylindrical actuator circuit [2].

388

CHAPTER 5 Electromechanical Devices

system of equations (5.86),

di(t) 1 "(i, x) : V (t) 9 i(t)(R ; R ) 9 v(t) dt L(x) Q x dv(t) 1 : F(i, x) m dt

(5.158)

dx(t) : v(t). dt To solve Eq. (5.158) we need expressions for L(x), "(i, x)/x, and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). DJB From Eq. (5.139) we have W A (i, x) : DJB

"(i, x)i 2

:

L(x)i . 2

(5.159)

Inductance: We determine the inductance using Eq. (3.75), L:

1 i

B · H dv. (5.160) 4 As the core and plunger have a high permeability, the H-ﬁeld in these elements is negligible, and Eq. (5.160) reduces to an integration over the nonmagnetic sleeve regions, L:

1 i

B · H dv ;

1 i

B · H dv,

(5.161)

4 4 where the subscripts top and bot refer to the top and bottom sleeve regions, respectively. Notice that V dgh and V dg(h 9 x). We need to determine H and H . To this end, apply Eq. (3.141) to either of the dotted paths in Fig. 5.14. This gives g ; H g : ni. (5.162) In addition, as there is no ﬂux leakage : , or B A :B A , (5.163) where A : dh and A : d(h 9 x). From Eq. (5.163) and the fact that H

5.8 MAGNETIC CIRCUIT ACTUATORS

B

: H and B : H , we ﬁnd that h H :H . (h 9 x)

389

(5.164)

Substitute Eq. (5.164) into Eq. (5.162) and obtain H

:

ni (h 9 x) . g (2h 9 x)

(5.165)

Finally, substitute Eqs. (5.164) and (5.165) into Eq. (5.161), which gives the inductance ndh (h 9 x) . L(x) : g (2h 9 x)

(5.166)

Coenergy: Once we know the inductance, the coenergy is easily obtained from Eq. (5.159), indh (h 9 x) W A (i, x) : . DJB 2g (2h 9 x)

(5.167)

We use Eqs. (5.137) and (5.138) to obtain " and f , and ﬁnd that DJB W A (i, x) DJB ": i indh (h 9 x) : , g (2h 9 x)

(5.168)

and f

DJB

:

W A (i, x) DJB x

ind h :9 . (2h 9 x) 2g

(5.169)

As x h, Eq. (5.169) is negative, which implies that the direction of f is DJB opposite to the direction of increasing x (i.e., opposite to the direction of increasing air gap). The total force F(i, x) is a sum of the forces due to the ﬁeld and spring, F(i, x) : f where

DJB

(i, x) ; f (x). Q

(5.170)

f (x) : 9k(x 9 x ). (5.171) Q Here, x is the initial position of the plunger, which is assumed to be the equilibrium position for the spring. When x x (x x ), f (x) tends to move Q

390

CHAPTER 5 Electromechanical Devices

the mass into (out of) the solenoid. Finally, substitute Eqs. (5.168), (5.169), (5.170), and (5.171) into Eq. (5.158) and obtain

di(t) (2h9x) ind h g : V (t)9i(t)(R;R ); v(t) Q dt ndh (h9x) g (2h9x) dv(t) 1 ind h :9 ; k(x 9 x ) m dt 2g (2h 9 x)

dx(t) : v(t). dt

(5.172)

This nonlinear ﬁrst-order system is valid for 0 x h. It is solved subject to the initial conditions x(0) : x , v(0) : v and i(0) : i . ) EXAMPLE 5.8.3 Set up the equations of motion for the actuator shown in Fig. 5.15. Assume that all magnetic materials are linear with . The nonmagnetic sleeves of thickness s provide low friction support for the moving plunger [4]. SOLUTION 5.8.3 Set up a coordinate system with f positive in the direction DJB of increasing x as shown. As the motion is linear, we use the ﬁrst-order system of equation (5.86)

di(t) 1 "(i, x) : V (t) 9 i(t)(R ; R ) 9 v(t) dt L(x) Q x dv(t) 1 : F(i, x) dt m dx(t) : v(t). dt

FIGURE 5.15

(5.173)

Cross section of an actuator circuit [4].

391

5.8 MAGNETIC CIRCUIT ACTUATORS

To solve Eq. (5.173) we need expressions for L(x), "(i, x)/x and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). From DJB Eq. (5.139) we have "(i, x)i W A (i, x) : DJB 2 :

L(x)i . 2

(5.174)

Inductance: We determine the inductance using Eq. (3.75), L:

1 i

B · H dv. (5.175) 4 As the core and plunger have a high permeability, H is negligible in these elements and Eq. (5.175) reduces to an integration over the gap region and the nonmagnetic sleeve regions, L:

1 i

B · H dv ;

2 i

B · H dv, (5.176) 4E 4Q where V and V are the volumes of the gap and sleeve regions, respectively. The E Q factor of 2 in the second integral takes into account the identical integrations over each sleeve region. Notice that V : A x and V : A s, where A and A E E Q Q E Q are the cross-sectional areas of the gap and sleeve, respectively. We need to determine H and H . To this end, apply Eq. (3.141) to the dotted path in Fig. E Q 5.15. This gives H x ; H s : ni. (5.177) E Q Further, as there is no ﬂux leakage : 2 , or E Q B A : 2B A . (5.178) E E Q Q Combining Eqs. (5.177) and (5.178) with B : H and B : H gives E E Q Q ni H : , (5.179) E (x ; (A /2A )s) E Q and A ni H : E . (5.180) Q 2A (x ; (A /2A )s) Q E Q Next, substitute Eqs. (5.179) and (5.180) into Eq. (5.176), which gives the inductance L(x) :

nA E . (x ; (A /2A )s) E Q

(5.181)

392

CHAPTER 5 Electromechanical Devices

Coenergy: Once we know the inductance, the coenergy is obtained easily from Eq. (5.159) inA E W A (i, x) : . DJB 2[x ; (A /2A )s] E Q Use Eqs. (5.137) and (5.138) to obtain " and f . We ﬁnd that DJB W A (i, x) DJB ": i :

inA E , [x ; (A /2A )s] E Q

(5.182)

(5.183)

and f

DJB

:

W A (i, x) DJB x

inA E . (5.184) 2[x ; (A /2A )s] E Q Notice that f is negative, which implies that the direction of f is opposite to DJB DJB the direction of increasing x (i.e., opposite to the direction of increasing air gap). The total force F(i, x) is a sum of the forces due to the ﬁeld and the spring, :9

F(i, x) : f

DJB

(i, x) ; f (x), Q

(5.185)

where f (x) : 9k(x 9 x ). (5.186) Q Here, x is the initial position of the plunger, which is assumed to be the equilibrium position for the spring. When x x (x x ), f (x) tends to move Q the mass into (out of) the circuit. Finally, substitute Eqs. (5.181), (5.184), (5.185), and (5.186) into Eq. (5.173) to obtain

di(t) (x ; (A /2A )s) inA E Q E : V (t) 9 i(t)(R ; R ) ; v(t) Q dt nA [x ; (A /2A )s] E E Q dv(t) 1 inA E :9 ; k(x 9 x ) dt m 2[x ; (A /2A )s] E Q dx(t) : v(t). (5.187) dt

This nonlinear ﬁrst-order system is solved subject to the initial conditions x(0) : x , v(0) : v , and i(0) : i . )

393

5.9 AXIAL-FIELD ACTUATORS

5.9 AXIAL-FIELD ACTUATORS Axial-ﬁeld actuators are moving coil actuators. These actuators are well suited and widely used for applications that require controllable bidirectional rotation. For example, they are routinely used for positioning read/write heads in computer disk drives. These actuators consist of pie-shaped magnets that are positioned above (and/or below) a coil that is free to rotate as shown in Fig. 5.16. When the coil is energized, it experiences a torque and rotates either clockwise or counterclockwise depending on the direction of current. A mechanical component such as a spring is often used to provide a restoring torque, and to ﬁx the unenergized position of the coil (Fig. 5.17). Axial-ﬁeld actuators can be designed and optimized prior to fabrication using lumped-parameter analysis [9, 10]. We develop a model for performing such analysis in the following example. EXAMPLE 5.9.1 Determine the equations of motion for the axial-ﬁeld actuator shown in Fig. 5.16. Assume that the coil has n turns. SOLUTION 5.9.1 This device is a moving coil actuator and is governed by Eqs. (5.102),

di(t) 1 : V (t) 9 i(t)(R ; R ) ; dt L Q

[( ; r) ; B

· dl

1 d (t) : i(t) [r ; (dl ; B )] · z ; T ()

dt J K d(t) : (t). dt

FIGURE 5.16

Drive magnet and coil for an axial-ﬁeld actuator.

(5.188)

394

CHAPTER 5 Electromechanical Devices

FIGURE 5.17

Axial-ﬁeld actuator with spring mechanism.

Here, L and R are the inductance and resistance of the coil, and J is its K moment of inertia above the z-axis. We need to evaluate the induced voltage and torque integrals in Eq. (5.188). We use a cylindrical coordinate system at rest with respect to magnets as shown in Fig. 5.16. Before we begin, we state some simplifying assumptions. Assumptions: The ﬁrst assumption is that the magnetic ﬁeld due to the magnet is essentially uniform and constant across the coil, that is,

9B z 0

K (5.189) B z 9 0,

K where 9 and ; deﬁne the angular positions of the radial edges of the K K B :

395

5.9 AXIAL-FIELD ACTUATORS

FIGURE 5.18

Electrical connection for a single turn of the coil.

magnet. In reality, the ﬁeld distribution above the magnet varies with height, and from point to point at a ﬁxed height. However, an average value of B

that is adequate for this analysis can be obtained using three-dimensional FEA, closed-form analysis [11], or measured data if the fabricated magnet is available. The second assumption is that each of the coil’s two radial edges rotates over a separate magnetic pole throughout the entire range of motion. Induced voltage: We determine the voltage induced at the terminals of the coil by summing the voltage induced across its various segments. For reference purposes, Fig. 5.18 shows the electrical connection of a single turn. The coil has four segments: an upper and a lower segment, and two side segments (Fig. 5.19). We consider a single turn of the coil and evaluate the side (radial) segments ﬁrst. We need to determine ( ; r) ; B along these segments. First, we know that (t) : (t)z . Then from Fig. 5.19 we have r : rr . Thus, ; r : (t)r Note that B is in the z-direction, 9B z B : B z

(side segments).

(left side) (right side).

396

CHAPTER 5 Electromechanical Devices

FIGURE 5.19

Top view of the coil with the magnetic ﬁeld.

Therefore, ( ; r) ; B

:

9 (t)rB r (t)rB r

(left side) (right side).

(5.190)

We evaluate

[(;r);B ] · dl

along the left and right sides where dl : dr r

(side segments).

We obtain

[(;r);B ] · dl : 9 left side

0 (t)B r dr 0

R 9 R , : 9 (t)B 2

(5.191)

397

5.9 AXIAL-FIELD ACTUATORS

and

[( ; r) ; B ] · dl :

0 (t)B r dr 0

right side

R 9 R . : 9 (t)B 2

(5.192)

Now we consider the top and bottom segments. For these we have ( ; r) ; B r . Because dl , it follows that [( ; r) ; B ] · dl r · : 0

(top/bottom segments).

Therefore,

[( ; r) ; B ] · dl : 0, top/bottom segments

which shows that these segments do not contribute to the induced voltage. Finally, add Eqs. (5.191) and (5.192), and multiply by n (the number of turns) to obtain an expression for the induced voltage

[( ; r) ; B ] · dl : 9 (t)nB (R 9 R).

(5.193)

Torque: Only the side segments of the coil contribute to the torque. We make a brief argument to show that the upper and lower segments make no contribution. Consider the torque on an element of the upper or lower segment, dT : [r ; (dl ; B )] · z , X where r : rr and dl . . Notice that dl ; B

. ; B . r .

This gives dT . [r ; r ] · z : 0, X which shows that the upper and lower segments generate no torque. Now

398

CHAPTER 5 Electromechanical Devices

consider the left- and right side segments. For these, we have dl : drr , and

B dr dl ; B : 9B dr

(left side) (right side).

Therefore,

B r dr z r ; (dl ; B ) : 9B r dr z

(left side) (right side).

(5.194)

We evaluate

[r ; (dl ; B )] · z

along the left- and right side segments with the integration in the direction of the current ﬂow. This gives

[r ; (dl ; B )] · z :

0 B r dr 0

left side

:B and

(R 9 R) , 2

(5.195)

0 B r dr [r ; (dl ; B )] · z : 9 0 right side

:B

(R 9 R) . 2

(5.196)

The torque on the coil is ni times the sum of Eqs. (5.195) and (5.196),

[r ; (dl ; B )] · z : niB (R 9 R). (5.197) Equations of motion: We are ﬁnally ready to write the equations of motion. Substitute the integrals (5.193) and (5.197) into Eq. (5.188) and obtain di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 nB (R 9 R) (t)] dt L Q d (t) 1 : [i(t)nB (R 9 R) ; T ()]

dt J K d(t) : (t). dt

(5.198)

399

5.9 AXIAL-FIELD ACTUATORS

These equations are solved subject to the initial conditions of Eq. (5.89). Finally, from Eq. (5.198) we see that the electrical constant K equals the torque constant C K (see Eq. (5.100)). Speciﬁcally, R K : K : nB (R 9 R). C R Calculations: We apply Eq. (5.198) to an actuator with the mechanical mechanism shown in Fig. 5.17. Here, the mechanical torque is supplied by the spring. To perform the analysis we need expressions for the coil inductance L and the mechanical restoring torque T ().

Inductance: The inductance of the coil can be estimated by considering a short circular coil with a height h equal to that of the actuator coil, and with A a radius r , A R 9R . r : A 2

(5.199)

To ﬁrst order, the ﬂux through the center of the coil (through each turn) is (i)

inr A . (h ; (2r )) A A

(5.200)

Therefore, the ﬂux linkage is "(i) : n(i). Recall that the inductance is d"(i) , di

(5.201)

nr A . (h ; (2r )) A A

(5.202)

L: which gives L

This approximation is appropriate if the mean circumferential arc subtended by the coil is approximately equal to R 9 R , which is the case for many practical designs. However, if this is not the case the inductance can be determined empirically, or calculated using three-dimensional FEA. Mechanical Torque: The mechanical torque is provided by the spring mechanism shown in Fig. 5.17. The restoring force is F(Y) : k (Y 9 Y ) ; F , Q where k is the spring constant, Y is the initial spring length, F is the force Q

400

CHAPTER 5 Electromechanical Devices

FIGURE 5.20

Applied voltage V (t). Q

when Y : Y , and Y is the stretched length. The length Y is deﬁned by the angle #. From geometry we have Y : (X ; D 9 2XD cos(# ; ). The restoring torque is given by () : DF(Y()) cos(# ; ). (5.203)

Analysis: We substitute Eqs. (5.202) and (5.203) into Eq. (5.198) and perform the analysis. The applied voltage V (t) is shown in Fig. 5.20 and the remaining Q parameters are as follows: T

B : 0.5 T

R;R :3 n : 100 turns R : 10 mm R : 15 mm

401

5.9 AXIAL-FIELD ACTUATORS

FIGURE 5.21

Current i(t).

D : 10 mm # : 45° h : 1.5 mm A k : 1.5 N/mm Q J : 6 ; 10\ kg · m. K The moment of inertia is computed using Eq. (5.77), which gives J : (R 9 R) A ; (R 9 R) A , K where and A are the density and area of the wire and is the

angular span of the coil (left- to right side) in radians. The response of the actuator is computed for the initial conditions i(0) : 0A, (0) : 0°, and (0) : 0 rad/sec. The current proﬁle i(t) is shown in Fig. 5.21. The rotation angle (t) and angular velocity (t) are shown in Fig. 5.22. Notice that the angular velocity (t) passes through zero and turns negative as the angular position (t) peaks at approximately 6.5 ms. )

402

CHAPTER 5 Electromechanical Devices

FIGURE 5.22

Angular position (t) and velocity (t).

5.10 RESONANT ACTUATORS Resonant actuators are used as drive mechanisms for scanning elements in optical scanning systems [12]. An example of such an actuator is shown in Fig. 5.23. In this actuator a wedge-shaped magnet provides an axial bias ﬁeld, and a coil rotates above the magnet once it is energized. A torsional pivot mechanism provides the restoring torque. When power consumption is an issue, the torsional mechanism can be designed to render the actuator resonant at the scanning frequency. When this is the case, the actuator can be brought to a resonant oscillation with a sequence of low-energy pulse excitations, thus minimizing power consumption. We demonstrate this in the following example. EXAMPLE 5.10.1 Develop a model for the design of the axial-ﬁeld resonant actuator shown in Fig. 5.23. SOLUTION 5.10.1 The behavior of this device is governed by the equations for an axial-ﬁeld actuator as derived in Example 5.9.1. We repeat them here for

403

5.10 RESONANT ACTUATORS

FIGURE 5.23

Resonant actuator.

convenience, di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 nB (R 9 R) (t)] dt L Q d (t) 1 : [i(t)nB (R 9 R) ; T ()]

dt J K d(t) : (t). dt

(5.204)

Mechanical torque: The restoring torque is provided by a torsional pivot mechanism T () : 9K . (5.205)

N Here, K is the spring constant of the pivot. This simple linear relation is N typically only valid for limited rotations, for example, 912° 12°. The

404

CHAPTER 5 Electromechanical Devices

resonant frequency f for this mechanism is f:

1 2

K N. J K

(5.206)

Calculations: We demonstrate the use of Eq. (5.204) with some sample calculations. We numerically integrate these equations using the fourth-order Runge-Kutta method (Appendix C). The following parameters are used for the analysis: B

: 0.55 T

R;R :6 n : 100 turns R : 3.2 mm R : 11.5 mm L : 0.4 mH J : 0.3 ; 10\ kg · m K f : 100 Hz The ﬁeld B : 0.55 T can be achieved using sintered NdFeB magnets above and below the coil with soft-magnetic ﬂux plates attached to the outer surface of each magnet. The ﬂux plates enhance the ﬁeld across the coil. We apply a periodic step function voltage V (t) with a magnitude of <3 V and a period of Q 10 ms as shown in Fig. 5.24. It only requires two cycles to bring the system to resonance after which the circuit is assumed to be open (i(t) : 0 for t 20 ms). The actuator response is computed with the following initial conditions: i(0) : 0 A, (0) : 0°, and (0) : 0 rad/s. The circuit current is plotted in Fig. 5.25. The rotation angle (t) and back voltage V (t) are plotted in Fig. 5.26. The back voltage is the voltage induced in the coil as it moves. It is given by V (t) : nB (R 9 R) (t). Note that V (t) passes through zero as the angular displacement peaks ( (t) : 0), as it should. Because there are no damping terms, the actuator continues in a resonant mode with an angular oscillation of : <12°. This type of performance is consistent with many such systems where the only appreciable dissipation results from windage and eddy current losses. )

405

5.10 RESONANT ACTUATORS

FIGURE 5.24

Applied voltage V (t). Q

FIGURE 5.25

Current i(t).

406

CHAPTER 5 Electromechanical Devices

FIGURE 5.26

V (t) and (t) vs t.

5.11 MAGNETOOPTICAL BIAS FIELD ACTUATOR Magnetooptical (M-O) recording was discussed in Section 4.10. The basic elements of an M-O system are shown in Fig. 5.27. Recall that M-O systems require oppositely oriented bias ﬁelds for the recording and erasure processes [13—15]. In this section we discuss an actuator that serves the dual function of providing the bias ﬁeld and implementing the bias ﬁeld reversal. Speciﬁcally, we analyze the rotary permanent magnet actuator shown in Fig. 5.28. This actuator consists of a cylindrical bipolar bias magnet (Fig. 5.29), surrounded by a stationary concentric conductive shell and a drive coil. When the coil is energized it rotates the magnet to one of two equilibrium positions, thereby producing a bias ﬁeld of desired polarity across the recording media. Rotation of the magnet gives rise to eddy currents in the conductive shell that impart a drag torque and reduce oscillations of the magnet about its equilibrium positions. We develop a model for this actuator in the following example [16, 17].

5.11 MAGNETOOPTICAL BIAS FIELD ACTUATOR

FIGURE 5.27

Basic elements of a magnetooptic recording system.

FIGURE 5.28

Cross section of a rotary bias ﬁeld actuator.

FIGURE 5.29

Cylindrical bipolar bias magnet.

407

408

CHAPTER 5 Electromechanical Devices

EXAMPLE 5.11.1 Set up the equations of motion for the rotary bias actuator shown in Fig. 5.28. The magnet has a radius a and length h, with a h. The coil has n turns and has a length l . A SOLUTION 5.11.1 This is a moving magnet actuator and is governed by the equations for rotational motion (5.87), which we repeat here for convenience:

di(t) 1 "(i, ) : V (t) 9 i(t)(R ; R ) 9 (t) dt L Q

d (t) 1 : T(i, , ) J dt K d(t) : (t). dt

(5.207)

These equations contain nonlinear terms and can be solved numerically once expressions for L, "(i, )/, and T(i, , ) have been determined. We have added an dependence in T(i, , ) to account for the effects of eddy currents in the conductive shell. Inductance: The inductance of the coil can be estimated using the result of Eq. (3.87) for two parallel wires (Example 3.2.10). Taking into account the fact that there are n turns, the inductance is approximated by

s nl , (5.208) L A ln r U where s is the mean separation of the coils, l is the length of each side of the A coil (into the page), and r is the radius of the wire. U Induced voltage: A voltage V is induced in the coil as the magnet rotates. It is given by V

(t) :

"(i, ) (t).

(5.209)

The ﬂux linkage "(i, ) can be written as the sum of two terms, "(i, ) : " (i) ; " ().

The ﬁrst term " (i) is the ﬂux linkage due to the current through the coil itself. The second term " () is the ﬂux linkage due to the magnet. Notice that

only the second term has an angular dependence and, therefore, "(i, ) " () : .

409

5.11 MAGNETOOPTICAL BIAS FIELD ACTUATOR

To determine this term we need an expression for the ﬁeld of the magnet. For this actuator, the length of the magnet is much greater than its diameter, and a two-dimensional analysis sufﬁces (Fig. 5.27). A two-dimensional ﬁeld solution for the magnet was derived in Example 3.6.3. The vector potential was found to be M a A (r, ) : Q sin()z

2 r

(a r).

(5.210)

Recall that Eq. (5.210) applies when the magnet is positioned with its north pole aligned with : 0 as deﬁned by Eq. (3.264). If the magnet is rotated by an angle then Eq. (5.210) is modiﬁed by substituting ; 9 , that is, A

M a (r, , ) : Q sin( 9 )z .

2 r

(5.211)

The ﬂux linkage due to the magnet is "

() : n

:n

1

B (r, , ) · ds

A

(r, , ) · dl.

Therefore, " ()

:n

A (r, , ) · dl

\JA :n A (R , 0, ) dz ; A (R , , ) dz

A

A \JA a : n M l cos(), (5.212) Q AR A where is the angular position of the magnet relative to the x-axis, and R is A the radial position of the sides of the coil. The coil has n side elements (into the page) that are positioned at r : R , and : 0 and , respectively. We A substitute Eq. (5.212) into Eq. (5.209) and obtain

V (t) : K ((t)) (t), C where K is the electrical constant C a K ((t)) : n M l cos((t)). C Q AR A

(5.213)

(5.214)

410

CHAPTER 5 Electromechanical Devices

Torque: The torque on the magnet can be written as a superposition of two terms, T(i, , ) : T

(i, ) ; T ( ). (5.215) The terms T (i, ) and T ( ) are the torques due to the coil and eddy currents induced in the surrounding conductive shell, respectively. Drive torque: We consider the torque imparted to the coil by the magnet. This is given by Eq. (5.79), T:i

r ; (dl ; B ), (5.216) where r is the distance from the current element dl to the axis of rotation and integration is in the direction of current ﬂow. Here, r : R r , and dl : dzz . A Therefore, r ; (dl ; B ) : R B dz z , A P where B is given by Eq. (3.274), P M a B (r, , ) : Q cos( 9 ). P 2 r

(5.217)

In Eq. (5.217), (r, ) is the observation point and is the rotational position of the magnet with respect to the x-axis. The coil has n pairs of wires positioned as described in the foregoing. Therefore, the axial component of the torque is

M a \JA T () : ni(t) Q cos(0 9 ) dz ; cos( 9 ) dz X 2 R \JA A a : 9ni(t) M l cos((t)). (5.218) Q AR A This is the torque imparted to the coil by the magnet. The torque on the magnet is of equal magnitude but in the opposite direction. Therefore, the torque imparted to the magnet by the coil is a (i(t), (t)) : ni(t) M l cos((t)). Q AR A By comparing Eq. (5.214) with Eq. (5.219) we ﬁnd that T

(i(t), (t)) : K ((t))i(t), R where K ((t)) is the torque constant and K ((t)) : K ((t)). R R C T

(5.219)

(5.220)

5.11 MAGNETOOPTICAL BIAS FIELD ACTUATOR

411

Eddy currents: As the magnet rotates it induces eddy currents in the conductive shell that give rise to a drag torque. To analyze this, we choose a reference frame that is at rest with respect to the shell, and with its origin on the axis of the magnet. We assume that the shell is thinner than the skin depth . associated with the desired ﬁeld reversal time. For example, . 25 mm for a ﬁeld reversal of 10 ms. Based on these assumptions, the magnetic ﬁeld induced in the shell is negligible compared to the magnet’s ﬁeld, which is obtained from the vector potential equation (5.211). Therefore, the electric ﬁeld induced at a position (r, ) in the shell is E(r, , (t)) : 9 :9

A

(r, , (t))

t

A

(r, , (t))

(t).

(5.221)

This gives rise to an induced current density in the shell, J(r, , (t)) : E(r, , (t)),

(5.222)

where is the conductivity of the shell. Substituting Eqs. (5.211) and (5.221) into Eq. (5.222) gives M a J (r, , (t)) : Q cos( 9 (t)) (t). X 2 r

(5.223)

Thus, the current through an inﬁnitesimal cross section r dr d of the shell is di(t) : J (r, , (t))r dr d. (5.224) X The drag torque is obtained by combining Eqs. (5.216), (5.217) and (5.224) and then integrating over the cross-sectional area of the shell (we assume that the shell is as long as the magnet). This gives T

M L 0 1 Q (t) cos( 9 (t)) dr d 2 r 0 M R : ha Q ln (t). (5.225) 2 R

( (t)) : ha

We write this as T

( (t)) : 9K (t),

(5.226)

where K : Sgn( (t))ha

M R Q ln , 2 R

(5.227)

412

CHAPTER 5 Electromechanical Devices

and Sgn( (t)) :

1 91

(t) & 0 (t) 0.

The term Sgn( (t)) takes into account the fact that the torque due to the eddy current acts to oppose the rotation. Notice that we have ignored the current induced in the conductive shell by the drive coil, and the voltage induced in the coil by the eddy currents within the shell. These effects are second order [17]. Equations of motion: Finally, substitute Eqs. (5.213), (5.220), and (5.226) into Eq. (5.207) and obtain the equation of motion for the actuator, di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 K ((t)) (t)] C L Q dt d (t) 1 : [K ((t))i(t) 9 K (t)] R dt J K d(t) : (t), dt

(5.228)

where L is given by Eq. (5.208). Calculations: To demonstrate the theory we consider an M-O system that requires a bias ﬁeld in the range 0.02—0.04 T along a radial length of 100 mm, and a ﬁeld-reversal time of 12 ms. A magnet for this system was designed in Example 4.10.1. Speciﬁcally, we use a sintered NdFeB magnet (M : 7.2;10 Q A/m) with a radius R : 1 mm, and a length h : 108 mm. This magnet needs to be rotated : <85° in 12 ms. Preliminary analysis: Before we design the actuator we need to estimate nominal values for several key parameters. First, we determine the moment of inertia of the magnet. This is given by J : 1/2hR, where : 7.52 ; 10 K kg/m is the density of sintered NdFeB, and R and h are as given in the preceding text. Using these values we obtain J : 1.275 ; 10\ kg/m. Second, K we estimate the number of amp turns required to reverse the ﬁeld in 12 ms. This is a three-step process: First, we estimate the angular acceleration : 2/t required to rotate the magnet through an angle of : in less than half of the reversal time, t : 5 ; 10\ s. We use this reduced time because the acceleration must be great enough to compensate for the drag due to the conductive shell. We calculate : 2.5 ; 10 rad/s. Next, we estimate the drive torque T : J , which yields T : 3.18 ; 10\ Nm. Finally, we estimate the amp K turns using Eq. (5.219). We use the following nominal values: (t) : /3, l : h, and R : 3 ; 10\ m. This value of R (radial position of the coil) places A A A

413

5.12 LINEAR ACTUATORS

the coil far enough away from the magnet to accommodate the conductive shell. Using these values we obtain T R A M ha cos(/3) Q : 20.

ni :

(5.229)

If we choose a nominal value of current i : 200 ma, then n : 100 turns. We are ﬁnally ready to perform the actuator design. Actuator design: Our goal is to determine the radial thickness of the conductive shell that renders the desired ﬁeld reversal time. We integrate Eq. (5.228) using the fourth-order Runge-Kutta method with the inner radius of the shell set to R : 2.0 mm (1.0 mm from the surface of the magnet) and the outer radius set to R : 2.5, 3.0 and 3.5 mm, respectively. Further, we assume that the shell is made from aluminum, which has a conductivity : 3.82 ; 10 m\. For these calculations, the radial position of the coil is 0.5 mm beyond the outer surface of the shell, that is, R : 3.0, 3.5 and 4.0 mm, A respectively. The excitation voltage is held constant at V : 12 V and 36-gauge Q wire is used with n : 100 turns. The response of the actuator is computed subject to the following initial conditions: i(0) : 0 A, (0) : 985°, (0) : 0 rad/s. The resistance and inductance of the coil compute to 32.6 and 2.0 mH, respectively. A parametric plot of (t) vs t with R : 2.5, 3.0 and 3.5 mm is shown in Fig. 5.30. Notice that as R increases, the eddy current damping increases and the magnitude of the oscillations about the equilibrium position of 90° decrease. When R : 3.5 mm, the magnet stabilizes at 90° within the desired reversal time of 12 ms. Thus a shell with R : 2.0 mm and R : 3.5 mm will sufﬁce for this application. Additional plots of i(t) and V (t) : K((t)) (t) for R : 3.0 mm are shown in Fig. 5.31. Last, we note that in practice, the equilibrium position of the magnet is offset from 90° (e.g., 85°) so that it can be rotated by the coil. This offset is usually accomplished using magnetic detent in which a soft-magnetic element is ﬁxed in proximity to the magnet. The presence of this element gives rise to second-order effects, and will not alter the analysis appreciably. )

5.12 LINEAR ACTUATORS Linear actuators are used for applications that require smooth and controllable bidirectional motion. For example, they are used as radial access actuators in optical and magnetic disk drives. In this section we

414

CHAPTER 5 Electromechanical Devices

FIGURE 5.30

Parametric plot of (t) vs t with R : 2.5, 3.0 and 3.5 mm.

FIGURE 5.31

Current and induced voltage vs t with R : 3.0 mm.

415

5.12 LINEAR ACTUATORS

FIGURE 5.32

Linear actuator: (a) perspective of actuator; and (b) cross-sectional

view.

study the moving coil linear actuator shown in Fig. 5.32a. This actuator consists of a stator and a moving coil that executes bidirectional motion. The stator consists of a closed soft magnetic core with a permanent magnet mounted on the inside surface of the base (Fig. 5.33a). The coil encloses the top segment of the stator and is free to move along it. The operation of the actuator is straightforward. The magnet produces a ﬁeld B in the gap region of the stator, and when the coil is E energized, the wires that pass through the gap experience a Lorentz force that causes the coil to move (Fig. 5.32b). This is a moving coil actuator and its behavior is governed by the equations for linear motion (5.98), di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt dx(t) m : i(t) dt

(u ; B ) · dl

(dl ; B ) · x ; F (x).

(5.230)

416

CHAPTER 5 Electromechanical Devices

Linear actuator stator: (a) perspective of stator; and (b) crosssectional view.

FIGURE 5.33

Here m, R , and L are the mass, resistance, and inductance of the coil. The terms m and R are easily determined given the number of turns n of the coil and the wire gauge. To solve this system we need expressions for L, the induced voltage (u ; B ) · dl, and the Lorentz force i (dl ; B ) · x . We derive these under the assumption that the stator is operating in a linear region of its B-H curve with B : H . First, we determine the inductance L. Consider the total ﬂux (i, x) through the top segment of the stator. As the magnetic circuit is linear, (i, x) is the sum of the ﬂux due to the coil (i), and the ﬂux due to the magnet (x),

(i, x) : (i) ; (x).

The ﬂux linkage can also be separated into two components, "(i, x) : " (i) ; " (x).

(5.231)

417

5.12 LINEAR ACTUATORS

The self inductance of the coil is L:

d" (i) . di

(5.232)

The magnetic circuit is essentially a coil wrapped around a closed core and was analyzed in Example 3.5.1. Thus, if we assume negligible leakage ﬂux, L is given by Eq. (3.164), L:

nA A, l A

(5.233)

where and A are the permeability and cross-sectional area of the core A and l is the average path length around the core. A Next, we determine the induced voltage (u ; B ) · dl. We use the reference frame shown in Fig. 5.33a. Speciﬁcally, we have u : u(t)x , B : B y , and dl : dzz . Only the wires in the gap contribute to the E induced voltage. Therefore,

(u ; B ) · dl : n

\UA u(t)B dz E

: 9nw B u(t), A E

(5.234)

where w is the width of the core (Fig. 5.33a). It remains to determine B . A E This follows from the results of Example 3.5.3. Speciﬁcally, we apply the analysis presented there to one half of the stator circuit (dotted line in Fig. 5.33b), assuming that there is no leakage ﬂux and that the magnet has a linear demagnetization curve of the form, B :B ; H . K P K K From Eq. (3.179) we have B l P K, B : E ( ; l /g) g K K

(5.235)

where : B /H and A : A . Substitute Eq. (5.235) into Eq. (5.234) K P A E K and obtain

l nw B K u(t). A P (u ; B ) · dl : 9 ( ; l /g) g K K

(5.236)

This is the desired expression for the induced voltage. If : , Eq. K

418

CHAPTER 5 Electromechanical Devices

(5.236) reduces to

nw B A P u(t). (u ; B ) · dl : 9 (g/l ; 1) K Last, we determine the Lorentz force. Again, we have dl : dz z and B : B y . Therefore, E \UA B dz, i (dl ; B ) · x : 9i n E :inw B . (5.237) A E Substitute Eq. (5.235) for B in Eq. (5.237) and obtain E nw B l A P K. (5.238) i (dl ; B ) · x : i ( ; l /g) g K K This is the Lorentz force on the coil. We are ﬁnally ready to write the equations of motion. Substitute Eqs. (5.233), (5.236), and (5.238) into Eq. (5.230). This gives

V (t) : i(t)(R ; R ) ; Q

nA di(t) dx(t) A ;K dt l dt A

(5.239)

dx(t) m : Ki(t), dt where we have used u(t) : dx(t)/dt and n w B l A P K. (5.240) ( ; l /g) g K K Equations (5.239) are solved subject to initial conditions i(0):i , x(0):x , and u(0) : dx(t)/dt : u . R The coupled equations (5.239) can be rewritten as a ﬁrst-order system and solved numerically using techniques such as the Euler or the Runge-Kutta method (Appendix C). Rapid parametric simulations are possible once drive voltage V (t) and resistance R are speciﬁed. Q The response time of a linear actuator is limited by the time constant of the coil : L/R . However, the response time can be reduced by putting a conductive sleeve around a segment of the stator as shown in Fig. 5.34. The sleeve is referred to as a shorted turn [18]. Consider an equivalent magnetic circuit for the coil-stator-sleeve system (Fig. 5.35). This is basically a transformer with the drive coil as the primary, and the shorted turn as the secondary. The electrical equations for this system K:

419

5.12 LINEAR ACTUATORS

FIGURE 5.34

Linear actuator with shorted turn.

are di di ; M QR V(t) : i R ; L dt dt di di 0 : i R ; L QR ; M , QR QR QR dt dt

(5.241)

where R and L are the resistance and self-inductance of the coil, R QR and L are the resistance and self-inductance of the shorted turn, and M QR is the mutual inductance. The mutual inductance can be expressed as M : (L L , (5.242) QR where 1 is a coupling constant. We desire an expression for the

FIGURE 5.35

shorted turn.

Equivalent magnetic circuit for the stator with the drive coil and

420

CHAPTER 5 Electromechanical Devices

response time of this system, that is, the time dependence of i . Take the Laplace transform of Eq. (5.241), which gives V(s) : I (s)R ; sL I (s) ; sMI (s) QR (5.243) 0 : I (s)R ; sL I (s) ; sMI (s), QR QR QR QR where all initial conditions have been set to zero. We solve Eq. (5.243) for I (s) and obtain V(s) s ; 1 QR I (s) : , (5.244) R s (1 9 ) ; s( ; ) ; 1 QR QR where : L /R and : L /R are the time constants for the coil QR QR QR and shorted turn, respectively. Since 1 we have

(1 9 ) ( ; ). QR QR The denominator in Eq. (5.244) can be simpliﬁed using the approximation sa ; sb ; 1 (sa/b ; 1)(sb ; 1), which applies when a b. Using this approximation reduces the denominator to s (1 9 ) ; s( ; ) ; 1 (s ; 1)(s( ; ) ; 1), (5.245) QR QR P QR where (1 9 ) : QR . P ( ; ) QR Substitute Eq. (5.245) into Eq. (5.244) and obtain s ; 1 V(s) QR I (t) : . R (s ; 1)(s( ; ) ; 1) P QR We examine the response to a step function voltage

0 t0 V 0 t. /s. Transform Eq. (5.246) back to the time domain and V(t) :

Thus, V(s) : V obtain I

(5.246)

1 V (t) : R ( ; 9 ) QR P ; [( 9 )(1 9 e\ROP) ; (1 9 e\RO >OQR)]. QR P

421

5.13 AXIAL-FIELD MOTORS

Since ( ; ), the time response is dominated by (1 9 e\ROP) and P QR the initial current surge can be approximated by V ( 9 ) P (1 9 e\ROP). (t) QR R ( ; ) QR /n we have I

Since L L QR

(1 9 ) . P QR 1 ; R /nR QR In practice, 1 and, therefore, 9 . Thus, the initial current QR P QR surge through the coil when a shorted turn is present is V R (t) (1 9 e\ROP) (shorted turn). R R ; nR QR The response of the coil without the shorted turn is I

I

V (t) : (1 9 e\RO ) R

(no shorted turn).

(5.247)

(5.248)

However, R (1 9 ) , R ; nR P QR and R (1 9 ) 1. R ; nR QR Therefore, . By comparing Eqs. (5.247) and (5.248) we ﬁnd that P the response time of an actuator with a shorted turn is much shorter than without one. Finally, an alternate linear motor conﬁguration with a shorted turn is shown in Fig. 5.36. The response time of this actuator has been analyzed by Wagner [18].

5.13 AXIAL-FIELD MOTORS Permanent magnet axial-ﬁeld motors are used for low-torque, servo and speed control applications. They are also referred to as brushless dc motors, and are commonly used as spindle motors for both disk and cassette tape drives [19]. A typical motor conﬁguration is shown in Fig. 5.37. The motor consists of a set of equally spaced stator coils, ﬁxed

422

CHAPTER 5 Electromechanical Devices

Linear actuator: (a) perspective of core; and (b) cross-sectional view with drive coil.

FIGURE 5.36

beneath a cylindrical multipole rotor magnet with ﬂux return plates positioned above the magnet and below the coils to enhance the ﬁeld in the motor cavity. During normal operation, the coils are energized in a synchronous fashion as they pass under the transition region between adjacent rotor poles. They are synchronized so as to impart a continual steady torque to the rotor. Hall effect sensors are mounted on the stator near the coils to sense the rotor’s magnetic ﬁeld and synchronize the activation of the coils. The numbers of rotor poles and coils are

423

5.13 AXIAL-FIELD MOTORS

FIGURE 5.37

Permanent magnet brushless dc motor.

chosen to ensure proper start up and steady drive torque with no dead zones. The behavior of the motor is relatively straightforward [20]. When a voltage V is applied, the rotor accelerates and eventually reaches a ﬁnal steady-state angular velocity . The equations governing the steady state are V : R i ; K , C C

(5.249)

T : K i, R

(5.250)

and

where R and i are the operating resistance and current and K and K C C R are electrical and torque constants. If there is substantial drag torque, it can be taken into account by modifying Eq. (5.250) as follows: T:K i9T . R B

(5.251)

Drag torque T is due to various loss mechanisms such as friction, air B resistance, etc.

424

CHAPTER 5 Electromechanical Devices

FIGURE 5.38

Performance curves of an axial-ﬁeld motor [19].

We combine Eqs. (5.249) and (5.250) and obtain RT V 9 C . (5.252) K K K C R C Notice that for a ﬁxed voltage V the speed is a linear function of T as shown in Fig. 5.38. There are two parameters that deﬁne this line, the no-load speed , and the stall-torque T , where ,* Q V : , (5.253) ,* K C and :

KV T : R . (5.254) Q R C We will show how to compute these parameters in text that follows. The motor efﬁciency is P T ) : : . P Vi As shown in Fig. 5.38, the efﬁciency depends dramatically on the operating point of the motor. A good design must take this into account. The design of an axial ﬁeld motor is usually an iterative process. The designer usually starts with some basic system requirements such as operating speed and voltage, and then determines motor parameters compatible with these speciﬁcations. While a complete description of

425

5.13 AXIAL-FIELD MOTORS

this process is beyond the scope of this book, nevertheless a brief review of a typical design cycle is instructive. Assume that a motor is needed and that the operating voltage V, speed , and torque T have been speciﬁed along with some basic size constraints. As a ﬁrst step, we decide on a rotor magnet and drive circuitry. We choose an eight-pole magnet (material and size), and a Y-connected three-phase motor driver circuit with an operating voltage V and resistance R . We choose a total C of six equally spaced drive coils with two coils per phase. Two phases will be active at a time to ensure proper start up and smooth drive. Therefore, the operating voltage V is applied across four coils in series. Next, we design the coils. We start with the operating resistnce R . C This is the total (series) resistance of all active coils. In our case, two phases are active at a time so there are four active coils (two per phase). Once R is known, the coils can be speciﬁed (number of turns and wire C gauge). In particular, the number of turns and length of each coil are related to the operating resistance as follows: N N l R : ?A U , C A U

(5.255)

where N is the number of active coils, N is the number of turns in ?A each coil, l is the length of the wire in each coil, and and A are the U U conductivity and cross-sectional area of the wire. In our case there are 4 active coils, N : 4. Notice that Eq. (5.255) constitutes one equation with ?A three unknowns: N , l , and A . Therefore, additional information is U U needed. One approach to obtaining the unknowns is to specify the orientation, shape, and dimensions of the coil. To this end, we specify the motor cavity height h, which puts an upper limit of h 9 t on the K height of the coils (Fig. 5.37). We choose the cross-sectional area A and A mean length per turn l/ (Fig. 5.39). Given the coil dimensions, we write R the following additional relations, N A A : U A f N

(5.256)

and l : f N l/ , (5.257) U E R where f is a wire packing factor and f is a geometric factor. For N E precision wound coils f 0.69. The three unknowns N , l , and A U N U are obtained via the simultaneous solution of Eqs. (5.255), (5.256), and (5.257).

426

FIGURE 5.39

CHAPTER 5 Electromechanical Devices

Coil geometry: (a) top view of the stator; and (b) perspective of

coil.

Once the orientation and geometry of the coils are known we compute the torque constant K . To this end, consider the force on an R inﬁnitesimal segment of coil wire dl. We assume that the rotor ﬁeld is in the z direction and obtain dF : idl ; B : i(drr ; r d ) ; B z E : iB (r dr 9 dr ), (5.258) E where B is the rotor ﬁeld in the gap region. The torque is given by E dT : r ; dF, and if we substitute Eq. (5.258) and take the z-component we ﬁnd that dT : 9iB r dr. X E This is the torque imparted to the segment of the coil by the rotor magnet. Notice that only current in the radial direction contributes to the axial torque. If we consider a typical coil geometry, there are two sides that have an effective length l in the radial direction (Fig. 5.39b). The C torque on the rotor due to an active coil can be estimated as T

: 2iN B R/ l , E C

(5.259)

427

5.13 AXIAL-FIELD MOTORS

where factor 2 takes into account the two sides of the coil and R/ is the mean radius of the magnet, R ;R . R/ : 2 The effective length l can be determined from the coil geometry. The C total torque is simply Eq. (5.259) times the number of active coils, T : 2iN N B R/ l . ?A E C Therefore, the torque constant is K : 2N N B R/ l . (5.260) R ?A E C Furthermore, in the MKS system K : K . At this point, all the terms in C R Eq. (5.260) are known except for B , the ﬁeld in the gap region. This can E be determined using the results of Examples 5.13.1 or 5.13.2 [21—23]. Once we know R , V, K , and K we use Eqs. (5.253) and (5.254) to C R C determine the no-load speed and stall torque T . We then plot the ,* Q speed-torque curve vs T (Fig. 5.38). As the operating speed is given, we can determine both the operating point of the motor and its efﬁciency. EXAMPLE 5.13.1 Develop a two-dimensional model for the ﬁeld in the gap region of an axial-ﬁeld motor (Fig. 5.37). Assume the magnet has a linear second quadrant demagnetization curve of the form B : H ; M , Q

(5.261)

where B : P . (5.262) H A SOLUTION 5.13.1 We reduce the three-dimensional motor geometry to an approximate two-dimensional geometry [21]. This is done by introducing a cylindrical cutting plane at the mean radius of the magnets. This cylinder is then imagined to be unrolled into a two-dimensional surface of inﬁnite extent into and out of the page, as illustrated in Fig. 5.40a. This two-dimensional geometry can be reduced further by exploiting the symmetry that results from the repeating magnetic structure. Speciﬁcally, the ﬁeld is symmetric about the vertical center lines of each pole. The ﬁeld solution for the reduced geometry is similar to that of the periodic geometry of Example 3.6.2. Therefore, we follow closely the solution procedure presented there. As depicted in Fig. 5.40b, there are three regions to consider. In this ﬁgure, l is the arc length from the center

428

FIGURE 5.40

CHAPTER 5 Electromechanical Devices

Motor geometry: (a) approximate 2D geometry; and (b) reduced

geometry.

of one pole to the next at the mean radius, and x is the distance along the ‘‘straightened’’ arc. Therefore, if R and R are the inner and outer radii of the magnet, respectively, and N : the number of poles, then (R ; R ) . l: N The variable x is related to the angular measure along the arc as follows: x 2 . lN The magnetization for the three regions is :

M y Q M : 9M y Q Q 0

(5.263)

(Region 1) (Region 2) (Region 3).

(5.264)

429

5.13 AXIAL-FIELD MOTORS

Because M is constant in each region, the equation for the magnetic scalar Q potential reduces to : 0 K

(5.265)

for all three regions. The three regions under consideration are collectively bounded by four lines that form a rectangular area. The vertical sides of the rectangle, which are deﬁned by the y-axis and the line x : l, represent lines of symmetry for the ﬁeld, along which the normal component H is zero. Moreover, L the boundaries that form the top and bottom of the rectangle are bordered by soft-magnetic materials (ﬂux return plates). For the sake of simplicity, we assume that the ﬂux plates have inﬁnite permeability ( : -). Consequently, the tangential component of the ﬁeld vanishes along these boundaries. When we impose these boundary conditions we ﬁnd that the general solutions for the magnet and gap regions are

n(y 9 h) nx (x, y) : C sinh cos

L l l L

(5.266)

and

ny nx (x, y) : C sinh cos , L l l L

(5.267)

respectively, where C and C are constants that need to be determined.

L L At the interface between the magnet and the gap, the tangential component of H must be continuous: H

R

(x, g) : H (x, g) R

(0 x l).

(5.268)

This condition, along with the fact that the functions sin(nx/l) are orthogonal on (0, l), implies that sinh(ng/l) C :C .

L L sinh(n(g 9 h)/l)

(5.269)

In addition, the normal component of B must also be continuous B

L

(x, g) : B (x, g) L

(0 x l).

(5.270)

430

CHAPTER 5 Electromechanical Devices

As B

L

:

H ; M

L Q H 9 M

L Q

(0 x l/2) (l/2 x l),

(5.271)

then Eq. (5.270) reduces to

M Q

(0 x l/2)

nx n C K(n, h, g, l, ) cos : L l l L 9 M Q where

(l/2 x l)

ng ng n K(n, h, g, l, ) : 9 cosh ; sinh coth (g 9 h) . (5.272) l l l Again, by exploiting the orthogonality of the functions cos(nx/l) on (0, l), it follows that C : L

(91)L\ 4l M Q (n)K(n, h, g, l, )

(n : 1, 3, 5, . . .).

(5.273)

Substituting Eq. (5.273) into Eq. (5.267) gives the complete solution for the gap region (x, y) :

4l M Q

(91)L\ ny nx sinh cos . l l (n)K(n, h, g, l, ) L

Therefore, the vertical or axial component of the ﬁeld in this region is B (x, y) : W

4M Q

(91)L\ (n)K(n, h, g, l, ) L

;cosh

ny nx cos , l l

(5.274)

where x equals the horizontal distance along a circumferential arc at the mean radius of the magnet, y equals the vertical distance above the lower ﬂux plate, g is the gap height, t is the thickness of the magnet, h : g ; t , l : K K (R ; R )/N , and R and R are the inner and outer radii of the magnet.

431

5.13 AXIAL-FIELD MOTORS

Calculations: We apply Eq. (5.274) to a motor with the following parameters: M : 4.0 ; 10 A/m Q g : 6.0 mm t : 4.0 mm K 40 R : mm 120 R : mm l : 20 mm N : 8. (5.275) Field values are computed with : 1.0 , 1.5 , and 2.0 , and with the vertical position in the gap set to y : 2 mm. Ten values are computed for each value of . Speciﬁcally, ﬁeld values are computed at : 0, 5, 10, . . . , 45° (i.e., from the center of one pole to that of its neighbor). These data are shown in Fig. 5.41. Notice that the ﬁeld decreases with increasing . )

FIGURE 5.41

Axial ﬁeld B vs . X

432

CHAPTER 5 Electromechanical Devices

FIGURE 5.42

Motor geometry for 3D analysis.

EXAMPLE 5.13.2 Develop a three-dimensional ﬁeld solution for the gap region of an axial-ﬁeld motor (Fig. 5.42). Assume that the magnet has a second quadrant demagnetization curve B : (H < M z ), Q

(5.276)

where the < term takes into account the alternating polarity of adjacent poles [22, 23]. SOLUTION 5.13.2 For the three-dimensional model we approximate the magnetic structure of the motor in terms of the magnetic circuit shown in Fig. 5.43. Speciﬁcally, we assume that the upper and lower ﬂux plates have inﬁnite permeability ( : -), and are of inﬁnite extent in both the vertical and horizontal directions. The ﬁeld in the gap region of this idealized geometry can be determined using the method of images (Section 3.7). That is, the ﬁeld in the gap can be expressed as a superposition of the ﬁelds from a double inﬁnite sum of image magnets. Let B (r, , z) denote the ﬁeld due to the motor magnet

X in free space. We studied this geometry in Example 4.2.8. The axial ﬁeld B (r, , z) is given by Eq. (4.163). Notice that B (r, , z) can be

X

X considered to be a function of the spatial coordinates (r, , z), and the axial position of the magnet, B (r, , z) : B (r, , z; z , z ),

X

X

(5.277)

5.13 AXIAL-FIELD MOTORS

FIGURE 5.43

433

Cross section of a motor model with inﬁnite ﬂux plates.

where z : z (1) and z : z (2) are the positions of the bottom and top of the Q Q magnet on the z-axis, respectively. Image magnets: To obtain a ﬁeld solution we need to take into account the ﬁelds from the image magnets. These are located in the regions occupied by the ﬂux plates. For example, the images for an inﬁnitesimal magnetic charge Q on the bottom surface of the motor magnet are shown in Fig. 5.44. Each image magnet has a ﬁeld component B (r, , z; z , z ), and the positions z and z

X are chosen so that the tangential component of H is zero and the normal component of B is continuous at the surface of the ﬂux plates. Each image magnet in one ﬂux plate gives rise to a mirror image magnet in the other ﬂux plate. Therefore, there is a double inﬁnite set of image magnets. We write the

434

CHAPTER 5 Electromechanical Devices

FIGURE 5.44

Image charges for Q.

total ﬁeld as a superposition of all the magnets, real and image, B (r, , z) : X

[B (r, , z; Z (n), Z (n))

X L

;B

X

(r, , z; Z (n), Z (n))].

(5.278)

5.13 AXIAL-FIELD MOTORS

435

The axial positions are given by the following recursive relations: Z (n ; 2) : Z (n) ; 2(h ; t ) K Z (n ; 2) : Z (n ; 2) ; 2t K and Z (n) : 9Z (n) Z (n) : 9Z (n), where Z (1) : h, Z (1) : h ; 2t (n : 1, 3, 5, . . .). According to these rela K tions, the ﬁrst few terms are as follows: Z (3) : h ; 2(h ; t ) K Z (3) : (3h ; 2t ) ; 2t K K Z (5) : (3h ; 2t ) ; 2(h ; t ) K K Z (5) : (5h ; 4t ) ; 2t K K and Z (1) : 9(h ; 2t ) K Z (1) : 9h Z (3) : 9(3h ; 4t ) K Z (3) : 9(3h ; 2t ) K Z (5) : 9(5h ; 6t ) K Z (5) : 9(5h ; 4t ). K The ﬁrst and second terms in Eq. (5.278) correspond to the images in the upper and lower ﬂux plates, respectively. If the lower ﬂux plate is absent, the ﬁeld reduces to B (r, , z) : B (r, , z; h, h ; 2t ). (5.279) X X K Calculations: We apply Eq. (5.278) to a motor geometry with the following parameters: M : 7.2 ; 10 (A/m) Q N :4 R : 0.5 cm R : 2.5 cm.

436

CHAPTER 5 Electromechanical Devices

FIGURE 5.45

B vs h. X

First, we compute B at the midpoint of the gap (r : (R ; R )/2, : 0°, and X z : h/2) for a range of gap heights h : 4, 5, . . . , 10 mm, and magnet heights t : 4, 6, 8 and 10 mm. These data are shown in Fig. 5.45. Notice that the ﬁeld K drops off sharply as the gap h increases. This analysis is performed with N : 40 and N : 20, and there is no improvement in accuracy beyond the ﬁrst P ﬁve terms in Eq. (5.278). From these initial calculations, one can select a gap height h and magnet thickness t that render the desired ﬁeld strength. For K example, h : 6 mm and t : 10 mm, which yields a ﬁeld of 0.428 T at the K midpoint of the gap. Next, we ﬁx h : 6 mm and t : 10 mm, and then perform K an analysis to determine the variation of B in the radial direction. Speciﬁcally, X B is computed along two different radial lines located midheight in the gap X (z : h/2) and spanning the surface of the magnets from R r R (Fig. 5.46). These lines are positioned at : 0° (center line of a pole) and : 30°, respectively. Notice that at : 0, B & 0.35 T for 10 mm r 23 mm, and X that it drops off rapidly outside of this interval. There is also a considerable drop in ﬁeld strength as one moves in an angular sense from the center of the pole : 0. To see this, the ﬁeld is computed along three different circumferential arcs spanning an angular measure of 90°, that is, from the center of one pole to the center of the neighboring pole at radii r : 10, 15 and 20 mm, respectively,

437

5.14 STEPPER MOTORS

FIGURE 5.46

B vs r. X

with (z : h/2) (Fig. 5.47). This analysis shows that the ﬁeld drops off sharply for 30°. In summary, we ﬁnd that the region of maximum ﬁeld strength for this motor geometry is 10 mm r 23 mm with 930° 30°. )

5.14 STEPPER MOTORS Stepper motors execute rotational motion in a sequence of discrete angular steps. Well suited for open-loop digital control, they are widely used for precision motion and positioning. These motors can be found in numerous industrial and consumer products including computers and peripherals, disk drives, printers, robotic devices, and in electronic cameras where they are used for lens focusing. Stepper motors come in a variety of conﬁgurations and designs. In this section we study a permanent magnet stepper motor known as a ‘‘can-stack’’ motor. This motor consists of two separate bobbin-wound drive coils that enclose axial sections of a cylindrical radially polarized multipole permanent magnet rotor (Fig. 5.48). The rotor is usually made from an isotropic grade of ceramic ferrite, Sm Co or Nd Fe B.

438

CHAPTER 5 Electromechanical Devices

FIGURE 5.47

FIGURE 5.48

B vs . X

Exploded view of a permanent magnet stepper motor.

5.14 STEPPER MOTORS

439

This is a two-phase motor with each coil comprising a separate drive phase. Each coil is mounted on a soft-magnetic stator element that consists of two opposing toothed components and a ﬂux return ring. The number of teeth on a stator equals the number of poles on the rotor. The drive coils are assembled by ﬁrst sliding the coil and ﬂux return ring on one of the toothed components and then inserting the remaining toothed element into the assembly in such a way that the stator teeth are equally spaced. A cross section of the interior of an assembled four-pole stator is shown in Fig. 5.49a. An ‘‘unrolled’’ perspective of an assembled stator element (without the coil) is shown in Fig. 5.50.

Cross section of stepper motor stator: (a) unenergized stator (rotor not shown); (b) energized stator (rotor not shown); (c) rotor at position of maximum drive torque; and (d) rotor rotated to position of zero drive torque.

FIGURE 5.49

440

CHAPTER 5 Electromechanical Devices

FIGURE 5.50

Magnetic circuit of stator element.

In the can-stack motor each rotation step is initiated by activating a single phase. Speciﬁcally, when one phase is activated (with the other phase off) the rotor experiences a torque that rotates it into magnetic alignment with that phase. The stator teeth of the two phases are offset from one another by half-a-tooth (or pole) pitch in an angular sense. Therefore, by sequentially activating the two phases the rotor can be repetitively stepped an angular measure equal to half a pole pitch. The torque developed by the motor can be understood by considering the activation of a single phase as depicted in Fig. 5.49. The activation sequence is as follows: (a) the coil is unactivated (rotor not shown); (b) the coil is activated and a magnetic ﬁeld is created in the stator cavity (rotor not shown); (c) the rotor is initially in a position where it experiences maximum drive torque that is due to the interaction of the radially directed stator ﬁeld and equivalent current sheets of the rotor (surface currents exist at the transitions between neighboring poles); and (d) the rotor has rotated into magnetic alignment with the phase where it experiences zero torque. This occurs when the equivalent current sheets of the rotor are centered with respect to the gap as shown. When

441

5.14 STEPPER MOTORS

the rotor is in this position there is no drive torque because the stator ﬁeld across the equivalent current sheets is parallel to the direction of rotation as shown in Fig. 5.49b. The alignment of the rotor to a given phase is enhanced by the lower reluctance of the rotor/stator circuit in the aligned position (Fig. 5.49d). When the rotor aligns with one phase it is in a position maximum torque for the other phase because the stator teeth for that phase are offset by half-a-tooth (or pole) pitch. Thus, the rotor can be continuously rotated (stepped) by activating the two phases in sequential fashion. In the following example, we derive an expression for the drive torque of a ‘‘can-stack’’ stepper motor. EXAMPLE 5.14.1 Derive an expression for the drive torque of the stepper motor shown in Fig. 5.48. Assume that the rotor is a radially polarized multipole cylinder with a magnetization M : <M r , (5.280) Q where < takes into account the alternating polarity of adjacent poles. Further assume that each stator is operating below saturation in a linear region of its B-H curve with . SOLUTION 5.14.1 The torque is derived in two steps. First, we obtain a ﬁeld solution in the interior of an energized stator with no rotor present. Second, we reduce the rotor to an equivalent current distribution and then compute the torque on it using a Lorentz force approach with the stator ﬁeld as an external ﬁeld. Stator ﬁeld: Let L and R denote the length and inner radius of the stator, Q respectively. Each stator has N teeth, which equals the number of poles on the rotor. The stator is enclosed by an n-turn coil and a ﬂux return ring. A cross section of the interior of a four-pole stator is shown in Fig. 5.49a. Let R denote the angular span of each tooth and denote the angular expanse of the E gap between each tooth. These angles are related by the following equation, 2 : 9 . (5.281) E N R We seek a ﬁeld solution for the interior of the stator. To obtain an analytical solution, we assume that the stator is inﬁnitely long. This assumption is not as radical as it might ﬁrst seem because the stator tends to conﬁne the ﬂux and minimize leakage etc. In the interior of the stator, ; H : 0.

442

CHAPTER 5 Electromechanical Devices

Thus, we can represent H in terms of a scalar potential, H : 9 (Section K 3.4). The ﬁeld problem reduces to a boundary value problem (BVP) in cylindrical coordinates (Section 3.6.2). Speciﬁcally, we solve : 0. (5.282) K The general solution to Eq. (5.282) that is well behaved at r : 0 is of the form (r, ) : a rN sin(p) ; b rN cos(p). (5.283) K N N N We choose a coordinate system in which : 0 coincides with the center of a stator tooth (Fig. 5.51a). In this coordinate system (r, ) must be an even K function of . Thus, we discard the sin(n) terms in Eq. (5.283) and obtain (r, ) : b rN cos(p). (5.284) K N N We determine the coefﬁcients b by matching the boundary conditions at the N

Stepper motor stator: (a) reference frame; and (b) proﬁle of scalar potential along the stator surface.

FIGURE 5.51

443

5.14 STEPPER MOTORS

interior surface of the stator. We assume that the potential (R , ) along this K Q surface is constant across each tooth and alternates from to 9 on

successive teeth. As the gap between neighboring teeth is small ( ), we E R assume that the potential is linear between the teeth and this implies a proﬁle for (R , ) as shown in Fig. 5.51b. To solve the BVP we need a Fourier series K Q representation for (R , ). First, we derive an expression for in terms K Q

of the motor’s physical parameters. Let g denote the gap between adjacent stator teeth, and let H denote the ﬁeld in the gap region (Figs. 5.49b and 5.50). We E assume that the potential is linear between adjacent teeth. Therefore, 1 K H :9 E R Q 2 :9 , g

(5.285) (5.286)

where : and g : R . We apply Eq. (3.138) from Section 3.5 to the E Q E dotted path in Fig. 5.50 and obtain ni H : . E g

(5.287)

Therefore,

:9

ni . 2

(5.288)

Now that we know , we can represent the potential along the stator

(R , ) in terms of the Fourier series K Q (R , ) : A cos( ), (5.289) K Q I I I

where

8

[cos(k) 9 1] cos R A :9 I I kN 2 E 4 ni : [cos(k) 9 1] cos R , (5.290) I 2 kN E and : kN /2. We evaluate Eq. (5.284) at r : R , compare this with Eq. I Q (5.289), and obtain b : A R\N N I Q

p:

kN . 2

444

CHAPTER 5 Electromechanical Devices

Thus, the potential in the interior of the stator is

r ?I A cos( ). (5.291) I R I I

Q The ﬁeld components are obtained from B : 9 . For example, K I A r? \ cos( ). B (r, ) : 9 (5.292) P I R?I I I I

Q Drive torque: To compute the torque we represent the rotor as a distribution of equivalent currents. Recall that the rotor is radially polarized with (r, ) : K

M : <M r . Q Let R and R denote its inner and outer radius, respectively. Assume that it is initially positioned with one of its north poles centered about 0° (i.e., each pole is initially centered with respect to a stator tooth). The equivalent current densities are given by Eq. (3.95). We ﬁnd that the volume current density is zero, J : ; M : 0, and that the surface current density is K j : M ; n K : <M z . Q These surface currents are located at the edges of each sector at angular positions

1 2 : n; (n : 0, 1, 2, . . . , N 9 1). L 2 N It is important to note that there are two such currents at these positions, one along the surface of each of the neighboring sectors. If the rotor is rotated by an angle , the positions of the surface currents change to

1 2 (n : 0, 1, 2, . . . , N 9 1). () : ; n ; L 2 N The surface current densities at these locations are

M z (n : 0, 2, 4, . . .) Q j (n) : K 9M z (n : 1, 3, 5, . . .). Q The torque is given by Eq. (3.100) from Section 3.3, that is, T() : 2L

, \ 0 r ; [j (n) ; B (r, ())] dr. K L L 0

(5.293)

(5.294)

445

5.14 STEPPER MOTORS

Factor 2 takes into account the contribution of each surface of the neighboring sectors. From Eq. (5.294) we ﬁnd that the component of torque about the z-axis is

, \ 0 (91)L B (r, ())r dr. P L L 0 Finally, substituting Eq. (5.292) into Eq. (5.295) gives T () : 2LM X Q

, \ (91)L L I

(R?I> 9 R?I>) I ;A cos[ ()], I ;1 I L R?I Q I where L and R are the length and interior radius of the stator and Q 4 ni A : [cos(k) 9 1] cos R . I kN I 2 E Here,

(5.295)

T () : 2 LM X Q

(5.296)

: I

kN , 2

and

1 2 () : ; n ; . L 2 N Calculations: We demonstrate Eq. (5.296) with some sample calculations. The following parameters are used in the analysis: M : 4.3 ; 10 A/m Q N : 10 n : 100 turns i : 0.25 A R : 6.0 mm R : 8.0 mm R : 8.1 mm Q L : 2.0 mm. The angular span of each stator tooth is taken to be : 27° (75% of the span R

446

CHAPTER 5 Electromechanical Devices

of a rotor pole). The angular span of the gap between neighboring teeth is : 9° ( : 2/N 9 ). There are two phases to consider, which are E E R labeled 1 and 2, respectively. Recall that the stator teeth of the two phases are offset from one another by an angular span equal to one half of the pole pitch, which in this case is 18°. The drive torque T () for each phase is shown in Fig. X 5.52. This plot shows the torque on the rotor as it rotates the angular span of one pole (036°). When :0, the rotor is positioned as in Fig. 5.49c and d for phase 1 and phase 2, respectively. In this initial position, phase 1 exerts almost maximum torque, and phase two exerts zero torque. The peak torque for phase 1 is 1500 N · m, which occurs at : 12°. When the rotor is rotated to : 18° (half a pole pitch) as shown in Fig. 5.49d, the torque of phase 1 is zero and the torque of phase 2 is near a maximum. For 18 36°, the torque of phase 1 reverses direction as shown. )

5.15 HYBRID ANALYTICAL-FEM ANALYSIS In the previous sections we derived analytical lumped-parameter models for the analysis of various electromechanical devices. Such models can be developed for some devices, but certainly not all. Geometric asymmetries and/or material nonlinearities often preclude a full analytical treatment. However, in such cases it is still possible to perform a lumped-parameter analysis using a hybrid analytical-FEM approach. In this section we describe the approach and demonstrate it via application to a practical device. In the hybrid analytical-FEM method we set up the equations of motion and identify which terms, if any, can be obtained in analytical form. We derive expressions for these terms using the methods presented in the previous sections. Next, we consider the remaining terms. We obtain expressions for these using parametric FEA. The method is as follows. First, specify a discrete set of values for the independent variables i and x or that span the working range of the device. Next, perform a series of FEA calculations to determine values for the unknown terms for each value of the independent variables. The computed data is then ﬁt to a smooth curve, which renders the unknown terms in analytical form. For example, suppose we are analyzing a rotary actuator and need an expression for the torque T() as a function of rotation angle . We ﬁrst specify a discrete set of angular values ( , , . . . , , . . . , ) G , that span the range of motion of the rotor. Next, we perform FEA calculations and determine T() at these values. This results in a series

5.15 HYBRID ANALYTICAL-FEM ANALYSIS

FIGURE 5.52

Drive torque vs : (a) phase 1; and (b) phase 2.

447

448

CHAPTER 5 Electromechanical Devices

Rotary permanent magnet actuator: (a) actuator perspective; and (b) cross-sectional view with reference frame.

FIGURE 5.53

of data ( , T( )) (i : 1, 2, . . . , N). This data is then ﬁt to a ﬁnite polyG G nomial of the form, ) T() : a I. I I This renders T() in a form suitable for a lumped-parameter analysis. We demonstrate the hybrid analytical-FEM approach via application to the rotary actuator shown in Fig. 5.53. This type of actuator is used to implement rotational motion that is typically 45°. The actuator consists of a drive coil that is wrapped around a soft-magnetic core and a

449

5.15 HYBRID ANALYTICAL-FEM ANALYSIS

cylindrical bipolar magnet mounted for rotation in the gap of the core. When the coil is energized, the ﬁeld generated by the coil imparts a torque to the magnet that causes it to rotate clockwise. As it rotates, the magnet experiences an additional counterclockwise torque due to the core itself. This is called reluctance torque. When the coil is turned off, the reluctance torque rotates the magnet back to its initial position where it is held in place against a mechanical stop mechanism. We develop a model for analyzing this actuator in the following example [24]. EXAMPLE 5.15.1 Derive the equations of motion for the actuator shown in Fig. 5.53. SOLUTION 5.15.1 This is a moving magnet actuator and its behavior is governed by rotational motion equations (5.87):

di(t) 1 "(i, ) : V (t) 9 i(t)(R ; R ) 9 (t) , dt L Q d (t) 1 : T(i, ), dt J K d(t) : (t). dt

(5.297)

To solve Eq. (5.297) we need expressions for "(i, ) , L:

"(i, ) , i

(5.298) (5.299)

and T(i, ).

(5.300)

Assumptions: In order to simplify the analysis we make the following assumptions. First, we assume that the core is operating in a linear region of its magnetization curve with a permeability . B : H (core). (5.301) Thus, the magnetic circuit is linear, and the effects of coil and magnet can be separated, B:B ;B ,

"(i, ) : " (i) ; " (),

(5.302) (5.303)

450

CHAPTER 5 Electromechanical Devices

and T(i, ) : T

(i, ) ; T (),

(5.304)

where B : ﬁeld due to the coil B : ﬁeld due to the magnet

T (i, ) : torque on the magnet due to the coil T () : torque on the magnet due to the core " () : ﬂux linkage due to the magnet

" (i) : ﬂux linkage due to the current. (5.305) Notice that " () and T () are functions of only. If the magnetic circuit

is nonlinear, then a nonlinear analysis needs to be performed for each time stepped integration of Eq. (5.297). The second assumption is that the magnet has a linear second-quadrant demagnetization curve of the form B : (H ; M ), (5.306) Q where M is uniform and ﬁxed. Q Flux linkage: The ﬂux linkage is given by Eq. (5.303). Therefore, Eqs. (5.298) and (5.299) reduce to "(i, ) d" () : , d

(5.307)

and L:

d" (i) , di

(5.308)

respectively. Flux linkage due to the magnet: The ﬂux linkage due to the magnet is determined using FEA, FEA $ "

(). (5.309)

Speciﬁcally, FEA calculations are performed for a discrete set of angles I (k : 1, 2, . . .) that span the anticipated rotation of the magnet. The ﬂux linkage " ( ) is computed (with the coil turned off) for each value of ,

I I " ( ) :

I

B 1

( ) · ds.

I

integration over coil

5.15 HYBRID ANALYTICAL-FEM ANALYSIS

451

Typical FEA ﬂux analysis: (a) rotor at : 0°; (b) rotor at : 45°; (c) rotor at : 90°; and (d) normalized ﬂux vs .

FIGURE 5.54

Typical FEA calculations are shown in Fig. 5.54. Speciﬁcally, ﬂux plots are shown for (a) : 0°, (b) : 45°, and (c) : 90°. Notice that the maximum ﬂux linkage occurs when : 0°. A normalized plot of () (the average ﬂux through the coil) is shown in (d). From these calculations we obtain the data ( , " ( )) and then ﬁt it to a polynomial, I I , " () : a I.

I I

(5.310)

452

CHAPTER 5 Electromechanical Devices

From this we obtain () ,

: a kI\ (FEA), I d I which is the desired analytical expression for Eq. (5.307). d"

(5.311)

Flux linkage due to the coil: The ﬂux linkage due to the coil can be determined using FEA or magnetic circuit analysis. As the circuit is linear, the ﬂux linkage is proportional to the current and the constant of proportionality C can be computed using FEA, "

(i) : niC (FEA). (5.312) On the other hand, if the gap g in the circuit is small relative to the dimensions of the cross-sectional area, then the ﬂux (i) through each turn of the coil can be determined using magnetic circuit analysis as in Example 3.5.1. This gives niA , (i) : g

(5.313)

where g and A are the length and area of the gap. As the coil has n turns, the ﬂux linkage is "

(i) : n (i) niA . : g

In either case, from Eq. (5.308) we have L:

nC nA g

(FEA), (5.314) (circuit analysis).

Torque: As noted already, the torque on the magnet can be separated into two components T(i, ) : T

(i, ) ; T (). (5.315) The term T (i, ) is the torque on the magnet due to the current through the coil. This is referred to as the drive torque. The other term, T () is the torque on the magnet due to the presence of the magnetic circuit (core). It is independent of the current, and is referred to as the reluctance torque.

5.15 HYBRID ANALYTICAL-FEM ANALYSIS

453

Drive torque: The drive torque can be computed using the formula for the torque on a bipolar cylinder in an external ﬁeld (Example 3.4.4), : M hR B sin()z , Q are the length and radius of the magnet. In addition, T

where h and R

ni B : g

(5.316)

is the ﬁeld in the gap due to the coil (Example 3.5.1). Therefore, ni sin(). T (i, ) : M hR Q g

(5.317)

This result is valid when magnetic circuit analysis applies. When this is not the case, an analytical expression for T (i, ) can be determined using FEA. Speciﬁcally, choose a reference current i and perform a single FEA to compute at a series of points (R , ) that span the ﬁeld components B and B

H P the circumference of the magnet, (R , ) : B (R , )r ; B (R , ) . (5.318) H P H H This data is then substituted into Eq. (3.134), which in turn can be evaluated numerically for a sequence of angles that span the angular rotation of the I magnet. The resulting data ( , T ( )) is then ﬁt to a ﬁnite polynomial I I i , T (i, ) : d I, (5.319) I i I where the multiplier i/i reﬂects the linearity of the circuit. Reluctance torque: The reluctance torque is computed using FEA B

FEA $ T

(). (5.320) This torque is computed for a set of angular values that span the angular I rotation of the magnet. The computed data ( , T ( )) is then ﬁt to a ﬁnite I I polynomial, , () : c I. (5.321) I I A sample plot of T () (obtained from FEA) is shown in Fig. 5.55; it is negative because it acts to rotate the magnet counterclockwise (clockwise rotation is taken to be positive). Notice that T () is maximum at : 45°. T

454

CHAPTER 5 Electromechanical Devices

FIGURE 5.55

Plot of reluctance torque T

() vs .

Calculations: We demonstrate the hybrid approach with some sample calculations. The following parameters are used in the analysis: R : 3.0 n : 500 turns M : 4.3 ; 10 A/m Q R : 4.0 mm

g : 10.0 mm h : 25.0 mm J : 6 ; 10\ kg · m. K The width of the core is 10 mm. Therefore, A : 250 mm. First, determine Eqs. (5.311), (5.314), (5.317), and (5.321) and then substitute these into the

455

5.16 MAGNETIC MEMS

FIGURE 5.56

Normalized coil current vs t.

equations of motion (5.297). We apply a step function voltage pulse of the form V (t) : Q

V 0

(0 t T ) (T t -),

where V :12 V and T :2.0 ms. The initial conditions are i :0 A, :0.175 radians (10°), and : 0 rad/s. We solve Eq. (5.297) numerically using Euler’s method. A normalized plot of i(t) is shown in Fig. 5.56. Notice that the current begins decaying after T when the applied voltage becomes zero. Normalized torque proﬁles are shown in Fig. 5.57. Finally, a plot of (t) is shown in Fig. 5.58. As expected, (t) reaches a maximum and then returns to its initial value due to the reluctance torque. )

5.16 MAGNETIC MEMS In the previous sections we have studied conventional electromechanical devices. These are macroscopic in size, typically greater than one cubic

456

CHAPTER 5 Electromechanical Devices

FIGURE 5.57

Normalized torque proﬁles vs t.

FIGURE 5.58

Rotation angle (t) vs t.

457

5.16 MAGNETIC MEMS

centimeter in volume. In this section we study micro-electromechanical devices. These are typically less than one cubic millimeter in volume, and can have features that are on the order of microns. Such devices fall under the general category of micro-electromechanical systems (MEMS) [25]. These MEMS devices are fabricated using techniques derived from the microelectronics industry. A typical fabrication cycle includes steps such as mask design and generation, photolithography, and bulk and surface micromachining [26]. These methods enable a high degree of integration, and a low manufacturing cost with tens to hundreds of devices realized on a single silicon wafer. Advances in microsystems technology have spawned the rapid development of a variety of devices for both research and commercial use. These include accelerometers, light modulators, microﬂuidic devices, micromotors, molecular ﬁlters, and various actuators and sensors [25, 27]. To date, the majority of MEMS transducers have been electrostatically driven. There are at least two reasons for this. First, electrostatic actuators lend themselves to microelectronic fabrication methods. Second, the electrostatic force scales relatively well at the micro domain. We demonstrate this using a model of a parallel plate capacitor. The force of attraction between the plates is

+ A F : E, C 2 where + is the permittivity of free space, A is the area of each plate, and E is the electric ﬁeld between the plates. If we keep the E-ﬁeld constant, F scales as L where L is the characteristic dimension of the capacitor. C This follows from the L dependency of A. Thus, if the dimensions of a capacitor are reduced by a factor of ten, the electrostatic force decreases by a factor of one hundred. The magnetostatic force for current-driven actuators does not scale as well as the electrostatic force. To see this, consider the force between a pair of parallel wires, s F : I I . A 2 d Here, I and I are the currents in each wire, s is the length of the wires, and d is the separation distance. The current is I : JA where J is the current density and A is the area of the wire. The force can be rewritten as s F : J J A A . A 2 d

(5.322)

458

CHAPTER 5 Electromechanical Devices

From Eq. (5.322) we see that if the current density is kept constant, the force scales as L, where L is the characteristic dimension of the circuit. Thus, if the dimensions of a circuit are reduced by a factor of 10, the force decreases by 10,000, which is two orders of magnitude weaker than the electrostatic case. If we allow the current density to increase, but keep the heat ﬂow through the surface of the wires constant, the force scales as L. Or, if we allow the current density to increase, but keep the maximum temperature rise constant, the force varies as L [28]. This last case scales as well as the electrostatic case, but it will likely give rise to deleterious thermal effects. We can perform the same analysis for the force between a magnet and a conductor. Consider a conductor carrying a current I with a length L perpendicular to a B-ﬁeld. The force on the conductor is : ILB.

The B-ﬁeld of the magnet is independent of size. That is, if we change all the linear dimensions of a permanent magnet, the ﬁeld strength at all the rescaled observation points remains constant. However, the current I scales as L for the constant current density case, L for the constant heat ﬂow case, and L for the constant temperature rise case [28]. Therefore, F scales as L for the constant current density

case, L for the constant heat ﬂow case, and L for the constant temperature rise case. Again, the last case scales as well as the electrostatic force, but there could be substantial issues regarding the management of thermal energy. It is instructive to consider how the ﬁeld of a current source scales at the micro domain. Consider an inﬁnite straight wire. The B-ﬁeld at a distance r from the wire is F

I B(r) : . 2r As I : JA we have JA B(r) : , 2 r which scales as L (B L). Thus, if we reduce the dimensions of a circuit, the current density J must increase by 1/L if ﬁeld strength at the rescaled observation points is to remain constant. By comparison, if we change all the linear dimensions of a permanent magnet, the ﬁeld strength at all the rescaled observation points remains constant (assuming that the magnetization is constant). Moreover, there is no power consumption.

459

5.16 MAGNETIC MEMS

Therefore, permanent magnets have advantages over conventional current sources in terms of the ﬁelds they provide at the micro domain. In spite of the relatively poor scaling of the magnetic force, there is a substantial and growing interest in magnetic microactuators. Over the last few years several research groups have demonstrated viable devices [29—31]. There has also been recent progress towards the commercialization of such devices [32]. However, most of these devices are not fabricated in a purely batch mode. Instead, they typically require a modest amount of manual assembly. It is likely that magnetic microactuators will gain in importance as methods for depositing rare-earth materials such as NdFeB onto microstructures improve. In the remainder of this section we discuss the operation of a batch-fabricated magnetic microactuator. A batch-fabricated magnetic microactuator is shown in Fig. 5.59. It consists of a soft magnetic (NiFe) plate supported on one side by a narrow torsion beam that is anchored at either end to a silicon substrate. The fabrication process for this structure is shown in Fig. 5.60a—e [30]. The operation of the actuator is as follows: When a uniform ﬁeld H is applied, it induces a magnetization in the plate and a magnetic torque develops that acts to rotate the plate up, away from the substrate. This rotation is resisted by a restoring torque that results from the torsion of the support beam. The plate rotates to an equilibrium deﬂection angle where the magnetic and restoring torques are in balance (Fig. 5.61b). To determine we need to compare the two torques. We consider the restoring torque ﬁrst. This is due to the torsion of the support beam. Since the midportion of the beam is attached to the plate, it does not twist. Therefore, the restoring torque is due to the torsion of the two end portions of the beam (between the plate and the anchor points). Each of these can be considered as a separate beam with length l and cross-sectional dimensions w and t. We assume that both

FIGURE 5.59

Magnetic MEMS actuator.

460

CHAPTER 5 Electromechanical Devices

FIGURE 5.60

Fabrication process for actuator: (a)—(e) [30].

beams obey Hooke’s law T () : k, K where T is the mechanical restoring torque and k is the torsional K stiffness. This is given by k : 2

KG ; J , l

(5.323)

461

5.16 MAGNETIC MEMS

FIGURE 5.61

A MEMS actuator: (a) top view; and (b) side view [30].

where K is a cross-section dependent factor, G is the shear modulus, is the residual stress in the beam, and J is the polar moment of inertia. The term J in Eq. (5.323) is usually negligible compared to KG. The formulas for K, G, and J are as follows: K:

1 nb tanh n 2a L

E G: , 2(1 ; v)

ab 192a 19 3 b

,

(5.324) (5.325)

and J:

1 (wt ; wt), 12

(5.326)

where E is Young’s modulus, v is Poisson’s ratio, a : min(w, t), and b : max(w, t). To get a sense of the magnitude of these terms we analyze a polysilicon beam with E:170 GPa, l : 400 m, w : 2.2 m, and t : 2.2 m. We evaluate Eqs. (5.324)—(5.326) and ﬁnd that K : 3.3 ; 10\ m, J : 3.9 ; 10\ m, and k : 2 nN-m/rad [30].

462

CHAPTER 5 Electromechanical Devices

We now consider the magnetic torque. If there is no external ﬁeld (H : 0) the plate is unmagnetized, and it remains at rest in a down position. When an external ﬁeld is applied (H 0), it induces a magnetization M in the plate, which couples to H and this gives rise to a torque T . The torque is given by Eq. (3.133), which in this case & reduces to T : V M ; H

& : V MH sin(), (5.327)

where V is the volume of the magnet and is the angle between M

and H . Before we can determine T we need an expression for the & magnetization M. The magnetization of the plate depends on H as speciﬁed by its M vs H plot (Section 1.11). However, H (the ﬁeld inside the plate) depends on M. Recall from Section 1.8 that the ﬁeld inside a magnetized specimen is H:H ;H ? B : H 9 N M, ? + where H is the applied ﬁeld and H : 9N M is the demagnetization ? B + ﬁeld (N is the demagnetization factor). Thus, M is functionally depend+ ent on itself, which makes it difﬁcult to determine. Often, we obtain M by making simplifying assumptions. For example, if the material is below saturation, we sometimes assume that it is linear (Example 1.8.1), or that it has an inﬁnite permeability - (Example 3.2.8). On the other hand, if the material is saturated we can assume that it has a ﬁxed magnetization M . For this application we model the magnetization of Q the plate below saturation with an anisotropy model, and above saturation we use a uniform magnetization M . Q Below saturation, the magnetization model for the plate is based on uniaxial anisotropy (Section 1.9). Speciﬁcally, we assume that the plate has a single easy axis. When an external ﬁeld is applied it induces a magnetization M and also rotates M an angle away from the easy axis. Consequently, the plate acquires an energy density w : K sin(). ? ? The constant K can have contributions from stress, crystalline and shape ? anisotropy (Section 1.9). In this case, the plate is a low-stress polycrystalline thin ﬁlm, and therefore the stress and crystalline anisotropy are negligible compared to the shape anisotropy. The shape-induced easy

463

5.16 MAGNETIC MEMS

axis runs the length of the plate, parallel to its surface. The constant K is ? M K : Q (N 9 N ), ? R J 2 where N and N are the anisotropy constants along thickness and length R of the plate, respectively [33]. There is a shape-induced torque on M which is given by w ? T () : 9V

? : 9V K sin(2), (5.328)

? where V is the volume of the plate. The negative sign in Eq. (5.328)

indicates that T () acts to oppose an increase in . That is, T () acts to ? ? align M with the easy axis. There is an equal and opposite torque 9T () ? on the plate itself. This acts to rotate the plate towards M, away from the substrate (Fig. 5.62). We use Eq. (5.328) to determine the equilibrium rotation of the magnetization. However, to determine this we still need an expression for M. We obtain this by considering the components of H along the direction of M. Speciﬁcally, we have H0 : H0 ; H0 , ? B where H0 and H0 are the applied and demagnetization ﬁeld components ? B parallel to M (Section 1.8). Let denote the ﬁxed angle between H and the initial (unactivated) direction of the easy axis, and let denote the equilibrium rotation of the easy axis (i.e., of the plate) (Fig. 5.62). The

FIGURE 5.62

Torque diagram with reference frame [30].

464

CHAPTER 5 Electromechanical Devices

angle between M and H is : 9 9 . Therefore, H0 can be written as ? H0 : H cos( 9 9 ). ? The component H0 is given by B H0 : 9N ()M, B + where N () is the demagnetization factor that accounts for the rotation + of M. Speciﬁcally, if M is at an angle with respect to the easy axis then Therefore,

N () : N cos() ; N sin(). + J R

H0 : 9(N cos() ; N sin() M. B J R At this point, we have expressions for H0 and H0 , but still no ? B expression for M. We obtain this by making a further assumption: When the plate is subjected to H , the domain structure within it changes in such a way as to minimize the magnetostatic energy. This gives H 0 inside the plate, which implies that H0 : 9H0 , ? B

or

H cos( 9 9 ) M : . (5.329) N () + We use Eq. (5.329) for M when the plate is below saturation. Above saturation we assume that M : M . Therefore, the magnetization can be Q written as H cos( 9 9 ) M : min ,M . (5.330) Q (N cos() ; N sin() J R We are ﬁnally ready to determine the equilibrium rotation angle of the plate. Equilibrium occurs when the magnetic and restoring torques are in balance. We consider the torque on both M and the plate (Fig. 5.62). The torque on M is

where and

T (, , M) : T (, , M) 9 T (), + & ? T : V MH sin( 9 9 )

&

(5.331)

T () : V K sin(2).

? ?

(5.332)

465

5.16 MAGNETIC MEMS

FIGURE 5.63

Rotation angle vs applied ﬁeld H [30].

We use Eq. (5.330) for M in Eq. (5.331). The torque on the plate is (, ) : T () 9 T (). ? K In equilibrium, the net torque on both M and the plate is zero. This occurs when T

T (, ) : T () & ?

(5.333)

and T () : T (). ? K Combining Eqs. (5.333) and (5.334) gives

(5.334)

T (, , M) : T () : T (). (5.335) & ? K The relation (5.335) represents a set of transcendental equations for , , and M. These equations can be solved numerically for the three variables. A typical response curve for the actuator is shown in Fig. 5.63 [30]. If a time-varying ﬁeld is applied the plate can be made to oscillate up and down [34].

466

CHAPTER 5 Electromechanical Devices

References 1. Brown, D. and Hamilton III, E. P. (1984). Electromechanical Energy Conversion, New York: Macmillan Publishing Co. 2. Fitzgerald, A. E., Kingsley, C., and Umans, S. D. (1983). Electric Machinery, 4th ed., New York: McGraw-Hill. 3. Krause, P. C. and Wasynczuk, O. (1989). Electromechanical Motion Devices, New York: McGraw-Hill. 4. Woodson, H. H. and Melcher, J. R. (1968). Electromechanical Dynamics, Part I: Discrete Systems, New York: John Wiley & Sons. 5. Woodson, H. H. and Melcher, J. R. (1985). Electromechanical Dynamics, Part II: Fields, Forces and Motion, Malabar, FL: John E. Krieger Publishing Co. 6. Taylor, E. F. and Wheeler, J. A. (1992). Spacetime Physics: Introduction to Special Relativity, 2nd ed., New York: W. H. Freeman and Co. 7. Moon, F. C. (1984). Magneto-solid Mechanics, New York: Wiley. 8. Lowther, D. A. and Silvester, P. P. (1986). Computer-aided Design in Magnetics, New York: Springer-Verlag. 9. Furlani, E. P. and O’Brien, M. (1994). Analysis of axial-ﬁeld actuators. IEEE Trans. Mag. 30(6): 4323. 10. Furlani, E. P. and O’Brien, M. (1996). Predicting the dynamic behavior of axial-ﬁeld actuators. IEEE Trans. Mag. 32(2): 302. 11. Furlani, E. P. (1994). A three-dimensional ﬁeld solution for axially-polarized multipole disks. J. Magn. Magn. Mat. 135(2): 205. 12. Furlani, E. P. (1996). An analytical method for predicting the transient behavior of resonant electromagnetic actuators. Int. J. Appl. Elec. Mech. 9: 101. 13. Marchant, A. B. (1990). Optical Recording: A Technical Overview, Reading, MA: Addison-Wesley Pub. Co. 14. Mansuripur, M. (1995). The Physical Principles of Magneto-optical Recording, Cambridge, England: Cambridge Univ. Press. 15. McDaniel, T. W. and Victora, R. H. (1997). Handbook of Magneto-Optical Recording; Materials, Subsystems, Techniques, Westwood, NJ: Noyes Pub. 16. Furlani, E. P. and Williams, C. (1995). Formulas for optimizing the bias ﬁeld for magneto-optic recording. J. Magn. Magn. Mat. 148: 475. 17. Furlani, E. P. (1997). Field analysis and simulation of a permanent magnet bias ﬁeld device for magneto-optic recording. J. Phys. D: Appl. Phys. 30: 1846. 18. Wagner, J. A. (1982). The shorted turn in the linear actuator of a high performance disk drive. IEEE Trans. Magn. 18(6): 1770. 19. Kenjo, T. and Nagamori, S. (1984). Permanent Magnet and Brushless DC Motors, Oxford: Oxford Univ. Press. 20. Campbell, P. (1974). Principles of a permanent magnet axial-ﬁeld dc machine. IEE Proc. 121: 1489. 21. Furlani, E. P. (1994). Computing the ﬁeld in permanent-magnet axial-ﬁeld motors. IEEE Trans. Mag. 30(5): 3660.

5.16 MAGNETIC MEMS

467

22. Furlani, E. P. and Knewtson, M. A. (1997). A three-dimensional ﬁeld solution for permanent-magnet axial-ﬁeld motors. IEEE Trans. Mag. 33(3): 2322. 23. Furlani, E. P. (1997). Field analysis and optimization of NdFeB axial-ﬁeld permanent-magnet motors. IEEE Trans. Mag. 33(5): 3883. 24. Furlani, E. P., Lee, J. K., and Dowe, D. (1993). Predicting the dynamic behavior of moving magnet actuators. J. Appl. Phys. 73(7): 3555. 25. Kovacs, G. T. A. (1998). Micromachined Transducers Sourcebook, Boston: McGraw-Hill. 26. Madou, M. (1997). Fundamentals of Microfabrication, Boca Raton, FL: CRC Press. 27. Trimmer, W. S. (1997). Micromechanics and MEMS; Classical and Seminal Papers to 1990, New York: IEEE Press. 28. Trimmer, W. S. N. (1989). Microrobots and micromechanical systems. Sensors and Actuators 19: 267. 29. Wagner, B., Benecke, W., Engelmann, G., and Simon, J. (1992). Microactuators with moving magnets for linear, torsional or multiaxial motion. Sensors and Actuators A (Physical) 32: 598. 30. Judy, J. W. and Muller, R. S. (1996). Magnetic microactuation of torsional polysilicon structures. Sensors and Actuators A (Physical) 53: 392. 31. Ruan, M., Shen, J., and Wheeler, C. B. (2000). Latching micro electromagnetic relays. Tech. Dig. Solid-State Sensor and Actuator Workshop (Hilton Head 2000) 146. 32. Dewa, A. S., Orcutt, J. W., Hudson, M., Krozier, D., Richards, A., and Laor, H. (2000). Development of silicon two-axis micromirror for an optical crossconnect. Tech. Dig. Solid-State Sensor and Actuator Workshop (Hilton Head 2000) 93. 33. Osborn, J. A. (1945). Demagnetizing factors of the general ellipsoid. Phys. Rev. 67: 351. 34. Judy, J. W. and Muller, R. S. (1996). Batch-fabricated, addressible, magnetically actuated microactuators. Tech. Dig. Solid-State Sensor and Actuator Workshop (Hilton Head 1996) 187.

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Appendix A: Vector Analysis

Vector analysis provides a succinct and elegant description of electromagnetic ﬁeld theory and is essential for the analysis of permanent magnet and electromechanical devices. In this appendix we review the fundamentals of vector analysis. We start with a discussion of vector calculus in the Cartesian, cylindrical and spherical coordinate systems. Next, we review the basics of vector integration. Following this, we state some key theorems and identities of vector analysis. Last, we discuss coordinate transformations. The interested reader can ﬁnd comprehensive discussions of vector analysis in numerous texts [1—3].

A.1 CARTESIAN COORDINATES For problems with rectangular geometries we use the Cartesian coordinates x, y, and z with unit vectors x , y , and z (Fig. A.1). Throughout the text, italicized symbols such as x and f denote scalars (numbers, functions etc.), while boldface symbols such as x, A, and B denote vectors. Thus, a vector A with components A , A and A has the representation V W X A : A x ; A y ; A z , V W X and magnitude A : (A ; A ; A, (A.1) V W X which is a scalar. We work with scalar-valued functions such as f(x, y, z), and vector-valued functions such as A(x, y, z) : A (x, y, z)x ; A (x, y, z)y ; A (x, y, z)z . V W X 469

470

APPENDIX A Vector Analysis

Cartesian coordinates: (a) reference frame; (b) constant coordinate surfaces; (c) unit vectors; and (d) inﬁnitesimal volume element [3].

FIGURE A.1

Note that each component of A(x, y, z) is itself a scalar-valued function. Such functions are also referred to as scalar and vector ﬁelds. Given two vectors A and B, we form the sum/difference A < B : (A < B )x ; (A < B )y ; (A < B )z V V W W X X and dot product, A·BYA B ;A B ;A B (dot product). V V W W X X Notice that the dot product gives a scalar. Two vectors are said to be orthogonal if their dot product is zero, A · B : 0 $ A orthogonal to B.

471

A.1 CARTESIAN COORDINATES

The unit vectors are orthonormal, x · x : y · y : z · z : 1, x · y : x · z : y · z : 0. In addition to the dot product, there is also the vector (cross) product, which is given by

x

y

z

A;B: A A A V W X B B B V W X : (A B 9 A B )x ; (A B 9 A B )y W X X W X V V X ; (A B 9 A B )z (cross product). (A.2) V W W V Notice that the vector product gives another vector with an orientation as shown in Fig. A.2. We use vectors to evaluate the distance between two points. Speciﬁcally, let p : (x , y , z ) and p : (x , y , z ) be arbitrary points, and let x : x x ; y y ; z z and x : x x ; y y ; z z be vectors from the origin to p and p , respectively. The distance between the two points is given by x 9 x , where

x 9 x Y ((x 9 x ) ; (y 9 y ) ; (z 9 z ) (distance from p : (x , y , z ) to p : (x , y , z )).

(A.3)

Inﬁnitesimal displacements give rise to the following expressions for

FIGURE A.2

Vector product.

472

APPENDIX A Vector Analysis

differential length dl, area ds, and volume dv, dl : dxx ; dyy ; dzz ,

(A.4)

dy dz

(constant x)

ds : dx dz

(constant y)

dx dy

(constant z),

and

(A.5)

dv : dx dy dz,

(A.6)

respectively. Next we review vector calculus. Let f(x, y, z) and A(x, y, z) be scalar and vector-valued functions, respectively. We introduce the differential (del) operator Y

x ; y ; z , x y z

(A.7)

and deﬁne the gradient, divergence, and curl f Y

f f f x ; y ; z x y z

(gradient),

(A.8)

A A A V; W; X x y z

(divergence),

(A.9)

·AY and

x y z ;AY x y x A A A V W X A A A A X 9 W x ; V 9 X y : y z z x

;

A A W 9 V z x y

(curl).

(A.10)

We will also make use of the Laplacian operator , which is deﬁned on scalar-valued functions as follows: f : · f :

f f f ; ; x y z

(Laplacian).

(A.11)

A.2 CYLINDRICAL COORDINATES

473

This operator can also be applied to vector-valued functions. Since x , y and z have no functional dependence, the spatial derivatives in apply to the vector components only (leaving the unit vectors unchanged). For example, A : (A )x ; (A )y ; (A )z . (A.12) V W X This operation is more complicated in other coordinated systems in which the unit vectors are themselves spatially varying functions. These include the cylindrical and spherical coordinate systems that we study next.

A.2 CYLINDRICAL COORDINATES For cylindrical geometries we use cylindrical coordinates r, , and z with unit vectors r , , and z as shown in Fig. A.3. The unit vectors for this system are orthonormal r · r : · : z · z : 1, r · : r · z : · z : 0. It is important to note that r and have different orientations at different points in space. Speciﬁcally, from Table A.2 we have r : cos()x ; sin()y and : 9sin()x ; cos()y . Thus, we ﬁnd that r : and : 9r .

(A.13)

Because of this, care is needed when performing vector calculus in this coordinate system. A vector is written as A : A r ; A ; A z , P X with A < B : (A < B )r ; (A < B) ; (A < B )z . P P X X

474

APPENDIX A Vector Analysis

Cylindrical coordinates: (a) reference frame; (b) constant coordinate surfaces; (c) unit vectors; and (d) inﬁnitesimal volume element [3].

FIGURE A.3

The dot and cross products are given by and

A · B : A B ; A B ; A B X X P P

r

z

A ; B : A A A P X B B B P X

: (A B 9 A B)r ; (A B 9 A B ) ; (A B 9 A B )z , (A.14) X X X P P X P P respectively.

475

A.2 CYLINDRICAL COORDINATES

We determine the distance between two points p : (r , , ) and p : (r , , ) using the Cartesian formula (A.3). First, transform these points into Cartesian vectors x : x x ;y y ;z z and x : x x ;y y ;z z using Table A.1. This gives x : r cos( ), y : r sin( ) for i : 1, 2. G G G G G G Next, substitute these relations into Eq. (A.3) and obtain

x 9 x : (r ; r 9 2r r cos( 9 ) ; (z 9 z ) (distance from p : (r , , z ) to p : (r , , z )).

(A.15)

The differential length, area, and volume are given by dl : drr ; r d ; dzz ,

r d dz

(constant r) (constant )

ds : dr dz

and

(A.16)

r dr d

(A.17)

(constant z),

dv : r dr d dz,

(A.18)

respectively. The operations (A.8), (A.9), and (A.10) take the form f : ·A:

f 1 f f r ; ; z , r r z

(A.19)

1 (rA ) 1 A A P ; ; X, r r r z

(A.20)

and

r r z ;A: r z A rA A P X 1 A A A A X 9 r ; P 9 X : r z z r

;

1 (rA) A 9 P z . r r

(A.21)

476

APPENDIX A Vector Analysis

The Laplacian equation (A.11) is f :

1 f 1 f f r ; ; . r r r r z

(A.22)

We apply to a vector ﬁeld A taking into account the functional dependency of the unit vectors and obtain

1 2 A r A : A 9 A 9 P r P r

; A 9

1 2 A P ; A z , A ; X r r

(A.23)

where A and A are given by Eq. (A.22) with f replaced by the P respective vector component.

A.3 SPHERICAL COORDINATES For spherical geometries we use spherical coordinates r, , and with unit vectors r , , and . These unit vectors are orthonormal r · r : · : · : 1 r · : r · : · : 0. They are also vector-valued spatially varying functions (Fig. A.4). From Table A.3 we have r : sin() cos()x ; sin() sin()y ; cos()z , : cos() cos()x ; cos() sin()y 9 sin()z , : 9sin()x ; cos()y ,

(A.24)

and, therefore, : 9r sin() 9 cos() r :

: 9r

: cos()

r : sin().

(A.25)

A.3 SPHERICAL COORDINATES

477

Spherical coordinates: (a) reference frame; (b) constant coordinate surfaces; (c) unit vectors; and (d) inﬁnitesimal volume element [3].

FIGURE A.4

We must be mindful of Eq. (A.25) when performing vector calculus in this system. Vectors have representations A : A r ; A ; A , P F with A ; B : (A ; B )r ; (A ; B ) ; (A ; B) . P P F F The dot and cross products are given by A · B : A B ; A B ; A B P P F F

478

APPENDIX A Vector Analysis

and

r

A;B: A P B P

A A F B B F

: (A B 9 A B )r ; (A B 9 A B) ; (A B 9 A B ) , F F P P P F F P

(A.26)

respectively. We evaluate the distance between two points p : (r , , ) and p : (r , , ) using Eq. (A.3). We transform p and p into Cartesian vectors x : x x ; y y ; z z and x : x x ; y y ; z z using Table A.1. This gives x : r sin( ) cos( ), y : r sin( ) sin( ), and z : r cos( ) for G G G G G G G G G G G i : 1, 2. These are substituted into Eq. (A.3), which yields

x 9 x : (r ; r 9 2r r sin( ) sin( ) cos( 9 ) (distance from p : (r , , ) to p : (r , , )).

(A.27)

The differential length, area, and volume are given by dl : drr ; r d ; r sin() d ,

ds :

r sin() d d

(constant r)

r sin() dr d

(constant )

r dr d

(constant ),

(A.28)

(A.29)

and dv : r sin() dr d d,

(A.30)

respectively. The vector operations (A.8), (A.9), and (A.10) are as follows: f :

·A:

f 1 f 1 f r ; ; , r r r sin()

1 (rA ) 1 (sin()A ) 1 A P ; F ; , r r r sin() r sin()

(A.31)

(A.32)

479

A.4 INTEGRALS OF VECTOR FUNCTIONS

and

r r r sin() ;A: r A rA r sin()A P F (A sin()) A 1 A 1 1 (rA) P9 : 9 F r ; r sin() r sin() r

;

1 (rA ) A F 9 P . r r

(A.33)

The Laplacian equation (A.11) is given by f :

f 1 f 1 f 1 r ; sin() ; . (A.34) r r sin() r sin() r r

We apply the operator to a vector ﬁeld A taking into account the functional dependency of the unit vectors (A.25) and obtain

2 2 A 2 cos() 2 A F9 A : A 9 A 9 A 9 r P r P r r sin() F r sin()

1 2 A 2 cos() A P9 A ; ; A 9 F r sin() F r r sin() ; A 9

1 2 A 2 cos() A P; F , (A.35) A ; r sin() r sin() r sin()

where A , A , and A are given by Eq. (A.34) with f replaced by P F the respective vector component.

A.4 INTEGRALS OF VECTOR FUNCTIONS In our study of magnetic structures and devices we frequently encounter integrals with vector-valued integrands. In this section we review such integrals, and the procedures for evaluating them. The ﬁrst integral we study is a line integral of the form

f dl. (A.36) ! In this integral f(x, y, z) is a scalar-valued function, C is the path of

480

APPENDIX A Vector Analysis

FIGURE A.5

Line integration path and line element.

integration, and dl is a differential line element along C (Fig. A.5). Notice that this integral yields a vector. For example, in Cartesian coordinates we use the line element (A.4), which gives

f dl :

!

f(x, y, z)[dxx ; dyy ; dzz ]

!

:

f(x, y, z) dx x ;

!

f(x, y, z) dy y ;

!

f(x, y, z) dz z , (A.37)

!

which is obviously a vector. If the integral is to be evaluated from point P to point P we write . f dl. If integration path C is closed we write . w f dl. Notice that in the second step in Eq. (A.37) we have taken the unit ! vectors outside the integral because they are constant in both magnitude and direction. This is true only in the Cartesian system. Care must be taken when evaluating such integrals using cylindrical and spherical coordinates because the direction of the unit vectors in these systems varies from point to point. EXAMPLE A.4.1 Evaluate the line integral of f(x, y) : x ; y along the three paths shown in Fig. A.6. SOLUTION A.4.1 Along path 1 we have

!

W V y dy y ; (x ; y) dx x x y ; y x ; y . :x 3 3

(x ; y) dl :

481

A.4 INTEGRALS OF VECTOR FUNCTIONS

FIGURE A.6

Line integration paths.

To integrate along path 2 we need to relate y and x. Along this path y y: x x and y dy : dx. x Therefore dl : dxx ; (y /x ) dxy . The integration reduces to

(x ; y) dl :

!

V

x ;

y x x

y dxx ; dxy x

y x y y x x ; 1 ; y . x 3 x 3

: 1; Last, along path 3 we have

(x ; y) dl : !

V x dx x ;

W

(x ; y) dy y

x y : x ; y x ; y . 3 3

)

482

APPENDIX A Vector Analysis

The second integral we consider is a line integral of the form

A · dl. (A.38) ! Here, A : A (x, y, x)x ; A (x, y, x)y ; A (x, y, x)z is a vector-valued funcV W X tion and A · dl is the projection of A along the path of integration C. This integral results in a scalar. For example, in Cartesian coordinates we have

A · dl :

[A (x, y, x) dx ; A (x, y, x) dy ; A (x, y, x) dz]. V W X

! ! EXAMPLE A.4.2 Evaluate the line integral of A(x, y) : 2xyx ; xy along the three paths shown in Fig. A.6. SOLUTION A.4.2 Along path 1 we have

!

A(x, y) · dl :

W

0 dy ;

: xy .

V 2xy dx

Along path 2 we have y y: x x and y dy : dx, x with dl : dxx ; (y /x ) dxy . Thus, the integration reduces to V V y A(x, y) · dl : 2xy dx ; x dx x ! y V y V :2 x dx ; x dx x x : xy . Last, along path 3 we obtain

!

V W 2x0 dx ; x dy : xy .

A(x, y) · dl :

)

483

A.4 INTEGRALS OF VECTOR FUNCTIONS

The third integral we study is a volume integral of the form

A dv.

(A.39)

4

We integrate the vector ﬁeld A over a volume V here, which results in a vector. To see this, consider the integral in Cartesian coordinates,

A dv : 4

4

:

[A (x, y, x)x ; A (x, y, x)y ; A (x, y, x)z ] dx dy dz V W X

4

;

A (x, y, z) dx dy dz x ; V

4

4

A (x, y, z) dx dy dz y W

A (x, y, z) dx dy dz z . X

The last integral we study is a surface integral of the form

A · n ds.

(A.40)

1

In this integral A is a vector ﬁeld, n is the unit normal to a surface S, and the projection A · n is integrated over S. The integral (A.40) is also written as

A · ds,

(A.41)

1

where ds : n ds. To evaluate Eq. (A.40) we need to know the sense of n . The following convention is adopted: If S is a closed surface (enclosing a volume V) then the positive direction of n is in the outward direction from the volume (i.e., pointing away from V). If S is an open surface bounded by a curve C, then the positive direction of n depends on the direction in which C is traversed. This is speciﬁed in terms of the right-hand rule (Fig. A.7). Speciﬁcally, if the ﬁngers of the right hand follow the direction of travel around the path C, then the thumb points in the direction of positive n . EXAMPLE A.4.3 Evaluate the integral w A · ds where A(x, y) : rr ; #zz , and 1 S is the closed surface of the cylinder shown in Fig. A.8. Here, and # are constants.

484

APPENDIX A Vector Analysis

FIGURE A.7

Orientation of a normal vector for a bounded surface.

SOLUTION A.4.3 We use cylindrical coordinates. First, we determine the unit normals on the surface of the cylinder. We ﬁnd that

n :

FIGURE A.8

z

z:L

r

r:a

9z

z : 0.

Cylindrical surface of integration.

485

A.5 THEOREMS AND IDENTITIES

Next, we evaluate A · ds over S using Eq. (A.17) and obtain

#Lr dr d

(ds : r dr d)

z:L

A · ds : a d dz

(ds : a d dz)

r:a

(ds : r dr d)

z : 0.

Therefore,

0

A · ds : #L

1

L ?

r dr d ; a

: #La ; a2L

* L

d dz

: (2 ; #)aL.

)

A.5 THEOREMS AND IDENTITIES In this section we review some important theorems, propositions, and identities of vector calculus. There are two key integral theorems that are fundamental to the study of electromagnetic ﬁeld theory. The ﬁrst is Stokes’ theorem. This theorem equates the integral of the curl of a vector ﬁeld over an open surface S to the line integral of the vector ﬁeld around the curve C, which forms the boundary of S (Fig. A.9),

A · dl : !

Integral around closed curve C bounding surface S

( ; A) · n ds

(Stokes’ theorem).

(A.42)

1 Integral over surface S

We digress brieﬂy to discuss the terms in this equation. Notice that to evaluate Eq. (A.42), we need to know the orientation of the unit normal n relative to the direction taken for the line integral around curve C. This is speciﬁed in terms of the right-hand rule (Fig. A.7). Speciﬁcally, if the ﬁngers of the right hand follow the direction of travel around the path C, then the thumb points in the direction of positive n . EXAMPLE A.5.1 Verify Stokes’ theorem for the vector ﬁeld A(x, y) : (2x 9 y)x 9 yzy ; yzz where S is the half-surface of the sphere x ; y ; z : a and C is its bounding curve in the x-y plane (Fig. A.10).

486

APPENDIX A Vector Analysis

FIGURE A.9

Geometry for Stokes’ theorem.

SOLUTION A.5.1 First, we evaluate ( ; A) · n ds. It is easy to check that 1 ; A : z . In this case, the open surface S is deﬁned by x ; y ; z : a with z & 0. The unit vector n on S is given by n :

xx ; yy ; zz . a

Therefore, z ( ; A) · n : . a

FIGURE A.10

Hemisphere geometry with reference frame.

487

A.5 THEOREMS AND IDENTITIES

We want to evaluate (z/a) ds over the hemisphere. We use spherical coordi1 nates in which z : r cos(), and ds : r sin() d d. On S we have r : a and, therefore,

z L L ds : a cos() sin() d d a 1 : a.

(A.43)

Next, we evaluate w A · dl using cylindrical coordinates. We transform A(x, y) ! into cylindrical coordinates using Table A.1. This gives A(r, ) : (2r cos() 9 r sin())(cos()r 9 sin() ) 9 r sin()z(sin()r ; cos() ) ; r sin()zz . On the curve C, r : a, z : 0, and dl : a d . Therefore, A(r, ) · dl : (92a cos() sin() ; a sin()) d and

!

A · dl :

L

(92a cos() sin() ; a sin()) d

: a.

(A.44)

Notice that Eq. (A.44) equals Eq. (A.43), which veriﬁes Stokes’ theorem.

)

The second integral theorem that we study is the Divergence theorem. Before we discuss this, we deﬁne the ﬂux of a vector ﬁeld A through a surface S, Flux of A : through surface S

A · n ds. 1 The Divergence theorem equates the integral of the divergence of a vector ﬁeld over a volume V to the total ﬂux passing through the closed

488

APPENDIX A Vector Analysis

Geometry for Divergence theorem.

FIGURE A.11

surface S that forms the boundary for V (Fig. A.11),

· A dv :

4

A · n ds

(Divergence theorem).

(A.45)

1

Integral over volume V

Integral over surface S enclosing V

In Eq. (A.45) n is the outward unit normal to the surface S. EXAMPLE A.5.2 Verify the Divergence theorem for the vector ﬁeld A(x, y) : rr ; #zz where V and S are the volume and surface of the cylinder shown in Fig. A.8. Here, and # are constants. SOLUTION A.5.2 We have already evaluated w A · n ds in Example A.4.3. 1 Speciﬁcally, we have

A · ds : #La ; a2L

1 : (2 ; #)aL.

(A.46)

It remains to evaluate · A dv. We use cylindrical coordinates and obtain 4 ·A:

1 (r) (#z) ; r r z

: 2 ; #.

489

A.5 THEOREMS AND IDENTITIES

Therefore,

· A dv : (2 ; #)

4

dv 4 : (2 ; #)aL, )

which is the same as Eq. (A.46). Throughout this text we make use of the following deﬁnitions: DEFINITION A.5.1 A vector ﬁeld V is said to be solenoidal if ·V:0

(solenoidal ﬁeld).

DEFINITION A.5.2 A vector ﬁeld V is said to be irrotational if ;V:0

(irrotational ﬁeld).

PROPOSITION A.5.1 A solenoidal vector ﬁeld V can be written as the curl of a vector ﬁeld, that is, ·V:0$V:;A for some A. PROPOSITION A.5.2 An irrotational vector ﬁeld V can be written as the gradient of a scalar-valued function, that is, ; V : 0 $ V : 9 for some . The following theorems are important in our study of Maxwell’s equations. THEOREM A.5.1 A vector is uniquely speciﬁed by giving its divergence and curl within a region, and its normal component over the boundary of the region. Let be a region in space and let f and K be given scalar and vector-valued functions, respectively. Then a vector ﬁeld V is uniquely speciﬁed if ·V:f ;V:K

(A.47)

and the normal component V is speciﬁed on the boundary of . L THEOREM A.5.2 Helmholtz’s Theorem: A vector ﬁeld V satisfying Eq. (A.47) with both f and K vanishing at inﬁnity may be written as the sum of two parts,

490

APPENDIX A Vector Analysis

one that is irrotational, and another that is solenoidal, that is, V : 9 ; ; A.

(A.48)

Given Eqs. (A.47) and (A.48), it can be shown that

(x) :

1 4

f(x) dv

x 9 x

(A.49)

A(x) :

1 4

K(x) dv.

x 9 x

(A.50)

and

Last, we list some useful identities:

Vector Algebra A ; B : 9B ; A (A ; B) · C : A · (B ; C) : (C ; A) · B

(A.51)

A ; (B ; C) : B(A · C) 9 C(A · B),

Differential Identities · ( ; A) : 0 · (A ; B) : B · ( ; A) 9 A · ( ; B) ; (A ; B) : (B · )A 9 (A · )B ; A( · B) 9 B( · A) ( ; A) ; A : (A · )A 9 (A · A) ; ; A : ( · A) 9 A · (fA) : f · A ; (A · )f ; (f) : 0 (fg) : fg ; gf,

(A.52)

491

A.6 COORDINATE TRANSFORMATIONS

and Integral Identities

f(g) · ds :

1

[fg ; (f) · (g)] dv 4

[fg 9 gf)] · ds :

1

(A.53) [fg 9 gf] dv.

4

A.6 COORDINATE TRANSFORMATIONS This section contains transformation relationships for the Cartesian, cylindrical and spherical coordinate systems. A vector A has the following representations: A : A x ; A y ; A z V W X

(Cartesian)

: A r ; A ; A z P X

(cylindrical)

: A r ; A ; A P F

(spherical).

We compute the components of A in a given coordinate system by taking the inner product with the appropriate unit vectors. For example, given the Cartesian representation for A we obtain the radial component in cylindrical coordinates as follows: A : A · r P : A x · r ; A y · r ; A z · r . V W X We summarize the transform relations in the tables that will follow. To avoid confusion, we use r and r to denote the cylindrical and spherical A Q radial coordinate variable, respectively. Table A.1 is used to convert from cylindrical or spherical coordinates to Cartesian coordinates. EXAMPLE A.6.1 Consider a vector A : A r ; A ; A z at a point (r , , z) P A X A in cylindrical coordinates. Determine its Alocation and representation in Cartesian coordinates.

492

APPENDIX A Vector Analysis Coordinate transformations for Cartesian coordinates

TABLE A.1

Cartesian

Cylindrical

x

: cos()r 9 sin() A

y z x y z A V

: : : : : :

A W

:

A X

:

Spherical

: sin() cos()r ; cos() cos() Q 9sin() sin()r ; cos() : sin() sin()r ;cos() sin() ;cos() A Q z : cos()r 9 sin() r cos() : r sin() cos() A Q r sin() : r sin() sin() A Q z : r cos() Q A cos()9A sin() : A sin() cos() ; A cos() cos() PA PQ F 9 A sin() A sin()9A cos() : A sin() sin() ; A cos() sin() PA PQ F ; A cos() A : A cos() 9 A sin() X PQ F

SOLUTION A.6.1 From Table A.1 the location in Cartesian coordinates is x : r cos(), y : r sin(), z : z. Similarly, the vector representation is A A A : (A cos() 9 A sin())x ; (A sin() 9 A cos())y ; A z . PA PA X A V

)

A W

Table A.2 is used to convert from Cartesian or spherical coordinates to cylindrical coordinates. TABLE A.2

Coordinate transformations for cylindrical coordinates

Cylindrical

Cartesian

r A z r A

: : : :

: tan\

z A P AA A X

: : : :

cos()x ; sin()y 9sin()x ; cos()y z (x ; y

y x

z A cos() ; A sin() V W 9A sin() ; A cos() V W A X

Spherical : : : :

sin()r ; cos() Q cos()r 9 sin() Q r sin() Q

: : : : :

r cos() Q A sin() ; A cos() P F AQ A cos() 9 A sin() PQ F

493

A.6 COORDINATE TRANSFORMATIONS TABLE A.3

Coordinate transformations for spherical coordinates

Spherical

Cartesian

r Q r Q

: : : :

: cos\

: tan\

A PQ A F A

sin() cos()x ;sin() sin()y ;cos()z cos() cos()x ;cos() sin()y 9sin()z 9sin()x ; cos()y (x ; y ; z

z

(x ; y ; z

y x

: A sin() cos() ; A sin() sin() V W ; A cos() X : A cos() cos() ; A cos() sin() V W 9 A sin() X : 9A sin() ; A cos() V W

Cylindrical : : : :

sin()r ; cos()z A cos()r 9 sin()z A (r ; z A z : cos\ (r ; z A

: : A sin() ; A cos() PA X : A cos() 9 A sin() P X : A

EXAMPLE A.6.2 Consider a vector A : A r ; A ;A at a point (r , , ) PQ Q F Q in spherical coordinates. Determine its location and representation in cylindrical coordinates. SOLUTION A.6.2 From Table A.2 the location of (r , , ) is r : r sin(), Q A Q :, and z : r cos(). The vector is Q A : (A sin() ; A cos())r ; A ; (A cos() 9 A sin())z . PQ PQ F A F A PA

)

A X

Table A.3 is used to convert from Cartesian or cylindrical coordinates to spherical coordinates.

EXAMPLE A.6.3 Consider a vector A : A x ; A y ; A z at a point (x, y, z) V W X in Cartesian coordinates. Determine its location and representation in spherical coordinates. SOLUTION A.6.3 From Table A.3 the location of (x, y, z) is r : (x;y;z, Q

494

APPENDIX A Vector Analysis

: cos\ z/(x ; y ; z !, and : tan\(y/x). The vector is A : (A sin() cos() ; A sin() sin() ; A cos())r V W X Q A PQ

; (A cos() cos() ; A cos() sin() 9 A sin()) V W X A F

; (9A sin() ; A cos()) . V W

)

A

References 1. Arftgen, G. (1985). Mathematical Methods for Physicists, 3rd ed., New York: Academic Press. 2. Zahn, M. (1987). Electromagnetic Field Theory: A Problem Solving Approach, Malabar, FL: R. E. Krieger Pub. Co. 3. Edminister, J. A. (1979). Theory and Problems of Electromagnetics, Schaum’s Outline Series, New York: McGraw-Hill.

Appendix B: Green’s Functions

In this section we review the basic properties of the free-space Green’s function for Poisson’s equation : f.

(B.1)

This equation plays a key role in both the current and charge models of magnetic materials as discussed in Sections 3.3 and 3.4, respectively. The Green’s function G(x, x) for Poisson’s equation satisﬁes G(x, x) : .(x 9 x), V where indicates differentiation with respect to the unprimed coordiV nates. The Dirac delta function .(x 9 x) satisﬁes

.(x 9 x) dv : 1,

\ where the integration is over all space. Physically, .(x 9 x) corresponds to a point source located at the point x. Given G(x, x), we obtain a solution to Eq. (B.1) using the following relation,

(x) :

G(x, x)f(x) dv.

(B.2)

\ In Eq. (B.2) x is an observation point, and x is a source point. 495

496

APPENDIX B Green’s Functions

For 3D problems it is well known that G(x, x) : 9

1 1 . 4 x 9 x

From Eqs. (A.3), (A.15), and (A.27) we obtain explicit formulas for the free space Green’s function in the Cartesian, cylindrical and spherical coordinate systems, respectively. Speciﬁcally, we ﬁnd that 1 1 , 4 ((x 9 x) ; (y 9 y) ; (z 9 z)

(B.3)

1 1 , 4 (r ; r 9 2rr cos( 9 ) ; (z 9 z)

(B.4)

1 1 , 4 (r ; r 9 2rr sin() sin() cos( 9 )

(B.5)

G(x, x) : 9 G(x, x) : 9 and G(x, x) : 9

respectively. These functions play a key role in obtaining the ﬁeld solutions for free-standing permanent magnet structures. We demonstrated this in various examples in Chapter 4.

Appendix C: Systems of Equations

In this section we review various numerical methods for solving systems of ﬁrst-order, time-dependent differential equations. Such systems naturally arise as the equations of motion for electromechanical devices (Section 5.6). We brieﬂy discuss the Euler, improved Euler, and fourth-order Runge-Kutta methods. Each of these can be demonstrated using an initial-value problem consisting of a single ﬁrst-order equation, dy : f(y, t), dt

(C.1)

and an initial condition y(t ) : y . (C.2) We seek a solution to Eqs. (C.1) and (C.2) for a given time interval (t , t ). As a ﬁrst step, divide the interval (t , t ) into a sequence of discrete values t , t , t , . . . , t , t , . . . , t , as shown in Fig. C.1. Next, L L>

integrate Eq. (C.1) over a typical interval, say (t , t ), L L> RL> dy RL> dt : f(y, t) dt. dt RL RL The left side can be simpliﬁed as follows:

y(t ) 9 y(t ) : L> L

RL>

f(y, t) dt.

(C.3)

RL 497

498

APPENDIX C Systems of Equations

Discretization of time interval.

FIGURE C.1

The various solution methods entail different estimates of the integral in Eq. (C.3). The accuracy of a method improves with the accuracy of the estimate.

C.1 EULER’S METHOD Euler’s method is a very basic solution method and is easy to implement. In this method, the area under the curve in Eq. (C.3) is estimated using the value of the function at the left endpoint of the interval (Fig. C.2),

RL>

RL

f(y, t) dt t f(y(t ), t ). L L

(C.4)

We substitute Eq. (C.4) into Eq. (C.3) and obtain y(t ) : y(t ) ; t f(y(t ), t ). L> L L L

(C.5)

This can be rewritten as y(t

) 9 y(t ) L> L : f(y(t ), t ). L L t

(C.6)

By comparing Eqs. (C.1) and (C.6) we see that Euler’s method amounts

499

C.1 EULER’S METHOD

FIGURE C.2

Integral estimate for the Euler method.

to estimating the derivative by a forward difference quotient, dy y(t ) 9 y(t ) L . L> dt t

(C.7)

This is equivalent to keeping only the ﬁrst term in the Taylor series expansion for y(t ; t), which is of the form y(t ; t) : y(t) ; ;

dy(t) 1 dy(t) t ; t dt 2! dt

1 dy(t) 1 dy(t) t ; t ; · · · . 4! dt 3! dt

(C.8)

It is convenient to rewrite Eq. (C.5) using index notation, y : y ; t f(y , t ). L> L L L

(C.9)

This relation forms the basis for the following solution procedure: 1. Given an initial-value problem (C.1) and (C.2), choose a time increment t and discretize the time interval (t , t ) for which the solution is desired. 2. Start with the initial value point (t , y ) and use Eq. (C.9) to compute the solution point (t , y ). 3. Given (t , y ), use Eq. (C.9) to compute the next solution point (t , y ).

500

APPENDIX C Systems of Equations

FIGURE C3.

Flow chart for implementation of the Euler method.

4. Repeat step 3 using the solution value (t , y ) to compute the L L subsequent solution values (t , y ), for n : 2, 3, 4, . . . . L> L> Continue in this fashion until solution values are obtained for the time interval (t , t ). A ﬂow chart for this procedure is shown in Fig. C.3. Although Euler’s method is easy to implement, it is somewhat limited in terms of its accuracy and stability. Regarding accuracy, the error of this method in any one interval is proportional to t, but the global error is proportional to t. For stability, Euler’s method may

501

C.1 EULER’S METHOD

become unstable if the time constant of the differential equation is negative. For example, consider the equation dy : 9ay, dt

(C.10)

with initial condition y(0) : y 0, and a 0. This equation has an analytical solution y(t) : y e\?R. However, if we apply Eq. (C.5) to (Eq. (C.10) we obtain ) : y(t )(1 9 at). L> L Notice that if at 1, the computed solution is diminishing and positive. However, if at 1, the solution becomes unstable with alternating positive and negative values, and if at 2, the magnitude of the solution increases (in a negative sense) after each step. Thus, the choice of step size t can be crucial to the stability of Euler’s solution procedure. Euler’s method is easily generalized to a system of linear ﬁrst-order equations as shown in the following example: y(t

EXAMPLE C.1.1 Apply Euler’s method to the equations of motion (5.86) for an electromechanical device. SOLUTION C.1.1 We reduce Eq. (5.86) to a set of algebraic equations using Euler’s method in which derivatives are approximated by the forward difference quotient (C.7). This results in the following system of equations: i(t ; t) : i(t) ;

"(i, x) t V(t) 9 i(t)(R ; R ) 9 x(t) L x

v(t ; t) : v(t) ;

t F(i, x) m

x(t ; t) : x(t) ; t v(t).

(C.11)

Equations (C.11) can be written using index notation,

(i , x ) t L L v V 9 i (R ; R ) 9 i :i ; L> L L L L L x t v : v ; F(i , x ) L> L m L L

x : x ; t v . (C.12) L> L L The solution procedure for Eq. (C.12) is similar to the preceding steps 1 through 4. Speciﬁcally, start with the initial conditions [i , v , x ] and use Eq. (C.12) to

502

APPENDIX C Systems of Equations

obtain [i , v , x ], which gives the solution at t : t. Repeat this process to obtain [i , v , x ], [i , v , x ], . . . , [i , v , x ], . . . , until solution values are L L L obtained for the entire interval (t , t ). )

C.2 IMPROVED EULER METHOD The accuracy of Euler’s method can be improved by using a more accurate estimate of the integral in Eq. (C.3). In the improved Euler method the integral is estimated using the trapezoidal rule as shown in Fig. C.4. This gives

RL>

t [f(y(t ), t ) ; f(y(t ), t )]. L> L> L L 2

(C.13)

t ) : y(t ) ; [f(y(t ), t ) ; f(y(t ), t )]. L> L L L L> L> 2

(C.14)

f(y, t) dt

RL We substitute Eq. (C.13) into Eq. (C.3) and obtain y(t

Notice that the unknown y(t ) appears on both sides of Eq. (C.14). L> Thus, it cannot be obtained explicitly. However, Eq. (C.14) can be transformed into an explicit relation for y(t ) by using an estimate on L> the right side. Speciﬁcally, y(t ) can be estimated using the Euler L> relation (C.5). Let y (t ) denote the estimate, L> y (t ) : y(t ) ; t f(y(t ), t ). L> L L L

FIGURE C.4

Integral estimate for the improved Euler method.

503

C.3 RUNGE-KUTTA METHODS

Substitute this in the right side of Eq. (C.14) and obtain t y(t ) : y(t ) ; [f(y(t ), t ) ; f(y (t ), t )]. L> L L L L> L> 2

(C.15)

This can be rewritten using index notation as follows: t y : y ; [f(y , t ) ; f(y , t )], L> L L L L> L> 2

(C.16)

where y : y ; t f(y , t ). (C.17) L> L L L Equations (C.16) and (C.17) can be solved iteratively as described in steps 1 through 4 given in the preceding text. A ﬂow chart of the solution procedure is shown in Fig. C.5. The improved Euler method is more accurate and stable than the standard Euler method. Speciﬁcally, the error of this method in any one interval is proportional to t and global error is proportional to t.

C.3 RUNGE-KUTTA METHODS Runge-Kutta methods are widely used for integrating ﬁrst-order equations. The fourth-order Runge-Kutta method is perhaps the most popular. In this method, the estimate of Eq. (C.3) is based on Simpson’s 1/3 rule [1]. This leads to the following set of formulas: k : t f(y , t ) L L

k t k : t f y ; , t ; L 2 L 2 k t k : t f y ; , t ; L 2 L 2 k : t f(y ; k , t ; t) L L y : y ; [k ; 2k ; 2k ; k ]. L> L

(C.18)

A ﬂow chart of the solution procedure is shown in Fig. C.6. The fourth-order Runge-Kutta method is accurate to the fourth-order term in the Taylor expansion. Thus, the error in any interval is proportional to

504

FIGURE C.5

APPENDIX C Systems of Equations

Flow chart for implementation of the improved Euler method.

C.3 RUNGE-KUTTA METHODS

FIGURE C.6

method.

505

Flow chart for implementation of the fourth-order Runge-Kutta

506

APPENDIX C Systems of Equations

t. The following example demonstrates the use of this method for the solution of a ﬁrst-order system of equations. EXAMPLE C.3.1 Apply the fourth-order Runge-Kutta method to the equations for rotational motion (5.87). SOLUTION C.3.1 The equations governing rotational motion are

di(t) 1 "(i, ) : V(t) 9 i(t)(R ; R ) 9 (t) , L dt

(C.19)

d(t) : (t), dt

(C.20)

and d (t) T(i, ) : . (C.21) J dt K Note that there are three dependent variables i(t), (t), and (t). We need a set of Runge-Kutta coefﬁcients for each variable. We denote these coefﬁcients by (k , k , k , k ), (m , m , m , m ), and (p , p , p , p ), respectively. To implement the Runge-Kutta method, it is convenient to rewrite Eqs. (C.19)—(C.21) as follows: di(t) : f(V(t), i(t), (t), (t)), dt d(t) : g( (t)), dt and d (t) : h(i(t), (t)), dt where f(V(t), i(t), (t), (t)) :

1 "(i(t), (t)) V(t) 9 i(t)(R ; R ) 9 (t) , L

g( (t)) : (t), and h(i(t), (t)) :

T(i(t), (t)) . J K

507

C.3 RUNGE-KUTTA METHODS

The Runge-Kutta coefﬁcients are determined from the following relations: k : t f(V(n), i(n), (n), (n)) m : t g( (n)) p : t h(i(n), (n)), k m p k : t f V(n), i(n) ; , (n) ; , (n) ; 2 2 2

(C.22)

p m : t g (n) ; 2

k m p : t h i(n) ; , (n) ; , 2 2 k m p k : t f V(n), i(n) ; , (n) ; , (n) ; 2 2 2

(C.23)

p m : t g (n) ; 2

k m p : t h i(n) ; , (n) ; , 2 2

(C.24)

and k : t f(V(n), i(n) ; k , (n) ; m , (n) ; p ) m : t g( (n) ; p ) p : t h(i(n) ; k , (n) ; m ). (C.25) Once the coefﬁcients have been computed, they are used to determine the subsequent solution values as follows: i : i ; [k ; 2k ; 2k ; k ] L> L : ; [m ; 2m ; 2m ; m ] L> L : ; [p ; 2p ; 2p ; p ]. (C.26) L> L The solution procedure for Eqs. (C.19)—(C.21) is similar to steps 1 through 4 of Euler’s method. Speciﬁcally, start with the initial conditions [i , , ] and use Eqs. (C.22)—(C.25) to determine the coefﬁcients (k , k , k , k ), (m , m , m , m ), and (p , p , p , p ). Once these are known, use Eq. (C.26) to

508

APPENDIX C Systems of Equations

obtain [i , , ], which gives the solution at t : t. Repeat this process to obtain [i , , ], [i , , ], . . . , [i , , ], . . . , until solution values are L L L obtained for the interval (t , t ). ) References 1. Nakamura, S. (1991). Applied Numerical Methods with Software, Englewood Cliffs, NJ: Prentice-Hall.

Appendix D: Units

In this section we discuss three systems of units that are commonly used in magnetism. These are the CGS or Gaussian system, and two SI systems (International System of Units or Système international d’unités). The SI systems are referred to as the Kennelly and Sommerfeld conventions, respectively. The CGS system is based on the fundamental units of centimeter, gram and second, and the SI systems are based on the MKSA units of meter, kilogram, second, and ampere. In this book we use SI system units in the Sommerfeld convention. The basic entities and units in the Sommerfeld and CGS systems are listed in Table D.1. Maxwell’s equations in these two systems are as follows: Maxwell’s Equation SI ;H:J;

CGS (Gaussian) D t

·B:0 B t ·D:

;E:9

4 1 D J; c c t ·B:0 1 B ;E:9 c t · D : 4.

;H:

(D.1)

509

510

APPENDIX D Units TABLE D.1

Fields and units in the SI (Sommerfeld) and CGS systems

Symbol

Description

SI

CGS

H B M

Magnetic ﬁeld strength Flux density Magnetization Flux

A/m Tesla A/m Weber

Oe Gauss emu/cm Maxwell

The constitutive relations are as follows:

Constitutive Relations SI (Sommerfeld)

CGS (Gaussian)

B : (H ; M) D:+ E;P : 4 ; 10\ 10 + : 4c

B : H ; 4M

(D.2)

D : E ; 4P :1 + : 1.

To convert between the Sommerfeld and CGS systems we use Tables D.2 and D.3. For example, from Table D.1 we ﬁnd that 1 A/m : 4/10 Oe and that 1 J/m : 40 GOe and so on. Similarly, from Table D.3 we ﬁnd that 1 Oe : 10/4 A/m and that 1 MGOe : 10/4 kJ/m and so on.

TABLE D.2

Conversion from SI to CGS units

SI H in A/m B in Tesla M in A/m in Webers (BH) in J/m

(BH) in kJ/m

to ; ; ; ; ; ;

4/10 10 10\ 10 40 4/10

CGS : : : : : :

H in Oe B in Gauss M in emu/cm in Maxwells (BH) in GOe

(BH) in MGOe

511

UNITS TABLE D.3

Conversion from CGS to SI units

CGS H in Oe B in Gauss M in emu/cm in Maxwells (BH) in GOe

in MGOe (BH)

to ; 10/4 ; 10\ ; 10 ; 10\ ; 10\/4 ; 10/4

SI : : : : : :

H in A/m B in Tesla M in A/m in Webers (BH) in J/m

(BH) in kJ/m

Other useful relations are 1 T:1

N , A·m

and 1 Wb : 1 T m. The Kennelly and Sommerfeld conventions differ in the deﬁnition of magnetization. In the Kennelly convention the constitutive relation is B: H;J (Kennelly convention). Here, J is called the magnetic polarization or intensity of magnetization and is measured in tesla. It is also referred to as the intrinsic ﬂux density and is denoted by B . Notice that J : M. The fundamental element in G magnetism is the magnetic dipole m. In the Sommerfeld convention m is measured in A · m. In the Kennelly and CGS systems it is measured in Wb · m and emu, respectively (1 emu : 4 ; 10\ Wb · m). If a magnetic dipole is subjected to an external B-ﬁeld, it acquires an energy E : 9m · B,

(D.3)

T : m ; B.

(D.4)

and experiences a torque

In both the Kennelly and CGS systems these are given by E : 9m · H and T : m ; H. In the remainder of this section we give conversion factors for various mechanical units. There are three systems that are commonly used for mechanical analysis. These are the MKS, CGS and British systems. In the MKS system the fundamental units of length, mass, and

512

APPENDIX D Units TABLE D.4

Conversion of mechanical variables from MKS

to CGS units MKS

to ; ; ; ;

mass in kg force in newtons (N) torque in N · m energy in joules

10 10 10 10

CGS : : : :

mass in grams force in dynes torque in dyne · cm energy in ergs

time are the meter (m), kilogram (kg), and second. The unit of force is the newton (N), and the unit of energy is the joule (J). We use the MKS system throughout this text. In the CGS system, the fundamental units are the centimeter (cm), gram (g), and second. In this system the units of force and energy are the dyne and erg, respectively. Last, in the British system the units of mass, length, and time are the slug, foot, and second, respectively. For example, acceleration due to gravity is g : 32 ft/s, whereas in the MKS system g : 9.8 m/s. In the British system the unit of force is the pound (lb), and a commonly used unit of energy is the British thermal unit (Btu). Tables D.4 and D.5 give conversion factors for various mechanical units. Other useful relations are as follows: 1. Length: 1.0 in : 2.54 cm, 1.0 ft : 0.3048 m. 2. Mass density: 1.0 lb/ft : 16.02 kg/m. 3. Torque: 1.0 lb · in:0.113 N · m, 1.0 oz · in:72 gm · cm, 1.0 N · m : 1.019 ; 10 g · cm.

Conversion of mechanical variables from British to MKS units

TABLE D.5

British mass in slugs force in lbs torque in ft · lb energy in Btu

to ; ; ; ;

14.59 4.45 1.3558 1055

MKS : : : :

mass in kg force in newtons torque in N · m energy in joules

Index A Actuators axial-ﬁeld, 393—401 linear, 413—421 magnetic circuit, 383—392 magnetooptical bias ﬁeld, 406—413 moving magnet, 449 plunger, 383, 387, 390 resonant, 402—406 rotary bias ﬁeld, 322 solenoid, 136 Alnico, 48—50 Ampere’s circuital law, 78, 88, 338 Angular momentum, orbital versus spin, 10—11 Anisotropic materials, 5, 77 Anisotropy magnetocrystalline, 27—31 shape, 31—33 Antiferromagnetic material, 6, 7 Atomic magnetic moments, 7—9 multielectron atoms, 14—17 single electron atoms, 9—14 Axial coupling, 286, 288, 297—306 Axial-ﬁeld actuators, 393—401 Axial-ﬁeld motors, 421—437 no-load speed, 424 stall-torque, 424 Axially polarized cylindrical shell, 245—253 Axially polarized multipole cylindrical disk magnet, 253—264 B Back emf constant, 368 Barium ferrite, 46 Barkhausen, 36 Bar magnet, 211—217 rectangular, 133—135 Bearing conﬁgurations, radial and axial, 273 Bessel function, 176 B-H hysteresis loop, 36—39 Bias ﬁeld actuators, 406—413 magnetooptical recording and, 323 Bias magnets cylindrical, 218—264 rectangular, 208—218

Biot-Savart law, 104 Bipolar cylindrical magnet, 218—224 Bipolar sphere, 182—185 Bloch walls, 34—35 Bohr magneton, 11 Boltzmann’s constant, 17 Boltzmann statistics, 17 Bonded ferrite magnets, 54 Bonded magnets, 53—55 Boundary conditions, 79—81 Boundary-value problems, 161 Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 magnet between plates, 165—169 spherical coordinates, 180—185 Bound charge, 75 Br, 38 Brillouin function, 19, 22 Brushless dc motors. See Axial-ﬁeld motors C Can-stack motors, 437—446 Carrier particles, 315 Cartesian coordinates, 469—473 boundary-value problem, 162—174 Laplace’s equation, 162—174 transformations, 491—494 Ceramic magnets, 46—47 CGS/Gaussian unit, 2—5, 509—512 Chain rule, 345 Charge bound, 75 free, 75 Charge density, free, 75 Charge model, 131—134 force, 135—140 torque, 141—143 Circuits current sources, 144—153 equivalent, 157—159 inductance, 151 magnet sources, 153—161 moving, 358—361, 366—376 no ﬂux leakage, 145, 150 stationary, 354—358, 363—366 Coenergy, 379 Coercive force, 31

513

514 Coercivity, 31, 38, 39 intrinsic, 29, 32 Conduction current, 75 Constitutive relations, 75—77 magnetostatic, 99 Contact latching, 268—269 Continuity equation, 75 Convection current, 75 Conversion factors, 3, 509—512 Coordinate transformations, 491—494 Coulomb gauge condition, 86, 103, 127 Coupling, synchronous applications, 286 axial, 286, 288, 297—306 radial, 286, 288—297 Russell-Saunders, 14—15 spin-orbit, 14 Cross product, 471 Curie’s law, 19 Curie temperature, 20 Curl, 472 Current conduction, 75 convection, 75 electrolytic, 75 equivalent, 128 free, 75 line, 104—106 sources, 144—153 surface, 107 Current density displacement, 75 upper limit of, 58 Current model, 126—131 force, 131 torque, 131 Cylindrical coordinates, 473—476 boundary-value problem, 175—180 Laplace’s equation, 175—180 transformations, 491—494 Cylindrical magnets axially polarized multipole disk magnet, 253—264 axially polarized shell, 245—253 axial polarization, 128—130 bipolar, 218—223 radially polarized multipole, 224—245 D Del operator, 343, 472 Demagnetization curve, 44, 66—67 for ceramic, 67 for samarium-cobalt, 67 Demagnetization ﬁeld, 23—27 Diamagnetic material, 6 Die-upsetting, 53 Differential Maxwell equations, 74, 79, 338

INDEX Dipole-dipole force, 140, 316 Dipole/dipole moment, 3, 108 Energy in external ﬁeld, 4, 511 magnetic, 140 of sphere in external ﬁeld, 26 torque in external ﬁeld, 4, 511 Direct current motors. See Axial-ﬁeld motors Dirichlet boundary-value problem, 162 Disk magnets, levitation, 273—281 Dislocations, 34 Displacement current density, 75 Divergence, 472 Divergence theorem, 78, 114, 487—489 Domains, 6, 20, 33—36 Domain walls, 34 motion of, 35—36 Dot product, 470 Drag torque, 423 Drive torque, 453 Dyadic, 113 E Earnshaw’s theorem, 273 Eddy currents, 350—353, 411—412 drag force, 352, 353 lift force, 352, 353 Electrical constant, 368 Electrical equations moving coils, 358—361 stationary circuits, 354—358 Electrolytic current, 75 Electromagnets, conventional, 308, 310 Electromechanical devices actuators, axial-ﬁeld, 393—401 actuators, linear, 413—421 actuators, magnetic circuit, 383—392 actuators, magnetooptical bias ﬁeld, 406—413 actuators, resonant, 402—406 basics, 335—337 electrical equations, 354—361 electromechanical equations, 363—376 energy analysis, 376—383 examples of, 335 hybrid analytical-FEM analysis, 446—455 mechanical equations, 361—363 micro, 455—465 motors, axial-ﬁeld, 421—437 motors, stepper, 437—446 quasi-static ﬁeld theory, 337—353 Electromechanical equations moving coils, 366—376 stationary circuits, 363—366 Electron spin, 10 Electrophotography, 312—318 Electrostatic theory, 89—91

515

INDEX emu, 3, 511 Energy analysis, 376—383 balance equation, 377 co-, 379 contours, 46 density, 117 exchange, 22 magnetostatic, 23, 116—117 point of maximum, 44, 153 product, 157 self-energy equation, 23, 117 Equivalent charge, 132 Equivalent circuit, 157—159 Equivalent currents, 128 Euler’s method, 363, 365—366, 418, 498—503 Exchange energy, 22 Exchange integral, 22 Expectation value, 10 F Faraday’s law of electromagnetic induction, 78, 85, 347 FEA. See Finite element analysis FEL. See Free-electron lasers FEM analysis, hybrid analytical, 446—455 Ferric oxide, 47 Ferrimagnetic material, 6, 7 Ferrites, 46—47 bonded, 54 Ferromagnetic material, 6 Ferromagnetism, 19—22 Finite bar magnet, 211—217 Finite difference method, 200—204 backward, 202 central, 202—203 forward, 202 Finite element analysis (FEA), 60, 190—200 Finite radially polarized multipole cylinder, 235—245 Flux density, 2, 3, 98 line, 62, 108—109 linkage, 119—120, 341 magnetic, 144 remanent, 31 vector ﬁeld, 78 Force between cylindrical magnets, 139—140 between dipoles, 140 between two wires, 111 charge model, 135—140 coercive, 31 current model, 131 on dipole in external ﬁeld, 317 Lorentz, 82, 110 magnet above plate, 188, 189

on steady currents, 110—112 on wire above plate, 188 Free charge density, 75 Free current density, 75 Free-electron lasers (FELs), 325—331 Fringe ﬁelds, 309 G Galilean transform, 344 Gamma ferric oxide, 47 Gauss’ law, 78, 90, 338 Gears, 282—286 Gradient, 472 Green’s function, free-space, 90, 103, 495—496 H Hall effect, 265, 422 Hard magnetic materials, 44—45 alnico, 48—50 bonded, 53—55 ferrites, 46—47 neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Helmholtz’s theorem, 489—491 Henry, 118 High ﬁeld structures, 265—268 Homogeneity, magnetic resonance imaging and, 309 Homogeneous material, 5, 77 Hund’s rules, 15 Hybrid analytical-FEM analysis, 446—455 Hysteresis loop, 36—39 I Images, method of. See Method of images Impulse magnetizers, 58—59 Inclusions, 34 Inductance, 341 external, 119 internal, 119 magnetostatic analysis, 118—126 mutual, 120—121 parallel wires, 122—123 self, 118 solenoid, 123—124, 125—126 toroid, 124—125 wire, 121—122 Induced voltage, 370, 372 Inertial reference frame, 341—342 Inﬁnitely long cylindrical magnet, 224—235 Inﬁnite rectangular magnet, 208—211 Inhomogeneous material, 5, 77 Initial magnetization curve, 37 Integral Maxwell equations, 77—79, 338 Integrals, of vector functions, 479—485 Intrinsic coercivity, 29 shape anisotropy and, 32

516 Iron oxides, 47 Iron-silicon alloys, 39, 43 Irreversible changes, 67, 70 Isotropic materials, 5, 77 K Kennelly units, 2, 509—512 Kerr effect, 320—321 Kirchhoff’s laws current, 146, 338 voltage, 89, 146—147, 339, 341 voltage (moving coil), 358—361 voltage (stationary circuit), 354, 357 L Land g factor, 14 Langevin function, 18 Laplace operator, 472—473 Laplace’s equation Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 spherical coordinates, 180—185 Larmor frequency, 307, 308 Larmor relation, 309 Latching, magnetic, 268—273 Legendre polynomials, 182—185 Legendre’s equation, 181 Lenz’s law, 78 Levitation, passive, 273—281 Linear actuators, 413—421 Linear materials, 5, 77 Line integral, 479—482 Load line, 44—46, 155—156 stability, 66—70 Lodestone, 47 Lorentz force, 82, 110, 362 Lorentz gauge condition, 83 Lorentz transformation, 343—345 M Magic cylinder, 265, 266 Magic sphere, 265—268 Magnequench, 52—53 Magnetic bearing. See Magnetic levitation Magnetic brush technique, 315—318 Magnetic circuit actuators, 383—392 Magnetic circuit analysis, 144—161 current sources, 144—153 equivalent circuits, 157—159 magnet sources, 153—161 Magnetic coupling. See Coupling Magnetic dipole. See Dipole moment Magnetic ﬁeld bar magnet, 211—217 bipolar sphere, 182—185 current loop, 108 cylindrical magnets. See Cylindrical magnets

INDEX line current, 104—107 magnet between plates, 165, 168 rectangular magnets. See Rectangular magnets solenoid, 100—101 strength, 2, 3, 98 structures of high, 265—268 surface current, 107 toroid, 101—102 wire, 99—100 Magnetic gears, 282—287 Magnetic latching, 268—273 Magnetic levitation, 273—281 Magnetic moments. See Atomic magnetic moments Magnetic quantum number, 10 Magnetic resonance imaging (MRI), 306—307 ﬁeld strengths, 308 ﬁeld systems for, 308, 310 fringe ﬁelds, 309 homogeneity, 309 hybrid, 310, 312 patient access, 309—310 stability, 309 yoked, 310, 311—312 yokeless, 310—311 Magnetic viscosity, 70 Magnetite, 47 Magnetization process, 55—66 Magnetizing ﬁeld, 56 Magnetizing ﬁxture design, 58—59 Magnetocrystalline anisotropy, 27—31 Magnetomotive force (mmf), 148 Magnetooptical bias ﬁeld actuators, 406—413 Magnetooptical (M-O) recording, 318—325 Magnetostatic analysis Biot-Savart law, 104 energy, 116—117 ﬁeld equations, 97—102 force and torque, 110—112 inductance, 118—126 Maxwell stress tensor, 112—116 vector potential, 102—109 Magnetostatic energy, 23 Magnetostatic theory, 87—89 Magnets equivalent circuit, 157—159 cylindrical. See Cylindrical magnets rectangular. See Rectangular magnets sources, 153—161 Magnets, structures of cylindrical, 218—264 rectangular, 208—218

517

INDEX Materials anisotropic, 5, 77 classiﬁcation of, 6—7 hard, 44—55 homogeneous, 5, 77 inhomogeneous, 5, 77 isotropic, 5, 77 linear, 5, 77 nonlinear, 5, 77 soft, 39—44 Maxwell, James Clerk, 76 Maxwell’s equations, 74, 338 boundary conditions, 79—81 constitutive relations, 75—77 differential, 74, 79 force and torque, 82 integral, 77—79 Maxwell stress tensor, 112—116 force, 115—116, 362 Mechanical equations, 361—363 Melt spinning, 51, 52 MEMS (micro-electromechanical systems), 455—465 Method of images, 185—190 magnet on plate, 188—189 wire above plate, 188 Micro-electromechanical systems (MEMS), 455—465 Mixed boundary-value problem, 162 MKS system, 368, 509—512 Molecular ﬁeld, 20 Moment of inertia, 361, 375 Motion, electromechanical equations moving coils, 366—376 stationary circuits, 363—366 Motion, mechanical equations, 361—363 linear, 361 rotational, 361 Motors axial-ﬁeld, 421—437 stepper (can-stack), 437—446 Moving coils, 358—361, 366—376 Moving reference frames, 341—353 MRI. See Magnetic resonance imaging Multipolar cylinders, radially polarized, 235—245 Multipole disk magnet, axially polarized, 253—264 Mutual inductance, 120—121 N Neodymium-iron-boron, 51—53 Neumann boundary-value problem, 162 Neumann functions, 176 Newton’s second law of motion, 361 Nickel-iron alloys, 39, 43 Noncontact latching, 269

Nonlinear materials, 5, 77 O Ohm’s law, 356 Operating point, 153—161 Orbital angular momentum, 10—11 P Paramagnetic material, 6 Paramagnetism, 17—19 Pauli exclusion principle, 14, 15 Permanent magnet structures. See Magnets Permanent magnets, 44—46 alnico, 48—50 bonded, 53—55 ferrites, 46—47 ﬁelds of. See Magnetic ﬁeld magnetic resonance imaging and, 308, 310 magnetization of. See Magnetization process neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Permeability absolute, 41 differential, 41 free space, 5, 76 initial, 41 maximum relative, 41 relative, 41 Permeance angular gap, 149 radial gap, 149 rectangular gap, 149 Permittivity, free space, 76 Planck’s constant, 10 Poisson’s equation, 200—201 Potential(s) difference, 90, 340 energy, 23 Maxwell’s equations and, 82—84 scalar, 82 vector, 82, 83, 102—109 Q Quadrupole magnet, 218 Quantum mechanics, 10 Quasi-static ﬁeld theory, 85—86, 337 moving reference frames, 341—353 stationary reference frame, 338—341 R Radial coupling, 286, 288—297 Radially polarized magnets, 224—245 ﬁnite, 235—245 inﬁnitely long, 224—235 Rapid quenching, 51, 52

518 Rare-earth magnets, 58 neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Rectangular magnets, 208—218 Reference frames moving, 341—353 stationary, 338—341 Relative permeability, 41 Reluctance, 145 torque, 453 Remanence, 31, 38 Remanent ﬂux density, 31 Remanent induction, 38 Resonant actuators, 402—406 Reversible changes, 67 Reversible temperature coefﬁcients, 67 Runge-Kutta method, 363, 418, 503—508 Russell-Saunders coupling, 14—15 S Samarium-cobalt, 50—51 Saturation induction, 38 Scalar-valued functions, 345, 469—473 Schrödinger’s equation, 9 Self-energy equation, 23, 117 Self-inductance, 118 Separation of variables Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 spherical coordinates, 180—185 Shape anisotropy, 31—33 Shorted turn, 418—421 SI (Système international d’unités), 2—5, 509—512 Soft ferrites, 39, 43—44 Soft iron, 39, 42—43 Soft magnetic materials, 39—44 Solenoid B-ﬁeld, 100—101 inductance, 123—124, 125—126 Sommerfeld units, 2—5, 509—512 Sphere in external ﬁeld, 26 Spherical coordinates, 476—479 boundary-value problem, 180—185 Laplace’s equation, 180—185 transformations, 491—494 Spin, 10 Spin angular momentum, 10—11 Spin-orbit coupling, 14 Stability, 66—70 magnetic resonance imaging and, 309 Static theory electrostatic, 89—91 magnetostatic, 87—89 Stationary circuits, 354—358, 363—366 Stationary reference frame, 338—341 Stepper motors, 437—446

INDEX Stokes’ theorem, 77, 356, 485—487 Stoner and Wohlfarth model, 28 Strontium ferrite, 46 Structural changes, 67 Superconducting coils, 308, 310 Surface current, 107 Surface integral, 483—485 Susceptibilities absolute, 42 differential, 42 initial, 42 Suspension, 273—281 Synchronous coupling. See Coupling, synchronous T Tesla, 2—3, 510—511 Tesla per amp (TPA) ratio, 60 Toroid B-ﬁeld, 101—102 inductance, 124—125 Torque, 82, 374 on bipolar cylinder, 142—143 charge model, 141—143 constant, 368 current model, 131 drag, 423 drive, 453 reluctance, 453 on steady currents, 110—112 U Undulator, 326 Units CGS/Gaussian, 2—5 conversion factors, 3 Kennelly, 2 SI (Sommerfeld), 2—5 summary of, 509—512 V Vector ﬁeld ﬂux, 78 Vector potential, 83, 102—109 line current, 104—106 surface current, 107—108 Vector-valued functions, 74, 345—346, 469— 473 Virtual displacement, 376—377 Volume integral, 483 W Webers, 2 Weiss molecular/mean ﬁeld theory, 20—22 Wigglers, 326—328 Halbach, 329—330 helical ﬁeld, 330 hybrid, 330—331