ONS MANUAL
INCLUDES:
--'.I-Step Solutions to 25% of the text's
""'I"I""'''-Chapter Problems 'r
I!lMIi tions in the same two-column format as the worked examples in the text and the Study Guide •
Carefully rendered art to help you visualize each ...r()DI�em
STUDENT SOLUTIONS MANUAL TO ACCOMPANY
PHYSICS FOR SCIENTISTS AND ENGINEERS FIFTH EDITION
This solutions manual doesn't just give you the answers. It shows you how to work your way through the problems. The Solutions Manual includes: • •
•
Detailed step-by-step Solutions to 25% of the text's end-of-chapter Problems Solutions in the same two-column format as the worked examples in the text and the Problems and Solutions in the Study Guide Carefully rendered art to help you visualize the Problem and Solution
Other great resources to help you with your course work: Study Guide Gene Mosca, United States Naval Academy Todd Ruskell, Colorado School of Mines Vol. 1,0-7167-8332-0,Vol. 2,0-7167-8331-2 •
Begins with review of Key Ideas and Equations for each chapter
•
Tests your knowledge of the chapter's material with True and False Exercises and Short Questions and Answers
•
Provides additional Problems and Solutions to help you master your understanding of the chapter content
Be sure to visit the Tipler/Mosca Student Companion Web site at: www.whfreeman.com/tipler5e Accessible FREE OF CHARGE, the site offers: •
Online Quizzing
•
IIMaster the Concept" Worked Examples
•
Concept Tester Interactive Simulations
•
Concept Tester Quick Questions
•
Solution Builders
•
Applied Physics Video Clips
•
Demonstration Physics Video Clips
Student Solutions Manual for Tipler and Mosca's
Physics for Scientists and Engineers Fifth Edition Volume 2
DAVID MILLS Professor Emeritus College of the Redwoods with
CHARLES L. ADLER Saint Mary's College of Maryland EDWARD A. WHITTAKER Professor of Physics Stevens Institute of Technology GEORGE ZOBER Yough Senior High School PATRICIA ZOBER Ringgold High School
W. H. Freeman and Company
New York
Copyright
© 2004 by W. H. Freeman and Company
All rights reserved. Printed in the United States of America ISBN: 0-7167-8334-7 First printing 2003 W. H. Freeman and Company
41 Madison Avenue
New York, NY 10010 Houndrnills, Basingstoke RG21 6XS, England
Contents To the Student, v Acknowledgments, vii About the Authors, ix Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
The Electric Field I: Discrete Charge Distributions,
1
The Electric Field II: Continuous Charge Distributions, Electric Potential,
51
Electrostatic Energy and Capacitance,
71
Electric Current and Direct-Current Circuits, The Magnetic Field,
135 189
29
97
159 219
Sources of the Magnetic Field, Magnetic Induction,
Alternating-Current Circuits,
Maxwell's Equations and Electromagnetic Waves,
275 291
Properties of Light, Optical Images,
Interference and Diffraction,
315
253
337 353
Wave-Particle Duality and Quantum Physics, Applications of the Schrodinger Equation, Atoms,
363 379 407
Molecules,
Solids and the Theory of Conduction, Relativity,
Nuclear Physics,
429
393
Elementary Particles and the Beginning
of the U ni verse,
447
To the Student This solution manual accompanies Physics
for Scientists and Engineers, 5e, by
Paul Tipler and Gene Mosca. Following the structure of the solutions to the Worked Examples in the text, we begin the solutions to the back-of-the-chapter numerical problems with a brief discussion of the physics of the problem, represent the problem pictorially whenever appropriate, express the physics of the solution in the form of a mathematical model, fill in any intermediate steps as needed, make the appropriate substitutions and algebraic simplifications, and complete the solution with the substitution of numerical values (including their units) and the evaluation of whatever physical quantity is called for in the problem. This is the problem-solving strategy used by experienced learners of physics, and it is our hope that you will see the value in such an approach to problem solving and learn to use it consistently. Believing that it will maximize your learning of physics, we encourage you to create your own solution before referring to the solutions in this manual. You may find that, by following this approach, you will find different, but equally
valid, solutions to some of the problems. In any event, studying the solutions
contained herein without having first attempted the problems will do little to help you learn physics. You'll find that nearly all problems with numerical answers have their answers given to three significant figures. Most of the exceptions to this rule are in the solutions to the problems on Significant Figures and Order of Magnitude and the problems dealing with nuclear physics. When the nature of the problem makes it desirable to do so, we keep more than three significant figures in the answers to intermediate steps and then round to three significant figures for the final answer. Some of the Estimation and Approximation Problems have answers to fewer than three significant figures.
Physics for Scientists and Engineers, 5e includes numerous spreadsheet problems. Most of them call for the plotting of one or more graphs. The solutions to these problems were generated using Microsoft Excel and its "paste special" feature, so that you can easily make changes to the graphical parts of the solutions.
v
Acknowledgments Charles L. Adler (Saint Mary's College of Maryland), Ed Whittaker (Stevens Institute of Technology, George Zober (Yough Senior High School) and Patricia Zober (Ringgold High School) are the authors of the new problems appearing in the Fifth Edition. Chuck, Ed, George, and Patricia saved me (dm) many hours of work by providing rough-draft solutions to these new problems, and I thank them for their help. Gene Mosca (United StatesNaval Academy and the co-author of the Fifth Edition) helped me tremendously by reviewing my work, helping me clarify many of my solutions, and providing solutions when I was unsure how best to proceed. It was a pleasure to collaborate with Gene in the creation of this solutions manual. All of us who were involved in the creation of this solutions manual hope that you will find the solutions useful in learning physics. We want to thank LayNam Chang (Virginia Polytechnic Institute), Brent A. Corbin (UCLA), Alan Cresswell (Shippensburg University), Ricardo S. Decca (Indiana University-Purdue University), Michael Dubson (The University of Colorado at Boulder), David Faust (Mount Hood Community College), Philip Fraundorf (The University of Missouri-Saint Louis), Clint Harper (Moorpark College), Kristi R. G. Hendrickson (University of Puget Sound), Michael Hildreth (The University ofNotre Dame), David Ingram (Ohio University), James J. Kolata (The University ofNotre Dame), Eric Lane (The University of Tennessee Chattanooga), Jerome Licini (Lehigh University), Laura McCullough (The University of Wisconsin-Stout), Carl Mungan (United StatesNaval Academy), Jeffrey S. Olafsen (University of Kansas), Robert Pompi (The State University of New York at Binghamton), R. J. Rollefson (Wesleyan University), Andrew Scherbakov (Georgia Institute of Technology), Bruce A. Schumm (University of Chicago), Dan Styer (Oberlin College), Daniel Marlow (Princeton University), Jeffrey Sundquist (palm Beach Community College-South), Cyrus Taylor (Case Western Reserve University), and Fulin Zuo (University of Miami), for their reviews of the problems and their solutions. Jerome Licini (Lehigh University), Michael Crivello (San Diego Mesa College), Paul Quinn (University of Kansas), and Daniel Lucas (University of Wisconsin-Madison) error-checked the solutions. Without their thorough and critical work, many errors would have remained to be discovered by the users of this solutions manual. Their assistance is greatly appreciated. In spite of their best efforts, there may still be errors in some of the solutions, and for those I (dm) assume full responsibility. Should you find errors or think of alternative solutions that you would like to call to my attention, I would appreciate it if you would communicate them to me by sending them to
[email protected].
Vll
It was a pleasure to work with Brian Donnellan, Media and Supplements Editor for Physics, who guided us through the creation of this solution manual.
Our thanks to Amanda McCorquodale and Eileen McGinnis for organizing the
reviewing and error-checking process. September
2003
David Mills
Professor Emeritus College of the Redwoods Charles
L. Adler
Saint Mary's College of Maryland Edward A. Whittaker
Professor of Physics Stevens Institute of Technology George Zober
Yough Senior High School Patricia Zober
Ringgold High School
\ YIll
About the Authors David Mills,
Professor Emeritus, College of the Redwoods, retired in May of
2000 after a teaching career of 42 years. He earned his bachelor's degree at
Humboldt State College, his master's degree at California State University-Hayward, and his doctoral degree at the University of Northern Colorado. His teaching career included experience with the Physical Science Study Committee materials, the Harvard Project curriculum, the Personalized System of Instruction, Microcomputer-Based Laboratory instruction, and the interactive-engagement movement in physics education. A 1996 NSF.lLI grant allowed him to transform instruction in physics at the College of the Redwoods from a traditional lecture-laboratory delivery system to one that was microcomputer based, eliminate the distinction between lecture and laboratory, and utilize interactive-engagement teaching and learning strategies. He authored the Test Bank to accompany Physics for Scientists and Engineers, 3e and 4e. He now lives in Henderson, NY and is an Adjunct Professor at the Community College of Southern Nevada.
Charles L. Adler is a professor of physics at St. Mary's College of Maryland. He received his undergraduate, masters, and doctoral degrees in physics from Brown University before doing his postdoctoral work at the Naval Research Laboratory in Washington, D.C. His research covers a wide variety of fields, including nonlinear optics, electrooptics, acoustics, cavity quantum electrodynamics, and pure mathematics. His current interests concern problems in light scattering, inverse scattering, and atmospheric optics. Dr. Adler is the author of over 30 publications. Edward A Whittaker
has been a professor of physics at Stevens Institute of Technology since 1984. His research interests include laser spectroscopy, quantum optics, and optical communications. In 2003 he was named an American Institute of Physics State Department Science Fellow.
George Zober
is a teacher of Advanced Placement Physics at Yough High School in western Pennsylvania. He serves as a Physics Consultant with the College Board and teaches Advanced Placement Physics Workshops at Wilkes University and Manhattan College during his summers. Patricia J. Zober
teaches Advanced Placement Physics at Ringgold High School in Monongahela, Pennsylvania and is a Physics Consultant for the College Board. Patricia presents Advanced Placement Workshops in Physics and during her summers is a faculty physics project leader with the Governor's School for the Sciences at Carnegie Mellon University.
IX
Chapter 21 The Electric Field 1 : Discrete Charge Distributions Conceptual Problems *1
••
Discuss the similarities and differences in the properties of electric charge
and gravitational mass. Similarities:
Differences:
The force between charges and
There are positive and negative charges but
masses varies as lIr2.
only positive masses.
The force is directly proportional to
Like charges repel; like masses attract.
the product of the charges or masses. The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. *5
••
Two uncharged conducting spheres with their conducting surfaces in contact
are supported on a large wooden table by insulated stands. A positively charged rod is brought up close to the surface of one of the spheres on the side opposite its point of contact with the other sphere.
(a) Describe the induced charges on the two conducting
spheres, and sketch the charge distributions on them. (b) The two spheres are separated far apart and the charged rod is removed. Sketch the charge distributions on the separated spheres. Determine the Concept Because the spheres are conductors, there are free electrons on
them that will reposition themselves when the positively charged rod is brought nearby.
(a ) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram.
+
(b) When the spheres are separated and far
+
+
+
+
o
apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the
1
)
o
Chapter 21
2
diagram. *7
A positive charge that is free to move but is at rest in an electric field E will
(a) accelerate in the direction perpendicular to E . (b) remain at rest. (c) accelerate in the direction opposite to E. Cd) accelerate in the same direction as E. ( e) do none of the above. Determine the Concept The acceleration of the positive charge is given by ii
=
F m
=
E. Because qo and
!l.!!.. m
direction as the electric field. *8
•
m
are both positive, the acceleration is in the same
I (d) is correct. I
If four charges are placed at the corners of a square as shown in Figure 21-
33, the field E is zero at
(a) all points along the sides of the square midway between two charges. (b) the midpoint of the square. (c) midway between the top two charges and midway between the bottom two charges. (d) none of the above. -q
Figure 21-33 Problem 8
,0
}------{:+ +q
+q t;'-}: ------{, -
-q
Determine the Concept E is zero wherever the net force acting on a test charge is zero.
At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the square will be zero.
I (b) is correct. I
*11
•
Two charges +q and -3q are separated by a small distance. Draw the electric
field lines for this system.
The Electric Field 1: Discrete Charge Distributions
3
Determine the Concept We can use the
rules for drawing electric field lines to draw the electric field lines for this system.
-3q
In the field-line sketch to the right we've
+q
assigned 2 field lines to each charge q.
*12
•
Three equal positive point charges are situated at the comers of an equilateral
triangle. Sketch the electric field lines in the plane of the triangle. Determine the Concept We can use the
rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we've assigned 7 field lines to each charge q.
*14
·
The electric field lines around an electrical dipole are best represented by
which, if any, of the diagrams in Figure 2 1-34?
(c)
(d)
Figure 21-34 Problem 1 4 Determine the Concept Electric field lines around an electric dipole originate at the
positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. *15
··
I (d) is correct. I
A molecule with electric dipole moment
p
is oriented so that
p
makes an
angle e with a uniform electric field E that is in the direction of increasing x. The dipole is free to move in response to the force from the field. Describe the motion of the dipole. Suppose the electric field is nonuniform and is larger in the x direction. How will the motion be changed?
4
Chapter 21
Determine the Concept Because 8* 0, a dipole in a unifonn electric field will
experience a restoring torque whose magnitude is pEx sin e . Hence it will oscillate about its equilibrium orientation, 8= o. If 8« 1 , sin8� e, and the motion will be simple hannonic motion. Because the field is nonunifonn and is larger in the x direction, the
force acting on the positive charge of the dipole (in the direction of increasing x) will be greater than the force acting on the negative charge of the dipole ( in the direction of decreasing x) and thus there will be a net electric force on the dipole in the direction of increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about 8 =0. *18
••
A metal ball is positively charged. Is it possible for it to attract another
positively charged ball? Explain. Determine the Concept Yes. A positively charged ball will induce a dipole on the metal
ball, and if the two are in close proximity, the net *19
force can be attractive.
··
A simple demonstration of electrostatic attraction can be done simply by tying a small ball of tinfoil on a hanging string, and bringing a charged wand near it. Initially, the ball will be attracted to the wand, but once they touch, the ball will be repelled violently from it. Explain this behavior. Determine the Concept Assume that the wand has a negative charge. When the charged
wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand. When it touches the wand, some of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand. Estimation and Approximation *23
··
A popular classroom demonstration consists of rubbing a "magic wand" made of plastic with fur to charge it, and then placing it near an empty soda can on its side ( Figure 2 1 -35). The can will roll toward the wand, as it acquires a charge on the side nearest the wand by induction. Typically, if the wand is held about 1 0 cm away from the can, the can will have an initial acceleration of about 1 rnIs2• If the mass of the can is 0.0 1 8 kg, estimate the charge on the rod.
Soda can
Figure 21-35 Problem 23
The Electric Field 1: Discrete Charge Distributions
5
Picture the Problem We can use Coulomb's law to express the charge on the rod in terms of the force exerted on it by the soda can and its distance from the can. We can apply Newton's 2nd law in rotational form to the can to relate its acceleration to the electric force exerted on it by the rod. Combining these equations will yield an expression for Q as a function of the mass of the can, its distance from the rod, and its acceleration.
Use Coulomb's law to relate the force on the rod to its charge Q and distance r from the soda can: Solve for Q to obtain: ( 1)
L reenter
Apply can:
of mass
=fa to the
Because the can rolls without slipping, we mow that its linear acceleration a and angular acceleration a are related according to: Because the empty can is a hollow cylinder: Substitute for I and Fto obtain:
a and
FR=fa a
a=R where R is the radius of the soda can.
f=MR2
where M is the mass of the can.
solve for
Q� r7Q
Substitute for Fin equation ( 1 ):
Substitute numerical values and evaluate Q:
Q=
= I 14l nC I
Electric Charge *27
·
How many coulombs of positive charge are there in 1 kg of carbon? Twelve
grams of carbon contain Avogadro's number of atoms, with each atom having six protons and six electrons. Picture the Problem We can find the number of coulombs of positive charge there are in 1 kg of carbon from
Q = 6nce,
where
nc is the number of atoms in
1 kg of carbon and the factor of 6 is present to account for the presence of 6 protons in
6
Chapter 21
each atom. We can find the number of atoms in lkg of carbon by setting up a proportion relating Avogadro's number, the mass of carbon, and the molecular mass of carbon to nco Express the positive charge in terms of the electronic charge, the number of protons per atom, and the number of atoms in 1 kg of carbon: Using a proportion, relate the number of atoms in 1 kg of carbon
nc, to Avogadro's number and the
nc
-
N
A
m
-�-n ---.' c
M
-
-
NA mC M
molecular mass M of carbon: Substitute to obtain:
Substitute numerical values and evaluate Q:
Q
=
6(6.02 X 1023 atoms/mol)(1 kg)(1.6 x 10-19 C) = 1 4.82 X 107 C 1 0.012 kg/mol
Coulomb's Law *32
••
A point charge of -2.5 jiC is located at the origin. A second point charge of 6
jiC is at x = 1 m, Y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium. Picture the Problem The positions of the
charges are shown in the diagram. It is
Y,m
apparent that the electron must be located along the line joining the two charges. Moreover, because it is negatively charged, it must be closer to the -2.5 jiC than to the 6.0 jiC charge, as is indicated in the figure. We can find the x and y coordinates of the electron's position by equating the two electrostatic forces acting on it and solving for its distance from the origin. We can use similar triangles to express this radial distance in terms of the x and y coordinates of the electron.
The Electric Field 1: Discrete Charge Distributions
7
Express the condition that must be satisfied if the elech-on is to be in equilibrium: Express the magnitude of the force that ql exerts on the electron: Express the magnitude of the force that q2 exerts on the electron: ql
Substitute and simplify to obtain:
_
l q2 1
�+JL25mY -7 Substitute for ql and q2 and simplify: Solve for r to obtain:
(-L4m-2) r2 + (22361m-1) r +125m = 0 r=2_036m and
r=-OA386m Because r < 0 is unphysical, we'll consider only the positive root Use the similar triangles in the diagram to establish the proportion involving the y coordinate of the
2_036m O_5m L12m
�
electron: Solve for
Ye:
Use the similar triangles in the diagram to establish the proportion involving the x coordinate of the
Ye =O_909m Xe 2_036m 1m L12m _
electron: Solve for
Xe:
The coordinates of the electron's position are:
Xe =L82m (xe,yJ=1 (-L82m,-O.909m) I
8
Chapter 2 1
*33
··
A charge o f -1.0 ;...tC is located at the origin; a second charge o f 2.0 ;...tC is
located atx = O,y
=
0. 1 m; and a third charge of 4.0;...tC is located atx = 0.2 m, y = O.
Find the forces that act on each of the three charges. Picture the Problem Let ql represent the
charge at the origin,
Y,m qz
q2 the charge at (0, 0. 1
m), and q3 the charge at ( 0.2 m, 0). The diagram shows the forces
F1.2
acting on each of the charges. Note the action-and-reaction pairs. We can apply
=
"
2 p.C
"
"
FZ•1
"
Coulomb's law and the principle of
"
"
"
superposition of forces to find the net force acting on each of the charges. Express the net force acting on q1 : Express the force that q2 exerts on ql:
Substitute numerical values and evaluate
F",
�
F2,1 :
(8.99xI0' N ·m'/C' )(2pC) tO.ll�! m (-O.lm)] (1.80N)] �
i:
Express the force that q3 exerts on ql:
Substitute numerical values and evaluate
F",
�
3,1
=
kq3ql r r3,13 3,1
F3 1 :
(8.99x10' N ·m'/C' )(4pC) tO.2IpCm J (-0.2m)i (0.899N)i F; : i; =\ (O.899N)i (1.80N)) I �
Substitute to find
Express the net force acting on q2:
+
-
-
-
F2 = F3,2 + F;,2
The Electric Field
1:
Discrete Charge Distributions
9
because F.,,2 and F2,I are action-and-reaction forces. Express the force that
q3 exerts on q2:
F32, =kq33q2 r3,2 r32, =kq�q2 r32,
[(-O.2m)i (O.lm)J]
Substitute numerical values and evaluate
F",
+
F3,2 :
(8,99 10' N, )(4 jiC) (0.(222jiC4m))' [(- 0,2m)i (0, lm)j1 =(-1.28N)i (0.640N)J
�
m 'IC '
X
+
+
Find the net force acting on
q2:
-(1.80N)J =(-1.28N)i (0.640N)J -(1.80N)J (-1.28N)i-(1.16N)J 1
F2 =F32, =1
+
F3, I are an action-and-reaction pair, as are F23, and F3,2 ' express the net force acting on q3:
Noting that
-
F., ,
3
and
-(0.899N)i -l(-1.28N)i (0.640N)JJ
F3 =FI3, F23, =-F3,1 - F32, = 1 =1 +
(0.381N)i -(0.640N)J
+
The Electric Field
*37
·
A charge of 4.0 J-LC is at the origin. What is the magnitude and direction of
the electric field on thex axis at (a) x = 6 m, and (b)x = - 10 m?
(c) Sketch the function Ex versusx for both positive and negative values ofx. (Remember that Ex is negative when E points in the negativex direction.) Picture the Problem Let
q
represent the charge at the origin and use Coulomb's law for
E due to a point charge to find the electric field atx = 6 m and -10 m. (a) Express the electric field at a point P located a distancex from a charge
q:
-() kq =-2 rpo
E
X
A
x
'
10 h C apter 21 09N .m2/C2 )(4,LlC ) �
E-(6m)= (8.99xl
Evaluate this expression for x=6 m:
=
(6mY
I
I (999N/C)i I
(b) Evaluate Eatx=-lO m:
(c) The following graph was plotted using a spreadsheet program:
500
250
� «5'
a
-250
-500
*38
·
-2
-1
x
a
(m)
2
Two charges, each +4 pC, are on thex axis, one at the origin and the other at
x = 8 m. Find the electric field on thex axis at (a)x=-2 m, (b) x = 2 m, (c) x = 6 m, and
(d) x = 10 m. x.
(e) At what point on thex axis is the electric field zero? if) Sketch Ex versus
Picture the Problem Let q represent the charges of +4 fI2 and use Coulomb's law for
Edue to a point charge and the principle of superposition for fields to find the electric field at the locations specified. Noting that ql =q2, use Coulomb's law and the principle of superposition to express the electric field due to the given charges at a point P a distancex from the origin:
The Electric Field 1 : Discrete Charge Distributions
)
()
(1
kql kq, 1 r + kql E(x) =Eq, (x +Eq, x = 2 rq"p+ q � r r "p = fJ " p 8m x 2 x x2 ( 8m-xY -
( ) 2)(71 rq"p + ( 1 x rq"p = (36kN'm 8m -Y J -
-
�
IC
�
�
�
fJ"P
�
(a) Apply this equation to the point atx = -2 m:
ii,(-
2m) (36kN �
l[ (2�)' (-i)+ (1O�)' (-il] I (-936kN/C)i I
m' IC
�
(b) Evaluate Eatx = 2 m:
E(2m)� (36kN.m'/cl[ (2�)' i)+ ( (6�)' (-I)H (S.oOkN/c)i I (c) Evaluate E atx
=
6 m:
E(6m)� (36kN ·m'/Ct6�)' �)+ (2�)' (-I)�] I (-S.oOkN/c)i I (d) Evaluate E atx
=
1 0 m:
E(10m)� (36kN m'/ctl 0�)' (I) + (2�)J)] I (935kN/C)i I �
(e) From symmetry considerations:
E(4m)= @]
(f) The following graph was plotted using a spreadsheet program:
11
J
12
Chapter 21
·4
*42
••
x
4
8
(m)
12
A point charge of +5.0 j.1C is located at x = -3.0 cm, and a second point
charge of -8.0 j.1C is located at x = +4.0 cm. Where should a third charge of +6.0 j.1C be
placed so that the electric field at x = 0 is zero?
Picture the Problem If the electric field at x = 0 is zero, both its x and y components must
be zero. The only way this condition can be satisfied with the point charges of +5.0 j.1C and -8.0 j.1C are on the x axis is if the point charge of +6.0 j.1C is also on the x axis. Let the subscripts 5, -8, and 6 identify the point charges and their fields. We can use Coulomb's law for Edue to a point charge and the principle of superposition for fields to determine where the +6.0 j.1C charge should be located so that the electric field at x = 0 is zero. Express the electric field at x = 0 in terms of the fields due to the charges
E()O = Es,£ E_8,£ E6,£ +
+
=0
of +5.0 j.1C, -8.0 j.1C, and +6.0 j.1C : Substitute for each of the fields to obtain: or
kqs kq6 ( ) kq_8 ( r_8
-
2 �
l� +
-
2 �
-l� +
� )-0
- -l 2
-
Divide out the unit vector i to obtain: Substitute numerical values to obtain:
5
6
_-_8_ =0
(3cmY (4cmY r6 =12.38cm I
T he Electric Field 1: Discreteh C arg e Distributions *45
··
A 5-p.C point charge is located at x = 1 m, Y = 3 m; and a
-
4 ,uC point charge -
is located at x = 2 m, Y = -2 m. (a) Find the magnitude and direction of the electric field
at x = -3 m, Y = 1 m. (b) Find the magnitude and direction of the force on a proton at
x = -3 m, Y = I m.
Picture the Problem The diagram shows the electric field vectors at the point of interest
P due to the two charges. We can use Coulomb's law for E due to point charges and the superposition principle for electric fields to find p . We can apply F to find the
E
E q=
force on a proton at ( -3 m, 1 m). Y,m 3
./
/q2 = SI-'C
2
--�---+��r----r---+-x,m 2 -3 -2 �'-.. '-..
-1
'-.
'-..
-2
'-..
'-..
'--. ,/, = -4 f.LC
(a) Express the electric field at
l and q
( -3 m, 1 m) due to the charges
q2 :
Evaluate
EI :
C) qk rl p = (8.99X109N .m 2C/ 2)(-4,u EI =l , 2 2 (S) m + (3) m p
]
[
m m +(3)) (-S)1 �(SY m +(3Y m 1'j� 5 7 + 0.514 = (-1.06kN/ C)(-0.81 ))= (0.908kN/ c)1 ( 0 544 kNC)) / ?
�
+
E 2:
Evaluate
-
13
.
14 h C apter 21 Substitute and simplify t o find
Ep:
Ep =(0.908kN/ C)i + (-0.544kN/ C)} (-2.0lkN/ C)i (-1.0lkN/ C)) C)) =(-1.1OkN/ C)i (-1.55kN/ +
+
+
The magnitude of
Ep is:
The direction of Ep is:
Y C + (1.55kN/ Ep = �(1.10kN/ Y C C ! = ! 1.90kN/ e
E
(
)
C =! 2350 ! =tan- I -1.55kN/ -1.10kN/ C
(
)
Note that the angle returned by your calculator for
C . tan_1 -1.55kN/ -1.10kN/ C
IS
the
reference angle and must be increased by
1800 to yield BE.
(b) Express and evaluate the force on a proton at point P:
C))J c)1+ (-1.55kN/ F q=Ep =(1.6x10-19C )l(-1.10kN/ =(-1.76x10-'6i N) + (-2.48X10-16} N) The magnitude of F is:
The direction of F is:
eF
]
(
! =tan- I -2.48X10-16N =! 2350 -1.76x10-16N '
where, as noted above, the angle returned
(
by your calculator for
tan- I -2 ·48X10-16N -1.76x10-16N
)
is the reference
angle and must be increased by yield BE.
1800 to
The Electric Field 1: Discrete Charge Distributions *48
•••
15
Two positive point charges + q are on the y axis at y = +a and y = -a as in
Problem 44. A bead of mass m can-ying a negative charge -q slides without friction along a thread that runs along the x axis. (a) Show that for small displacements of x « a, the bead experiences a restoring force that is proportional to x and therefore undergoes simple harmonic motion. (b) Find the period of the motion. Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to
equal positive charges located at ( 0, a) and (O,-a), is given by
Ex
=
2kqx(x2 + a2 t/2. We can use T = 2Jr�m/ k' to express the period of the motion
in terms of the restoring constant Ii. (a) Express the force acting on the on the bead when its displacement from the origin is x: Factor a2 from the denominator to obtain:
Fx =
For x « a: i.e., the bead experiences a linear restoring force.
(b) Express the period of a simple harmonic oscillator:
T
=
2Jr 1m
fk7
Obtain Ii from our result in part (a):
Substitute to obtain:
Motion of Point Charges in Electric Fields *50
·
(a) Compute elm for a proton, and find its acceleration in a uniform electric
field with a magnitude of 100 N/C. (b) Find the time it takes for a proton initially at rest in such a field to reach a speed of O.Ole ( where e is the speed of light).
16 C a er h pt
21
Picture the Problem We can use Newton's 2nd law of motion to find the acceleration of
the proton in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of O.Ole and the distance it travels while acquiring this speed.
1.6 10-19 C 1.67x10-27 kg = 19.58x107 Clkg 1 Fnet eE a=--=e=
(a) Use data found at the back of your text to compute elm for an electron:
X
------
mp
Apply Newton's 2nd law to relate the
mp
acceleration of the electron to the electric field:
mp
10-19 C){l00N/C) a= (1.6x1.67 x10-27 kg =19.58 109 m1s2 1 The direction of the acceleration of a proton is in the direction of the electric field.
Substitute numerical values and evaluate a:
X
(b)
Using the definition of a
acceleration, relate the time required for an electron to reach O.Ole to its
a
acceleration: Substitute numerical values and evaluate f..t: *54
••
A particle leaves the origin with a speed of 3 x 10 6 m/s at 35 ° to the x axis. It -
moves in a constant electric field E =
Eyj. Find Ey such that the particle will cross the �
x
axis atx = 1.5 em if the particle is (a) an electron, and (b) a proton.
Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the particle in terms of the parameter t and Newton's 2nd law to express
the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m that we can solve for Ey. Express thex and y coordinates of the particle as functions of time:
x {vcosB)t =
h T e Electric Field 1: Discreteharg C e Distributions and y
Apply Newton's 2nd law to relate the acceleration of the particle to the net
a
={vsinB)t
-
-t ai2
Ey Fnet,y q =--=- y m m
force acting on it: Substitute in the y-coordinate equation to obtain: Eliminate the parameter t between the two equations to obtain: Set y = 0 and solve for
Y
q t =(vsin) B t E ---y 2 2m
y
E x2 = ( tan)x B - q 2mv2 cos2 B Y
mv2 sin2B = Ey x q
Ey :
Substitute the non-particle specific data to obtain:
sin 7 0° E y - m(3qx106(O.mist OIS) m _
=(S.64x1014m /s2) m q
(a) Substitute for the mass and charge of an electron and evaluate
Ey :
(b) Substitute for the mass and charge of a proton and evaluate
*58
·
Ey:
kg s ) 9.11x10-31 E y =(S.64x1014m 12 1.6x10-1C9 =13.21kN/C I - 7 kg 1.67 x102 ml 2 14 s ) = (S 64x 10 . Ey 1.6x10-1C9 =IS.89MN/C I
A dipole of moment 0.5 e·nm is placed in a uniform electric field with a
magnitude of
4.0x104N/C. What is the magnitude of the torque on the dipole when (a)
the dipole is parallel to the electric field, (b) the dipole is perpendicular to the electric field, and (c) the dipole makes an angle of 30° with the electric field? (d) Find the potential energy of the dipole in the electric field for each case. Picture the Problem The torque on an electric dipole in an electric field is given by
. i = PxE and the potential energy of the dipole by U -pE. =
Using its definition, express the torque
i=pxE
17
18
Chapter
21
on a dipole moment in a uniform
and
electric field:
T
E p = sin8
where () is the angle between the electric dipole moment and the electric field. (a) Evaluate ,for () = 0°:
T
(b) Evaluate ,for ()
90°:
T
30°:
T
(c) Evaluate dor ()
=
=
(d) Using its definition, express the potential energy of a dipole in an
E p = si n 0° =� =(O.Se · n)( m 4.0 x10 4N/ )sin C 90° =13.20x10-24N'm 1 = (0.Se . n)( m 4.0xl0 4N/ ) C sin30° = 11.60 xl 0-24N .m 1
U
=-p·i =p E - cos8
electric field: 0°:
U
90°:
U
Evaluate U for () = 30°:
U
Evaluate U for ()
Evaluate U for ()
*59
··
=
=
=-(O.Se . n)( m 4.0x104N/ )cosoo C =1-3.20xl0-24J 1 =-(O.Se . n)( m 4.0x104N/ )cos90° C =0 )cos30° =-(O.Se · n)( m 4.0x104N/ C =1-2.77x10-24 J I
For a dipole oriented along the x axis, the electric field falls off as
lIx3 in the
lIi in the y direction. Use dimensional analysis to prove that, in any direction, the field far from the dipole falls off as llr3•
x direction and
Picture the Problem We can combine the dimension of an electric field with the
dimension of an electric dipole moment to prove that, in any direction, the dimension of
[ L 1/ 3] and, hence, the electric field far from the dipole falls
the far field is proportional to off as
lIr3.
Express the dimension of an electric field:
] [ E ] = [ kQ 2 [ L]
The Electric Field
1: Discrete Charge Distributions
Express the dimension an electric dipole moment: Write the dimension of charge in terms of the dimension of an electric dipole moment: Substitute to obtain: This shows that the field E due to a dipole 3 p falls off as lIr . General Problems
*63 · ( a) What mass would a proton have if its gravitational attraction to another proton exactly balanced out the electrostatic repulsion between them? ( b) What is the true ratio of these two forces? Picture the Problem We can equate the gravitational force and the electric force acting on a proton to find the mass of the proton under the given condition.
( a) Express the condition that must be satisfied if the net force on the proton is zero: Use Newton's law of gravity and Coulomb's law to substitute for Fg
and Fe:
Solve for m to obtain:
m= e
H;
Substitute numerical values and evaluate m:
N·m 2Cf 2 = i 1.86xlO-9kg i. m= (1.6xlO-19C ) 6.68.99xl09 7 x 10-11 N·m 2fkg2 . ( b) Express the ratio of Fe and Fg:
ke2
Gm� � r
2
ke2
Gm�
19
20 h C apter 21 Substitute numerical values to obtain:
(8.99xl09N .m 2C/ 2)(1.6xlO-19C) =1 1.24xl036 g )(1.67 x 10-27 k) g (6.67 x 10-11N · m 2 /k2 *66
••
In copper, about one electron per atom is free to move about. A copper penny
has a mass of 3 g. (a) What percentage of the free charge would have to be removed to give the penny a charge of 15 fLC? (b) What would be the force of repulsion between two pennies carrying this charge if they were 25 cm apart? Assume that the pennies are point charges. Picture the Problem We can find the percentage of the free charge that would have to
be removed by finding the ratio of the number of free electrons ne to be removed to give the penny a charge of 15 fLC to the number of free electrons in the penny. Because we're assuming the pennies to be point charges, we can use Coulomb's law to find the force of repulsion between them. (a) Express the fractionJof the free charge to be removed as the quotient
f =N ne
of the number of electrons to be removed and the number of free electrons: Relate Nto Avogadro's number, the mass of the copper penny, and the molecular mass of copper:
N -=-:::: ::>N=NA M NA M m
m
Relate ne to the free charge Q to be removed from the penny:
Q
f= -e = N� A M
Substitute numerical values and evaluate!
(-15,uC )(63.5g /m ) ol =3.29xlO-9 =1 3.29x 10- 7 0/0 I f =- ()3( g 1.6xl 0-1C9 )(6 .02 x 1023 m orl)
h T e Electric Field 1: Discreteh C arg e Distributions
21
(b) Use Coulomb's law to express the force of repulsion between the two pennIes: Substitute numerical values and evaluate F:
F = (8.99xl0 *69
••
9
N·m C/ 2
)(9.38x 1013 Y(1.6xlO-1C9 Y 1 = 32. 4 N I (0 25 m 2)
2
.
A positive charge Q is to be divided into two positive charges ql and q2.
Show that, for a given separation D, the force exerted by one charge on the other is greatest ifql q2
= =1Q·
Picture the Problem We can use Coulomb's law to express the force exerted on one
charge by the other and then set the derivative of this expression equal to zero to find the distribution of the charge that maximizes this force. Using Coulomb's law, express the force that either charge exerts on the other: Express q2 in terms of Q and ql: Substitute to obtain:
Differentiate F with respect to ql and set this derivative equal to zero for extreme values:
-dF = k -d[ql (Q-ql )] d d l D2 ql q = 2 [ql (- l) + Q - q]l =0
; for extrem a
Solve for ql to obtain: To determine whether a maximum or a minimum exists at ql
=1Q ,
differentiate F a second time and evaluate this derivative at ql
= 1Q:
d -2 l d2F k - [ 2 2 d D qd l Q q] ql = k2 (-2) D 0 _
<
independently of
ql .
22
C hapter 21
Q
I .. q) = q2 = t m axim izesF. 1 ·
*70
••
A charge Q is located atx = 0, and a charge 4Q is atx = 12.0 cm. The force
on a charge of -2;.£ is zero if that charge is placed atx = 4.0 cm, and is 126.4 N in the
positivex direction if placed atx = 8.0 cm. Determine the charge Q.
Picture the Problem We can apply Coulomb's law and the superposition of forces to
relate the net force acting on the charge q = -2 ;.£ tox. Because Q divides out of our equation when F(x) = 0, we'll substitute the data given for
x = 8.0 cm.
Using Coulomb's law, express the net force on q as a function ofx:
F{x)
Simplify to obtain:
kq Solve for Q:
Q=
=
[� _
X
[
2 +
4 (12cm-xY
F{x) 1
4
kq - _ 2 + ..,..-___ X (12cm-x)2
]Q ]
Evaluate Q forx = 8 cm:
*78
··
Two small spheres of mass m are suspended from a common point by threads
of length L. When each sphere carries a charge q, each thread makes an angle e with the
�
vertical as shown in Figure 21-42. (a) Show that the charge q is given by mgtane . q= 2L sm e k
where k is the Coulomb constant. (b) Find q if m = 1 ° g, L = 50 cm, and e= 10°.
h T e Electric Field I: Discreteh C arg e Distributions
23
Figure 21-42 Problem 78 y
Picture the Problem Each sphere is in
static equilibrium under the influence of -
-
the tension T , the gravitational force F ,
E
and the electric force F . We can use
g
----x ----�--.-
Coulomb's law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the
�=mg
charge on each sphere. (a) Apply the conditions for static equilibrium to the charged sphere:
2
IF, =FE -TsinB= kc; r
and
IFy =TcosB-mg= 0
Eliminate Tbetween these equations to obtain: Solve for q:
Referring to the figure, relate the separation of the spheres r to the
2 tanB= kq 2 -
q=r
r
�
mgr
mgtanB k
2 = LsinB
length of the pendulum L: Substitute to obtain:
q=
2L .
�
sm B
mgtanB k
-TsinB =
0
24
Chapter
21
(b) Evaluate q for m q
*83
··
=
1 0 g, L
=
5 0 cm, and
= 2(O.5 m)sin 1 0°
B=
1 0°:
-'-------=-'---'---:--'-
=
I O.24 1,liC I
An electron ( charge -e, mass m) and a positron (charge +e, mass m ) revolve
around their common center of mass under the influence of their attractive coulomb force. Find the speed of each particle v in terms of e, m, k, and their separation r. Picture the Problem The forces the electron and the proton exert on each other
constitute an action-and-reaction pair. Because the magnitudes of their charges are equal and their masses are the same, we find the speed of each particle by finding the speed of either one. We'll apply Coulomb' s force law for point charges and Newton's 2nd law to relate v to e, m, k, and r. Apply Newton' s 2nd law to the positron:
v = 1Tk7l �
Solve for v to obtain:
*90 ••• Two neutral polar molecules attract each other. Suppose that each molecule has a dipole moment p, and that these dipoles are aligned along the x axis and separated
by a distance d. Derive an expression for the force of attraction in terms of
P and d.
Picture the Problem We can relate the force of attraction that each molecule exerts on
the other to the potential energy function of either molecule using
F = -dV/dx. We can
relate U to the electric field at either molecule due to the presence of the other through U
= -pE. Finally, the electric field at either molecule is given by E Express the force of attraction between the dipoles in terms of the spatial derivative of the potential energy function of
PI :
Express the potential energy of the dipole
PI :
Express the electric field at
P2:
P I due to
F = - ddxVI
= 2kp /x3 •
( 1)
VI = -PI EI where EI is the field at PI due to P2. EI
=
2 kp2 X
3
The Electric Field 1 : Discrete Charge Distributions
25
where x is the separation of the dipoles. Substitute to obtain:
Substitute in equation ( 1) and differentiate with respect to x : Evaluate
F for PI = P2 = P and
x = d to obtain: *93 ••• In Problem 92, there was a description of the Millikan experiment used to determine the charge on the electron. In the experiment, a switchable power supply is used so that the electrical field can point both up and down, but with the same magnitude, so that one can measure the terminal speed of the microsphere as it is pushed up ( against the force of gravity) and down. Let Vu represent the terminal speed when the particle is moving up, and Vd the terminal speed when moving down. (a) If we let v = Vu + Vd, show that v = qE /31r1]r , where q is the microsphere's net charge. What advantage does measuring both Vu and Vd give over measuring only one? (b) Because charge is quantized, v can only change by steps of magnitude LlV. Using the data from Problem 92, calculate Llv. y
Picture the Problem The free body diagram shows the forces acting on the microsphere of mass m and having an excess charge of q = Ne when the electric field is downward. Under terminal-speed conditions the sphere is in equilibrium
I
under the influence of the electric force Fe ' its weight mg, and the drag force
2 nd
F
d•
m, Ne
We
can apply Newton's law, under terminal-speed conditions, to relate the number of excess charges N on the sphere to its mass and, using Stokes' law, to its terminal speed. (a) Apply
IFy
=
may to the
microsphere when the electric field is downward: Substitute for obtain:
Fe and FcI,terminal to
mg
Fe
- mg - Fd = may or, because ay = 0, F. - mg - F:t terminal 0 =
qE - mg - 61rTJrvu
=
0
or, because q = Ne,
NeE - mg
-
61rTJrvu
=
0
26
Chapt er 2 1
Solve for
Vu
to obtain:
With the field pointing upward, the electric force is downward and the application of
I Fy
=
NeE - mg 6mJr
vu =
may to the
microsphere yields: Solve for Vd to obtain:
Add equations ( 1) and (2) to obtain:
(1)
----
Fd, temlinal - Fe - mg - O or
6Jr7]rvd - NeE - mg = 0 NeE + mg 6Jr7]r
vd =
---=
V = Vu
_
NeE- mg -
+ vd
NeE 3TrTJr
_
(2)
�
6 TrTJr NeE + mg + ------=6TrTJr
I 1 qE 3TrTJr
This has the advantage that you don' t need to know the mass of the microsphere. (b) Letting �V represent the change in the terminal speed of the microsphere due to a gain ( or loss) of one electron we have: Noting that �V will be the same whether the microsphere is moving upward or downward, express its terminal speed when it is moving upward with N electronic charges on it: Express its terminal speed upward when it has N + 1 electronic charges: Substitute and simplify to obtain:
vN -
VN + l
NeE - ma b 6TrTJr
=
1 ) eE - mg (N +� -'--
--
6Jr7]r
The Electric Field
1 : Discrete Charge Distributions
Substitute numerical values and evaluate
�v:
=
1 5.15
X
10-5 m/s
1
27
Chapter 22 The Electric Field 2: Continuous Charge Distributions Conceptual Problems *1
True or false:
••
(a) Gauss's law holds only for symmetric charge distributions.
(b) The result that E = 0 inside a conductor can be derived from Gauss's law. (a) False. Gauss's law states that the net flux through any surface is given =
by ¢net
4EndA
=
4nkQnside . While it is true that Gauss's law is easiest to apply to
symmetric charge distributions, it holds for any surface.
(b) True *5
•
True or false:
(a) If there is no charge in a region of space, the electric field on a surface surrounding the region must be zero everywhere.
(b) The electric field inside a uniformly charged spherical shell is zero. (c) In electrostatic equilibrium, the electric field inside a conductor is zero. (d) If the net charge on a conductor is zero, the charge density must be zero at every point on the surface of the conductor. (a) False. Consider a spherical shell, in which there is no charge, in the vicinity of an infinite sheet of charge. The electric field due to the infinite sheet would be non-zero
---
everywhere on the spherical surface.
(b) True ( assuming there are no charges inside the shell). (c) True. (d) False. Consider a spherical conducting shell. Such a surface will have equal charges on its inner and outer surfaces but, because their areas differ, so will their charge densities. *9
••
Suppose that the total charge on the conducting shell of Figure 22-36 is zero.
It follows that the electric field for r < RJ and r > R 2 points (a) away from the center of the shell in both regions.
29
30
Chapter 22
( b) toward the center of the shell in both regions. (c) toward the center of the shell for r < R, and is zero for r > R2 . (d) away from the center of the shell for r < R , and is zero for r > R 2 . Determine the Concept We can apply Gauss's law to determine the electric field for r R , and
r
> R 2 • We also know that the direction of an electric field at any point is
<
determined by the direction of the electric force acting on a positively charged object located at that point. From the application of Gauss's law we know that the electric field in
E
n
= kQ r
2
both of these regions is not zero and is given by: A positively charged object placed in either of these regions would experience an attractive force from the charge -Q located at the center of the shell. *10
••
I (b) is correct. I
If the conducting shell in Figure 22-36 is grounded, which of the following
statements is then correct? (a) The charge on the inner surface of the shell is +Q and that on the outer surface is -Q.
(b) The charge on the inner surface of the shell is +Q and that on the outer surface is zero. (c) The charge on both surfaces of the shell is +Q. (d) The charge on both surfaces of the shell is zero. Determine the Concept We can decide what will happen when the conducting shell is
grounded by thinking about the distribution of charge on the shell before it is grounded and the effect on this distribution of grounding the shell. The negative point charge at the center of the conducting shell induces a positive charge on the inner surface of the shell and a negative charge on the outer surface. Grounding the shell attracts positive charge from ground; resulting in the outer surface becoming electrically neutral.
I (b) is correct. I
Estimation and Approximation *14
·· Given that the maximum field sustainable in air without electrical discharge is approximately 3 x l 0 6 N/C, estimate the total charge of a thundercloud. Make any assumptions that seem reasonable.
Picture the Problem We'll assume that the total charge is spread out uniformly (charge
density = 0-) in a thin layer at the bottom and top of the cloud and that the area of each
The Electric Field 2 : Continuous Charge Distributions
31
surface of the cloud is 1 km2. We can then use the definition of surface charge density and the expression for the electric field at the surface of a charged plane surface to estimate the total charge of the cloud. Express the total charge Q of a thundercloud in terms of the surface area A of the cloud and the charge density () : Express the electric field just outside the cloud: Solve for () : Substitute for (} to obtain:
Q = E O EA
Substitute numerical values and evaluate Q:
Remarks: This charge is in reasonably good agreement with the total charge transferred in a lightning strike of approximately 30 C.
Calculating E From Coulomb's Law * 17 =
0 to x = 5
A uniform line charge of linear charge density A. = 3 .5 nC/m extends from x m.
(a) What is the total charge? Find the electric field on the x axis at (b) x =
6 m, (c) x = 9 m, and (d) x = 250 m. (e) Find the field at x = 250 m, using the
approximation that the charge is a point charge at the origin, and compare your result with that for the exact calculation in Part (d). Picture the Problem We can use the definition of A. to find the total charge of the line of
charge and the expression for the electric field on the axis of a finite line of charge to evaluate Ex at the given locations along the x axis. In part (d) we can apply Coulomb's law for the electric field due to a point charge to approximate the electric field at x = 250 m. (a) Use the definition of linear charge density to express Q in terms of A.: Express the electric field on the axis of a finite line charge:
Q = AL
(
)( )
I
= 3 . S nC/m S m = 1 7.S nC
I
32
Chapter
22
(b) Substitute numerical values and
evaluate Et at x = 6 m:
. m 2 /C 2 )(1 7.5 )n C 0 9x N (8.9 9 I = , m (6 ) E (6) m (6 m - 5) m =1 26.2N/C I .
(c) Substitute numerical values and evaluate Ex at x = 9 m:
9N·m 2 /C 2 )(1 7 .5 nC) E (9 m ) = (8 . 99 x I(90m )(9 m - 5 m) x
= 1 4 . 3 7 N/C I
(d) Substitute numerical values and evaluate Ex at x = 250 m:
(e) Use Coulomb' s law for the electric field due to a point charge to obtain:
kQ E)X ) =2 X
Substitute numerical values and evaluate ExC250 m):
) =\8 .99 10' .mm' / ' )(17 .5n) C =! 2 52 mN/C ! E, (250m 250 x
7 �
.
Note that this result agrees to within 2% with the exact value obtained in (d) . *25 ·· (a) Using a spreadsheet program or graphing calculator, make a graph of the electric field on the axis of a disk of radius r = 30 cm carrying a surface charge density (J = 0.5 nC/m2 . (b) Compare the field to the approximation E = 2 Jrkcr. At what distance does the approximation differ from the exact solution by 1 0 percent?
Picture the Problem (a) The electric field on the x axis of a disk of radius r carrying a surface
charge density (J is given by:
(b) The electric field due to an
infinite sheet of charge density (Jis
independent of the distance from the plane and is given by:
Eplate =2 nk<J
The Electric Field 2 : Continuous Charge Distributions
33
A spreadsheet solution i s shown below. The fOlll1ulas used to calculate the quantities in
the columns are as follows: Cell B3 B4 B5 A8 A9 B8
ContentiFormula 9.00E+09 5 .00E- I 0 0.3 0 0.01 2*PIO*$B$3*$B$4*( I -A81 (A81\2+$B$51\2Y'2)1\0.5)
C8
2*PIO*$B$3*$B$4 ",1'-
"'"1
2 3 4 §
< '��' '",A i.; ".
,
� k1� 1-
-
12 13 [ 4 ;�" 15
�.
1�· .3)
("
k=
74 ';�' 7' 5 ,--� 76 ?"'. 77 78
.;;,
k
() r
xo
xo
+ 0.01
2nkcr(l -
�
.!'I"B
I"',�.�
�
¥;;iI';
x 0.00 0.01 0 .02 0.03 0.04 0.05 0.06 0.07
E(x) 28 .27 27.33 26.39 25 .46 24.54 23 .63 22.73 2 1 .85
E plate 28.3 28.3 28.3 28.3 28.3 28.3 28.3 28.3
0 .65 0 .66 0.67 0.68 0.69 0.70
2.60 2.53 2.47 2.41 2 . 34 2.29
28.3 28.3 28.3 28.3 28.3 28.3
.
)
.,
�"
Nml\2/CI\2 Clml\2 m
r=
x 2 + r2 x
27rku
9.00E+09 5 .00E- I 0 0.3
slgma=
,6 ;11 �) 7 w� 8 16 U
i;.'
Algebraic Form
-i��"" "'';
The following graph shows E as a function of x. The electric field from an infinite sheet with the same charge density is shown for comparison - the magnitude of the electric fields differ by more than 1 0 percent for x = 0.03 m.
34
Chapter 22
f:2 b '-<.l
15 10 5 0
*30
•••
0.0
0.1
0.2
0.3
x
(m)
0.4
0. 5
0. 6
0.7
A hemispherical thin shell of radius R carries a uniform surface chargeO".
Find the electric field at the center of the hemispherical shell (r = 0). z
Picture the Problem Consider the ring with its axis along the z direction shown in the diagram. Its radius is z = rcos B and its width is rdB. We can use the equation for the field on the axis of a ring charge and
.F------;)l- y
then integrate to express the field at the center of the hemispherical shell. x
Express the field on the axis of the ring charge:
dE =
kzdq l (r 2 sin 2 a + r 2 60 s 2 a ) 2 kzdq r3
where z = rcos B Express the charge dq on the ring:
Substitute to obtain:
dq
=
adA = O"{2nr sin a }rd a
=
21CO"r 2 sin Bda
2 sin Bd a dE = k{r cos a )21CO"r r3 = 21lk0" sin a cos ad a
The Electric Field 2 : Integrate dE from e = 0 t o Tel2 to obtain:
Continuous Ch arge Distributions
E
=
2nka-
=
2nk
35
:r/ 2 fsin e cos ed e o
o-[t sin 2 e ] �/2
=
I nka- I
Gauss's Law •
*32
A single point charge q
=
+2 j.iC is at the origin. A spherical surface of radius
3 .0 m has its center on the x axis at x = 5 m. (a) Sketch electric field lines for the point
charge. Do any lines enter the spherical surface? (b) What is the net number of lines that
cross the spherical surface, counting those that enter as negative? (c) What is the net flux
of the electric field due to the point charge through the spherical surface?
Determine the Concept While the number of field lines that we choose to draw radially outward from q is arbitrary, we must show them originating at q and, in the absence of
other charges, radially symmetric. The number of lines that we draw is, by agreement, in proportion to the magnitude of q . (a) The sketch of the field lines and of the sphere is shown in the diagram to the right.
Given the number of field lines drawn from q, 3 lines enter the sphere. Had we chosen to draw 24 field lines, 6 would have entered the spherical surface. (b) (c)
The net number of lines crossing the surface is zero.
I The net flux is zero. I
*36
•
Since Newton' s law of gravity and Coulomb' s law have the same inverse-
square dependence on distance, an expression analogous in form to Gauss's law can be found for gravity. The gravitational field g is the force per unit mass on a test mass mo .
Then, for a point mass m at the origin, the gravitational field g at some position r is
36
Chapter 22 g _
= -
Gm
-r-
,- r
�
Compute the flux of the gravitational field through a spherical surface of radius r centered at the origin, and show that the gravitational analog of Gauss' s law is
¢net = -4 nGm inside Picture the Problem We'll define the flux of the gravitational field in a manner that is analogous to the definition of the flux of the electric field and then substitute for the gravitational field and evaluate the integral over the closed spherical surface. Define the gravitational flux as:
¢g =
g
¢g =
Substitute for
and evaluate the
integral to obtain:
{g.iidA {(-�� } ( ��) ) r
= -
iidA = -
1
7{
4nr 2 = - 4nGm
dA
I
Spherical Symmetry * 42
••
Consider two concentric conducting spheres (Figure 22-3 8). The outer sphere
is hollow and initially has a charge -7 Q deposited on it. The inner sphere is solid and has a charge +2Q on it. (a) How is the charge distributed on the outer sphere? That is, how
much charge is on the outer surface and how much charge is on the inner surface? (b) Suppose a wire is connected between the inner and outer spheres. After electrostatic
equilibrium is established, how much total charge is on the outside sphere? How much charge is on the outer surface of the outside sphere, and how much charge is on the inner surface? Does the electric field at the surface of the inside sphere change when the wire is connected? If so, how? (c) Suppose we return to the original conditions in Part (a),
with +2Q on the inner sphere and -7Q on the outer. We now connect the outer sphere to ground with a wire and then disconnect it. How much total charge will be on the outer sphere? How much charge will be on the inner surface of the outer sphere and how much will be on the outer surface?
---
-7Q
,.--- +2Q
Figure 22-38 Problem 42
The Electric Field
2:
Continuous Charge Distributions
37
Dete,·mille the Concept The charges on a conducting sphere, in response to the repulsive
Coulomb forces each experiences, will separate until electrostatic equilibrium conditions
exit. The use of a wire to connect the two spheres or to ground the outer sphere will cause additional redistribution of charge. (a) Because the outer sphere is conducting, the field in the thin shell must vanish. Therefore, -2Q, uniformly distributed, resides on the inner surface, and -5Q, uniformly distributed, resides on the outer surface. (b) Now there is no charge on the inner surface and -5Q on the outer surface of the spherical shell. The electric field just outside the surface of the inner sphere changes from a finite value to zero.
(c) In this case, the -5Q is drained off, leaving no charge on the outer surface and -2Q
on the inner surface. The total charge on the outer sphere is then -2Q. *46
··
Repeat Problem 44 for a sphere with volume charge density p = C// for r <
R; p = 0 for r > R.
Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r=R. By
symmetry, the electric fields must be radial. To find Er inside the charged sphere we
choose a spherical Gaussian surface of radius
r
<
R. To find Er outside the charged sphere
we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is
�
constant. Gau s ' s law then relates Er to the total charge inside the surface. \ \
(a ) Express the charge dq in a shell of thickness dr and volume 4nl dr:
Integrate this expression from r = 0 to R to find the total charge on
C dq 4m-2pdr 4m-- 2dr r 4nCdr =
=
=
Q
the sphere:
=
R
4nC fdr [4nCr] � =
o
(b) Apply Gauss's law to a spherical surface of radius r > R that is
concentric with the nonconducting sphere to obtain:
?
or
4m- 2 Er Qinside =
Eo
38
Chapter 22
E (r > R )
Solve for Er:
r
=
�
1
Qinside J 4 1C Eo r -
_ _
k4nCR r2
=
�I
kQinside ?
r-
CR Eo r
Apply Gauss' s law to a spherical
2
I
surface of radius r < R that is
concentric with the nonconducting
or
sphere to obtain:
4nr
2
E
r
=
Qnside Eo
Solve for Er:
��
The graph of Er versus rlR, with Er in units of C l Eo R , was plotted using a spreadsheet program.
10 8 6 Er 4 2 0
0.0
0.5
1 .0
1 .5
2.0
2.5
3.0
rlR
Cylindrical Symmetry * 52
••
Consider two infinitely long, concentric cylindrical shells. The inner shell
has a radius R1 and carries a uniform surface charge density of 0'] , and the outer shell has a radius R2 and carries a uniform surface charge density of 0'2' (a) Use Gauss's law to find the electric field in the regions r < R], R1
<
r < R2, and r > R2•
(b) What is the ratio of the surface charge densities 0'2/0'1 and their relative signs if the
The Electric Field 2 : Continuous Charge Distributions
39
electric field i s zero at r > R2 ? What would the electric field between the shells be in this
case? (c) Sketch the electric field lines for the situation in Part (b) if CJ, is positive.
Picture the Problem From symmetry; the field tangent to the surfaces of the shells must vanish. We can construct a Gaussian surface in the shape of a cylinder of radi us r and length L and apply Gauss' s law to find the electric field as a function of the distance from
the centerline of the infinitely long, uniformly charged cylindrical shells. (a) Apply Gauss's law to the cylindrical surface of radius r and length L that is concentric with the
or
cylindrical shell:
2trrLEn
infinitely long, uniformly charged
=
Qnside Eo
where we've neglected the end areas because no flux crosses them. Solve for En :
For r < R j , Qinside
En
=
2kQnside Lr
(1)
= 0 and:
Express Qinside for Rj
<
r < R2:
Substitute in equation ( 1 ) to obtain:
Express Qinside for r > R2:
Qnside
= =
Substitute in equation ( 1 ) to obtain:
En
(
r
O"j A, + 0"2 Az
21rO"jRj L + 2tr0"2 R 2 L
> R2 )
= 2k (21r0", R,L + 21r0"2 R2 L) Lr
0", R, + 0"2 R2 Eo r
(b) Set E
= 0 for r > R2 to obtain: or
40
Chapter 22
Solve for the ratio o f (JI to
(J2 :
Because the electric field is determined by the charge inside the Gaussian surface, the field under these conditions would be as given above: (c) Assuming that (J1 is positive, the field lines would be directed as shown to the right.
Charge and Field at Conductor Surfaces *58
·
A penny is in an external electric field of magnitude 1 . 6 kN/C directed perpendicular to its faces. (a) Find the charge density on each face of the penny, assuming the faces are planes. (b) If the radius of the penny is 1 cm, find the total charge on one face. Picture the Problem Because the penny is in an external electric field, it will have charges of opposite signs induced on its faces. The induced charge (J is related to the electric field by E = atco . Once we know 0; we can use the definition of surface charge density to find the total charge on one face of the penny.
(a) Relate the electric field to the
charge density on each face of the
penny: Solve for and evaluate
a:
(J' = E o = =
(b) Use the definition of surface charge density to obtain:
E
(8.85
x
)(
1 0-1 2 C2 /N . m2 1 . 6 kN/C
I 1 4.2 nC/m2 I
)
The Electric Field 2: Continuous Charge Distributions
Solve for and evaluate Q:
Q = (TW2
41
= ;;r(14.2 n Ch 2 )(0.01 mY n
= I 4.45 p C I
A positive point charge of magnitude 2.5 j1C is at the center of an uncharged spherical conducting shell of inner radius 60 cm and outer radius 90 cm. (a) Find the charge densities on the inner and outer surfaces of the shell and the total charge on each surface. (b) Find the electric field everywhere. (c) Repeat Part (a) and Part (b) with a net charge of +3.5 j1C placed on the shell.
*63
••
Picture the Problem Let the inner and outer radii of the uncharged spherical conducting
shell be a and b and q represent the positive point charge at the center of the shell. The positive point charge at the center will induce a negative charge on the inner surface of the shell and, because the shell is uncharged, an equal positive charge will be induced on its outer surface. To solve part (b), we can construct a Gaussian surface in the shape of a sphere of radius r with the same center as the shell and apply Gauss's law to find the electric field as a function of the distance from this point. In part (c) we can use a similar strategy with the additional charge placed on the shell. (a) Express the charge
the inner surface:
density on
Express the relationship between the positive point charge q and the charge induced on the inner surface Substitute for qinner to obtain:
(Y.mner
= qilmer A
q +qinner =
(Yinner = -q
4;;ra 2
Substitute numerical values and evaluate Oinner:
(Yinner =
Express the charge density on the outer surface:
(Y
Because the spherical shell is uncharged: Substitute for qouter to obtain:
0
outer =
-2.5;£)2 4n (0.6m
qouter
A
q outer
+ qinner = 0
(Y outer
= - qinner ntl
4
=
I - 0.553 ,uC/m 2 I
42 Chapter 22 Substitute numerical values
evaluate (JOUler:
and
(b) Apply Gauss's law to a spherical surface of radius r that is concentric with the point charge:
O'oulcr
2.5f-LC 1 = 41Z' (0.9m )2 = 0.246 pC/m 21
{ En dA _1 =
Eo
Qinside
or 4nr2En = Qinside Eo
Solve for En:
(1)
For r < a = 0.6 m, Qinside = q. Substitute in equation (1) and evaluate En(r < 0.6 m) to obtain:
(2.2 5x104 N.m2/C)� r-
Because the spherical shell is a conductor, a charge -q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:
Qinside = 0
and En (0.6m
For r > 0.9 m, the net charge inside the Gaussian surface is q and: EJr > 0.9m)= k; = (2.2 5x104 N.m2/C)� r r
(c) Because E = 0 in the conductor:
qinner = -2.5 J-LC and O'inner = 1 - 0.553 pC/m 2 1 as before.
Express the relationship between the charges on the inner and outer surfaces of the spherical shell:
qouter + qinner = 3.5 J-LC and qOllter = 3.5 J-LC - qinner = 6.0 J-LC
The Electric Field 2: Continuous Charge Distributions (JOllIer is now
given by:
For r < a = 0.6 m, Qinside = q and En(r < 0.6 m) is as it was in (a): Because the spherical shell is a conductor, a charge -q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:
(JOUler
En
JiC y t 4Jr 0.9m
(
r < a
Qinside
and En
=
=
43
I 0.5 89 }lC/m 2 I
) = ( 2.25 x 104 N· m 2 /C)� r
=0
(0.6m
< r <
0.9m ) = [I]
For r > 0.9 m, the net charge inside the Gaussian surface is 6 J.iC and: I
EJr
>
0.9m) = k; = (8 .99x109N .m 2 /C2 )(6JC i )� r
r
=
( 5.3 9x1 04 N .m 2 /C)� r
General Problems
*69 ·· A thin nonconducting uniformly charged spherical shell of radius r (Figure 22-41a) has a total charge of Q. A small circular plug is removed from the surface. (a) What is the magnitude and direction of the electric field at the center of the hole? (b) The plug is put back in the hole (Figure 22-41b). Using the result of Part (a), calculate the force acting on the plug. (c) From this, calculate the "electrostatic pressure" (force/unit area) tending to expand the sphere.
Plug
Figure 22-41a
Problem 69
Figure 22-41b
Problem 69
Picture the Problem If the patch is small enough, the field at the center of the patch comes from two contributions. We can view the field in the hole as the sum ofthe field from a uniform spherical shell of charge Q plus the field due to a small patch with surface charge density equal but opposite to that of the patch cut out.
(a) Express the magnitude of the electric field at the center of the hole:
E
= Espherical shell + Eho1e
44 Chapter 22 Apply Gauss's law to a spherical gaussian
surface just outside the
given sphere: Solve for Espherical shell to obtain:
The electric field due to the small hole (small enough so that we can treat it as a plane surface) is: Substitute and simplify to obtain:
Espherical
Assuming that the patch has radius a, express the proportion between its charge and that of the spherical shell:
nr 2
=
=
Eo
Q
Q EO
2
4 7r EO r
-(j
Ehole = 2- E
o
=
Express the force on the patch:
(4 ) = Qenclosed
EsPherical shell
=
(b)
shell
Q
Q
47r E o r2 - 2 E o (47r r2 )
ISff;' r' I
F=qE
where q is the charge on the patch. or
Substitute for q and E in the expression for F to obtain: The pressure is the force divided by the area of the patch: (c)
*73 ·· An infinitely long cylindrical shell is coaxial with the y axis and has a radius of 15 cm. It carries a uniform surface charge density a = 6 j.1Clm2 . A spherical shell of radius 25 cm is centered on the x axis at x = 50 cm and carries a uniform surface charge density a = -12 j.1Clm2 . Calculate the magnitude and direction of the electric field at (a) the origin; (b) x = 20 cm, y = 10 cm; and (c) x = 50 cm,Y = 20 cm. (See Problem 48.) Picture the Problem We can find the electric fields at the three points of interest by
adding the electric fields due to the infinitely long cylindrical shell and the spherical
The Electric Field 2 : Continuous Charge Distributions 45 it was established that, for an infinitely long cylindrical shell of radiusR, Er (r
R)= uRIEo r. We know that, for a spherical shell ofradiusR, Er(rR)= uR 2 / Eo r 2 . shell. In Problem 42
y,cm
(- ---<... ,
, I , I I
- -
(T =
'\
-_ .......
10
-
-:::
f..I-C/m2
(20,10) / ./
-......
15 20 25
o
" ......
, , I I I I
6
....... __
""
./
\
I I I I
Express the resultant electric field as the sum of the fields due to the cylinder and sphere:
\
-......
"
/'
----
(T
= -12 f..I-C/m2
-- ...... •
(50,20)
-......
r_---...... 50
...... _--_ .......
"
-
E
......
--- ./ -
-
= E y + ESPh e ]
"\
\ -......" \ ./ x,m /
/'
./
/
/
(1)
Eey] (0, 0)= 0
(a) Express and evaluate the electric
field due to the cylindrical shell at the origin:
because the origin is inside the cylindrical shell.
Express and evaluate the electric field due to the spherical shell at the origin: E sph (0' 0) =
2 (- i)= UE R r2 o
- 12 ,uClm2 (0.25mY (- i)=(339kN/c ) i 8.85 1 0-]2C2/N .m2 (0.5mY
Substitute in equation (1) to obtain:
x
E(O,O)= 0 +(339kN/C) i = I (339kN/C) i I or E(O,O) = I 339kN/C I and e= �
46 Ch pt r 22 a
e
(b) Express and evaluate the electric field due to the cylindrical shell at (0.2 m,0.1 m):
-
o..,....;. - (0.2 m, 0. 1m ) = EoR r i = ,...-_-'------,-----,-L._...O..- ,..._
E cyl
___.
�
- (r) = -� o-R 2 E r2
Express the electric field due to the charge on the spherical shell as a function of the distance from its center:
Esph
o
�
r
where r is a unit vector pointing from (50 cm, 0) to (20 cm, 10 cm).
r = 0.316m
Referring to the diagram shown above, find rand r:
and r
= -0.949i+0.316 }
Substitute to obtain:
- 12)1Clm2 (0.2Sm Y ( 0.949i+0.316]':) 8.8Sx10-l2C2/N'm2 (0.316m Y = (- 849kN/C )(- 0.949i+0.316 }) = (806 kN/c )i+(- 268kN/C) }
Esph (O.2m, O.lm ) =
Substitute in equation (1) to obtain:
or and
E(0.2m,0.lm ) = (S08kN/C) i+(806kN/C) i+(- 268kN/C) } = 1 (1310kN/C) i+(-268kN/C) ) 1 E(0.2m,0.lm) = �(1310kN/CY +(- 268kN/CY = 1 13 40kN/C 1
(
e = tan-l - 268 kN/C = 1 3480 I 13 10 kN/C
J
(c) Express and evaluate the electric field due to the cylindrical shell at (0.5 m, 0.2 m):
The Electric Field 2 : C ont i nu ous Charge Distributions
Express and evaluate the electric field due to the spherical shell at (0.5 m,0.5 m):
ESPh (0.5m,0.2m) = ° because (0.5 m, 0.2 m) is inside the spherical shell.
Substitute in equation (1) to obtain:
E(0 . 5 m 0 .2 m ) = (203 kN/C) i + ° = I (203 kN/C ) i I or E (0.5 m,0 . 2 m ) = ! 203 kN/C ! and
47
,
B=
@:J
*76 ·· Using the results of Problem 75, if we placed a proton above the nucleus of a hydrogen atom, at what distance r would the electric force on the proton balance the gravitational force mg acting on it? From this result,explain why even though the electrostatic force is enonnously stronger than the gravitational force, it is the gravitational force we notice more. Picture the Problem We will assume that the radius at which they balance is large
enough that only the third tenn in the expression matters. Apply a condition for equilibrium will yield an equation that we can solve for the distance r.
--k -2 (-k 2 e2
Apply IF = ° to the proton:
a2
rla
e
-mg = °
To solve for r, isolate the a 2 e2 n I r exponential factor and take the 2 mga 2 natural logarithm of both sides of the equation: Substitute numerical values and evaluate r: r=
J ]
0.0529 nm 2 8 . 99 x lO N ·m - I C2 1 . 60 x lO C 2 = 1 .16 nm ln 2 1 . 6 7 x lO 2 kg 9.81 rnls 2 0 .0 529nm
[
-
9 7
?
(
-19
Y !
I
Thus, ev en though th e unscr een ed electrostatic forc eis 40 ord ers of magn it ud e larg er than th e g rav itationa lforc e, scr eening r educ es it to sma llerthan th e gravitat iona lforc e within a few nanom et ers. Remarks: Note that the argument of the logarithm contains the ratio between the gravitational potential energy of a mass held a distance ao above the surface of the
48 Chapter 22 earth and the electrostatic potential energy for two unscreened charges a distance
no
apart. ••
A ring of radius R that lies in the yz plane carries a positive charge Q uniformly distributed over its length. A particle of mass that carries a negative charge of magnitude q is at the center of the ring. (a) Show that if x « R, the electric field along the axis of the ring is proportional to x. (b) Find the force on the particle of mass m as a function ofx. (c) Show that if m is given a small displacement in the x direction,it will perform simple harmonic motion. Calculate the period ofthat motion. *84
m
Picture the Problem Starting with the equation for the electric field on the axis of ring
charge,we can factor the denominator ofthe expression to show that,for x « R, Ex is proportional to x. We can use Fx = qEx to express the force acting on the particle and apply Newton's 2nd law to show that,for small displacements from equilibrium,the particle will execute simple harmonic motion. Finally,we can find the period ofthe motion from its angular frequency, which we can obtain from the differential equation of motion. (a) Express the electric field on the axis of the ring of charge: Factor R2 from the denominator of Ex to obtain:
Express the force acting on the particle as a function of its charge and the electric field: (b)
(c) Because the negatively charged particle experiences a linear restoring force, we know that its motion will be simple harmonic. Apply Newton's 2nd law to the negatively charged particle to obtain:
or the differential equation of simple harmonic motion.
Th e Electric Fi eld 2: Contin uo us Charg e Distri butions
Relate the period T of the simple harmonic motion to its angular frequency w:
T
49
= 2J[ OJ
From the differential equation we have: Substitute to obtain:
T�
2tr�mR' kq Q
A dipole p is located at a distance r from an infinitely long line charge with a uniform linear charge densityX Assume that the dipole is aligned with the field due to the line charge. Determine the force that acts on the dipole. *94
•••
Picture the Problem We can find the field due to the infinitely long line charge from
E = 2kA/ r and the force that acts on the dipole using F pdE/d r . E Express the force acting on the F = pd dr dipole: =
The electric field at the location of the dipole is given by: Substitute to obtain:
E = 2kA r
[
F = p� 2kA = dr r
1_ 2kr�p 1
J where the minus sign indicates that the dipole is attracted to the line charge.
Chapter 23 Electric Potential Conceptual Problems •
A positive charge is released from rest in an electric field. Will it move toward a region of greater or smaller electric potential?
*1
Determine the Concept A positive charge will move in whatever direction reduces its
potential energy. The positive charge will reduce its potential energy if it moves toward a region of lower electric potential. ••
Figure 23-26 shows a metal sphere carrying a charge -Q and a point charge +Q. Sketch the electric field lines and equipotential surfaces in the vicinity of this charge system. *7
-Q +Q
Figure 23-26 Problems 7 and 8 Picture the Problem The electric field
lines, shown as solid lines, and the equipotential surfaces (intersecting the plane of the paper), shown as dashed lines, are sketched in the adjacent figure. The point charge +Q is the point at the right, and the metal sphere with charge -Q is at the left. Near the two charges the equipotential surfaces are spheres, and the field lines are normal to the metal sphere at the sphere's surface. Two equal positive point charges +Q are on the x axis. One is at x = -a and the other is at x =+a. At the origin, (a) E = ° and V= 0, (b) E = ° and V = 2kQla, (c) E (2kQ /a )i and V = 0, (d) E = (2kQ / a ) i and V = 2kQla, or ( e) none of the above is correct. *11 ·
-
=
2
2
A
-
2
2
A
Picture the Problem We can use Coulomb's law and the superposition of fields to find E
at the origin and the defmition of the electric potential due to a point charge to find V at 51
52 Chapter 23 the origin. Apply Coulomb's law and the superposition of fields to find the electric field E at the origin: Express the potential Vat the origin:
-
E
-
-
= E+Qal-a + E +Qala
v= =
V+Qal-a
kQ
a
+
+ V +Qala
kQ
a
=
2kQ
a
and I (b) is co rrect. I Two charged metal spheres are connected by a wire, and sphere A is larger than sphere B (Figure 23-27). The magnitude of the electric potential of sphere A is (a) greater than that at the surface of sphere B; (b) less than that at the surface of sphere B; (c) the same as that at the surface of sphere B; (d) greater than or less than that at the surface of sphere B, depending on the radii of the spheres, or (e) greater than or less than that at the surface of sphere B, depending on the charge on the spheres.
*16
·
Figure 23-27
Problem 16
When the two spheres are connected, their charges will redistribute until the two-sphere system is in electrostatic equilibrium. Consequently, the entire system must be an equipotential. I (c) is co rrent. I Determine the Concept
Estimation and Approximation Problems *18 · Estimate the potential difference across the spark gap in a typical automobile spark plug. Because of the high compression of the gas in the piston, the electric field at which the gas sparks is roughly 2xl0 7 Vim.
The potential difference between the electrodes of the spark plug is the product of the electric field in the gap and the separation of the electrodes. We'll assume that the separation of the electrodes is 1 mm. Picture the Problem
Express the potential difference between the electrodes of the spark
V = Ed
Electric Potential of their separation d and electric field E between them:
plug as a function
V:
Substitute numerical values and evaluate
v
53
(2 107 V /m )(1O-3 m ) = i 20.0kV i =
X
Potential Difference
electron gun fires electrons at the screen of a television tube. The electrons start from rest and are accelerated through a potential difference of 30,000 V. What is the energy of the electrons when they hit the screen (a) in electron volts and (b) in joules? (c) What is the speed of impact of electrons with the screen of the picture tube? *26
··
An
The work done on the electrons by the electric field changes their kinetic energy. Hence we can use the work-kinetic energy theorem to find the kinetic energy and the speed of impact of the electrons.
Picture the Problem
Use the work-kinetic energy theorem to relate the work done by the electric field to the change in the kinetic energy of the electrons: (a) Substitute numerical values and evaluate Kr: (b) Convert this energy to eV:
or
Kr
= ef... V
K,
= (3 x10' eV 1.6
(1)
l(
�
Xe
"
-
J
= 1 4 .80xl0-J5 J I (c) From equation (1) we have:
tmv;
= ef... V
Solve for Vr to obtain: Substitute numerical values and evaluate V(
vf
= 2 1.6xl0-J9OC3 (30k V ) 9.11xl - Jkg = 1 1.03xl08m 1s I
Remarks: Note that this speed is about one-third that of light.
J
54
Chapter 23
Potential Due to a System of Point Charges
Two point charges q and q' are separated by a distance a. At a point a/3 from q and along the line joining the two charges the potential is zero. Find the ratio qlq'. *32
•
We can use the fact that the electric potential at the point of interest is the algebraic sum of the potentials at that point due to the charges q and q' to find the ratio qlq'. Picture the Problem
Express the potential at the point of interest as the sum of the potentials due to the two charges: Simplify to obtain:
kq + kq' = 0 a/3 2a/3 , q+!L=O 2
Solve for the ratio qlq': A point charge of +3e is at the origin and a second point charge of 2e is on the x axis at x = a. (a) Sketch the potential function Vex) versus x for all x. (b) At what point or points is Vex) zero? (c) How much work is needed to bring a third charge +e to the point x= t a on the x axis? *34
··
-
For the two charges, r = Ix-al and Ixl respectively and the electric potential at x is the algebraic sum of the potentials at that point due to the charges at x = a and x O. We can use the graph and the function found in part (a) to identify the points at which Vex) = o. We can find the work needed to bring a third charge +e to the point x= t a on the x axis from the change in the potential energy of this third charge. Picture the Problem
=
Express the potential at x:
V(x) = k(3e) + k(-2e) Ixl Ix-al
The following graph of Vex) for ke = 1 and a = 1 was plotted using a spreadsheet program.
Electric Potential
55
25 20 15
�
:::...
10 5
0
-5
-10
-15
-3
-2
-1
x
0
(m)
2
V(x) 0 when:
x=l±ool
Examining the function, we see that V(x) is also zero provided:
3 ---=0 2 Ixl Ix-al
For x> 0, V(x) = 0 when:
x=�
For 0 <x < a, V(x) = 0 when:
x=1 0.6a I
(c) Express the work that must be
W
Evaluate the potential at x = 1 a :
V ( 2 a) = k (3ea) + k (-2e) 11 l 11 a-al 4ke 2ke 6ke --=a a a
(b)
From the graph we can see that =
done in terms of the change in potential energy of the charge:
3
=!1U =qV(1a)
1.
Substitute to obtain: Computing the Electric Field from the Potential
The potential due to a particular charge distribution is measured at several points along the x axis, as shown in Figure 23-28. For what value(s) in the range 0 <x < 10 m is Ex = O? *36
·
56 Chapter 23 VCr), volts
•
• •
Figure 23-28
•
• • •5 •
•
•
•
•
•
•
• •
•
10
x, m
Problem 36
Because Ex = -dVldx, we can find the point(s) at which 0 by identifying the values for x for which dVldx = O.
Picture the Problem
Ex
=
Examination of the graph indicates that dVldx = 0 at x = 4.5 m. Thus Ex Oat:
=
x= ! 4.5m !
*42 An electric field is given by Ex = 2.0X 3 kN/C. Find the potential difference between the points on the x axis at x = 1 m and x = 2 m.
Because Vex) and Ex are related through Ex = - dVldx, we can find V from E by integration.
Picture the Problem
Separate variables to obtain: Integrate V from VI to V2 and x from 1 m to 2 m:
Simplify to obtain:
V2
X2
fd V = -(2.0kN/C) fx3 dx
V2 -� = ! - 7.50kV !
Calculations of V for Continuous Charge Distributions
A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with its center at the origin. (a) Find the potential as a function of position along the x-axis. (b) Show that the result obtained in Part (a) reduces to V = kQlx for x» L. *47
• •
Electric Potential z
Let the charge per unit length be A =QIL and dy be a line element with charge Ady We can express the potential dVat any point on the x axis due to My and integrate of find Vex, 0). Picture the Problem
.
y
kJe y dV=-d
(a) Express the element of potential dV due to the line element dy:
r
where r = �X2 + y2
LIf2 dy k Q = V(x,o) L -L12 �X2 + y2
Integrate dV from y = -Ll2 to y =Ll2:
R; 2 1 + - + L 2 x V(x,O) = kQ In --,-==4X= =-_2_ L + L' .!:... 4x 2 2x
(b) Factor x from the numerator and denominator within the parentheses to obtain:
�I
. Use In-a = In a -Inbto obtam:
_
b
�J - ( r;:rz �2xJ}
+ + V(x,o) = kQ In r;:rz Vl � 2x L
{(
Let &
-
_
In
+ Vl �
...
2 .. 2 R;L L 1 2 2 --2 and use (1 + & ) / - 1 + "2 & - "8 & + to expand + --2 4x 4x I
_
I
--
1 L2 -1 ( L2 J2 L2 JI/2 =1 + --(1 + -+ ... � 1 forx» L. 4X2 2 4x2 - 8 4x2
Substitute to obtain:
i HI + �) - ln( l- :J}
V(x,o)� k
57
58
Chapter 23
( )
Let 5= � and use In(1+5)=5-t52 + . . to expand In l± � : .
2x
In( 1+-) L
2x
L
�-2x
2x
( )
L2
2
L L L and In 1-�-----2 forx» L. 2 4x 2x 2x 4x
--
Substitute and simplify to obtain:
V(x,o)= kQ{� _�_(_�_ J} L
2x
4x
2x
L2 4x 2
= 1kQl
W
A disk of radius R carries a charge density +0"0 for r < a and an equal but opposite charge density -0"0 for a < r < R. The total charge carried by the disk is zero. (a) Find the potential a distance x along the axis of the disk. (b) Obtain an approximate expression for Vex) when x» R. *51
••
The potential at any location on the axis of the disk is the sum of the potentials due to the positive and negative charge distributions on the disk. Knowing that the total charge on the disk is zero and the charge densities are equal in magnitude will allow us to find the radius of the region that is positively charged. We can then use the expression derived in the text to find the potential due to this charge closest to the axis and integrate dV from r = R/ .fi to r = R to find the potential at x due to the negative charge distribution. Picture the Problem
x
(a) Express the potential at a distance x along the axis of the disk as the sum of the potentials due to the positively and negatively charged regions of the disk: We know that the charge densities are equal in magnitude and that the total charge carried by the disk is zero. Express this condition in terms
or
Electr ic Potent ial
of the charge in each of two regions of the disk: Solve for a to obtain: Use this result and the general expression for the potential on the axis of a charged disk to express V +(x): Express the potential on the axis of the disk due to a ring of charge a distance a from the axis of the nng:
r>
r=
r
Integrate this expression from R/12 to = R to obtain:
59
-
a = 12R
= r' r'= r2 . x( ) = -21lk(Jo f 2r 2 dr -Jx +r ' �� 2nka{ N +R' � �x' +� J
x( ) -21lk(Jo !...- dr where -Jx 2 + dV_
V_
Substitute and simplify to obtain:
R
R/.fi
{� �' + 2nka{�x, +R' � �x' +�' J � 2nk
(b)
x
expansion:
and
60
Chapter 23
Substitute to obtain:
Show that for x > > R the potential on the axis of a disk charge approaches kQlx, where Q = o'1rR2 is the total charge on the disk. [Hint: Write (x2 + R2)1/2 = x(l + R21x2)112 and use the binomial expansion.]
*58
··
The potential on the axis of a disk charge of radius R and charge density 0" is given by V = 2nko- l(x2 + R2 r - x1-
Picture the Problem
Express the potential on the axis of the disk charge: Factor x from the radical and use the binomial expansion to obtain:
Substitute for the radical term to obtain:
provided x > > R. Find the greatest surface charge density O"max that can exist on a conductor before dielectric breakdown of the air occurs. *63
·
Electric Potent ial
61
We can solve the equation giving the electric field at the surface of a conductor for the greatest surface charge density that can exist before dielectric breakdown of the air occurs. Picture the Problem
Relate the electric field at the surface of a conductor to the surface charge density: Solve for CTunder dielectric breakdown of the air conditions:
(J"max
Substitute numerical values and evaluate CTmax:
(J"max
=Eo Ebreaddown = (8.85 10-12C2/N . m2)(3 MV /m ) = 1 26.6 j1C/m2 1 X
*66 Calculate the potential relative to infinity at the center of a uniformly charged sphere of radius R and charge Q. Picture the Problem We can find the potential relative to infinity at the center ofthe sphere by integrating the electric field for 0 to We can apply Gauss's law to find the electric field both inside and outside the spherical shell. •••
00.
The potential relative to infinity the center of the spherical shell is: Apply Gauss's law to a spherical surface of radius r < R to obtain:
'" R (1) JErRd r o R d = Er
V =
+
Using the fact that the sphere is uniformly charged, express Qinside in terms of Q: Substitute for Qinside to obtain: Solve forEr R: <
Apply Gauss's law to a spherical surface of radius r > R to obtain: Solve for Er>R to obtain:
Er>R = 4
Q
:r
=
kQ
Eo 2 r2 r
62
Chapter 23
Substitute for E,.R in equation (1) and evaluate the resulting integral:
V
R fr dr R C
kQ )
= -
0
kRq [ 20]
=
R
+
kQ
+
f-drr 2 R
'"
[ �J'" r
f3kQl R llRJ
kQ -
=
General Problems ·· Two infinitely long parallel wires carry a unifonn charge per unit length A and -A respectively. The wires are in the xz plane, parallel with the z axis. The positively charged wire intersects the x axis at x = - a, and the negatively charged wire intersects the x axis at x +a. (a) Choose the origin as the reference point where the potential is zero, and express the potential at an arbitrary point (x y) in the xy plane in tenns ofx, y, A, and a. Use this expression to solve for the potential everywhere on the y axis. (b) Use a spreadsheet program to plot the equipotential curve in the xy plane that passes through the point x = i a, y = O . Use a 5 cm and A = 5 nC/m. *71
=
,
=
The geometry of the wires is shown to the right. The potential at the point whose coordinates are (x, y) is the sum of the potentials due to the charge distributions on the wires. Picture the Problem
(a) Express the potential at the point whose coordinates are (x, y):
(x,y) A
-a
{ y)
V x,
a
r2 �{x a Y y2 : =
-
+
On the y-axis, x = 0 and:
+ y2 and
A
-x
-
+ �2kA{;" )+2k(-,l){;") 2kA[{;" ) {;" )] - 2:E, m(;,) =
Vwire at-a
�
Because lj = �{x + a ) 2
y
where V(O)
Vwireata
_
=
o.
Electric Potential
63
(b) Evaluate the potential at
(�a,a) (1 . 2 5 a) : =
em,
) (
3 - -- In 2" Eo 5
Equate V(x,y) and V(� a,a):
A.
Solve for y to obtain:
y=±.J2 1.25x-x2 -25
A spreadsheet program to plot y=±.J21.25x-x2 -25 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell ContentIFormula Algebraic Form A2 1 .25 .La A3 A2 + 0.05 x+& B2 SQRT(21 .25*A2 - A2"2 - 25) y .J21 .25x -x2 -25 B4 -SQRT(21 .25*A2 - A2"2 - 25) y=-.J2 1 .25x-x2 -25 4
=
1·;i4 1 . Pb2 ,3 4 ,:�: 5,.' ;;?6 7 370 1"371 1;'372 '):3'73 374 3 �5 1,376
1'-"
1 .25 1 .30 1 .35 1 .40 1 .45 1 .50
B y--pos 0.00 0.97 1.37 1 .67 1 .93 2.15
C y neg 0.00 -0.97 -1 .37 -1 .67 -1 .93 -2.15
1 9.65 1 9.70 19.75 1 9.80 1 9.85 1 9.90 1 9.95
2.54 2.35 2.15 1.93 1.67 1 .37 0.97
-2.54 -2.35 -2.15 -1.93 -1 .67 -1 .37 -0.97
A x
'"
The following graph shows the equipotential curve in the xy plane for
V(� a,a) = _ A._In(�)5 . 2" Eo
64 Chapter 23 10
8
6
E
�
""
4
2
0
-2
-4 -6 -8
-10
0
5
x
10
(em)
15
20
A Van de Graaff generator has a potential difference of 1 .25 MY between the belt and the outer shell. Charge is supplied at the rate of 200 flCls. What minimum power is needed to drive the moving belt? *76
••
We can use the definition of power and the expression for the work done in moving a charge through a potential difference to find the minimum power needed to drive the moving belt.
Picture the Problem
Relate the power need to drive the moving belt to the rate at which the generator is doing work:
p= dW dt
Express the work done in moving a charge q through a potential difference L1 V:
W =q/lV
Substitute to obtain:
P=�[q/lV]= /lV dq dt dt
Substitute numerical values and evaluate P:
P= (1 .25 MV )(200j£ls) = !250W!
A metal sphere centered at the origin carries a surface charge of charge density (J'= 24.6 nC/m2• At r = 2.0 m, the potential is 500 V and the magnitude ofthe electric field is 250 V1m. Determine the radius of the metal sphere. *82 ··
We can use the definition of surface charge density to relate the radius of the sphere to its charge Q and the potential function V(r) = kQ/r to relate Q Picture the Problem R
Electric Potential 65 to the potential at 2 m. Use its definition, relate the surface charge density erto the charge Q on the sphere and the radius R of the sphere: r =
Solve for R to obtain:
R - [(2 ��
Relate the potential at r = 2.0 m to the charge on the sphere:
V(r) r rV(r) =
kQ
Q k
Solve for Q to obtain:
=
Substitute to obtain:
Substitute numerical values and evaluate R:
x O-1 R = 8 . 8 5 l 2 C 2/N ·m 2 ?(2m )(500V ) = 1 O .600m 24.6nC/m -
•••
.
1
.
A point charge q is a distance d away from a grounded conducting plane of infinite extent (Figure 23-3 1a). For this configuration the potential V is zero, both at all points infinitely far from the particle in all directions, and at all points on the conducting plane. Consider a set of coordinate axes with the particle located on the x axis at x = d. A second configuration (Figure 23-31 b) has the conducting plane replaced by a particle of charge -q located on the x axis at x = -d. (a) Show that for the second configuration the potential function is zero at all points infinitely far from the particle in all directions, and at all points on the yz plane-just as was the case for the first configuration. (b) A theorem, called the uniqueness theorem, shows that throughout the half-space x> 0 the potential function V-and thus the electric field E -for the two configurations are identical. Using this result, obtain the electric field E at every point in the yz plane in the second configuration. (The uniqueness theorem tells us that in the first configuration the electric field at each point in the yz plane is the same as it is in the second configuration.) Use this result to find the surface charge den$ity erat each point in the conducting plane (in the first configuration). *87
66
Chapter 23 (1. "', ",
+
+
to
,.
+�
V=60V
p
(b)
(a)
Problem 87 Picture the Problem We can consider the relationship between the potential and the electric field to show that this arrangement is equivalent to replacing the plane by a point charge of magnitude -q located a distance dbeneath the plane. In (b) we can first find the field at the plane surface and then use (J = Eo E to find the surface charge density. In (c) the work needed to move the charge to a point 2d away from the plane is the product of the potential difference between the points at distances 2d and 3d from -q multiplied by the separation Llx ofthese points. Figure 23-31
(a)
Th epot entia lan ywh er eon th ep lan eis 0 in eith er arrang em ent and th e el ectric field is p erp endicu lar to th ep lan ein both arrang em ents, so th ey must giv e th esam epot entia l ev erywh er ein th exy p lan e. Also, becaus e th en et charg eis zero, th epot entia lat in finit y is zero.
The surface charge density is given by: (b)
At any point on the plane, the electric field points in the negative x direction and has magnitude:
(1)
E
=
d
2
kq +r
2
cos e
where e is the angle between the horizontal and a vector pointing from the positive charge to the point of interest on the xz plane and r is the distance along the plane from the origin (i.e., directly to the left of the charge).
Electric Potential
Because cos e = .J d
d2 + r2
67
E= 2 kq 2 d d +r .Jd2 + r2 kqd
4
7r Eo
(d2 + r2 )3/2
Substitute for E in equation (1) to obtain: Consider two concentric spherical metal shells of radii Q and b, where b > Q. The outer shell has a charge Q, but the inner shell is grounded. This means that the inner shell is at zero potential and that electric field lines leave the outer shell and go to infinity, but other electric field lines leave the outer shell and end on the inner shell. Find the charge on the inner shell. *90
•••
\ Q \
The diagram shows a cross-sectional view of a portion of the concentric spherical shells. Let the charge on the inner shell be q. The dashed line represents a spherical Gaussian surface over which we can integrate E . ndA in order to find Er for r� b. We can find V(b) from the integral of Er between r= co and r= b. We can obtain a second expression for V(b) by considering the potential difference between Q and b and solving the two equations simultaneously for the charge q on the inner shell. Picture the Problem
\ , / /
Apply Gauss's law to a spherical surface of radius r� b: Solve for Er to obtain: Use Er to find V(b):
Er
= k (Q +2 q ) r
f� V(b) = -k(Q + q )bd ",r
68
Chapter 2 3
We can also determine V(b) by considering the potential difference between a, i.e., 0 and b:
( �)
Equate these expressions for V(b) to obtain:
k(Q +q ) = a � k b a b
Solve for q to obtain:
q= - Q �
_
I I
*94 ••• Problem 93 can be modified to be used as a very simple model for nuclear fission. When a 235U nucleus absorbs a neutron, it can fission into140the fragments 14094Xe 94 and Sr, plus 2 neutrons ejected. The 235U has 92 protons, while Xe has 54 and Sr has 38. Estimate the energy liberated by this fission process (in MeV), assuming that the mass density of the nucleus is constant and has a value p� 4xl017 kg/m3. Picture the Problem We can use the definition of density to express the radius R of a nucleus as a function of its atomic mass N. We can then use the result derived in Problem 91 to express the electrostatic energies of the 235U nucleus and the nuclei of the 140 94 fission fragments Xe and Sr.
The energy released by this fission process IS: Express the mass of a nucleus in terms of its density and volume:
(1)
Nm=}ptrR3
where N is the nuclear number.
�
Solve for R to obtain:
R = 3Nm 47rp
Substitute numerical values and evaluate R as a function of N:
1 R = 3 3 1 .66 0 x 0 -2 7 kg NI/3 47r 4 x l 017 kg/m 3 = � .97 x 1 0-16 m )N1/3
The 'radius' of the 235U nucleus is therefore:
Ru = (9.97 x lO-IGm )(2 3 5Y/3 =6. 1 5 x 1 0-15 m
From Problem 91 we have:
3Q2U = --=2 07r Eo R
Substitute numerical values and evaluate the electrostatic energy of the 2 35U nucleus:
Electric Potential
69
3 (nx1.6 x lO -'9C 2 20 Jr8 .8 5x1 0 -'2C2/N'm2 6. 1 5x lO -') m leV 1 1 89 MeV = 1 .9 I x I 0 -10 J x 1 .6 x 1 0-1 9 J/eV _
Um
=
'U
=
Proceed as above to find the electrostatic energy of the fission fragments 140Xe and 94Sr: U 140X
e
and U94
Sr
3 54 x 1 .6 x 1 0 - 1 9 C 2 - 20Jr 8.8 5x 1 0 -1 2 C 2 IN·m 2 leV =6. 57 x 1 0 -1 1 J x = 41 0 MeV 1 .6 x l0 - 1 9 J/eV
_
3 38 x 1 .6 x 1 0 -19 C 2 20 Jr8.8 5x 1 0 - 1 2 C 2 IN. m 2 leV = 3 .2 5x 1 0 - 1 1 J x = 203 MeV 1 .602 x 1 0 - 1 9 J/eV =
Substitute for U u' U 140 ,and U94 in equation (1) and evaluate sr 2J5
M:
Xe
M
= 1 1 89 MeV - (4 1 0 MeV + 203 MeV ) = 576 MeV
1
I
Chapter 24 Electrostatic Energy and Capacitance Conceptual Problems
If the voltage across a parallel-plate capacitor is doubled, its capacitance (a) doubles. (b) drops by half. (c) remains the same. *1
•
The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the voltage across the capacitor. I (c) is co rrect. I Determine the Concept
A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor (a) quadruples. (b) doubles. (c) remains unchanged. (d) drops to half its previous value. (e) drops to one-fourth its previous value. *5
••
The energy stored in a capacitor is given by U = tQV and the capacitance of a parallel-plate capacitor by C Eo Ald. We can combine these relationships, using the definition of capacitance and the condition that the potential difference across the capacitor is constant, to express U as a function of d. Picture the Problem
=
Express the energy'stored in the capacitor:
U
Use the definition of capacitance to express the charge of the capacitor:
Q= CV
-l.QV 2 -
Substitute to obtain:
C= A d where A is the area of one plate.
Express the capacitance of a parallel-plate capacitor in terms of the separation d of its plates:
Eo
Substitute to obtain: Because U rx ..!.. , doubling the d separation of the plates will reduce 71
72
Ch apter 24
the energy stored in the capacitor to 112 its previous value:
I (d) is correct. I
*10 ·· Two capacitors half-filled with a dielectric are shown in Figure 24-28. The area and separation of each capacitor is the same. Which has the higher capacitance, that shown in Figure (a) or in Figure (b)?
Figure 24-28
Problem 10
Picture the Problem We can treat the configuration in (a) as two capacitors in parallel and the configuration in (b) as two capacitors in series. Finding the equivalent capacitance of each configuration and examining their ratio will allow us to decide whether (a) or (b) has the greater capacitance. In both cases, we'll let CI be the capacitance of the dielectric-filled capacitor and C2 be the capacitance of the air capacitor.
In configuration (a) we have: Express CI and C2:
c
I
=
K Eo AI
�
=
K Eo
d
tA
=
K Eo A
and C2 - Eo Az - Eo t A - Eo A T, -d- U -
_
_
Substitute for CI and C2 and simplify to obtain: In configuration (b) we have: Express CI and C2:
c
I
= Eo AI
and
d
I
=
Eo A
l.2d
=
2 Eo A d
2d
Electrostatic Energy and Capacitance c =
2
73
A = 2K Eo A d td
K Eo � = K Eo
d2
Substitute for C) and C2 and simplify to obtain:
=
Because (
4K
K+l
)2
<
1
(
2 Eo
A �
d
K+l
)
for K > l :
Estimation and Approximation
*14 ·· To create the high-energy densities needed to operate a pulsed nitrogen laser, the discharge from a high-capacitance capacitor is used. Typically, the energy requirement per pulse (i.e., per discharge) is 100 J. Estimate the capacitance required if the discharge is applied through a spark gap of 1 mm width. Assume that the dielectric breakdown of nitrogen occurs at E 3 x l 06 V1m. �
Picture the Problem
The energy stored in a capacitor is given by U = t CV2 .
Relate the energy stored in a capacitor to its capacitance and the potential difference across it:
U
= lCV 2 2
Solve for C: The potential difference across the spark gap is related to the width of the gap d and the electric field E in
V = Ed
74
Chapte 24 r
the gap: Substitute for V in the expression for C to obtain: Substitute numerical values and evaluate C:
c=
2(100 J ) (3 106 V/m) (0.001m)2 = 1 22.2# 1 x
Electrostatic Potential Energy
What is the electrostatic potential energy of an isolated spherical conductor of 1 0 cm radius that is charged to 2 kV?
*19 ·
V = kQ/r = QV
The potential of an isolated spherical conductor is given by ,where Q is its charge and r its radius, and its electrostatic potential energy by U 1- . We can combine these relationships to find the sphere's electrostatic potential energy. Picture the Problem
Express the electrostatic potential energy of the isolated spherical conductor as a function of its charge Q and potential V: Express the potential of the spherical conductor: Solve for Q to obtain: Substitute to obtain: Substitute numerical values and evaluate U:
u = t QV
V = kQ Q = rVk u = .l(r v )v = rV2 2 k 2k r
u - (0.1m)(2kVY 2(8.99 109 N · m2/C2) = 1 22.2,uJ 1 _
x
Electrostatic Energy and Capacitance
75
Capacitance
An isolated spherical conductor of 10 cm radius is charged to 2 kY. (a) How much charge is on the conductor? (b) What is the capacitance of the sphere? (c) How does the capacitance change if the sphere is charged to 6 kV? *22
·
The charge on the spherical conductor is related to its radius and potential according to V = kQlr and we can use the definition of capacitance to find the capacitance ofthe sphere. Picture the Problem
(a) Relate the potential V ofthe spherical conductor to the charge on it and to its radius:
V = krQ
Solve for and evaluate Q:
rV Q= k = (O . 1 m)(2 kV2)
8.99 x 1 09 N · m /C 2
(b) Use the definition of capacitance to relate the capacitance of the sphere to its charge and potential: (c)
1
= 22.2 nC
1
c = Q = 22 .2 nC = 1 1 1 . 1 PF 1 V 2 kV
1 Itdo esn't. Th ecapacitanc eof a sph er eis a function of its radius. 1
The Storage of Electrical Energy
Find the energy per unit volume in an electric field that is equal to 3 MV1m, which is the dielectric strength of air. *28
·
The energy per unit volume in an electric field varies with the square of the electric field according tou = Eo E2 /2 .
Picture the Problem
Express the energy per unit volume in an electric field: Substitute numerical values and evaluate u :
U
= -'-2 E E 2
U
= 1(8.85 X 1 0- 1 2 C 2 /N . m 2 )(3 MV/mY = 39 . 8 J/m3
0
!
!
76
Chapter 24
A parallel-plate capacitor with plates of area 500 cm2 is charged to a potential difference Vand is then disconnected from the voltage source. When the plates are moved 0.4 cm farther apart, the voltage between the plates increases by 100 V. (a) What is the charge Q on the positive plate of the capacitor? (b) How much does the energy stored in the capacitor increase due to the movement of the plates? *3 1
We can relate the charge Q on the positive plate of the capacitor to the charge density of the plate O"using its definition. The charge density, in tum, is related to the electric field between the plates according to (J =Eo E and the electric field can be found from E = i1 Vli1d. We can use !::,. U = t Q!::" V in part (b) to find the increase in the energy stored due to the movement of the plates. Picture the Problem
(a) Express the charge Q on the positive plate of the capacitor in terms of the plate's charge density 0" and surface area A : Relate O"to the electric field E between the plates of the capacitor: Express E in terms of the change in V as the plates are separated a distance i1d: Substitute for 0" and E to obtain:
E=
!::,. V !::"d
!::,. V Q =Eo EA = Eo A !::"d
Substitute numerical values and evaluate Q:
(b) Express the change in the electrostatic energy in terms of the change in the potential difference:
!::,. U - .l2 Q!::,. V
Substitute numerical values and evaluate i1 U:
!::,. U = t (1 1 . 1 nC)(1 00 V) = 0 . 553
I
fll l
El ectrostatic En erg y and Capacitanc e 77 Combinations of Capacitors
*35
Three capacitors are connected in a triangle as shown in Figure 24-29. Find
•
Figure
24-29 Problem 3 5
c
Because we're interested in the equivalent capacitance across terminals a and c, we need to recognize that capacitors C1 and C3 are in series with each other and in parallel with capacitor C2 . Picture the Problem
Find the equivalent capacitance of C1 and C3 in series: Solve for C1 +3:
C1+3 C1 C3 C1 + C3 =
Find the equivalent capacitance of C1+3 and C2 in parallel: Three identical capacitors are connected so that their maximum equivalent capacitance is 1 5 ;if. (a) Describe how the capacitors are combined. (b) There are three other ways to combine all three capacitors in a circuit. What are the equivalent capacitances for each arrangement? *38 ··
We can use the properties of capacitors connected in series and in parallel to find the equivalent capacitances for various connection combinations. Picture the Problem
(a)
If
th eir capac itanc e is to bea max imum, th eymust beconn ect ed in para llel.
Find the capacitance of each capacitor:
=
=
l iP C 3 C 1 5 f-U. and eq
78 Chapter 24 (1) Connect the three capacitors in series:
_1_= 3 Ceq S JlF
(2) Connect two in parallel, with the third in series with that combination:
Ceq, two in parallel
(b)
and Ceq 1 1 . 6 7 ,uP 1 =
_ _
and
=
1
1
=
2(S JlF) 1 0 ,uP 1
+ -- = -1 0 JlF S ,uP Ceq
Solve for Ceq: (3) Connect two in series, with the third in parallel with that combination:
Ceq
(1 0 ,uP)(S ,uP) 1 0 ,uP + 5 ,uP
=
1
Ceq, two in series
Find the capacitance equivalent to 2.5 liF and 5 J..iF in parallel:
Ceq
=
1 3 33 ,uP 1 .
2 S ,uP
Ceq, two in series
or
=
=
2 . S ,uP
2.S ,uP + S ,uP
=
1 7.S0 JlF 1
Find all the different possible equivalent capacitances that can be obtained using a I-J..iF, a 2-J..iF, and a 4-liF capacitor in any combination that includes all three, or any two, of the capacitors. *44
••
We can connect two capacitors in parallel, all three in parallel, two in series, three in series, two in parallel in series with the third, and two in series in parallel with the third. Picture the Problem
Connect 2 in parallel to obtain: or Ceq
=
1 ,uP + 4 ,uP
=
or Ceq
Connect all three in parallel to obtain:
= 2 ,uP + 4,uP =
1 S ,uP I I 6 ,uP I
Electrostatic Energy and Capacitance
79
Connect two in series: or
Connect all three in series:
Connect two in parallel, in series with the third:
or
Connect two in series, in parallel with the third: or
Parallel-Plate Capacitors
Design a 0.1-,uF parallel-plate capacitor with air between the plates that can be charged to a maximum potential difference of 1000 V. (a) What is the minimum possible separation between the plates? (b) What minimum area must the plates of the capacitor have? *49 ··
80
C hap ter 24
The potential difference across the capacitor plates V is related to their separation d and the electric field between them according to V = Ed. We can use this equation with Emax = 3 MV/m to find dmin. In part (b) we can use the expression for the capacitance of a parallel-plate capacitor to find the required area of the plates.
Picture the Problem
�
1 000 V 3 MV/m
(a) Use the relationship between the potential difference across the plates and the electric field between them to find the minimum separation of the plates:
d
(b) Use the expression for the capacitance of a parallel-plate capacitor to relate the capacitance to the area of a plate:
c
=
Eo A d
A
=
(O. 1 ,uF)(O. 33 3 mm) 8 . 85 x lO-1 2 C 2/N · m2
.
mm
=
Emax
=
=
1 O. 333 mm 1
Solve for A : Substitute numerical values and evaluate A :
=
1 3 .7 6 m 2
I
Cylindrical Capacitors *53 ·· A goniometer is a precise instrument for measuring angles. A capacitive goniometer is shown in Figure 24-34. Each plate of the variable capacitor consists of a flat metal semicircle with inner radius Rl and outer radius R 2• The plates share a common rotation axis, and the width of the air gap separating the plates is d. Calculate the capacitance as a function of the angle Band the parameters given.
0":::::::6�
, - -lIIiiiitril
Figure 24-34a Problem 53
Rl
�-----------
�
Figure 24-34b Problem 53
Electrostatic Energy and Capacitance 8 1 We can use the expression for the capacitance of a paraIlel plate capacitor of variable area and the geometry of the figure to express the capacitance of the goniometer. Pictu re the Problem
I I!
"" - - - - ...... -...... � f) --
The capacitance of the parallel-plate capacitor is given by: The area of the plates is: If the top plate rotates through an angle !1e, then the area is reduced by: Substitute for A and M in the expression for C to obtain:
Spherical Capacitors
*55 ·· A spherical capacitor consists of two thin, concentric spherical shells of radii RI and R2• (a) Show that the capacitance is given by C 4Jr E o RI R2 /(R2 - RJ (b) Show that when the radii of the shells are nearly equal, the capacitance is given approximately by the expression for the capacitance of a parallel-plate capacitor, C = EoA/d, where A is the area of the sphere and d = R2 - RI• =
Picture the Problem We can use the definition of capacitance and the expression for the
potential difference between charged concentric spherical shells to show that C 4 Jr E o R1 R2 /(R2 - Rl ) . =
(a) Using its definition, relate the capacitance of the concentric spherical shells to their charge Q and the potential difference V between their surfaces:
82
Chapter 24
Express the potential difference between the
conductors:
Substitute to obtain:
(b)
Because R2 = R) + d:
R) R2 = R) (R) + d) = R)2 + R) d R2 - R2 � �
)
-
because d is small. Substitute to obtain: Disconnected and Reconnected Capacitors
Two capacitors, C) 4 � and C2 12 �, are connected in series across a 12-V battery. They are carefully disconnected so that they are not discharged and they are then reconnected to each other, with positive plate to positive plate and negative plate to negative plate. (a) Find the potential difference across each capacitor after they are connected. (b) Find the initial energy stored and the [mal energy stored in the capacitors. *60
··
=
=
When the capacitors are reconnected, each will have the charge it acquired while they were connected in series across the 12-V battery and we can use the definition of capacitance and their equivalent capacitance to find the common potential difference across them. In part (b) we can use U = t CV 2 to find the initial and final energy stored in the capacitors. Picture the Problem
Using the definition of capacitance, express the potential difference across each capacitor when they are reconnected:
where Q is the charge on each capacitor before they are disconnected.
Find the equivalent capacitance of the two capacitors after they are connected in parallel:
Ceq = C) + C2 = 4 .uF + 1 2 ,uF = 1 6 ,uF
(a)
V=
2Q Ceq
(1)
Electrostatic Energy and Capacitance
83
Express the charge Q on each capacitor before they are disconnected: Express the equivalent capacitance of the two capacitors connected in senes: Substitute to find Q:
Q = (3 ,LLF)(12 V ) = 36;£
Substitute in equation (1) and evaluate V: (b) Express and evaluate the energy stored in the capacitors initially: Express and evaluate the energy stored in the capacitors when they have been reconnected:
Vi
Vf
= tC'el'; 2 = t (3 ,uF)(12VY = 1 216,uJ 1 = tceq V/ = t (16 ,LLF)(4 .5VY = 1 162,uJ I
••
A 20-pF capacitor is charged to 3 kV and then removed from the battery and connected to an uncharged 50-pF capacitor. (a) What is the new charge on each capacitor? (b) Find the initial energy stored in the 20-pF capacitor, and find the final energy stored in the two capacitors. Is electrostatic potential energy gained or lost when the two capacitors are connected? *64
Let the numeral 1 refer to the 20-pF capacitor and the numeral 2 to the 50-pF capacitor. We can use conservation of charge and the fact that the connected capacitors will have the same potential difference across them to find the charge on each capacitor. We can decide whether electrostatic potential energy is gained or lost when the two capacitors are connected by calculating the change /). U in the electrostatic energy during this process. Picture the Problem
(a) Using the fact that no charge is lost in connecting the capacitors, relate the charge Q initially on the 20pF capacitor to the charges on the two capacitors when they have been connected:
(1)
84
Chapter 2 4
Because the capacitors are in parallel, the potential difference across them is the same:
v.I
=
V2 => Q Q2 C C2 I
=
Solve for QI to obtain: Substitute in equation (1) and solve for Q2 to obtain:
(2)
Use the definition of capacitance to find the charge Q initially on the 20pF capacitor:
Q C1 V = (20 pF)(3 kV ) = 60nC
Substitute in equation (2) and evaluate Q2 :
Q2 = 1 + 20pF/50pF = I 42 . 9 nC I
Substitute in equation ( 1 ) to obtain:
QI = Q - Q2
=
60nC
I
= 60nC - 42.9nC = 17.lnC
(b) Express the change in the electrostatic potential energy of the system when the two capacitors are connected:
Substitute numerical values and evaluate 11 U:
I
/::" U = Ur - Uj
--Q2 - --Q2
/::" U =
(60nCY
2 = -64.3 jiJ
(
1 1 __ _ __ 70pF 20pF
)
Because /::" U < 0, electrostatic energy is lost when the two capacitors are connected.
A capacitor of capacitance C has a charge Q. A student connects one terminal of the capacitor to a terminal of an identical uncharged capacitor. When the remaining two terminals are connected, charge flows until electrostatic equilibrium is reestablished and both capacitors have charge QI2 on them. Compare the total energy initially stored in the one capacitor to the total energy stored in the two after the second has been charged. Where did the missing energy go? This energy was dissipated in the connecting wires via Joule heating, which is discussed in Chapter 25. *66 ··
85
Electrostatic Energy and Capacitance Picture the P roblem We can use the express i on for the energy stored in a capacitor
express the ratio of the energy stored in the system after the discharge of the first capacitor to the energy stored in the system prior to the discharge. 2 Express the energy U initially Q stored in the capacitor whose U = 2C capacitance is C:
to
The energy U stored in the two capacitors after the first capacitor has discharged is: Express the ratio of U to U:
Q2
� = �� = �
=>
I
U' = t U
I
2C
Dielectrics *72 ·· Two capacitors, each consisting of two conducting plates of surface area A, with an air gap of width d. They are connected in parallel, as shown in Figure 24-37, and each has a charge Q. A slab of width d and area A with dielectric constant K is inserted between the plates of one of the capacitors. Calculate the new charge Q' on that capacitor.
Figure 24-37
Problem 72
Let the charge on the capacitor with the air gap be QI and the charge on the capacitor with the dielectric gap be Q2 . If the capacitances of the capacitors were initially C, then the capacitance of the capacitor with the dielectric inserted is C' = KG. We can use the conservation of charge and the equivalence of the potential difference across the capacitors to obtain two equations that we can solve simultaneously for QI and Q2 . Picture the Problem
Apply conservation of charge during the insertion of the dielectric to obtain: Because the capacitors have the same potential difference across
(1)
Q1 C
_
Q2 K.C
(2)
Chapter 24
86
them: Solve equations ( 1 ) and (2) simultaneously to obtain: What is the dielectric constant of a dielectric on which the induced bound charge density is (a) 80 percent of the free-charge density on the plates of a capacitor filled by the dielectric, (b) 20 percent of the free-charge density, and (c) 98 percent of the free-charge density? *75
••
The bound charge density is related to the dielectric constant and the free charge density according to = 1 Picture the Problem
Solve the equation relating eJi" and K for Kto obtain:
O'r,
O'r
(a) Evaluate this expression for Oil = 0.8:
(b) Evaluate this expression for OilO'r = 0.2:
O'r
(c) Evaluate this expression for
Oil
*77
=
0.98:
••
CTb
(
:)CTr .
---1 - = 5.00 1 K=1 - 0.8 1 K = _I_ = [ill 1 - 0.2
1 K = l - CTb/CT
r
.
1 = 50.0 K = 0.98 1 1 1-
Find the capacitance of the parallel-plate capacitor shown in Figure 24-3 8.
Figure 24-38
Problem 77
Picture the Problem
We can model this parallel-plate capacitor as a combination of two
Electrostatic Energy and Capacitance 87 capacitors C, and C2 in series with capacitor C3 in parallel. Express the capacitance of two series-connected capacitors in parallel with a third:
=
5
Express each of the capacitances C" C2, and C3 in terms of the dielectric constants,plate areas, and plate separati ons:
(1)
C C3 + Cs where C C,C2 C, + C2 C,
=
C
=
and C3
=
2
(2)
=
K, Eo
(1 A)
l. d 2
K2 Eo
(1 A)
l. d 2
K3 Eo
d
(1 A)
=
=
=
K, Eo
A
d K2 Eo
A
d K3 Eo
' '
A
2d
Substitute in equation (2) to obtain:
Substitute in equation (1) to obtain:
General Problems
Three capacitors have capacitances of 2 ;iF, 4 ;iF, and 8 ;iF. Find the equivalent capacitance if (a) the capacitors are connected in parallel and (b) ifthe capacitors are connected in series.
*81
•
We can use the equations for the equivalent capacitance of three capacitors connected in parallel and in series to find these equivalent capacitances.
Picture the Problem
Express the equivalent capacitance of three capacitors connected in parallel:
(a)
88
Chapter 24
Substitute numerical values and evaluate Cq :
= 2.0 ,LlF + 4. 0 1 1 4 . 0 1LF I
Ceq
=
liF +
8. 0 ,uP
(b) Express the equivalent capacitance of the three capacitors connected in series: Substitute numerical values and evaluate Ceq:
••
Figure 24-40 shows four capacitors connected in the arrangement lmown as a *85 capacitance bridge. The capacitors are initially uncharged. What must the relation between the four capacitances be so that the potential between points c and d is zero when a voltage V is applied between points a and b? a.
Figure 24-40
Problem 85
Note that with V applied between a and b, C1 and C3 are in series, and so are C2 and C4 • Because in a series combination the potential differences across the two capacitors are inversely proportional to the capacitances, we can establish proportions involving the capacitances and potential differences for the left- and right hand side of the network and then use the condition that Vc = Vd to eliminate the potential differences and establish the relationship between the capacitances. Picture the Problem
Letting Q represent the charge on capacitors 1 and 2, relate the potential differences across the capacitors to their common charge and capacitances:
and
V; CQ3 =
-
Electrostatic Energy and Capacitance
he first of these equations by the second to obtain:
Divide
t
(1)
Proceed similarly to obtain: Divide equation ( I ) by equation (2) to obtain:
89
(2) V; V4 V3 V2
_
C3 C2 C1 C4
(3)
If Vc Vd then we must have: =
Substitute in equation (3) and rearrange to obtain: *90 ·· A parallel-plate capacitor is constructed from a layer of silicon dioxide of thickness 5 x l 0-6 m between two conducting films. The dielectric constant of silicon dioxide is 3.8 and its dielectric strength is 8xl06 Vim. (a) What voltage can be applied across this capacitor without dielectric breakdown? (b) What should the surface area of the layer of silicon dioxide be for a 10-pF capacitor? (c) Estimate the number of these capacitors that can fit into a square 1 cm by 1 cm. Picture the Problem The maximum voltage is related to the dielectric strength of the medium according to Vmax = Emax d and we can use the expression for the capacitance of a parallel-plate capacitor to determine the required area of the plates.
(a) Relate the maximum voltage that can be applied across this capacitor to the dielectric strength of silicon dioxide:
)
Substitute numerical values and evaluate Vmax :
Vmax = (8 X 106 V/m (S x 10-6 m
(b) Relate the capacitance of a parallel-plate capacitor to area A of its plates:
C = K ED A
Solve for A to obtain:
\
= 40.0V
d
\
)
90
Chapter 24
Substitute numerical values and evaluate A : = 1 . 49 x I 0-6 m 2 = I . 49 mm2
I
I
(c) Express the number of capacitors n in terms of the area of a square 1 cm by 1 cm and the area required for each capacitor:
(l cm Y = -'--------'
11
A
1 00 mm2 1 .49 mm 2
--- �
@] 67
A parallel-plate capacitor is filled with two dielectrics of equal size, as shown in Figure 24-42. (a) Show that this system can be considered to be two capacitors of area t A connected in parallel. (b) Show that the capacitance is increased by the factor (KJ + *94
••
K2)/2 .
d Figure 24-42
Problem 94
We can express the ratio of Ceq to Co to show that the capacitance with the dielectrics in place is (KJ + K2)/2 times greater than that of the capacitor in the absence of the dielectrics. Picture the Problem
(a)
Because the capacitor plates are conductors, the potentials are the same across the entire upper and lower plates. Hence, the system is equivalent to two capacitors, each of area An, in parallel.
(b) Relate the capacitance Co, in the absence of the dielectrics, to the plate area and separation:
c
o=
Eo A d
91
Electrostatic Energy and Capacitance Express the equivalent capacitance of capacitors C1 and C2, each with plate area A12, connected in paral1el:
Express the ratio of Ceq to Co and simplify to obtain:
*97
•••
An electrically isolated capacitor with charge Q is partly filled with a
dielectric substance as shown in Figure 24-43. The capacitor consists of two rectangular plates of edge lengths is inserted is x.
(a)
a and b separated by distance d.
The distance which the dielectric
What is the energy stored in the capacitor?
(Hint: the capacitor can be
thought of as two capacitors connected in parallel.) (b) Because the energy of the
capacitor decreases as x increases, the electric field must be doing positive work on the dielectric, meaning that there must be an electric force pul1ing it in. Calculate the force by examining how the stored energy varies with x.
(c)
Express the force in terms of the
capacitance and voltage. (d) What is the origin of this force?
Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel, one with an air gap and other filled with a dielectric of constantK. Let the numeral 1 denote the capacitor with the dielectric material whose constant is K and the numera12 the air-filled capacitor.
(a)
Using the hint, express the
energy stored in the capacitor as a function of the equivalent capacitance Ceq:
(
)
capacitors are:
Eo a a -x K Eo ax = �---'-----'CI = -----"-- and C2 d d
Because the capacitors are in
c
The capacitances of the two
parallel, Ceq is the sum of C1 and C2:
�
(
= c + C = K Eo ax + Eo a a - x I 2
d
d
)
92
Chapter 24
Substitute for Ceq in the expression for U and simplify to obtain:
(b) The force exerted by the electric field is given by:
2a [(K-l)x+a] 2 Eo
(c) Rewrite our result in (b) to obtain:
(K-l)Q' F
=
('7)
{ a; ) [(K-l)x+al' (K-l)Q2 ('7) '
'
Note that this expression is independent of x.
This force originates from the fringing fields around the edges of the (d) capacitor. The effect of the force is to pull the dielectric into the space between the capacitor plates.
*102
You are asked to construct a parallel-plate, air-gap capacitor that will store
•••
100 kJ of energy. capacitor?
(b)
(a)
What minimum volume is required between the plates of the 8 Suppose you have developed a dielectric that can withstand 3xl0 Vim and
has a dielectric constant of K= 5. What volume of this dielectric, between the plates of the capacitor, is required for it to be able to store 100 kJ of energy?
Picture the Problem Recall that the dielectric strength of air is 3 MV1m. We can express the maximum energy to be stored in terms of the capacitance of the air-gap capacitor and the maximum potential difference between its plates. This maximum potential can, in tum, be expressed in terms of the maximum electric field (dielectric strength) possible in
Electrostatic Energy and Capacitance
93
the air gap. We can solve the resulting equation for the volume of the space between the plates. In part (b) we can modify the equation we derive in part (a) to accommodate a dielectric with a constant other than
1.
(a) Express the energy stored in the
Umax
-
c=
Eo
I 2"
CV2 max
capacitor in terms of its capacitance and the potential difference across it:
Express the capacitance of the air gap parallel-plate capacitor:
d
A
Relate the maximum potential difference across the plates to the maximum electric field between them:
Substitute to obtain:
=.1 2
where u
=
E0
vE2
Ad is the volume between the
plates.
Solve for u:
V=
2
-Umax Eo
Substitute numerical values and
V-
_
evaluate u:
(1)
---'-"""?
E�lllx
2(100kJ)
(8.85
x
10-12 C2/N. m2
I
)(3 MV/mY
= 2.51x103 m3
(b) With the dielectric in place
Evaluate equation (2) with K = 5 and Emax
=
3xl08 Vim:
2Unlllx V= ---"'-= '--? K Eo
equation (1) becomes:
I (2)
E�lllx
2(100kJ) - --;------�--..!.,-,�-----;-::V5 8.85 x 10-12 C2/N . m2 x 108 Vim
)(3
(
)
1
= 5.02x10-2m3
*106 CI
=
•••
0.4 j.iF
The two capacitors shown in Figure 24-45 have capacitances and C
2
=
1.2 j.iF. The voltages across the two capacitors are VI and V ,
2
respectively, and the total stored energy in the two capacitors is 1.14 m1. If terminals b and
c are connected together,
the voltage is Va
- Vd
=
80 V;
if terminal a is connected to
94
Chapter 24
terminal b, and terminal c is connected to terminal d, the voltage Va - Vd initial voltages VI and
V2.
=
20 V.
Find the
c Figure 24-45 Problem 106 Picture the Problem We can express the two conditions on the voltage in terms of the charges to find
Q,
Q,
and
and
Q2
Q2.
voltages V, and
and the capacitances
C,
and
C2 and solve the equations simultaneously
We can then use the definition of capacitance to find the initial
V2.
Express the condition for the series connection:
or
, Q C,
Substitute numerical values to
+
2 Q C2
=
80V
=80V
obtain: or
Use the definition of capacitance to expre·ss the condition for the parallel connection:
Because the capacitors are now connected in parallel:
Substitute to obtain:
(2)
Solve equations
(1)
and (2)
Electrostatic Energy and Capacitance
95
simultaneously to obtain:
�
Use the definition of capacitance to obtain:
*110
=
Q C1
=
32;£
0.4 /-IF
\
= 80 0 Y .
\
inclined at a small angle, as shown in Figure 24-48. The plate separation varies from d =
A capacitor has rectangular plates oflength a and width b. The top plate is
•••
Yo at the left to d = 2yo at the right, where Yo is much less than a or b. Calculate the
capacitance using strips of width dx and length b to approximate differential capacitors of area b dx and separation d = Yo + (yo/a)x that are connected in parallel.
.... ... 1-----.
Figure 24-48
a
------i.�
Problem 1 1 0
Picture the Problem
Choose a coordinate system in which the positive x direction is the
dC and then integrate this expression to find C for this capacitor.
right and the origin is at the left edge of the capacitor. We can express an element of capacitance
dC of length b, width dx and
Express an element of capacitance d = Yo+
(yo/a)x:
separation
These elements are all in parallel, so the total capacitance is obtained by integration:
]
1 dx = Eo ab In2 b C = Eo Yo----' Yo ol+x/a '------"-
Chapter 25 Electric Current and Direct-Current Circuits Conceptual Problems *1
•
In our study of electrostatics, we concluded that there is no electric field within a
conductor in electrostatic equilibrium. How is it that we can now discuss electric fields inside a conductor? Determine the Concept
When current flows, the charges are not in equilibrium. In that
case, the electric field provides the force needed for the charge flow. *5
••
A metal bar is to be used as a resistor. Its dimensions are 2 by 4 by 10 units.
To get the smallest resistance from this bar, one should attach leads to the opposite sides that have the dimensions of (a) 2 by 4 units. (b) 2 by 10 units. (c) 4 by 1 0 units. (d) All connections will give the same resistance.
(e) None of the above is correct. Picture the Problem
The resistance of the
metal bar varies directly with its length and inversely with its cross-sectional area. Hence, to minimize the resistance of the 2
bar, we should connect to the surface for which the ratio of the length to the contact
a
4
area is least. Denoting the surfaces as a, b, and c,
L
A
LlA
a
10
8
0.8
b
4
20
0.2
c
2
40
0.05
Surface
complete the table to the right:
I (c) is correct. I
Because connecting to surface c minimizes R:
*10 · Two resistors with resistances RI and R2 are connected in parallel. If RI » R2 , the equivalent resistance of the combination is approximately (a) RI• (b) R2 . (c) o. (d) infinite Picture the Problem We can find the equivalent resistance of this two-resistor combination and then apply the condition that RI » R2 .
97
98
Chapter 25
Express the equivalent resistance of
RJ and R2 in parallel:
Solve for Reg to obtain:
Factor RJ from the denominator and simplify to obtain:
Req = Reff R2 and I (b) is correct I �
*13
·
Two resistors are connected in series across a potential difference. Resistor A
has twice the resistance of resistor B . If the current carried by resistor A is I, then what is
the current carried by resistor B? (a) I (b) 21 (c) 112 (d) 41 (e) 114
Determine the Concept
In a series circuit, because there are no alternative
pathways, all resistors carry the same current. The potential difference across each resistor, keeping with Ohm's law, is given by the product of the current and the resistance and, hence, is not the same across each resistor unless the resistors are identical. *17
•
(a) infinite (b) zero
I (a) is correct. I
An
ideal ammeter should have ---- internal resistance.
ideal ammeter would have zero resistance. An ammeter consists of a very small resistance in parallel with a galvanometer movement. The small resistance accomplishes two purposes: 1) It protects the galvanometer movement by shunting most of the current in the circuit around the galvanometer movement, and 2) It minimizes the loading of the circuit by the ammeter by minimizing the resistance of the Determine the Concept An
ammeter. *21
··
I (b) is correct. I
A battery is connected to a series combination of a switch, a resistor, and an
initially uncharged capacitor. The switch is closed at statements is true?
t
=
o. Which of the following
(a) As the charge on the capacitor increases, the current increases. (b) As the charge on the capacitor increases, the voltage drop across the resistor Increases. (c) As the charge on the capacitor increases, the current remains constant.
El ectric Cunent and Direct-Current Circuits
99
(d) As the charge on the capacitor increases, the voltage drop across the capacitor decreases. (e) As the charge on the capacitor increases, the voltage drop across the resistor decreases.
=
Applying Kirchhoff's loop rule to the circuit, we obtain 0, where VR is the voltage drop across the resistor. Applying Ohm's law to
Determine the Concept
[; - VR - Vc
the resistor, we obtain VR decreases with time. *24
·
=
JR. Because J decreases as the capacitor is c�arged, VR
I (e ) is co
rre c t.
I
All voltage sources have some internal resistance, usually on the order of 1 00
n or less. From this fact, explain the following statement that appears in some
electronics textbooks: "A voltage source likes to see a high resistance." Determine the Concept
R
The potential difference across an external resistor of resistance
V, where r is the internal resistance and V the voltage supplied by the r +R source. The higher R is, the higher the voltage drop across R. Put differently, the higher the resistance a voltage source sees, the less its own resistance will change the circuit.
R is given by
--
Estimation and Approximation *30
••
Compact fluorescent lightbulbs cost $6 each and have an expected lifetime of
8000 h. These bulbs consume 20 W of power, but produce the illumination equivalent to 75-W incandescent bulbs. Incandescent bulbs cost approximately $ 1 .50 each and have an expected lifetime of 1 200 h. If the average household has, on the average, six 75-W incandescent lightbulbs on constantly, and if energy costs 1 1 .5 cents per kilowatt-hour, how much money would a consumer save each year by installing the energy-efficient fluorescent lightbulbs? Picture the Problem
We can find the annual savings by taking into account the costs of
the two types of bulbs, the rate at which they consume energy and the cost of that energy, and their expected lifetimes. Express the yearly savings:
.6.$ =
Express the annual cost with the
Costincandescent = Costbulbs + Costenergy
incandescent bulbs:
Costincandescent - Costfluorescent
Express and evaluate the annual cost of the incandescent bulbs:
(1)
100
Chapter 25
= [
Cost bulbs = number of bulbs in use x annual consumption of bulbs x cost per bulb
365.24d x 24 h d ($1.50) $65.74 (6) 1200h
J
=
Find the cost of operating the incandescent bulbs for one year: Cost
energy = energy consumed x cost per unit of energy = 6(75W)(365.25d)(24h1d)($O.115/kW .h) = $453.64 Costnuorescenl = Cost bulbs + Costenergy
Express the annual cost with the fluorescent bulbs:
Express and evaluate the annual cost of the fluorescent bulbs: Cost
bulbs
=
1
[
number of bulbs in use x annual consumption of bulbs x cost per bulb
365.24d x 24h d ($6)= $39.45 = (6) 000h 8
Find the cost of operating the fluorescent bulbs for one year:
Costenergy = energy consumed x cost per unit of energy 6(20W) 365.24dX 2 h $0.1l5/kW .h) =
(
:}
= $120.97 Substitute in equation (1) and evaluate the cost savings �$:
�$ = Costincalldescent - Costnuorescenl = ($65.74 + $453.64) - ($39.45 + $120.97) = 1 $358.96 1 ••
A laser diode used in making a laser pointer is a highly nonlinear circuit element. For a voltage drop across it less than approximately 2.3 V, it behaves as if it has effectively infinite internal resistance, but for voltages across it higher than this it has a very low internal resistance-effectively zero. (a) A laser pointer is made by putting two 1.55 V watch batteries in series across the laser diode. If the batteries each have an internal resistance between 100 Q and 1 50 n, estimate the current in the laser diode. *32
'Electric Current and Direct-Current Circuits
101
(b) About half of the power delivered to the laser diode goes into radiant energy. Using this fact, estimate the power of the laser diode, and compare this to typical quoted values of about 3 mW. (c) If the batteries each have a capacity of20-mA hours (i.e., they can deliver a constant current of20 mA for approximately one hour before discharging), estimate how long one can continuously operate the laser pointer before replacing the batteries. Picture the Problem
Let r be the internal resistance of each battery and use Ohm's law
to express the current in laser diode as a function of the potentia! difference across r. We
can find the power of the laser diode from the product of the potential difference across the internal resistance of the batteries and the current delivered by them Iand the time-to discharge from the combined capacities of the two batteries and I. (a) Use Ohm's law to express the
current in the laser diode:
The potential difference across the internal resistance is:
ntemal resistance
V;ntemal resistance I
Substitute to obtain:
Assuming that r = 125 0:
I
(b) The power delivered by the batteries is given by: The power of the laser is half this value: Express the ratio of Plaser to
= V; 2r
I
Pquoted:
P
= £ - 2.3V
= £ -2r2.3V )- 2.3V = 1 3.20mA I = 2(1.25(1525Q ) = = (3.2mA)(2. 3V) = 7.36mW IV
�aser
=1P=1(7. 36mW) = 1 3.68mW I 8mW = 1.23 �aser = 3.36mW �uoted � = 1 123%�uoted = 0mA· h = 1 1 2. 5 h I = 43.20mA
or
aser
(c) Express the time-to-discharge:
Because each battery has a capacity of20 mA·h, the series combination has a capacity of 40 mA·h and:
/).t
M
Capacity I
102 Chapter 25 Current and the Motion of Charges *37
A 10-gauge copper wire and a 14-gauge copper wire are welded together end
••
to end. The wires carry a current of 15 A. If there is one free electron per copper atom in each wire, find the drift velocity of the electrons in each wire. The current wj]] be the same in the two wires and we can relate the drift velocity of the electrons in each wire to their current densities and the cross
Picture the Problem
sectional areas of the wires. We can find the number density of charge carriers n using n= pN A IM, where pis the mass de�sity, NA Avogadro' s number, and Mthe molar mass. We can find the cross-sectional area of 1 0- and 14-gauge wires in Table 25-2 . Relate the current density to the drift
1 10 gauge
velocity of the electrons in the 1 0-
Alogauge
= nevd
gauge WIre: Solve for Vd:
The number density of charge carriers n is related to the mass
Vd,IO =
IIOoauoe neA10 gauge /:)
/:)
-M
N n= p A
density p, Avogadro's number NA, and the molar mass M: 3 For copper, p= 8 .93 g/cm and M = 63.5 g/mol. Substitute and evaluate
n:
(8.93 glem3 )(6.02 x 1023 atoms/mol) 63.5 glmol = 8.47 x 1028 atoms/m3
n=
Use Table 25-2 to find the cross sectional area of 10-gauge wire: Substitute numerical values and evaluate Vd.JO:
Express the continuity of the current
1 10 gauoe = 1 14oauoe
in the two wires:
or
0
0
0
neVd,JOA logaUge = nevd,I4A I4gaUge
Solve for
Vd,14
to obtain:
'Electric Current and Direct-Current Circuits 1 03 Vd,14 Vd,IO AJooauoe 0
=
0
A14 gauge
Use Table 25-2 to find the cross sectional area of 14-gauge wire: Substitute numerical values and evaluate
*39
Vd,14:
··
Vd,14 = (0.21 0 ) S.261 2.081mm = I 0.S31mm1s I mm1s
rnrn2 2
In a proton supercollider, the protons in a 5-mA beam move with nearly the
speed of light. (a) How many protons are there per meter of the beam? (b) If the cross sectional area of the beam is 1 0-6 m2 , what is the number density of protons? Picture the Problem
We can relate the number of protons per meter Nto the number n of
free charge-carrying particles per unit volume in a beam of cross-sectional area A and then use the relation between current and drift velocity to relate n to 1. (a) Express the number of protons
N= nA
(1)
per meter Nin terms of the number
n of free charge-carrying particles per unit volume in a beam of cross sectional area A: Use the relation between current and drift velocity to relate I and n: Solve for n to obtain:
Substitute to obtain:
Substitute numerical values and evaluate N:
f = enAv f n= - eAv
� fA = N= eAv ev
- (1.60xlO-1SmA 9C)(3x108m/s) = 1 1.04x 1 08m-1
N-
104 Chapter
25
(b) From equation (1) we have:
m-I
n=
N 1 .04 x 108 - = ----10-6 m2 A
1
= 1 .04xl014
m-3
Resistance and Ohm' s Law *44 · The third (current-carrying) rail of a subway track is made of steel and has a cross-sectional area of about 55 cm2. The resistivity of steel is n·m. What is the resistance of km of this track?
10-7
10
Picture the Problem
We can use R= pL/ A to find the resistance of the track.
(a) Relate the resistance of the track
to its resistivity p, cross-sectional
L R= p A
area A , and length L:
Substitute numerical values and
evaluate R: *49
··
R=
(10-70. m ) 0 m2 = 1 0. 1 820 I 55cm 1 k
A copper wire and an iron wire with the same length and diameter carry the
same current!. (a) Find the ratio of the potential drops across these wires. (b) In which
wire is the electric field greater?
We can use Ohm's law to express the ratio of the potential differences across the two wires and R= pL/ A to relate the resistances of the wires to
Picture the Problem
their lengths, resistivities, and cross-sectional areas. Once we've found the ratio of the potential differences across the wires, we can use E = V / L to decide which wire has the greater electric field.
(a) Apply Ohm's law to express the potential drop across each wire:
VCu
=
1 Rcu
and
VFe = IRFe Divide the first of these equations
( 1)
by the second to express the ratio of the potential drops across the wires: Relate the resistances of the wires to their resistivity, cross-sectional area, and length:
L RCu - pcu cu Acu and
'Electric CUlTent and Direct-CulTent Circuits 105
)(Fe
=
Divide the first of these equations
� Fe PFe A Fe Pcu
by the second to express the ratio of
A -"'-c,,-u �Cu
Pcu PFe
�Fe PFe A Fe because L cu = LFe and Acu AFe, _ _
the resistances of the wires:
_
=
Substitute in equation ( 1) to obtain:
Substitute numerical values (see
VCu VFe
Table 2625-1 for the resistivities of copper and iron) and evaluate the
=
1.7 x10-8 Q, m = 0.170 I 10x10-8 Q·m
I
ratio of the potential differences: (b) Express the electric field in each
ECu
conductor in terms of its length and the potential difference across it:
=
VCu �Cu
and
EFe = VFe LFe
Divide the first of these equations by the second to obtain:
or
E EFe = � = 5.88ECu 0.17
I E is greater in the iron wire. I
Because EFe = 5,88E cu:
= I0 \e V 125 mY - 1) , where 10 - 2x lO
*54 ]
··
(
A diode is a circuit element with a very nonlinear IV curve. In a diode, -9
· A. Usmg a spreadsheet program, make a graph of
1(V) for a typical diode, for both forward biasing ( V> 0) and back-biasing ( V < 0). Show that if you plot In Ivs V for forward biasing (using V> 0.3 V or so), you get nearly a straight line, What is the slope of the line?
106
Chapter 25
Picture the Problem A spreadsheet program to plot 1 as a function of V is shown below. The formulas used to calculate the quantities in the columns are as follows:
Cell B1 AS A6 B5
Algebraic Form 10 V (mV)
ContentlF onnula 2 -200 AS +25 $B$l * (EXP(A5/25) - 1 )
V (mV) -200.0 - 1 7S.0
V+t..V
Jo(eVI25I1lV -1)
I (nA) -2.00 -2.00
7S.0 1 00.0 The following graph was plotted using the data in spreadsheet table shown above. 110 90
�
5
....
70 50 30 10 -10
-200
-150
-100
-50
V (mY)
0
50
100
A spreadsheet program to plot In(1) as a function of V for V> 0.3 V follows. The formulas used to calculate the quantities in the columns are as follows: Cell Bl AS A6 BS
ContentIFormula 2 300 AS + 1 0 LN($B$l *(EXP(AS/2S) - 1 ))
Algebraic Form 2 nA V
-l)J InlJo(eVl25mv V + t..V
'Electric Current and D irect-Current Ci r cuits -
A I 0=
"
;-
Y (mY) 300 310 320 330 340 350 ,
'
.
970 980 990 1000
,
B
107
C nA
2 In(I) 12,69 13.09 1 3 .49 1 3 .89 14.29 14.69 -,;-
-,il:
39.49 39.89 40.29 40.69
Ie' -,,'
'.;:',�'!
A graph of ln(J) as a function of V follows. Microsoft Excel' s Trendline feature was used to obtain the equation of the line. 45 40 35 30 25 � ..E 20 �
15 10 5 0
300
400
500
For V» 25 mY:
600
700
900
1000
eV 125 mY -1 eV 125mY >:::
and
1 Take the natural logarithm of both sides of the equation to obtain:
800
V(mV)
� �
1oeVI2
5mY
In (J ) = In (Ioev/25mY ) 1 V = In (Jo ) + 25rnV
which is of the form y = mx + b, where m =
1
25 rnV
=
I 0.04(rnVt I
.L--� ---'----'-----'-
in agreement with our graphical result.
108 Chapter *58
25
The space between two concentric spherical-shell conductors is filled with a
material that has a resistivity of 1 0 Q·m. If the inner shell has a radius of l.5 cm and the outer shell has a radius of 5 cm, what is the resistance between the conductors? (Hint: Find the resistance of a spherical-shell element of the material of area 4m,2 and length dr,
9
and integrate to find the total resistance of the set of shells in series.) The diagram shows a cross-sectional view of the concentric Picture the Problem
spheres of radii a and b as well as a spherical-shell element of radius r. Using the Hint we can express the resistance dR of the spherical-shell element and then integrate over the volume filled with the material whose resistivity p is given to find the resistance between the conductors. Note that the elements of resistance are in senes.
-
_P � dR -- P dr 4nr2 A
Express the element of resistance dR: Integrate dR from r
=
a to r = b to
obtain:
R
Substitute numerical values and evaluate R:
_ 109 n . m 4Jr
=
( 1.5cm1 _5_cml_J
I 3 .7 1 109 n I x
Temperature Dependence of Resistance *60
·
A
tungsten rod is 50 cm long and has a square cross-sectional area with sides
of l .0 mm. (a) What is its resistance at 20°C? (b) What is its resistance at 40°C? Picture the Problem We
can use R = pLIA to find the resistance of the rod at 20°e.
Ignoring the effects of thermal expansion, we can we apply the equation defining the temperature coefficient of resistivity,
a,
to relate the resistance at 40°C to the resistance at
20°e. (a) Express the resistance of the rod at 20°C as a function of its
R20 = P20L
A
' El ectric Current and Direct-Current C i rcuits
109
resistivity, length, and cross sectional area:
)
evaluate R20:
0'S R20 =(S.SXIO-SQ .m) Imm 2 ( = / 27.SmQ /
(b) Express the resistance of the rod at 40°C as a function of its
R40 =P40 AL
Substitute numerical values and
resistance at 20°C and the temperature coefficient of resistivity
a:
=P2o[1+ a(tc - 20Co)].:f. A =P20 AL +P20 AL a(tc -20Co) =R20 [1+a(tc -20Co)]
Substitute numerical values (see Table 25- 1 for the temperature coefficient of resistivity of tungsten) and evaluate R40:
*65
•••
A wire of cross-sectional area A, length
L" resistivity Ph and temperature
al is connected end to end to a second wire of the same cross-sectional area, length L2, resistivity 0., and temperature coefficient a2, so that the wires carry the same current. (a) Show that if PILIal+0.L2a2= 0, the total resistance R is independent of coefficient
temperature for small temperature changes. (b) If one wire is made of carbon and the other is copper, find the ratio of their lengths for which R is approximately independent of temperature. Picture the Problem
Expressing the total resistance of the two current-carrying (and
hence warming) wires connected in series in terms of their resistivities, temperature coefficients of resistivity, lengths and temperature change will lead us to an expression in
PILlal+0.L2a2= 0, the total resistance is temperature independent. In part (b) we can apply the condition that PILIal + 0.L2a2 = ° to find the ratio of the lengths of the
which, if
carbon and copper wires.
(a) Express the total resistance of these two wires connected in series:
R =RI +R2 LI (1+alL\T)+P2 L2 ( 1+a 2L\T)+.l[PILI(1+alL\T)+P2 L2(1+a 2L\T)] = PI A A A
110
Chapter
25
Expand and simplify this expression to obtain:
the te mperature. (b) Apply the condition for temperature independence obtained in (a) to the carbon and copper WIres:
Solve for the ratio of Leu to L e:
Leu Le
_
Peae Peuaeu
Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of carbon and copper) and evaluate the ratio of Leu to Le:
- 0.5 10-3 K-1 = I 64 I --8-Q�3�-1 . -7-x-10- . m .9- -x-10---3 -K---'
---L
Energy in Electric Circuits Find the power dissipated in a resistor connected across a constant potential difference of 120 V if its resistance is (a) 5 Q and (b) 10 O. *68
·
Picture the Problem
We can use P =
two resistors.
V 2 /R to find the power dissipated by the
Express the power dissipated in a resistor as a function of its resistance and the potential difference across it: (a) Evaluate P for V= 120 V and R = 5 0: (b) Evaluate P for V= 120 V and R = 10 0:
P=
1 ( 20VY 1 1.44kW I. IOn
=
·
i
' E lectri c Current and Direct-Current C rcuits *73
•
111
(a) How much power is delivered by the emf of the battery in Problem 72
when it delivers a current of20 A? (b) How much of this power is delivered to the starter? (c) By how much does the chemical energy of the battery decrease when it
delivers a current of 20 A to the starter for 3 min? (d) How much heat is developed in the battery when it delivers a current of 20 A for 3 min? Picture the Problem
We can find the power delivered by the battery from the
product of its emf and the current it delivers. The power delivered to the battery can be found from the product of the potential difference across the terminals of the starter (or across the battery when current is being drawn from it) and the current being delivered to it. In part (c) we can use the definition of power to relate the decrease in the chemical energy of the battery to the power it i s
delivering and the time during which current is drawn from it. In part (d) w e can use conservation of energy to relate the energy delivered by the battery to the heat develope the battery and the energy delivered to the starter (a) Express the power delivered by the battery as a function of its emf
P = EI = (12V)(20A) = 1240W I
and the current it delivers: (b) Relate the power delivered to the starter to the potential difference across its terminals: (c) Use the defmition of power to express the decrease in the chemical energy of the battery as it delivers
�tarter = Vstarter = (1 l.4V)(20A) = I 228W I = PM = (240W)(3min) = I 43.2kJ I I
M
current to the starter: (d) Use conservation of energy to relate the energy delivered by the
Edelivered by battery = Etransfomled into heat =
battery to the heat developed in the battery and the energy delivered to the starter: Express the energy delivered by the battery and the energy delivered to the starter in terms of the rate at which this energy is delivered: Solve for Q to obtain :
Pl1t = + �l1t Q
+ Edelivered to starter
Q + Edelivered to starter
112 Chapter 25 Substitute numerical values and
Q
evaluate Q: *77
··
(240 W - 228 W)(3min) . = 12.16kJ I =
A lightweight electric car is powered by ten 1 2-V batteries. At a speed of 80
kmlh, the average frictional force is 1 200 N. (a) What must be the power of the electric motor if the car is to travel at a speed of 80 kmlh? (b) If each battery can deliver a total charge of 160 A·h before recharging, what is the total charge in coulombs that can be delivered by the 1 0 batteries before charging? (c) What is the total electrical energy
delivered by the 1 0 batteries before recharging? (d) How far can the car travel at 80 km/h
before the batteries must be recharged? (e) What is the cost per kilometer ifthe cost of recharging the batteries is 9 cents per kilowatt-hour? Picture the Problem
We can use P = fo to find the power the electric motor must
develop to move the car at 80 km/h against a frictional force of 1 200 N. We can find the total charge that can be delivered by the 1 0 batteries using fl.Q NIfl.t The total
= .
electrical energy delivered by the 1 0 batteries before recharging can be found using the definition of emf. We can find the distance the car can travel from the definition of work and the cost per kilometer of driving the car this distance by dividing the cost of the required energy by the distance the car has traveled. (a) Express the power the electric motor must develop in terms of the speed of the car and the friction
p
= tv = (1200N)(80 = I 26.7kW I
force: (b) Use the definition of current to express the total charge that can be
fl.Q
delivered before charging:
kmIh
)
= 1 O(160A. h) ( 36�OS ) = 15.76MC I NIM
=
where N is the number of batteries.
(c) Use the definition of emf to express the total electrical energy available in the batteries: (d) Relate the amount of work the
= (5.76MC)(1 2V) = 1 69.1MJ I W = fd
W
=
Q&
batteries can do to the work required to overcome friction: Solve for and evaluate d:
9. 1MJ = I 57.6km I d fW = 61200N =
'Electric Current and Direct-Current Circuits
1 13
(e) Express the cost per kilometer as the ratio of the ratio of the cost of the energy to the distance traveled before recharging:
Cost/km =($O.09/kW.h)Elt =($O.09IkW.hX 120V)(160A·h) =I $O.03/km I d 57.6km Combinations of Resistors *79
(a) Find the equivalent resistance between point a and point b in Figure 25-
·
49. (b) If the potential drop between point a and point b is 12 V, find the current in each resistor.
3Q a
Figure 25-49
6Q
b
Problem 79
Picture the Problem
We can either solve this problem by using the expression
for the equivalent resistance of three resistors connected in parallel and then using Ohm's law to find the current in each resistor, or we can apply Ohm's law first to find the current through each resistor and then use Ohm' s law a second time to find the equivalent resistance of the parallel combination. We'll follow the first procedure.
of the three resistors in parallel and
1 1 1 =1 +-+Req 4Q 3Q 6Q
solve for
and
(a) Express the equivalent resistance
Req:
(b) Apply Ohm's law to each of the resistors to find the current flowing through each:
Req = 1 1.33Q I
=� =12V =1 3.00A .I 4Q 4Q · 12 I3 =� = V =1 4.00A I . 3Q 3Q · I
4
and
114
Chapter
25 1
6
=
�
l 2V
6Q 6Q = 1 =
2 00A .
I
Remarks: You would find it instructive to use Kirchhoff's junction rule (conservation of charge) to confirm our values for the currents through the three resistors.
*83
··
A 5-V power supply has an internal resistance of 50 n. What is the smallest resistor that we can put in series with the power supply so that the voltage drop across the resistor is larger than 4.5 V? -�
Let r represent the resistance of the internal resistance of the power supply, 8 the emf of the power supply, R the resistance of the external resistor to be placed in series with the power supply, and I the current drawn from the power supply. We can use Ohm' s law to express the potential difference across R and apply Kirchhoffs loop rule to express the current through R in terms of 8, r, and R.
Picture the Problem
1
R
(1)
Express the potential difference across the resistor whose resistance is R: Apply Kirchhoffs loop rule to the
& -Ir-IR
=
0
circuit to obtain: Solve for I to obtain:
Substitute in equation (1) to obtain:
I=� r+ R VR =
R (� r+R )
Solve for R to obtain:
Substitute numerical values and evaluate R:
R = (4.5V)(50Q) = I 450Q I
5V -4.5V
' Electric ClllTent and Direct-Current Circuits *89
••
1 15
A length of wire has a resistance of 120 Q. The wire is cut into N identical
pieces that are then connected in parallel. The resistance of the parallel arrangement is 1.875 Q. Find N.
We can use the equation for N identical resistors connected in parallel to relate N to the resistance R of each piece of wire and the equivalent resistance
Picture the Problem
Express the resistance of the N pieces connected in parallel:
N
R
1 Req
where R is the resistance of one of the N pIeces.
Relate the resistance of one of the N pieces to the resistance of the wire: Substitute to obtain:
R
=
Rwire N
1
Solve for N:
Substitute numerical values and
evaluate N:
N=
1 20n 1.875n
=
f8l �
Kirchhoff's Rules *93
·
In
Figure 25-56, the emf is 6 V and R = 0.5 Q. The rate of joule heating in R
is 8 W. (a) What is the current in the circuit? (b) What is the potential difference across R? (c) What is r?
R
Figure 25-56
Problem 93
116
Chapter
25
Picture the Problem We can relate the current provided by the source to the
=12 R
and use Ohm's law and Kirchhoffs rules to rate of louIe heating using P . find the potential difference across R and the value of r. (a) Relate the current Iin the circuit to rate at which energy is being
or
dissipated in the form of louIe heat:
I=
�
r
evaluate I:
I = W =\4.00A \ 0.50
(b) Apply Ohm's law to find VR:
VR
(c) Apply Kirchhoffs loop rule to obtain:
E -1r -1R =0
Solve for r:
r =E -IR =E -R I I
Substitute numerical values and
r
Substitute numerical values and
evaluate r: *97
••
=IR =(4A)(0.5n) =\2.00V \
=
:� -0.50
=
\1.000 \
In the circuit shown in Figure 25-59 the batteries have negligible internal
resistance. Find (a) the current in each resistor, (b) the potential difference between point
a and point b, and (c) the power supplied by each battery.
12V
g
Fi ure 25-59
+
60
+
12V
b
Problem 97
Picture the Problem
Let II be the current delivered by the left battery, 12 the current
delivered by the right battery, and 13 the current through the 6-0 resistor, directed down.
We can apply Kirchhoffs rules to obtain three equations that we can solve simultaneously for 12 , and 13• Knowing the currents in each branch, we can use Ohm's law to find the
I},
'Electric Current and Direct-Current Circuits
117
potential difference between points a and b and the power delivered by both the sources. (a) Apply Kirchhoffs junction mle at junction a: Apply Kirchhoffs loop rule to a
12 V -
loop around the outside of the
or
(4£1)11 (3£1)12 -12 V +
+ (3 £1)12
=
- (4£1)1,
Apply Kirchhoffs loop rule to a
12V -(4£1)1 -(6£1)13
loop around the left-hand branch of
0
0
circuit to obtain:
1
=
=
0
the circuit to obtain: Solve these equations simultaneously to obtain:
= I 0.667 A I, 12 I 0.889 A I, 11
=
and 13
(b) Apply Ohm's law to find the potential difference between points
a and b:
(c) Express the power delivered by the 12-V battery in the left-hand branch of the circuit: Express the power delivered by the 12-V battery in the right-hand
=
Vab
1 1.56A I
=
(6£1)13
=
�eft
1 9.36V 1
= Ell =
Prighl
= (6£1)(1.56
(12V)(0.667
A) = 1 8.OOW I
= E12 = (1 2V)(O 889A) = I IO 7 W I .
branch of the circuit:
A)
.
The circuit fragment shown in Figure 25-6 1 is called a voltage divider. (a) If R10ad is not attached, show that VOUI = V(R2/(RI + R2». (b) If RI= R2 = 10 kn, what is the smallest value of Rload that can be used so that VOUI drops by less than 10 percent from its unloaded value? (VOUI is measured with respect to ground.) *100 ··
1 1 8 Chapter
25
v
+
f------r- VOLIl I
�RI()ad I I I I
Figure 25-61
Problem 100
Let the current drawn from the source be 1. We can use Ohm's law in conjunction with Kirchhoffs loop rule to express the output voltage as a function of V, Rt, and R 2. In (b) we can use the result of (a) to express the condition on the output voltages in terms of the effective resistance of the loaded output and the resistances R l and R 2• Picture the Problem
(a) Use Ohm's law to express VOU! in terms of R2 and I: Apply Kirchhoffs loop rule to the circuit to obtain: Solve for I:
Substitute for I in the expression for VOU! to obtain:
(b) Relate the effective resistance of the loaded circuit Reff to R2 and R1oad:
1
--
Rerr
=1+ -
R2
1
--
R10ad
Solve for R1oad:
Letting V'OU! represent the output voltage under load, express the condition that VOll! drops by less than 10 percent of its unloaded value: Using the result from (a), express V' OU! in terms of the effective output load Reff:
(1)
VOU! - V'out
=
Vout
V'out
1- V'out 0. 1 <
VOU!
=( V
Rerr R1
+ Rerr
)
(2)
'Electric Current and Direct-Current Circuits
Substitute for Vout and V'Olit in equation (2) and simplify to obtain:
119
1 - _R-,--, R_+_Re:-=.f.:...-T O. 1 <
2
or
1 - RefT ( R, +R2 ) 0.1 R2 ( R I +Re ) fT <
Solve for Reff:
0.9(10kQ)(10kQ) efT IOkQ+0.I(10kQ) 8.l8kQ (IOkQ)(8.18kQ) 1 44.9kQ I R 10kQ-8.l8kQ >
Substitute numerical values and evaluate RefT:
R
Finally, substitute numerical values in equation (1) and evaluate Rload:
=
load
<
=
Ammeters and Voltmeters ••
A digital voltmeter can be modeled as an ideal voltmeter with an infinite internal resistance in parallel with a 10-MO resistor. Calculate the voltage measured by the voltmeter in the circuit shown in Figure 25-64 when (a) R = 1 ill, (b) R = 10 ill, (c) R =1 MO, (d) R =10 MO, and (e) R =100 MO. if) What is the largest value of R possible if we wish the measured voltage to be within 10 percent of the true voltage (i.e., the voltage drop without the voltmeter in place)? *105
lOV
+
R
V
Voltmeter
2R
Figure 25-64
Problem 105
Let Ibe the current drawn from source and Req the resistance equivalent to R and 10 MO connected in parallel and apply Kirchhoffs loop rule to express the measured voltage V across R as a function of R.
Picture the Problem
The voltage measured by the voltmeter is given by: Apply Kirchhoffs loop rule to the circuit to obtain:
(1)
lOY -1Req -I (2R ) 0 =
1 20
Chapter
2S
Solve for 1:
lOY Req +2R
1=
eq
Express R in tenns of R and lO-MQ resistance in parallel with it:
eq
Solve for R :
Substitute for lin equation (1) and simplify to obtain:
eq
Substitute for R and simplify to obtain:
(a) Evaluate equation (2) for R = 1 kQ: (b) Evaluate equation (2) for R =10 kG: (c) Evaluate equation (2) for R=1MQ: (d) Evaluate equation (2) for R =10MQ: (e) Evaluate equation (2) for R=100MQ:
1 = 1 +-1 Req 10MQ R (10MQ)R Req = R+lOMQ v
=
V R ( R10+2R eq eq J
=
10V 2R 1+_ Re q
= (lOV )(SMQ) (2) R+1SMQ = (lOV )(SMQ ) = 1 3.33 V I V 1kQ+1SMQ
v
v
= (lOV )(SMQ) = 1 3.33 V 10kQ+1SMQ
= (lOV)(SMQ ) = 1 3.13V I . 1MQ+1SMQ · = (lOV )(SMQ) = 1 2.OOV I V . 10MQ+1SMQ · = (lOV )(SMQ) = I 0.43SV I V . 100MQ+1SMQ · v
if) Express the condition that the measured voltage to be within 10 percent of the true voltage Vtrue: Substitute for Vand VinIC to obtain:
I
(10V)(SMQ )
1- R+1SMQ 0.1 IR or, because 1= 10 V/3 R, (lOV)(SMQ) 1- R+1SMQ 0.1 10 V 3 <
<
' E lectric Current and Direct-Current Circuits
Solve for R to obtain:
R<
12 1
1.50.MQ I L67MQ I 9 =
•• *109 Show that the meter movement in Problem 106 can be converted into a voltmeter by placing a large resistance in series with the meter movement, and find the resistance needed for a full-scale deflection when 10 V are placed across it.
The circuit diagram shows a fragment of a circuit in which a resistor of resistance r is connected in series with the meter movement of Problem 106. The purpose of this resistor is to limit the current through the galvanometer movement to 50 JlA and to produce a deflection of the galvanometer movement that is a measure of the potential difference V. We can apply Kirchhoffs loop rule to express r in terms of Vg, Ig, and R.
Picture the Problem
Apply Kirchhoffs loop rule to the circuit fragment to obtain:
a
V -r1g -RJg
=
v
0
Solve for r:
(1)
Use Ohm's law to relate the current Ig through the galvanometer movement to the potential difference Vg across it: Use the values for Vg and Ig given in Problem 106 to evaluate R: Substitute numerical values in equation (1) and evaluate r:
Vg Va 1g = _o �R = R 1g
0.25V = 5000Q 50jJA 10V -5000Q = 1 1 5kQ 9 I r= 50jJA
R=
Remarks: The total series resistance is the sum of rand R or 200 ill.
RC Circuits *113
••
In the cilicuit previously shown in Figure 25-40, emf 8
=
50 V and
C = 2.0 Jff!; the capacitor is initially uncharged. At 4 s after switch S is closed, the voltage drop across the resistor is 20 V. Find the resistance of the resistor. Picture the Problem
We can find the resistance of the circuit from its time constant and
use Ohm's law and the expression for the current in a charging RC circuit to express -ras a function of time, Vo, and V(t).
122
Chapter
25
Express the resistance of the resistor
R =�
(1)
C
in terms of the time constant of the circuit:
V(t) = I(t)R
Using Ohm's law, express the voltage drop across the resistor as a function of time: Express the current in the circuit as a function of the elapsed time after the switch is closed: Substitute to obtain:
t
Take the natural logarithm of both sides of the equation and solve for
r
to obtain:
R =-
Substitute in equation (1) to obtain:
t
cm[V::l]
Substitute numerical values and evaluate R using the data given for
=4 s: t Qo
A O.12-,uF capacitor is given a charge Qo. After What is the effective resistance across this capacitor?
••
*114
is
t
.
Picture the Problem
4 s, the capacitor's charge
We can find the resistance of the circuit from its time constant and
Qo, and Q(t) .
use the expression for the charge on a discharging capacitor as a function of time to express ras a function of time, Express the effective resistance across the capacitor in terms of the time constant of the circuit: Express the charge on the capacitor as a function of the elapsed time after the switch is closed:
R =�
C
(1)
' Electric CUlTent and D i rect-CulTent C ircuits
123
Take the natural logarithm of both sides of the equation and solve for
r
to obtain: Substitute in equation (1) to obtain:
Substitute numerical values and evaluate R:
*120
•• •
R =
R =
4s
I
o (O. 12 ,uF)ln z Q Qo
= 1 48 JMQ I
A photojournalist's flash unit uses a 9-V battery pack to charge a
0 .15-,uF capacitor, which is then discharged through the flash lamp of 10.5-0 resistance when a switch is closed. The minimum voltage necessary for the flash discharge is 7 V. The capacitor is charged through an 18-kO resistor. (a) How much time is required to
charge the capacitor to the required 7 V? (b) How much energy is released when the lamp flashes? (c) How much energy is supplied by the battery during the charging cycle and what fraction of that energy is dissipated in the resistor?
We can find the time-to-discharge by expressing the voltage across 2 the capacitor as a function of time and solving for t. We can use U{t) C V (t) to find Picture the Problem
=t
the energy released/stored in the capacitor when the lamp flashes. In part (c) we can
integrate dUbat = &dI(t) to find the energy supplied by the battery during the charging
cycle.
(a) Express the voltage across the capacitor as a function of time:
Solve for t to obtain:
Substitute numerical values and evaluate t:
t = -(I S k.G)(O. 15 1'l') ln =
(b) Express the energy stored in the capacitor as a function of time:
1 4 .06 ms I
( ��J l-
124
Chapter
25
U(t) t cv/ (1 - e-l/ RC r
Substi tute for Vet) to obtain: Substitute numerical values and evaluate
=
U(4.06 ms):
U{4.06ms) = t(0.15 ,LlF)(9V Y (I - e-4.06ms/(1 8H2)(O.15f'!') Y = 1 3.67Jil l Ubat (t) & fI(t')dt' = � fe -I'/Rc dt' o & 2 [R C(I - e -t/RC )] R = C& 2 (1 - e-I/RC )
(c) Relate the energy provided by
=
the battery to its emf and the current
I
?
I
0
it delivers:
=
Substitute numerical values and evaluate
Ubat(4.06 ms):
Ub•t (4 . 06 ms) = {0.1 5 ,Llf){9 V Y (1 - e-4.o6ms/(18kn)(0 15f'!') )= I 9.45 Jil I Express the fractionJof the energy supplied by the battery during the charging cycle that is dissipated in
f = UR Ubat
the resistor: Use conservation of energy to relate the energy supplied by the battery to the energy dissipated in the resistor and the energy released when the
Ub•t = UR + Uflash
or
UR Ubat - Ufl•sh =
lamp flashes: Substitute to obtain:
Substitute numerical values and evaluate! *121
•••
For the circuit shown in Figure
f = Ubat - Uflash = 1 - Uflash Ubat Ubat Jil f = 1 _ 3..467Jil = 1 6 1 .2% I 9 5 25-68, (a) what is the initial battery current
immediately after switch S is closed? (b) What is the battery current a long time after switch S is closed? (c) What is the current in the
600-0 resistor as a function of time?
' E lectri c CUlTent and Direct-CulTent Circuits
/ - � S 50lV
1 25
200 Q
5 ,uF
r-----� I�------I 600 Q Figure 25-68
Problem 121
200 n, R2 = 600 n, II and 12 their currents, and 13 the current into the capacitor. We can apply Kirchhoffs loop rule to find the initial battery Picture the Problem
Let RI
=
current 10 and the battery current 100 a long time after the switch is closed. In part (c) we can apply both the loop and junction rules to obtain equations that we can use to obtain a linear differential equation with constant coefficients describing the current in the 600-n resistor as a function of time. We can solve this differential equation by assuming a solution of a given form, differentiating this assumed solution and substituting it and its derivative in the differential equation. Equating coefficients, requiring the solution to hold for all values of the assumed constants, and invoking an initial condition will allow us to find the constants in the assumed solution.
(a) Apply Kirchhoffs loop rule to the circuit at the instant the switch is closed: Because the capacitor is initially uncharged : Solve for and evaluate 10:
I
a
= __ c_ =
200 n
50 Y
200 n
=
I 0.250A I
(b) Apply Kirchhoffs loop rule to the circuit after a long time has passed: Solve for 100 to obtain:
I
00
(c) Apply the junction rule at the junction between the 200-n resistor
=
50 Y
800 n
=
1 62.5 mA I ( 1)
126
Chapter
25
and the capacitor to obtain: Apply the loop rule to the loop containing the source, the 200-0
£ - R/I - -QC = O
(2)
QC - Ro-I2 = 0
(3 )
resistor and the capacitor to obtain: Apply the loop rule to the loop containing the 600-0 resistor and the capacitor to obtain: Differentiate equation (2) with
�dt [£ - RI1I - QC ] = O - RI ddt11 lC ddtQ = -RI ddt11 _ lC 3 = 0 (4) RI ddt11 = - ll C3 �dt [ QC - R2 2 ] = lC ddtQ - R2 ddt12 = 0 R2 ddt12 = lC l3 d12 = _1 (II -12) dt R2 C 1I = £ - 0C = £ - �4 _
respect to time to obtain:
I
or
I
Differentiate equation (3 ) with respect to time to obtain:
or
(5)
Using equation (1), substitute for in equation (5) to obtain: Solve equation (2) for
II 12:
Substitute for
13
II :
in equation (6) and
simplify to obtain the differential equation for
(6)
_ _ _
�
�
d12 = _1_ ( £-R212 2 ) dt R2 C RI - RI R£2C - ( RRII R+2RC2 )12 12 (t) = -I
_
To solve this linear differential equation with constant coefficients we can assume a solution of the form:
a + be-f/r
(7)
' E l ectric Current and Direct-Current Circuits d 2 J dt
Differentiate 12(t) with respect to time to obtain:
=
� [a dt
+
be-I/r
]
=
1 27
- � e-Ilr T
Substitute for 12 and d121dt to obtain: lr Equate coefficients of e -t to obtain:
Requiring the equation to hold for all values of a yields:
If
12
is to be zero when
t
=
O
0:
=
or b
=
a
+
-a
b = -
[;
R, +R2
Substitute in equation (7) to obtain:
(2000---'-)-'---(6000 (- 5 ,uP-R1 R2 C = --'--------'-)..:... ...:.. )-'R, +R2 2000+6000 0.750ms 50V (1 - ) 2 (t) 2000+6000 )I = I (62.5 mA)(l -
where T
=
=
J
Substitute numerical values and evaluate
12(t):
e-tlo.750ms
=
e -tjO.750 ms
General Problems *125
··
In
Figure 25-71,
RI
=
currents through these resistors by
13• (c) 13> 12. (d)
4 0,
R2
=
6 0, and
1" 12, 13, and
none of the above is correct.
R3
=
12 0. If we denote the
respectively, then (a)
1z
11 > >
h (b)
12
=
1 28 Chapter 2 5
Figure 25-71
Problems 125 and 127
Determine the Concept
Because all of the current drawn from the battery passes through
R], we know that II is greater than 12 and 13. Because R2 Because R3 > R2, 13
< 12
and so (c) is false.
7:-
R3, h 7:- 13 and so (b) is false.
I (a) is correct. I
•• A closed box has two metal terminals a and b. The inside of the box *131 contains an unknown emf c in series with a resistance R. When a potential difference of 2 1 V is maintained between terminal a and terminal b, there is a current of 1 A between the terminals a and b. If this potential difference is reversed, a current of2 A in the reverse direction is observed. Find c and R.
Picture the Problem
We can apply Kirchhoff's loop rule to the circuit that includes the
box and the 2 1 -V source to obtain two equations in the unknowns
c and
solve simultaneously. R
II
a
. AAA V VV
1 1 6'
I
---
Apply Kirchhoff's loop rule to the circuit when the polarity of the
+
21 V
b
-
21V+E-
(l A)R
=
0
2 1 -V source and the direction of the current are as shown in the diagram: Apply Kirchhoff' s loop rule to the circuit when the polarity of the source is reversed and the current is 2 A in the opposite direction:
- 21V + E +
(2 A)R
=
0
R that we can
'Electri c Current and Direct-Current Circuits R
Solve these equations simultaneously to obtain: * 1 33
••
=
129
1 14. 0 Q 1 and E = 1 -- 7.00V 1
The circuit shown in Figure 25-73 is a slide-type Wheatstone bridge. This
bridge is used to determine an unknown resistance R" in terms of the known resistances R b R2, and Ro. The resistances R1 and R2 comprise a wire 1 m long. Point a is a sliding
contact that is moved along the wire to vary these resistances. Resistance R1 is
proportional to the distance from the left end of the wire (labeled 0 cm) to point a, and R2
is proportional to the distance from point a to the right end of the wire (labeled 100 cm). The sum of R1 and R2 remains constant. When points a and b are at the same potential,
there is no current in the galvanometer and the bridge is said to be balanced. (Because the
galvanometer is used to detect the absence of a current, it is called a null detector.) If the fixed resistance Ro = 200 [2, find the unknown resistance Rx if (a) the bridge balances at
the 18-cm mark, (b) the bridge balances at the 60-cm mark, and (c) the bridge balances at
the 95-cm mark.
""
O cm
R2
_
100
1\ _---'/
cm
P (J
Figure 25-73
Problems 13 3 and 13 4
Let the current flowing through the galvanometer by IG. By applying Kirchhoff's rules to the loops including 1) R ] , the galvanometer, and Rx, and 2)
Picture the Problem
R2, the galvanometer, and Ro, we can obtain two equations relating the unknown
resistance to R], R2 and Ro. Using R = pLiA will allow us to express Rx in terms of the
length of wire L1 that corresponds to R1 and the length of wire L2 that corresponds to R2. Apply Kirchhoff's loop rule to the
(1)
loop that includes R], the galvanometer, and Rx to obtain: Apply Kirchhoff's loop rule to the
(2)
loop that includes R2, the galvanometer, and Ro to obtain: When the bridge is balanced,
(3 )
1 30 Ie
=
Chapter
25
0 and equations ( 1 ) and (2) (4)
become: Divide equation (3 ) by equation (4)
(5)
and solve for x to obtain:
Express RJ and R2 in terms of their lengths, cross-sectional areas, and the resistivity of their wire:
RJ
(a) When the bridge balances at the 18-cm mark, L J = 18 cm, L2 = 82 cm and: (b) When the bridge balances at the
60-cm mark, L J = 60 cm,
-
95-cm mark, L J = 95 cm,
L2 = 5 cm and:
L
p
L2 A
Rx R0 � =
L2
=
(2000) 18cm 1 43.9 0 1 82cm
Rx
=
(2000) 60cm 1 3000 1 40cm
Rx
=
Rx
L2 = 40 cm and:
(c) When the bridge balances at the
p�
and R2
Substitute in equation (5) to obtain:
=
=
=
(2000) 95cm 1 3.80kO 1 5cm =
*140 ·· (a) Show that a leaky capacitor (one for which the resistance of the dielectric is finite) can be modeled as a capacitor with infinite resistance in parallel with a resistor. (b) Show that the time constant for discharging this capacitor is r = EopK. (c) Mica has a dielectric constant K = 5 and a resistivity p = 9x l 0 J 3 n·m. Calculate the time it takes for the charge of a mica-filled capacitor to decrease to 1 0 percent of its initial value.
We'll assume that the capacitor is fully charged initially and apply Kirchhoffs loop rule to the circuit fragment to obtain the differential equation describing the discharge of the leaky capacitor. We'll show that the solution to this equation is the familiar expression for an exponential decay with time constant Picture the Problem
r = EoPK.
- +
r-------� �------� C
R
--..
I
'Electri c Current and Direct-Current Circui ts
13 1
If we think of the leaky capacitor as a resi stor/capacitor combination,
(a)
the voltage drop across the resistor must be the same as voltage drop
across the capacitor. Hence, they must be in parallel.
(b) Assuming that the capacitor is initially fully charged, apply Kirchhoffs loop rule to the circuit fragment to obtain:
Q - RJ = O C or, because
J=_
dQ , dt
Q dQ =0 +R dt C equation to obtain:
dQ 1 - = - - dt RC Q
From Problems l 3 8 and l 3 9 we
RC =E O pK
Separate variables in this differential
have:
equation to obtain:
dQ Q
Integrate this equation from
Q = Qoe-I/r
Substitute for RC in the differential
1 --
Q' = Qo to Q to obtain:
where
(c) Because QIQo = 0. 1:
e-I/r = 0. 1
Solve for t by taking the natural
t - - = In 0.1
logarithm of both sides of the
r
dt
-----' = '--1 E o pK 1
r
=>
t = - Eo pK In O.1
equation: Substitute numerical values and evaluate t:
*144
•••
Capacitors C1 and C2 are connected in parallel by a resistor and two switches,
as shown in Figure 25-76. Capacitor C1 is initially charged to a voltage Vo, and capacitor C2 is uncharged. The switches SI and S2 are then closed. (a) What are the final charges on C1 and C2? (b) Compare the initial and final stored energies of the system. (c) What caused the decrease in the capacitor-stored energy?
1 32
Chapter 25 R
s Figure 25-76
�C2
Problems 144 and 145
Picture the Problem
Let QI and Q2 represent the final charges on the capacitors C1 and
C2. Knowing that charge is conserved as it is redistributed to the two capacitors and that the final-state potential differences across the two capacitors will be the same, we can obtain two equations in the unknowns QI and Q2 that we can solve simultaneously. We can compare the initial and final energies stored in this system by expressing and simplifying their ratio. We can account for any difference between these energies by considering the role of the resistor in the circuit.
(a) Relate the total charge stored initially to the final charges QI and Q2 on C1 and C2: Because, in their final state, the potential differences across the two capacitors will be the same: Solve equation (2) for Q2 and substitute in equation (1) to obtain: Solve for Q2 to obtain:
Substitute in either (1) or (2) and solve for QI to obtain:
(1)
(2)
'Electri c Current and D irect-Current Circuits
133
(b) Express the ratio of the initial and final energies of the system:
Simplify this expression further to I obtain:
=
Ur �
I
C1 l + C,
I
or U. is greater than Ur by a factor of 1
(c)
+ C2/C1 ,
The decrease in energy equals the energy dissipated as Joule heat in the resistor connecting\the two capacitors. \
\ " The differential r� istance of a nonohmic circuit element is defined as Rd d VldI, where Vis the voltage acr6ss the element and Iis the current through the element. Show that for V > 0.6 V, the differential resistance of a diode (Problem 54) is
*147
=
•••
approximately Rd
=
(25 mY)!I, and for V< 0, Rd increases exponentially with
this result to justify the approximation given in Problem 55.
IVI.
Use
We can use the definition of differential resistance and the expression for the diode current given in problem 54 to express R d and establish the required results.
Picture the Problem
The differential resistance Rd is given by: From Problem 54, the current in the diode is given by:
1 = 10 (e
V/25m Y
- 1)
Substitute for Ito obtain:
= For V> 0.6 V, equation (1) becomes:
1
� �
25 mV e-V/25mV
1o e
10
V/25mV
(1)
1 34
Chapter 25 I -V/25mV � 10 e V/25 mV � - => e
Solve for the exponential factor to obtain:
10
Substitute in equation (2) to obtain:
Rd �
Examination of equation (2) shows that, for V This result, together with that for V
>
Problem 5 5 .
25 mV <
Io
10 I
=1
I
-
25 mV I
I
0, Rd increases exponentially.
0.6 V, justi fies the assumptions made in
Calculate the equivalent resistance between points a and b for the infinite *151 ••• ladder of resistors shown in Figure 25-80, where RJ and R2 can take any value.
a _-�I
R}
b ...... Figure 25-80 Problem 151
R}
-'-
-
'"'--
-
Let Req be the equivalent resistance of the infinite ladder. If the resistance is finite and non-zero, then adding one or more stages to the ladder will not change the resistance of the network. We can apply the rules for resistance combination to the diagram shown to the right to obtain a quadratic equation in Req that we can solve for the equivalent resistance between points a and b.
-
-
-
-
Picture the Problem
The equivalent resistance of the series combination of RJ and (R2 11 Req ) is Req, so: Simplify to obtain: Solve for the positive value of Req to obtain:
a
b .-------�--�
Chapter 2 6 The Magnetic Field Conceptual Problems *1
•
When a cathode-ray tube is placed horizontally in a magnetic field that is directed
vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 26- 3 0. The correct path is (a) 1. (b) 2. (c) 3 . (d) 4. (e) 5 .
Figure 26-30
Problem 1
Determine the Concept
Because the electrons are initially moving at 90° to the magnetic
field, they will be deflected in the direction of the magnetic force acting on them. Use the
F = qv x jj , remembering that, for a negative charge, the force is in the direction
right-hand rule based on the expression for the magnetic force acting on a moving charge opposite that indicated by the right-hand rule, to convince yourself that the particle will follow the path whose terminal point on the screen is 2.
I (b) is correct. I
*5 · Ajlicker bulb is a lightbulb with a long, thin filament. When it is plugged in and a magnet is brought near the lightbulb, the filament is seen to oscillate rapidly back and forth. Why does the filament oscillate, and what is the frequency of oscillation? Determine the Concept The alternating current running through the filament is changing direction every 1/60 s, so in a magnetic field the filament experiences a force which alternates in direction at that frequency. • The north-seeking pole of a compass needle located on the magnetic equator *8 is the end of the needle that points toward the north, and the direction of any magnetic field B is specified as the direction that the north-seeking pole of a compass needle points when the needle is aligned in the field. Suppose that the direction of the magnetic field B were instead specified as the direction of a south-seeking pole of a compass needle aligned in the field. Would the right-hand rule shown in Figure 26-2 then give the direction of the magnetic force on the moving positive charge, or would a left-hand rule be required? Explain.
1 35
136
Chapter
26
Determine the Concept The direction in which a particle is deflected by a magnetic field
will be unchanged by any change in the definition of the direction of the magnetic field. Since we have reversed the direction of the field, we must define the direction in which particles are deflected by a "left-hand" rule instead of a "right-hand" rule. *12
·
An
electron moving with speed
v
to the right enters a region of uniform
magnetic field that points out of the paper. When the electron enters this region, it will be (a) deflected out of the plane of the paper. (b) deflected into the plane of the paper. (c)
deflected upward. (d) deflected downward. (e) undeviated in its motion. Determine the Concept
Application of the right-hand rule indicates that a positively
charged body would experience a downward force and, in the absence of other forces, be deflected downward. Because the direction of the magnetic force on an electron is opposite that of the force on a positively charged object, an electron will be deflected upward.
I (c) is correct. I
Estimation and Approximation CRTs used in monitors and televisions commonly use magnetic deflection to steer the electron beams. A schematic diagram is shown in Figure 26-32. The electron beam is accelerated through a potential difference and the electron beam is then accelerated through a magnetic field that deflects the electron beam, as shown in the figure. Given the following parameters, estimate the magnitude of the magnetic field needed for maximum deflection: accelerating voltage, V = 15 kV; distance over which electron is in magnetic field, d = 5 cm; length, L = 50 cm; diagonal of CRT, r = 19 in. *16
Figure 26-32
Problem 16
Picture the Problem If the electron enters the magnetic field in the coil with speed v, it will travel in a circular path under the influence of the magnetic force acting on. We can d apply Newton's 2n law to the electron in this field to obtain an expression for the magnetic field. We'll assume that the deflection of the electron is small over the distance it travels in the magnetic field, but that, once it is through the region of the magnetic field, it travels at an angle e with respect to the direction it was originally traveling.
Apply
I F = mac to the electron
in the magnetic field to obtain:
2 v evB = m r
Solve for B:
B K
The kinetic energy of the electron is:
=
mv er
= eV = l.2 mv
v=
Solve for v to obtain:
The Magnetic Field
2
= � 2e v ! 2m V = er m r e
�
�
B
Because 8 « 1:
d r sin e ';::!
B
For maximum deflection, 8 ';::! 45°. Substitute numerical values and evaluate B:
B
2
� ::
Substitute for v in the expression for r:
Substitute for r in the expression for B to obtain:
1 37
=
sin e
d
=>
�
= sin 4 S o O.OS m
r � �
sin e
2m v
e
2
9. 1 1 x l O-31 kg (l S kV) 1 .60 x l O-19 C
= I S.84mT I
Force Exerted by a Magnetic Field A straight wire segment IL = (2.7 A)(3 cm i + 4 cm j ) is in a uniform -
*22
•
magnetic field B -
=
1.3 T i . Find the force on the wire.
Picture the Problem
segment.
�
We can use F
Express the force acting on the wire segment: Substitute numerical values and evaluate F :
*26
··
�
=
-
IL
-
x B to find the force acting on the wire -
-
-
F = IL x B
i = (2 . 7 A)[(3 cm)i + (4cm)}]x (1.3 T )i = 1 - (0. 1 40 N)k
1
A simple gaussmeter for measuring horizontal magnetic fields consists of a
stiff 50-cm wire that hangs from a conducting pivot so that its free end makes contact with a pool of mercury in a dish below. The mercury provides an electrical contact without constraining the movement of the wire. The wire has a mass of 5 g and conducts a current downward. (a) What is the equilibrium angular displacement of the wire from
vertical if the horizontal magnetic field is 0.04 T and the current is 0.20 A? (b) If the
1 38
Chapter
26
current is 20 A and a displacement from vertical of 0.5 mm can be detected for the free end, what is the horizontal magnetic field sensitivity of this gaussmeter?
Picture the Problem The
diagram shows
/ / / / /
the gaussmeter displaced from equilibrium under the influence of the gravitational and magnetic forces acting on it. We can apply the condition for translational equilibrium in the x direction to find the equilibrium
--- ---
angular displacement of the wire from the
vertical. In part (b) we can solve the equation derived in part (a) for B and
.... Ft"---.;��
I / ,--- --- � x / , / , / ,
evaluate this expression for the given data to find the horizontal magnetic field sensitivity of this gaussmeter.
(a) Apply
I Fx = O to the wire to
___
mgsin e -Fcos B 0 =
obtain: Substitute for F and solve for e to obtain:
mgsin B -IeB cosB 0 =
and
(1)
(:;J tan-I [ (0.2A)(0.S m)(0.04 T )] ( kg)(9.81m1s2)
e tan -' =
Substitute numerical values and evaluate e :
B
=
=
(b) Solve equation (1) for B to obtain: For a displacement from vertical of 0.5 mm:
B
=
mg tan B Ie �
B
evaluate B:
1 4'-.66---' 01
tan B sm. B O.S mm 0.001 O.S m and
Substitute numerical values and
O.OOS
=
=
=
0.001 rad
(0.00Skg)(9.81m1s2 )(0. 001 rad) B (20A)(0.Sm) - -' 1 4'-. 91 j.1T 1 =
=
The Magneti c Field
1 39
Mo tio n o f a Point Charge in a Magnetic Field *31
A proton moves in a circular orbit of radius 65 cm perpendicular to a uniform
·
magnetic field of magnihlde 0.75 T. (a) What is the period for this motion? (b) Find the speed of the proton. (c) Find the kinetic energy of the proton. Picture the Problem
We can apply Newton's 2nd law to the orbiting proton to relate its
speed to its radius. We can then use T= 2 m-Iv to find its period. In Part (b) we can use the
relationship between T and v to determine v. In Part (c) we can use its definition to find
the kinetic energy of the proton. (a) Relate the period T of the
motion of the proton to its orbital speed v: Apply
L Fradial = mac to the proton
to obtain:
Solve for vir to obtain:
T = 2nr v
(1)
v2 qvB = m r
v = qB m r
Substitute to obtain:
T = 2mn qB
Substitute numerical values and evaluate T: (b) From equation ( 1 ) we have:
2nr = ----'----"2 n(O.65m) v = -T 87. 4ns = 1 4.67x107 rnIs I
(c) Using its definition, express and evaluate the kinetic energy o f the proton:
=
*37
field
iJ
··
I U .4MeV I A beam of particles with velocity v enters a region of uniform magnetic that makes a small angle B with v . Show that after a particle moves a distance 2Jr
140
Chapter 26
(m/qB)v cos e, measured along the direction of B , the velocity of the particle i s in the same direction as it was when the particle entered the field. Picture the Problem
The particle's velocity has a component VI parallel to B and a
component V2 normal to B .
VI = V
cos e and is constant, whereas V2 = v sin e , being
normal to B , will result in a magnetic force acting on the beam of particles and circular motion perpendicular to B . We can use the relationship between distance, rate, and time and Newton' s 2nd law to express the distance the particle moves in the direction of the field during one period of the motion. (1)
Express the distance moved in the direction of B by the particle during one period: Express the period of the circular motion of the particles in the beam: Apply
I Fradial = mac to a particle
in the beam to obtain:
v2 - m_ qv2 B r 2
qBr v2 - --
Solve for V2:
m
Substitute to obtain:
T
=
2m-
qBr m
Because
VI
becomes:
=
2mn qB
= V cose, equation ( 1 )
The Velocity Selector *41
·
A velocity selector has a magnetic field o f magnitude 0.28 T perpendicular to
an electric field of magnitude 0.46 MV/m. (a) What must the speed of a particle be for the
particle to pass through undeflected? What energy must (b) protons and (c) electrons have to pass through undeflected? Picture the Problem
Suppose that, for positively charged particles, their motion is from
left to right through the velocity selector and the electric field is upward. Then the magnetic force must be downward and the magnetic field out of the page. We can apply
the condition for translational equilibrium to relate v to E and B. In (b) and (c) we can use
The Magnetic Field
141
the de finition o f kinetic energy to find the energies of protons and electrons that pass through the velocity selector undeflected. (a) Apply
I Fy = 0 to the particle or qE - qvB
to obtain:
=
0
E
Solve for v to obtain:
V=-
B
Substitute numerical values and evaluate
v:
(b) Express and evaluate the kinetic energy of protons passing through
V=
0.46 MV/m = 1.64 x 106 m1s 0.28 T
1
K = -L m v2 p 2 P
= 1 1 .67 X 10-27 kg 1 .64 x l 06
)(
(
the velocity selector undeflected:
= 2 .26 x 10-1 5 J x
I
= 14 .0 keV
(c) The kinetic energy of electrons passing through the velocity selector undeflected is given by:
Ke
I
l eV
-L2 me v2
=
1 9 . 11 x 1O-3 \g 1 .64 x 106
)(
= 1 .23 x 10 -1 8 J x = 1 7 .66 eV
mls ')
1 .6 0 x 10-1 9 J
=
(
I
l eV
mls ')
1 .60 x 10-1 9 J
I
Thomson's Measurement of q/m for Electrons and the Mass Spectrometer *43
••
The plates of a Thomson q/m apparatus are 6.0 em long and are separated by
l.2 em. The end of the plates is 30.0 em from the tube screen. The kinetic energy of the
electrons is 2.8 keY. (a) If a potential of 25 V is applied across the deflection plates, by how much will the beam deflect? (b) Find the magnitude of the crossed magnetic field that will allow the beam to pass between the plates undeflected. Picture the Problem
Figure 26- 1 8 is reproduced below. We can express the total
deflection of the electron beam as the sum of the deflections while the beam is in the field between the plates and its deflection while it is in the field-free space. We can, in turn,
1 42
Chapter 26
use constant-acceleration equations to express each of these deflections. The resulting
equation is in terms of Vo and E. We can find Vo from the kinetic energy of the beam and E from the potential difference across the plates and their separation. In part (b) we can equate the electric and magnetic forces acting on an electron to express B in terms of E and Vo .
I �-- xl-I"----·--X2 F
(a) Express the total deflection �y of the electrons:
------
�y = �Yl
+ .0.Y2
"i (1)
where �Y l i s the deflection o f the beam while i t is in the electric field and �Y2 is the deflection
of the beam while it travels along a straight-line path outside the electric field. Use a constant-acceleration equation to express �Yl :
(2)
where �t = X l /VO is the time an electron is in the electric field between the plates.
Apply Newton' s 2nd law to an electron between the plates to obtain: Solve for ay and substitute into
equation (2) to obtain:
a
Y
=
and
qE
m
(3)
Express the vertical deflection �Y2 of the electrons once they are out of the electric field:
(4)
The Magnetic Field
Use a constant-acceleration equation to find the vertical speed of an electron as it leaves the electric field: Substitute in equation (4) to obtain:
Substitute equations (3) and (5) in equation ( 1 ) to obtain: or A
LlY =
Use the definition of kinetic energy to find the speed of the electrons:
K
(2 )
qExl XI 2 - + X2 mvo
--
(6)
= ..L2 mv20
9.11 10-3 1 kg = 3.14 x107 m1s X
Express the electric field between the plates in tenns of their potential
E = dV
difference: Substitute numerical values and evaluate E:
E
25 V = 2.08kV/m = Vd 1.2cm =
Substitute numerical values in equation (6) and evaluate Lly: 1'1
y
=
1.60x10-19 C (2.08kV/m)(6cm) ( 6cm + 30 c 9.11x 10-31 kg (31AMmlsY 2
(b) Because the electrons are
Fmag
deflected upward, the electric field
or
must be downward and the magnetic field upward. Apply electron to obtain:
L Fy = O to an
-
F.lec =
qvB = qE
0
m
)
=
1 7 . 34mm I
1 43
144
Ch apter
26
Solve for B:
B=
Substitute numerical values and
evaluate B: *46
B
E v
2 . 08k\/m I = 66.2 ,uT I 3 . 1 4 x l O m/s
=
··
A beam of 6Li and 7Li ions passes through a velocity selector and enters a magnetic spectrometer. If the diameter of the orbit of the 6Li ions is 7 15 cm, what is the diameter of the orbit for Li ions?
Picture the Problem
We can apply Newton's 2nd law to an ion in the magnetic field of
the spectrometer to relate the diameter of its orbit to its charge, mass, velocity, and the
magnetic field. If we assume that the velocity is the same for the two ions, we can then express the ratio of the two diameters as the ratio of the masses of the ions and solve for
the diameter of the orbit o eLi. Apply
L Fradial = mac to an ion in
the field of the spectrometer: Solve for r to obtain:
2 v B = qv m r mv r=qB 2 mv
Express the diameter of the orbit:
d=
Express the diameters of the orbits for 6Li and 7Li:
d6 -
Assume that the velocities of the two ions are the same and divide the
2nd of these diameters by the first to
obtain:
d7
d6
=
qB
2 m6v qB
and d7
2m 7 v
m7 qB 2 m6v - m6 qB
Solve for and evaluate d7 :
The Cyclotron *49
··
A cyclotron for accelerating protons has a magnetic field of 1 .4 T and a
radius of 0.7 m. (a) What is the cyclotron frequency? (b) Find the maximum energy of
The Magnetic Field
145
the protons when they emerge. (c) How will your answers change if deuterons, which have the same charge but twice the mass, are used instead of protons? Picture the Problem We can express the cyclotron frequency in tenns of the maximum orbital radius and speed of the protons/deuterons. By applying Newton's 2nd law, we can
relate the radius of the particle's orbit to its speed and, hence, express the cyclotron frequency as a function of the particle's mass and charge and the cyclotron's magnetic
field. In part (b) we can use the definition of kinetic energy and their maximum speed to find the maximum energy of the emerging protons. (a) Express the cyclotron frequency in tenns of the proton's orbital speed
f-
and radius: Apply
I Fradial
=
mac to a proton in
the magnetic field of the cyclotron: Solve for r to obtain:
Substitute to obtain:
Substitute numerical values and evaluate!
(b) Express the maximum kinetic energy of a proton: Solve equation (1) for Vrnax to obtain:
Substitute to obtain:
1
1
_ ----
T
2m/v
-_
v2 qvB = m
2w
(1)
r
r=
v
--
mv
-
f-
qB qBv
qB
_ -----
2mnv
2mn
(2)
146
Chapter 26
Substitute numerical values and
K=
evaluate K:
1.((1.60X 1 Y(1.4 J( 1.67 kg 2
TY
0-19 C
x 10-27
= 7.36x10-12 Jx
1
= 46.0M eV
0.7 mY
leV
1.60xl0-19 J
I
heuterons = 1- fprotons = 1 10.7MHz I
(c) From equation (2) we see that
doubling m halves f
Kdeuterons = 1- Kprotons = I 23.0MeV I
From equation (3) we see that doubling m halves K:
Torques on Current Loops and Magnets
*55
•
A current-carrying wire is bent into the shape of a square of edge-length L =
6 em and is placed in the xy plane. It carries a current 1= 2.5 A. What is the magnitude of
the torque on the wire if there is a uniform magnetic field of 0.3 and (b) in the x direction?
Picture the Problem
We can use i=
orientations of the magnetic field. Express the torque acting on the coil: Express the magnetic moment of the coil:
(a) Evaluate i for B in the z direction: (b) Evaluate i for B in the x
direction:
jl x
B
T
(a) in the z direction
to find the torque on the coil in the two
i=
jl x B
jl = ±IA k = ±IL2 k
k k = ±IL2B (k x k) = 0
i= ±IL2 xB
k t (k xi) = ±(2.5 A)(0.06m )2 (0.3T )J = I ± (2.70x 10-3 N . m)J I
i = ±IL2 xB = ±IL2B
The Magnetic F ield 147 Magnetic Moments A small magnet of length 6.8 cm is placed at an angle of 60° to the direction *60 •• of a uniform magnetic field of magni tude 0.04 T. The observed torque has a magnitude of
0 . 10 N·m. Find the magnetic moment of the magnet. Picture the Problem
Because the small magnet can be modeled as a magnetic dipole; we
can use the equation for the torque on a current loop to find its magnetic moment. Express the magnitude of the torque
T
= jlBsin ()
acting on the magnet: Solve for J.1 to obtain:
T J..i. = - B sin ()
Substitute numerical values and
evaluate J.1:
*64
•••
J..i. =
O.ION·m = 1 2.89A . m2 (0.04 T ) sin60°
1
A hollow cylinder has length L and inner and outer radii Ri and Ro,
respectively (Figure 26-3 8). The cylinder carries a uniform charge density p. Derive an
expression for the magnetic moment as a function of (j), the angular velocity o f rotation of
the cylinder about its axis.
Figure
26-38 Problem 64
Picture the Problem
We can express the magnetic moment of an element of charge dq
in a cylinder of length L, radius r, and thickness dr, relate this charge to the length,
radius, and thickness of the cylinder, express the current due to this rotating charge, substitute for A and dI in our expression for f.1. and then integrate to complete our
derivation for the magnetic moment of the rotating cylinder as a function of its angular velocity. Express the magnetic moment of an
dJ..i. = Ad!
148
Chapter
26
element of charge dq in a cylinder
where
m·2.
of length L, radi us r, and thic kness
A
Relate the charge dq in the cylinder
dq = 2nLprdr
dr:
=
to the length of the cylinder, its radius, and thickness: Express the current due to this rotating charge:
dI
=
�
27r
dq
=
�
27r
(2nLprdr) LOJprdr =
Substitute to obtain: Integrate r from Ri to Ro to obtain:
Ro
fl LOJp7r Jr3dr =
Ri
=
I *LOJp7r(R� -Rt) I
*69 ••• A unifonn disk of mass m, radius R, and surface charge (5 rotates about its center with angular velocity OJ as shown in Figure 26-40. A unifonn magnetic field of
magnitude B threads the disk, making an angle e with respect to the rotation axis of the disk. Calculate (a ) the net torque acting on the disk and (b ) the precession frequency of the disk in the magnetic field. (See pages 3 1 6-3 1 7 for a discussion of precession. ) I�otilti()n nxis �
Figure
26-40 Problem 69
Picture the Problem We can use its definition to express the torque acting on the disk and the defmition of the precession frequency to tind the precession frequency of the disk.
(a ) The magnitude of the net torque acting on the disk is:
From example 26 -1 1 :
r
=
flBsinB
where JI. is the magnetic moment of the disk.
1
4
fl =-4 7rO"r OJ
The Magnetic Field 149 S ubstitute for fl in the expression for T to obtain:
'f
= 41
-
.D SIll · e 1[01' 4 UJn
(b ) The p recession frequency Dis equal to the ratio of the to rque divided by the spin angular momentum:
I =-21 mr 2
For a solid disk, the moment of inertia is given by: S ubstitute for
r
1[Qr2B . e sm
and I to obtain:
---
2m
Remarks: It's interesting that the precession frequency is independent of OJ.
The Hall Effect ••
*72
A coppe r s trip (n
=
8.47xl022 electrons per cubic centimete r) 2-cm wide and
OJ-cm thick is used to measure the magnitudes of unknown magnetic fields that are
perpendicular to the strip. Find the magnitude of B when 1= 20 A and the Hall voltage is
(a) 2 00 flY, (b ) 5 .25 flY, and (c) .
Picture the Problem
8.00 ;N.
We can use
VH =
vd
1= nqvdA to eliminate the drift velocity
Bw to express B in te rms of VH and
Vd
and de rive an expression for B in terms of VH,
n, and t. Relate the Hall voltage to the drift velocity and the magnetic field: Solve for B to obtain:
Exp ress the current in the metal strip in terms of the drift velocity of the electrons: Solve for Vd to obtain:
I Vd = nqA
--
150
Chapter 26
Substitute and simplify to obtain:
B
VH
nqAVH
nqwtVH
-- w
Iw
Iw
=
I
nqA
=
nqt I
VH
Substitute numerical values and simplify to obtain:
(a) Evaluate B for VH = 2.00 Jl.V:
B = (6.7 8 X 105 s/m2 )(2.00 .uV)
\
= 1 36
(b) Evaluate B for VH = 5 .25 Jl.V:
.
T\
B = (6.7 8 x 105 s/m2 )(5.25 .uV)
\
= 3 . 56T (c) Evaluate B for VH = 8.00 Jl.V:
B
I
(6.7 8 x 105 s/m2 )(8.00 .uV) =1 5.42T \
=
3 3 Aluminum has a density of 2.7x l0 kg/m and a molar mass of 3 27 g/mol. The Hall coefficient of aluminum is R = -0.3 x l O-1O m /C. (See Problem 74 for
*75
··
the definition of R. ) Find the number of conduction electrons per aluminum atom. Picture the Problem
We can determine the number of conduction electrons per atom
from the quotient of the number density of charge carriers and the number of charge carriers per unit volume. Let the width of a slab of aluminum be wand its thickness t. We can use the definition of the Hall electric field in the slab, the expression for the Hall
voltage across it, and the definition of current density to find n in terms of R and q and na = pNA M , to express n•.
/
(1)
Express the number of electrons per atom N:
where n is the number density of charge carriers and n. is the number of atoms per unit volume. From the definition of the Hall coefficient we have:
15 1
The Magnetic Field Express the Hall electric field in t he
E = y
slab: Express the current densi ty in the
VH W
I
Jx- =- = nqvd wt
slab:
VH W
Substitute to obtain: R=
nqvdBz
VH nqvdwBz
Express the Hall voltage in terms of Vd, B,
and w:
Substitute and simplify to obtain:
Solve for and evaluate
R=
vdBzw nqvdwBz
=
1_ _
nq
1
n:
n=Rq
Express the number of atoms na per unit volume: Substitute equations (2) and (3 ) in e quation (1 ) to obtain:
(2)
N na =p�
(3 )
M
N=
M
qRpNA
Substitute numerical values and evaluate N: N
27g/mol - (-1.60 x 10-19 C)(- 03 x 10-10 m3/C)(2.7 x 103 kg/m3 )(6.02 x 1023 atoms/mol) = 1 3.46 1 _
General Problems
*79
··
A particle of mass m and charge q enters a region where there is a uniform
magnetic field
B along the
x
axis. The initial velocity of the particle is v=vo
so the particle moves in a helix. (a) Show that the radius of the helix is
r=
)
+ VOy
)
,
mvoy/qB.
(b) Show that the particle takes a time t = 2mn/qB to make one orbit around the helix.
F = qv x B to show that motion of the particle in the x direction is not affected by the magnetic field. The application of Newton's 2nd law to Picture the Problem
We can use
motion of the particle in yz plane will lead us to the result that r = mVOy /qB. By expressing
the period of the motion in terms of VOy we can show that the time for one complete orbit
15 2
Chapter 26
around the helix is t = 27rrI'l/qB.
(a) Express the magnetic force acting on the particle: Substitute for v and B and simplify
to obtain:
i=q (vo} + VOy} )x Bi
= qvoxB (i xi )+ qVoyB (J xi ) =0 qVoyBk=-qvoyBk A
A
-
i.e., the motion in the direction of the magnetic field (the x direction) is not affected by the field.
L FradiaJ =mac to the motion
Apply
of the particle in the plane
perpendicular to i (i.e., the yz
2
Voy qvOyB=mr
(1 )
plane):
Solve for r:
2nr t=VOy
(b) Relate the time for one orbit around the helix to the particle's orbital speed: Solve equation (1 ) for
qBr Voy =-m
VOy:
Znr zmn t qBr - qB m _
Substitute and simplify to obtain:
-
*80
··
-
I I
A metal crossbar of mass m rides on a pair of long, horizontal conducting
rails separated by a distance L and connected to a device that supplies constant current I
to the circuit, as shown in Figure 26-42. A uniform magnetic field B is established, as
shown. (a) If there is no friction and the bar starts from rest at t = 0, show that at time t the bar has velocity (b) In which direction will the bar move? (c) If the
V=(BIL/m)t
.
coefficient of static friction is f.is, find the minimum field B necessary to start the bar movmg.
The Magnetic Field 153 y
x
Bin
._---
x
x x
X
X
x
x
x
x
I
t
x
x x
Figure
x
x
x
26-42 Problems 80 and 81
Picture the Problem
We can use a constant-acceleration equation to relate the velocity
of the crossbar to its acceleration and Newton's 2nd law to express the acceleration of the
=
crossbar in terms of the magnetic force acting on it. We can determine the direction of -
motion of the crossbar using a right-hand rule or, equivalently, by applying F
-
-
If!. x B .
We can find the minimum field B necessary to start the bar moving by applying a condition for static equilibrium to it.
(a) Using a constant-acceleration equation, express the velocity of the bar as a function of its acceleration and the time it has been in motion: Use Newton's 2nd law to express the acceleration of the rail:
v = va +at
V=
or, because Va = 0, at
F m
a =where F is the magnitude of the magnetic
force acting in the direction of the crossbar's motion. Substitute to obtain:
Express the magnetic force acting on the current-carrying crossbar: Substitute to obtain:
(b)
F m
v = -t
F=ILB V=I I� I t
Apply to conclude that the magnetic force is to the right and so the motion of the crossbar will also be to the right.
26 (c) Apply IF. Oto the c rossbar: 154
Chapter
=
ILBmin - Is,max or
ILBmi n Solve for BOlin to obtain:
Bmin
-
-
=
0
f.1smg=0
I f.1ILsmg 1
A stiff, straight horizontal wire of length 25 cm and mass 20 g is supported *85 ••• by electrical contacts at its ends, but is otherwise free to move vertically upward. The wire is in a uniform, horizontal magnetic field of magnitude
0.4 T perpendicular to the
wire. A switch connecting the wire to a battery is closed and the wire flies upward, rising to a maximum height h. The battery delivers a total charge of 2 C during the short time it makes contact with the wire. Find the height h. Picture the Problem
We can use a constant-acceleration e quation to express the height
to which the wire rises in terms of its initial speed and the acceleration due to gravity. We can then use the impulse-change in momentum equation to express the initial speed of the wire in terms of the impulsive magnetic force acting on it. Finally, we can use the definition of current to relate the charge delivered by the battery to the time during which the impulsive force acts. Using a constant-acceleration equation, relate the height h to the initial and final speeds and the acceleration of the wire: Solve for h:
Use the impulse-momentum e quation to relate the change in momentum of the wire to the
v2 =v� +2ayh or, because
v=
o=v� -2gh
=
g,
2
(1 )
!::.P. =F!::.t. or Pr - Pi=FM and, because Pi 0, mvo =FM =
F=IRB
force acting on the wire: Substitute to obtain:
ay
h=� 2g
impulsive force accelerating it: Express the impulsive (magnetic)
0 and
mvo
=
IRB!::..t
The Magnetic Field 155 Solve for Va and substitute in equation (1):
Use the definition of current to relate the charge delivered by the
6Q=IM
battery to the time during which it delivers the current: Substitute to obtain:
Substitute numerical values and evaluate h:
h
[(O = .25m){0.4T){2C)2r =i5.lOm i 2{O.02kgY(9.81m1s )
*88 ••• The special theory of relativity tells us that a particle's mass depends on its speed through the formula:
m(v) =
m = (v)m g 2 y v 1
c2
2 2
where m is the particle's mass and y(v)= 11 �1- (v / c ). (a) Taking into account the special theory o/relativity, what is the radius and period of a particle's orbit if it has speed v and is moving in a magnetic field with magnitude B that is perpendicular to the
jj).
direction of the velocity? Assume the force on the particle is given by F =q(v x The particle has mass m and charge q. (b) Using a spreadsheet program, make graphs of the radius and period of the orbit of an electron in a lO-T magnetic field versus rev) for speeds between v O.1c and v = O.999c. Use a logarithmic scale to display rev). =
We can apply Newton's 2nd law to the particle to derive an expression for the radius of its orbit and then express its period in terms of its orbital speed and radius.
Picture the Problem
(a) Because B is perpendicular to v , the magnitude of force on the particle is given by: Apply
IF= ma to the orbiting
particle to obtain: Solve for r:
F=qvB v2 = v m v2 qvB = m{v)y{ ) r
r
�
I y{qBv)mv I
r
Chapter 26
156
The period T of the particle's motion is related to the radius rof its orbit and its orbital speed v:
2m
T= v
Substitute for rand simplify to obtain:
(b)
T
�
12nr(v)m
qB
I
A spreadsheet program to calculate rand T as functions of In(y) follows.
The formulas used to calculate the quantities in the columns are given in the table. Cell Bl B2 B3 B4 A7 A8 B7 C7 D7
ContentIFormula 9.11E-31 1.60E-19 10 3.00E+08 0.100 0.101 1/SQRT(1 - (A7Y'2) LN(B7) B7*$B$1*A7*$B$4/($B$2*$B$3)
Algebraic Form
E7
D7*10"8 (2*PIO*A7*$B$1I($B$2*$B$3»*10"12
106r
F7
1 2 3 4 5 6" 7 8 9 .., 10 '11 903 904 905 906
A
B 9.11E-31 1.60E-19 10 3.00E+08
C
"..
m= e= B= c=
vic 0.100 0.101 0.102 0.103 0.104 0.996 0.997 0.998 0.999
gamma 1.0050 1.0051 1.0052 1.0053 1.0055
11.1915 12.9196 15.8193 22.3663
kg C T rn/s In(gamma) 0.005 0.005 0.005 0.005 0.005 2.415 2.559 2.761 3.108
,.."
m e
B
c vic vic + 0.001 y
In(y) ymv
qB
2nym
qB
xlO12
D
'" .,E .J. .','
r
r (microns) 17.2 17.3 17,5 17.7 17.9
1.72E-05 1.73E-05 1.75E-05 1.77E-05 1.79E-05
1.90E-03 2.20E-03 2.70E-03 3.82E-03
1904.0 2200.2 2696.7 3816.6
P',
F
"
T
(ps)
0.358 0.361 0.365 0.368 0.372
3.563 3.567 3.570 3.574
The following graph of r as a function of In(y) was plotted using the data in columns C and E.
The Magnetic Field 157 4000
1" ,"" .. "0 ,�-,'" ..
3000
r
--. '" c::
-
I,
1
c'
i'_,:""._,,: . : �'-I
2000 1:-- - ' T; "lIT .1 "'. gl2
.�
_
'-
1000
I";,. "
.:1:::_
o��" 0.5 0.0
k-I-I.
'I -
'
l
1.0
1.5
2.0
In(gamma)
2.5
3.0
3.5
The following graph of T as a function of In(r) was plotted using the data in columns C and F. 4 ,"
3 --. '"
�2
h
o 0.0
0.5
1.0
1.5
2.0
In (gamma)
2.5
3.0
3.5
Chapter 27 Sources of the Magnetic Field Conceptual Problems Compare the directions of the electric force and the magnetic force between
*1
two positive charges, which move along parallel paths (a) in the same direction, and (b)
in opposite directions. Picture the Problem
The electric forces are described by Coulomb's law and the laws of
attraction and repulsion of charges and are independent of the fact the charges are
moving. The magnetic interaction is, on the other hand, dependent on the motion of the
charges. Each moving charge constitutes a current that creates a magnet field at the location of the other charge.
(a) The electric forces are repulsive; the magnetic forces are attractive (the two charges
moving in the same direction act like two currents in the same direction).
(b) The electric forces are again repulsive; the magnetic forces are also repulsive.
*6
•
A wire carries an electrical current straight up. What is the direction of the
magnetic field due to the wire a distance of 2 m north of the wire? ( a) North (b) East (c)
West (d) South (e) Upward Determine the Concept
Applying the right-hand rule to the wire to the left we see that
the magnetic field due to the current points to west at all points north of the wire.
I (c) is correct. I
*9 Make a field-line sketch of the magnetic field due to the currents in the pair of coaxial coils (Figure 27-43). The currents in the coils have the same magnitude but are opposite in direction in each coil. The field-line sketch is shown below. An assumed direction for the current in the coils is shown in the diagram. Note that the field lines never begin or end and that they do not touch or cross each other. Because there are an uncountable infinity of lines, only a representative few have been shown.
Picture th e Problem
159
160
Chapter 27
*12
·
If the magnetic susceptibility is positive, (a) paramagnetic effects or
ferromagnetic effects must be greater than diamagnetic effects.
(b)
diamagnetic effects
must be greater than paramagnetic effects. (c) diamagnetic effects must be greater than ferromagnetic effects. (d) ferromagnetic effects must be greater than paramagnetic effects. (e) paramagnetic effects must be greater than ferromagnetic effects. Determine the Concept
-
The magnetic susceptibility Xm is defined by the
-
equation M = X m � , where M is the magnetization vector and
B
flo
-
Bapp is the applied
magnetic field. For paramagnetic materials, Xm is a small positive number that depends
on temperature, whereas for diamagnetic materials, it is a small negative constant independent of temperature. *18
·
I (a) is correct. I
When a current is passed through the wire in Figure 27-44, will the wire tend
to bunch up or form a circle?
I
Figure
27-44 Problem 18
Determine the Concept
The force per unit length experienced by each segment of the
wire, due to the currents in the other segments of the wire, will be equal. These equal forces will result in the wire tending to form a circle.
Sources of the Magnetic Field 161 Estimation and Approximation *21 ·· Estimate the transient magnetic field 100 m away fi·om a lightning bolt if a charge of about 30 C is transferred from cloud to ground and the average velocity of the charges is 106 rn/s.
Picture the Problem We can model the lightning bolt as a current in a long wire and use the expression for the magnetic field due to such a current to estimate the transient magnetic field 100 m from the Jightning bolt.
B 4Jr f.-lo 21 r where r is the distance from the wire.
The magnetic field due to the current in a long, straight wire is:
Assuming that the height of the cloud is 1 lan, the charge transfer will take place in roughly 10-3 s and the current associated with this discharge is:
Substitute numerical values and evaluate
=
B:
7 2 x l04 A) B 4Jrx lO-4JrN/A 2(3100m 1 60 0f.-lT I =
=
.
*22 ·· The rotating disk of Problem 125 (page 896) can be used as a model for the magnetic field due to a sunspot. lf the sunspot radius is approximately 107 m rotating at an angular velocity of about 10-2 rad/s, calculate the total charge Q on the sunspot needed to create a magnetic field of order 0.1 T at the center of the sunspot. What is the electrical field magnitude just above the center of the sunspot due to this charge? A rotating disk with total charge Q and surface charge density a is shown in the diagram. We can find Q by deriving an expression for the magnetic field at the center of the disk due to its rotation. We'll use Ampere's law to express the field at the center of the disk due to the element of current dI and then integrate over to find Picture the Problem
B
dB r
B.
Applying Ampere's law to a
circular current loop of radius obtain : The
r we
B field at the center of an annular ring dB flo d1 r and thickness 2r
on a rotating disk of radius
dr is:
=
(1)
162
Chapter 27
If o-represents the surface charge density, then the current in the annular ring is given by:
Because
T=
Substitute for
obtain:
dI = o-(2n r) dr' T
where
(J'=
2n :
dI=(J'(j) rdr
dI in equation (1) to
dB = f.1o (j) rdr= f.1o(J'(j) dr 2 2r (J'
-
(j)
Integrate from r = ° to R to obtain: Substitution for o-yie1ds:
0
--=-
JrR2
R
B= f.1o(J'(j) fdr= f.1o(J'(j)R 2 2 0
B
�
p,(�2 )aill
2n R
Solve for Q to obtain: Substitute numerical values and evaluate Q:
The electric field above the sunspot is given by:
Substitute numerical values and evaluate E:
5.00 x 1014C 2n(8.85 x 10-12 C2 IN· m2 )(107 m Y =I 90.0 GN/C I
E=
The Magnetic Field of Moving Point Charges
*27
••
Two equal charges q located at (0, 0, 0) and (0, b, 0) at time zero are moving
with speed v in the positive x direction (v« c). Find the ratio of the magnitudes of the magnetic force and the electrostatic force on each charge.
Picture the Problem
We can find the ratio of the magnitudes of the magnetic and
electrostatic forces by using the expression for the magnetic field of a moving charge and Coulomb's law. Note that v and r, where r is the vector from one charge to the other,
Sources of the Magnetic Field
163
are at right angles. The field B due to the charge at the origin at the location (0, b, 0) is
perpendicular to
v
and r .
Express the magnitude of the
magnetic force on the moving
charge at (0, b, 0):
FB
=
qvB =
2
2
f.10 � b2
47r
and, applying the right hand rule, we find
that the direction of the force is toward the
charge at the origin; i.e., the magnetic force
between the two moving charges is Express the magnitude of the
attractive.
repulsive electrostatic interaction
between the two charges:
Express the ratio of FB to FE and
simplify to obtain:
47rco
b2
where c is the speed of light in a vacuum. The Magnetic Field of Currents: The Biot-Savart Law
* 30
For the current element in Problem 28, find the magnitude of
•
4
dB
and
indicate its direction on a diagram at (a) x = 2 m, Y = m, Z = ° and (b) x =2 m,
y=0,z=4m.
Picture the Problem
(dB 47r Idre2x =
f.10
f
We can substitute for v and q in the Biot-Savart relationship
) , evaluate r and f for the given points, and substitute to find
Express the Biot-Savart law for the given current element:
4
f
dB 47r Ide x / 2 x (10-7 N/A2 2 A){2mm)k r ( 0.400nT ·m2 )-kr2x{2m)i {4m)J, 2.J5m, =
=
f.1 0
r2
f
=
(a) Find r and f for the point whose coordinates are (2 m, m, 0):
dB.
r=
r
=
+
f
164 Chapter 27 and �
r =
Evaluate
dB
at (2 m, 4 m, 0):
1�I + 152-: I = + 15 15 15 2 2 2
�
2�) kx (1 -�i+-j
dii{Zm,4m,0)� (0.400nT .m') (215mr
4
J
J
-:
�
�
1- {17.9pT)i + {8.94pT)j 1 y
The diagram is shown to the right :
d8
� (b) Find r and r for the point whose coordinates are (2 m, 0, m):
4
x
=(215m)i + (4m)k, r=2 m, = 2152�i+ 2154 k= 151 i+ 152 k
r
and
�
r
Evaluate
dB
�
�
at (2 m, 0, 4 m):
�kx (1 � 2�k ) dii{zm,0,4m)� (0.400nT .m') (215mr 1{8.94pT)j 1 -
i
+
-
�
The diagram is shown to the right:
y
B Due to a Current Loop *32
·
A single-tum circular loop of radius 10.0 cm is to produce a field at its center
that will just cancel the earth's magnetic field at the equator, which is 0.7 G directed
north. Find the current in the loop and make a sketch showing the orientation of the loop and the current.
1 65
Sources of the Magnetic Field .
Picture the Problem We can solve Bx
=
/-L 2n:R21 0 _ 4n (x- +R- )W2 ?
?
.
for I wIth x
=
.
0 and substltute
the earth's magnetic field at the equator to find the current in the loop that would produce a magnetic field equal to that of the earth. Express B on the axis of the current loop:
Solve for I with x
=
0:
1=
4
7r
/-Lo
�
2n
Bx
Substitute numerical values and evaluate I:
North
The orientation of the loop and current is shown in the sketch:
*38 ••• Anti-Helmholtz coils are used in many physics applications, such as laser cooling and trapping, where a spatially inhomogeneous field with a uniform gradient is desired. These coils have the same construction as a Helmholtz coil, except that the currents flow in opposite directions, so that the axial fields subtract, and the coil separation is r .J3 rather than r. Graph the magnetic field as a function of x, the axial distance from the center of the coils, for an anti-Helmholtz coil using the same parameters as in Problem 36. Picture t h e Problem
centered at
x
Let the origin be midway between the coils so that one of them is
= -r .J3 /2
and the other is centered at
denote the coil centered at x = -r .J3 x
=
r
x
= r .J3
/2.
Let the numeral 1
12 and the numeral 2 the coil centered at
.J3 12 . We can express the magnetic field in the region between the coils as the
difference of the magnetic fields B, and B2 due to the two coils. Express the magnetic field on the x axis due to the coil centered at x
= -r .J3 12:
where N is the number of turns.
1 66
Chapter
27
Express the magnetic field on the x
axis due to the coil centered at x=
rJ3
J
2J 7]3/2 Nr o f1 2 [[rJ3 2 +r-
B (X ) =
12:
2
-- - x 2
Subtract these equations to express the total magnetic field along the x axis:
The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Algebraic Form
Cell Bl B2 B3 B3 B5
Formula/Content 1.26xl0-6 0.30 250 15 0.5*$B$1*$B$3*($B$2/\2)*$B$4
A8 B8
-0.30 $B$5*(($B$2*SQRT(3)/2+A8Y2 +$B$2/\2Y( -3/2)
C8
$B$5* (($B$2*SQRT(3)/2-A8Y2 +$B$2/\2Y(-3/2)
D8
10/\4*(B8-C8)
·i� '" 'B '" .,'1��i;;,A · " ,� r· I" \; : : � mu 1.26E-06 0= ';�� i '
li�I2�
1;';¥:3;-;'� 4', .
r=
N= 1= 5,/t, Coeff= "" (,f '
x ·7 , .. 8, -0.30
0.3 250 15 2.13E-04 B 1 5.63E-03
"'"
J.lo r
N
I
Coeff =
J-loNr2I 2
p'Nr'I[[r� --+xJ' +r' r [[---x p,Nr'I r� ' 'f" +r 2 2 -r
2
2
J
Bx =B1 -B2
:C. ';:..: ,
N/A/\2 m turns A
B 2 1.34E-03
f)
'�
B(x) 68.4
Sources of the Magnetic Field 1 67 9 10 'Il 12 1 '16';1
-0.29 -0.28 -0.27 -0.26 -0.25 -0.24 -0.23
w/�
" "
, ' /'
14'· 1·�.1
0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30
6 ��� 62 63'3 64.: "\65,�:1 66 I'P67r g�,
5.86E-03 6.08E-03 6.30E-03 6.52E-03 6.72E-03 6.92E-03 7.10E-03 �"
1.87E-03 1.78E-03 1.70E-03 1.62E-03 1.55E-03 1.48E-03 1.4lE-03 1.34E-03
1.41E-03 1.48E-03 1.55E-03 1.62E-03 1.70E-03 1.78E-03 1.87E-03
.:;; .
.,
68.9 69.2 69.2 68.9 68.4 67.5 66.4
' ""
7.10E-03 6.92E-03 6.72E-03 6.52E-03 6.30E-03 6.08E-03 5.86E-03 5.63E-03
..
-66.4 -67.5 -68.4 -68.9 -69.2 -69.2 -68.9 -68.4
The following graph of Bx as a function of x was plotted using the data in the above table. 60 40 20
S
<:>:l
0 -20 -40 -60
-0.2
-0.3
-0.1
0.1
0.0 x
(m)
0.2
0.3
Straight-Line Current Segments
*43
.
27-45 are in the negative x direction, find -3 cm, (b) y = 0, (c)y +3 cm, and (d)y = +9 cm.
If the currents in Figure
points on the y axis at (a)y Picture the Problem
=
=
B at the
Let +denote the wire (and current) aty = +6 cm and - the wire (and
current) aty = -6 cm. We can use
JL 21 0 - to find the magnetic field due to each of B = 4n R _
the current carrying wires and superimpose the magnetic fields due to the currents in the
wires to find B at the given points on they axis. We can apply the right-hand rule to find
the direction of each of the fields and, hence, of B .
168
Chapter 27
(a) Express the resultant magnetic field aty = -3 cm:
jj(- 3em) jjJ-3em)+ jjJ- 3em) =
Find the magnitudes of the magneti c fields at y = -3 cm due to each wire:
=
44.4jlT
and
(20A) B_(-3em) = (10-7 T . m1Af0.03m
Apply the right-hand rule to find the -
directions of
B
+
-
and B
_
:
jj+ (-3em) (44.4jlT)k =
and
jj_ (-3em) -(13 3jlT )k =
jj(- 3em) = (44.4jlT)k - (133jlT)k 1- (88.6jlT)k 1
Substitute to obtain:
=
(b) Express the resultant magnetic field at y = Because
0:
jjJO)
=
i
-
t
(0):
jj(O)=�
(c) Proceed as in (a) to obtain:
B+ (3em) = (133jlT)k, jjJ3em) -(44.4jlT)k, =
and
B(3em) (133jlT)k - (44.4jlT)k 1- (88.6jlT)k 1 =
=
(d) Proceed as in (a) withy to obtain:
=
9 cm
jj+ (9em) -(133jlT)k, jj_(gem) -(26.7jlT)k, =
=
and
B(9em) -(133jlT)k - {26.7 jlT)k 1-{160jlT)k 1 =
=
Sources of the Magnetic Field *52
1 69
Three long, parallel strai ght wires pass tlu-ough the corners of an equilateral
••
triangle of sides 10 cm, as shown i n Figure
27-47, where a dot means that the current is
out of the paper and a cross means that the current is i nto the paper. If each current is 15 A, find (a ) the force per unit length on the upper wire and (b) the magnetic field
B at the
upper wire due to the two lower wires.
10 Figure
em
27-47 Problems 5 2 a nd 53
Picture t h e Problem
Note that the forces on the upper wire are away from a nd directed
along the lines to the lower wire and that their horizontal components cancel. We can use
F = 2 f.10 1 2 to find the resultant force in the upward direction (the y direction) I!
4n-
R
acting o n the top wire. In part (b) we can use the right-hand rule to determine the
directions of the magnetic fields at the upper wire due to the currents i n the two lower .
WIres a nd use
. f.1 21 to find the magnItude B = _0 of the resultant field due to these 4n- R -
currents . (a ) Express the force per unit length each of the lower wires exerts on the upper WIre:
F
_
--
I!
2
f.1 0
12
--
4n-
R
Noting that the horizontal compone nts add up to zero, express the net upward force per u nit length on the upper wire:
Substitute numerical values and evaluate
I FI!
�:
IFI!y = 4(1O-7 T . m1 A )(150.1AmY cos30° = 17.79 x 1 0-4 N/m I
170 Chapter 27 (b) Noting, from the geometry of the wires, the magnetic field vectors
B=
I'
21
�
2_0 -cos300i 4/'r R
both are at an angle of 30° with the horizontal and that their y
components cancel, express the
resultant magnetic field: Substitute numerical values and evaluate
B:
B 2� 0-7 T . m/A) 2O(15.lmA) cos 30° =
= *57
1 52.0 I'T 1
Four long, straight parallel wires each carry current 1. In a plane
••
perpendicular to the wires, the wires are at the comers of a square of side a. Find the force per unit length on one of the wires if (a) all the currents are in the same direction and (b) t he currents in the wires at adjacent comers are oppositely directed. Picture the Problem
Choose a coordinate system with its ori gin at the lower left-hand
comer of the square, the positive x axis to the right and the positive y axis upward. Let
the numeral 1 denote the wire and current in the upper left-hand comer of the square , the numeral
2 the wire and current in the lower left-hand comer (at the origin) of the square,
and the numeral 3 the wire and current in the lower ri ght-hand comer of the square. We can use
B
=
1'0
4Jr
21 R
and the right-hand rule to find the magnitude and direction of the
magnetic field at, say, the upper right-hand comer due to each of the currents, superimpose these fields to find the resultant field, and then use F =
HB to find the
force per unit length on the wire.
(a) Express the resultant magnetic
(1 )
field at the upper right-hand comer:
�B'
When all the currents are into the paper their magnetic fields at the upper right-hand comer are as shown to the right: Express the magnetic field due to the current 1\:
jj
__�
\-
l
' B\
21 � 4/'r a J
Sources of the Magnetic Field Express the magnetic field due to
the curre nt 12:
Express the magnetic field due to the curre nt 13: Substitute in equation (1) and simplify to obtain:
B-
/-Lo 21 -: /-Lo 21 ( -: ) +--1 /-Lo 21 = -/-Lo 21 4n a 4n 2a 4n a 4n a /-Lo 21 +- 1+ - -- ] - 3/-Lo1 [1-]-:] --4n a 2 2 4mz =---]+-- 1-] �
_
[(1 1 ) ( 1 1 ) ] �
Using the expression for the magnetic force on a current carrying wire, express the force per
-:
�
_
--
( -: 1 ( 2
-: )
�
F= Bl
(2)
e
right-hand comer:
F
-=
e
3/-Lo12 [1-]-:] �
4mz
and
: (�J ( 3:: J +
�
'
3J2/-LoI2 4mz
(b) When the current in the upper righ�-hand co mer of the square is out of the page, and the currents in the wires at adjacent comers are oppositely directed, the magnetic fields at the upper right-hand are as shown to the right:
�
�
unit length on the wire at the upper
Substitute to obtain:
8,
)
-]+- 1-] +1
\-., ., I
17 1
172
Chapter 27
Express the magnetic field at the
B-2
upper right-hand comer due to the current 12:
21 o( � ) = fl o -COS 45 - I +} 47r a 12 flo 21 (-�l +}-: ) = -:
-
--
47r 2a
-
-
Using BJ and B3 from (a), substitute in equation (1) and simplify to obtain:
flo 21
flo 21 -: = flo 21 flo 21 (-I-: +},: ) +--1
B =---}+--
-
':
[(1 ) ( ) ,:]
47r a flo 21 = 47r a --
--
(
': 1 ( -: +}) +1-:
- } +- -I
,:
47r a 47r a 2 47r 2a 1 1 � flo! r � -:] flo 21 � 1 - - 1 + - 1 +- } =-- -l--} = -- L' - } 2 2 47r a 2 2 4mz
Substitute in equation (2) to obtain:
[1
-:]
)
! 2 r-: -:] flo - = -- � - } -
F f!.
4mz
and F f!.
B Due to a Current in a Solenoid *60
•
A solenoid 2.7-m long has a radius of 0.85 cm and 600 turns. It carries a
current 1 of 2. 5 A. What is the approximate magnetic field B on the axis of the solenoid? Picture the Problem
We can use
Bx = flon! to find the approximate magnetic field on
the axis and inside the solenoid. Express Bx as a function of n and Substitute numerical values and evaluate Bx:
I: Bx = (47rXIO-7N/A2 )
= 1 O 69 8mT 1 .
[ 600 J 2.7m
(2.SA)
Ampere's Law
*63
•
A long, straight, thin-walled cylindrical shell of radius R carries a current
Find B inside the cylinder and outside the cylinder.
I.
Sources of the Magnetic Field
173
Picture the Problem We can apply Ampere's law to a circle centered on the axis of the
cylinder and evaluate this expression for r Rto find B inside and outside the cylinder.
Apply Ampere's law to a circle
centered on the axis of the cylinder: Evaluate this expression for
{B. if = f.1oIc
Note that, by symmetry, the field is the
same everywhere on this circle.
r
Solve for Binside to obtain:
Binside = [2J
Evaluate this expression for r >R:
Solve for Boutside to obtain:
*67
••
Boutside
-I 1 -
f.101
2trR
Show that a uniform magnetic field with no fringing field, such as that shown
in Figure 27-50, is impossible because it violates Ampere's law. Do this by applying
Ampere's law to the rectangular curve shown by the dashed lines.
Figure
27-50 Problem 6 8
Determine the Concept
portions for which
The contour integral consists o f four portions, two horizontal
{B . if = 0 , and two vertical portions. The portion within the
magnetic field gives a nonvanishing contribution, whereas the portion outside the field
gives no contribution to the contour integral. Hence, the contour integral has a finite
value. However, it encloses no current; thus, it appears that Ampere's law is violated.
What this demonstrates is that there must be a fringing field so that the contour integral does vanish.
1 74
Chapter 27 The xz plane contains an infinite sheet of current in the positive z di rectio n.
* 72
The current per unit length (alon g the x direction) is A. Figure 27-52a shows a point P
above the sheet (y > 0 ) a nd two portions of the current sheet labeled I, and 12 ,
(a) What is
the direction of the magnetic field B at P due to the two portions of the current shown?
(b) What is the direction of the magnetic field
B at point P due to the entire sheet?
What is the direction of B at a point below the sheet (y < O)? the rectangular curve shown in Fi gure -
above the sheet is given by B =
-
(c)
(d) Apply Ampere's law to
27-52b to show that the magnetic field at any point �
1 f.1oAi .
(a) i.- w --oj
(b)
Figure
)(
)(
x
x
)(
x
x
27-52 Problem 72
Picture the Problem In
parts (a),
(b), and (c) we can use a right-hand rule to determine
the direction of the magnetic field at points above and below the infinite sheet of current.
In part
(d) we can evaluate {B . if around the specified path and equate it to We and
solve for B.
(a )
its in the j direction JIsince At P the magnetic field points to the right li.e., ' \' vertical components cancel.
(b)
Because the sheet is infinite, the same argument used in (a) applies; B is in the i direction.
l
-
A
-
(c) Below the sheet the magnetic field points to the left, i.e., in the i direction. The vertical components cancel. (d) Express iii
0
if , in the
counterclockwise direction, for the given path:
{ii o de = 2 fii o de + 2 fii o de parallel
..l.
Sources of the Magnetic Field
fB . cif = o
For the paths perpe ndicular to the -
-
sheet, B and
l 75
d.e are perpendicular
.1
to each other and:
fB . d£ = Bw parallel
For the paths parallel to the sheet, -
B and
d.e are in the same direction -
and: Substitute to obtain:
Solve for
paral el B = .l rIlo A
B:
2
and
Babove = 1 - t floAf I Magnetization and Magnetic Susceptibility
*79
••
A cyli nder of magnetic material is placed in a long solenoid of n turns per
unit length and current I. The values for magnetic field given below. Use these values to plot B versus
B withi n the material versus nI are
Bapp and Km versus nI.
nI, AIm
0
50
100
150
200
500
1000
10,000
B, T
0
0.04 0.67
1.00
1.2
1.4
1.6
1.7
Picture the Problem
We can use the data in the table and
Bapp. We can fi nd Km using B = Km Bapp . We can find the applied field
Bapp = flo nI to plot B versus
Bapp
for a long solenoid using: Km can be found from
Bapp and B
usmg:
B Km = -Bapp
The following graph was plotted using a spreadsheet program. The abscissa values for the graph were obtained by multiplying nl by J.1o.
B initially rises rapidly, and then becomes
nearly flat. This is characteristic of a ferromagnetic material.
1 76
Chapter
27 2.0 ����----'-�--���'---��--�-r-----�
E
'l:l
1 .6
1
12
16
0.8
II! '
6· ""'_ .
..
d
I !
.
'·. .';,,"-" 1
I
" r"
!-
iI •
- I
L
'. Y .
0.4 0. 0
.'
0.000
'''1'
0.002
,
1
0.004
'I
0.006
I
0.008
B app (T)
· r
0.0 1 0
0.01 2
0.014
The graph of Km versus nI shown below was also plotted using a spreadsheet program. Note that Km becomes quite large for small values of nI but then diminishes. A more revealing graph would be to plot BI(n!), which would be quite large for small values of nI and then drop to nearly zero at nI = 10,000 Aim, corresponding to saturation of the magnetization.
6000 5000 4000 :..:5: 3000 2000 1 000 o
r" "' . o
2000
4000 III
6000
(AIm)
8000
"'?'T
1 0000
Atomic Magnetic Moments
*82
••
Nickel has a density of
8.7 g/cm3 and a molecular mass of 58.7 glmol.
Nickel ' s saturation magnetization is given by !JoMs = 0.61 T. Calculate the magnetic moment of a nickel atom in Bohr magnetons.
Sources of the Magnetic Field
1 77
Picture the Problem We can find the magnetic momen t of a nickel atom J-l from its relationship the saturation ma gnetization Ms using Ms = n I' where 11 is the number of
molecules.
11,
in tum, can be found from Avogadro ' s number, the density of nickel, and
its molar mass using n
= NAMP .
- nr Ms il
Express the saturation magnetic field in terms of the number of molecules
or
Ms I' = _ n
per unit volume and the magnetic moment of each molecule: Express the number of molecules per unit volume in terms of Avo gadro ' s number NA, the molecular mass M, and the density
p:
Substitute and simpli fy to obtain:
Substitute numerical values and evaluate
J-l:
-- --'-- ----: :...� ..- '_rr_ --____:_'i T)--'---58.7 x 10,--3 kg/mol ----,,...,-!:-: ---== I' -r-4;r x 10-7--=-N/(0.61 A 2 6.02 1023 atoms/mol =
Express the value of 1 Bohr magneton: Divide
J-l by J-lB to obtain:
X
I'B = 9.27x lO-24 A · m 2 I' 5.44xlO-24 A · m 2 = 0.587 I'B 9.27 xlO-24 A · m 2 =
or
I' = I 0.587 1'B I
Paramagnetism
*86
··
Assume that the magnetic moment of an aluminum atom is 1 Bohr magneton. The density of aluminum is 2.7 g/cm3 , and its molecular mass is 27 g/mo1. (a) Calculate Ms and JioMs for aluminum. (b) Use the results of Problem
84 to calculate Xm at T
=
300
K. (c) Explain why the result for Part (b) is larger than the value listed in Table 27- 1 .
1 78
Chapter
27
Picture t h e Problem
In (a) we can express the saturation magnetic field in terms o f t he
number of molecules per unit volume and the magnetic moment of each molecule and use n = NA P 1M to express the number of molecules per unit volume in terms of Avogadro's
number NA, the molecular mass M, and the density p. We can use XI1l from Problem 84 to calculate Xm .
=
J.-loJ.-LMs/3kT
(a ) Express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule: Express the number of molecules per unit volume in terms of Avogadro' s number NA, the molecular mass M, and the density
p:
Ms
Substitute to obtain:
_
-
NA P
M
J...lB
Substitute numerical values and evaluate Ms : Ms =
_
-
(6.02 x 1 023 atoms/mol)(2 .7 x 103 kg/m3 )(9 .27 x 10-24 A · m2 )
! S.S8x 105 AIm !
27g/mol
and
(b) From Problem 84 we have:
Xm
=
J...loJ.-lMs 3kT
Substitute numerical values and evaluate Xm: Xm
(c)
=
1 0-7 N/A 2 9.27 x10-""'--24IKA--'r;105 AIm--L ! S .23 x 10-4 ! ·(m-2LiS.S8x -"4/'Z"x --.,. -------,3-.1.3L--'-8-1-x1 0-23 J 3 00 K ) =
! In calculating Xm in (b)we neglected any diamagnetic effects. !
179
Sources of the Magnetic Field Ferromagnetism For annealed iron, the relative permeability Km has its maximum value of approximately 5 500 at Bapp 1 .57x 10-4 T. Find M and B when Km is maximum.
*90
·
=
Picture the Problem
We can use
B Km Bapp =
to find B and M
=
(Kill - l)Bapp/ Jl.o
to
find M. Express B in tenns of M and Km: Substitute numerical values and evaluate B:
B
= =
(5500)(1.57 x l 0-4 T ) 1 0.864T 1
Relate M to Km and Bapp :
Substitute numerical values and evaluate M:
*96
··
M
(5500)(1.57 x 10-4 T ) 47r x 10-7 N/A 2 = 16.87 x 105 Aim 1
=
Two long straight wires 4-cm apart are embedded in a unifonn insulator that
has a relative penneability of Km = 1 20. The wires carry 40 A in opposite directions. (a) What is the magnetic field at the midpoint of the plane of the wires?
(b) What is the force
per unit length on the wires? Picture the Problem
Because the wires carry equal currents in opposite directions, the
magnetic field midway between them will be twice that due to either current alone and
will be greater, by a factor of Km, than it would be in the absence of the insulator. We can
use Ampere ' s law to find the field, due to either current, at the midpoint of the plane of
dF -
the wires and
=
-
-
Id.e x B to find the force per unit length on either wire.
(a) Relate the magnetic field in the insulator to the magnetic field in its absence : Apply Ampere' s law t o a closed circular path a distance
r
from a
current-carrying wire to obtain:
180
Chapter 27
Solve for Bapl' to obtain:
Because there are two current carrying wires, with their currents in opposite directions, the fields are additive and: Substitute numerical values and evaluate B:
_
B-
=
(b) Express the force per unit length experienced by either wire due to the current in the other: Apply Ampere' s law to obtain:
F f!
120(4Jr 10-7 N/A 2 )(40 A) Jr(0.02 m) I 96.0mT I
=
X
B1
1.8 . ie
=
B (2 nr)
=
Ji0 1c
=
Ji0 1
where r is the separation of the wires. Solve for B:
Substitute to obtain:
Substitute numerical values and evaluate
F f!
F _ 120(4Jr x 10-7 N/A 2 )(40 A Y £ 2Jr(0.04m) =
I 0.960 N/m I
General Problems
*101
•
In Figure 27 -5 5 , find the magnetic field at point P, which is at the common
center of the two semicircular arcs.
Sources of the Magnetic Field
Figure
27-55 Problem
Picture the Problem
181
1 01
Let out of the page be the positive x direction. Because point P is
on the line connecting the straight segments of the conductor, these segments do not
contribute to the magnetic field at P. Hence, the resultant magnetic field at P will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the expression for the magnetic field at the center of a current loop to find
B
p .
Express the resultant magnetic field at P: Express the magnetic field at the center of a current loop:
B
=
f..Lo1 2R
where R is the radius of the loop.
Express the magnetic field at the center of half a current loop: -
Express
-
B\ and B 2 : and
jj
2
=
_
f..LoI i 4R 2
Substitute to obtain:
* 1 04
••
A power cable carrying 50 A is 2 m below the earth' s surface, but the cable's
direction and precise position are unknown. Show how you could locate the cable using a compass. Assume that you are at the equator, where the earth's magnetic field is north. Picture the Problem
0.7 G
Depending on the direction of the wire, the magnetic field due to
its current (provided this field is a large enough fraction of the earth' s magnetic field)
1 82
Chapter 27
will either add to or subtract from the earth' s field and moving the compass over the ground in the vicinity of the wire will indicate the di rection of the current. Apply Ampere ' s law to a circle of radius r and concentric with the center of the wire:
BlVire = f.10
Solve for B to obtain:
1
2w
Substitute numerical values and evaluate BlVire :
Express the ratio of Bwi re to Bearth:
(47r x lO-7 N/A 2 )(50A) 2 7r(2 ) = 0.0500G _
Bwire -
Bwire = 0.05G 7 Bearth 0 . G
m
::::>
7%
Thus, the field of the current-carrying wire should be detectable with a good compass. If the cable runs east-west, its magnetic field is in the north-south direction and thus either adds to or subtracts from the earth's field, depending on the current direction and location
of the compass. Moving the compass over the region one should be able to
detect the change. If the cable runs north-south, its magnetic field is perpendicular to that of the earth, and moving the compass about one should observe a change in the direction of the compass needle. A very long straight wire carries a current of 20 A. An electron 1 cm from the center of the wire is moving with a speed of 5 .Ox 1 06 mls. Find the force on the
* 108
••
electron when it moves (a) directly away from the wire, (b) parallel to the wire in the
direction of the current, and (c) perpendicular to the wire and tangent to a circle around the wire . Picture the Problem
Chose the coordinate system shown t o the right. Then the current i s
in the positive z direction. Assume that the electron i s a t ( 1 cm, 0, 0). W e can use
F-
=
- -
-
-
q v x B to relate the magnetic force on the electron to v and B and B
=
1 21
f. 0 - j to _
47r
r
express the magnetic field at the location of the electron. We'll need to express v for each of the three situations described in the problem in order to evaluate F = q v
�
xB.
Sources of the Magnetic Field '\
,.
-
-
-
\
I Y' \
/
cm
B
\
q
-
/
183
/
(1,0,0) -
I
/
-
-
x, cm
F = qv x B
Express the magnetic force acting on the electron: Express the magnetic field due to the current in the wire as a function of distance from the wire:
/-Lo 2 1 ': = 2q/-LoI (v- X ] ) F- = qv x--] 4Jr 4nr
Substitute to obtain:
,:
-
r
(a) Express the velocity of the electron when it moves directly away from the wire:
( X ] ) = 2q/-LoIv k� F- = 2q/-LoI 4nr 4nr ; VI
Substitute to obtain:
Substitute numerical values and evaluate
,:
F:
2(4Jr X 10-7 N/A2 )(- 1.6 x 10-19 C)(5 X 106 rnIs) (20 A)k 4Jr(0.01m) = 1 (- 3.20 x 10-16 N)k 1 _
F-
(b) Express
v when the electron is
traveling parallel to the wire in the direction of the current: Substitute in equation (1) to obtain:
( k� X ] ) = - 2q/-LoIv ; F- = 2q/-LoI 4nr 4nr V
,:
I
(1)
184
Chapter
27
Substitute numerical values and evaluate F :
F=
_
2 (47r 10-7 N/A 2 )(-1. 6 10-19 C )(5 106 m/s)(20A)i = I (3 . 20 10-1 6 N)i I 47r(0.01m) X
x
(c) Express v when the electron is
traveling perpendicular to the wire and
X
v
x
= vj
tangent to a circle around the wire: Substitute in equation
*1 1 1
••
Figure
(1)
27-60
to obtain:
I (vJ� x J�) = fQl F- = 2q4f-lo � nr
shows a bar magnet suspended by a thin wire that provides a
0.8
restoring torque -K8. The magnet is 1 6 em long, has a mass of kg, a dipole moment of f1 = 0. 1 2 Am2 , and it is located in a region where a uniform magnetic field B can be
established. When the external magnetic field is
0.2
T and the magnet is given a small
angular displacement /18, the bar magnet oscillates about its equilibrium position with a period of 0.500 s. Determine the constant K and the period of this torsional pendulum when B = O .
B
Figure
27-60 Problem
Picture the Problem
111
We can apply Newton' s 2nd law for rotational motion to obtain the
differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for K when there is an external field and for the period T in the absence of an external field. Apply
IT = I
a
to the bar magnet
when B *- 0 to obtain the differential equation of motion for the magnet:
-
K() -
d 2 () j.1Bsin() = 1 dt 2
where I is the moment of inertia of the magnet about an axis through its point of suspenSIOn.
Sources of the Magnetic Field For small displacements from equilibrium (8
«
1 ):
Rewrite the differential equation as:
d 2e + ( K+ j.iB )e 0 I dt2
or
=
Because the coefficient of the linear
(1/
term is the square of the angular
=
K + j.iB I
(1)
frequency, we have: Express the moment of inertia (see
L2 1 = ...L 12 m
Table 9-1) of the bar magnet about an axis through its center: Substitute to obtain:
())
2 K+ j.iB L2
= --'--- I
Ti m
Solve for K to obtain:
Substitute numerical values and evaluate K :
Substitute B = 0 and equation
0) =
(I ) to obtain:
2mT in
Solve for T:
Substitute numerical values and evaluate T:
T
=
=
7r(0.16 m) 3(0.2460.8kg N . mlrad) I 0. 523s I
1 85
1 86 Chapter 27 An iron bar of length 1 .4 m has a diameter of 2 em and a uniform magnetization of 1 .72x 1 06 Aim directed along the bar ' s length. The bar is stationary in *] 1 6
••
space and is suddenly demagnetized so that its magnetization disappears. What is the rotational angular velocity of the bar if its angular momentum is conserved? (Assume that Equation 27-27 holds where m is the mass of an electron and q = -e.)
Picture the Problem
We can use the definition of angular momentum and Equation 27-
27, together with the definition of the magnetization M of the iron bar, to derive an expression for the rotational angular velocity of the bar j ust after it has been demagnetized.
Assuming its angular momentum to
be conserved, use the definition of L
L
= 10)
to express the angular momentum of the iron bar j ust after it has been demagnetized: Solve for the angular velocity
cu:
Assuming that Equation 27-27 holds yields:
L
0) = 1 L
= 2m J1 = 2me MV 2me Mm- 2 ,e q
=
e
e
where r is the radius of the bar and ,e its length.
Modeling the bar as a cylinder, express its moment of inertia with respect to its axis: Substitute to obtain:
0) = Substitute numerical values (see Table 1 3 - 1 for the density of iron) and evaluate
* 1 18
••
cu:
A relatively inexpensive ammeter, called a tangent galvanometer, can be
made using the earth ' s field. A plane circular coil of N tums and radius R is oriented such that the field Be it produces in the center of the coil is either east or west. A compass is
Sources of the Magnetic Field
1 87
placed at the center of the coil. When there is no cw-rent in the coil, the compass needle
points north. When there is a current J, the compass needle points in the direction of the
resultant magnetic field B at an angle e to the north. Show that the current J is related to e and to the horizontal component of the earth's field Be by
Picture the Problem
Note that
Be and Be are perpendicular to each other and that the
resultant magnetic field i s at an angle e with north. We can use trigonometry to relate Be and Be and express Be in terms of the geometry of the coil and the current flowing in it.
Express Be in terms of Be:
Be = Be tanB
where e i s the angle of the resultant field from north.
\
Express the field Be due to the current in the d-oil: \
Solve for J:
•••
2R
where N is the number of turns.
Substitute t6- obtain:
*125
N I Be = Jio
NJi I 0_ = Be tanB 2R
_ _
1=
2RBe
__
Jio N
tanB
A disk of radius R carries a fixed charge density
a
and rotates with
angular velocityw. (a) Consider a circular strip of radius r and width dr with charge dq. Show that the current produced by this strip dI (oi2 7r) dq = war dr. (b) Use your result from Part (a) to show that the magnetic field at the center of the disk is B = t Jio CJ"OJR . ( c) =
Use your result from Part (a) to find the magnetic field at a point on the axis of the disk a distance x from the center.
188 Chapter 2 7 Picture the Problem
The diagram shows
the rotating disk and the circular strip of radius r and width dr with charge dq. We can use the definition of surface charge density to express dq in terms of r and dr and the definition of current to show that dI = OJCJr
dr. We can then use this current and
expression for the magnetic field on the axis of a current loop to obtain the results caned for in (b ) and (c). (a) Express the total charge dq that
dq = CJ"dA = 2 1CCJrdr
passes a given point on the circular strip once each period: Using its definition, express the current in the element of width dr:
(c) Express the magnetic field dBx at a distance x along the axis of the disk due to the current loop of radius r and width dr:
Integrate from r = 0 to r = R to
obtain:
(b) Evaluate Bx for x = 0 :
d 1C dr dI = q = 2 CJr = I (j)ardr I 2 1C dt (j)
Chapter 28 Magnetic Induction Conceptual Problems
*1
•
A conducting loop lies in the plane of this page and carries a clockwise
induced current. Which of the following statements could be true? (a) A constant magnetic field is directed into the page. (b) A constant magnetic field is directed out of
the page. (c) An increasing magnetic field is directed into the page. (d) A decreasing
magnetic field is directed into the page. (e) A decreasing magnetic field is directed out of
the page. Determine the Concept
We lmow that the magnetic flux (in this case the magnetic field
because the area of the conducting loop is constant and its orientation is fixed) must be changing so the only issues are whether the field is increasing or decreasing and in which direction. Because the direction of the magnetic field associated with the clockwise current is into the page, the changing field that is responsible for it must be either increasing out of the page (not included in the list of possible answers) or a decreasing field directed into the page.
*6
·
I (d) is correct . I
If the current through an inductor were doubled, the energy stored in the
inductor would be (a) the same. (b) doubled. (c) quadrupled. (d) halved.
(e) quartered.
Determine the Concept
Um
=
The magnetic energy stored in an inductor is given by
-t LI 2 . Doubling I quadruples Um . I (c) is correct. I
A pendulum is fabricated from a thin, flat piece of metal. At the bottom of its *1O · arc, it passes between the poles of a strong permanent magnet. In Figure 28-42 a, the metal sheet is continuous, whereas in Figure 28-42b, there are slots in it. The pendulum with slots swings back and forth many times, but the pendulum without slots comes to a stop in no more than one complete oscillation. Explain why.
N
(b)
(a) Figure
28-42 Problem 1 0
1 89
190
Chapter 28
Determine the Concept
In the configuration shown in (a), energy i s dissipated by eddy
currents from the emf induced by the pendulum movement. In the configuration shown
in (b), the slits inhibit the eddy currents and the braking effect is greatly reduced. Estimation and Approximation
* 1 3 ·· A physics teacher attempts the following emf demonstration. She has two of her students hold a long wire connected to a voltmeter. The wire is held slack, so that there is a large arc in it. When she says "start", the students being rotating the wire in a large vertical arc, as if they were playing jump-rope. The students stand 3 .0 m apart, and the sag in the wire is about 1 . 5 m. (You may idealize the shape of wire as a semi-circular arc of diameter d = 1 . 5 m.) The induced emf from the jump rope is then measured on the voltmeter. (a) Estimate a reasonable value for the maximum angular velocity which the students can rotate the wire. (b) From this, estimate the maximum emf induced in the wire. The magnitude of the earth' s magnetic field is approximately 0.7 G. (c) Can the students rotate the jump-rope fast enough to generate an emf of 1 V? (d) Suggest modifications to the demonstration that would allow higher emfs to be generated. Picture the Problem We
can use Faraday' s law to relate the induced emf to the angular velocity with which the students turn the jump rope.
(a) It seems unlikely that the students could tum the "jump rope" wire faster than 5 revolutions per second. This corresponds to a maximum angular velocity of: (b) The magnetic flux ¢Jm through the rotating circular loop of wire varies sinusoidally with time according to: Because the average value of the cosine function, over one revolution, is Yz, the average rate at which the flux changes through the circular loop is: From Faraday' s law, the magnitude of the induced emf in the loop is: Substitute numerical values and evaluate E:
OJ = 5
rev s
x 2 n rad = I 3 1 .4 rad/s. I rev
¢Jm = BAsin OJt d¢m = BAOJcosOJt dt d¢m l = l.BAOJ = l.nr 2 BOJ 2 dt 2
and
av
Magnetic Induction 191 (c)
No. To generate an emf of 1 V, t he studenls would have to o ta te thejump rope about 500 times fast er.
(d)
The use of multiple strands of lighterwire(so that the compositewire could be rotated at the same angular speed) looped several times around wouldincrease the induced emf.
r
Magnetic Flux
*17 ·
A circular coil has 25 turns and a radius of 5 cm. It is at the equator, where
the earth' s magnetic field is 0.7 G north. Find the magnetic flux through the coil when its
plane is (a) horizontal, (b) vertical with its axis pointing north,
(c) vertical with its axis pointing east, and (d) vertical with its axis making an angle of 3 0° with north. Picture the Problem Because the coil defines a plane with area A and B is constant in
magnitude and direction over the surface and makes an angle 8 with the unit normal vector, we can use NBA to find the magnetic flux through the coil.
rpm =
cos e
Substitute for N, B, and A to obtain:
(
)
n(5 10-2 mY cose rpm = NBn r2 cos e = 25 0.7 G· + 10 G = (1 .37 lO-s W b)cos e X
X
(a) When the plane of the coil is horizontal, 8= 90° :
(b) When the plane of the coil is vertical with its axis pointing north, ()= 0° :
(c) When the plane of the coil is vertical with its axis pointing east, 8= 90°:
(d) When the plane of the coil is vertical with its axis making an angle of 30° with north, 8= 3 0° :
rpm = (l.3 7 10-s W b )cos90° =0 X
rpm = (1 .3 7 10-s W b)cos 00 = 11 .37 x lO-s W b I X
rpm = (1 .37 x lO-s W b)cos90° =0 rpm = (1 .3 7 x lO-s W b)cos30° = 11 . 19x lO-s W b I
192 Chapter 28 *24 ·· A long straight wire carries a current 1. A rectangular loop with two sides parallel to the straight wire has sides a and b, with its near side a distance d from the straight wire, as shown in Figure 28-45 . (a) Compute the magnetic flux through the rectangular loop. (Hint: Calculate the flux through a strip of area dA b dx and integrate from x = d to x = d + a.) (b) Evaluate your answer for a 5 cm, b = 1 0 cm, d = 2 cm, and 1= 20 A. =
=
n
b
Figure 28-45 Problem 24 Picture the Problem We can use the hint to set up the element of area dA and express the flux d
¢m through it and then carry out the details of the integration to express rAn.
(a) Express the flux through the strip of area dA: Express B at a distance x from a long, straight wire:
dCPm = BdA
where dA = bdx.
B
Substitute to obtain:
Integrate from x = d to x = d + a:
(b) Substitute numerical values and evaluate
¢m :
flo 21
flo 1
4Jr
2Jr X
_ ------
X
Magnetic Induction 193 Induced EMF and Faraday's Law •
A uniform magnetic field B is established perpendicular to the p1ane of a loop of radius 5 cm, resistance 0.4 n, and negligible self-inductance. The magnitude of *27
B is increasing at a rate of 40 mT/s. Find (a) the induced emf [; in the loop, (b) the
induced current in the loop, and (c) the rate of joule heating in the loop.
Picture the Problem We can find the induced emfby applying Faraday' s law to the loop. The application of Ohm ' s law will yield the induced current in the loop and we can find the rate of joule heating using P (a) Apply Faraday' s law to express the induced emf in the loop in tenns
= 12 R .
d AB)= AdB = ;rrR2 dB ( lEI= dq)mt = !£ dt dt dt
of the rate of change of the magnetic field: Substitute numerical values and evaluate
lEI:
(b) Using Ohm's law, relate the induced current to the induced
lEI= Jr(0.OSmY(40mT/s)= 1 0.3 14 mV 1
1= ER = 0.3 14mV = 1 0.78S mA 1 O.4n
voltage and the resistance of the loop and evaluate J:
(c) Express the rate at which power is dissipated in a conductor in terms of the induced current and the resistance of the loop and evaluate
p
= 12 R = (0.78S mAY(0.4n) = 1 0.247 flW 1
P: *31
••
A 1 00-tum circular coil has a diameter of 2 cm and resistance of 50 n. The
plane of the coil is perpendicular to a uniform magnetic field of magnitude
1 T. The direction of the field is suddenly reversed. (a) Find the total charge that passes
through the coil. If the reversal takes 0. 1 s, find (b) the average current in the coil and (c) the average emf in the coil. Picture the Problem We can use the definition of average current to express the total charge passing through the coil as a function of Jay. Because the induced current is
proportional to the induced emf and the induced emf, in tum, is given by Faraday' s law,
we can express �Q as a function of the number-of turns of the coil, the magnetic field, the resistance of the coil, and the area of the coil. Knowing the reversal time, we can find the average current from its definition and the average emf in the coil from Ohm' s law.
194
Chapter 28
(a) Express the total charge
that passes through the coil in terms of the induced current: Relate the induced current to the induced emf:
£ 1 = 1av = R
Using Faraday's law, express the induced emf in terms of rPnl: Substitute and simplify to obtain:
l1¢m
I1Q = £ I1t = M I1t = 2 ¢m R R R _
=
=
2 NBA
- R NBtrd2 2R
=
-
2 NB( : d2
)
-'----<-
--
R
where d is the diameter of the coil. Substitute numerical values and evaluate tlQ:
I1Q= (100)(lT)n-(0.02 mY 2(50Q) 1 - 1.26 mC I =
(b) Apply the definition of average current to obtain: (c) Using Ohm's law, relate the average emf in the coil to the average current:
Iav =
I1Q 1.26 mC 1 12 .6 mA I I1t O.ls =
=
£av = 1av R = (12 .6 mA)(50Q) =
1 630 mV I
Motional EMF
A rod 30 cm long moves at 8 mls in a plane perpendicular to a magnetic field of 5 00 G. The velocity of the rod is perpendicular to its length. Find (a) the magnetic force on an electron in the rod, (b) the electrostatic field E in the rod, and (c) the potential difference Vbetween the ends of the rod. *36
·
Picture the Problem We
can apply the equation for the force on a charged particle
Magnetic Induction 195 moving in a magnetic field to find the magnetic force acting on an electron in the rod. We can use
E =vx jj to find E and
V = Ef. , where f.
is the length of the rod, to find the
potential difference between its ends. (a) Relate the magnetic force on an electron in the rod to the speed of the rod, the electronic charge, and
-
F
=
-
qvxB
and
=
F qvBsin B
the magnetic field in which the rod IS .
movmg: .
Substitute numerical values and evaluate
F:
F = (1.6xl 0-19 C)(8 m/s)(0.05 T ) sin90° =I 6.40x1 0-20 N I -
-
(b) Express the electrostatic field
E =vxB
E in the rod in terms of the
and
magnetic field B :
E
Substitute n umerical values and
E
=vBsin B = (8 m/s)(0.05T )sin90° =I 0.400V/m I
evaluate B:
(c) Relate the potential difference between the ends of the rod to its length f! and the electric field E: Substitute numerical values and evaluate *42
··
V:
V = Ef.
V =(0.4 V/m)(0.3 m)=I 0. 120V I
In Example 28-9, find the total energy dissipated in the resistance and show
that it is equal to
t mv� .
Picture the Problem In Example 28-9 it is shown that the speed of the rod is given
byv = voe-(B2e2/mR)t. We can use the definition of power and the expression for a motional emf to express the power dissipated in the resistance in terms of B, f. ,
v,
and
R. We can
then separate the variables and integrate over all time to show that the total energy dissipated is equal to the initial kinetic energy of the rod. Express the power dissipated in terms of Eand
R:
Express Eas a function of B, f. and ,
= Bf.v
E
196
Chapter 28
v:
p
Substitute to obtain:
=
(B£vY R
The total energy dissipated as the rod comes to rest is obtained by integrating dE = Pdt:
Evaluate the integral (by changing variables to
u = -
2B 2 £2 ) to mR
obtain: *45
••
In Figure 28-50, a conducting rod of mass
m
and negligible resistance is free
to slide without friction along two parallel rails of negligible resistance separated by a distance
£ and connected by a resistance R. The rails are attached to a long inclined
plane that makes an angle Bwith the horizontal. There is a magnetic field B directed upward. (a) Show that there is a retarding force directed up the incline given by
F = (B2 £2VCOS 2 e)/R .
(b) Show that the terminal speed of the rod is
vt (mgR sin e)/(B 2 £2 Cos2 e) . =
Figure 28-50 Problem 45
Magnetic Induction
1 97
Picture the Problem The free-body
/Y
diagram shows the forces acting on the rod as it slides down the inclined plane. The retarding force is the component of
Fm
acting up the incline, i.e., in the -x direction. We can express
Fm using the '"
expression for the force acting on a conductor moving in a magnetic field.
/
Recognizing that only the horizontal
/
'"
'"
x� '"
component of the rod' s velocity v produces
'"
mg
an induced emf, we can apply the expression for a motional emf in
conj unction with Ohm ' s law to find the
-
--
--
)e'\.
--- - .:::..
induced current in the rod. In part (b) we
can apply Newton' s 2nd law to obtain an
expression for dv/dt and set this expression equal to zero to obtain Vt. (a) Express the retarding force
F = Fm cosB
acting on the rod:
where
(1)
Fm = IRB and J is the current induced in the rod as a consequence of its motion in the magnetic field. Express the induced emf due to the motion of the rod in the magnetic
[;
= BRvcosB
1
BR. = [; = vcosB
field: Using Ohm ' s law, relate the current J in the circuit to the induced emf:
R
R
Substitute in equation ( 1 ) to obtain:
(b) Apply
L Fx = max to the rod:
. - B 2 R. 2v cos 2 B = mdv mgsmB R dt --
and
198
Chapter 28 dv -
=
dt
2£2V 2 B . gsmB- -cos B mR
When the rod reaches its tenninal velocity
Vb
dvldt
=
° and:
Solve for Vt to obtain:
*51
•••
At time t
m RsinB
vt = 2g£2 2 B cos B
The loop in Problem 24 moves away from the wire with a constant speed v.
0, the left side of the loop is a distance d from the long straight wire. (a) Compute the emf in the loop by computing the motional emf in each segment of the loop that is parallel to the long wire. Explain why you can neglect the emf in the segments that =
are perpendicular to the wire. (b) Compute the emf in the loop by first computing the flux through the loop as a function of time and then using
& =
-dr/>rr/dt and compare your
answer with that obtained in Part (a). Picture the Problem We can use the expression for a motional emf and Ampere' s law to express the net emf induced in the moving loop. We can also use express the magnetic flux through the loop and apply Faraday' s law to obtain the same result. (a) Express the motional emf induced in the segments parallel to
c
= B(x)vb
the current-carrying wire: Using Ampere' s law, express
B(d + vt) and B(d + a + vt):
B(d+vt) = and
JL o I 2rc(d+vt )
B(d + a+ vt) = Substitute to express wire and
&1 for the near
&2 for the far wire:
JL o I 2rc(d+ a + vt)
JLo lvb = 1 2rc(d+vt)
c
and c
1
o lvb
JL - 2rc(d+ a+vt)
Magnetic Induction c
Noting that the emfs both point
=
C, - c2
199
f.1ol-.,."- vb� = -:... 2n(d+vt) 2n(d+a+vt)
upward and hence oppose one another, express the net emf induced in the loop:
The motion of the segments perpendicular to the long wire does not change the flux through the rectangular loop. Consequently, these segments do not contribute to the the induced emf.
= _ drpmdt drpm = B(x)dA= B(x)bdx B(x)= f.12m01 lb f d rpm = fB(x)dx=� 2n x = f.12nolb In[d+d+a+vt vt]
(b) From Faraday' s law we have:
c
Express the magnetic flux in an area of length b and width vdt:
where, from Ampere' s law,
Substitute and integrate from
x= d+vt to d+a + vt :
d+a+vt
d+a+vt
d+vt
d+vt
�
Differentiate with respect to time and simplify to obtain: c
= _ �dt [f.12nolb In d+d+a+vt vt] _ f.12nolb�dt [In d+d+a+vt vt] )[(d+vt)v-(d+2a+vt)v]J =_f.12:rr0lb[( d+d+vt a+vt (d+vt) f.lolbv [(d+vt)- (d+a+vt)]= f.1olbv [ 1 1] =-� (d+vt)(d+a+vt) -� d+a+vt-d+vt f.1olbv [1 1] =� �-d+a+vt =
Inductance *54
1=10
·
A coil with self-inductance
L carries a current I, given by
sin 2;ift. Find and graph the flux rpm and the self-induced emf as functions of time.
200
Chapter 28
Picture the Problem We can apply
rpm
=
LI to find
rPm and Faraday' s law to find the
self-induced emf as functions of time.
rpm
Use the definition of self-inductance to express rPm:
=
LI
=
I LIo
sin
2ift I
The graph of the flux rPm as a function of time shown below was plotted using a spreadsheet program. The maximum value of the flux is LIo and we have chosen 21if = 1 rad/s.
1.0 "..,.....".,..,.,...�
-1.0
::; .p.:.J:.!�::.:;..:�4����E:;L 2
o
3 t
Apply Faraday' s law to relate and dI/dt:
&,
L,
4
6
5
(s)
dI dt
d [ . d.] I 8m2'':!t dt
& = -L- = -L-
0
The graph of the emf & as a function of time shown below was plotted using a
spreadsheet program. The maximum value of the induced emf is 21ljLIo and we
have chosen 21if= I rad/s.
Magnetic Induction
201
-1.0 -f""'''''--'-�--+-''--' �-+' '' � �"'--I'��="4=-' ---' =--1 ''=---''----' 'F'''''' o
2
t
*57
••
4
3
5
6
(s)
A long insulated wire with a resistance of
18 Wm is to be used to construct a
resistor. First, the wire is bent in half, and then the doubled wire is wound in a cylindrical form as shown in Figure 28-53. The diameter of the cylindrical form is 2 cm, its length is
25 cm, and the total length of wire is 9 m. Find the resistance and inductance of this wire wound resistor.
Figure 28-53 Problem 57 Picture the Problem Note that the current in the two parts of the wire is in opposite directions. Consequently, the total flux in the coil is zero. We can find the resistance of the wire-wound resistor from the length of wire used and the resistance per unit length. Because the total flux in the coil is zero: Express the total resistance of the WIre:
L=0 R = (18 �)L= (18 �}9m)= 1 162Q 1
202 Chapter 28 Magnetic Energy
*61
··
In a plane electromagnetic wave, such as a light wave, the magnitudes of the
electric fields and magnetic fields are related by E
=
cB, where
c =
1/�Co 110 is the speed
of light. Show that in this case the electric energy and the magnetic energy densities are equal. Picture the Problem We can examine the ratio of c
=
Urn to
UE with E
= cB and
1/�Co 110 to show that the electric and magnetic energy densities are equal. B2
Express the ratio of the energy
urn
density in the magnetic field to the energy density in the electric field:
Substitute E
=
uE
cB:
_
2 110 .1 G o E2 2 C'
B2 ""'0 "0 E2 /I C'
1
Substitute for c:
*64 ·· You are given a length d of wire which has radius a, and told to wind it into an inductor in the shape of a cylinder with a circular cross-section of radius r. The windings are to be as close together as possible without overlapping. Show that the selfm . ductance 0f th·IS m . ductor IS .
L
(rdj
= 110 4a).
Picture the Problem The wire of length d and radius a is shown in the diagram, as is the inductor constructed with this wire and whose inductance L is to be found. We can use the equation for the self-inductance of a cylindrical inductor to derive an expression for
L.
The self-inductance of an inductor with length .e, cross-sectional area A,
(1)
Magnetic Induction and number of turns per unit length
203
n IS:
The number of turns N is given by:
N-2a 2a1 ;rr N(2;rr ) (�)2;rr 2a a
N= �
The number of tums per unit length
n=
n IS:
Assuming that a « r, the length of the wire d is related to nand r:
d
=
e
=-
r =
r=
r
e
Solve for .e to obtain:
Substitute for e, A, and n in equation ( 1 ) to obtain:
RL Circuits *69
••
In the circuit of Figure 28-29, let &0 = 1 2 V, R O. At time t
=
3 n, and L = 0.6 H. The
0.5 s, find (a) the rate at which the battery supplies power, (b) the rate of joule heating, and (c) the rate at which energy is being switch is closed at time t
=
=
stored in the inductor. Picture the Problem We can find the current using
I = If (1 - e -tlr 1
where
If = &oIR ,and ,= LIR , and its rate of change by differentiating this expression with respect to time.
Express the dependence of the current on If and Evaluate Ir and
r:
r:
and 'f
Substitute to obtain:
=
R 3Q O.2s L 0.6H = =
204
C hapte r
Express
28
dl/dt:
J(O.S s) = (4 A)(l - e-s(o.s
(a) Find the current at t = 0.5 s :
s)s-J
= 3.67 A
)
p(O.Ss) = J(0.5s) & = (3.67 A) (12 V) =144. 0 W I
The rate a t which the battery supplies power at t = 0.5 s is:
�(O.Ss) = [J(O.Ss)Y R = (3.67 Ay(30) = 1 40.4W I
(b) The rate ofjoule heating is:
dUL = � [1.LJ2 ]= LJ dJ dt dt dt 2
(c) Using the expression for the magnetic energy stored in an inductor, express the rate at which energy is being stored:
dUL = � [1.LJ2 ]= LJ dJ dt dt 2 dt
Substitute for L , 1, and dJ/dt to obtain:
Substitute numerical values and evaluate
Evaluate this expression for
t = 0.5 s :
dU dt
L :
dUL = ( 8W)(l- e-s(O.5s)s-J )e-S(o.ss)s-J dt 4 = (48W)(l - e-2.S )e-2.S =13.62W I
Remarks: Note that, to a good approximation, dUJdt = P - PJ•
*75 ••• Given the circuit shown in Figure 28-56, assume that the switch S has been closed for a long time so that steady currents exist in the inductor, and that the inductor L has negligible resistance. (a) Find the battery current, the current in the 100 n resistor
Magnetic Induction
205
and the current tlu'ough the inductor. (b) Find the initial voltage across the inductor when the switch S is opened. (c) Using a spreadsheet program, make graphs of the CUlTent and voltage across the inductor as a function of time.
10V
+
100 Q
2H
Figure 28-56 Problem 75 Picture the Problem The self-induced emf in the inductor is proportional to the rate at which the current through it is changing. Under steady-state conditions, dlldt = 0 and so
the self-induced emf in the inductor is zero. We can use Kirchhoff s loop rule to obtain the current through and the voltage across the inductor as a function of time. (a) Because, under steady-state
10V
conditions, the self-induced emf in
and
the inductor is zero and because the inductor has negligible resistance,
I
=
-
(IOQ)I
=
0
10 V I 1. 00 A I . =
IOQ ·
we can apply Kirchhoff s loop rule to the loop that includes the source, the 10-0 resistor, and the inductor to find the current drawn from the battery and flowing through the inductor and the 10-0 resistor: By applying Kirchhoff s j unction rule at the junction between the
II oo-n resistor = Ibattery Iinductor = @] -
resistors, we can conclude that: (b) When the switch is closed, the current cannot immediately go to zero in the circuit because of the inductor. For a time, a current will circulate in the circuit loop between the inductor and the 100-0 resistor. Because the current flowing through this circuit is initially 1 A, the voltage drop across the 100-0 resistor is initially
1100 V . I Conservation of energy (Kirchhoff s loop rule) requires that the voltage drop across the inductor is also 1100 V. I
206
Chapter 28
(c) Apply Kirchhoffs loop rule to the RL circuit to obtain: R
1
1(t)- 1 e-[.' - 1 e--;: L 2H
The solution to this differential
-
equation is:
where
-
0
T =
-
R
=
0
--
lOOn
=
0.02s
A spreadsheet program to generate the data for graphs of the current and the voltage across the inductor as functions of time is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 A6 B6
),Jl:,:r, L"
2
,3'"
[�4/
56 .7 ", 8" 9' 10 Iv 1 1 12 32 33 34 :35 36
Formula/Content 2 1 00 1 0 $B$3 *EXP((-$B$2/$B$ 1)* A6) A �I"" {." B L= 2 R= 1 00 1 0= 1 t 0.000 0.005 0.0 1 0 0.0 1 5 0.020 0.025 0.030 ""ill,
0. 1 3 0 0. 1 35 0. 1 40 0. 1 45 0. 1 50
H
C
Algebraic Form L R 10
to
1 0e
R --I
L
,,_
ohms A
let) 1 .00E+00 7 .79E-01 6.07E-01 4.72E-01 3 . 68E-01 2. 87E-01 2 .23E-01
Vet) 1 00.00 77.88 60.65 47.24 3 6.79 28.65 22.3 1
1 .5 0E-03 1 . 1 7E-03 9 . 1 2E-04 7 . 1 0E-04 5 .53E-04
0. 1 5 0. 1 2 0.09 0.07 0.06
,., '�
The following graph of the current in the inductor as a function of time was plotted using the data in columns A and B of the spreadsheet program.
Magnetic Induction
207
'\-��+--'=������ -':'=�-=--1
0.0
0.00
0.03
0.06
t
0.09
0.12
0.15
(5)
The following graph of the voltage across the inductor as a function of time was plotted using the data in columns A and C of the spreadsheet program. 100
80
� ;".
60
40
20
0 0.00
0.03
0.06
t
0.09
0.12
0.15
(5)
For the circuit of Example 2 8- 1 3 , find the time at which the power *80 •• dissipation in the resistor equals the rate at which magnetic energy is stored in the inductor. Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by
I = If (1 e-t/r 1 where If = eo!R and T = LIR is the time constant -
for the circuit. We can find the time at which the power dissipation in the resistor equals the rate at which magnetic energy is stored in the inductor by equating expressions for these rates and using the expression for 1 and its rate of change. Express the rate at which magnetic energy is stored in the inductor: Express the rate at which power is
dUL = � [.lLI 2 ]= LI dI dt dt 2 dt
208
Chapter 28
dissipated in the resistor: Equate these expressions to obtain :
Simplify to obtain:
(1)
1= Ir(l- e-t/r )
Express the current and its rate of change:
and
dI = dt =
(1 - e-t/r )= -I e-t/r ( _!) If !!..dt f I e-t/r f r
r
Substitute in equation (1) to obtain: or
1- e-t/r = e-t/r 1 = 2 e-t/r t - -dnl. -
Solve for t: Using
T=
=>
3 3 3 f.1s from Example 28-
1 1, evaluate t to obtain:
2
t = -(333 ,us)ln1 = I 23 1f.1s I
General Problems
*85
••
Figure 28-58 shows an ac generator. It consists of a rectangular loop of
dimensions a and b with N turns connected to slip rings. The loop rotates with an angular velocity OJ in a uniform magnetic field B. (a) Show that the potential difference between the two slip rings is
E=
NBabOJ sin
at what angular frequency value is 110 V?
OJ must
CtJt.
(b) If a = 1 cm, b = 2 cm, N = 1000, and B = 2 T,
the coil rotate to generate an emf whose maximum
Magnetic Induction 209 I�·---b (JJ
1
�I
x-x B
x
x
X
x
x
X
N turns
Figure 28-58 Problems 85 and 86 Picture the Problem We can apply Faraday's law and the definition of magnetic flux to derive an expression for the induced emf in the coil (potential difference between the slip rings). In part (b) we can solve this equation for wunder the given conditions. (a) Use Faraday' s law to express the induced emf: Using the definition of magnetic flux, relate the magnetic flux
C =_ drpm dt rpm ( t)= NBAcos (f)t
through the loop to its angular velocity: Substitute to obtain:
c
sin{f)t = 1
(b) Express the condition under which & = &max: Solve for and evaluate condition:
w under
= - !!:...-[NBAcos{f)]t dt = - NBabm(- sinmt) = I NBab{f)sinmt I
this
{f)= Cmax NBab
--
1 l 0V - (1 000)(2T)(0.01m)(0.02 m) = I 275 radls I
*88 ·· Show that the effective inductance for two inductors LJ and L 2 connected in parallel, so that none of the flux from either passes through the other, is given by
21 0 C hapter 28
Picture the Problem We can use the common potential difference across the parallel combination of inductors and the fact that the current into the parallel combination is the sum of the currents through each inductor to find an expression of the equivalent inductance. Define L eff by:
�
=
Leff
dI/ dt
or
dI dt Relate the common potential difference across the inductors to
£
1
=
their inductances and the rate at
and
which the current is changing in
£
each:
2
1
(1)
dI1 -'-1 dt
(2)
=
=
£ __ Leff
T
L
2
dI2 dt
=
Because the current divides at the
I I) +12
parallel junction:
and
dI dt
dI1 dt
-=-
Solve equations (2) and (3) for dI/dt and dI21dt and substitute to obtain: Express the relationship between an emf & applied across the parallel
combination of inductors and the emfs &1 and
&2 across the individual
inductors:
Substitute to obtain:
Substitute in equation ( 1 ) and solve for II Leff:
(3)
dI2 dt
+-
Magnetic Induction
211
*89 ·· Figure 28-59a shows an experiment designed to measure the acceleration of gravity. A large plastic tube is encircled by a wire, which is arranged in single loops separated by a distance of 1 0 cm. A strong magnet is dropped through the top of the loop. As the magnet falls through each loop, the voltage rises, then rapidly falls through 0 to a large negative value as the magnet passes through the loop, and the returns to O. The shape of the voltage signal is shown in Figure 28-59(b). (a) Explain how this experiment works. (b) Explain why the tube cannot be made of a conductive material. (c) Qualitatively explain the shape of the voltage signal in Figure 28-59b. (d) The times at which the voltage crosses 0 as the magnet falls through each loop in succession are given in the table in the next column. Use these data to calculate a value for g. (a)
_�...-.Magnet
Tube
Oscilloscope
(b)
ovr-------�r_--�
TIme
Figure 28-59 Problem 89 Loop Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Zero crossing time (s) 0.01 1 1 89 0.063 1 33 0. 1 0874 0. 1 4703 0. 1 805 2 0.2 1 025 0.23 85 1 0.26363 0.28853 0.3 1 1 44 0.33494 0.3 5476 0.37592 0.39 1 07
212
Chapter 28
Picture the Problem
(a)
As the magnet passes through the coil, it induces an emf because of the changing flux through the coil. This allows the coil to" sense" when the magnet is passing through it.
(b)
One cannot use a cylinder made of conductive material because eddy currents induced in it by a falling magnet would slow the magnet.
( c)
As the magnet approaches the loop, the flux increases, resulting in the increasing voltage signal. When the magnet is passing the coil, the flux goes from increasing to decreasing, so the induced emf becomes zero and then negative. The time at which the induced emf is zero is the time at which the magnet is at the center of the coil.
(d) Each time represents a point when the distance has increased by 1 0 cm. The following graph of distance versus time was plotted using a spreadsheet program. The regression curve, obtained using Excel ' s "Add Trendline" feature, is shown as a dashed line. 1.4
1.2 1.0 ...-- 0.8
S �
0.6
0.4 0.2 0.0 0.1
0
0.2 t
0.4
(5)
The coefficient of the second-degree term is
g
0.3
t g.
= 2(4.9257m1s2 ) = 1 9.85m1s2
Consequently,
I
Magnetic I *91
n du
ct ion 213
Suppose the coil of Problem 90 is rotated about its vertical centerline at
constant angular velocity of2 radls. Find the induced current as a function oftime. Picture the Problem We can apply F araday ' s law and the definition of magnetic flux to
derive an expression for the induced emf in the coil. We can then apply Ohm's law to find the induced current as a function of time. Note that only halfofthe loop is in the magnetic field.
I( t) C:R( t)
Apply Ohm ' s law to relate the
=
induced current to the induced emf:
(1)
Use Faraday' s law to express the induced emf:
rpm ( t)=
Using the definition of magnetic
NBAcos wt
flux, relate the magnetic flux through the loop to its angular velocity:
C:( t)= - � [NBAcos wt] dt = - NBAw(- sinwt) = NBAw sin wt
Substitute to obtain:
. Wt I( t) NBAw Slll R
Substitute in equation ( 1 ) to obtain:
Substitute numerical values and evaluate
=
I(t):
. (2 radJ)s t I( t)= ( SO)(1. 4T)( O . 252m )(Q O.15 m)(2radls)SIll 4 I (0.350A )sin(2rad/s)t I =
*96 ••• Figure 2 8-62 shows a rectangular loop of wire, 0.30 m wide and 1 .50 m long, in the vertical plane and perpendicular to a uniform magnetic field B = 0.40 T, directed inward as shown. The portion of the loop not in the magnetic field is 0. 1 0 m long. The resistance of the loop is 0.20 n and its mass is 0.05 kg. The loop is released from rest at
t
=
O. (a) What is the magnitude and direction of the induced current
when the loop has a downward velocity v? (b) What is the force that acts on the loop as a result of this current? (c) What is the net force acting on the loop? (d) Write the equation of motion of the loop. (e) Obtain an expression for the velocity of the loop as a function
214 Chapt er 28 o f time. ( f) Integrate the expression obtained in Part (e) t o find the displacementy a s a function of time. (g) Using a spreadsheet program, make a graph of the positiony of the loop as a function of time for values ofy between 0 m and 1.4 m (i.e., when the loop
leaves the magnetic field). At what time t doesy = 1.4 m? Compare this to the time it
would have taken if B = O .
fx----x----x- - - x-: ;x
x
x
;x
x
x
:x
x
x
x;
x
x;
,
:x
x
:x
x
:x
x
,
x
x;
x:
Figure 28-62 Problems 96 and 97 Picture the Problem We can use 1= c/R and & = Bv f to find the current induced in the loop and Lenz' s law to determine its direction. We can apply the equation for the force
on a current-carrying wire to find the net magnetic force acting on the loop and then sum the forces to find the net force on the loop. Separating the variables in the differential equation and integrating will lead us to an expression for v(t) and a second integration to an expression foryet). We can solve the latter equation fory =
1.40 m to find the time it
takes the loop to exit the magnetic field and our expression for v(t) to find its exit speed.
Finally, we can use a constant-acceleration equation to find its exit speed in the absence of the magnetic field. (a) Relate the magnitude of the induced current to the induced emf
1=
£
R
and the resistance of the loop: Relate the induced emf to the
£= Bvl!
motion of the loop: Substitute for & to obtain:
1=
I �I! I v
As the loop falls, the flux into the page decreases. The direction of the induced current is such that its magnetic field opposes this decrease, i.e., clockwise. (b) Express the velocity-dependent force that acts on the loop in terms
Magnetic Induction
215
of the current in the loop: Substitute for I to obtain:
Apply dF = Idf!. x B to the horizontal portion of the loop that is in the magnetic field to conclude that the net magnetic force is upward. Note that the magnetic force on the left side of the loop is to the left and the magnetic force on the right side of the loop is to the right. (c) The net force acting on the loop is the difference between the downward gravitational force and the upward magnetic force: (d) Apply Newton' s 2nd law of motion to the loop to obtain its equation of motion:
or
Factor g to obtain an alternate form of the equation of motion:
(e) Separate the variables to obtain:
dv
----
gor
B2f2
--
mR
= dt
v
dv = dt a-bv
--
where
Integrate v ' from 0 to v and t ' from o to t:
Vf-dv' S I ' = dt
o
B2f!.2
a= g and b = - mR
a -bv,
0
=>
1( )
a-bV --In -- = t b a
Chapter 28
216
Transform from logarithmic to exponential fom1 and solve for v to
ob tain : .
Notmg that
vt
V{t)= I (1- e-'/r ) I
a
= - , we have: b
vt
where
if) Write
v
as
r
v
v
a
g
=--.t... =--.t....
dyldt and separate
variables to obtain: Integrate
y' from 0 to y and t' from 0 to t:
fo dy'= Vt f(l- e-"/r }it'
Y
I
0
y{t)= I VI [t - (1 - e-l/r )] I
and
r
y
(g) A spreadsheet program to generate the data for graphs of position as a function of time is shown below. The formulas used to calculate the quantities in the columns are as follows:
t
Cell Bl B2 B3 B4 BS
F ormulalContent O.OS 0.2 0.4 0.3 $B$ 1 *$B$7*$B$2/($B$}I'2*$B$4"2)
Algebraic Form
B6 B7 AlO B I0 CI0
$B$SI$B$7 9.8 1 0.00 $B$S*(AI0-$B $6*( 1 -EXP(-AI01$B$6))) 0.S*$B$7*AI0"2
r
R= B= L= vt=
0.2 0.4 0.3 6.8 1 3 0.694 9.8 1
m mls s mls"2
m
R B L
VI t y
g I
t2
-zg
Magnetic Induction r
9 ..10
,11
;, 1 2 ,,1 3
. 't, 14 15
"? , ;1 6-,;"
'':''
'
"
"
','
""
,
17
18 \;1 9 ',,"'10 21" 22 1'"/ ;;,' 2 3 OJ, 24 25 26
;'27-
.
,
28 29
3'0
' ''' ,'f
,.;
t 0.00 0.05 0.10 0. 1 5 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1 .00
Y
Y (no B)
0.000 0.0 1 2 0.047 0 . 1 03 0.179 0.273 0.384 0.51 1 0.654 0.809 0.978 1 . 1 59 1 .3 51 1 .553 1 .764 1 .985 2 .2 1 4 2 .451 2 .695 2.946 3 .202
0.000 0.0 1 2 0.049 0.1 10 0. 1 96 0.307 0.44 1 0.60 1 0.785 0.993 1 .226 1 .484 1 .766 2.072 2.403 2.759 3 . 1 39 3 .544 3 .973 4.427 4.905
Examining the table, we see that y = 1 .4 m when t :::::
217
I 0.60 I s.
The following graph shows y as a function of t for B 7:- 0 (solid curve) and B = 0 (dashed curve).
0.0
0.2
0.4
I
0.6
(5)
0.8
1.0
Chapter 29 Alternating-Current Circuits Conceptual Problems *1
•
As the frequency in the simple ac circuit in Figure 29-27 increases, the rms
current through the resistor (a) increases. (b) does not change. (c) may increase or decrease depending on the magnitude of the original frequency. (d) may increase or decrease depending on the magnitude of the resistance.
(e) decreases.
R
Figure 29-27 Problem 1 Determine the Concept Because the rms current through the resistor is given by
ImlS *5
=
cmlS/ R
•
and both
&rms
and R are independent of frequency,
I (b) is correct. I
If the frequency in the circuit in Figure 29-29 is doubled, the capacitive
reactance of the circuit will (a) increase by a factor of 2 . (b) not change. (c) decrease by a factor of2. (d) increase by a factor of 4. (e) decrease by a factor of 4.
-
c
Figure 29-29 Problem 5 Determine the Concept The capacitive reactance of an capacitor varies with the frequency according to Xc
*9
••
=
1/ OJC.
Hence, doubling
(i)
will halve Xc.
I (c) is correct. I
Making LC circuits with oscillation frequencies of thousands of hertz or
more is easy, but making LC circuits that have small frequencies is difficult. Why?
219
220
Chapter
29
Determine the Concept To make an LC circuit with a small resonance frequency
requires a large inductance and large capacitance. Neither is easy to construct. *1 2
·
Are there any disadvantages to having a radio tuning circuit with an
extremely large Q factor? Determine the Concept Yes; the bandwidth must be wide enough to accommodate the modulation frequency. Estimation and Approximation *18
··
The impedances of motors, transformers, and electromagnets have inductive
reactance. Suppose that the phase angle of the total impedance of a large industrial plant is 25° when the plant is under full operation and using 2.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 60 Hz rms line voltage
at the plant is 40,000 V. The resistance of the transmission line from the substation to the plant is 5.2 Q. The cost per kilowatt-hour is 0.07 dollars. The plant pays only for the
actual energy used. (a) What are the resistance and inductive reactance of the plant' s total
load? (b) What is the current in the power lines and what must be the rms voltage at the substation to maintain the voltage at the plant at 40,000 V? (c) How much power is lost in transmission? (d) Suppose that the phase angle of the plant' s impedance were reduced to 1 8° by adding a bank of capacitors in series with the load. How much money would be saved by the electric utility during one month of operation, assuming the plant operates at full capacity for 1 6 h each day? (e) What must be the capacitance of this bank of capacitors? Picture the Problem We can find the resistance and inductive reactance of the plant's total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the rms voltage at the substation by applying Kirchhoff's loop rule to the substation-transmission wires-load circuit. The power lost in transmission can be found from Ptrans = I!lSRtrans. We can find the cost savings by finding the difference in the power lost in transmission when the phase angle is reduced to 1 8°. Finally, we can find the capacitance that is required to reduce the phase angle to 1 8° by first finding the
capacitive reactance using the definition of tan 8 and then applying the definition of
capacitive reactance to find C.
Rtr•n,
=
5.2.Q
Z = 1� + iXt.
(irms
=
40 kV
0=25°
Alternating-Current Circuits (a) Relate the resistance and inductive reactance of the pla nt's total load to
Z and 5:
Express Z in terms of the current I in the power lines and voltage &""s at
the plant: Express the power delivered to the plant in terms of CmlS' 1m1s and J and ,
solve for 1ffi1S:
R
=
Zcosb
and
XL = Zsinb Z EI
mlS
=
P.v = E
mlS
ImlS
cos b
and
Pay I m1S = ----
(1)
E cos b mlS
Substitute to obtain:
Substitute numerical values and evaluate Z:
Substitute numerical values and evaluate R and XL:
Z= Z
=
&�
cosb
P.v
( 40kVY cos 25° 630n 2.3 MW =
R (630n)cos25° =
=
and
XL (630n)sin25° =
(b) Use equation ( 1 ) to find the current in the power lines:
I
=
Ztot
=
ffi1S
1 571n 1
=
2.3 MW ( 40kV)cos250
1 266n I
=
I 63: 4A I
Apply Kirchhoff s loop rule to the circuit: Solve for Csub: Evaluate
Ztot:
Substitute numerical values and evaluate Csub:
Esub
=
= =
�R2 + X�
�(57 1nY +(266ny 630n =
(63 .4AX5.2n + 630n) 1 40.3kV I
22 1
222
Chapter
29
(c) The power lost in transmission IS:
2 R trans = (63 . 4 A)2 (S . .Ptrans = 1 mlS =
1 20.9 kW I
') Q)
(d) Express the cost savings ;j.C in
terms of the difference in energy consumption per-unit cost
(P25° - PI 8o);j.t and the
U
of the energy:
Express the power list in transmission when 0 = 1 8°: Find the current in the transmission lines when 0 = 1 8° : Evaluate
11 80 =
2.3 MW = 60.S A (40 kV )cos1 8°
� 80 :
Substitute numerical values and evaluate ;j.C:
!1C = (20.9kW - 1 9.0kW)(16h1d)(30dlrnonth)($0.07 / kW . h) = 1 $63.84 1 Relate the new phase angle 0 to the
inductive reactance XL, the reactance
tan 0 = xL
- XC
R
due to the added capacitance Xc, and the resistance of the load R: Solve for and evaluate Xc:
Xc
=
XL
- R tan 0
= 266Q - (S71 Q)tan1 8° = 80.S Q
Substitute numerical values and evaluate C: Alternating Current Generators *21
·
A 2-cm by 1 .S-cm rectangular coil has 300 turns and rotates in a magnetic
field of 4000 G. (a) What is the maximum emf generated when the coil rotates at 60 Hz? (b) What must its frequency be to generate a maximum e mf of 1 1 0 V? Picture the Problem We can use the relationship
Emax
= 21iNBAf to relate the
maximum emf generated to the area of the coil, the number of turns of the coil, the
Alternating-Current Circuits magnetic field in whic h the coil is rotating, and the frequency at which i t rotates.
emax =
(a) Relate the induced emf to the
=
NBA (1) 2TiNBAf
(1)
magnetic field in which the coil is rotating: Substitute numerical values and evaluate
Cmax :
(b) Solve equation ( 1 ) forf
f-
emax
2TiNBA
Substitute numerical values and evaluate f
1 l0V - 2 JZ" (300)(0. 4 T)(2 10-2 )(1 .5 10-2 )
f-
x
m
m
x
-
-
I 486Hz I
Alternating Current in a Resistor * 23
•
A l OO-W light bulb is plugged into a standard 1 20-V (rms) outlet. Find (a)
Inns, (b) Imax, and (c) the maximum power. Picture the Problem We can use and
Pmax = 1max cmax to fmd Pmax.
P.v = emlJmlS to find Inns, Imax = .J21nns to find Imax,
(a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it: Solve for and evaluate Inns:
Inns
=
P.v
emlS
=
lOO W 1 20V
=
I 0.833A I
(b) Express Imax in terms of Inns: Substitute for Inns and evaluate Imax :
( c) Express the maximum power in terms of the maximum voltage and
1max
=
.J2(0 . 8 33A )= I l.I8A I
223
224 Chapter 29 maximum current: Substitute numerical values and evaluate
Pmax =
(1 . 18A)J2(1 20V) 1 200 W I =
Alternating Current in Inductors and Capacitors *29
•
An emf of 1 0 V maximum and frequency 20 Hz is applied to a 20-J.LF
capacitor. Find (a) Imax and (b) ImlS •
Picture the Problem We can use lmax = &maxlXC and Xc = 1 1 we to express Imax as a function of &max,j, and C. Once we've evaluate Imax, we can use
Im1S = ImaJ J2 to find Im1S.
Express Imax in terms of &max and Xc:
max Imax = & XC
Express the capacitive reactance:
Xc
= _1_ = _1_ OJC 2JifC
Substitute to obtain: (a) Substitute numerical values and evaluate
lmax:
(b) Express Im1S in terms ofImax :
= 2Jr(20s -I )(20,uF)(lOV) = I 25. 1 mA I lmax = 25 . 1 mA = I i7 . 8 mA I IJlllS =
Imax
J2
J2
.
.
LC and RLC Circuits without a Generator *32
•
Show from the definitions of the henry and the farad that
unit s-I . Picture the Problem We can use XL = (f)L and Xc = 1 1 we to show the
1/.JLC 1/.JLC
has the
has the
unit S- I . Alternatively, we can use the dimensions of C and L to establish this result. Substitute the units for L and C in the expression
1/.JLC
to obtain:
Alternating-Current Circuits
[e] = [Q] [V ]
A l ternatively, use the defining
equation (C
=
225
QI V) for capacitance
to obtain the dimension of C: Solve the defining equation = dJ/ dt ) for inductance to
(V L
obtain the dimension of L :
Le :
Express the dimension of 1/..J
1 [V ][T]2 [Q] [Q] [V ]
Le has units of �
Because the SI unit of time is the second, we've shown that 1/..J
*37 ·· An inductor and a capacitor are connected, as shown in Figure 29-30. With the switch open, the left plate of the capacitor has charge Qo. The switch is closed and the charge and current vary sinusoidally with time. (a) Plot both Q versus t and J versus t and explain how to interpret these two plots to illustrate that the current leads the charge by 90°. (b) Using a trig identity, show the expression for the current (Equation 29-3 8) leads the expression for the charge (Equation 29-39) by 90° . That is, show that
J = -Jpeaksin OJ{ = Jpeakcos (OJt + 1C ).
2
Figure 29-30 Problem 3 7 Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect to time, an expression for the current as a function of time. We' ll use a spreadsheet program to plot the graphs.
226
Chapter
29
Apply Kirchhoff s loop rule to a clockwise loop just after the switch is closed: Because 1 =
dQ/dt : Q(t) = Qo c os (cut - 8)
The solution to this equation is:
where
cu =
JLC1
Because Q(O) = Qo, 8= 0 and:
Q(t) = Qo cos cut
The current in the circuit is the derivative of Q with respect to t:
1
=
dQ = !!.- [Qo cos cut] = -cuQo sin cut dt dt
(a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Qo were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°.
o
-1 o
2
4
6 t
8
10
(s)
(b) The equation for the current is:
1 = -cuQo sin cut
The sine and cosine functions are related through the identity:
- sin B = cos B + ;
Use this identity to rewrite equation (1):
1
( )
=
(1)
( )
-cuQo sin cut = cuQo cos cut + %
showing that the current leads the charge by 90°.
Alternating-Current Circuits
227
RL Circui ts w ith a Generatol" *44 ·· A resistor and an inductor are connected in parallel across an emf C = Cmax as shown in Figure 29-32 . Show that (a) the current in the resistor is lR = (cmax /R) cos 0Jl, (b) the current in the inductor is h ( Cmax /XL) cos (wt - 90°), and (c) 1 = lR + h = Imax cos (wt - 5), where tan 8= R/XL and Imax = &max/Z with =
Z-2 = R-2 + XZ2 .
R
L
Figure 29-32 Problem 44 Picture the Problem We can apply Kirchhoff s loop rule to obtain expressions for lR and
lL and then use trigonometric identities to show that 1 = lR + h
tan 8 = R/XL and Imax = cma.jZ with
Z-2 = R-2 + XZ2 .
= Imax cos (wt - 5), where
(a) Apply Kirchhoff s loop rule to a clockwise loop that includes the source and the resistor: Solve for h
E IR - �coS - R OJt
(b) Apply Kirchhoff s loop rule to a
Emax
clockwise loop that includes the
because the current lags the potential
source and the inductor:
difference across the inductor by 90°.
Solve for h:
IL =
cos(OJt - 900)- IL XL = 0
�laXL cos(cot -90°)
(c) Express the current drawn from the source in terms of lmax and the phase
constant 8:
Use a trigonometric identity to expand cos(wt - 5):
1 = Imax (cos OJt cos 6' + sin OJtsin 8) = Imax cos OJt cos 8 + Inlax sin OJt sin 6'
228
Chapter 29
From our results in (a):
1
=
1R + 1L
= £max cos cut
£max XL
cos(cut 9 0°)
+
=£
� COS
R
A useful trigonometric identity is:
R
-
cut + £max sm. cut XL
A cos cut + B sin cut = .JA 2 + B 2 cos(cut -8)
where
8 = tan -1 B A
Apply this identity to obtain:
{ ) (�' ) co,("" -o)
I � . &;'
'
'+
(1)
and
8= Simplify equation ( 1 ) and rewrite equation (2) to obtain:
£max XL tan-I I
£max R
=
tan
-l
( J R
XL
(2)
{ ) (�) co,("" - o) G) +UJ co,("" - o) �G), co,("" - o)
I � . &;'
'
� &="
'
+
'
� &=,
=1
£max
where
Z
tano �
cos(cut -8)
I l R XL
and
_1_2 = _1_2 + _1_2 Z
R
XL
Filters and Rectifiers *48 ·· The circuit shown in Figure 29-3 5 is called an RC high-pass filter because it transmits signals with a high-input frequency with greater amplitude than low-frequency
Alternating-Current Circuits signals. If the input voltage is �n
VOUI ( t )= VH cos (OJt - 5) where Vpeok VH = ----;======= 1+
229
( t )= VpCOk cos OJt, show that the output voltage is
(_)2 1
OJRC
C
----f l------r-Figure 29-35 Problem 48 Picture the Problem We can use Kirchhoff s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We'll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. Repeating this process for the output side ofthe filter will yield the desired equation. Apply Kirchhoff s loop rule to the input side of the filter to obtain:
Substitute for Yin and I to obtain:
Because Q
=
C V:
Substitute for dQldt to obtain:
V;n - V - IR = 0
where V is the potential difference across the capacitor.
dQ Vpeak co s OJt - V - R -= 0 dt dQ � dV = [CV]= C dt dt dt dV Vpeak cOS OJt - V - RC-= 0 dt the differential equation describing the potential difference across the capacitor.
Assume a solution of the form:
V= � cos OJt + � sin OJt
Substitution of this assumed solution and its first derivative in the differential equations, followed by equating the coefficients of the sine and cosine terms, yields two coupled linear equations:
Vc + OJR C� = Vpeak and
� - OJRC� = 0
230 Chapter 29 Solve these equations simultaneously to obtain:
vc
1
=
Va l + (cuRCY pe k
and
cuRC Vpeak 1 + (cuRCY
Vs =
Note that the output voltage is the voltage across the resistor and that it is phase shifted relative to the input voltage:
VH COS{cut - 5) where VH is the amplitude of the signal.
Assume that VH is of the form:
VH (t) = Vc cos cut + Vs sin cut
The input, output, and capacitor voltages are related according to:
=
VOllt
VH (t ) = �n (t ) - V(t )
Substitute for VH (t ) , Vpeak (t ) , and V {t ) and use the previously established values for Vc and Vs to obtain:
Vpeak - Vc
Vc
=
Vs
= -
and
�
Substitute for Vc and Vs to obtain:
vc - (cuRCY Vpeak 1 + (wRCY and cuRC vs = Vpeak 1 + (cuRCY
VH, vc, and Vs are related according to the Pythagorean relationship:
VH = Jv� + v�
_
Substitute for Vc and Vs to obtain:
cuRC V Jl + (wRCY peak Vpeak =
VH =
_rC"�c)'
Show that the average power dissipated in the resistor of the high-pass filter of Problem 48 is given by
*53
··
(
Vp�ak {wRCY P,ve = 2R 1 + {wRCY Picture the Problem
J
We can express the instantaneous power dissipated in the resistor
Alternating-Current Circuits
23 1
and then use the fact that the average value of the square of the cosine function over one cycle is 1'2 to establish the given result. The instantaneous power pet) dissipated in the resistor is:
PCt) =
VO�,t R
The output voltage Vout is: From Problem 4 8 :
Substitute i n the expression for pet) to obtain:
V.H
=
(_1)2 Vpea k
---;==��=
1+
v. 2 R [
OJRC
PCt) = --1:L COS2 (OJt
-
0)
( )2
V,;.'
1 R 1+ _ OJRC
rS' (wt - O)
Because the average value of the square of the cosine function over one cycle is 1'2:
Simplify this expression to obtain:
p
ave
- 2R ( 1 +(wRCY (OJRCY J -
Vp�ak
*57 ·· Using a spreadsheet program, make a graph of VL versus/ rd27r and 5 versus/for the low-pass filter of Problem 55. Use R = 1 0 kQ and C = 5 nF. =
Picture the Problem We can use the expressions for VL and 5 derived in Problem 56 to plot the graphs of VL versus/and 0 versus/for the low-pass filter of Problem 55. We'll simplify the spreadsheet program by expressing both VL and 5 as functions ofh dB . From Problem 56 we have:
L = �1 + (OJRCY
v
Vpeak
and
0 = tan-l (OJRC)
Rewrite each of these expressions in terms ofh dB to obtain:
2 ( L J 1+ h
dB
232 Chapter 29
( ]
and 8
= tan- J (2;ifR C) tan- J L =
J3dB
A spreadsheet program to generate the data for graphs of VL versus/and 5 versus/for
the low-pass filter is shown below. Note that Vpeak has been arbitrarily set equal to 1 V. The formulas used to calculate the quantities in the columns are as follows: Cell Bl B2 B3 B4 B8
F ormula/Content 2 .00E+03 5 .00E-09 1 (2 *PIO * $B$ l * $B$2)",- 1 $B$3/SQRT( 1 +((A8/$B$4)",2))
C8
ATAN(A8/$B$4)
D8
C8 * 1 80IPIO
A graph of Vout as a function of / follows:
Algebraic Form R C Voeak
h dB
Vpeak
( ) tan-' ( J
�
1+
J hdB
L hdB
'
5 in degrees
Alternating-Current Circuits
23 3
1 .0
0.8
�
-:::.g..
0.6
0.4
0.2
0.0 0
10
20
30
40
50
30
40
50
f (kHz)
A graph of 6 as a function of f follows:
';;) '" f! OJ) '" 2� ., '0
60
40
20
o
10
20
f (kHz)
*59
•••
frequency
Show that the trap filter, shown in Figure 29-3 7, acts to reject signals at a
0) 1/..JL C =
.
How does the width of the frequency band rejected depend on
the resistance R?
\(in
R
Figure 29-37 Problem 5 9 Picture the Problem We can apply Kirchhoff s loop rule to both the input side and output side of the trap filter to obtain an expression for the impedance of the trap. Requiring that the impedance of the trap be zero will yield the frequency at which the circuit rejects signals . Defining the bandwidth as
/),.0) 10) =
-
{j)lrap
1 and requiring that
234 Chapter 29
IZlrap 1 = R will yield an expression for the bandwidth and reveal its dependence on R. Apply Kirchhoff s loop rule to the output of the trap circuit to obtain:
VOtlt - JXL - JXc = 0
Solve for VOtlt:
= J (XL + Xc ) = JZtrap where Ztrap = XL + Xc Vout
(1)
Apply Kirchhoffs loop rule to the input of the trap circuit to obtain:
�n �n = -----""--J = ------"-'--R + XL + Xc R + Ztrap
Solve for I:
Substitute for I in equation ( 1 ) to obtain:
Because
Vout
= VIn
Ztrap R + Ztrap
XL = i wL and
-i
Xc = - : wC Note that Ztrap = 0 and Vout = 0 provided: Let the bandwidth /j. OJ be: Let the frequency bandwidth to be defined by the frequency at which
IZtrap I = R . Then:
Because
wtrap =
Solve for
2 2
m - mtrap :
OJ ;::: OJtrap, m - trap 2 mtrap :
Because W
.1 .JLC
�
(2)
1 mL - - = R mC or
o/LC- 1 = mRC
(�J2 mtrap
- 1 = mRC
Alternating-Current Circuits Substitute in equation (2) to obtain :
tJ.
W W = I W W.mp 1 = R C ,�,"P
2 I 2L I
_
=
R
23 5
LC Circuits with a G enerator *64
•• •
L C circuit
One method for measuring the compressibility of a dielectric material uses an with a parallel-plate capacitor. The dielectric is inserted between the plates and
the change in 'resonance frequency is determined as the capacitor plates are subjected to a compressive stress. In such an arrangement, the resonance frequency is 1 20 MHz when a dielectric of thickness 0 . 1 em and dielectric constant
K =
6.8 is placed between the
capacitor plates. Under a compressive stress of 800 atm, the resonance frequency decreases to 1 1 6 MHz. Find Young's modulus of the dielectric material. Picture the Problem We can use the definition of the capacitance of a dielectric-filled capacitor and the expression for the resonance frequency of an LC circuit to derive an expression for the fractional change in the thickness of the dielectric in terms of the resonance frequency and the frequency of the circuit when the dielectric is under compression. We can then use this expression for /).t/t to calculate the value of Young' s modulus for the dielectric material. Use its definition to express Young' s modulus of the dielectric material: Letting t be the initial thickness of the dielectric, express the initial capacitance of the capacitor: Express the capacitance of the capacitor when it is under compreSSIOn: Express the resonance frequency of the capacitor before the dielectric is
Y
-
_
tJ.P
(1)
_ - --
strain
tJ.t/t
Co KEo A =
t
Cc KEo A =
W
0
compressed: When the dielectric is compressed:
e
s tr s s --
-
t - tJ..t
)CoL - �KE� AL 1
1
--- - --= = = =
c - )CcL - �KEo AL
W
1
1
= - -- - --= = =
t - tJ..t
236 Chapter 29 Express the ratio of We to
eva
and
we Wo
simplify to obtain:
�KEOt �1- l:J.t �K Eo AL
_
-
_
AL
-
t
t - l:J.t
Expand the radical binomially to
We Wo
obtain:
=
( 1 )1/2 _
l:J.t t
(
I1t 2 1 - We W t =
1
_
l:J.t 2t
t.
provided I1t « Solve for I1tlt:
�
o
J
Substitute in equation ( 1 ) to obtain:
y
Substitute numerical values and evaluate Y:
-
_
=
(800 atm) (1 0 1 . 325 kPaJatm) 2 1 1 16 MHZ 120 MHz ! 1.22 x 109 N/m 2 !
(
J
RL C Circuits with a Generator *69
Show that the formula �v
··
=
R&�, j
Z2
gives the correct result for a circuit
containing only a generator and (a) a resistor, (b) a capacitor, and (c) an inductor. Picture the Problem The impedance of an ac circuit is given by
Z
=
�R 2
+
(XL - Xc Y
. We can evaluate the given expression for Pay first for
XL = XC = ° and then for R
(a) For X= O, Z = R and:
(b), (c) If R
=
=
0.
0, then:
Remarks: Recall that there is no energy dissipation in an ideal inductor or capacitor.
Alternating-Current Circuits
237
*74 " FM radio stations have carrier frequencies that are separated by 0.20 MHz. When the radio is tuned to a station, such as 1 00 . 1 MHz, the resonance width of the receiver circuit should be much smaller than 0.2 MHz, so that adjacent stations are not received. Iffo = 1 00. 1 MHz and 11/= 0.05 MHz, what is the Q factor for the circuit? Picture the Problem We can use its definition, circuit.
Q = fo /l1f to find the Q factor for the
Express the Q factor for the circuit:
Substitute numerical values and evaluate Q: *79
"
Q=
1 00. l MHz 0.05 MHz
�
=�
In the circuit shown in Figure 29-42, the ac generator produces an rms
voltage of 1 1 5 V when operated at 60 Hz. What is the rms voltage across points (a) AB,
(b) BC, (c) CD, (d) AC, and (e) BD?
137 mH
SO Q
Figure 29-42 Problem 79 Picture the Problem We can find the rms current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations.
(a) Express the potential difference between points A and B in terms of
VAB = IrmsXL
(1)
Irms and XL: Express Irms in terms of c and Z:
£
ImlS = Z
£
�R 2 + (XL - XC Y
238 Chapter 29 Eva luate XL and Xc to
obtain:
XL
and
XC
= 27ifL = 2Jr(60s-1 )(1 3 7 mH) = 5 1 .6 0 1 1 _ -_ - 27ifC 2Jr(60s-1 )(25 JLF) = 1 06. 1 0 _
_
. 1 15V �(500Y + (5 1 .6 0 - 1 06. 1 0Y = 1 .55 A
Substitute numerical values and evaluate Irms:
I =
Substitute numerical values in equation (1) and evaluate VAB :
VAB (1 .55 A)(S 1 . 6 0) 1 80.0V I
mlS
=
=
(b)
VBC = IrrnsR = (1 .55 A)(500)
(c) Express the potential difference
VCD = Irms Xc = (1 .55 A)(1 06. 1 0)
Express the potential difference between points B and C in terms of Irms and R: between points C and D in terms of Irrns and Xc: (d) The voltage across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right: Use the Pythagorean theorem to find
VAC :
(e) The voltage
across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right:
= 1 77.5V I = 1 1 64V I V;IC
VAC = �V1B + V�c
= �(80.0 VY + (77.5 VY = l l l 1 V I
Altemating-CUlTent Circuits
239
Use the Pythagorean theorem to find VBD:
*84
••
Show that Equation 29-48 can be written as
Picture the Problem We can substitute for XL and Xc in Equation 29-48 and simplify the resulting equation to obtain the given equation for Imax. Equation 29-48 is:
Imax
=
.
�R 2 + (Cmax ) XL - X 2 C
Substitute for XL and Xc to obtain:
Simplify algebraically to obtain:
Imax
cmax
(
)
= ---;========== = ====
1 R 2 + o i L2 1 - / o Le
2
Cmax
Cmax
*88
••
A method for measuring inductance is to connect the inductor in series with a
known capacitance, a known resistance, an ac ammeter, and a variable-frequency signal generator. The frequency of the signal generator is varied and the emf is kept constant until the current is maximum. (a) If C = 1 0 fiF, cmax = 1 0 V, R = 1 00 fl, and I is maximum at 0)= 5 000 rad/s, what is L? (b) What is lmax ? Picture the Problem We can use the fact that when the current is a maximum, XL = Xc, to find the inductance of the circuit. In (b), we can find Imax from �ax and the impedance of the circuit at resonance.
240 Chapter 29 (a) Relate XL and Xc at resonance:
Solve for L to obtain:
Substitute numerical values and evaluate L :
(b) Noting that, at resonance,
L=
1
(S O O Os-I )(lOflF)
Imax
0, express Imax in terms of the applied emf and the impedance of
X=
=
&max Z
=
10 V l oo n
=
=
I ·
4 00 mH
I
1 O.lO O A 1
the circuit at resonance: *90
••
In the circuit shown in Figure 29-44, R
= 1 0 Q, RL
=
30 Q, L =
1 50 mH, and C = 8 flF; the frequency of the ac source is 1 0 Hz and its amplitude is 1 00 V. (a) Using phasor diagrams, determine the impedance of the circuit when switch S is closed. (b) Determine the impedance of the circuit when switch S is open. (c) What are the voltages across the load resistor RL when switch S is closed and when it is open? (d) Repeat Parts (a), (b), and (c) with the frequency of the source changed to 1 000 Hz. (e) Which arrangement is a better low-pass filter, S open or S closed?
a
R
b
Figure 29-44 Problem 90 Picture the Problem Because we'll need to use it repeatedly in solving this problem, we 'll begin by using complex numbers to derive an expression for the impedance Zp of the parallel combination of C with L and RL in series. The total impedance of the circuit is then Z = R + Zp. We can apply Kirchhoff s loop rule to obtain expressions for the voltages across the load resistor with S either open or closed.
Use complex numbers to relate Zp to
RL, XL, Xc: and
Alternating-Current Circuits 241 1 ---1 1 = --+ Zp -iXc RL+ iXL L+i(XL- Xc) _ RXCXL - iRLXC -iRLXC Zp = RXCXL L+i(XL- Xc) -iRLXC RL-i(XL-Xc) Zp = RXCXL L+i(XL- XC) RL-i(XL-Xc)
or
Multiple the numerator and denominator o f this fraction by the complex conjugate o f
RL i(XL- Xc): +
Simplify to obtain: (1)
(a)
S is closed.
Because L is
shorted: Evaluate
Xc:
Substitute numerical values in
equation ( 1 ) and evaluat e Zp, and 0:
Z, IZI,
1 _ _- ( 1 Xc _- _ 2ife 27r lOs-1)(8,uP) = 1.99kO Zp = 300-i(0.4520), Z = 400-i(0.4520) , IZI = �(400/ +(0.4520/ = 140.00 1
and
In Problem 29-77 we showed that for a parallel combination o f a resistor and capacitor, the phase angle 0 is given by:
Substitut e numerical values and
evaluate 0:
5 = tan -1 ( - 0.400 4 520 ) = 1-89.40 1 No phasor diagram is shown because it is
242
Chapter
29 impossible to represent it to scale.
( b)
S is open; i.e., the inductor is
in the circuit.
Find XL :
Substitute numerical values in
equation (1) and evaluat e Zp, Z, and 5:
IZI,
XL = = 27ifL = 2Jr(1 0s-1 )(0.15H) = 9.42£1 zp = 30. 3 £1+ i (9. 0 1£1) , = 40.3 £1+i(9.0H2), IZI = �(40.3 £1)2 + (9.01£1)2 = 1 41. 3 £1 1 9.0 1£1J () = tan-I ( X) = tan-I ( 40.3£1 = 1 12.60 1 wL
z
and
R
The phasor diagram for this case is shown to the right .
( c)
S is closed.
Apply Kirchhoff s
loop rule to a loop including the
£-IR -VR L
=0
source, R, and RL:
VR = 1= Z L
Express the current I in the circuit :
£-IR
£
Substitut e and simplify to obt ain:
From (a) we have:
Zp = 30£1 -i(0.452£1) , = 40£1 -i(0.452£1) , IZI = 40. 0 £1, {) = tan-I (-0��52 ) = _0.647o::::0: 0 Z
and
i c s 243 =(1-�}lOOV)COS[(20S-' )m] 1 (7 5V)cos[(20s-l)m] 1 -IR - IXL- RL = 0 Alternatin g-Current C
Substitute numer ical values to obtain :
r u it
V"
=
S
is open.
Apply Kirchhoffs loop
rule to a loop including the source, R, L, and Solve for
RL
&
V
when S is open:
VR : I.
Express the current I in the circuit:
1= & Z
Substitute to obtain :
Substitute numerical values and
evaluate Zp and Z:
=
=
and s: u =
Substitute numerical values and evaluate
VRL :
30. 30 +i(9 . 0 10), 40.30 +i(9 . 0 10), 41. 30, tan -I ( XL ) = tan _ 1 ( 9.420 ) R+R L 40. 30
Zp Z IZI=
. 420 ) = (1 _ 100+9 41. 30 (100 V) cos[(20 S-I )m - 13.2°] =1 (53.0V)cos[(20s-l)m-13.2°] I XL 2Jr(1 000s-1 )(0.15H) 9420 Xc = 2 Jr(1000s1 -I )(8,LlF) = 19 . 90
VRL
X
(d) Find XL and Xc when f= 1 000 Hz:
S
is closed. XL
to:
=
and
= 0, and Zp simplifies
=
244 Chapter 29 Substitute numerical values in
equ ati on 1 ( ) and evaluate Zp, and 8:
Z, IZI,
=9. 170 -i (13. 8 0) , = 19.170 - i (13.80) , = �(19.17 oy + (13.80Y =1 23.6 0 1 o = tan-, ( -19.113.780° ] = 1 _ 3s.70 1
Zp Z \ZI
and
A phasor diagram for this circuit is shown to the right,
S is o pe n.
Substitute numerical
values in equation ( 1 ) and evaluate
Zp, Z, IZI, and 8:
=0.01400-i (20.30) , = 10.00 -i(20.30) , Izl = �(1O· 00r + (20. 30r = 1 22.6 0 1 Zp
Z
and Find the total impedance, its magnitu de, and phase angle f or the circuit:
= 10. 0 0 -i (20AO) , �(1O.00r + (20.40Y = 1 22. 7 0 I 8�tan-' ( - ���(} 1 - 63. 9 0 I
Z Z
=
and
The phasor diagram is shown to the right.
(e)
\
The load voltage at the higher frequency is much more attenuated with S open, while opening S does not reduce the low frequency load voltage significantly. Therefore, S open is the better arrangement for a low - pass filter.
*97
•••
(a) Show that Equation 29-47 can be written as
tan 5
= Q(o/(j)(j)-o (j)2 ) 0
(b) Show that near resonance
Alternating-Current Circuits
245
())
(c) Sketch a plot of 6'v ersus x, where x = wi %, for a circuit with high Q and for one with
low Q .
Picture the Problem
We can manipulate Equation 29-47 into a form that has the ratio of
L to R in it and then use the definition of Q to eliminate L and R. In ( b) we can approximate
u/ - ()); , near resonance, as 2OJofj,0J and substitute in the result from ( a) to
obtain the desired result. ( a) From Equation 29-47:
tan 8
OJL -ljOJC = OJ2 L R OJR 2 ))2 = L(OJ -1/ L ) = L(( - OJ; ) OJR OJR
-ljC
=
C
Express Q in terms of
%,
L and R :
Solv e for LlR to obtain:
Substitute to obtain:
( b) Near resonance:
Q = OJoRL L Q R OJo
-=
tan 8 =
Q(OJ2 - OJ;) ())())o
( 1)
OJ2 - OJ; = (OJ+OJo)(())-OJo) � 2())0fj,())
Substitute in equation ( 1) to obtain:
(c) A following graph of 6' as a function of x = wi % w as plotted using a spreadsheet program. The solid curve is for a high-Q circuit and the dashed curve is for a low-Q circuit.
246 Chapter 29 1.5 1.0
.g �
0.0
-1.0 -1.5
0.0
*102
0.5
1.0 x
2.0
1.5
One method for measuring the magnetic susceptibility of a sample uses an
•••
LC c ircuit consisting of an air-core solenoid and a capacitor. The resonant frequency of the c ircuit w ithout the sample is determined and then measured again w ith the sample inserted in the solenoid. Suppose the solenoid is 4 cm long, 0 . 3 cm in diameter, and has 400 turn s of fine wire. Assume that the sample that is inserted in the solenoid is also long and fills the air space. Neglect end effects. (In practice, a test sample oflrnown
4
cm
susceptibility of the same shape as the unknown is used to calibrate the instrument.) ( a) What is the inductance of the empty solenoid?
(b)
What should be the capacitance of the
capacitor so that the resonance frequency of the c ircuit without a sample is 6.0000 MHz?
(c) When a sample is inserted in the solenoid, the resonance frequency drops to 5.9989
MHz. Determine the sample 's susceptibility. Picture the Problem
We can use L
=
JLon2 Af to determine the inductance of the empty
solenoid and the resonance condition to find the capacitance of the sample-free c ircuit when the resonance frequency of the circuit is 6.0000 MHz. By expressing L as a function offo and then evaluating dfoldL and approximating the derivative with !1foiM ,
w e can evaluate X from its definition. ( a) Express the inductance of an air core solenoid:
Substitute numerical v alues and evaluate L: L
=
2 1Z' (0. 003m)2(0.04m) I 35.5,uH I (41Z' 10-7 N/A2)( 0.400 ) 04m 4 X
=
247
Alt ernating-Current Circuit s (b) Express the condition for resonance in the LC circuit :
or
1_ 2..riL0 = _ 2lifoC
( 1)
/1:/
Solve for C to obt ain:
Substitut e numerical values and evaluate C:
(c) Express the sample ' s
susceptibility i n terms of L and M: Solve equation ( 1) forfo:
Differentiat e 10 with respect to L:
1 C - 47r2(6 MHz)(3S. S,uH) --/ 119 =M X
;if
/
( 2)
L
1 27r-JLC dlo 1 d 1 dL 27r.,Jc dL 47r.,Jc 1 = _ 10 47rL-JLC 2L 10 =
r-J/2 =
=
_
L-3/2
=
Approximate dfoldL by f:.folM :
M -=-- or-=--
!110 M
Substitute i n equation ( 2) t o obt ain:
x=
Substitute numerical values and
x=
evaluate x:
10
2L
!110 10
2L
-2 !11100 _2( S.99 896.M0HZ000-6.MHz0000o MHZ ) = 1 3. 67 xlO-4 1
The Transformer *105
•
An ac voltage of 24 V is required for a device whose impedance is 1 2 Q.
( a) What should the turn ratio of a transformer be, so the device can be operated from a 1 20-V line?
(b)
Suppose the transformer is accidentally connected reversed ( i.e., with the
secondary winding across the 1 20-V line and the 1 2-0 load across the primary). How much current will then flow in the primary winding?
Chapter 29
248
Picture the Problem Let the subscript I denote the primary and the subscript 2 the
secondary. We can use
V2N1
=
V;N2 and N/l NJ2 to find the tum ratio and the =
primary current when the transformer connections are reversed. (1)
(a) Relate the number of primary and secondary turns to the primary and secondary voltages: Solve for and evaluate the ratio
N2INI:
(b)
Relate the current in the primary
to the current in the secondary and
N2 NI I1
_
=
24 V -I! 1 V; 120V 5
V2
_
N2
I N 2 1
to the turns ratio: Express the current in the primary winding in terms of the voltage across it and its impedance:
I
2
=
V2 2
Z
Substitute to obtain: Substitute numerical values and evaluate II:
*110
••
An audio oscillator (ac source) with an internal resistance of2000 Q and an
open-circuit rms output voltage of 12 V is to be used to drive a loudspeaker with a resistance of 8 Q. What should be the ratio of primary to secondary turns of a
transformer, so that maximum power is transferred to the speaker? Suppose a second identical speaker is connected in parallel with the first speaker. How much power is then supplied to the two speakers combined? Picture the Problem Note:
In a simple circuit maximum power transfer from source to
load requires that the load resistance equals the internal resistance of the source. We can use Ohm's law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the speaker(s).
Alternating-Current Circuits 249 Express the effective loudspeaker resistance at the primary of the
R
efT
=
�
1
I
transformer:
Express
I] in terms of h N], and N2:
1] 12 N 2 N] =
Substitute to obtain:
(1 )
(2)
Express the power delivered to the two speakers connected in parallel: Find the equivalent resistance Rsp of the two 8-0 speakers in parallel:
1 1 1 2 1 Rsp 80 80 80 40 Rsp = 40
-=-
+ -=- = --
and
Solve equation (1) for RefT to obtain:
Substitute numerical values and evaluate RefT:
Find the current drawn from the source:
V = 12V = 4.00mA I] =Rtot 20000 9990 +
250
Chapter 29 �p
Substitute numerical values in equation (2) and evaluate the power
=
(4mAy(999Q)= 1 16.0mW I
delivered to the parallel speakers: General Problems *115
Figure29-47 shows the voltage Vversus time t for a square-wave voltage.
••
If V = 12 V, (a) what is the rms voltage of this waveform? (b) If this alternating
o
waveform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what now is the rms voltage of the rectified waveform?
v
Va
t
Figure 29-47
Problem 115
Picture the Problem
The average of any quantity over a time interval f:...T is the integral
of the quantity over the interval divided by f:...T. We can use this definition to find both the
average of the voltage squared,
(V 2 t and then use the definition of the rms voltage.
(a) From the definition of Vrn1s we have: Noting that
- Vo2 V0 2 , evaluate =
(b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval
+f:...T:
Express the square of the voltage during this half cycle:
Alternating-Current Circuits 251 Calculate
(V2 t by integrating V2
from t = 0 to t =
by 6.T:
t t1T
and dividing
(V2)
av
=
V02
t1T
r,L\7 dt 1
=
V02 [t]1t.T
t1T
0
=
� 2
V2 0
Substitute to obtain:
*120
••
Repeat Problem 1 1 9 if the resistor R is replaced by a 2-j.iF capacitor.
Picture the Problem
We can apply Kirchhoff's loop rule to obtain an expression for
charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find Imax and Imin by
considering the conditions under which the time-dependent factor in I will be a maximum
or a minimum. We can use the maximum value of the current to find IrnlS• Apply Kirchhoff's loop rule to obtain: Substitute numerical values and solve for q(t):
Differentiate this expression with respect to t to obtain the current as a function of time:
Express the condition that must be
E1 + E2
q(t) (2 ,uF(2,uF)(20)(18V )V)cos(1 131 ) = (40flC)cos(1 131s-l)t 361lC I = ddtq = �[ dt (40flC)cos(1 131s-l)t 36 flC] =1-(45. 2 mA)sin(1 131s-l)t 1 sin(1 131 )t = 1 1- 45. 2 m A 1 sin(1 131s-l)t = - 1 = 1 45. 2 m A 1 =
mlmmum:
Imin
and
maXImum:
Imax
average value of the sine function over a period is zero:
Iav
=
-I
+
=
satisfied if the current is to be a
capacitor as an open circuit and the
+
-I S
and
Because the dc source sees the
S
+
satisfied if the current is to be a
Express the condition that must be
- qC(t) = 0
@]
t
252 Chapter 29 Because the peak CUlTent is 45 .2 rnA:
Inns
=
Ji 45 �A I 32. 0 mA I =
=
Chapter 30 Maxwell's Equations and Electromagnetic Waves Conceptual Problems *1
•
True or false:
(a) Maxwell's equations apply only to fields that are constant over time. (b) The wave equation can be derived from Maxwell's equations.
(c) Electromagnetic waves are transverse waves.
(d) In an electromagnetic wave in free space, the electric and magnetic fields are in phase.
(e) In an electromagnetic wave in free space, the electric and magnetic field vectors E and B are equal in magnitude.
(j) In an electromagnetic wave in free space, the electric and magnetic energy densities are equal. (a) False. Maxwell's equations apply to both time-independent and time-dependent fields. (b) True
(c) True (d) True
(e) False. The magnitudes of the electric and magnetic field vectors are related according toE= cB. (j) True *4
•
Are the frequencies of ultraviolet radiation greater or less than those of
infrared radiation? Determi ne the Concept
The frequencies of ultraviolet radiation are greater than those of
infrared radiation (see Table 30-1). *8
• A helium-neon laser has a red beam. It is shone in tum on a red plastic filter (of the kind used for theater lighting) and a green plastic filter. (A red theater-lighting filter transmits only red light.) On which filter will the laser exert a larger force?
A red plastic filter absorbs all the light incident on it except for the red light and a green plastic filter absorbs all the light incident on it except for the green light. If the red beam is incident on a red filter it will pass through, whereas, if it is
Determine the Concept
253
254
Chapt er
30 absorbs more a greater force on the green filter.
incident on the green filter it will be absorbed. Because the green filter energy than
does
the
red filter, the laser beam will
exert
Estimation and Approximation *12 ·· Estimate the radiat ion pressure force exerted on the earth by the sun, and compare the radiation pressure force to the gravit at ional attract ion of the sun. At the earth's orbit the intensity of sunlight is 1 .37 kW/m2 •
We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton' s law of gravit ation to find the gravitational force the sun exerts on the earth. Picture the Problem
The radiation pressure exerted on the earth is given by:
� =�
Fr
=>
= �A
where A is the cross- sectional area of the earth. Express the radiation pressure in terms ofthe intensity of the radiation I from the sun: Substituting for Pr and A yields:
= In R2 7 m2 )(6370kmY Fr = n(13 0W/ 3 xl08 mls = ! 5. 82x108 N ! Fr
Substitute numerical values and evaluate
Fr:
The gravitational force exerted on the earth by the sun is given by:
Gmsun me arth r2 where r is the radius of the earth's orbit .
F
Substitute numerical values and evaluate F
=
c
=
F:
10-11 . m2 I kg2 )(1.99 1030 kg)(5.98 1024 kg) = 3. 5 3 x 1022 N (1. 5 1011 m) Fr 5. 82 108 N = 1. 6 5 10-14 Fr F 3 . 5 3 1022 N F: The gravitational force is greater by a factor of approximately 1 014.
(6. 6 7
X
N
Express the ratio of the force due radiation pressure t o the gravitational force
X
X
X
X
X
X
Maxwell's Equations and Electromagnetic Waves 255 'k13 Repeat Problem 12 for the planet Mars. Which planet has the larger ratio of radiation pressure to gravitational attraction? Why? Picture the Problem We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton's law of gravitation to find the gravitational force the sun exerts on Mars.
The radiation pressure exerted on Mars is given by:
� =�
�
Fr
= �A
where A is the cross-sectional area of Mars. Express the radiation pressure on Mars in terms of the intensity of the radiation IMars from the sun: Substituting for Pr and A yields:
Express the ratio of the solar constant at the earth Iearth to the solar constant IMars at Mars:
I = ( J2 I = I I Mars
rearth
earth
rMars
�
Mars
earth
( J2 rearth
rMars
Substitute for IMars to obtain:
Substitute numerical values and evaluate Fr: Fr
1 I m J 2 = 1 7.09 x1 0 7 N I = 7r(1370W/3x m1028)ml(33s95km)2 ( 2.1. 520X10 9x 1011 m (0 ) =
The gravitational force exerted on Mars by the sun is given by:
F
Gmsun mMars
=
GmSW1
.1 1 mearth
r2 r2 where r is the radius of Mars' orbit.
Substitute numerical values and evaluate F: F
= (6.67 X10-1 N· m2 Ikg2(2.)(1.2999X101x 1031 m0 kg) )(0.11)(5. 9 8 x 1024 kg) = 1. 66 x 102 1 N due 9 X10 7 N = 4. 27 X10-1 4 _ 1.7 .606x1021 N
Express the ratio of the force radiation pressure Fr to the gravitational force F:
Fr F
256
Chapter Because the ratio of these forces is for the earth and 4.2 7 for Mars, Mars has the larger ratio. The reason that the ratio is higher for Mars is that the dependence of the radiation pressure on the distance from the Sun is the same for both forces whereas the dependence on the radii of the planets is different. Radiation pressure varies as , whereas the gravitational force varies as (assuming that the two pla nets have the same density, an assumpti on that is nearly true). Consequently, the ratio of the forces goes as Because Mars is smaller than earth, the ratio is larger. 30
1.65 x l 0-14
x 1 0-14
(r - 2 ) ,
R2
R3
1 R 2 I R 3 = R- •
*14 ·· In the new field of laser cooling and trapping, the forces associated with radiation pressure are used to slow down atoms from thermal speeds of hundreds of meters per second at room temperature to speeds of just a few meters per second or slower. An isolated atom will absorb radiation only at specific resonant frequencies. If the frequency of the laser-beam radiation is one of the resonant frequencies of the target atom, then the radiation is absorbed via a process called resonance absorption. The effective cross-sectional area of the atom for resonant absorption is approximately equal to A?, where A is the wavelength of the laser beam. (a) Estimate the acceleration of a rubidium atom (atomic mass 85 glmol) in a laser beam whose wavelength is 780 nm and intensity is 10 W/m2 • (b) About how long would it take such a light beam to slow a rubidium atom in a gas at room temperature (300 K) down to near-zero velocity? We can use Newton' s 2nd law to express the acceleration of an atom in terms of the net force acting on the atom and the relationship between radiation pressure and the intensity of the beam to find the net force. Once we know the acceleration of an atom, we c an use the definition of acc eleration to find the stopping time for a rubidium atom at room temperature. Picture the Problem
(a) Apply obtain:
I F=ma to the atom to
F=ma
where beam.
F is the force exerted by the laser
The radiation pressure Pr and intensity of the beam I are related acc ording t o: Solve for
F to obtain: F
Substitute for in the expression of Newton' s 2nd law to obtain: Solve for a:
F=
fA = e
f}}
-=ma e
f},}
a= me
IA} e
Maxwell's Equations and Electromagnetic Waves Substitute numerical values and evaluate a: a=
(b)
257
lOW/m2 ) (780nm)2 = 1 1.44xI Osm/s2 1 1 mol (85� mol x 6. 02 xl023 particles J (3X 108 m/s) = (
U sing the definition of
acceleration, express the stopping
Vfinal
I1t
a
time I':..t of the atom: Because Vfinal � 0:
I1t
Using the rms speed as the initial
�
speed of an atom, relat e Vini!ial to the t emperature of the gas:
Vini!ial
Substitut e in the expression for the
I1t
st opping time to obt ain:
- vini!ial
- v ini!ial a
= = �3:T VmlS
= _� �3kT a
In
Substitut e numeric al values and evaluate I':.. t:
I1t
= - 1.44x 1105 m /s2 31. 38x10-23 lmol(300K) = 1 2. 06ms I (85� molX 6. 02 x 1023 particles J JIK
___ ____
Maxwell's Displacement Current
*19 ··
Current of lOA fl ows into a capacitor having plates with areas of 2 0.5 m . (a) What is the displacement current between the plates? What is dEldt
(b) B· de
between the plates for this current? (c) What is the line integral of
around a circle
of radius 10 cm that lies within the plates and parallel to the plates? Picture the Problem
We can use the conservation of charge to find fct, the definitions of
the displacement current and electric flux to find dEldt, and Ampere' s law to evaluate
B· de
around the given path.
(a) From conservation of charge we know that :
Id
= = I IO.OA I I
258
Chapter
30
( b) Express the displa cemen t cutTent
h
Substitute for
e A elE d =E 0 leltrpe =E0 �[EA]=E 0 elt elt
1
elE elt
dEldt:
Substitute numerical values and evaluate dEldt:
�
Eo A
dtelE = (8. 8 5 X10-1 2 C2lOAI N · m2 )(0. 5 m2 ) = 1 2.26X101 2 � I {B if = !lol enclosed .
(c) Apply Ampere' s law to a circular path of radius r between the plates and parallel to their surfaces to obtain: Assuming that the displacement current is uniformly distributed:
Ienclosed Id Tr
r2 = A
=>
Ienclosed _
7r:
r2 Id
A
where R is the radius of the circular plates. Substitute for
lenclosed to obtain:
Substitute numerical values and evaluate
*23
{B. if :
•••
Show that the generalized form of Ampere ' s law (Equation 30-4) and the Biot Savart law give the same result in a situation in which they both can be used. Figure 30-13 shows two charges +Q and -Q on the x axis at x = -a and x = +a, with a current
1
=
- dQldt along the line between them. Point P is on the y axis at y = R. (a) U se the
Biot-Savart law to show that the magnitude of B at point P is B
(b)
= !l27r:R.ola .JR2 1 a2 +
Consider a circular strip of radius r and width dr in the yz plane with its center at the
origin. Show that the flux of the electric field through this strip is
Maxwell's Equations and Electromagnetic Waves 259 ExdA = il a r2 Eo
(
+
a2
t/2 rdr
(C) Use your result from Part ( b) to find the total flux CPe through a circular area of radius
R. Show that
( d) Find the displacement current h and show that
(e) Finally, show that Equation 30-4 gives the same result for B as the result found in Part ( a) .
Figure 30-13
y
p
Problem 23
Picture the Problem
We can follow the step-by-step instructions in the problem
statement to show that Equation 30-4 gives the same result for B as that given in Part ( a). ( a) Express the magnetic field at P using the expression for B due to a straight wire segment:
where
. eI sm
Substitute for sinel and sin� to obtain:
B
=
Ji
s
. e2 m
=
I
.,j
a
R2 +a2
p - 4JZ'o R .,jR2 +a2 _
Jiola
2a
2JZ'R .,j
1
R 2 + a2
260
Chapter
30
(b) Express the electric fl ux through
the circular strip of radius rand width dr in the yz plane:
The electric field due to the dipole IS:
Substitute for Ex t o obtain:
d¢e = ExdA =
=
2kQa
(2
2 W2 (2rcrdr )
r +a J
2Qa
(2
2 W2 (2rcrdr)
4rc Eo r +a J
=
(c) Multiply both sides of the expression for ¢e by Eo: Integrate r from 0 to R to obt ain: , Eo
- Rf (2
¢ e - Qa
o
r
rdr
-
(d) The displacement current is defined to be:
The t ot al current is the sum of I and
h
(e) Apply Equation 30-4 (the generalized form of Ampere's law) to obtain:
(
-1
2 - Qa -JR 2 +a 2 +a 2 )1
1
+a
J
-
Q
(1 - -J 2
a
R +a
2
J
Maxwell's Equations and Electromagnetic Waves 261 B=�(I+I) 2Jr R B-- � [ I a 2Jr R .JR2+a2 J
Solve [or B:
d
Substitut e for I + ld from (d) to obt ain:
= f.1oIa 1 W.JR2+a2
Maxwell's Equations and the Electromagnetic Spectrum
*25
·
Find the wavelength for (a) a typical AM radio wave with a frequency of
1000 kHz and (b) a typical FM radio wave with a frequency of 100 MHz. Picture the Problem
We can use c =fA to find the wavelengths corresponding to the
given frequencies. Solve c =fA for A :
(a) Forf= 1000 kHz :
A=
3 X 108m1s = 300 mI 1000 x103 S-I I
(b) Forf= 100 M Hz :
*26
•
What is the frequency of a 3-cm microwave?
Picture the Problem
wavelength. Solve
c =fA
We can use c =fA t o find the frequency corresponding to the given
forf
Substitute numerical values and evaluatef
f= !:...A 3x m1s =10 I oHz = 1 10.0 H f= 3x108 . G zI 102_ m .
262 Chapter
30
Elect.-ic Dipole Radiation *32
At a distance of 30 Ian from a radio station broadcasting at a frequency of 0.8 3 MHz, the intensity of the electromagnetic wave is 2 x 10-1 W/m2. The transmitting •••
antenna is a vertical dipole. What is the total power radiated by the station? Picture the Problem
The intensity of radiation from an electric dipole is given by
lo( sin2 B)/r2, where B is the angle between the electric dipole moment and the position
vector r. We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate 10. Knowing 10 we can find the total power radiated by the
station. From the definition of intensity we
dP = IdA
have:
and
Ptot = ffI(r, e)dA
where, in polar coordinates,
sinededrp 21"[ = �ot f f I(r, e) r2 sin ededrp o0
dA = r2
Substitute for dA to obtain:
Express the intensity of the signal as
(1)
a function of rand B: Substitute for lr( , 8):
�ot = 10
2""
f fSin3 eded¢
o0
From integral tables we find that:
Substitute and integrate with respect to ¢ to obtain:
From equation ( 1) we have:
Substitute to obtain:
2" 4 8n 2" =-1 Io f drp=-Io[] rpo �ot=0 3 0 4
Io =
3
I(r,e)r2 e
sin2
8n I(r,e)r2 tot = 3 e
p
sin2
or, because B= 90°,
3
Maxwell's Equations and Electromagnetic Waves 263 ( ?"Ol =�I ) 2 ?"ot = ¥ (2X10- 13 W/m 2 )(30km )2 = / 1. 5 1mW / 8Jr .:>
r r
Substitute numerical values and
evaluate Ptot:
Energy and Momentum in an Electromagnetic Wave
*38
·
The rms value of the magnitude of the magnetic field in an electromagnetic
wave is Brms
=
0.245 ,uT . Find (a) EmlS, ( b) the average energy density, and (c) the
intensity. Picture the Problem
Given BmlS, we can find Erms using Erms = cBrms. The average energy
density of the wave is given by Uav = E mlSBrmslf.1oC and the intensity of the wave by 1= UaVC
(a) Express Enns in terms of BmlS: Substitute numerical values and
Erms
evaluate Erms:
= ( x 108 mls)(0.245,uT) = 1 Vim 1 3
73.5
(b) The average energy density Uav is given by: Substitute numerical values and evaluate Uav:
Uav
= (4Jr(x10-Vi7 N/mA) (20.)(3245x10,uT8 )mls) =1 1 73.5
47.8nllm3
(c) Express the intensity as the product of the average energy density and the speed of light in a vacuum: Substitute numerical values and
evaluate 1:
)( x 10 8 m/s ) 1=( = 1 14. 3 W /m2 I 47.8nllm3 3
264 Chapter 30 *41
··
An AM radio station radiates an isotropi c sinusoidal wave with an average
power of 50 kW. What are the amplitudes of Emax and Bmax at a distance of (a) 500 m, ( b) 5 km, and (c) 50 km?
Picture the Proble m
of
Pav
We can use 1=
Pavl4trr2
and the distance r from the station .
and 1 = ErmsBnn/f.1o to express Errns in terms
Express the intensity I of the
radiation as a function of its average power and the distance r from the station: The intensity is also given by:
I
=
EmlSBrms E� E�13X =
flo
_ 4rc r 2 E
�
Equate these expressions to obtain:
Solve for Emax:
=
cflo
2cflo
�x IIo 2cr
�(!)r
Emax �� =
(a) Substitute numerical values and evaluate Emax for r = 500 m: Emax
(SOOm)
=
U se Bmax = Emaxlc to evaluate Bmax:
4rc 10 -7 N/A2 (SOkW) ( 1 !3.46V/m I 500m J 3.46V/m 1 11. Sn T I B 3X 108 rnIs x
=
2rc
max
=
=
(b) Substitute numerical values and evaluate Emax for r = 5 km:
Use Bmax = Emaxlc to evaluate Bmax:
Bmax
=
0.346 VIm /l.lSn T / 3X 108 rnIs
(c) Substitute numerical values and evaluate Emax for r = 50 km:
=
Maxwell's Equations and Electromagnetic Waves 265 Use Bmax
*43
=
EmaJc to evaluate Bmax:
••
Bmax
. 0 346 Vim = I 0.115nT I = 03xl0 8 mls
Inst ead of sending power by a 750-kV, 1 000-A transmission line, one desires
to beam this energy via an electromagnetic wave. The beam has a uniform intensity within a cross-sectional area of 50 m2• What are the rms values of the electric and the magnetic fields? Picture the Problem
We can use 1
of 1. We can then use Brms
=
=
ErmsBmlSlJ.1o and Bm1s = EmlS/c to express Erms in t erms
EmlS/c t o find BmlS .
Express the intensity 1 of the
radiation as a function of its average
EmlS BmlS
I
=
I
= =
_
E!lS Cllo
f-Lo
power and the distance r from the station:
U se the definition of intensity t o relat e the intensity o f the electromagnetic wave to the power
p A
Itrans.line V A
in the beam: Substitute for 1 to obtain:
Cf-Lo1trans.line V
EmlS
=
Brms
7S. 2kV/m = 1 0.2S1mT I = 3xl0 8 m 1s
A
Substitut e numerical values and evaluate EmlS:
E rms
=
266 Chapter 30 *45
The electric fi eld of an electromagnetic wave oscillates in the y direction and
••
the Poynting vector is given by
where x is in meters and t is in seconds. ( a) What is the direction of propagation of the wave?
(b)
Find the wavelength and the frequency. ( c) Find the electric and magnetic
fields. Picture the Problem We
can determine the direction of propagation of the wave, its
wavelength, and its frequency by examining the argument of the cosine function. We can find E from
lSi
= E2
/f.1oc and B from B
=
Elc. Finally, we can use the definition of the
Poynting vector and the given expression for S to find E and B . -
-
-
Because the argument of the cosine function is of the [onn kx the wave propagates in the positive direction. 2lr 10m-1 (b)
( a)
-
illt,
x
Examining the argument of the
cosine function, we note that the
k=
wave number k of the wave is: Solve for and evaluate A:
A
=
A
=
2�_1 I 0.628m I 10 =
Examining the argument of the cosine function, we note that the angular frequency
OJ of the wave
is:
Solve for and evaluatef to obtain:
f = 3xl02lr9s-1 1477 MHz I =
(c) Express the magnitude of Sin terms of E: Solve for E: Substitute numerical values and evaluate E:
E=
�f.1oCISI
Maxwell's Equations and Electromagnetic Waves Because
S(x,t) =(100W/m2)cos2[lOx-(3x109)t]i and S = _1- Ex B: Jlo
U se B = Elc to evaluat e B: -
Because S with
B
267
=
1
-
Jlo
B=
194 Vim = 0.6 7 Jl 4 3x108m/S
T
E x B , the direction of B must be such that the cross product of E
is in the positive x direction:
The Wave Equation for Electromagnetic Waves
*52
•••
( a) Using argument s similar to those given in the text, show that for a plane
wave, in which E and B are independent ofy and z,
aEz
__
ax
aBy
= --
at
and
aBy
--=Jl 0 E 0 aEz ax
--
at
(b) Show that Ez and By also satisfy the wave equation. Picture the Problem
We can use Figures 30-10 and 30- 1 1and a derivation similar to that
in the text to obtain the given results. In Figure 30- 1 1, replace Bz by Ez• For i'll" small:
Eva luate the line integral of
Earound the rectangular area i'll"fu:
Express the magnetic flux through the same area :
- 1JE · de;:::
aE z ax
&!:J.z
__
( 1)
268
Chapter 30
Apply Faraday's law to obtain:
B = o(B && ) at y at
a E ·d.e�--
1-
-
i
n
dA
--
Substitute in equation ( 1 ) to obtain:
aEaxz _ aBaty
or
In Figure 3 0- 1 0, replace
Ey By by
and evaluate the line integral of
B
around the rectangular area &�:
Evaluate these integrals to obtain:
(b)
provided there are no conduction curre nts.
aBaxy _ aEatz
-- - flo Eo --
Using the first result obtained in
Ez
(a), find the second partial derivative of
with respect to x:
Use the second result obtained in (a) to obtain:
2 1 / c , aax2E2 c21 aat2E2z aax (aBaxy)_- 0 0 axa (aEz) -at
or, bec ause f..1oEo = z
__
Using the sec ond result obtained in
By
(a), find the second partial derivative of
with respect to x:
Use the second result obtained in (a) to obtain:
=
---
___
II E-
Maxwell's Equations and Electromagnetic Waves 269 ?
or, because Po Eo = l/c-, (iBy
--
8x2
1
82 By
c2 8t2
---
General Problems
*57 ··
A circul ar l oop of wire can be used to detect electromagnetic waves. Suppose
a 1 00-MHz FM radio station radiates 50 kW uniformly in all directions. What is the maximum rms voltage induced in a l oop of radius 30 cm at a distance of 1 05 m from the station? Picture the Problem
The maximum rms voltage induced in the loop is given by
Crms = AwBo / J2, where A is the area of the loop, Bo is the amplitude of the magnetic
field, and
OJ
is the angular frequency of the wave . We can use the definition of density
and the expression for the intensity of an electromagnetic wave to derive an expression
for Bo.
The maximum induced rms emf occurs when the plane of the l oop is perpendicular to
B:
From the definition of intensity we have:
Crms =
AwBo
J2
JiR2wBo
(1)
where R i s the radius of loop of wire .
I =� 47r r2 where r is the distance from the transmitter.
The intensity is al so given by:
Substitute to obtain:
Solve f or Bo:
Substitute in equation ( 1 ) to obtain:
1=
EoBo B�c = 2J.10 2J.10
270
30
Chapter
Substitute numerical values and evaluate E
*61
-
_
nns
··
given by E)
6fms:
(0.3m-=Y( 100MHz) "':" -J22 (1Os- -m"""T")
-'--
---"-
The electric fields of two harmonic waves of angular frequency =
E)O, cos(k)x - w/)J
and
w)
and
OF.!
are
E2 E2,0 COS(k2X - w2t + 5)J . Find (a) the =
instantaneous Poynting vector for the resultant wave motion and (b) the time-average Poynting vector. If the direction of propagation of the second wave is reversed so E2 =
E2 0 COS(k2X + w2t + 5)J , find (c) the instantaneous Poynting vector for the
resultant wave motion and (d) the time-average Poynting vector. Picture the Problem
We can use the definition of the Poynting vector and the -
relationship between Band
-
E to find the instantaneous Poynting vectors for each of the
resultant wave motions and the fact that the time average of the cross product term is zero for
lV) :;t OF.!,
vectors.
and 12 for the square of cosine function to find the time-averaged Poynting -
(a) Because E) and the
-
E2 propagate in
x direction:
-
-
"
E x B = fLoSi
=>
-
,..
B = Bk
Expre ss B in terms of E) and E2: Substitute for E) and E2to obtain:
Express the instantaneous Poynting vector for the resultant wave motion:
X =
=
�(E),o cos(k)x - w)t)+ E2,0 COS(k2X - w2t + 8))k
2 _1_ (E)'0 cos(k)x - w)t)+ E2 0 COS(k2X - w2t + 5)\ (J x k) ,
flo c
_1_ �c
J
[E)�o cos2(k)x - w)t)+ 2E) OE20 cos(k)x - w/) '
X
,
2 COS(k2X - w2t + 5)+ E�,o cos (k2X - w2t + 5)] i
Maxwel l's Equations and Electroma gnetic Wave s
(b)
The time average of the cross product term is zero f or lVj :;r 0>.1, and the time aver age of the square of the c osine ter ms is 1'2:
1 Sav = - --
2 /-ioC
[2 2]:E].o + E2.0 I
(c) In this case B2 = Bk bec ause the wave with k = k2 prop agates in the The magnetic field is then: -
2 71
�
-
�
-
i direction.
Express the instantaneous Poynting vector for the resultant wave motion:
s=
1 (E]0 cos(k]x - OJ]t)+ E20 COS(k2X - OJ2t + ))J x �(E1,0 cos(k]x - OJ]t)- E2,0 COS(k2X + OJ2t + o))ic
_ _
f-Lo
'
0
,
(d) The time average of the square
Sav
of the cosine terms is 1'2:
1 [E 20 - E220]:-I 1 ,
=
--
2/-ioC
'
*62
·· At the surface of the earth, there is an approximate average solar flux of 0 .75 2 kW/m . A family wishe s to c onstruct a solar energy conversion system to p ower their home . If the conversion system is 30 percent efficient and the family needs a maximum of 25 kW, what effecti ve surface area is needed for perfectly absorbing collectors? Picture the Problem We can use
the definitions of power and intensity to express the
area of the surface as a function of P, 1, and the efficiency
Use the definition of power to relate
E & IA& = t
the requ ired surface area to the
p=
intensity of the solar radiation:
where
Solve for A to obtain:
G
A =� 1&
Substitute numerical values and evaluate A:
G.
is the efficiency of the system.
272 Chapter *65
30
A long cylindrical conductor of length L, radius
•••
and resistivity p carries a
a,
steady CUlTent I that is uniformly distributed over its cross-sectional area. ( a) Use Ohm's law to relate the elect ric field E in the conductor to I, p, and a.
(b)
Find the magnetic field
B just outside the conductor. (c) Use the results for Part (a) and Part (b) to compute the
(i jj);f.1o at (the edge of the conductor). In what direction is S ? (d) Find the flux 1S n dA through the surface of the conductor into the conductor,
Poynting vector
S=
x
r= a
and show that the rate of energy flow into the conductor equals iR, where R is the
resistance of the cylinder. (Here Sn is the inward component of surface of the conductor.) Picture the Problem
-
S perpendicular to the
We can use Ohm's law to relate the electric field E in the conductor
to I, p, and a and Ampere's law to find the magnetic field B j ust outside the conductor.
Knowing
- E and B we can find S and, using its normal component, show that the rate of
energy flow into the conductor equals iR, where R is the resistance . (a) Apply Ohm ' s law to the cylindrical conductor to obtain:
v = IR = IpL = IpL2 = EL A TCa
Solve for E:
(b)
Apply Ampere 's law to a
circular path of radius a at the
fji . ie
=
B{2TCa) = f.1oIenclosed = f.101
surface of the cylindrical conductor: Solve for B to obtain:
(c) The electric field at the surface of the conductor is in the direction of the current and the magnetic field at the surface is tangent to the surface. Use the results of (a) and
(b)
and the right-hand rule to
evaluate
S:
- 1 1(
- S=-ExB f.1o Ip f.101 "tangent = - -- "parallel 2TCa f.1o TCa 2
J
�
X(
J
�
where f is a unit vector directed radially outward from the cylindrical conductor.
(d) The flux through the surface of the conductor into the conductor is:
273
Maxwell's Equations and Electromagnetic Waves Substitute for 5n, the in ward
component of S and simplify to obtain : ,
pL pL . Smce R = = : A T[{1 2 -
*67
•••
--
fSll dA = �
Small particles might be blown out of solar systems by the radiation pressure
of sunlight. Assume that the particles are spherical with a radius r and a density of 1 g/cm3 and that they absorb all the radiation in a cross-sectional area of ;rr2 . The particles are a distance R from the sun, which has a power output of 3 .83 x l 0 26 W. What is the radius r for which the radiation force of repulsion just balances the gravitational force of attraction to the sun? Picture the Problem
We can use a condition for translational equilibrium to obtain an
expression relating the forces due to gravity and radiation pressure that act on the particles. We can express the force due to radiation pressure in terms of the radiation pressure and the effective cross sectional area of the particles and the radiation pressure in terms of the intensity of the solar radiation. We can solve the resulting equation for r. Apply the condition for translational equilibrium to the particle: ( 1) The radiation pressure Pr depends on the intensity ofthe radiation I: The intensity of the solar radiation at a distance R is:
Substitute to obtain:
Substitute for Pr, A, and m in equation ( 1) : Solve for R t o obtain:
Substitute numerical values and evaluate r:
r = 16rc p3Pc GM
-----
s
274 Chapter
30
= I 0. 5 74 *69
3 108 m/s 6.67 x 10-1 1 N · m2 2 I kg
X
JLll1
I
1 .99
x 1030
kg
An intense point source of light radiates 1 MW isotropically. The source is
•••
located 1 m above an infinite, perfectly reflecting plane. Detenrune the force that acts on the plane. Picture the P roblem
ring of radius
r
Let the point source be a distance a above the plane. Consider a
and thickness
dr
in the plane and centered at the point directly below the
light source. Express the force of force on this ring and integrate the resulting expression to obtain
F.
The intensity anywhere along this infinitesimal ring is PI4Jr (r2 + a 2) and the element of force
on this ring of area 2Jr
dF
rdr
is given by:
F
d
P rdr r 2 +a 2
_
-
(
c
a 2 r +a 2
) .J
Pardr 2 r 2 +a 2
= ---;---� -,-
c
)1
{
where we have taken into account that only the normal component of the incident radiation contributes to the force on the plane, and that the plane is a perfectly reflecting plane.
Integrate
dF r from
=
r
0 to =
00 :
1
From integral tables:
a
Substitute to obtain:
F = (!) F = 3x110M8W = 1 3 . 3 3mN I Pa c
Substitute numerical values and evaluate
F:
a
=
P
mls
e
Chapter 3 1 Properties of Light Conceptual Problems
*5
••
The density of the abnosphere decreases with height, as does the index of
refraction. Explain how one can see the sun after it has set. Why does the setting sun appear flattened? Determine the Co ncept
The change in abnospheric density results in refraction of the
light from the sun, bending it toward the earth. Consequently, the sun can be seen even after it is just below the horizon. Also, the light from the lower portion of the sun is refracted more than that from the upper portion, so the lower part appears to be slightly higher in the sky. The effect is an apparent flattening of the disk into an ellipse.
*10
••
We learned in Chapter 30, Section 30-3 , that an oscillating electric dipole
produces electromagnetic radiation (see Figure 3 0-8). Assuming that the light reflected off and refracted into the surface of a piece of transparent material is caused by such dipoles, show that the condition for Brewster's angle (Equation 3 1 -2 1 ) is exactly the same as saying that the refracted ray is perpendicular to the axis of the radiating dipoles for light polarized in the plane of incidence. Determine the Co ncept
The diagram shows that the radiated intensity for a dipole is
zero in the direction of the dipole moment. Because the dipole axis is in the same direction as the polarization, for light polarized parallel to plane of incidence, the dipole axis will point in the same direction as the reflected wave, i.e., in the direction described by Brewster's law. As the diagram indicates, there is zero field in the direction of the refracted ray. On the other hand, if the incoming wave is polarized perpendicular to the plane of incidence, the dipole axis will never point along the direction of propagation for the reflected or refracted wave. Unpolarized incident ray Polarized reflected
roy
Slightly polarized refracted ray
27 5
276
Chapter 3 1
*1 4
It is a common experience that on a calm, sunny day one can hear voices of
persons in a boat over great distances. Explain this phenomenon, keeping in mind that sound is reflected from the surface of the water and that the temperature of the air just
above the water' s surface is usually less than that at a height of 1 0 m or 20 m above the water. Picture the Problem
The sound is reflected specularly from the surface of the water (we
assume it is calm). It is then refracted back toward the water in the region above the water because the speed of sound depends on the temperature of the air and
is greater at the
higher temperature. The pattern of the sound wave is shown schematically below.
Sources of Light *24
··
Singly ionized helium is a hydrogen-like atom with a nuclear charge of 2e. Its energy levels are given by Ell = -4Eofn2 , where Eo = 1 3 .6 eV. If a beam of visible white light is sent through a gas of singly ionized helium, at what wavelengths will dark lines be found in the spectrum of the transmitted radiation? Determine the Concept
The energy difference between the ground state and the first
excited state is 3Eo = 40. 8 eV, corresponding to a wavelength of 30.4 nm. This is in the far ultraviolet, well outside the visible range of wavelengths. There will be no dark lines in the transmitted radiation. The Speed of Light
*29 ··
In Galileo's attempt to determine the speed of hght, he and his assistant were
located on hilltops about 3 km apart. Galileo flashed a light and received a return flash
from his assistant. (a) If his assistant had an instant reaction, what time difference would
Galileo need to be able to measure for this method to be successful?
(b)
How does this
time compare with human reaction time, which is about 0.2 s? Picture the Problem
We can use the distance, rate, and time relationship to find the time
difference Galileo would need to be able to measure the speed of light successfully.
(a) Relate the distance separating Gahleo and his assistant to the speed of light and the time required for it travel to the assistant and back to Galileo:
D
=
cf:.t
Propeliies Solve for t1t:
Substitute numerical values and
evaluate t1t:
(b) Express the ratio of the human
reaction time to the transit time for ,
the light:
!:'t
=
!:' t
=
277
D
-
c
2 3 m ( k ) x 3 l 08 mls
!:, t rcaction
or
of Light
!:'t
!:'treaction
=
=
0.2 s
20 ,us
11
=
=
04 !:'t
I 20.0 ,us I 1 04
I
Reflection and Refraction *31
••
A ray of light is incident on one of a pair of mirrors set at right angles to each
other. The plane of incidence is perpendicular to both mirrors. Show that after reflecting off each mirror the ray will emerge in the opposite direction, regardless of the angle of incidence.
Picture the Problem The diagram shows
ray 1 incident on the vertical surface at an
angle 81> reflected as ray 2, and incident on
the horizontal surface at an angle of
incidence � . We'll prove that rays 1 and 3
are parallel by showing that 81 = 84 , i.e., by showing that they make equal angles with the horizontal . Note that the law of
reflection has been used in identifying
equal angles of incidence and reflection.
We know that the angles of the right
triangle formed by ray 2 and the two
mirror surfaces add up to 1 80°: The sum of fh. and � is 90°: Because 81
=
82 :
The sum of 84 and � is 90°: Substitute for � to obtain:
3
278
Chapter
*37
31
Light is incident nonnally on a slab of glass with an index of refraction n =
1 .S . Reflection occurs at both surfaces of the slab. Approximately what percentage of the
incident light energy is transmitted by the slab? Picture the Problem Let the subscript
1
refer to the medium to the left (air) of the
rt l
first interface, the subscript 2 to glass, and
=
1
1"13
= 1
II,l
the subscript 3 to the medium (air) to the
-
right of the second interface. Apply the
equation relating the intensity of reflected
---
II
light at nonnal incidence to the intensity of
the incident light and the indices of
13
refraction of the media on either side of the interface to both interfaces. We'll neglect
multiple reflections at glass-air interfaces. Express the intensity of the
transmitted light in the second medium:
Express the intensity of the
transmitted light in the third
medium:
Substitute for 12 to obtain :
Solve for the ratio VII :
Substitute numerical values and evaluate 1311) :
13 I)
5 )2 (1. 5 -1)2 -[1 (1-1. ][ ] 1 1+1.5 1. 5 +1
=
0.922 I 92.2% I =
Properties of Light 279 *40
Figure 3 1 -5 6 shows a beam of light incident on a glass plate of thickness d
and index of refraction
(a ) Find the angle of incidence so that the perpendicular
11.
separation between the ray reflected from the top surface and the ray reflected from the bottom surface and exiting the top surface is a maximum. (b) What is this angle of incidence if the index of refraction of the glass is 1 .60? What is the separation of the two beams if the thiclmess of the glass plate is 4.0 cm?
Figure 31 -56
Air Problem 40
Picture the Problem
l __ e � 1 I I I I I I
Let x be the
perpendicular separation between the two rays and let .e be the separation between the points of emergence of the two rays on
AiI
the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of .e,
d, er,
d
Glass
and �. Setting the derivative of the resulting equation equal to zero will yield Air
the value of � that maximizes x. (a) Express .e in terms of angle of refraction er:
d
and the
Express x as a function of .e, en and �:
d,
Differentiate x with respect to �:
f.
=
2dtanBr
Chapter 31 dx = 2e1 --d (tan Br cos Bj ) 2d( - tan Br sm. Bj sec 2 Br cos Bj dBr d8 J d8 d8 n J sin Bj = n2 sin Br nJ n2 n, sinBj = nsinBr
280
=
-
I
I
+
Apply Snell's law to the air-glass interface:
or, since
Differentiate implicitly with respect to� to obtain:
__
I
= 1 and
(1)
(2)
=
or
dBr = dBj
- --
Substitute in equation ( 1 ) to obtain:
1 - sin 2 Bj 1 . B sm. Br cos 2 B -sm n
Substitute j
for
and
j
for
to
obtain:
cos 2 Br /cos 2 Br dx = 2d( 1 - sin 2 Bj sin 2 Bj cos 2 Br = 2d (1 - sm. 2 B. - sm. 2 B. cos 2 Br ) dBj ncos3 Br ncos3 Br J ncos3 Br 1 - sin 2 Br cos 2 Br : dx dB. 1 n Bj sin Br -si n 2d [1 -sm. 2 B. - sm. 2 B. (1 - -sm 1 . 2 B. )] = dBj ncos3 Br n2 lIn2,
Multiply the second term in parentheses by
and simplify to obtain:
-
--
I
Substitute
I
for
I
Substitute
for
dx
--
Factor out
to obtain:
I
I
simplify, and set equal to zero to obtain:
I
Properties of Light 281
dx 2d3 [si e - 2n2 sin2 e; +n2 ] = O for extrema de; n 3 cos er n4
;
If dxld81 = 0, then it must be true
that:
Solve this quartic equation for (), to
obtain:
(b) Evaluate Bt for
n
=
1 .60:
B; sin -(1.6 1 48 . 5 ° I 2d taner cose; =
=
In (a) we showed that: Solve equation (2) for
x =
8r:
Substitute numerical values and
evaluate
8r:
Substitute numerical values and
evaluate x:
er sin-I ( _1.I-6 sin48 . 5 0 ) 27. 9° 2(4cm)tan2 7.9°cos48. 5 ° 1 2. 8 1cm I =
=
x =
=
Total Internal Reflection
*46
••
An optical fiber allows rays of light to propagate long distances through total
internal reflection. As shown in Figure 3 1 -57, the fiber consists of a core material with index of refraction n2 and radius b, surrounded by a cladding material of index
n3
< nz.
The numerical aperture of the fiber is defined as sin 8J , where 81 is the angle of incidence
of a ray of light impinging the end of the fiber that reflects off the core-cladding interface at the critical angle. Using the figure as a guide, show that the numerical aperture is
given by
��
n - n� assuming the ray is incident from air. (Hint: Use of the Pythagorean
theorem may be required. )
282
Chapter 3 1 Incident ray f13 --c;:::"""'"".,....,.,-:"7;:-"'7;-:-�-=:=:��
r-=-���""""-""
11,
Figure
31-57 Problems 46, 47, and 48.
Picture the Problem We can use the geometry of the figure, the law of refraction at the
air-n ) interface, and the condition for total internal reflection at the n)-n2 interface to show that the numerical aperture is given by
Referring to the figure, note that:
�n� - n� .
n3
.
SIn ec
n2
and
the right triangle to obtain:
or
a2
=-
c
b2
-+- =
c
b
2
c
1
a
!:
Substitute for !!:.- and � to obtain: c c
sin B
c
2
�l- : �l-
Solve for - : c
�
c
b
SIn . e2 Apply the Pythagorean theorem to
a
= -=
c
2
�
n
;
n 22
Use the law of refraction to relate 8) and �:
Substitute for sin � and let n) = 1
(air) to obtain:
Dispersion
*51
••
A beam of light strikes the plane surface of silicate flint glass at an angle of
incidence of 45°. The index of refraction of the glass varies with wavelength, as shown in
the graph in Figure 3 1 -26. How much smaller is the angle of refraction for violet light of wavelength 400 nm than that for red light of wavelength 700 nm?
Properties of
Light
283
Picture the Problem We can apply Snell's law of refraction to express the angles of
refraction for red and violet light in silicate flint glass. \
Express the difference between the angle of refraction for violet light
11
=
(1)
B Br red - Br violet ,
,
and for red light:
Apply Snell ' s law of refraction to the interface to obtain:
sin 45 0 n sin Br =
B
SIn. -, ( J21 n )
AB
= SIn. , ( J21nred J - SIn-, ( J2n1violet J
r =
Substitute in equation ( 1 ):
Substitute numerical values and
evaluate 11 8 :
Ll
MJ
.
-
Sin -' ( )2(: 60)) -Sin-' ( )2(:.66)) = 26.23 0 - 25. 2 1 0 1 1. 020 1 =
=
Polarization
*60 ··
A stack of N + 1 ideal polarizing sheets is arranged with each sheet rotated
by an angle of Jr/(2N) rad with respect to the preceding sheet. A plane, linearly polarized light wave of intensity
10
is incident normally on the stack. The incident light is polarized
along the transmission axis of the first sheet and is therefore perpendicular to the
transmission axis of the last sheet in the stack. (a) Show that the transmitted intensity through the stack is given by the expression
10 COS2 N(�).
(b) Using a spreadsheet or
graphing program, plot the transmitted intensity as a function of N for values of N from 2
to 1 00. (c) What is the direction of polarization of the transmitted beam in each case? Picture the Problem Let In be the intensity after the nth polarizing sheet and use
1 = 10 cos2 B
to find the ratio of 1n+' to In.
(a) Find the ratio of
1,,+1
to In:
I
-.!!±!.. =
In
COS2 2N 1r
284 Chapter
31
1
N+I
Because there are N such reductions
11
of intensity:
=
1N+I 10
=
) COS2 N(� 2N
and
IN+I
10
=
cos 2N (2N) n
IN+/lo
as a function of N is shown below. The (b) A spreadsheet program to graph formulas used to calculate the quantities in the columns are as follows:
�.
I,
•
1 2 3 4 5,
co '
"'�§
96 97 98 99
WO
A graph of
1110
Algebraic Form N N+ 1
ContentIForrnula 2 A2 + 1 (cos(PIO/(2* A2)Y'(2* A2)
Cell A2 A3 B2
COS2N ( 2:V) B
A N 2 3 4 5
0.250 0.422 0.5 3 1 0.605
95 96 97 98 99 1 00
0.974 0 .975 0.975 0.975 0.975 0.976
1110
."
as a function of N follows.
t lf
c
1 .0
".,-,=-r .,.....,.,...,
0.6
-· H!.o---f��=:!.�����f;::;:';,....;:;,p�-+-�-H'-"""4
0.5
H'---1I--'-':!";!-'''''::':-*
o
10
20
30
40
50 N
60
70
80
90
1 00
(c)
*62
Properties of Light In each case, the pol arization of the transmitted beam is perpendicular to that of the incident beam. ••
285
Show that a linearly polarized wave can be thought of as a superposition of a
right and a left circularly polarized wave.
Picture the Problem A circularly polarized wave is said to be
right circularly polarized
if the electric and magnetic fields rotate clockwise when viewed along the direction of
propagation and left circularly polarized if the fields rotate counterclockwise.
Ex = Eo cos OJt and Ey = Eo sin OJt or Ey = -Eo sin OJt
For a circularly polarized wave, the x
and y components of the electric
field are given by:
for left and right circular polarization,
respectively. For a wave polarized along the x
Eright + EJeft -
aXIS:
*64
x
••
-
= Eo cos OJt i +Eo cos OJt i = I 2Eo cos OJt i I ,..
"
Show that the electric field of a circularly polarized wave propagating in the
direction can be expressed by
i Eo sin(kx mt)J + Eo cos(kx -OJt)k =
-
Picture the Problem We can use the components of E to show that E is constant in -
-
time and rotates with angular frequency
w.
Express the magnitude of E in terms of its components:
Substitute for Ex and Ey to obtain:
E = �[Eo sin(kx - OJt)Y + [Eo cos(kx -OJt)] 2 = �E� [sin 2 (kx -OJt) + cos 2 (kx -OJt)] = Eo and the i vector rotates in the yz plane with angular frequency w.
Chapter
2 86
31
General P roblems *68
••
Figure 3 1 -5 8 shows two plane minors that make an angle e with each other.
Show that the angle between the incident and retlected rays is 2 e.
Figure
31-58 Problem 68
Picture the Problem
Angle ADE is the
angle between the direction of the incoming ray and that reflected by the two minor surfaces. Note that triangle ABC is
A
isosceles and that angles CAB and ABC are equal and their sum equals e. Also from the law of reflection, angles CAD and CBD equal angle ABC. Because angle BAD is twice BAC and angle DBA is twice CBA, angle ADE is twice the angle e.
*71
··
A swimmer at the bottom of a pool 3 m deep looks up and sees a circle of
light. If the index of refraction of the water in the pool is 1 .3 3 , find the radius of the circle. Picture the Problem
We can apply Snell' s law to the water-air interface to express the
critical angle ee in terms of the indices of refraction of water (nl) and air (n2) and then relate the radius of the circle to the depth Air,
n2
Water,
=
1
n,
=
d
1 .33
of the swimmer and ee.
r--
1
r
--.. 1
Properties of Light Relate the radius of the circle to the depth
d
of the point source and the
r =
287
d tan Be
critical angle ee:
Apply Snell's law to the water-air interface to obtain: Solve for ee:
Substitute for ee to obtain:
Substitute numerical values and evaluate r:
*76 ·· A Brewster window is used in lasers to preferentially transmit light of one polarization, as shown in Figure 3 1 -59. Show that if is the polarizing angle for the
/
/
Bp I nl n2 interface, then Bp 2 is the polarizing angle for the n2 n I interface.
Figure
31-59 Problem 76
Picture the Problem
Let the angle of refraction at the first interface by el and the angle
of refraction at the second interface be fh.. We can apply Snell's law at each interface and eliminate el and
n2 to show that fh. = Bt>2.
Apply Snell' s Brewster' s law at the
nl-n2 interface:
tan B
PI
-
_
2
n nl
_
288
Chapter 3 1
Draw a reference triangle consistent with Brewster's law:
Apply Snell 's law at the n l -n2 interface: Solve for B, to obtain:
Referring to the reference triangle we note that:
8I
e,
.
(
sin -,
(-:-: -�r=n=�n=:==n=; J
nl . 8 = sm _I �sln �
PI
J
i.e., B, is the complement of �" Apply Snell's law at the n2-n, interface: Solve for (h. to obtain:
Refer to the reference triangle again to obtain:
Equate these expressions for 8, to obtain:
n2 sin
*79
••
(a) For a light ray inside a transparent medium that has a planar interface
with a vacuum, show that the polarizing angle and the critical angle for internal reflection
satisfy tan Bp
=
sin Be. (b) Which angle is larger?
Properties of Light 289 P i c t u r e the Problem
angle to show that tan
We can apply Snell's law at the critical angle and the polarizing
Bp Be. = sin
(a) Apply Snell's law at the medium-vacuum interface:
sin Be sin 90° 1 tan BP tan Bp 1 n
For
B1
= Bp, n 1
=
n, and n2 = 1 :
=
=
n
= _ 2 =
n
1
n
=>
n
=
1
Because both expressions equal one: (b) For any value of
*85
••
B:
Suppose rain falls vertically from a stationary cloud 1 0,000 m above a
confused marathoner running in a circle with constant speed of 4 rnIs. The rain has a terminal speed of 9 rnIs. (a) What is the angle that the rain appears to make with the vertical to the marathoner? (b) What is the apparent motion of the cloud as observed by the marathoner? (c) A star on the axis of the earth's orbit appears to have a circular orbit of angular diameter of 4 1 .2 seconds of arc. How is this angle related to the earth's speed in its orbit and the velocity of photons received from this distant star? (d) What is the speed of light as determined from the data in Part (c)? Picture the Problem
The angle that the rain appears to make with the vertical, according
to the marathoner, is the angle whose tangent is the ratio of Vrunner to Vrain . The circular motion of the star is analogous to the circular motion of the cloud with Vrutmer = Vearth and Vrain = c.
(a) The angle that the rain appears to make with the vertical to the
B tan-I [ =
marathoner is given by: Substitute numerical values and evaluate
B:
(b) The cloud moves in a circle whose radius is given by:
B
=
R
=
J tan -I [ 4 mJS J Vrwmer vram
9 mJs
HtanB
=
1 24 . 0 ° 1
290
Ch er ap t
31
!
Substitute numerical values and
R = (1 0 km)tan 24° = 4.45 km
evaluate R : (c) Here Vrunner = Vearth and
(1)
Vrain = c :
where e = 1 (angular (d) From equation (1):
c
Convert 20.6" to degrees:
= vearth
tan e
=
evaluate c:
c:
diameter)
2 nRearth-sun T.arth
20.6" = 20.6" x
Substitute numerical values and
Substitute numerical values and evaluate
I
tan
� 60"
e
x � = 5 .722 x 1 0-3 0 60'
Chapter 32 Optical Images Conceptual Problems· *4
Under what condition will a concave mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? ••
Determine the Concept Let s be the object distance andfthe focal length of the mirror.
(a) If
s
< f, the image is virtual, upright, and larger than the object.
(b) If
s
< f, the image is virtual, upright, and larger than the object.
(c) If s > 2f, the image is real, inverted, and smaller than the object. (d) Iff< s < 2f, the image is real, inverted, and larger than the object. *9
•
Under what conditions will the focal length of a thin lens be (a) positive and
(b) negative? Consider both the case where the index of refraction of the lens is greater than and less than the surrounding medium. Determine the Concept
(a) The lens will be positive if its index of refraction is greater than that of the surrounding medium and the lens is thicker in the middle than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be positive if it is thinner at its center than at the edges. (b) The lens will be negative if its index of refraction is greater than that of the surrounding medium and the lens is thinner at the center than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be negative if it is thicker at the center than at the edges. *14
•
If an object is placed 2 5 cm from the eye of a farsighted person who does not
wear corrective lenses, a sharp image is formed (a) behind the retina, and the corrective lens should be convex. (b) behind the retina, and the corrective lens should be concave. (c) in front of the retina, and the corrective lens should be convex. (d) in front of the retina, and the corrective lens should be concave.
291
292
Chapter 32
Determine the Concept The
eye muscles ora farsighted person lack the ability to shorten the focal length of the lens in the eye sufficiently to form an image on the retina of the eye. A convex lens (a lens that is thicker in the middle than at the circumference)
will bring the image forward on to the retina.
I (a) is correct. I
*17 · The image of a real object formed by a convex mirror (a) is always real and inverted. (b) is always virtual and enlarged. (c) may be real. (d) is always virtual and diminished.
Determine the Concept Referring to the ray diagram show below we note that the image
is always virtual and diminished.
I (d) is correct. I \
\
\ --
-)\--
-
I I
F
-_
c
Explain the following statement: A microscope is an object magnifier, but a telescope is an angle magnifier.
*21
·
Determine the Concept Microscopes ordinarily produce images (either the intermediate
one produced by the objective or the one viewed through the eyepiece) that are larger than the object being viewed. A telescope, on the other hand, ordinarily produces images that are much reduced compared to the object. The object is normally viewed from a great distance and the telescope magnifies the angle subtended by the object. Estim ation and Approximation
Estimate the maximum value that could be usefully obtained for the magnification of a simple magnifier, using Equation 32-20. (Hint: Think about the smallest f ocal length lens that could be made from glass and still be used as a magnifier.) *24
••
Picture the Problem Because the focal length of a spherical lens depends on its radii of
curvature and the magnification depends on the focal length, there is a practical upper limit to the magnification. Use equation 32-20 to relate the magnification M of a simple
Xnp M =-
f
Optical Images 293 magnifier to its focal lengthf Use the lens-maker's equation to relate the focal length of a lens to its radii of curvature and the index of refraction of the material from which it is constructed: For a plano-convex lens, Hence:
r2
1 ( - l)
-= n
f
(
I
1
--rl
r2
J
= 00.
Substitute in the expression for M and simplify to obtain:
1)
Note that the smallest reasonable value for rl will maximize M. reasonable smallest value for the radius of a magnifier is 1 cm. Use this value and n = 1.5 to estimate
A
( )( ) Mrnax 1.5-1lcrn25 cm = @}] =
Plane Mirrors
Two plane mirrors make an angle of 90°. The light from an object point that is arbitrarily positioned in front of the mirrors produces images at three locations. For each image location draw two rays from the object that, upon one or two reflections, appear to come from the image location. *27
·
Determine the Concept Draw rays of light
from the object that satisfy the law of reflection at the two mirror surfaces. Three virtual images are formed, as shown in the adjacent figure. The eye should be to the right and above the mirrors in order to see these images.
I ��� ;±;;;;;;;l I
J I
//
1/
if
/
/
/
/ /
/ I I I I I I I I I' I' •
Spherical Mirrors
concave spherical mirror has a radius of curvature of 24 cm. Draw ray diagrams to locate the image (if one is formed) for an object at a distance of (a) 55 cm, (b) 24 cm, (c) 12 cm, and Cd) 8 cm from the mirror. For each case, state whether the *30
··
A
294
Chapter 32
image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. The easiest rays to use in locating the image are 1) the ray parallel to the principal axis and passes through the focal point of the mirror, the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and 2) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. Picture the Problem
(a) The ray diagram is shown to the right. The image is real, inverted, and reduced.
(b) The ray diagram is shown to the
The image is real, inverted, and reduced.
right.
( c) The ray diagram is shown to the right. The object is at the focal point of the mirror.
The image is real, inverted, and the same size as the object.
The emerging rays are parallel and do not form an image.
Optical (d) The ray diagram is shown to the right.
Images 29S ./
-7+
J J J ---L
?
_
The image is virtual, erect, and enlarged . dentist wants a small mirror that will produce an upright image with a magnification of 5.5 when the mirror is located 2.1 cm from a tooth. (a) What should the radius of curvature of the mirror be? (b) Should the mirror be concave or convex? *35
.
A
Picture the Problem We can use the mirror equation and the definition of the lateral
magnification to find the radius of curvature of the mirror. (a) Express the mirror equation:
1 = -1 =-2 -s1 + s' r f
Solve for r:
r = 2ss'
The lateral magnification of the mirror is given by:
m = --s'
Solve for s':
s' = -ms
Substitute for s' in equation (1) to obtain:
ms r = -2 I-m
--
s'+s
Substitute numerical values and evaluate r: (b)
(1)
s
---
)(2.1 cm ) = l s .13cm r = -2(S.S I I-S.S
The mirror must be concave. A convex mirror always produces a diminished virtual image.
concave mirror has a radius of curvature 6 crn. Draw rays parallel to the axis at 0.5 cm, 1 cm, 2 cm, and 4 cm above the axis, and find the points at which the reflected rays cross the axis. (Use a compass to draw the mirror and a protractor to find the angle of reflection for each ray.) (a) What is the spread ox of the points where these *39
••
A
296
Chapter 32
rays cross the axis? (b) By what percentage could this spread be reduced if the edge of the milTor were blocked off so that parallel rays more than 2 cm from the axis could not strike the mirror? Picture the Problem
(a) The figure to the right shows the mirror and the four rays drawn to scale. Using a calibrated ruler, the spread of the crossing points is & � 1.0 cm. Note that the triangles formed by the center of curvature, the point of reflection on the mirror, and the point of intersection of the reflected ray and the mirror axis are isosceles triangles. Express the equal angles of the isosceles triangles:
,
,
,
,
,
, ,
,
, ,
,
,
,
,
/
.:{;��;;;==::::-=--_-__:�-- �_----rc -
C
Or
=
Sin-I(�)
where y is the distance of the incoming ray from the mirror axis andR is the radius of curvature of the mirror. Using the law of cosines, the distance between the point of intersection and the mirror is given by: Evaluate d for y/R = 2/3:
= 1 .975 cm Evaluate d for y/R = 1112:
= 2.990cm Express the spread &:
5x = 2.990cm-1 .975 cm = l l .Ol cm I
in good agreement with the result obtained above.
(b) Evaluate d for y/R
Optical Images 297
= 1/3: =
Express the new spread &':
Ox'
Express the ratio of &' to &:
&'
2.81 8cm 2.990cm-2.81 8cm 0 . 172 cm
=
=
=
Ox
0 . 172 cm = 17.0% 1 .01cm
By blocking off the edges of the mirror so that only paraxial rays within 2 cm of the mirror axis are reflected, the spread is reduced by 83.0%. Images Formed by Refraction *44
··
A very long glass rod of3.5-cm
diameter has one end ground to a convex
spherical surface of radius cm. Its index of refraction is 1.5. (a) A point object in air is on the axis of the rod35 cm from the surface. Find the image and state whether the image
7. 2
is real or virtual. Repeat (b) for an object 6.5 cm from the surface and (c) an object very far from the surface. Draw a ray diagram for each case. Picture the Problem We can use the equation for refraction at a single surface to find the
images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances: Here we have n] = and n2 n 1 .5. Therefore:
1
=
=
Solve for s':
(a) Substitute numerical values (s =35 cm and r = 7. 2 cm) and evaluate s':
n]
-+
s
n2
-
s'
n
1
-+- =
s s'
s,
=
s' = =
n? - n 1
= --=--
...!..
_
r
(1)
n-1
--
r
nrs s(n-1 )-r
--,__.
(1.5 )(7.2 cm )(35cm ) (35 em)(1.5-1)-(7.2 em)
1 3 6 .7 em I
where the positive distance tells us that the
298 Chapter 32 image is 36.7 cm in back of the surface and is
(b) Substitute numerical values
(s 6.5 cm and r = 7.2 cm) and evaluate Sf: =
I real. I
(1 .5 )(7.2 em )(6.5 em) 6.5 em 1 . 5 - 1 - 7.2 em = 1 -17. 8 em I
:""" --' )""" s/ = -;-���-----!" )�(....,..--( )(
where the minus sign tells us that the image is 17.8 cm in front of the surface and is
I virtual. I
(c) When s = 00, equation (1) becomes: Solve for s/:
Substitute numerical values and evaluate Sf:
n n-1
s/
r
nr s/ =--
n-1 (1 . 5 ) (7 .2 em) = I· 21.6 em I. s/ = 1 .5-1
i.e., the image is at the focal point, is
I real, I and of zero size. •
F
Optical Images 299 A glass rod 96 cm long with an index of refraction of 1.6 has its ends ground *49 to convex spherical surfaces of radii 8 cm and 16 cm. A point object is in air on the axis of the rod 20 cm from the end with the 8-cm radius. (a) Find the image distance due to refraction at the first surface. (b) Find the final image due to refraction at both surfaces. (c) Is the final image real or virtual? ••
Picture the Problem We can use the equation for refraction at a single surface to find the
images due to refraction at the ends of the glass rod. The image formed by the refraction at the first surface will serve as the object for the second surface. The sign of the final image distance will tell us whether the image is real or virtual. (a) Use the equation for refraction at a single surface to relate the image and object distances at the first surface:
n2
nl
n2 - nl
-+ - = ----"-_...!... r s Sf
(1)
Solve for Sf:
Substitute numerical values and evaluate Sf:
(1. 6 )(Scm )(20em) s f -,----)--"--'-<-(----; --'(20-'---em--) (1.-77-6 -l - S cm)
____
=
(b) The object for the second surface is 96 cm - 64 cm = 32 em from the surface whose radius is 16 cm. Substitute numerical values and evaluate Sf:
(c)
I 64.0em I
(1)(-1 6 em) (32 em) (32 em)(1-1. 6) - (1. 6) (-1 6 em) = I -SO.Oem I
sf =
--',--'-,--'- ----,----!-+--,-..,..-'-' ---,-
--;--
The final image is 9 6 em -SO cm =1 6 em from the surface whose radius is S em and is virtual .
Thin Lenses *53
·
A
double concave lens of index of refraction 1.45 has radii of magnitudes30
em and 25 cm. An object is located 80 cm to the left of the lens. Find (a) the focal length of the lens, (b) the location of the image, and (c) the magnification of the image. (d) Is the image real or virtual? Is the image upright or inverted? Picture the Problem We can use the lens-maker's equation to find the focal length of the
300 Chapter 32 lens and the thin-lens equation to locate the image. We can use
m = -�s to find the
lateral magnification of the image.
( J
� ( )� � = n-1 -
(a) The lens-maker's equation is:
f
r2
fj
where the numerals 1 and 2 denote the first and second surfaces, respectively. Substitute numerical values to obtain:
-1 = (1.45-1)
Solve forf
f=
(b) Use the thin-lens equation to relate the image and object
1 = -1 -1 + '
f
1 1 -(-30cm 25 cm
1 3 0.3 cm 1
J
-
s s
f
distances: Solve for s':
fs s , =-s- f
Substitute numerical values and
s' =
evaluate S':
s' s
(c) The lateral magnification of the image is given by:
m = --
cm = 1 0.275 1 m =--22 80cm
Substitute numerical values and evaluate
m:
(d)
*55
Because s' < 0 and •
m >
(-30.3 cm)( 80 cm ) = 1 - 22.0cm 1 80cm - (- 3 0.3 cm)
0, the image is 1 virtual and upright. 1
An object3 cm high is placed 25 cm in front of a thin lens of power 10 D.
Draw a precise ray diagram to find the position and the size of the image, and check your results using the thin-lens equation. Picture the Problem The parallel and central rays were used to locate the image in the
diagram shown below. The power P of the lens, in diopters, can be found from P = l/j and the size of the image from
y' = - s-. m= s
'
y
Optical Images 301
The image is real, inverted, and diminished. The thin-lens equation is:
Solve for s ':
Use the definition of the power of the lens to find its fotal length: Substitute numerical values and evaluate s': Use the lateral magnification equation to relate the height of the image
y' to the height y of the object
-I +-I =I s s' I Is s, =-s-I 1 -- 1 I = 0 . 1 m = 1 0 em 10mP (10cm)(25 em ) = s' 25cm-10cm = 1 1 6.7 cm I
I-
y' = --s' m =y s
and the image and object distances: Solve for
y' = --s's y
y':
Substitute numerical values and evaluate
y':
cm (3 cm ) = 1 _2.00cm I y, =_16.7 25cm
>
Because s' 0 and y' = -2.00 em, the image is real, inverted, and diminished in agreement with the ray diagram. *59
··
Two converging lenses, each of focal length 10 cm, are separated by35 cm.
An object is 20 cm to the left of the first lens. (a) Find the position of the final image using both a ray diagram and the thin-lens equation. (b) Is the image real or virtual? Is the image upright or inverted? (c) What is the overall lateral magnification of the image?
302
Chapter
32
We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.
Picture the Problem
(a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.
__
______
L
Apply the thin-lens equation to express the location of the image
(1)
formed by the first lens: Substitute numerical values and evaluate SI' : Find the lateral magnification of the first image:
S' = 1
{IO cm){20 em) = 20cm 20cm-IOcm
s/ 20cm ml= --= - --- =- I 20cm s
Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm 20 cm 1 5 cm. Equation -
=
( 1 ) applied to the second lens is: Substitute numerical values and evaluate S2':
/ (IOcm s =
){I5 cm) = 30cm
15cm-IOcm
and the final image is object. Find the lateral magnification of the second image:
sz'
! 85 . 0cm i from the
30cm
m2 = --= - --- = -2 1 5 cm s
Because S'2 0 and m = m1m2 = 2, the image is real, erect, and twice the size of the obj ect. >
Optical Images 303 The overall lateral magnification of the image is the product of the magnifications of each image: An object is 1 5 cm in front of a positive lens of focal length 15 cm. A second positive lens offocal length 1 5 cm is 20 cm from the first lens. Find the final image and draw a ray diagram. *64
••
We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.
Picture the Problem
Apply the thin-lens equation to express the location of the image formed by the first lens: Substitute numerical values and evaluate s/: With
SI
'
=
ro,
the thin-lens equation
applied to the second lens becomes:
(1)
s'
=
15cm-15 cm
1
� S2
(15 cm)(15 cm)
=
f
l
2
=>
s
z
'
=
=
<Xl
1 1 1 5.0 cm I 2 =
A ray diagram is shown below:
The final image is 50 cm from the object, real, inverted, and the same size as the object. Aberrations
Chromatic aberration is a common defect of (a) concave and convex lenses. (b) concave lenses only. (c) concave and convex mirrors. (d) all1enses and mirrors. *70
·
Determine the Concept
Chromatic aberrations are a consequence of the differential
refraction of light of differing wavelengths by lenses.
I (a) is correct. I
304
Chapter 32
The Eye
The Model Eye 1: A simple model for the eye is a lens with variable power P located a fixed distance d in front of a screen, with the space between the lens and the screen filled by air. Refer to Figure 32-60. The "eye" can focus for all values of s such that xnp � s � Xfp. This "eye" is said to be normal if it can focus on very distant objects. (a) Show that for a normal "eye," the minimum value of P is *73
••
Pmi
n
1
=d
(b) Show that the maximum value of P is Pmax
1 +1 d
=xnp
(c) The difference
=
Pmax - Pmin is
called the accommodation. Find the minimum power and accommodation for a model eye with d = 2.5 cm and Xnp = 25 cm.
A
p
Screen_
d-----+l Model eye ,
Figure 32-60 Problems 73, 74, and 75 Picture the Problem The thin-lens equation relates the image and object distances to the
power of a lens.
)
(a Use the thin-lens equation to
relate the image and object distances to the power of the lens: Because s' = d and, for a distance object, s = 00:
Prrlln
=
1 lL] ;r m =
O t
p ical Image s 305
(b) If xnp is the closest distance an object could be and still remain in clear focus on the screen, equation ( 1 ) becomes: (c) Use our result in (a) to obtain : Use the results of (a) and (b) to express the accommodation of the model eye: Substitute numerical values and evaluate A:
= OD 2.5cm 1 40 . I 1 1 --1 = -1 A = P -po =--+d d Xnp Xnp
Pmin
=
1
max
m1l1
1 1 4.00D I _= A =_ 25cm
If two point objects close together are to be seen as two distinct objects, the images must fall on the retina on two different cones that are not adjacent. That is, there must be an unactivated cone between them. The separation of the cones is about 1 JLID. Model the eye as a uniform 2.S-cm-diameter sphere with a refractive index of 1 .34. (a) What is the smallest angle the two points can subtend? (See Figure 32-6 1 .) (b) How close together can two points be if they are 20 m from the eye?
*79 ·
Figure 32-61
Problem 79
We can use the relationship between a distance measured along the arc of a circle and the angle subtended at its center to approximate the smallest angle the two points can subtend and the separation of the two points 20 m from the eye. Picture the Problem
(a) Relate Bmin to the diameter of the eye and the distance between the activated cones: Solve for
Bmin:
Bmin
= -2,liIll d-eye
Chapter 32
306
Substitute numerical values and evaluate �"ill:
2 JLI11 = I 80.0 Jifad I emil = 2.5cm
(b) Let D represent the separation of
D = Remin
the points R = 20 m from the eye to obtain:
= (20m )( 80 Jifad) = 1 1. 60mm 1
The Simple Magnifier
person with a near-point distance of 30 cm uses a simple magnifier of power 20 D. What is the magnification obtained if the final image is at infinity? *85
·
A
Picture the Problem We
(M =
can use the definitions of the magnifying power of a lens xnp /I ) and of the power of a lens (P = 1/1 ) to find the magnifying power of the
given lens.
The magnifying power of the lens is given by:
Xnp =Pxnp M=j where P is the power of the lens.
Substitute numerical values and evaluate M: *90
••
M = (20m- 1 )(0.3m) 1 6 .00 1 =
(a) Show that if the final image of a simple magnifier is to be at the near
point of the eye rather than at infinity, the angular magnification is given by
x M =�+ 1 I
(b) Find the magnification of a 20-D lens for a person with a near point of 30 cm if the final image is at the near point. Draw a ray diagram for this situation. Picture the Problem
We can use the definition of the angular magnification of a lens
x
and the thin-lens equation to show thatM =� +
I
(a ) Express the angular
magnification of the simple magnifier in terms of the angles subtended by the object and the image:
1. ( 1)
Optical Images 307 Solve the thin-lens equation for s: Because the image is virtual:
s = s'fs' -
--
f
S = -xnp ,
Substitute to obtain:
Express the angle subtended by the object: where y is the height of the object. Express the angle subtended by the Image: Substitute for s to obtain:
Substitute in equation ( 1) and simplify:
(b) In terms of the power of the magnifying lens: The magnification of a 20-D lens for a person with a near point of 30 cm and the final image at the near point IS:
A ray diagram for this situation is
shown to the right:
M = (0.3 m)(20m-1)+ 1 = �
308
Chapter 32
The Microscope
A microscope has an objective of focal length 8.5 mm and an eyepiece that *93 gives an angular magnification of 1 0 for a person whose near point is 25 cm. The tube length is 1 6 cm. (a) What is the lateral magnification of the objective? (b) What is the magnifying power of the microscope? ••
Picture the Problem
The lateral magnification of the objective is
magnifying power of the microscope is M = m M o e' (a) The lateral magnification of the
objective is given by: Substitute numerical values and evaluate mo:
(b) The magnifying power of the microscope is given by:
mo - L/ fa and the =
L m =-a fa m o = - 16c m = 1 -1 .881 8.5 mm
M = m oMe where Me is the angular magnification of the lens.
Substitute numerical values and evaluate M:
M = (-1.88)(10) = 1 - 18 .8 1
A compound microscope has an objective lens with a power of 45 D and an eyepiece with a power of 80 D. The lenses are separated by 28 cm. Assuming that the final image is formed 25 cm from the eye, what is the magnifying power? *95
••
Picture the Problem The
magnifying power of a compound microscope is the product of the magnifying powers of the objective and the eyepiece. (1)
Express the magnifying power of the microscope in terms of the magnifying powers of the objective and eyepiece: From Problem 82, the magnification of the eyepiece is given by: The magnification of the objective is given by:
L m o =-fo
Optical Images where 111
Substitute to obtain:
o
L
=
309
D- fo - fe
o =_ D - J+-J: fo
e
Substitute for me and mo in equation ( 1 ) to obtain: Substitute numerical values and evaluate M: M
]
(
=[ (80D )(0.2 5 m)+ 1 ] - 28c m-2.22c m-l .25c m = 1_ 2321 2 . 22c m
The Telescope
The 200-in (5 . 1 -m) mirror of the reflecting telescope at Mt. Palomar has a •• focal length of 1 .68 m. (a) By what factor is the light-gathering power increased over the 40-in (l.OI6-m) diameter refracting lens of the Yerkes Observatory telescope? (b) If the focal length of the eyepiece is 1 .25 cm, what is the magnifying power of this telescope?
*99
Because the light-gathering power of a mirror is proportional to its area, we can compare the light-gathering powers of these mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths of the objective and eyepiece lenses to find the magnifying power of the Palomar telescope. Picture the Problem
(a) Express the ratio of the light gathering powers of the Palomar and Yerkes mirrors:
P.Palomar PYerkes
_
APalomar mirror AYerkes mirror
Jr
2
Jr
2
- 4 d Palomar mirror d "4 Yerkes mirror
_ d;alomar mirror d�erkes mirror Substitute numerical values and evaluate PPaloma/PYerkes:
Ppalomar PYerkes
or Ppalomar
(b) Express the magnifying power of the Palomar telescope:
= (200inr = 25.0 (40 in) = I (25.0)PYerkeS
310
Chapter 32
Substitute numerical values and evaluate M:
M=-
1 .68 m = 1 -134 1 1.25c m
General Problems *105
•
A camera uses a positive lens to focus light from an object onto film. Unlike
the eye, the camera lens has a fixed focal length, but the lens itself can be moved slightly to vary the image distance to the image on the film. A telephoto lens has a focal length of 200 mm. By how much must it move to change from focusing on an object at infinity to an object at a distance of 30 m? We can express the distance lis that the lens must move as the difference between the image distances when the object is at 30 m and when it is at infinity and then express these image distances using the thin-lens equation.
Picture the Problem
Express the distance lis that the lens must move to change from focusing on an object at infinity to one at a distance of 3 0 m: Solve the thin-lens equation for s': Substitute and simplify to obtain:
Js s , = -s- f jj,s = Js30 _
Substitute numerical values and evaluate lis:
*110
••
jj,s
S30
-f
S30
-f
Js30
30m -1] [30m-0 .2 m
=
(200mm)
=
1 1 .3 4mm 1
A scuba diver wears a diving mask with a face plate that bulges outward
with a radius of curvature of 0.5 m. There is thus a convex spherical surface between the water and the air in the mask. A fish is 2.5 m in front of the diving mask. (a) Where does the fish appear to be? (b) What is the magnification of the image of the fish?
Optical I mages 3 1 1 Picture the Problem We can use the equation for refraction at a single surface to locate the image of the fish and the expression for the magnification due to refraction at a spherical surface to find the magnification of the image.
(a) Use the equation describing refraction at a single surface to relate the image and object distances:
!2+� ' s
s
=
nz -nl r
Solve for s':
Substitute numerical values and evaluate s':
s
'
m) = (1 - 1 .33(1 ) (0.5 m)(2.5 ) (2.5 m) - (1 .33 ) (0.5 m) = 1 -0. 839m 1
Note that the fish appears to be much closer to the diver than it actually is.
(b) Express the magnification due to refraction at a spherical surface: Substitute numerical values and evaluate m:
n I s' m = - -n2 s m =
_
(1 .33 ) (- 0 . 83 9 m) 1 0.446 1 (1)(2.5 m) =
Note that the fish appears to be smaller than it actually is. (a) Find the focal length of a thick, double convex lens with an index of refraction of l.5, a thickness of 4 cm, and radii of +20 cm and -20 cm. (b) Find the focal length of this lens in water. *115
••
Here we must consider refraction at each surface separately. To find the focal length we imagine the object at s = 00, and find the image from the first refracting surface at s' I. That image serves as the object for the second refracting surface. We'll find that this is a virtual image for the second refracting surface, i .e., S2 is negative. Using the equation for refraction at a single surface a second time, we can locate the image formed by the second refracting surface by the virtual object at S2' The location of that image is then the focal point of the thick lens. We'll let the numeral 1 denote the first surface and the numeral 2 the second surface. In part (b) we can proceed as in part (a) (except that now nl 1.33 for the first refraction and nz = 1.33 for the second refraction) to determine the focal length in water, which we denote byfw. Picture the Problem
=
312
Chapter
32
(a) Use the equation for refraction
at a single surface to relate s, and
s,':
For s,
!i + � =
s, s,'
112
-
11,
1)
= 00:
Solve for
21] S/ = -11-"--'112 -11,
SI':
Substitute numerical values and evaluate
SI' :
The object distance second lens is:
S2 for the
m) = 60.0c m S ' = (1 .5)(20c 1 .5-1 1
S2 =-(s,' -4c m) = -(60c m-4c m) =-56c m
Solve the equation for refraction at a single surface for
S2'- -112r2s2 - (112 111 )S2 - 11lr2
Substitute numerical values and evaluate
5'
S2' :
52' :
_
= 5z' + 2c m = 1 9.3c m + 2c m = I 21 .3c m I
f
(b) Substitute numerical values in
m) = 17 6c m 5, = (1 .5)(20c 1.5- 1 .33
The object distance second lens is:
S2 for the
I
52 =-(5/ -4c m) =-(17 6c m-4c m) =-17 2c m
Substitute numerical values in equation (2) and evaluate
5z' =
S'2:
(1.33 ) (- 20c m) (- 17 2c m) =77 .2c m (1 .33 -1 .5)(-172c m)-(1 .5)(- 20c m)
Because fw is measured from the center of the lens:
(2)
(1 ) (- 20c m)(- 56c m) 2 - (1-1 .5)( - 56c m)- (1 .5)(- 20c m) = 19.3c m
Because f is measured from the center of the lens:
equation (1) and evaluate 5/ :
(1)
fw
= 5z' + 2c m =77.2c m + 2c m = 1 7 9.2c m I
Optical Images 313 Rem arks: Note that if we use the expression given in Problem 114 we obtain
j.v = 83.3 em, in only moderate agreement with the exact
result given above.
When a bright light source is placed 30 em in front of a lens, there is an upright image 7.5 e m from the lens. There is also a faint inverted image 6 em in front of the lens due to reflection from the front surface of the lens. When the lens is turned around, this weaker, inverted image is 1 0 cm in front of the lens. Find the index of refraction of the lens. *120
•••
The mirror surfaces must be concave to create inverted images on reflection. Therefore, the lens is a diverging lens. Let the numeral 1 denote the lens in its initial orientation and the numeral 2 the lens in its second orientation. We can use the mirror equation to find the magnitudes of the radii of the lens' surfaces, the thin-lens equation to find its focal length, and the lens. maker' s equation to find its index of refraction. Picture the Problem
Solve the mirror equation for
I'll:
Substitute numerical values and evaluate I'll: Solve the mirror equation for
2 1'1 1= s] S]S/ , +S] I'll 2(30c mX 6c m) = 1O.Oc m
6c m +30c m
=
hi:
Substitute numerical values and evaluate h�: Solve the thin-lens equation for!
Substitute numerical values and evaluate! Solve the lens-maker' s equation for n to obtain:
i mXIOc m) = 1 5. 0c m h 2(30c Oc =
I m +30c m
j= SS' s' + s j= (30cm ) (-7 .5cm ) =-lO.Oc m -7 .5c m + 30c m
314
Chapter 32
Because the lens is a diverging lens, rJ - 1 0 cm and r2 = 1 5 cm. Substitute numelical values and evaluate n:
=
n
=
(
1 1 1 _ _ l_ (- IOcm ) - 10cm IS cm
=�
J
+
The lateral magnification of a spherical mirror or a thin lens is given by m = -s' ls. Show that for objects of small horizontal extent, the longitudinal magnification is approximately _m2. (Hint: Show that ds'/ds = - s,2/i.) *125
•••
Picture the Problem We
due to a change in s.
examine the amount by which the image distance s' changes
Solve the thin-lens equation for s':
Differentiate s' with respect to s:
ds' � dsd[(1 -1J-I -; ] �- (� 1- 1 �-7�-m H I The image of an object of length will have a length m2/j,s. 1 ds
s'
f
/j,s
S
'2
2
-
Chapter 33 Interference and Diffraction Conceptual Problems *1
When destructive interference occurs, what happens to the energy in the light
•
waves? The energy is distributed nonuniforrnly in space; in some regions the energy is below average (destructive interference), in others it is higher than average (constructive interference). Determine the Concept
*6 • A loop of wire is dipped in soapy water and held so that the soap film is vertical. (a) Viewed by reflection with white light, the top of the film appears black. Explain why. (b) Below the black region are colored bands. Is the first band red or violet? ( c) Describe the appearance of the film when it inriewed by transmitted light.
(a) The phase change on reflection from the front surface of the film is 1 80°; the phase
change on reflection from the back surface of the film is 0°. As the film thins toward the top, the phase change associated with the film's thickness becomes negligible and the two reflected waves interfere destructively.
(b) The first constructive interference will arise when t = Al4. Therefore, the first band will be violet (shortest visible wavelength). (c) When viewed in transmitted light, the top of the film is white, since no light is reflected. The colors of the bands are those complementary to the colors seen in reflected light; i.e., the top band will be red.
A double-slit interference experiment is set up in a chamber that can be evacuated. Using monochromatic light, an interference pattern is observed when the chamber is open to air. As the chamber is evacuated, one will note that (a) the interference fringes remain fixed. (b) the interference fringes move closer together. (c) the interference fringes move farther apart. (d) the interference fringes disappear completely. *10
·
Determine the Concept
Ym
= m
J..L
d
,
The distance on the screen tomth bright fringe is given by
where L is the distance from the slits to the screen and d is the separation of
the slits. Because the index of refraction of air is slightly larger than the index of refraction of a vacuum, the introduction of air reduces A, to Aln and decreases Ym. Because the separation of the fringes is Y", - Ym-), the separation of the fringes decreases
315
3 16 Chapter 33 and I (b) is correct. I Estimation and Approximation *12 It is claimed that the Great Wall of China is the only human object that can be seen from space with the naked eye. Make an argument in support of this claim based on the resolving power of the human eye. Evaluate the validity of your argument for observers both in low-earth orbit (�400 km altitude) and on the moon. Picture the Problem We'll
assume that the diameter of the pupil of the eye is
5 mm and that the wavelength of light is 600 nm. Then we can use the expression for the
minimum angular separation of two objects than can be resolved by the eye and the relationship between this angle and the width of an object and the distance from which it is viewed to support the claim. Relate the width w of an object that can be seen at a height h to the critical angular separation ac:
tanac = -h
Solve for w:
W = htanac
The minimum angular separation a.: of two point objects that c<:tn just be resolved by an eye depends on the diameter D of the eye and the wavelength A of light:
ac = 1 .22-A
w
D
Substitute for a.: in the expression for w to obtain: In low-earth
orbit:
w
(
( )
= htan 1 .22�
)
= 86 W = (400km ) tan 1 .22 6500nrn mm 5 . m Because the width of the Great Wall is about 5 m, a naked eye would not be able to see it from the moon. At a distance equal to that of the distance of the moon from earth:
(
)
' W = (3 oS 4 X 10 m) tan 1 .22 6�= = S 6 02km Because the width of the Great Wqll is about 5 m, a naked eye would not be able to see it from the moon.
Interference and Diffraction
317
Human hair has a diameter of approximately 70 ,um. If we illuminate a ha ir using a helium-neon laser with wavelength /L = 632.8 nm and intercept the light scattered from the hair on a screen 1 0 m away, what will be the separation of the first diffraction peak from the center? (The diffraction pattern of a hair with diameter d is the same as the diffraction pattern of a single slit with width a = d.) *16
••
diagram shows the hair whose diameter d = a, the screen a distance L from the hair, and the separation 6y of the first diffraction peak from the center. We can use the geometry of the experiment to relate L\y to L and a and the condition for diffraction maxima to express B in terms of the diameter of the hair and the wavelength of the light illuminating the hair. Picture the Problem The
1 : G T : a
��
-
l
'1
___
-
Relate Bto 6y: Solve for L\y: Diffraction maxima occur where:
-
-
-
-
-
-
T 1 /:;y
________ _
L----�
tan 8 L\Ly L\y L tan 8 =
=
asin8
(m+-t)A -1[( +-t)A] SIn m =
wherem = 1 , 2, 3, . . . Solve for B to obtain:
8=
.
a
Substitute for Bin the expression for to obtain:
L\y
For the first peak,m = 1 . Substitute numerical values and evaluate L\y:
318
Chapter 33
Phase Difference and Coherence
Two coherent microwave sources that produce waves of wavelength 1 .5 cm are in the � plane, one on the y axis at y = 1 5 cm and the other at x = 3 cm, y = 1 4 cm. If the sources are in phase, find the difference in phase between the two waves from these sources at the origin. *19
··
Picture the Problem The difference
in phase depends on the path difference according
r to 0 b. 360° . The path difference is the difference in the distances of (0, 1 5 cm) and =
A
(3 cm, 1 4 cm) from the origin. Relate a path difference !':.r to a phase shift 8.
b.r = 1 5c m - �(3c m Y + (14c mY 0.682c m
The path difference !':.r is:
=
0 = 0.682c m 3600 = I I64° I I .5c m
Substitute numerical values and evaluate 5: Interference in Thin Films
The diameters of fine fibers can be accurately measured using interference patterns. Two optically flat pieces of glass of length L are arranged with the wire between them, as shown in Figure 33-40. The setup is illuminated by monochromatic light, and the resulting interference fringes are detected. Suppose that L 20 cm and that yellow sodium light (A. � 5 90 nm) is used for illumination. If 1 9 bright fringes are seen along this 20-cm distance, what are the limits on the diameter of the wire? (Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all.) *21
··
=
L
Figure 33-40
�I
Problem 2 1
The condition that one sees m fringes requires that the path difference between light reflected from the bottom surface of the top slide and the top Picture the Problem
Interference and Diffraction 319 surface of the bottom slide is an integer multiple of a wavelength of the light.
m1 2
Themth fringe occurs when the path difference 2d equals m wavelengths:
2d =m1
Because the nineteenth (but not the twentieth) bright fringe can be seen, the limits on d must be:
(m-1) � < d «m +1) �
=>
d
= -
wherem 19 =
Substitute numerical values to obtain: or
1 5 . 46 Jll1l < d < 5.7 5 Jll1l 1
A film of oil of index of refraction n = 1 .45 floats on water (n 1.33 ). When illuminated with white light at normal incidence, light of wavelengths 700 nm and 500 *26
••
=
nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because the index of refraction of air is less than that of the oil, there is a phase shift of trrad (11) in the light reflected at the air-oil interface. Because
the index of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflected from the oil-glass interface. We can use the condition for constructive interference to determinem for It = 700 nm and then use this value in our equation describing constructive interference to find the thickness t of the oil film. Express the condition for constructive interference between the waves reflected from the air-oil interface and the oil-glass interface: Substitute for obtain:
l' and
solve for It to
Substitute the predominant wavelengths to obtain:
2t +11' = 1',21',31',... or
2t
- 1 1 I 3 1 I S '} I - 2"/\',2"/\' ,2" /\',
.
- (m + 2"/\' )
_ . .
1
1 I
(1 )
where It' is the wavelength of light in the oil andm = 0, 1, 2, . . .
1 =� m +l. 2
700nm
2nt 2nt and 500nm =-= -m +l. m +1. 2
2
320
Chapter 33
Di vide the first of these equations by the second to obtain:
700 nm 500nm m
Solve for m :
=
2nt m + -.L2
= __
_
2nt
m + l2
__
m + -.L 2
2 for A- 7 00 nm =
Solve equation (1) for t:
t - (2 + "21 ) 7 00nm) _- I 603 nm 1 2 (1 .45
Substitute numerical values and evaluate t: Newto n ' s Rings
A Newton's ring apparatus consists of a plano-convex glass lens with radius of curvatureR that rests on a flat glass plate, as shown in Figure 33-42. The thin film is air of variable thickness. The pattern is viewed by reflected light. (a) Show that for a thickness t the condition for a bright (constructive) interference ring is *27
··
t
=
A- m 0,1, 2,... (m+ t)-, 2
=
(b) Apply the Pythagorean Theorem to the triangle of sides r, R - t, and hypotenuseR to
show that for t < < R, the radius of a fringe is related to t by r
=
-J2tR
(c) How would the transmitted pattern look in comparison with the reflected one? (d) Use R = 1 0 m and a diameter of 4 cm. How many bright fringes would you see if the apparatus were illuminated by yellow sodium light (A :::::: 590 nm) and viewed by reflection? (e) What would be the diameter of the sixth bright fringe? if) If the glass used in the apparatus has an index of refraction n = 1 .5 and water (nw 1 .33) is placed =
between the two pieces of glass, what change will take place in the bright-fringe pattern?
R
Figure 33-42
Problem 27
I
I
Interference and Diffraction 321 This arrangement is essentially identical to a "thin fi lm" configuration, except that the "film" is air. A phase change of 1 800 ( 1-A ) occurs at the Picture the Problem
top of the flat glass plate. We can use the condition for constructive interference to derive the result given in (a) and use the geometry of the lens on the plate to obtain the result given in (b). We can then use these results in the remaining parts of the problem. (a) The condition for constructive interference is:
2t + 1-A = ,,1., 2,,1.,3,,1., ...
or
2t = 1- A, t A, -t A, .. . = (m + 1-),,1.
where A is the wavelength of light in air and m = 0, 1 , 2, . . .
Solve for t:
A = 0,1, 2, t = (m + 1- )-,m ... 2
(1)
(b) From Figure 33-39 we have: or
R2 = r 2 + R2 -2Rt + t 2
For t « R we can neglect the last term to obtain: Solve for r : (c)
r = I �2Rt I
(2)
The transmitted pattern is complementary to the reflected pattern.
(d) Square equation (2) and substitute for t from equation ( 1 ) to obtain: Solve for m :
r2 1 m = --RA 2
Substitute numerical values and evaluate m :
Y - ! = 67 m = (1 ° m(2cm )(S 90nm) 2
(e) The diameter of the mlh fringe is:
D = 2r
and so there will be � bright fringes. = 2�(m + 1-)RA
322
C
hapter 33
=
D = 2 �(5 + t)(lO m)(S 90nm) = l . 1 4cm
Noting thatm 5 for the sixth fringe, substitute numerical values and evaluate D:
I
1
The wavelength of the light in the film becomes Itair / 444 nm The separation between fringes is reduced and the number of fringes that will be seen is increased by the factor l .3 3 . n =
(j)
.
n=
Two-Slit Interference Pattern
Two narrow slits separated by 1 mm are illuminated by light of wavelength 600 nm, and the interference pattern is viewed on a screen 2 m away. Calculate the number of bright fringes per centimeter on the screen. *30
·
Picture the Problem The number of bright fringes per unit distance is the reciprocal of
the separation of the fringes. We can use the expression for the distance on the screen to themth fringe to find the separation of the fringes. Express the number N of bright fringes per centimeter in terms of the separation of the fringes:
I N=_
(1)
Lly
Express the distance on the screen to themth and (m + 1 )st bright fringe: Subtract the second of these equations from the first to obtain:
U ily =
Substitute in equation ( 1 ) to obtain:
N=�
Substitute numerical values and evaluate N:
N=
d
AL
X ) = 1 8 .33cm -1 I (600nm 2m I
White light falls at an angle of 30° to the normal of a plane containing a pair of slits separated by 2.5 J-lffi . What visible wavelengths give a bright interference maximum in the transmitted light in the direction normal to the plane? (See Problem 34.) *35
••
Picture the Problem Let the separation of the slits be d. We can find the total path
difference when the light is incident at an angle ¢ and set this result equal to an integer multiple of the wavelength of the light to relate the angle of incidence on the slits to the direction of the transmitted light and its wavelength.
Interference and Diffraction 323 !'::,. £ =
Express the total path difference:
d sin ¢ + d si n e
!'::,.£ = m A
The condition for constructive interference is:
where m is an integer.
d sin ¢ + d sin e = rnA
Substitute to obtain: Divide both sides of the equation by d to obtain:
. d.. • n Slll 'f' + Slll u
Set B= 0 and solve for A:
A
Substitute numerical values and simplify to obtain:
A =
=
=
rnA
d
-
d sin¢ m
(2.5 ,um)sin 30° 1 .25 ,um m
=
rn
Evaluate A for positive integral values of m :
From the table we can see that 625 the electromagnetic spectrum.
m
A (nm)
1 2 3 4
1250 625 417 3 13
run
and 4 1 7
run
are in the visible portion of
Diffraction Pattern of a Single Slit
*39 Measuring the distance to the moon (lunar ranging) is routinely done by firing short-pulse lasers and measuring the time it takes for the pulses to reflect back from the moon. A pulse is fired from the earth; to send it out, the pulse is expanded so that it fills the aperture of a 6-in-diameter telescope. (a) Assuming the only thing spreading the beam out to be diffraction, how large will the beam be when it reaches the moon, 382,000 km away? (b) The pulse is reflected off a retroreflecting mirror left by the Apollo 1 1 astronauts. If the diameter of the mirror is 20 in, how large will the beam be when it gets back to the earth? (c) What fraction of the power of the beam is reflected back to the earth? (d) If the beam is refocused on return by the same 6-in telescope, what fraction of the original beam energy is recaptured? Ignore any atmospheric losses. 00
Picture the Problem The diagram shows the beam expanding as it travels to the moon
and that portion of it that is reflected from the mirror on the moon expanding as it returns to earth. We can express the diameter of the beam at the moon as the product of the beam
324
Chapter
33
divergence angle and the distance to the moon and use the equation describing diffraction at a circular aperture to find the beam divergence angle. We can follow this same procedure to find the diameter of the beam when it gets back to the earth. In Parts (c) and (d) we can use the dependence of the power in a beam on its cross-sectional area to find the fraction of the power of the beam that is reflected back to earth and the fraction of the original beam energy that is recaptured upon return to earth.
I I
/ � --.. --..
\
11
v, 1\
dt,l,seo ,
\
--.. --..
�
\
\T \
/ -� \ " ----/- -
D >::: 8 L
(a) Relate the diameter D of the beam at the moon to the distance to
the moon L and the beam
divergence angle B :
The angle B subtended by the first diffraction minimum is related to the wavelength
/L sin 8 1 .22 --=
A of the light and the
dtelescope
diameter of the telescope openi�g
dtelescope by : Because B «
1 , sin B �
B and:
/L 8 >::: 1 .22 -dtelescope
Substitute for B in equation obtain:
( 1 ) to
D=
1.22L/L dtelescope
Substitute numerical values and evaluate D:
D = (3. 8 2 x 10 (b)
)-, _8 ) ___ 22--,-(S_0_0_ nm--,._ m 6 . 12.54em 1m 10 em
The portion of the beam reflected back to the earth will be that portion incident on the mirror, so the diffraction angle is:
mx
m
x
2
/L 8 � 1 .22-dmirror
=
I 1 . S 3km I
Interference and Diffraction 325
[
The beam will expand back to:
DI = L 1 .22 � dnllrror
]
Substitute numerical values and evaluate D' :
D' = (3.82 x l 08 m ) (c) Because the power of the beam is proportional to its cross-sectional area, the fraction of the power that is reflected back to the earth is the ratio of the area of the mirror to the area of the expanded beam at the moon:
1 .22(500 nm ) = 1 459m I 2.54em 1m :-20 m' x in x 2 10 em pi P
_
Jr
2
'4 dmirror
�llITOT
Jr
Abeam
4
D
Substitute for D to obtain: pi
dmirror 1 .22LA =
P
dtelescope
Substitute numerical values and evaluate P'IP:
2 =
(
( "; )
d,,=,d.,=op,
1 .22LA
(
)
' 2 5 cm ) ) ( (20 in 6 in � pi = 8 P 1 .22(3. 8 2 1 0 m)(500nm ) X
= 1 1.10x l 0-7 1 (d) The angular spread of the beam from reflection from the 20-in mirror is given by:
, ,
d
A 8 :::::; 1 .22 -dmirror
The diameter D' of the beam on return to earth will be:
D' :::::; 1 .22L -A
Letting pI! represent the power intercepted by the telescope, we have:
P /I
dmirror
pi
_
L1
-'telescope
f
Jr
_
2 dtelescope -4 _ __
---'-
J 2
(1)
32 6
Chapter
33
Substitute for D' and simplify:
p" = P'
[ dtelescopedmirror )2
(2)
1.22L?
Multiply equation (2) by equation ( 1 ) and simplify to obtain: p" P ' = p" = P' p P
2 [ dtelescopedrnirror ) [ dmirrordtelescope )2
1 .22L?
[ dmirrordtelescope )
1 .22L?
4
1 .22L?
4 ' ) 25 cm � (20in )(6 in) ( 8 1 .22(3. 82 10 m)(500nm )
Substitute numerical values and evaluate P"IP:
=
p" p
x
=
1 1 .21
X
1 10 - 4
Interference-Diffraction Pattern of Two Slits
* 43 •• Suppose that the central diffraction maximum for two slits contains 1 7 interference fringes for some wavelength of light. How many interference fringes would you expect in the first secondary diffraction maximum? Determine the ConceptPicture the Problem There are 8 interference fringes on each
side of the central maximum. The secondary diffraction maximum is half as wide as the central one. It follows that it will contain 8 interference maxima. Using Phasors to Add Harmonic Waves
* 46
•
Find the resultant of the two waves EI = 4 sin 0Jt and E
2
=
3 sin ( 0Jt + 600).
Picture the Problem Chose the coordinate system shown in the phasor diagram. We can
use the standard methods of vector addition to find the resultant of the two waves.
The resultant of the two waves is of the form:
E
=
R
sin (
mt +
6')
Interference and D i ffraction
-
Express the x component of
R:
R , = 4 + 3 cos 60° = 5 . 5 0
Express the y component of
R:
R = 0 + 3 si n 60° = y
Find the magnitude of
R:
( J R
2 . 60
I tan- ( 2. 60 ) = 25.3 ° 5.50
Find the phase angle 5between R and E1 :
0=
t
Substitute to obtain:
E=
1 6.08 sin(mt 25 .3 °) I
an -I
y
R x
=
327
+
Remarks: We could have used the law of cosines to find R and the law of sines to
find &
*52 ••• For single-slit diffraction, calculate the first three values of ¢ (the total phase difference between rays from each edge of the slit) that produce subsidiary maxima by (a) using the phasor model and (b) setting dJ/d¢ = 0, where l is given by Equation 33-19. Picture the Problem We can use the phasor diagram shown in Figure 3 3 -26 to determine
the first three values of ¢ that produce subsidiary maxima. Setting the derivative of Equation 33-19 equal to zero will yield a transcendental equation whose roots are the values of ¢ corresponding to the maxima in the diffraction pattern. (a) Referring to Figure 33-26 we see
that the first subsidiary maximum occurs when: A minimum occurs when: Another maximum occurs when: Thus, subsidiary maxima occur when:
(b) The intensity in the single-slit diffraction pattern is given by:
¢ = 5n ¢ = (2n + 1 )n; n = 1, 2, 3, ...
and the first three subsidiary maxima are at ¢ = 3 n; 5 n; and 71r.
1=1 o
( sin
1¢ 2 1¢
J
Set the derivative of this expression equal to zero for extrema:
3 28 Ch apter 33 d1 - = 210 d¢
( . "'J [ sm - If' I
2
i¢
I '" - If' cos 4
. ",]
- -) - sm -
( I
'" If'
i¢
I
2
2
I
2
If'
0 for relative maxima and minima
=
Simplify to obtain the transcendental equation: Solve this equation numerically (use the "Solver" function of your calculator) to obtain:
¢
=
I 2.86n; 4.92n; and 6.94n I
Remarks: Note that our results in (b) are smaller than the approximate values found in (a) by 4.80%, 1 .63 %, and 0.865% and that the agreement improves as n increases. Diffraction and Resolution
Two sources of light of wavelength 700 nm are separated by a horizontal *55 • distance x. They are 5 m from a vertical slit of width 0.5 mm. What is the least value of x for which the diffraction pattern of the sources can be resolved by Rayleigh's criterion? Picture the Problem We can use
Rayleigh's criterion for slits and the geometry of the diagram to express x in terms of 2, L, and the width a of the slit.
-, x L Referring to the diagram, relate a." L, and x: For slits, Rayleigh' s criterion is:
Equate these two expressions to obtain: Solve for x:
x i':j
a
C L
a
= a
C
x
L X =
A
A a
-AL a
Interference an d D i ffraction
Substitute numerical values and evaluate x :
3 29
(700 nm ) (5 m ) 1 7.00 mm 1 0.5 mm
x
The star Mizar in Ursa Major is a binary system of stars of nearly equal *60 magnitudes. The angular separation between the two stars is 14 seconds of arc. What is the minimum diameter of the pupil that allows resolution of the two stars using light of wavelength 550 nm? ••
We can use Rayleigh' s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum diameter D of the pupil that allows resolution of the binary stars. Picture the Problem
1 T
.... .. t-. -
Your pupil
\
L -----.� I
A
(a) Rayleigh's criterion is satisfied provided:
a c = 1 .22 -
Solve for D:
D = I .22�
D
ac
550 nm 14"x 610 x 8 d 3 00" 1 00 I = 1 9 . 89 rnm :::d cm
Substitute numerical values and evaluate D:
D l .22 =
7r fa
---
--
Diffr action Gratings
*62 · With the diffraction grating used in Problem 6 1 , two other lines in the firstorder hydrogen spectrum are found at angles 81 x l O-2 rad and fh. = l .32 l O 1 rad. Find the wavelengths of these lines. =
9.72
x
-
We can solve rn A for A. with m = 1 to express the location of the first-order maximum as a function of the angles at which the first-order images are found.
Picture the Problem
d sin e
=
330
Chapter
33
The interference maxima in a diffraction pattern are at angles 8 given by: Solve for A:
d sin B
=
inA
where d is the separation of the slits and In = 0, 1 , 2, . . . A
= d sin B In
Relate the number of slits N per centimeter to the separation d of the slits:
N = !...d
Let m = and substitute for d to obtain:
d = sin B N � sin(9.72x 10-2 rad) 1 = I = 48 5nm I 2000 cm� sin(1 .32 X 10- l rad) 1 = = 65 8 nm I 2000cm-1
1
Substitute numerical values and evaluate Al for 81 = 9.72 rad:
X 10-2
Substitute numerical values and evaluate Al for 8 2 = 1 .32 rad:
xl 0-1
A
*67 ·· A diffraction grating with 4800 lines per centimeter is illuminated at normal incidence with white light (wavelength range of 400 nm to nm). For how many orders can one observe the complete spectrum in the transmitted light? Do any of these orders overlap? If so, describe the overlapping regions.
700
d sin B = rnA, rn = 1 , 2, 3, .. . to
Picture the Problem We can use the grating equation
express the order number in terms of the slit separation d, the wavelength of the light A, and the angle e. The interference maxima in the diffraction pattern are at angles 8 given by:
d sin B = rnA, rn = 1, 2, 3, . ..
Solve for m :
d sin B m = --A
If one is to see the complete spectrum:
sin B � 1
and
d m �A
Interference and Diffraction Evaluate 117 ma, :
Because m max
1
lnmax
=
4800 cm-1
Am3x
----
331
1
4800 cm-1 = 2.98 700 nm
----
2.98, one can see the complete spectrum only for m = 1 and 2.
Express the condition for overlap:
Because 700 run < 2 x 400 run, there is no overlap of the second - order spectrum into the first - order spectrum; however, there is overlap of long wavelengths in the second order with short wavelengths in the third - order spectrum. *71 ·· Mercury has several stable isotopes, among them 198Hg and 202Hg. The strong spectral line of mercury, at about 546.07 nm, is a composite of spectral lines from the various mercury isotopes. The wavelengths of this line for 1 98Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be the resolving power of a grating capable of resolving these two isotopic lines in the third-order spectrum? If the grating is illuminated over a 2-cm-wide region, what must be the number of lines per I centimeter of the grating? Picture the Problem We can use the expression for the resolving power of a grating to
find the resolving power of the grating capable of resolving these two isotopic lines in the third-order spectrum. Because the total number of the slits of the grating N is related to width w of the illuminated region and the number of lines per centimeter of the grating and the resolving power R of the grating, we can use this relationship to find the number of lines per centimeter of the grating The resolving power of a diffraction grating is given by:
R
=
Substitute numerical values and evaluate R:
R
= =
Express n, be the number of lines per centimeter of the grating, in terms of the total number of slits N of the grating and the width w of the
n =
A = mN IflAI
-
(1)
546.07532 15 46.07532 - 546.0735 51
1 3.09 x l 05 I N
w
332
Chapter 3 3
grating: From equation ( 1 ) we have :
N= R
m
Substitute to obtain:
Substitute numerical values and evaluate n :
n =
n
_
-
R
--
mw
3.09 1 05 (3 )(2 cm) x
_
-
I 5 15 .
x
1 0 4 em
-I 1
General Problems
*77 · A long, narrow horizontal slit lies 1 mm above a plane mirror, which is in the horizontal plane. The interference pattern produced by the slit and its image is viewed on a screen 1 m from the slit. The wavelength of the light is 600 nm. (a) Find the distance from the mirror to the first maximum. (b) How many dark bands per centimeter are seen on the screen? Picture the Problem We can apply the condition for constructive interference to find the
angular position of the first maximum on the screen. Note that, due to reflection, the wave from the image is 1 800 out of phase with that from the source. (a) Because Yo « L, the distance from the mirror to the first maximum is given by: Express the condition for constructive interference: Solve for e :
For the first maximum, m = 0 and:
Substitute in equation ( 1 ) to obtain:
(1)
d sin e = (m + t)/l" m
=
0, 1,2, . . .
Interference and Di ffraction
Because the image of the slit is as far behind the mirror's surface as the slit is in front of it, d 2 mm. Substitute numerical values and evaluate yo :
Yo
The number of dark bands per centimeter is the reciprocal of the fringe separation: Substitute numerical values and evaluate n:
_
) -) [(1.- ) 6002 mmnm ] .
sm
?
= I 0 . 150 mm I
=
(b) The separation of the fringes on the screen is given by:
(1 - m
333
}.L L'-.y = d n
d 1 ==-
n=
�y }.L
2 mm = I 3.33x10 m (600nm) (1m ) 3
)
I
*82 •• A thin layer of a transparent material with an index of refraction of 1 .30 is used as a nonreflective coating on the surface of glass with an index of refraction of 1 .50. What should the thickness of the material be for it to be nonreflecting for light of wavelength 600 nm? A
Picture the Problem Note that reflection
at both surfaces involves a phase shift of 1r rad. We can apply the condition for destructive interference to find the thickness t of the nonreflective coating.
=
600 nm
Air Coating Glass
The condition for destructive interference is:
2t = (m + t)Acoatina = (m + t) �
Solve for t:
t = (m + t)
Evaluate t for m = 0:
t = (2 ) 600nm = 1 1 15 nm I 2(1 .30)
o
ncoating
Aair 2ncoating
1.
*87 ·· The Impressionist painter Georges Seurat used a technique called point illism, in which his paintings are composed of small, closely spaced dots of pure color, each about 2 mm in diameter. The illusion of the colors blending together smoothly is
334
Chapter 3 3
produced in the eye o f the viewer by diffraction effects. Calculate the minimum viewing distance for this effect to work properly. Use the wavelength of visible light that requires the greatest distance so that you're sure the effect will work for all visible wavelengths. Assume the pupil of the eye has a diameter of 3 mm . ,
We can use the geometry of the dots and the pupil of the eye and Rayleigh's criterion to find the greatest viewing distance that ensures that the effect will work for all visible wavelengths. Picture the Problem
f.------
Referring to the diagram, express the angle subtended by the adjacent dots: Letting the diameter of the pupil of the eye be D, apply Rayleigh's criterion to obtain: Set e =
ac
to obtain:
Solve for L: Evaluate L for the shortest wavelength light in the visible portion of the spectrum:
L
----i�
d () � -
L
ac
d L
= 1 .22-A D
=
L=
1 .22
� D
Dd
1( .22A )( ) 2 mm = 1 12.3m I L = 3 mm(400nm ) 1 .22
*88 ••• A Jamin refractometer is a device for measuring or for comparing the indexes of refraction of gases. A beam of monochromatic light is split into two parts, each of which is directed along the axis of a separate cylindrical tube before being recombined into a single beam that is viewed through a telescope. Suppose that each tube is 0 .4 m long and that sodium light of wavelength 589 nm is used. Both tubes are initially evacuated, and constructive interference is observed in the center of the field of view. As air is slowly allowed to enter one of the tubes, the central field of view changes to dark and back to bright a total of 1 9 8 times. (a) What is the index of refraction of air? (b) If
Interference and D i ffraction
335
the fringes can be counted to ± 0 .25 fringe, where one fringe is equivalent to one complete cycle of intensity variation at the center of the field of view, to what accuracy can the index of refraction of air be determi ned by this experiment? It is given that with one tube evacuated and one full of air at l -atm pressure, there are 1 98 more wavelengths of light in the tube full of air than in the evacuated tube of the same length. We can use this condition to obtain an equation that expresses this difference in terms of L, A,,, and Ao. We can obtain a second equation Picture the Problem
relating An, n, and Ao ( IL"
IL0 =_ ) and solve the two equations simultaneously to find n
n.
= 1L0 n
(a) The wavelengths are related by:
IL"
The number of wavelengths in length L is the length L divided by the wavelength. Thus:
L L -= 1 98 IL" 1L0
Substitute for A,,:
Solve for An to obtain: Substitute numerical values and evaluate n :
n = 1 + 1 9 81L0 L 89 = 1 1 .0002916 1 n = 1 + 1 9 8 5OA m
( run]
(b) Replace 1 98 with 1 9 8 ± 0.25 and assume that the uncertainties in L and Ao are
negligible:
n = 1 + � (1 98 ± 0.25) = 1 1 .0002916 ± 0.0000004 1
Chapter 34 Wave-Particle Duality and Quantum Physics Conceptual Problems
*1
•
The quantized character of electromagnetic radiation is revealed by
(a) the Young double-slit experiment. (b) diffraction of light by a small aperture. (c) the
photoelectric effect. (d) the J.J. Thomson cathode-ray experiment. Determine the Concept The Young double-slit experiment, the diffraction of light by a
small aperture, and the J.J. Thomson cathode-ray experiment all demonstrated the wave nature of electromagnetic radiation. Only the photoelectric effect requires an explanation based on the quantization of electromagnetic radiation.
I (e )is correct. I
*5 The work function of a surface is ¢. The threshold wavelength for emission of photoelectrons from the surface is (a) he/¢. (b) ¢/hI (c) hjl¢. (d) none of the answers are correct. Determine the Concept The threshold wavelength for emission of photoelectrons is
related to the work function of a metal through ¢ = he/At . Hence At
I ( )is correct. I a
=
he/¢ and
*11 · Explain why the maximum kinetic energy of electrons emitted in the photoelectric effect does not depend on the intensity of the incident light, but the total number of electrons emitted does depend on the intensity of the incident light. Determine the Concept In the photoelectric effect, an electron absorbs the energy of a
single photon. Therefore, Kmax = hf- ¢, independently of the number of photons incident on the surface. However, the number of photons incident on the surface determines the number of electrons that are emitted. Estimation and Approximation
*16 ·· Students in an advanced physics lab use X rays to measure the Compton wavelength, A.c. The students obtain the following wavelength shifts � - II, as a function of scattering angle () () � - II,
45° 0.647 pm
135° 90° 1 80° 75° 1 .67 pm 2.45 pm 3 .98 pm 4.95 pm
337
338
Chapter 3 4
Use their data to estimate the value for the Compton wavelength. Compare this number with the accepted value. Picture the Problem From
ILz - /l,
=
the Compton-scattering equation we have Ac (1 - cos 8), where Ac hi mec is the Compton wavelength. Note that this =
equation is of the form y = mx + b provided we let y = � - AI and x = 1 - cos e. Thus, we can linearize the Compton equation by plotting �A = ,,1,2 - A, as a function of 1 - cos 8 . The slope of the resulting graph will yield an experimental value for the Compton wavelength. (a) The spreadsheet solution is shown below. The formulas used to calculate the
quantities in the columns are as follows: Cell A3 B3 C3
Formula/Content Algebraic Form 45 e (deg) 1 - cos e 1 - cos(A3 *PIOI1 80) 6.47E"-13 �A = ILz - A, e (deg) 45 75 90 135 1 80
1- cose
�-AI
0.293 0.741 1 .000 1 .707 2.000
6.47E- 1 3 1 .67E- 1 2 2.45E- 1 2 3 .98E- 1 2 4.95E-12
The following graph was plotted from the data shown in the above table. Excel' s "Add Trendline" was used to fit a linear function to the data. The regression equation is
�A
=
2.48 x 1 0-1 2 (1 - cos 8) - 1 .03 x 1 0-13 6.0E- 1 2 5.0E-12 � 25 E .."!
f'! 0;
-0
4.0E-12 3.0E- 1 2 2.0E-12 1 .0E- 1 2 O.OE+OO
0.0
0.5
1 .0
l -cos(theta)
1 .5
2.0
Wave-Parti cle Dual i ty and Quantum Physics
From the regression line we note that the experimental value for the Compton wavelength AC,exp is:
.1c,exp =
I 2.48
X
1 0- 1 2
m
339
I
The Compton wavelength is given by: Substitute numerical values and evaluate k:
.1 = 1 2 40 eV �nm = 2 . 43 x l O -1 2 C 5 . 1 1 x 1 0' eV
Express the percent difference between k and k,exp:
% diff =
m
AC,exp - Aexp AC,exp _1 = Aexp Aexp 12 2 .48 x 1 0 - 1 = 1 2. 0 6 % = 2.43 x 1 0- 1 2 m
m
1
*17 Baseball, tennis, golf, and soccer are sports that involve placing a ball in play with a certain speed. Estimate which of these sports has a ball with the smallest de Broglie wavelength when the ball is moving with the highest speed typically created by a professional athlete. ••
The de Broglie wavelength of an object is given by A = hlp, where p is the momentum of the obj ect. Picture the Problem
The de Broglie wavelength of an object, in tenus of its mass m and speed v, is: The values in the following table were obtained using the internet:
The de Broglie wavelength of a baseball, moving with its maximum speed, is:
A=� mv
Type of ball Baseball Tennis Golf Soccer
m
(g) 142 57 57 250
Vmax (rn/s )
44 54 42 31
6 . 6 3 x 1 0-34 J . S 1 . 0 6 x 1 0-34 A= (0 . 1 42 kg )(44 m1s ) =
m
340
Chapter
34
Proceed as above to obtain the values shown in the table:
Type of ball
In
(g) (mls) 142 44 54 57 57 42 250 3 1
Baseball Tennis Golf Soccer
A
Vmax
(m) 1 .06 x l O -34 2 . 1 5 x 1 0-34 2.77 x l O -34 0.85 5 x l O-34
Examination of the table indicates that the soccer ball has the shortest de Broglie wavelength.
The Particle Nature of Light : Photons
*20 · Find the photon energy for light of wavelength (a) 450 nm, (b) 550 nm, and (c) 650 nm. Picture the Problem We can use E = he/A to find the photon energy when we are given
the wavelength of the radiation.
(a) Express the photon energy as a
function of wavelength and evaluate E for A = 450 nm:
E=
he A
= l 2 40 eV · nm = 1 2.76 eV 450 nm
(b) For A = 5 5 0 nm:
. E = 1 240 ev nm = 1 2 . 2 5 ev 550 nm
I
(c) For A
l E = 2 40 eV · nm = 1 1 .91 eV 650 nm
I
=
650 nm:
I
*23 • Lasers used in the telecommunications network typically have a wavelength near 1 .5 5 pm. How many photons per second are being transmitted if such a laser has an output power of 2.5 mW? Picture the Problem The number of photons per unit volume is, in turn , the ratio of the
power of the laser to the energy of the photons and the volume occupied by the photons emitted in one second is the product of the cross-sectional area of the beam and the speed at which the photons travel; i.e., the speed oflight. Relate the number of photons emitted per second to the power of
N
=
P
E
=
PA he
Wave-Particle Duality and Quantum Physics
341
the laser and the energy of the photons: Substitute numerical values and evaluate N:
N-
(2.
)(
)
1.55 ,um 34 6.63x10- J. s x108m/ s 5mW
( )(3 = 1 1.95xlOI6 s-1 I _
)
The Photoelectric Effect *28
••
When a surface is illuminated with light of wavelength 780 nm, the
maximum kinetic energy of the emitted electrons is 0.37 eV. What is the maximum kinetic energy if the surface is illuminated with light of wavelength 410 nm? Picture the Problem
We can use Einstein's photoelectric equation to find the work
function of this surface and then apply it a second time to find the maximum kinetic energy of the photoelectrons when the surface is illuminated with light of wavelength 365 nm. Use Einstein's photoelectric equation to relate the maximum
Kmax
=
he T-¢
kinetic energy of the emitted electrons to their total energy and the work function of the surface: Using Einstein's photoelectric equation, find the work function of the surface:
¢ = E - Kmax =
=
he K T - max
1240eV · nm
-O.37eV
780nm
=1.22eV
Substitute for ¢ and 2 and evaluate
Kmax
=
1240eV ·nm -1 .22eV 410nm
1
= 1.80eV
I
Compton Scattering *32
•
Compton used photons of wavelength 0.0711 nm. (a) What is the energy of
these photons? (b) What is the wavelength of the photon scattered at e = 180°? (c) What is the energy of the photon scattered at this angle?
342
Chapter 34
Picture the Problem We can use the Einstein equation for photon energy to find the energy find
of both
the incident and scattered photon and
the Compton scattering equation to
the wavelength of the scattered photon.
(a) Use the Einstein equation for photon energy to obtain:
E= he = 1240eV·run = 1 17 .4 keV I 0.0711run A,
(b) Express the wavelength of the scattered photon in terms of its pre scattering wavelength and the shift in its wavelength during scattering: Substitute numerical values and evaluate Az:
-3 4J·s = ,,1,2 = 0.0711run (9 11x610. 63xlO 1 -3 kg )(3x 10 m/s )(1-COS180 ) I0 .0760run I . +
(c) Use the Einstein equation for photon energy to obtain:
8
0
E= he =1240eV·run = 11 6 . 3 keV I ,,1,2 0.0760run
Electrons and Matter Waves *39
··
An electron, a proton, and an alpha particle (the nucleus of a helium atom)
each have a kinetic energy of 150 keY. Find (a) their momenta and (b) their de Broglie wavelengths. Picture the Problem
The momenta of these particles can be found from their kinetic
energies and speeds. Their de Broglie wavelengths are given by A= hlp. (a) The momentum of a particle p, in terms of its kinetic energy K, is
p =.J2mK
given by: Substitute numerical values and evaluate Pe:
p,
(
= � = 2�.1lx10-" kg ) 150 keVx 1. 6x��-" C
= 12 .09xlO-22N .s I
)
Wave-Particle Duality and Quantum Physics
34 3
Substitute numerical values and evaluate Pp:
Pp
(
= �2m pK = 2(1.67 X10-27 kg ) 150 kevx 1.6X10-19 C eV = IS.95x10-2IN.s I
J
Substitute numerical values and evaluate Pa:
P.
�
�2m.K
�
(
J(
2 4 ux 1 . 66x �O- " kg 150 kevx 1.6x:�-" C
= 11.79 x10-20 N·s I
J
(b) The de Broglie wavelengths of the particles are given by: Substitute numerical values and evaluate
Ap:
Ap
=
� Pp
J· s = 6.63x10-34 = 8.95 x10- 21 N . s 17.41x10-14 m I Substitute numerical values and evaluate Ae:
Ae
=� Pe
= 6.63x10-34 J· s = 13.17x10-12 m I 2.09x10-22 N· s Substitute numerical values and evaluate Aa:
Aa
=
� Pa
= 6.63x10-3 4J·s =1 3.70x10-14 m 1 1.79x lO-20 N· s *42
•
A proton is moving at v = O.003c, where c is the speed of light. Find the
electron's de Broglie wavelength. Picture the Problem
this proton.
We can use its defmition to calculate the de Broglie wavelength of
344
Chapter 34
Use its definition to express the de Broglie wavelength of the proton:
Substitute numerical
values and evaluate Ap:
3x10-34J·s = A= ( .67x10-6.6 27 kg)[0.003(3x 108 m/s)] 10·441 pm 1 1
*47
•
An electron microscope uses electrons of energy 70 keV. Find the
wavelength of these electrons.
Picture the Problem
We can use
A
=
1.}t/
nm,
where K is in eV, to find the
wavelength of 70-ke V electrons. Relate the wavelength of the electrons to their kinetic energy: A
Substitute numerical values and
evaluate Il:
=
1.226 �70xl03 eV
nm =
! 4.63 pm I
A Particle in a Box
*52
··
Use a spreadsheet program or graphing calculator to plot IjI(x) and the
probability distribution
.
PIcture the Problem
1f/2(X) of a particle in a box for the states
The wave functlOn for state n ISlf/ll .
.
n =
1, 2, and 3.
{2 sm. nTlX ' The () �L x =
following graphs were plotted using a spreadsheet program.
L
Wave-Particle Duality and Quantum Physics
The graph of If!(x) for
n =
1 is shown below:
;:::-"'----::---r--""'--:;;: '"l : , >�
1.0
c:
.9 <:) c:
<2C!) >
0.5
'"
�
I'"
.,
-;,§ I'
0.0 "
"�, !"
0.0
0.2
'. -
...,
! . ",. 0.4
'." I "'"
.. "'"1
!"
,
0.8
0.6
1.0
xlL
The graph of 1(/2 (X) for n
=
1 is shown below:
1.0
o 'in
iii
"0
....
p..
L
0.6
I
0.4
!..
0.2
I.,."; .....---"'"".:.��.
0.0 !,:l;ii!!"" 0.0
"'"
'/� -I
0.8
o
�15
...
'k 'o�;" !I -
.. ",Y
Li"
.
n "I
! • -
I
0.2
• I
xlL
I" .!..! "'"
I
0.6
0.4
-"",�") .1...'
1
'.... �
0.8
c�
1.0
345
346
Chapter
34
The graph of !f!(x) for n
1.0
t::
.2 U t::
.z
.., > '"
=
2
is shown below:
h
.",,"
OC:,
.1
0.5
0.0
�
-0.5
-1.0
1'"
'I- - ', "'"'
0.0
0.2
0.4
0.6
0.8
1.0
0.6
0.8
1.0
xlL
The graph of
1f/2(X) for n = 2 is shown below: 1.0.,.,.."..-.."....,�
0.8
o ';;:j
t:: �
� �
.D
2
P-<
0.6
0.4
0.2
0.0
1"""" .",c,,,,".·','IL"
0.0
0.2
""""'....''1' .. .. <. 0.4
xlL
Wave-Particle Duality and Quantum Physics 347 The graph of If'{x) for n = 3 is shO\\fl1 below:
'"
.9 U '"
'" > OJ
0.0
3': -0.5
0.0
0.2
0.4
0.6
0.8
1.0
0.6
0.8
1.0
x/L
The graph of
\lf2(X) for n = 3 is shown below:
0.8
b 'Vi "0
�
.f' :E
2
e 0...
0.6
0.4
0. 2
0.0
0.2
0.4
x/L
Calculating Probabilities and Expectation Values *56
••
Repeat Problem 55 for a particle in the first excited state (n = 2).
Picture the Problem
The probability of finding the particle in some range fu; is
The interval fu; = O.002L is so small that we can neglect the variation in
2 compute \If fu;.
\lf2dx.
If'{x) and just
348
Chapter
Express
the probability
34 of finding
the particle in the interval ill:
Express the wave function for a particle in its first excited state: Substitute
to obtain:
p p(x)fu: = If2 (x)fu: =
1f2
p
() H. x
= =
=
(a) Evaluate P at
x= Ll2:
P
=
=
(b) Evaluate P at
x=
2L/3:
P
= =
(c) Evaluate P at
*64
··
x=
L:
P=
=
2m sm -L L
-
2m sin2 2 L L
fu:
� (si n2 2�)0.00
2L )
2m . 2 -004 sm L
O.
2 0.004si n2 21iL 2L 0.004si n 7r @J =
0.004si n2 4 1iL 0.004si n2 47r3 3L \0.003 \ =
0.004si n2 27r 0.004si n2 21iL L =
(a) Use a spreadsheet program or graphing calculator to plot the expectation
value for position
(x) and the square of the position (X2 ) as a function of the quantum
number n for the particle in the box described in Problem 60, for values of n from 1 to
100. Assume L = 1m for your graph. Refer to Problem 63. (b) Comment on the
significance of any asymptotic limits that your graph shows.
()
2
2
X2 ) L - nL2 2 . A \ 3 2 2 spreadsheet program was used to plot the following graphs of <x> and <X2 > as a function
Picture the Problem
of n .
From Problem 63 we have
x
=
L
-
and /
=
-
7r
W ave-P ar ticle Du ali ty and Qu antum P hysics 349
0.25 +-�+---'--'I=:"'--+----"'-+--'---:'-+--+-'--1--i 10
19
28
37
46
n
55
64
73
82
91
100
General Problems *67
•
A light beam of wavelength 400 nm has an intensity of 100 W/m2 .
(a) What is the energy of each photon in the beam? (b) How much energy strikes an area of 1 cm2 perpendicular to the beam in 1 s? (e)How many photons strike this area in 1 s? Picture the Problem
We can use the Einstein equation for photon energy to find the
energy of each photon in the beam. The intensity of the energy incident on the surface is the ratio of the power delivered by the beam to its delivery time. Hence, we can express the energy incident on the surface in terms of the intensity of the beam. (a) Use the Einstein equation for photon energy to express the energy
Epholon
he = hf = -;:
of each photon in the beam: Substitute numerical values and
evaluate Epholon:
(b) Relate the energy incident on a
surface of area A to the intensity of
the beam:
1240eV·ru n = 3.10e � I 400run I
Epholon =
E=IAM
350
Chapte r 34
Substitute numerical values and evaluate E:
E
(100W/m2 )(10-4 m2 )(ls) = O.OIJx leV 1.60 xl 0-19 J = 1 6.25 1016eV 1 =
X
(c) Express the number of photons striking this area in 1 s as the ratio of the total energy incident on the surface to the energy delivered by
N= E 6.25x1016ev Ephoton 3.10eV = 1 2.02x10161
each photon: *73
••
Suppose that a 100-W source radiates light of wavelength 600 nm uniformly
in all directions and that the eye can detect this light if only 20 photons per second enter a dark-adapted eye with a pupil 7 mm in diameter.How far from the source can the light be detected under these rather extreme conditions? Picture the Problem
We can relate the fraction of the photons entering the eye to ratio of
the area of the pupil to the area of a sphere of radius R. We can find the number of
photons emitted by the source from the rate at which it emits and the energy of each photon which we can find using the Einstein equation. Letting r be the radius of the pupil,
Nentering eyethe number of photons per second entering the eye, and Nemitted
Nenteringeye Aeye Nemitted 47rR2
the number of photons emitted by
the source per second, express the fraction of the light energy entering
the eye at a distance R from the source:
Solve for R to obtain:
Find the number of photons emitted by the source per second: Using the Einstein equation, express the energy of the photons:
R
= !... NNemitted 2 enteringeye
p Nemitted= � photon
Ephoton = -he ;:
( 1)
W ave-P ar ticle Du ali ty and Qu an tum P hysics Substitute numerical values and
= 1 240 eV·
EphOlon
evaluate Epholon: Substitute and evaluate Nemitted:
-(
and evaluate R:
2.0 7 eV
lOOW
)(
_
3 .5mm
3 .0 2xl020 s -'
2
20 s -'
=
)
2.0 7 eV 1.60xl0-J9 J/eV 2 3.0 2 X 10 0 S-J
= R
=
600run
Nemitted
Substitute for Nemittcd in equation (1)
nm
35 1
= 1 6.80 x103 km I *79
••
The Pauli exclusion principle states that no more than one electron may
occupy a particular quantum state at a time. Therefore, if we wish to model an atom as a collection of electrons trapped in a one-dimensional box, each electron in the box must have a unique value of the quantum number n. Calculate the energy that the most
energetic electron would have for the uranium atom with atomic number 92, assuming
the box has a width of 0.05 nm. How does this energy compare to the rest-mass energy of the electron? Picture the Problem
We can use the expression for the energy of a particle in a well to
find the energy of the most energetic electron in the uranium atom. Relate the energy of an electron in
2
(
h2
E" = n 8mL 2
the uranium atom to its quantum number n:
J
Substitute numerical values and evaluate E92:
E 92 = ( )
922
[
? - 4 6 .6 3xl0 3 J· s 8 9 .11xlO-3I kg 0 .0 5run
(
y
X
l eV
9 1.6xl0-' J
]
=
I 1·28MeV I
The rest energy of an electron is:
2 Express the ratio of E92 to mec :
E 92 m G2 e
=
1.28MeV 0 .512MeV
=
2.50
Chapter
352
34
T he energy of the mostenerge tic elec tron is approxi mately 2.5 ti mes the rest - m assenergy of an elec tron. *83
••
and state 11
(a) Show that for large n, the fractional difference in energy between state +
1 for a particle in a one-dimensional box is given approximately by
n
(b) What is the approximate percentage energy difference between the states 111 =
1000 and 112 = 1001? (c ) Comment on how this result is related to Bohr's
correspondence principle.
Picture the Problem
We can use the fact that the energy of the nth state is related to the
energy of the ground state according to E n
=n
2
E1 to express the fractional change in
energy in terms of n and then examine this ratio as n grows without bound. (a) Express the ratio
(Ell
+
1-
En)/EII:
for n» 1 (b) Evaluate E]OO] - E 000 . E1000
1.
E100] - E]OOO E]OOO
l':::
I
_2_ = 0.2% 1000
ally,the e ne rgy isco ntinuous. Fo rve ry l arge values of n, the (c) eClneassic rgy diffe re nce betwee n adj ace ntlevelsisi nfi nitesi mal.
I
Chapter 35 Applications of the Schrodinger Equation Conceptual Problems *4
•
The Schrodinger Equation could be applied equally well to baseballs as to
electrons yet we would never analyze the motion of a baseball with a wave function. Explain why this is the case by estimating the quantum mechanically predicted lowest energy level of a baseball trapped inside a locker. You can treat the locker as if it were a one-dimensional infinite potential well. What value of the quantum number n would you need for a ball rolling around in the locker, after you toss it in, so that the kinetic energy is approximately equal to the quantum mechanically calculated energy? Picture the Problem
Assume a mass of 1 50 g for the baseball, 30 cm for the width of the
locker, and 1 cmls for the speed of the ball, and equate the kinetic energy of the ball and the quantum-mechanical energy and solve for the quantum number n. The allowed energy states of a particle of mass m in a
I-dimensional infinite potential well
of width L are given by: The kinetic energy of the ball is:
1 2 K=-mv 2
For EI/ = K:
Solve for the quantum number n:
Substitute numerical values and evaluate n:
n=
2 mvL
h
--
(
)(
)(
)
m/s 2 0. 15 kg 0.0- 1...". 0.3 m ..,----"--'--"n = ---'---...=:..<.-'-6.6 3xl 0-34 J S = l . 36x1030
��
.
The Harmonic Oscillator •• *8 Use the procedure of Example 35-1 to verify that the energy of the first excited state of the harmonic oscillator is El = t limo. (Note: Rather than solve for a
again, use the result
a = m(1)o/
(2 1i ) obtained in Example 35-1.) 35 3
354
Chapter
35
Picture the Problem We can differentiate IjI(x) twice and substitute in the Schrodinger
equation for the harmonic oscillator. Substitution of the given value for a will lead us to
an expression for EI.
The wave function for the first excited state of the harmonic oscillator is: Compute dlf/l(X)/dx:
Substitute in the Schrodinger equation:
Divide out
2
Ale-ax
to obtain:
or
Substitute for a to obtain:
-
( ) - -Ii
2
2m
4
mOJo 2 1i
2
fl.". 2 3 x + -6 2m
(--) mOJo 2 1i
J
2
x+ -mOJ0 x3 =Ex I 2
Solve for EI to obtain:
Reflection and Transmission of Electron Waves: Barrier Penetration *14
••
A particle of mass m with wave number kl is traveling to the right along the
negative x axis. The potential energy of the particle is equal to zero everywhere on the
A pplic ationsoft he Sc hrodinge r Equ ation
355
Uoeverywhere on the positive axis, Uo> O. (a) Show aUo, w here a � 1, wave number k2 in the region > 0 is
negative x axis, and is equal to that if the total energy isE= given by
k2 =
x
x
� a \J a-
(b) Using a spreadsheet program or graphing calculator, graph the reflection coefficient R and transmission coefficient T for 1 � Picture the Problem
a �5 .
We can use the total energy of the particle in the region
x> 0 to express k2 in terms of and kJ• Knowing k2 in terms of kJ, we can use (kJ - k 2 R= Y2 to find Rand T 1 - Rto determine the transmission coefficient T a
(kJ + k 2)
=
(a) Using conservation of energy, express the energy of the particle in the region
x> 0:
k2= From the equation for the total energy of the particle: Express the ratio of k2 to kJ:
(b) The reflection coefficient Ris given by:
-,-�_2m_U_o(_ a_-_1) Ii
�2m aUo
k 1 =�--Ii
356
Chapter 35
Factor k, from the numerator and
denominator to obtain:
Substitute our result from (a) for
k2Ik,:
-l FaI l
1 - -a 1+
FaI-a
The transmission coefficient is given by:
T=I-R=I-
A spreadsheet program to plot Rand T as functions of a is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 B2
ContentlFormula 1.0 (1-SQRT«A2-1)/A2))/ (1+SQRT«A2-1)/A2)Y'2
Algebraic Form a
1+ C2
�
1- a
I-B2 I-
2
� a
, �' � '-
1-
a-I -a
1+
-a "
A pplic ations ofthe Schr6dinger Equ ation I.
5.0
22
357
0.971
0.029
The following graph was plotted using the data in the above table:
1.0
0.8
0.6
0.4
0.2
0.0 2
2
3
4
3
4
5
5
alpha
*18
··
A 10-eV electron is incident on a potential barrier of height 25 eV and width
of 1 nm. (a) Use Equation 3 5-29 to calculate the order of magnitude of the probability that the electron will tunnel through the barrier. (b) Repeat your calculation for a width of 0.1 nm. Picture the Problem
The probability that the electron with a given energy will tunnel
through the given barrier is given by Equation 35-29. (a) Equation 35-29 is:
T
=
e-2an
where
a
/2m(Uo -E) �2m(Uo -E) h h2 =
=
�
Multiply the numerator and denominator of a by c to obtain:
where Using
mec2
=
511keV,evaluate T:
hc
=
e
1.974 x10-13 M V· m
358 Chapte r 35
{
( m )� 2(511keVX 25eV-Ioev) T= ex p - 210 1 .974 x10-13 MeV· -9
m
(b) Repeat with a = 10- 10 m: T
}
=
I 5.91xlO -18 1
}
{
( -10 m )� 2(511keVX 25ev -10eV) = I 1.89 xI 0 2 I = ex p - 2 10 1 1.974x l 0- 3 MeV·m
The Schrodinger Equation in Three Dimensions *23
·
Give the wave functions for the lowest ten quantum states of the particle in
Problem 22.
ASin(nLlI/C xJsin(n2/C2LI yJSin(n34L/CI zJ
Picture the Problem
If
=
The wave functions are of the form
The Schrodinger Equation for Two Identical Particles *28
•
What is the ground-state energy of ten noninteracting fermions, such as
neutrons, in a one-dimensional box of length L? (Because the quantum number associated with spin can have two values, each spatial state can hold two neutrons.) Picture the Problem
For fermions, such as neutrons for which the spin quantum number
is Y2, two particles can occupy the same spatial
state.
The lowest total energy for the 10 fermions is:
General Problems *33
··
Eight identical noninteracting fermions (e.g., neutrons) are confined to a two-
dimensional square box of side length L. Determine the energies of the three lowest states. (See Problem 26.) Picture the Problem We
can determine the energies of the state by identifying the four
lowest quantum states that are occupied in the ground state and computing their combined energies. We can then find the energy difference between the ground state and
A pplic ations ofthe Schrodinger Equ ation
359
the first excited state and use this information to find the energy of the excited state. Each
n,
m
state can accommodate
(1,1), (1,2), (2,1) and (2,2)
only 2 particles. Therefore, in the ground state of the system of 8
Note that the states (1,2) and (2,1) are
distinctly different states because the x and
fennions, the four lowest quantum
y directions are distinguishable.
states are occupied. These are: The energies are quantized to the values given by:
The energy of the ground state is the sum of the energies of the four lowest quantum states:
[ � J( (�J (
=2
=2
8mL
8mL
)
( � J(
f + 12 + 2
8mL
)
)
( J(
12 + 2 2 + 2
h2
( J
) � (2 2 + 2 2 ) 8mL
2 2 + 12 + 2
8mL2
2 +5 + 5+ 8
5h2 mL2
The next higher state is achieved by taking one fennion from the (2,2) state and raising it to the next higher
M =EI3 -E?? ,
=
unoccupied state. That state is the (1,3) state. The energy difference
=
-,-
�(12 + 32 ) _ �(2 2 + 22 ) 8mL2
8mL2
h2 1 0-8) = ( 4mL2 8mL2 h2
-
between the ground state and this
-
state is: Hence, the energies of the degenerate states (1,3) and (3,1) are:
E1,3 =E3,1 =Eo+ M =
5h2 --
mL2
+
21h2 =-4mL2 4mL2 h2
--
The three lowest energy levels are therefore: and two states of energy 21h' E1 �E � 2 4mL2
1 I
360 *41
Chap ter 35 In this problem you will derive the ground-state energy of the harmonic
oscillator using the precise form of the uncertainty principle,
tul1p 2 fi12, where I1x: and I1p are defined to be the standard deviations (�;xi = [(x - xavi]av and (I1Pi = [(P -Pavi]av (see Equation l7-35a). Proceed as follows: 1. Write the total classical energy in terms of the position x and momentum P using U(x) = mOJ�x2 and K = p2/2m. 2. Use the result of Equation 17-35 to write (�y l(x - xaJ2 j= (x2 t -x;v and (flpY l(p- PavY J= (p 2 t - p�v · 3. Use the symmetry of the potential energy function to argue that Xav and Pay must be zero, so that (tui = (x)av and (l1pi = (P2)av. 4. Assume that I1p = fil(2tu) to eliminate (P2)av from the average energy Eav = (P2)avl(2m) + t mOJ\X2)av and write Eav as Eav = fi2/(8mZ) + t mOJ2Z, where Z = (X2)av. 5. Set dEldZ = 0 to find the value of Z for which E is a minimum. 6. Show that the minimum energy is given by (Eav)min = +t fi i4J. =
=
Picture the Problem
We can follow the step-by-step procedure outlined in the problem
statement to show that (Eav)min = +
t n OJ.
1. The total classical energy is:
2. Express the standard deviation of I1p:
(flPY = [(P-Pav YL _ _2 -[P 2 PPav _ Pay2]av
Because pay = 0: 3. Express the standard deviation of
tu:
_ = rx2 l - 2xxav X2] av av
Because Xav = 0:
A pplic ationsofthe Sc hr6dinger Equ ation 4. Use the uncertainty principle I1 p =
P!l211.x to eliminate (p2)a,' from the
Eav _
-
J.. 2
m OJ
2
361
( 2)av + (/1p2) 2m X
average energy in equation (1):
_
I
- 2
Let Z =
(X2)av to obtain:
5. Differentiate
Eav with respect to Z
and set this derivative equal to zero:
Solve for Z to find the value of Z
that minimizes Eav (see the remark
Eav
=
mOJ2 ( 2)av + 8mti22 ( )av X
x
ti2 J..2 mOJ2Z -8mZ +
[
]
� J..m())2Z � = dZ dZ 2 8mZ ti2 = 0 for extrema = tmal --8mZ2
dEav
+
ti_ Z= _ 2m{))
below):
6. Evaluate Eav when Z = til2mw:
(EaJnlin = tm{))2
( 2�{)) ) :� ( 2�()) } +
= I tti{)) I Remarks: All we've shown is that Z =
nl2m w is an extreme value, i.e., either a
maximum or a minimum. To show that Z = n12m m minimizes Eay, we must either 1) show that the second derivative of Eay with respect to Z evaluated at
positive, or
Z= nl2mlU
Z = nl2m w is
2) confirm that the graph of Eay as a function of Z opens upward at
Chapter 36 Atoms Conceptual Problems *1
•
As n increases, does the spacing of adjacent energy levels increase or decrease?
Examination of Figure 35-4 indicates that as n increases, the spacing of adjacent energy levels decreases.
Determine the Concept
*7 For the principal quantum number n = 4, how many different values can the orbital quantum number.e have? (a) 4, (b) 3, (c) 7, (d) 16, or (e) 6. •
Determine the Concept
We can find the possible values of .e by using the constraints on
the quantum numbers n and .e . The allowed values for the orbital
n
.e
quantum number .e for n = 1, 2, 3,
1
°
and 4 are summarized in table
2
0, 1
shown to the right:
3
0, 1, 2
4
0, 1, 2, 3
I (a) is
From the table it is clear that .e can have 4 values. *10
••
correct.
I
Why is the energy of the 3s state considerably lower than the energy of the
3p state for sodium, whereas in hydrogen these states have essentially the same energy? Determine the Concept
The s state, with .e = 0, is a "penetrating" state in which the
probability density near the nucleus is significant. Consequently, the 3s electron in
sodium is in a region of low potential energy for a significant portion of the time. In the
state .e = 1, the probability density at the nucleus is zero, so the 2p electron of sodium is
shielded from the nuclear charge by the 1s electrons. In hydrogen, the 3s and 2p electrons
experience the same nuclear potential. *14
•
For the principal quantum number n = 3, what are the possible values of the
quantum numbers f. and me! Picture the Problem
We can apply the constraints on the quantum numbers f. and me to
find the possible values for each when n = 3. Express the constraints on the
quantum numbers n, .e , and me:
n I!
=
=
and
36 3
1, 2, 3, ... ,
O,1,2 , . . . ,n -1 ,
364 Chapter 36 me
So, for n = 3, the constraints on £
= 1 0,1, and 2 · 1
£
limit it to the values: me
can take on the values:
=- £, - £+1, . . . , £
me
=
I-
2, - 1, 0, 1, 2
I
*18 ·· The Ritz combination principle states that for any atom, one can find different spectral lines AI> �, A3, and A4, so that 1/Al + 1/A2 IIA3 + 1/A4. Show why this is true using an energy-level diagram.
=
Determine the Concept The Ritz combination principle is due to the quantization of energy levels in the atom. We can use the relationship between the wavelength of the emitted photon and the difference in energy levels within the atom that results in the emission of the photon to express each of the wavelengths and then the sum of the reciprocals of the first and second wavelengths and the sum of the reciprocals of the third and fourth wavelengths.
Express the wavelengths of the spectral lines AI> �, A3, and A4 in terms of the corresponding energy transitions:
and A
_
4-
Add the reciprocals of Al and � to obtain:
Add the reciprocals of A3 and A4 to obtain:
Because the right-hand sides of equations (1) and (2) are equal:
he EI- Eo
E3 - E2 + E 2 - Eo ---"'------"he he A, � _ E3 -E o he 1
1
-+-
1
=
E3 - EI ---'EI - E o -"+ he he � ,14 _ E3 - Eo he 1
-+-
=
(1)
-
(2)
Atoms One possible set of energy levels is shown to the right:
.-\,
.-\3
.-\4
.-\2
365
E3 E2 E]
Eo
Estimation and Approximation *20 ·· In laser cooling and trapping, atoms in a beam traveling in one direction are slowed by interaction with an intense laser beam in the opposite direction. The photons scatter off the atoms via resonance absorption, a process by which the incident photon is absorbed by the atom, and a short time later a photon of equal energy is emitted in a random direction. The net result of a single such scattering event is a transfer of momentum to the atom in a direction opposite to the motion of the atom, followed by a second transfer of momentum to the atom in a random direction. Thus, during photon absorption the atom loses speed, but during photon emission the change in speed of the atom is, on average, zero (because the directions of the emitted photons are random). An analogy often made to this process is that of slowing down a bowling ball by bouncing ping-pong balls off of it. (a) Given a typical photon energy used in these experiments of about 1 eV and a momentum typical for an atom with a thermal speed appropriate to a temperature of about 500 K (a typical temperature for an atomic beam), estimate the number of photon-atom collisions that are required to bring an atom to rest. (The average kinetic energy of an atom is equal to � kT, where k is the Boltzmann constant. Use this to
estimate the speed of the atoms.) (b) Compare this with the number of ping-pong ball bowling ball collisions that are required to bring the bowling ball to rest. (Assume the speeds of the incident ping-pong balls are all equal to the initial speed of the bowling ball.) (c) 85Rb is a type of atom often used in cooling experiments. The wavelength of the light resonant with the cooling transition is A. = 780.24 nm. Estimate the number of photons needed to slow down an 85Rb atom from a typical thermal velocity of 300 m/s to a stop. The number of photons need to stop a 85Rb atom traveling at 300 m/s is the ratio of its momentum to that of a typical photon. Picture the Problem
(a) The number N of photon-atom collisions needed to bring an atom to rest is the ratio of the change in the momentum of the atom as it stops to the momentum brought to the collision by each photon: The kinetic energy of an atom whose temperature is T is: Substitute for v to obtain:
N
=
!1Patol11 PphOIOIl
where
1.. 2
mv 2
N
=
m
=
mv E
=
mve E
e
is the mass of the atom.
= 1. 2
kT => v =
�
�
3kT m
--
me 3kT = �-J3mkT m E E
366
Chap ter 36
For an atom use mass is 50 u:
3x 108 m/s --N= ---x10-19 J ---1eVx 1.6 eV (b) The number N of ping-pong ball bowling ball collisions needed to bring the bowling ball to rest is the ratio of the change in the momentum of the bowling ball as it stops to the momentum brought to the collision by each ping-pong ball:
N= l',.Pbowling ball
_
Pping-pong ball
N= l',.Pbowling ball �
Provided the speeds of the approaching bowling ball and ping pong ball are approximately the same:
Pping-pong ball
mbb Vbb mppb Vppb
mbb mppb
� 6 kg � �. 4g
( c) The number of photons N needed to stop a 85Rb atom is the ratio of the change in the momentum of the atom to the momentum brought to the collision by each photon: Substitute numerical values and evaluate N:
N= 85(1.66
X
10-27 kg)(300 m/s)(780.24 nm ) 6.63 X 10-34 J . S
=
14.98
X
104
1
The Bohr Model of the Hydrogen Atom *26
••
Repeat Problem 25 for the Brackett series,
Picture the Problem
n2 =
4.
We can use Bohr's second postulate to relate the photon energy to
its frequency and use A
=
1240eV nm to find the wavelengths of the three longest ·
Ei -Ef
wavelengths in the Brackett series. (a) Use Bohr's second postulate to express the energy of the photons in the Paschen series: For the series limit:
n
=
00
and
Ei
=
0
Atoms Substitute to obtain:
367
(1)
Evaluate the photon energy for n2 =
4:
Express the wavelength of the radiation resulting from an energy
A.
. =1240eV run M
(2)
transition M = hf
provided the energies are expressed in eV.
Evaluate A"un for the transition
A..
n
= ex) to n2 = 4:
mm
nj
(b) For the three longest wavelengths:
.run =1240eV 0.850eV =1 1459run 1
=5, 6, and7
Equation (1) becomes:
Evaluate equation (3) for n = 5:
M5�4
1) �--25 =(13.6eV)(16 =1 0.306eV 1
A. 5�4
Evaluate equation (2) for
.
M = 0 306 eV: Evaluate equation (3) for n = 6:
M = 0. 472 eV: Evaluate equation (3) for n
M6�4
A. 6�4
Evaluate equation (2) for
=
7:
.run = 4052run =1240eV 1 0.306eV 1 = (13.6eV)(�--1 ) 16 36 =10.472eV 1
=1240eV .run =1 2627run 1 0.472eV
M7�4
�--1 ) =(13.6eV)(16 49
=1 0.572eV 1
368 Chap ter 36 It7-->4 =
Evaluate equation (2) for
M = 0. 572 eV:
1 2 40 e V
.
run
0.572 eV
=
I 2 168 I run
The positions of these lines on a horizontal linear scale are shown below with the wavelengths and transitions indicated.
7�4 6�4 1
5�4
1
1
--- --------- --------------------------------------------- -----
2 168 nm 2627 nm
4052 nm
*29 ••• In the center-of-mass reference frame of a hydrogen atom, the electron and nucleus have equal and opposite momenta of magnitude p. (a) Show that the total kinetic
energy of the electron and nucleus can be writtenK = p
2 /(2 j.l) where
j.l = meM / (M +me) is called the reduced mass, me is the mass of the electron, and M is the mass of the nucleus. (b) For the equations for the Bohr model of the atom, the motion of the nucleus can be take into account by replacing the mass of the electron with the reduced mass. Use Equation 36-14 to calculate the Rydberg constant for a hydrogen atom with a nucleus of mass M = mp. Find the approximate value of the Rydberg constant by letting M go to infinity in the reduced mass formula. To how many figures does this approximate value agree with the actual value. (c) Find the percentage correction for the ground-state energy of the hydrogen atom by using the reduced mass in Equation 36-16. Remark: In general, the reduced mass for a two-body problem with masses mJ and m2 is given by j.l =
m1m2 ml +m2
Picture the Problem
We can express the total kinetic energy of the electron-nucleus
system as the sum of the kinetic energies of the electron and the nucleus. Rewriting these kinetic energies in terms of the momenta of the electron and nucleus will lead to K =
p2l2mr.
(a) Express the total kinetic energy of the electron-nucleus system: Express the kinetic energies of the electron and the nucleus in terms of their momenta:
2
2
K e =J!...-and K = L
2me
n 2M
Ato ms 369 Substitute to obtain:
�I :�, I
provided we define Ji = meAt/eM + m )
e.
(b) From Equation 36-14 we have: (1)
where
2 C= k e_4 4JZ"C1i3 __
Use the Table of Physical Constants at the end of the text to obtain:
C =1.204663 1037 m -I / kg X
ForH:
-C RHSubstitute numerical values and evaluate RH:
RH
kg = (1.204663x 1037 m -I / kg ) 9.11xlO-31 9.11xlO-3I kg 1+1.67 x 10-27 kg
Let M -+ CXJ in equation (1) to obtain
RH,approx
RH,approx:
=Cme
Substitute numerical values and evaluate RH,approx:
RH,approx
=
(1.204663x 1037 m-I / kg )(9.11x10-31 kg ) = \1.097448 x 107 m-I \
370 Chapter 36 R H and RH,ap Tox agree to three significant figures. P
(c) Express the ratio of the kinetic energy K of the electron in its orbit about a stationary nucleus to the kinetic energy of the reduced-mass systemK':
1
Substitute numerical values and
K
evaluate the ratio of the kinetic
K'
1
1+
energIes:
3 9.11x10- 1 kg 1.67 X 10-27 kg
= 0.999455
or K
=
0.999455K'
and the correction factor is the ratio of the
I
masses or 0.0545%
I
Remarks: The correct energy is slightly less than that calculated neglecting the motion of the nucleus.
*30 ·· The Pickering series of the spectrum ofHe+ (singly-ionized helium) consists of spectral lines due to transitions to the n = 4 state ofHe+. Experimentally, every other line of the Pickering series is very close to a spectral line in the Balmer series for hydrogen transitions to n = 2. (a) Show that this is true. (b) Calculate the wavelength of a transition from the n = 6 level to the n = 4 level ofHe+, and show that it corresponds to one of the Balmer lines.
We can use Equation 36-15 with Z = 2 to explain how it is that every other line of the Pickering series is very close to a line in the Balmer series. We can use the relationship between the energy difference between two quantum states and the wavelength of the photon emitted during a transition from the higher state to the lower state to find the wavelength of the photon corresponding to a transition from the n = 6 to the n = 4 level ofHe+. Picture the Problem
(a) From Equation 36-15, the energy levels of an atom are given by:
ForHe+, Z= 2 and:
Eo
En
= _Z2
En
=-4n2
n2 where Eo is the Rydberg constant (13.6 eV).
Eo
Atoms 371 Because of this, an energy level with even principal quantum number 11. in l-k!- will have the same energy as a level with quantum number 11.12 in H. Therefore, a transition between levels with principal quantum numbers 2m and 2p in He+ will have almost the same energy as a transition between level m and pin H. In particular, transitions from 2m to 2p = 4 in He+ will have the same energy as transitions from In to n = 2 in H (the Balmer series). (b) Transitions between these energy levels result in the emission or absorption of a photon whose wavelength is given by:
(1)
E6
=
and
E4 Substitute for E6 and E4 in equation (1) and evaluate A:
/L
=
4 (13 ·:2eV )
_
eV ) · 4 (13 : 2
_
=
=
-1.51eV
-3.40eV
1240eV· nm
=
(
)
=
I 656nm I
-1.51eV - - 3.40eV which is the same as the n = 3 to transition in H.
n =
2
Quantum Numbers in Spherical Coordinates *35
(a)
··
Find the minimum value of the angle e between L and the z axis for and (c) £ =50.
£ l ( b) £ 4 =
,
=
,
Picture the Problem
The minimum angle between the z axis and L is the angle between
the L vector for m = i! and the z axis. Express the angle e as a function of
Lz and L:
Relate the z component of L to me
and i! :
Express the angular momentum L: Substitute to obtain:
B
=
cos
-l(�i!(i! l)n ] -l(�T+i IT] Hz
+
=
cos
372 Chap ter 36
(a) Fore = 1:
(b) For f! = 4:
e=cos-I(�� (50)=�
(c) For f! = 50:
Quantum Theory of the Hydrogen Atom *39
·
(a) If spin is not included, how many different wave functions are there
corresponding to the first excited energy level
n =
2 for hydrogen? (b) List these
functions by giving the quantum numbers for each state. Picture the Problem
We can use the constraints on n, £, and m to detennine the number
of different wave functions, excluding spin, corresponding to the first excited energy state of hydrogen.
= ol O r
For n = 2:
£
(a) For £ = 0, me= ° and we have:
1 state
For £= 1, me = -1, 0, +1 and we
3 states
have: Hence, for n = 2 we have: (b) The four wave functions are summarized to the right.
s s
14 t te 1 a
n
£
me
(n, £ , me)
2
°
°
(2,0,0)
2
1
-1
(2,1,-1)
2
1
°
(2,1,0)
2
1
1
(2,1,1)
*44 ·· Show that the ground-state hydrogen wave function (Equation 36-33) is a solution to SchrMinger's equation (Equation 36-21) and the potential energy function (Equation 36-26).
Atoms 373
.
solutlOn to
=
We wish to show that '1',, 0,0
11. 2 2 -0 ( r2 -0 ) (r --2mr or or
Picture the Problem
If/
+U
L ff
=
); (:, f' .,z'I"" Ce,z'I"" (r ) - --. kZe2 r =
where U
E If/ ,
=
is a
Because the
ground state is spherically symmetric, we do not need to consider the angular partial derivatives in Equation 36-21. The normalized ground-state wave function is:
II' '1' 1 0 0
, ,
r
-- ( J e - Ce I "1/ 7r
I O
r2:
Multiply both sides of this equation by
r
ao
-Zr/ao _
-Zr/ao
o '-- - C -0 [e ] - -C -Z e ----'-' or or ao r2 0 or = -C -Z r2e If/ , ,D
Differentiate this expression with respect to to obtain:
3/2
_1_ �
_
-zr/ao _
-Zr/ao
-Zr/ao
If/I,O,O
ao
Differentiate this expression with respect to to obtain:
2 Z Z r 2z 0 ( ) 2 2 C -J - -- r e - [--- +r ( J ]ce 2 2 2 e k Z r C 11. ] C 2z 2 C e e --- ----+r e 2 J ( r 2mr [
-o (r 2 &
O lf/I ,D,O &
_
� &
-zr/ao _
�
-Zr/ao
�
Substitute in Schrodinger' s equation to obtain: Z
ao
Solve for E:
Because
ao =
ao
-Zr/ao
-Zr/ao _
E
-Zr/ao
11.--2 2 : mke
= Because this is the correct ground state energy, we have shown that Equation 36-33, is a solution to Schrodinger's Equation 36-21 with the potential energy function Equation 36-26.
374 Chapte r 36 The Spin-Orbit Effect and Fine Structure *50
·
The potential energy of a magnetic moment in an external magnetic field is given by U = jL . B . (a) Calculate the difference in energy between the two possible -
-
orientations of an electron in a magnetic field B
= l.SO Tk. (b) If these electrons are �
bombarded with photons of energy equal to this energy difference, "spin flip" transitions can be induced. Find the wavelength of the photons needed for such transitions. This phenomenon is called electron spin resonance . Picture the Problem
The energy difference between the two possible orientations of an
electron in a magnetic field is 2J1B and the wavelength of the photons required to induce
a spin-flip transition can be found from helM. The magnetic moment JiB associated with
the spin of an electron is 5.79x l O-5 eVrr. (a) Relate the difference in energy
M
between the two spin orientations in terms of the difference in the potential energies of the two states:
= 2J1l3 = 2( S.79x10- 5eV/ T )(O.6 T) = 1 6 .9 SxlO- 5eV I
(b) Relate the wavelength of the photon needed to induce such a transition to the energy required: Substitute numerical values and evaluate A.:
A
. 6.9SxlO-)eV = 1 1.78 cm I
= 124 0eV :un =1.78xl07nm
The Periodic Table *56
•
Write the electron configuration of (a) carbon and (b) oxygen.
Determine the Concept We
can use the atomic numbers of carbon and oxygen to
determine the sum of the exponents in their electronic configurations and then use the rules for the filling of the shells to find their electronic configurations. (a) The atomic number Z of carbon is 6. So we must fill the subshells of the electronic
configuration until we have placed its 6 electrons. This is accomplished by writing
1 1s 2 2s 22p2 I
(b) The atomic number Z of oxygen is 8. So we must fill the subshells of the electronic
Atoms 375 configuration until we have placed its 8 electrons. This is accomplished by writing l s 2 2s 2 2 p 4
\
\
Optical Spectra and X-Ray Spectra *61
·
(a) Calculate the next two longest wavelengths in the K series (after the Ka
line) of molybdenum. (b) What is the wavelength of the shortest wavelength in this series? Picture the Problem When
an electron from state n drops into a vacated state in the n
=
1 shell, a photon of energy M = En - E ) is emitted. We can find the wavelength of this photon using A he/ M . The second and third longest wavelengths in the K series
=
correspond to transitions from n
=
3 to
n =
1 and n = 4 to
wavelength to the transition from n = 00 to n = 1. Express the wavelength of the emitted photon in terms of the
A
n =
he En - E)
=
1 and the shortest
1240eV .run En - E)
energy transition within the atom: Express the energy of the nth energy state:
En
=
-( Z-lY nE�
where n = 1, 2, . . . Substitute to obtain:
1240eV .run - ( Z-lY :� - (- ( Z -lY �o) 1240eV .run
(a) Evaluate this expression with n =
3 and Z = 2 to obtain:
4
�
= =
1240eV .run (42-1Y(13 . 6eV{1- 3\ )
I O.0610run I
376 Chapte r 36 Use n
=
4 and Z = 42 to obtain:
A,
=
Of
1240eV . 1) (42 -1Y(13.6eV) 1- 4"2
= I 0 .0578 nm I
1240eV · nm '" (42-lY( 1 3.6eV)(1-0) I 0.0542nm I
A
(b) The shortest wavelength in the series corresponds to the largest energy difference between the initial
=
=
and final states. Repeat the calculation in part (a) with n = 00 to
obtain:
General Problems *68
··
We are often interested in finding the quantity ke21r in electron volts when r
is given in nanometers. Show that ke2 = Picture the Problem
1 .44 eY-nm.
We can show that ki =
ground state energy of an atom for ke2•
1.44 eV ·nm by solving the equation for the
Express the ground state energy of an atom as a function of k, e, and ao: Solve for ke2 : Substitute for Eo and ao to obtain:
*71
••
ke2
2 (1 3 .6eV)(0.0529 nm) = Il .44eV . nm I =
The combination of physical constants a = e2kllic, where k is the Coulomb
constant, is known as the fine-structure constant. It appears in numerous relations in
atomic physics. (a) Show that a is dimensionless. (b) Show that in the Bohr model of
hydrogen v" = ca in, where Vn is the speed of the electron in the stationary state of quantum number n.
Picture the Problem
We can show that a is dimensionless by showing that it has no
units. In part (b) we can use Bohr's 3Td postulate and the expression for the radii of the
Bohr orbits, together with the definition of a, to show that the speed of the electron in a stationary state of quantum number n is related to a according to
VI1 =
ca In.
Atoms 377
(a) Express the units of a:
Because a i s unitless, it is also dimensio nless. (b) Apply the quantization of angular momentum postulate to
v"
=
obtain: The radii of the Bohr orbits are given by:
nh m rn
-
r" = n 2
v"
mke2 nh
mn 2
h2
m ke2
�
v" =
ke2
--
nh
--
ke2 a
Solve for v,,:
h2
=
Divide this expression by the definition of a to obtain:
m kZe2
or, because Z = 1 for hydrogen,
r" = n 2
Substitute and simplify to obtain:
h2
nh e2k he
e
n
� n
*74 · A Rydberg atom is one in which an outer-shell electron is placed into a very high excited state (n � 40 or higher.) S uch atoms are useful for experiments which probe the transition from quantum mechanical behavior to classical. Furthermore, these excited states have extremely long lifetimes (i.e., the electron will stay in this high excited state for a very long time.) A hydrogen atom is in the n = 45 state. (a) What is the ionization energy of the atom when it is in this state? (b) What is the energy level separation (in e V) between this state and the n = 44 state? (c) What is the wavelength of a photon resonant with this transition? (d) How large is the atom in the n = 45 state?
Picture the Problem The ionization energy of the electron is the magnitude of the energy of the atom in the given state. We can use E = -EoIn2, where Eo is the ground-state energy, to find the energy levels in the 44th and 45th states and, hence, the energy level separation between the states. The wavelength of a photon resonant with this transition
378
Chapt e r
36
can be found from A helM. We'll approximate the size of the atom in the by finding the radius of the outer-shell electron. =
(a) The energy of the atom in its nth state is: The energy of the atom in the n = 45 state is: The ionization energy is the negative of the energy in the state n = 45 : (b) The energy level separation between the n = 45 and n = 44 state IS:
n =
45 state
Eo EI/ = -n2 13.6eV E45 = - (45Y = -6.72meV Eionizing = -E45 = 16.72meV I E45�44 -
(13(45.6eV)2
__
=
13.6 ev
_
(44)2
13.09x10-4 eV I
)
(c) The photon wavelength is:
Substitute numerical values and evaluate A:
A=
.
1240 eV 01x106 run ! 3.09x10-4 eV = !4.
(d) The radii of the Bohr orbits are given by:
r =n 2Z
Substitute numerical values and evaluate the radius of the 45th Bohr orbit:
r = (45Y
run
ao
0.05�9 run = !107nm
!
Chapter 37 Molecules Conceptual Problems *1
•
Would you expect the NaCI molecule to be polar or nonpolar?
Determine the Concept Yes. Because the center of charge of the positive Na ion does
not coincide with the center of charge for the negative CI ion, the NaC I molecule has a permanent dipole moment. Hence, it is a polar molecule . *5
••
The elements on the far right column of the periodic table are sometimes
called noble gases because they virtually never react with other atoms to form molecules. However, this behavior is sometimes modified if the resulting molecule is formed in an electronic excited state. An example is ArF. When it is formed in the excited state, it is written ArF* and is called an excimer (for excited dimer). Refer to Figure 37- 1 3 and discuss how this diagram would look for ArF in which the ArF ground state is unstable but the ArF* excited state is stable. Remark: Excimers are used in certain kinds of lasers. Determine the Concept The diagram would consist of a non-bonding ground state with
no vibrational or rotational states for ArF (similar to the upper curve in Figure 37-4) but for ArF* there should be a bonding excited state with a definite minimum with respect to inter-nuclear separation and several vibrational states as in the excited state curve of Figure 37-1 3 . Estimation and Approximation *14
Repeat Problem 1 3, finding the quantum number
••
v
and spacing between
adj acent energy levels for a S-kg mass attached to lS00-N/m spring vibrating with an amplitude of 2 cm. Hint: Pick
v
so that the quantum energy formula (Equation 37-18)
gives the correct energy for the given system. Then find the energy increase for the next highest energy level. Picture the Problem We can solve Equation 37- 1 8 for v and substitute for the frequency of the mass-and-spring oscillator to estimate the quantum number vand spacing between adjacent energy levels for this system.
The vibrational energy levels are given by Equation 37-1 8 : Solve for
v:
Ev
= (v+t)hl
where v= 0, 1 , 2, . . .
1 Ev v=--hi 2 379
380
Chapte r
37 or, becaus e Ev v>::;
The vibrational energy of the object attached to the spring is:
Ev
=
v=
l..2 kA2
kA2
2hf l /k f=_ 2;r �;;
The frequency of oscillation f of the mass-and-spring oscillato r is given by:
Substi tute numerical values and evaluate
Set
hf
_
where A is the amplitude of its motion.
Substitute for E v in the expression for v to obtain:
v=
v» 1,
--
v:
n(0.02_�Y �(Skg)(lS00N/m) = I 1 .64 x 1032 I 6 . 63 x 1 0 J·s
v= 0 in Equation 37- 1 8 to
2
adjacent energy levels: Substitute numerical values and evaluate
Eov:
Remarks: Note that our value for
/k 4n � ;;
EOv = 2. hif = �
express the spacing between
_6.63 x 10-34J . s lS00N/m S kg 4n = 1 9.14 x 10-34J I
EOv -
vjustifies our assumption that v »
1.
Molecular Bonding ·
*18
The equilibrium separation of the HF molecule is 0.09 1 7 nm, and i ts
measured electric dipole moment is 6.40x ionic?
OO( J.
10-30 em. What percentage of the bonding is
Picture the Problem The percentage of the bonding that is ionic is given by
1
Pmeas PIOO
Molecules 381 Express the percentage of the bonding that is ionic:
Pe rcent
Express the dipole moment for
P IOO
1 00 % ionic bonding: Substitute to obtain:
=
io ni c bo ndi ng
=
( )
I OO
er
Pe rcent ioni c bonding =
Pmeas P IOO
100( P;as )
Substitute numerical values and evaluate the percent ionic bonding:
Pe rcent ion ic bonding *24
•••
=
'C m · [(1.60x6.4100_�/C0);0.0917nm )]
I OO
=
I 43.6% I
Assume that the potential energy associated with the core repulsion of the
two ions of a diatomic molecule with ionic bonding can be represented by a potential
energy of the form Urep = Clr", so the total potential energy is
U= Ue + Urep + M, where Ue
=
-ke2 /r.
/)ill is the energy of the two ions at infinite
separation less the energy of the two neutral atoms at infinite separation (see Figure 3 71).
Use dUidr= 0 at r = ro to show that n
=
U I e(r o)1 Urep (ro)
-'--'--.:......!.
Picture the Problem U(r) is the potential energy of the two ions as a function of separation distance r. U(r) is chosen so U(oo) = -M, where M is the negative of the
energy required to form two ions at infinite separation from two neutral atoms also at
infinite separation. Urep(r) is the potential energy of the two ions due to the repulsion of
the two closed-shell cores. Ed is the disassociation energy, the energy required to separate the two ions plus the energy M required to form two neutral atoms from the two ions at
infinite separation. The net force acting on the ions is the sum of Frep and Fe . We can find
Frep from Urep a nd Fe from Coulomb ' s law and then use dUldr = Fnet = 0 at r= ro to solve for n .
(1)
Express the net force acting on the IOns: Find Frep from Urep :
Frep
=
dUrep dr nC
=
!!:..-[Cr-"l dr
=
-nCr-n-1
382 Chapter 37 The electrostatic potential energy of the two ions as a function of separation distance is given by: Find the electrostatic force of attraction Fe from Ue:
Substitute for Frep and Fe in equation
( 1 ) to obtain:
Because dU/dr
=
Ue =
Fe =
FlIet
_r
ke2
dUe dr
=
_
=
nC
r
11+1
�[_ ] ke2
dr
+
r
=
ke2 r2
ke2 r2
Fnet = 0 at
Multiply both sides of this equation
by ro to obtain:
Solve for n to obtain:
n=
lue h)1 U rep (ro)
Energy Levels of Spectra of Diatomic Molecules
*27
·
The separation of the two oxygen atoms in a molecule of O is actually
2
slightly greater than the 0. 1 nm used in Example 37-3, and the characteristic energy of rotation Eor is 1 .78x 1 0 -4 eV, rather than the result obtained in that example. Use this value to calculate the separation distance of the two oxygen atoms.
Picture the Problem We can relate the characteristic rotational energy Eor to the moment
of inertia of the molecule and model the moment of inertia of the O molecule as two
2
point objects separated by a distance r. The characteristic rotational energy of a molecule is given by: Express the moment of inertia of the molecule: Substitute for 1 to obtain:
li2 EOr = 21
Molecules
383
Solve for r:
Substitute numerical values and evaluate r: r
�
=
*30
(1.055X�O-" J ·s ) I O.121nm I
The equilibrium separation between the nuclei of the LiH molecule is 0 . 1 6 nm. Determine the energy separation between the = and rotational levels o f this ••
f! 3
diatomic molecule.
f! = 2
Picture the Problem We can use the expression for the rotational energy levels of the
diatomic molecule to express the energy separation M between the
f! = 3 and
f! = 2 rotational levels and model the moment of inertia of the LiH molecule as two point
objects separated by a distance roo
The energy separation between the
f! 3 and R 2 rotational levels of =
=
this diatomic molecule is given by:
Express the rotational energy levels Ee=3 and Ee= in terms of Eor:
2
EM = 3(3 + I)Eor = 12EOr
and
E =2 2(2 + I)Eor 6EOr t
=
=
Substitute for Ee= 3 and Et=2 to
obtain:
The characteristic rotational energy of a molecule is given by:
h2 = .L/1E EOr = 21 6
�
-
/1E = 3h2 1
Express the moment of inertia of the molecule:
where f.1 is the reduced mass of the
molecule.
384 Chapter
37
Substi tute for 1 to obtain:
M=
2
2
3tz = -3tz ---2 mLimH mLi + mH 3tz2 (mLi + mH ) mLimH rO 2 JL
r0
Substitute numerical values and evaluate M: M=
*31
3 1 .055x lO-34J·s
(6.94u+1u) (6.94u)(1 u)(0.16nmY 1.602x10-19 J/eV 1.660 x 1 0-27 kg/u ••
=
1 5 . 6 1meV I
Derive Equations 37-14 and 37- 1 5 for the moment of inertia in tenns of the
reduced mass of a diatomic molecule. Picture the Problem Let the origin of coordinates be at the point mass m l and point mass
m2 be at a dis tance ro from the origin. We can express the moment of inertia of a diatomic molecule with respect to its center of mass using the definitions of the center of mass and
the moment of inertia of point particles. Express the moment of inertia of a
(1)
diatomic molecule: The r coordinate of the center of mass IS:
The distances of
m
center of mass are:
I
and
m2 from the
Substitute for r l and r 2 in equation
(1) to obtain:
Simplifying this expression leads to: or
M olecules
11
=
/-t
whe re
ro2
385
36-14 36-15
ml and m2 are attached to a spring of force constant k roo (a) Show that when ml is moved a distance L'lrl from the center of mass, the force exerted by the spring is m,;,m, )",r, -k ( *36
··
Two objects of mass
and equilibrium length F
�
(b) Show that the frequency of oscillation is
f
=
_12:r_ �k/
I-l, where J.L is the reduced
mass. Picture the Problem For a two-mass and spring system on which no external forces are
acting, the center of mass must remain fixed. We can use this condition to express the net force acting on either obj ect. Because this force is a linear restoring force, we can conclude that the motion of the object whose mass is angular frequency given by
OJ �
given in (b). (a) If the particle whose mass is
�km'ffl
ml
ml will be simple harmonic with an
. Substitution for leo" will lead us to the result
I1r2 _mmI I1lj from (or toward) the center 2
L'lrl from (or toward) the cente r of mass, then the particle whose mass is m2 must
of mass.
Express the force exerted by the
F
moves a distance
=
move a distance:
sprIng: Substitute for L'lrz to obtain:
F
=
�
�
(b) A displacement in a restoring force:
L'lr of m results 1
1
F
=
-kl1r = -k(l1lj +I1r2)
-{ +�, ) m,;,m } -k ( m +m 2 ) -k -k ( m "'r,
"'r,
r,
1
2
11r.1
=
efT 11r.1
386 Chapter
37 where
keff
=
k
(
m1 +m2 m2
Because this is a linear restoring
J
force, we know that the motion will be simple harmonic with:
Substitute for ke ff and simplify to
obtain:
or, because f.1 =
mm I 2 ml + m2
is the reduced
mass of the two-particle system,
f = \llkl .
lEkJ
General Problems *40
··
The effective force constant for the HF molecule
frequency of vibration for this molecule.
IS
970 N/m. Find the
Picture the Problem We can use the result of Problem 36 to find the frequency of
vibration of the HF molecule. In Problem 36 it was established
that:
l fI f =_ 2n � -;
The reduced mass is:
Substitute for f.1 to obtain:
Substitute numerical values and evaluate f
l f =_ 2n
k = 1 ----
Mole cule s 387
*47
··
For a molecule such as CO, which has a permanent electric dipole moment,
= 1
radiative transitions obeying the selection rule 6,.e ± between two rotational energy levels of the same vibrational level are allowed. (That is, the selection rule 1'1 v = ± does
1
not hold.) (a) Find the moment of inertia of CO and calculate the characteristic rotational
Eo r £= 5 o for £ = O .
£= 0 E
(in eV). (b) Make an energy level diagram for the rotational levels for energy to for some vibrational level. Label the energies in electron volts starting with Indicate on your diagram transitions that obey 6,.£
= -1 and calculate the
=
energy of the photon emitted (c) Find the wavelength of the photons emitted for each transition in (b). In what region of the electromagnetic spectrum are these photons?
Picture the Problem We can find the reduced mass of CO and the moment of inertia of a
CO molecule from their definitions. The energy level diagram for the rotational levels for = Finally, we can find the wavelength to can be found using Me,t-I =
£ 0 £= 5
2 £EOr'
1 . . d lor . C each transItIon usmg o f the ph otons emItte /l,e , (-I .
1-
(a) Express the moment of inertia of CO:
he = --M
r,2
e t- I
he 2 £MOr
/I 0 r
where f.1 is the reduced mass of the CO
molecule. Find f.1:
In Problem 39 it was established that
ro = 0. 1 1 3 nm. Use this result to evaluate I:
1 = (6.86u)(1.66x 10-27 kg/u)(0.113nrnY = 11.45 x 10-46 kg ·m2 1
2
EOr = li1 2
The characteristic rotational energy
Eor is given by:
Substitute numerical values and evaluate
Eor:
EOr = -'----r--L.�--::---L = 1 0.239 meV I
388 Chapter 37 f = 5, E
(b) The energy level diagram is shown to the right. Note that Me,c-I>
=
2.14 meV
-
the energy difference between adj acent levels for t>"e = - 1 , is
1iEe,c_1 = 2 .eEOro
e
=
e
f.
e
=
=
=
4, E
3, E
2, E
1, E
liEu_I
2.86 meV----,--'--
=
=
=
e
(c) Express the energy difference between energy levels in
4.76 meV-
=
1.43 . meV --,--'----
0.476 mev
=
O,E
=
, -
0-'-
liEeH , , = hieH
tenus of the frequency of the emitted radiation: Be cause
e= ieHAeH
he e,H
Ae,e-I = liE
:
he
Substitute numerical values to obtain:
Ae,e-I =
(40136x10-15 eV os)(3x108m/s)= 2 596,urn .e 2 .e(0.2 39meV)
For JI. = 1 :
29 �,O= 5 ,urn = 1 2 596 Jl111 1
For.e = 2:
Az,1 = 2 59 ,urn = 1 1298,urn I
For.e
�,2 = 2 59 ,urn = I 865,urn I
=
3:
�
�
�
-
Mole cule s 389 For e Fore
=
=
4:
A4.3
5:
A5,4
=
=
2596f1I11 4 2596f1I11 5
=
=
\ 649 \ f-trl1
\ 5 1 9 f1I11 \ .
The s e wave le ngths fa lli n the microwave re gion of the s pe ctrum. *48
•••
Use the results of Problem 24 to calculate the vibrational frequency of the
LiCl molecule. The dissociation energy of LiCl is 4.86 eV, and the equilibrium separation is 0.202 nm. The electron affinity of chlorine is 3.62 eV, and the ionization energy of lithium is 5 .3 9 eV. To do this, expand the potential about r = ro, where ro i s the
equilibrium separation, in a Taylor series. Retain only the term proportional to (r - ro)2.
Recall that the potential energy of a simple harmonic oscillator is given by USHO = t m(ix2. What is the wavelength resulting from transitions between adj acent harmonic oscillator levels of this molecule?
Picture the Problem The wavelength resulting from transitions between adjacent
harmonic o scillator levels of a Liel molecule is given by A expression for
OJ by
=
2ne . We can find an (()
following the procedure outlined in the problem statement.
The wavelength resulting from transitions between adj acent harmonic oscillator levels of this
A
=
he he 2ne M n(() (() =
=
(1)
molecule is given by: From Problem 24 we have:
U(r)
=
-
ke2
r + r� , where M i s constant.
The Taylor expansion of U(r) about r
=
ro i s:
Because U(ro) i s a constant, it can be dropped without affecting the physical results and because
(2)
390
Chapter
(dU )
dr .
37
= 0:
'0
Differentiate U(r) twice to obtain: B ecause dU/dr = r= ro:
Fuet
=
0 at
c= ker2 a nrO 2
Solving for C yields:
( d2d U2 )
11+1 _
kera 2 II-I n
S ubstitute for C and evaluate
r
ro
to obtain:
Substitute for equation (2) :
(�:�1.
in
B ecause the potential energy o f a s imple harmonic oscillator is given by
/
USHO = tmo (r - ro Y :
Solve for
w to
obtain:
S ubst itute }1LiCI for m to obtain:
(n -1 )ke2
OJ ';::;
3
mrO
(n -1)ke2
OJ ';::;
mLimCl r O mLi +mCI
3
(3)
(n -l ) (mLi + mcl )ke2 3
mLimClrO
From Problem 24:
Urep is related to Ue, Ed, and M
n=
IUe(ro) I Urep (ro )
Urep -(Ue + Ed + M )
=
(4) (5)
Molecules
391
according to: The energy needed to form Li+ and cr from neutral lithium and
M
=
Substitute ro and evaluate Ue:
Substitute numerical values in
e2
Ue
= -k
Ue
=
equation (5) and evaluate Urep:
equation (4) and evaluate
n:
E
electronaffinity
n=1
1 .44 eV .run = 7.1 3eV 0.202nm
= -(- 7 . 1 3eV + 4.86 eV + 1 .77 eV) = 0.500 eV - 7.1 3 eV I = 14.3 0.500 eV
Substitute numerical values in equation (3) and evaluate
(j);::::
-
1 .44 eY·run
-
Urep
Substitute for Urep(ro) and Ue(ro) in
ionization
= 5.3geY - 3.62 eY = l.77 eY
chlorine atoms is:
Ue(ro) is given by:
E
co:
(1 4.3 -1)(6 . 941u + 35.453u)(1 .44 eV .run) 1 .60 x 10-19 l/ eV (6.941 u)(35.453u) 1 .66 x 10-27 kglU (0.202runY
= 1 1 .96 x 1 014 S-I
Substitute numerical values in
equation (1) and evaluate A:
I
Chapter 38 Solids and the Theory of Conduction Conceptual Problems *2
•
When the temperature of pure copper is lowered from 300 K to 4 K, its
resistivity drops by a much greater factor than that of brass when it is cooled in the same way. Why? Determine the Concept The resistivity of brass at 4 K is almost entirely due to the
"residual resistance, " the resistance due to impurities and other imperfections of the crystal lattice. In brass, the zinc ions act as impurities in copper. In pure copper, the
resistivity at 4 K is due to its residual resistance, which is very low if the copper is very pure. How does the change in the resistivity of copper compare with that of silicon *8 when the temperature increases? Determine the Concept The resistivity of copper increases with increasing temperature; the resistivity of (pure) silicon decreas es with increasing temperature b ecause the number of charge carriers increases.
The Structure of Solids *18
· Find the value of n in Equation 3 8 -6 that gives the measured dissociation energy of74 1 kJ/mol for LiCl, which has the same structure as NaCl and for which ro = 0 .257 nm.
Picture the Problem W e can solve Equation 3 8 -6 for n.
Equation 3 8-6 is: and
Solve for n to obtain:
1
393
394 Chapter 38 Substitute numerical values and evaluate n: n=
(1
1 14.641 eV/ion pair (741kJ/mol) (0.2 7nm) 6.47 kJ/mol 5 9 1 (1.7476)(1 .44eV . nm)
_______________
-
J
=
�--:-:-----:----
---
A Microscopic Picture of Conduction
*22 ·· Silicon has an atomic weight of28 .09 and a density of 2 .4 1 x1 0 3 Each atom of silicon has 2 valence electrons and the Fermi energy of the material is 4.88 eV. (a) Given that the electron mean free path at room temperature is A = 27.0 nm, estimate the resistivity. (b) The accepted value for the resistivity of Silicon is 640 n-m (at room temperature). How does this accepted value compare to the value calculated in part (a)?
kg/m3.
Picture the Problem We can use Equation 3 8-14 to estimate the resistivity of silicon.
(a) From Equation 38-14:
(1)
The speed of the electrons is given by: Substitute numerical values and evaluate Vav:
The electron density of Si is given by:
v
av
1
2(4.88 eV) 1.60 x 1 0-19 J (9. 1 x 10-31kg) 1 eV = 1 .31x106m1s =
ne =
MN
A
Nalam
where Nalam is the number of electrons per atom.
Substitute numerical values and evaluate ne: ne
=
2X 1023atoms J(�) (2.41X 1 03 mkg3 )( 6.00.0280 atom 9 kg
=
1.0 3x1 029 e/m3
Substitute numerical values in equation (1 ) and evaluate p:
(b) The accepted resistivity of 640 n·m is much greater than the calculated value. We assume that valence electrons will produce conduction in the material. Silicon is a
Solids and the Theory of Conduction 395 semiconductor and a gap between the valence band and conduction band exi sts. Only electrons with sufficient energies will be fo und in the conduction band. The Fermi Electron Gas *26
Calculate the Fermi temperature for (a) AI, (b) K, and (c) Sn.
•
Picture the Problem The Fermi temperature
Fermi energy.
TF is defined by kTF EF, where EF is the =
E F = kF
The Fermi temperature is given by:
T.
(a) For AI:
T.
(b) For K:
T.
(c) For Sn:
T.
F = 8.62x10-�) eV/K = 1 1.36x105K I 11.7 V
F = 8.62x10-�) eVIK = 1 2.45x104 K I 2.11 V
F = 8.62x10-)eVIK = 1 1.18x105 K I 10.2 �V
*31 ·· (a) Assuming that gold contributes one free electron per atom to the metal, calculate the electron density in gold knowing that its atomic wei ght is 196.97 and its mass density is 1 9 .3x1 03 kg/m3. (b) If the Fermi speed for gold is 1.3 9x1 06 mis, what is the Fermi energy in electron volts? (c) By what factor is the Fermi energy hi gher than the energy at room temperature? (d) Explain the difference between the Fermi energy and kTenergy.
kT
Picture the Problem We can use n e =
pV =
pN Na om A
m
t
, where Natom is the number of
electrons per atom, to calculate the electron density of gold. The Fermi energy is given 2 - 1 bY - z mevF•
EF
(a) The electron density of gold i s given by:
n e = pV =
pNANatom m
S ubstitute numerical values and evaluate n e:
ne =
(
19. 3X103
� ) (6.02x1023atoms)(1atIeom )
k
m
0.197 kg
(b) The Fermi energy is given by:
1
= 5.90x1028e/m3
1
39 6 Chapter 38 Substitute numerical val ues and evaluate
EF:
leV - = I s . so ev I EF =� (9. 1 1 XIO3- 1kg )(1 . 39XI06m/S) 1.60xlO 19J 2
)
(
(c) The factor by which the Fermi energy is higher than the kT energy at room temperature is:
E f= kTF
At room temperature kT= 0.026 eV. S ubstitute numerical val ues and evaluatef
S.SOeV f= 0.026 eV= @IJ
EF
(d) is 2 1 2 times kT at room temperature. There are so many free electrons present that most of them are crowded, as described by the Pauli exclusion principl e, up to energi es far higher than they would be according to the classical model . *32
••
particles by
The pressure of an ideal gas is related to the average energy of the gas PV t , where N is the number of particles and is the average
Eav
= NEav
energy. Use this to calculate the pressure of the Fermi electron gas in copper in newtons per square meter, and compare your result with atmospheric pressure, which is about 1 0 5 N/m 2. (Note: The units are most easily handled by using the conversion factors 1 N/m 2 = 1 J/m 3 and 1 eV = 1 .6 l 0 1 J.)
x 9 -
Picture the Problem We can solve
express P in terms of NIV and Solve
PV= t NEav
B ecause
for P:
Eav = t EF :
EF•
PV= t NEav for P and substitute for Eav in order to
p=� N Eav 3 V
( )
p= � N 'i EF = � N EF 3V S SV
( )( ) ( )
Substitute numerical values (see Table 38-1) and evaluate P:
p= �S (8.47 x1 022 electrons/cm3 )(7.04eV) (1.60x1 0-19 JleV) = 1 3.82xl0ION/m2 1. = 3.82xl0ION/m2 x 101 .32Slatm x1 03 N/m2 = 1 3.77xl05 atm I
Solids and the Theory of Conduction 397 Heat Capacity Due to Electrons in a Metal *35 ·· Gold has a Femli energy of 5 .53 eV. Determine the molar specific heat at constant volume and room temperature for gold.
Picture the Problem We can use Equation 3 8-29 to find the molar specific heat of gold at constant volume and room temperature.
The molar specific heat is given by Equation 3 8-29:
e 'y
The Fermi energy is given by:
Substitute for TF to obtain:
Substitute numerical values and evaluate e 'y :
Jr2 (8.3IJ/molK)(1.38x 10-23 J/mol)
e'
y
(
l eV 1.60x 10
-19
J
]
(300K)
= --------------------�----��--------�----=
1 0.192J/mol·K 1
2(5.53
eV )
Remarks: The value 0.192 J/mol K is for a mole of gold atoms. Since each gold atom contributes one electron to the metal, a mole of gold corresponds to a mole of electrons. Quantum Theory of Electrical Conduction
*37
··
The resistivity of pure copper is increased approximately 1 x 1 0-8 n·m by the
addition of 1 percent (by number of atoms) of an impurity throughout the metal. The mean free path depends on both the impurity and the oscillations of the lattice ions according to the equation 1/ A = 1/At + 1/ A,. (a) Estimate A; from data given in Table 38-1 . (b) If r i s the effective radius of an impurity lattice ion seen by an electron, the scatterin g cross section i s
Equation 3 8- 1 6.
7U"2. Estimate this area, using the fact that
r
is related to Ai by
Picture the Problem We can solve the resistivity equation for the mean free path and
then substitute the Fermi speed for the average speed to express the mean free path as a function of the Fermi energy.
398 Chapter 38 (a) In terms of the mean free path and the mean speed, the resistivity IS:
Solve for /L to obtain:
AI
= meuF2
ne Pi
Express the Fenni speed UF in
tenns of the Fenni energy
Er::
Substitute to obtain:
Substitute numerical values (see Table 38-1) and evaluate�:
�2(9.11x10-31 kg)(7.04eV)(1.60X1�-91 l/eV) = 66.1n mI 1 (8.47 x1028 electrons/m3)(1.60 x10-91 C) (10-8 Q. m)
Ai =
(b) From Equation 3 8 -1 6 we have:
Solve for 7r?:
Substitute numerical values and
evaluate ;rr2:
nA rcr
1 2- r 2 - \8.47X10 8 m-3)(66.1nm) xl 0--4-- nm-:--'2 1 = 1.79x10-22m2= lr--l .-79--
Band Theory of Solids *39
·
You are an electron sitting at the top of the valence band in a silicon atom,
longing to jump across the 1 . 1 4-eV energy gap that separates you from the bottom of the conduction band and all of the adventures that it may contain. What you need, of course, is a photon. What is the maximum photon wavelength that will get you across the gap? Picture the Problem We can relate the maximum photon wavelength to the energy gag using
DE = hf = he/A.
Express the energy gap as a function of the wavelength of the photon:
DE =hf = he A
Solids and the Theory of Conduction
399
Solve for A: Substitute numerical values and evaluate A:
A=
1 240eV ·nm 1 . 1 4eV
=
1 1.09 1 J11ll
Semiconductors *44
••
When a thin slab of semiconducting material i s illuminated with
monochromatic electromagnetic radiation, most of the radiation i s transmitted through the slab if the wavel ength i s greater than 1 . 85 J.lill. For wavel engths less than 1 .85 J.lill,
most of the incident radiation i s absorbed. Determine the energy gap of this semiconductor.
Picture the Problem We can use E= hf to find the energy gap of this semiconductor.
The energy gap of the semiconductor is given by:
E g=
hI" =
where Substitute numerical values and evaluate Eg:
*46
••
E 1=
'J
he
A
he = 1240 eV·nm o 1240eV ·nm E " = 1 .85,urn = 0.670eV
1
1
The ground-state energy of the hydrogen atom i s given by e 2 me
8 E o2 h 2
Modify this equation in the spirit of Problem 45 by replacing Eo by KEo and me by an effective mass for the electron to estimate the binding energy of the extra el ectron of an impurity arsenic atom in (a) silicon and (b) germanium. Picture the Problem We can make the same substitutions we made in Problem 45 in the expression for E1 (= 13.6 eV) to obtain an expression that we can use to estimate the binding energy of the extra electron of an impurity arsenic atom in silicon and germalll urn.
Make the indicated substitutions in the expression for E1 to o btain:
e2 memeff
8meK 2 E� h 2
400
Chapter 38
(a) For silicon:
(b) For gennanium:
Semiconductor Junctions and Devices *51
··
In Figure 3 8-27 for the pnp-transistor amplifier, suppose
Rb= 2 kO and RL=
1 0 kO . Suppose further that a 1 0-I'A ac base current generates a 0.5-mA ac collector current. What is the voltage gain of the amplifier? Picture the Problem We can use its definition to compute the voltage gain of the amplifier.
The voltage gain of the amplifier is given by: Substitute numerical values and evaluate the voltage gain:
IcRL . = -IbR b (0. 5 mA)(10 kQ) V o 1tage gam (1 0.uA) ( 2kQ) = 1 25 0 1 Voltage gam .
_
*54 ·· A "good" silicon diode has the current-voltage characteristic given in Problem 49. Let kT= 0 .025 eV (room temperature) and the saturation c urrent 10 = 1 nA. (a) Show that for small reverse-bias voltages, the resistance is 25 MO. (Hint: Do a Taylor expansion o/ the exponential/unction or use your calculator and enter small values/or V )b . (b) Find the dc resistance for a reverse bias of 0.5 V. (c) Find the dc resistance for a 0.5-V forward bias. What is the current in this case? (d) Calculate the ac resistance dVld1for a 0.5 -V forward bias.
Picture the Problem We can use Ohm's law and the expression for the c urrerit from Problem 49 to find the resistance for small reverse-and-forward bias voltages.
(a) Use Ohm ' s law to express the resistance:
R=Vb I
From Problem 47, the current across a pn j unction is given by:
(2)
For eVb« kT:
S ubstitute to obtain:
(1)
e b I= IoV kT
Solids and the Theory of Conduction 4 0 I Substitute for fin equation simplify:
(1) and
VIJ kT _ eVIJ elo 1o
R=
kT
Substitute numerical val ues and evaluate R:
1
1
= 25.0 MO (b) S ubstitute equation (2) in equation ( 1 ) to obtain:
R-
eVb
eVb
kT
kT
Evaluate -- for V b = - 0.5V: Evaluate equation (3) for
Vb=-0.5V: ev' (c) Evaluate _b for
Vb= + 0.5V:
kT
Evaluate equation (3) for
Vb= +0.5V: (d) Eval uate Rae = dVldlto obtain:
R=
eVb
kT
R=
b
(
(3)
)
) )
19 C -O.5V = l.60x 10= -19. 8 l. 38x lO-23 JIK 293K
(
I
-O.5V = 5 oOMO 10 9 A -,98. _1
)(e
(
)
( ) = 19.8 l . 38x 10-23 J/K (293K )
1
O.5V = l.26 0 1 0-9 A el98 -1
=
)
H
(
R ae _
1
l.60x lO-'9 C O.5V
=
=
Substitute numerical values and evaluate Rae:
V,
e 10 e vb/kT -1
dV dI
_
( dI )-1 dV
{:v [I, (
{ (
e -'.1" -1
}
1
ljf
e1o eVb/kT -1 kT -eVb/kT e = e e 10 kT
)
1
Rac = 25 MO e-19.8 = 0.0629 0
1
The BCS Theory *57
·
Repeat Problem 56 for lead 3 of 2.73xl O- eV.
(Tc = 7 . 1 9 K), which has a measured energy gap
Picture the Problem We can calculate Eg using
Eg = 3.5k�
and find the wavel ength of
a photon having sufficient energy to break up Cooper pairs in tin at
T = 0 using
402
Chapter38
(a) From Equation 3 8 -24 we have: Substitute numerical values and evaluate Eg:
Express the ratio of Eg to Eg,measured:
10-5 eV/K)(7.19K) = 1 2.17meV I
Eg = 3.5(8.62 x
Eg Eg,measured or
Eg
:::::
2.17meV = 0.795 2.73xl O-3eV
I 0.8Eg,measured
(b) The wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 is given by: Substitute numerical values and evaluate A:
1
. V run = 4.54xl05 run = 1240e x 3 2.73 l O eV = I 0.454mmI
The Fermi-Dirac Distribution *60
··
(a) Use Equation 38-22a to calculate the Fermi energy for silver. (b) Determine the average energy of a free electron and (c) find the Fermi speed for silver.
Picture the Problem Equation 38-22a expresses the dependence of the Fermi energy EF on the number density of free electrons. Once we've determined the Fermi energy for silver, we can find the average electron energy from the Fermi energy for silver and then use the average electron energy to find the Fermi speed for silver.
(a) From Equation 38-22a we have:
2 3N 3/2 EF - h 8me "V _
Use Table 27- 1 to find the free electron number density NIV for silver:
-- (--)
= 5.86xl022 electrons V cm3 = 5.86xl 028 elect: ons m N
Solids and the Theory of Conduction 403 Substitute numerical values and evaluate EF:
3
6.63xlO-34J·s 2 [3(S.86xl028electrons/m3)]2/ [ leV EF = 89.11xlO-3Ikg 1.60xlO-19J 7[
= / S . S l eV /
(b) The average electron energy is given by:
3 Eav = -EF S
Substitute numerical values and evaluate Eav:
Eav =%(S.SleV)=/3.3lev /
]
(c) Express the Fermi energy in terms of the Fermi speed of the electrons: Solve for VF:
S ubstitute numerical val ues and evaluate VF:
*63
··
energy of
vF
2(3.3leV) [1.60X 10-19 J] 9.11xlO-3Ikg leV =11.08xl06m1s I -
What is the probability that a conduction electron in silver will have a kinetic
4.9 eV at T= 300
K?
Picture the Problem The probability that a conduction electron will have a given kinetic
energy is given by the Fermi factor. The Fermi factor is:
Because EF *70
•••
-
4.9 eV » 300k:
(a) Show that for E �
1_ = OJ f(4.geV)= _0+1 0, the Fermi factor may be written as
f(E) = l/(CeE/kT +1) (b) Show that if C» e-E1kT,f(E) Ae-E1kT « =
I; in other words,
A
show that the Fermi factor is a constant times the classical Boltzmann factor if « 1 . (c) Use
fn (E)dE = N and Equation 38-41 to determine the constant A. (d) Using the result
obtained in Part (c), show that the classical approximation is applicable when the electron concentration is very small and/or the temperature is very high.
(e) Most semiconductors
have impurities added in a process called doping, which increases the free electron
404
Chapter
38
concentration so that it is about 1 017/cm3 at room temperature. Show that for these systems, the classical distribution function is applicable. Picture the Problem We can follow the step-by-step procedure outl ined in the problem
statement to obtain the indicated results. (a) The Fermi factor is:
1
CeE/kT
=
provided C (b) If C»
e-E1kT:
1
e-EdkT
1 1 f ( E) - CeE/kT 1�CeE/kT - I Ae-E/kT 1 _
_
where A (c) The energy distribution function
=
+
+
=
l/C
n{E)dE g{E)dE f{E) =
IS:
S ubstitute for g(E)dE and j(E) in the expression for N to obtain: The definite integral has the value: S ubstitute to obtain:
Solve for A:
(d) Evaluate A at T= 300 K: A
.J2(6.63xl O-3 4J,sLn �4xl O-26n 2 - 81Z"3 /2�.l 1xl O-3 1 kgr/2[(1.38xl O-23JIK)3 { 00K)r/
-
where the units are S1.
Solids and the Theory of Conduction 4 05 The valence electron concentration is typically about 1039 m -3. To satisfy the condition that A « 1 at room temperature, n should be less than 1023 m -3, or
about one millionth of the valence electron concentration. Because A depends on T-3/2 , the electron concentration may be greater the higher the temperature.
(e)
1017 cm-3 = 1023 m-3. So, according to the criterion in
Cd), the classical
approximation is applicable.
General Problems *76 ·· Detennine the energy that has 1 0 percent free electron occupancy probability for manganese at T = 1 300 K.
Picture the Problem The Fenni factor gives the probability of an energy state being occupied as a function of the energy of the state E, the Fenni energy EF for the particular material, and the temperature T.
The Fenni factor is:
For 1 0 percent probability:
1 0.1 = (E-E )jkT F +1 e
Take the natural logarithm of both sides of the equation to obtain:
E -EF kT
Solve for E to obtain:
E = EF +kTln 9
From Table 37- 1 , EF(Mn)
(
=
1 1 .0 eV. Substitute numerical values and evaluate E:
)(
E = 11.0eV+ 1. 38X10-23JI K 1 300K *78
-,- = In 9
__
)
(
l eV 9
1. 60xlO-1 J
)
l
ln 9 = l1.2 eV
I
A 2 -cm2 wafer of pure silicon is irradiated with light having a wavelength of 775 nm. The intensity of the light beam is 4.0 W/m2 and every photon that strikes the •••
sample is absorbed and creates an electron-hole pair. (a) How many electron-hole pairs are produced in one second? (b) If the number of electron-hole pairs in the sample is 6.25x1 0" in the steady state, at what rate do the electron-hole pairs recombine? (c) If every recombination event results in the radiation of one photon, at what rate is energy radiated by the sample?
406
Chapter
38
Picture the Problem The rate of production of electron-hole pairs is the ratio of the
incident energy to the energy required to produce an electron-hole pair.
(a)The number of electron-hole pairs N produced in one second is:
N=
fA
he A
= fAA he
Substitute numerical values and evaluate N:
(b) In the steady state, the rate of recombination equals the rate of generation. Therefore: (c) The power radiated equals the power absorbed: Substitute numerical values and evaluate Prod:
Prad = 4( .0 W/m2 )(2 x 1 0-4 m2 ) = I 0.800mJ/s I
Chapter 39 Relativity Conceptual Problems *1
•
The approximate total energy of a particle of mass m moving at speed u « c
2 is (a) me2 +1mu2 . (b) 1mu • (c) emu. (d) 1me2 . (e) 1emu.
Picture the Problem The total relativistic energy E of a particle is defined to be the sum
of its kinetic and rest energies. The total relativistic energy of a particle is given by:
*2
•
and
I (a) is corr ect. I
A set of twins work in an office building. One twin works on the top floor
and the other twin works in the basement. Considering general relativity, which one will age more quickly? (a)
They will age at the same rate. (b) The twin who works on the
top floor will age more quickly. (c) The twin who works in the basement will age more quickly. (d) It depends on the speed of the office building. (e) None of these is correct. Determine the Concept The gravitational field of the earth is slightly greater in the
basement of the o ffice building than it is at the top floor. Because clocks run more slowly in regions of low gravitational potential, clocks in the basement will run more slowly than clocks on the top floor. Hence, the twin who works on the top floor will age more quickly.
I (b) is corr ect. I
Estimation and Approximation *7
The most distant galaxies which can be seen by the Hubble telescope are moving away from us with a redshift parameter of about z = 5 . (See Problem 30 for a definition of z) . (a) What is the relative velocity of these galaxies with respect to us (expressed as a fraction of the speed of light)? (b) Hubble's law states that the recession velocity is given by the expression v = Hx, where v is the velocity of recession, x is the distance, and H is the Hubble constant, H = 75 km/s/Mpc. ( 1 pc 3 .26 light-year). Estimate the distance of such a galaxy using the information given. ••
=
Picture the Problem We can use the result from Problem 30, for light that is Doppler
shifted with respect to an observer,
( : - 11)
v=c u
u- +
, where u = z + 1 and z is the red-shift
parameter, to find the ratio of v to c. In (b) we can solve Hubble's law for x and substitute our result from (a) to estimate the distance to the galaxy.
407
408 Chapter 39 (a) Use the result of Problem 30 to
express vic as a function o f z :
Substitute for z and evaluate vic:
� _ (z + lY - 1 (z + lY + l
c
: - 1 1 0.946 1 (5 + 1) + 1
� = (5 + 1)
c (b) Solve Hubble' s law for x:
Substitute numerical values and evaluate x:
=
v
X=H x
=
0.946c
H
=
(
0. 946 3 x 10 5
kmI s
kmI s )
75 --
Mp c
=
3.78 x 10 3 Mp c x
=
1 1 2 . 3 Gc . y 1
3.26 x 10 6 c · y
---- Mp c
Time Dilation and Length Contraction *10 ·· Unobtainium (Un) is an unstable particle that decays into Nonnalium (Nr) and Standardium (St) particles. (a) An accelerator produces a beam of Un which travel s t o a detector located 1 000 m away from the accelerator. The particles travel with a velocity of v = 0.866c. How long do the particles take (in the laboratory frame) to get to the detector? (b) By the time the particles get to the detector, half of them have decayed. What is the half-life of Un? JNote: Half-life as it would be measured in a frame moving with the particles). (c) A new detector is going to be used, which is located 10,000 m away from the accelerator. How fast should the particles be moving if half of them are to make it to the new detector?
Picture the Problem The time required for the particles to reach the detector, as measured in the laboratory frame of reference is the ratio of the distance they must travel to their speed. The half life of the particles is the trip time as measured in a frame traveling with the particles. We can find the speed at which the particles must move if they are to reach the more distant detector by equating their half life to the ratio of the distance to the detector in the particle' s frame of reference to their speed.
(a) The time required to reach the
detector is the ratio of the distance to the detector and the speed with which the particles are traveling:
Substitute numerical values and
evaluate /1{:
6t = & = v
&
0.866c
Relativity (b) The half life is the trip time as measured in a frame traveling with
6l
,
=
the particles: Substitute numerical values and evaluate 6.t' :
-;;
' 1- O.8 6C
( :
l1l' = 3.85,u s =
1 1. 93,u s 1
(c) In order for half the particles to reach the detector:
? Fffi) -
v
6t = 6l 1r
l1l ' = l1x ' rv
=
l1x '
)
Fffi v
where t:u' is the distance to the new detector.
Rewrite this expression to obtain:
fix '
v
Squaring both sides of the equation yields:
Substitute numerical values for t:u' and 6.t' and simplify to obtain:
_ _ = ( 104 2 V_
I-
m
'
m
93 s 1 ,u
J
2
= (17.3cY
Divide both sides of the equation by 2 c to obtain:
2 2
(17.3Y 1+ (17.3)-
Solve this equation for V /C :
Finally, solving for v yields:
v=
1
0.998c 1
?
=
0. 9967
409
410
Chapter
39
The Lorentz Transformation, Clock Synchronization, and Simultaneity *17
••
A spaceship of proper length Lp = 400 m moves past a transmitting station at
a speed ofO.76c. At the instant that the nose of the spaceship passes the transmitter, clocks at the transmitter and in the nose of the spaceship are synchronized to t = t'
=
O.
The instant that the tail o f the spaceship passes the transmitter a signal i s sent and subsequently detected by the receiver in the nose of the spaceship. (a) When, according to the clock in the spaceship, is the signal sent? (b) When, according to the clock at the transmitter, is the signal received by the spaceship? (c) When, according to the clock in the spaceship, is the signal received? (d) Where, according to an observer at the transmitter, is the nose of the spaceship when the signal is received? Picture the Problem Let S be the reference frame of the spaceship and S' be that of the
earth (transmitter station). Let event A be the emission of the light pulse and event B the
reception of the light pulse at the nose of the spaceship. In (a) and (c) we can use the
classical distance, rate, and time relationship and in (b) and (d) we can apply the inverse Lorentz transformations. (a) In both S and S' the pulse travels
at the speed c. Thus:
(c) The elapsed time, according to
L 400rn = 1 . 76,us 1 tA = -P = v 0.76c 1 tB
=
tpulse to travel length of ship
the clock on the ship is: Find the time of travel of the pulse to the nose of the ship:
tpulse to travel lcngth of ship
=
+ tA
400rn 2.998x1 08 rn1s
= 1 .33,us Substitute numerical values and evaluate tB : (b) The inverse time transformation is:
tB
= 1 .33 ,us + 1 .76,us = 1 3 . 09,us I
( ;)
tB ' = r t -
1 1 g = ---;= === = 1 .54 2 - (0.7 6c y -
where r
=
1- 2 c
1
c2
Relativity Substitute numerical values and evaluate t' B :
tB '
=
= =
(d) The inverse transfonnation for x
IS:
41 1
(1 .54 3. 0 9 JiS ( 0 . 7 6 (400m)
{
�
_
[
1
6) (400m) (1 . 5 4) 3.09 JiS - (- 0.7 3 x 08 m/s
1 6.3 2 JiS I
)
J
x ' = r (x - vt )
Substitute numerical values and evaluate x' :
*22
•••
An observer in frame S standing at the origin observes two flashes of colored
light separated spatially by &
=
2400 m. A blue flash occurs first, followed by a red flash
5 J..lS later. An observer in S' moving along the x axis at speed v relative to S also observes the flashes 5 f.1s apart and with a separation of 2400 m, but the red flash is observed first.
Find the magnitude and direction of v.
Picture the Problem We can use the inverse time dilation equation to derive an
expression for the elapsed time between the flashes in S' in tenns of the elapsed time
between the flashes in S, their separation in space, and the speed v with which S' is movmg.
From the inverse time transfonnation we have:
[
M' = r M -
;
&
]
where �t' is the time between the flashes in
S' and �t and & are the elapsed time
between the flashes and their separation in S.
Set �t'
=
-�t to obtain:
- �t v = l1t - - & 2 r
or - l1t
Square both sides of the equation:
c
1 -H c
2 2
v c2
= M --&
4 12 Chapter 39
Simplify to obtain:
Solve for v: v =
( 1L\�)2 L\t [ 1( J ]2 1 ( J 18 I I
1+
Substitute numerical values and evaluate v:
-
e
2
v =
+
2400m 5 f.1s
2400m
3 x 10' m ls
5 ps
= 2.697 x 0 m1s = 0.89ge Because v is positi ve, S' ism ovingin the positi ve x directi on . The Velocity Transformation
*24 ·· A spaceship, at rest in a certain reference frame S, is given a speed increment of O.50c (call this boost 1). Relative to its new rest frame, the spaceship is given a further 0.50c increment 1 0 seconds later (as measured in its new rest frame; call this boost 2). This process is continued indefinitely, at 1 0- second intervals, as measured in the rest frame of the ship. (Assume that the boost itself takes a very short time compared to 1 0 s .) (a) Using a spreadsheet program, calculate and graph the velocity of the spaceship in reference frame S as a function of the boost number for boost 1 to boost 1 0. (b) Graph the gamma factor the same way. (c) How many boosts does it take until the velocity of the ship in S is greater than 0.999c? (d) How far has the spaceship moved after 5 boosts, as measured in reference frame S? What is the average speed of the spaceship (between boost 1 and boost 5) as measured in S?
Picture the Problem We'll let the velocity (in S) of the spaceship after the ith boost be
and derive an expression for the ratio of v to c after the spaceship' s
Vi
( i + 1 )th boost a s a function o f N. W e can use the definition o f y, i n terms o f vic t o plot r as a function of N.
(a ) and (b) The velocity of the spaceship after the (i + 1 )th boost is given by relativistic velocity addition equation:
Vi+1
=
1
v. I
+
+ 0.5e 0.5e i
(
e2
)V
Relativity 4 1 3 Factor c from both the numerator and denominator to obtain:
v.
----'- + 0.5 C --== Vi+l Vi 1 + 0.5 c
1; is given by:
1
A spreadsheet program to calculate vic and r as functions of the number of boosts N is shown below. The formulas used to calculate the quantities in the columns are as follows: ContentlFormula 0 0 (B2+0.5)/( 1 +0.5 *B2) 1/(1 -B2/\2)/\0.5
Cell A3 B2 B3 C1
Algebraic Form N Vo Vi+ l r
,
B · A boost 1 vic 0 2 0.000 3 1 0.500 4 2 0.800 3 0.929 '.' " . 5 4 0.976 ;, 6 5 7 0.992 6 8 0.997 7 0 .999 9 :' 10 1 .000 8 ':' 1 1 9 1 .000 12 1 0 1 .000 '. A graph of vic as a functIOn of N IS shown below: . ;.
.. . ,.
rr c
gamma 1 .00 1.15 1 .67 2 .69 4.56 7.83 1 3 .52 23.39 40.5 1 70. 1 5 1 2 1 .50
1 .0 0.8 � �
0.6 0.4 0.2 0.0
a
2
4
N
6
8
10
4 14
Cha pter 39
A graph of y as a function of N is shown below: 1 20 1 00
'"
E E '"
0/)
80 60 40 20 0
(c)
0
2
4
N
6
8
10
E xa mi nati on of the sp read sheet or of the grap h of vic a s a fu ncti on of N i ndi cate s that, after 8b oost s , t he ve locity of t he sp a ce s hip i s greate rtha n 0.99g e .
(d) After 5 boosts, the spaceship has traveled a distance ill, measured in the earth frame of reference (S), given by:
!'!:,xH 2+ !'!:,x -+2 3 + !'!:,x 3-+4 + !'!:,x4-+S = (0. 5 e)(1 Os )r1-+ 2+ (0.8e)(1 Os )Y -+2 3 + (0.92ge)(1 Os)Y -+3 4 + (0.976e)(1 Os)Y4-+S + (0. 992e )(1 0S)YS-+ 6 = (0. 5 e)(1 Os )(1 1. 5 )+ (0. 8e)(1 Os )(1 .67) + (0.92ge)(1 Os )(2.69) + (0.976e)(1 0s) (4.5 6)+ (0. 992e)(1 Os )(7.83) = 1 1 66e .s I
!'!:,x =
The average speed of the spaceship, between boost 1 and boost 5 , as measured in S is given by:
!'!:,x vav = -
f..t
where !J.t is the travel time as measured in the earth frame of reference.
Express !J.t as the sum of the times the spaceship travels during each 1 0-s interval following a boost in its speed:
f..t = f..tl-+ 2+ f..t -+2 3 + f..t 3-+4 + f..t4-+S = ( 1 0S)YI-+ 2+ (1 0s)Y -+2 3 + (1 0S)Y 3-+4 + (1 0S)Y4-+S + (1 0S)YS-+ 6 = (l Os )( YH 2+ Y -+2 3 + Y 3-+4 + Y4-+S + YS-+ 6)
Substitute numerical values and evaluate !J.t: f..t =
(l Os )(1 1. 5 + 1 .67+ 2. 69+ 4.5 6+ 7.83) = 1 79s
Relativity 4 1 5 Substitute for t..,'( and /':,1 and evaluate Vav :
vav
=
1 66 c · s 1 79 s
= I O. 92 7c I
Remarks: This result seems to be reasonable. Relativistic time dilation implies that the spacecraft will be spending larger amounts of time at high speed (as seen in reference frame S). The Relativistic Doppler Shift *29
··
A clock i s placed in a satellite that orbits the earth with a period of
90 min. By what time interval will this clock differ from an identical clock on earth after 1 y? (Assume that special relativity applies and neglect general relativity.) Picture the Problem Due to its motion, the orbiting clock will run more slowly than the
earth-bound clock. We can use Kepler' s third law to find the radius of the satellite's orbit in terms of its period, the definition of speed to find the orbital speed of the satellite from
the radius of its orbit, and the time dilation equation to find the difference 0 in the
readings of the two clocks.
Express the time 0 lost by the clock:
(
!::"t = !::,.t 1 --I 0= !::"t - !::"t = !::,.t - P
r
Because v « c, we can use part (b)
r
J
of Problem 1 3 :
Substitute to obtain:
Express the square of the speed of the satellite in its orbit:
(1)
v'
=
e;r ) '
(2)
where T is its period and is the radius of
r
its (assumed) circular orbit.
Use Kepler' s third law to relate the period of the satellite to the radius of its orbit about the earth: Solve for r:
Substitute numerical values and evaluate r:
416
Chapter
39
4Jr2 (6.65 x 106mL (90min x 60s/min Y = 5.99 x 107 m 2 I S2
Substitute numerical values in equation (2) and evaluate v2 :
v2 =
Finally, substitute for v2 in equation ( 1 ) and evaluate 5:
(
)
1
5 .!. 5.99 x 107 m 2 1 S2 (1 y x 3 1 .56 Ms/ y ) = = 10 . 5ms 8 2 (3 x 10 m/sL
I
,
A particle moves with speed O.8c along the x" axis of frame S which moves with speed O.8c along the x ' axis relative to frame S '. Frame S ' moves with speed O.8c along the x "
axis relative to frame S. (a) Find the speed of the particle relative to frame S '. (b) Find
the speed of the particle relative to frame S. Picture the Problem We can apply the inverse velocity transformation equation to
express the speed of the particle relative to both frames of reference. (a) Express
u/ in terms of u/' :
u "+v ux ' = ---"x .....,.. vu " l + -xe2 _
where V of S ', relative to S', is O.8c. Substitute numerical values and evaluate u/ :
(b) Express
Ux in terms of u/ :
Substitute numerical values and evaluate
Ux:
u/ =
0 . 8e + 0.8e = 1 .6e = I 0.976c I (0.8e)2 1 .64 1+ e2
u '+v ux = ---"x,--_ vu ' l + --x e2 where v of S, relative to S ', is O.8c. 0.976e +0 . 8e = 1 .776e Ux 1 + (0.8e)(0 .976e) 1 .781 e2 = I O.997e I =
Relati vit y
41 7
Relativistic M omentum and Relativistic Energy *34
•
A proton (rest energy 938 MeV) has a total energy of 2200 MeV.
(a ) What is its speed? (b) What is its momentum? Picture the Problem We can use the relation for the total energy, momentum, and rest
energy to find the momentum of the proton and Equation 39-26 to relate the speed of the
proton to its energy and momentum. Relate the energy of the proton to its momentum: (b) Solve for p to obtain:
Substitute numerical values and evaluate p:
p= p=
�{2200 MeV Y - (938 MeV Y
1
= 1 .99 (a) From Equation 39-26 we have:
v
c
= pc
Solve for v to obtain: Substitute numerical values and evaluate v: *39
··
v=
;
GV
1
c
E
1 .99 GeV c = I O.905c I 2200 M eV
Two protons approach each other head-on at O.Sc relative to reference frame
S' . (a) Calculate the total kinetic energy of the two protons as seen in frame S' . (b)
Calculate the total kinetic energy of the protons as seen in reference frame S, which is moving with speed O.Sc relative to S' so that one of the protons is at rest.
Picture the Problem The total kinetic energy of the two protons in part (a) is the sum of
their kinetic energies and is given by K
= 2 {r - l)Eo . Part (b) differs from part (a) in
that we need to find the speed of the moving proton relative to frame S. (a) The total kinetic energy of the protons in frame S' is given by:
4 1 8 Chapte r 3 9 Substitute for r and Eo and evaluate
K:
K
1
=2 1-
(O.Scy c2
- 1 (93 8 . 2 8 MeV)
= 1 290 MeV I
(1)
(b) The kinetic energy of the moving proton in frame S is given by:
1
Express the speed u of the proton in frame S:
Substitute numerical values and evaluate
where
u:
Evaluate r
Substitute numerical values in equation (1) and evaluate K:
+ u = u 'vuv l + __2_x c
----'x', --_
.Sc + O. Sc = O. 00c 8 u = O(O .Sc)(O.Sc) + 1 c2 r
K
=
1
� 1-
(O . 8 c)(O 8 c) c2 .
= 1 . 67
= (1 .67 - 1)(93 8 . 2 8 MeV) = 1 62 9 MeV I
General Relativity *42 " Light traveling in the direction of increasing gravitational potential undergoes a frequency redshift. Calculate the shift in wavelength if a beam of light of wavelength A. = 63 2 . 8 nm is sent up a vertical shaft of height L 1 00 m. =
Picture the Problem Let m represent the mass equivalent of a photon. We can equate the change in the gravitational potential energy of a photon as it rises a distance L in the gravitational field to h;').j and then express the wavelength shift in terms of the frequency shift.
The speed of the photons in the light beam are related to their frequency and wavelength:
C = fA
=>
f = Ac
Relativity
41 9
e dl -2 = - ell = - 112 dll
Differentiate this expression with respect to A to obtain: Approximate dj1dA by f..j1f..1l and solve for f..J Divide both sides of this equation by
f..1 I
fto obtain:
e
Solve for f..1l :
(1) M = f.. U = mgL
The change in the energy of the photon as it rises a distance L in a gravitational field is given by: Because M = hf..f :
hf..1 = mgL
Letting m represent the mass
E = me2
Divide equation (2) by equation (3)
-- =
equivalent of the photon:
hf..1 hi
to obtain:
=
(2)
hi
mgL
--
me2
(3)
f..1 I
gL � - =e2
Substitute for f..j1fin equation ( 1 ):
(9.8 1 mJs2 )(1 00 m)(632. 8 nm) (3 x 1 08 mJs) = 1 - 6. 9 0x 10-1 2 I
Substitute numerical values and
f..1l =
evaluate f..1l :
_
Dm
General Problems *47
··
Frames S and S are moving relative to each other along the x and x' axes.
Observers in the two frames set their clocks to t
=
0 when the origins coincide. In frame
S, event 1 occurs at X I 1 .0 c·y and tl 1 y and event 2 occurs at X 2 2.0 c·y and t2 = 0.5 y. These events occur simultaneously in frame S . (a) Find the magnitude and =
=
=
direction of the velocity of S relative to S. (b) At what time do both these events occur as measured in S?
420
Chapter 3 9
Picture the Problem We can use Equation 39- 1 2, the inverse time transformation
equation, to relate the elapsed times and separations of the events in the two systems to the velocity of S' relative to S. We can use this same relationship in (b) to find the time at
which these events occur as measured in S' . (a) Use Equation 39-12 to obtain:
[(tz - tJ- ; (X2 J] = [ �t - :2 ]
�t' = t/ -t/
=
-x
&
r
Because the events occur
r
simultaneously in frame S' , /).t' = 0 and:
Solve for v to obtain:
Substitute for evaluate V: Because M
�t and /).x and
= t2
- tl
= -0.5 y :
e2 (0 . 5 y - l y) = 1 - 0.5e I V = 2.0e · y - 1 .0e · y
I
S' mo ves in
the negative xd ire ction .
I
(b) Use the inverse time transformation to obtain :
Substitute numerical values and evaluate tz' and tl ' :
*51
•••
t2 ' t ' = =
I
O
. 5 y (- 0.5e) (22.0e . y) e 1 - (- 0.52 e? e _
In a simple thought experiment, Einstein showed that there is mass
associated with electromagnetic radiation. Consider a box of length L and mass M resting
on a frictionless surface. At the left wall of the box is a light source that emits radiation of
energy E, which is absorbed at the right wall of the box. According to classical
electromagnetic theory, this radiation carries momentum of magnitude p
= Ele (Equation
30- 1 3). (a) Find the recoil velocity of the box so that momentum is conserved when the
light is emitted. (Since p is small and M is large, you may use classical mechanics.) (b)
Relativity 42 1 When the light is absorbed at the right wall of the box the box stops, so the total momentum remains zero. If we neglect the very small velocity of the box, the time it takes for the radiation to travel across the box is 111
Lie. Find the distance moved by the box in this time. (c) Show that if the center of mass of the system is to remain at the same place, the radiation must carry mass m = Ele2 • =
Picture the Problem We can use conservation of energy to express the recoil velocity of
the box and the relationship between distance, speed, and time to find the distance
traveled by the box in time 111 = Lie. Equating the initial and final locations of the center of mass will allow us to show that the radiation must carry mass m = Ele2 . (a) Apply conservation of momentum to obtain:
E -+MV = Pi = 0 e
Solve for v:
vL e
(b) The distance traveled by the box
d = v!::..t = -
Substitute for v from (a):
d=
in time 111 = Lie is:
(c) Let x = 0 be at the center of the box and let the mass of the photon be m. Then initially the center of
x
CM
( )I
L _� = LE e Me Me 2 �
= -M.l+mLm
I
----=2'----_
mass is at:
When the photon is absorbed at the other end of the box, the center of mass is at: Because no external forces act on the system, these expressions for
XCM must be equal: Solve for m to obtain:
-{mL M+m
[ -MeMEL2 + m("2 L EL )] 1
_
M+m
E
- ];k2
Chapter 3 9
422
Because and E
=
Mc2 is of the order of 1 0 16 J
hfis of the order of 1 J for
reasonable values off and: *55
EIMc2 «
1
When a projectile particle with kinetic energy greater than the threshold kinetic energy Kth strikes a stationary target particle, one or more particles may be created in the inelastic collision . Show that the threshold kinetic energy of the projectile is given by ...
Here Lmjn is the sum of the masses of the projectile and target particles, Lmfin is the sum of the masses of the final particles, and mtarget is the mass of the target particle. Use this
expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton-antiproton pair; compare your result with the result of Problem 40. Picture the Problem Let mj denote the mass of the incident (proj ectile) particle. Then Lmjn
=
j
m + mtarget and we can use this expression to determine the threshold kinetic
energy of protons incident on a stationary proton target for the production of a proton antiproton pair. Consider the situation in the center of mass reference frame. At threshold we have:
E 2 - p 2 2 = '"' � mfin c 2 C
Note that this is a relativistically invariant expreSSIOn.
In the laboratory frame, the target is at rest so: We can, therefore, write: For the incident particle:
E 2 - po2 2 E'20 .
1
I
C
=
I,
and
Ej = Ej O + Kth
where Kth is the threshold kinetic energy of the incident particle in the laboratory frame. Express Kth in terms of the rest energIes:
(Et,o + Ej•O ) 2 + 2Kth Et,0 = (I mfin c2 Y
where
Relativity 4 2 3 E1,0 + Ei. O =
and
L
n1fill c
2
Substitute to obtain: Solve for Kth to obtain:
mill +
For the creation of a proton - antiproton pair in a proton proton collision:
L mfin L mfin - L min
c2
2 m larget
L min = 2 mp L mfin = 4 mp
and
(
J(
J
2 mp + 4 m 4 mp - 2 m c2 Kth = --'------'-----'--'-----'----'--2 mp
Substitute to obtain:
�
(6rnp
��
)
rn, c' p
�
I 6rnpc' I
in agreement with Problem 40. *59
•••
For the special case of a particle moving with speed
u
along the y axis in
frame 5, show that its momentum and energy in frame S' , a frame that is moving along
the x axis with velocity v, are related to its momentum and energy in 5 by the transformation equations
( (
Px = r Px
,
E'
c
=r
VE
-
7
) )
E vPx
c
_
c
, Py
,
,
= Py ' pz = pz
.
Compare these equations with the Lorentz transformation for x ' , y' , Zl, and t' . These
equations show that the quantities px, Py, pz, and Elc transform in the same way as do x, y,
z,
and ct.
Picture the Problem We can use the expressions for
p and E in 5 together with the
relations we wish to verify and the inverse velocity transformation equations to establish
424
Chapter 3 9
the condition
U,2
=(
U/) 2 + (U/) 2 + (U/ ) 2
= �r2 v2 +
and then use this result to verify
the given expressions for px' , py' , p/ and Eric. In any inertial frame the momentum and energy are given by:
_
P=
g c2 mu
2
1 --
where Ii is the velocity of the particle and u
The components of
p in S are:
Because Ux = Uz = 0 and uy = u :
is its speed.
Px
= :-2 ' = -2 g 11c2 g c2
Px =
and
mu
pz =
m uy
Py
, and
0
mu
Substituting zeros for Px and pz in
the relations we are trying to show yields:
px'=r(o-::) = -r ::, p/ = pE/ = E E ' = r( c c )=r c g g 1-c2 1 - -c2 g 1 -c2 0 , and
_
In S' the momentum components are:
, Px =
P/ =
The inverse velocity transformations are:
mu x
0
,
U ,2
mu ' Z
PY '
U '2
' PY
,
=
mu y '
U, 2
, and
Relativity 425
u: = �
Uz
1-
Substitute obtain:
Ux Uz =
=
vu ,
c-,-
U = -v ' U y 1
0 and u) ' = u to
X
1
=
yu and
,
uz 1 = 0
U , 2 = (U/) 2 (U /) 2 + (u/Y u2 = v2 r2
Thus:
+
+
First we verify that
Next we verify that
pz' pz =
P; C
mu l -� �G1
= [£J Next we verify that
pz - m (O)1 2 = pz = fOl � u 1 c2
0:
, _
j
py' = py:
1 mu y I Py = 2 U 1 2
-
=
mu 2 v2 - u 1 r c 2 2c 2
r
g� C
V
( J ( J
u2 - v2 - v2 1-1 c2 c2 c2 = 2 PY � 1 - �2 - �2 1 - �2 1- 2 c c c c
(l -2 �)2(l - �J J -�( 1-� c2 c 2
p/ = r( Px - ::
� -::
1 mu U r 1 - 2 - u2 1 2 e2 e2 r 2-
)
:
426 Chapte r 3 9
Px - !n u x 2 U 1-C2
g I
I -
Y
�
-I 1 - u: y 2 c m cJ'V 2 - U2 c - 1 Uc 2 1 - c2 y 2c2 ' 2 - ;2 2 ;1-� 1 c2 C C J'V E 2c v 2 - �2 - v2 1 1-c2 c2 c2
zg
- mv v - U_2 1--C 2 Y 2 C2 ?
V
1 - :: J ( - J ( ( J ---,:: -;---- C 2 E v 2 u 2 v2 1 - 2 - 2 (1 - 2 ( J J = -;E 1 I vPx . E' = y( -E - _ E E' = yE : = y ) c c c c g U -I -' �1 - U 2' 1 y 2 2 2 yE v 2 c 2 E I = !n C I 2 = rm C 2 g ' � = U 1-u g - -2 - 1 - Uc 2 g 1 -1 2 2 c c y 2c2 C (1 - -Uc?-2 J(1 - �c2 J 1 - -vc22 - �c22 (1 - -cv22 J = yE I 2 2 v2 = yE v 2 2 v 2 u (1 - -J -�( --1 -� 1-2 c C2 c2 c2 c 2 1 c 2 J -
.
_
J'V
1
----,----
�
_ _
Fmally, we venfy that
,
or
Y
-
= 1 yE 1
The x, y, z, and t transformation equations are:
X ' = y {x - vt)
y' = y Z' = z
( ) = y ( x - � ct )
and
v t = Y t - 2x I
The x, y, are:
-
z,
and ct transformation equations
X'
y' = y z' = z
and
(
ct' = y ct -� x
)
, ( - -; -;;E) ,
The Px, PY' Pz, and £/c transfonnation equations are:
Px = r Px
v
Py = Py ' p PZ = z
and E' c
Note that the transfonnation equations for x, y, for Ph PY' pz, and £/c are identical. *62
000
Z,
=r
(
E c
Relativity 427
_'!.- Px c
)
and ct and the transfonnation equations
A long rod that is parallel to the x axis is in free fall with acceleration g
parallel to the -y axis. An observer in a rocket moving with speed v parallel to the x axis passes by and watches the rod falling. Using the Lorentz transfonnations, show that the observer will measure the rod to be bent into a parabolic shape. Is the parabola concave upward or concave downward? We can use the inverse Lorentz transfonnation for time to show that the observer will conclude that the rod is bent into a parabolic shape.
Picture the Problem
In frame S where the rod is not
moving along the x axis, the height of the rod at time t is:
The inverse Lorentz time transfonnation is: Express y' (t) in the moving frame of reference:
(1) B ecause equation (1)is th e equation of a parabola, w e'v e shown that th e moving obs erv er conclud e that th e rod is b ent into a parabolic shap e. B ecaus e th e co effici ent of x2 is n egativ e,th e parabola is concav e downward.
Evaluate y'(t) at t' = 0 to obtain:
will
Chapter 40 Nuclear Physics Conceptual Problems *4
The half-life of 14C is much less than the age of the universe, yet14C is found in nature. Why?
Determine the Concept14C
is found on earth because it is constantly being formed by cosmic rays in the upper atmosphere in the reaction 14N + n -7 14C + IH.
Write and balance equations for each of the following nuclear decays: *13 (a) beta decay of I�, (b) alpha decay of248Fm, (c) positron decay of 12N, (d) beta decay of81Se, (e) positron decay of6lCu, and if) alpha decay of228Th. Knowing the parent nucleus and one of the decay products, we can use the conservation of charge, the conservation of energy, and the conservation of the number of nucleons to identify the participants in the decay.
Determine the Concept
(a) beta decay of1�
214008Fm�244Cf+ 4He + Q I 98 2
(b) alpha decay of248Fm
(c) positron decay of 12N
I · + 013 + + Q 6291C U�61N I 28 1 + 1 0 22908Th�22884Ra+ 4He + Q .I 2
0 0813 S e�381B 5 r+_1 13 + 0 v � Q
(d) beta decay of81Se
4
(e) positron decay of61Cu if) alpha decay of 228Th
°
V
Write and balance reaction equations for each of the following: (a) 240pU undergoes spontaneous fission to form two fission fragments and three neutrons. One of the fission fragments is a 90Sr nucleus. (b) A 72Ge nucleus absorbs an alpha particle and ejects a photon. (c) A 1271 nucleus absorbs a deuteron and ejects a neutron. (d) A235U nucleus absorbs a slow neutron and fissions forming a113 Ag nucleus, two neutrons, and another fission fragment. (e) A 55Mn nucleus is struck with a high-energy 7Li nucleus, resulting in a triton, 3H, and a new nucleus. if)238U absorbs a slow neutron resulting in a compound nucleus that emits a beta particle. What is the resulting nucleus? *14
·
We can use the information regarding the daughter nuclei to write and balance equations for each of the reactions.
Determine the Concept
429
430
Chapter
40
Properties of Nuclei *17
238U.
40-1
·
Calculate the binding energy and the binding energy per nucleon from the masses given in Table for (a) 12C, (b) 56Fe, and (c) To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2 . To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus.
Picture the Problem
(a)For 12C, Z = 6 and N= 6. Add the mass of the neutrons to that of the protons:
6mp +6mn 6x1.007825u+6xl.008665u 12.098940u =
=
Subtract the mass of 12C from this result:
6mp +6m,J -m'2c 12.098940u -12u 0.098940u
(
=
=
Multiply the mass difference by c2 and convert to MeV:
Eb (�m )c2 (0.098940u)c2 x 931.5�ev/c2 I 92.2MeV I . . . Eb 92.2MeV I 7.68MeV I and the bmdmg energy per nucleon A 12 =
=
=
IS
=
=
(b)For56Fe, Z = 26 and N= 30. Add the mass of the neutrons to that of the
protons:
26mp +30mn 26x1.007825u+30x1.008665u 56.463400u =
Subtract the mass of 56Pe from this result:
=
Nuclear Physics 431 (26m p + 30m n ) -m" = 56.463400u -55.934942 u 0.528458u Multiply the mass difference by c2 and convert toMeV: (�m)c2 (0.528458u)c 2 931.51Meu V/ c2 = !. 492MeV ! and the binding energy per nucleon is = 49256MeV = 18.79Me · V I. (c)For238U, and Add the mass of the neutrons to that of the =
-c
E
b
=
=
x
.
E _b A
Z = 92
N= 146.
protons:
92mp +l46mn 92x1.007825u +146x1.008665u 239.984990 u =
=
Subtract the mass of238U from this result:
(92mp +l46mn)-mmu 239.984990u -238.050783u 1.934207u Multiply the mass difference by c2 and convert toMeV: (�m)c2 (1.934207u)c2 x 931.5�eV/ c2 11802MeV I 1802MeV I 7.57MeVI . and the binding energy per nucleon 238 · =
E b
=
=
=
=
.
1S
E _b A
=
=
e+
*21 ·· The neutron, when isolated from an atomic nucleus, decays into a proton, electron, and an antineutrino as follows: �n�:H+_� gv. The thermal energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In both joules and electron volts, calculate the energy of a thermal neutron at 25°C. (b) What is the speed of this thermal neutron? A beam of monoenergetic thermal neutrons is produced at 25°C with intensity 1. After traveling 1350 lan, the beam has an intensity of I12. Using this information, estimate the half-life of the neutron. Express your answer in minutes.
(c)
Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from 1 to 112, is the half life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed.
432 Chapter 40 (a) The thermal energy of the
neutron is:
Ethen11"1 = = (1.38 x 10-23 11K)(25 273)K 1 4.11x10-21J I l eV 4.1lxlO-21Jx 1.60xlO-1 9J = 1 25.7m eV I Ethermal = 2 2Ethermal kT
+
=
=
Ethermal
(b) Equate and the kinetic energy of the neutron to obtain: Solve for v to obtain:
I
2: mn V
v=
Substitute numerical values and evaluate v: (c) Relate the half-life, t1/ ' to the 2 speed of the neutrons in the beam: Substitute numerical values and evaluate tl/ : 2
2 4.11xlO-2I J 7 1.67 x10-2 I 2.22 s I
v=
kg
=
kml
1350 = 608s x Imin /12 2.22 60s s 1 10.1min I km km1
=
t
=
*24 ·· In 1920, twelve years before the discovery of the neutron, Rutherford argued that proton-electron pairs might exist in the confines of the nucleus in order to explain the mass number, A, being greater than the nuclear charge, Z. He also used this argument to account for the source of beta particles in radioactive decay. Rutherford's scattering experiments in 1910 showed that the nucleus had a diameter of about 10 fm. Using this nuclear diameter, the uncertainty principle, and given that beta particles have an energy range of 0.02MeV to 3 MeV, show why electrons cannot be contained within the nucleus.
.40
Picture the Problem
1 2
TheHeisenberg uncertainty principle relates the uncertainty in
position, Llx, to the uncertainty in momentum, I1p, by Solve theHeisenberg equation for I1p:
I1.P>::;j
Substitute numerical values and evaluate I1p:
11.
� � .p =
&l1.p;:::
- ti.
ti 2&
1.05xlO-34J·s 2�OxlO-15m) 5.25 x10-21 . m/s kg
Nuclear Physics 433 The kinetic energy of the electron is given by: Substitute numerical values and evaluate K:
K =pc
5.25x l O 1 kg .m/s )(3xl08rn /s ) leV 1.58xlO-12Jx 1.60xlO-1 9 9.88MeV
K= = =
1 1000
(
-
2
J
This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of to eV. Radioactivity *31 ·· Plutonium is a highly hazardous and toxic material to the human body. Once it enters the body it collects primarily in the bones, although it can also be found in other organs. Red blood cells are synthesized within the marrow of the bones. The isotope 239pU is an alpha emitter with a half-life of years. Since alpha particles are an ionizing radiation, the blood-making ability of the marrow is, in time, destroyed by the presence oe39pu. In addition, many kinds of cancers will also be initiated in the surrounding tissues by the ionizing effects of the alpha particles. (a) If a person accidentally ingested Jig of239pU and it is absorbed by the bones of the victim, how many alpha particles are produced per second within the skeleton? (b) When, in years, will the activity be alpha particles per second?
24,360
2.0 1000
A,
Picture the Problem Each 239pU nucleus emits an alpha particle whose activity, depends on the decay constant of239pU and on the number N of nuclei present in the ingested239PU. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of239pu. Finally, we can use the dependence of the activity on time to find the time at which the activity be alpha particles per second.
(a) The activity of the nuclei present in the ingested239pU is given by: Find the constant for the decay of 239pU:
A=AN
A
_
-
=
Express the number of nuclei present in the quantity of 239pU ingested: Substitute numerical values and evaluate N:
1000 (1)
1n 2 - 0.693 (24360y)(31.56Ms/y) tl/ 2 9.02xlO-13s-1 _
N= mPu
_ N
A_
M Pu
where Mpu is the atomic mass of 239PU. N= =
u lei/mol ] (2.0 j.Lg) ( 6.02xl239023gnlmol 5.04xlOl5nuclei c
434 Chapter 40 Substitute numerical values in
A
equa ti on(I) and evaluate A:
= =
(9.02xlO-13s-I)(5.04xlOI5a) 14.55xl03 a/s I
(b) The activity varies with time
according to:
Solve for t to obtain:
Substitute numerical values and evaluate t:
The fissile material 239pU is an alpha emitter. Write the equation of this reaction. Given that 239pU, 235U, and an alpha particle have respective masses of 239.052 156 u, 235.043 923 u, and 4.002 603 u, use the equations appearing inProblem 32 to calculate the kinetic energies of the alpha particle and the recoiling daughter nucleus. *33
··
A
We can write the equation of the decay process by using the fact that the post-decay sum of the Z and numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations fromProblem 32 is given by Q = -(f1m)c2• Picture the Problem
239pU undergoes alpha decay
according to:
The Q value for the decay is given by: Substitute numerical values and evaluate Q:
Q [ (239.052156u) -(23S.043923u + 4.002603U)](93 1.::eV ) IS.24MeV I FromProblem the kinetic energy ( A�4 )Q of the alpha particle is given by: Substitute numerical values and =( 2��4 )S.24MeV) evaluate = I S.lSMeV I =
=
32,
Ka:
K K
a
a
=
From Problem 32, the kinetic energy of the J39U IS ' given b y:
-
.
K
U
=
Nuclear Physics 435
4Q A
Substitute numerical values and evaluate Ku: *36 ·· Radiation has long been used in medical therapy to control the development and growth of cancer cells. Cobalt-60, a gamma emitter of 1.17 and 1.33MeV energies, is used to irradiate and destroy deep-seated cancers. Small needles made of 60Co of a specified activity are encased in gold and used as body implants in tumors for time periods that are related to tumor size, tumor cell reproductive rate, and the activity of the needle. (a) A 1.00 f.1.g sample of 60 Co, of half-life of 5.27 y, is prepared in the cyclotron of a medical center to irradiate a small internal tumor with gamma rays. In curies, determine the initial activity of the sample. (b) What is the activity of the sample after 1.75 years?
R = Roe-A.t
Picture the Problem
We can use
Ro =
AN to find the initial activity of the sample and
to find the activity of the sample after 1.75 y.
(a) The initial activity of the sample is the product of the decay constant 60 It for Co and the number of atoms N of 60 Co initially present in the sample: Express N in terms of the mass m of the sample, the molar mass M of 60Co, and Avogadro's number NA:
R
-AN ° -
N
=.!!!..M
(1)
NA
Substitute numerical values and evaluate N: N
=(1.6000Xg/10mol-6g ](6.02x1023nuclei/mol) 1.00x1016 nuclei =
The decay constant is given by:
Substitute numerical values and evaluate It:
Ao:
Substitute numerical values in equation (1) and evaluate
0.693 (5.27 y )(31.56Ms /y ) =4.l7x10-9S-1 Ro =(4.17x10-9 )(1.00x1016nuclei ) 1Ci =4.l7x107s-IX 3.7x1010s -1 =! 1.13mCi !
A
=
S
-I
436
Ca e
h pt r 40
(b) The activity varies with time according t o: Evaluate R at
I
=
R
=
1.75 y:
R
= =
*41
··
Roe-A1
=
( a.69JI)
. R0e 5 27y
mCi) e I0.898mCi I (1.13
( a.693xI.75Y ) 5.27y
The counting rate from a radioactive source is measured every minute. The
resulting counts per second are 1000, 8 20, 673, 552, 453, 371, 305, and 250. Plot the
counting rate versus time on semilogarithmic graph paper, and use your graph to find the half-life of the source. Picture the Problem
The following graph was plotted using a spreadsheet program.
Excel's "Add Trendline" feature was used to determine the equation of the line.
7.0 6.8 6.6 6.4 6.2
� ..=
6.0 5.8 5.6 5.4 5.2 5.0
0
2
3
5
4 t
6
7
(min)
The linearity and negative slope of this graph tells us that it represents an exponential decay.
The decay rate equation is:
R - Rae-AI
Take the natural logarithm of both
lnR
sides of the equation to obtain: This equation is of the form:
= =
lne-AI + InRa -At+lnRa
y = mx+b where y = In R,
x
=
I, m
= -A, and
Nuclear Physics 437 b
The half-life of the radioisotope is:
= InRo.
tl/2 =
In 2 }L,
-
=
In 2 I 3.50mm I 0.198mm .
_I
=
.
Nuclear Reactions *47
for
I� N
••
(a) Use the atomic masses
m = 14.003 242 u for I:C and m = 14.003 074 u
to calculate the Q value(inMeV) for the fJ decay
14C�14N +fJ- +v 6
7
e
(b) Explain why you do not need to add the mass of the fJ- to that of atomic N for this calculation. Picture the Problem
)
We can use Q = -(L'1m c2 to find the Q values for this reaction.
(a) The masses of the atoms are:
14.003242u 14.003074u
m '4C = m'4N =
Calculate the increase in mass:
14.003074u -14.003242u -0.000168u -(L'1m) -(-0.000168 )C' (931.5Me�1 c' J I 0.156MeV I
= =
Calculate the Q value:
Q= = =
c2
U
The masses given are foratoms,not nuclei,so fornuclear masses the masses are too large bythe atomic numbertimes the mass of an electron. For the given nuclearreaction, the mass of the carbon atom is too large by 6me and the mass of the nitrogen atom is too large by 7me. Subtracting 6me from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron)is mathematicallyequivalent to explicitlysubtracting 1me from the right side of the equation.
(b)
438 Chapter 40 Fission and Fusion *49
Assuming an average energy of 200MeV per fission, calculate the number of
•
fissions per second needed for a 500-MW reactor. Picture the Problem
The power output of the reactor is the product of the number of
fissions per second and energy liberated per fission. Express the required number N of
N=
fissions per second in terms of the power output P and the energy
released per fission Eper fission:
P ---
Eper fission
500MW N= 200 1 5x108 .:!.x 1.60xl019J 200 =1 1.56x1019 S-1
Substitute numerical values and evaluate N:
MeV
eV
s
MeV
1 9
*51
··
139L +2on1 + Q
Consider the following fission process:
2 + on --*425 M0+ 9235U
57
a
. The masses 0f the neutron, U,M0, and La are
1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u, respectively. Calculate the Q-value, inMeV, for this fission process. Compare the result to the result obtained in Problem 23. Picture the Problem
We can use Q
value. The Q value is given by:
=
-(t::.m)c2, where /)"m
Q= (
)
= mf - mi,
- �m c2
x
to calculate the Q
931.5 lu
MeV/c2
Calculate the change in mass /)"m:
=94.905842 u + 138.906348 u + 2(1.008665 u)-(235.043923 u 1.008665 u) =-0.223068 u Q =-(-0.223068 u)x 931.lu5 =1208 1
� m = mf - mi
+
Substitute for /)"m and evaluate Q:
MeV
MeV
Nuclear Physics
The ratio of Q to U found in Problem 23 is: *54
•••
Q U
=
208 MeV 1 88 236MeV =
.1%
439
1
The fusion reaction between 2H and 3H is
Using the conservation of momentum and the given Q value, find the final energies of both the 4He nucleus and the neutron, assuming the initial kinetic energy of the system is 1.00MeV and the initial momentum of the system is zero. We can use the conservation of momentum and the given Q value to find the final energies of both the 4He nucleus and the neutron, assuming that the Picture the Problem
initial momentum of the system is zero. Apply conservation of energy to obtain:
18.6MeV
tmHeV�e +tmnv� = KHe +Kn =
(1) (2 )
Apply conservation of momentum to obtain: Solve equation (2) for VHe:
Substitute for
v�e in equation (1):
Solve for Kn:
Kn
=
18.6MeV mn 1 +-mHe
Substitute numerical values for mn and mHe and evaluate Kn:
Kn
=
18.6MeV I 14.86MeV I 1 + 1.008665 4.002603u u
=
440
Chapter 40 KHe
Use equation (I) to find KHe:
=
18.6MeV - Kn
= 18.6MeV -14.86MeV =
1 3.74MeV I
General Problems *57
·
The counting rate from a radioactive source is 6400 counts/so The half-life of
the source is lO s.Make a plot of the counting rate as a function of time for times up to
1 min. What is the decay constant for this source? Picture the Problem
We can use the given information regarding the half-life of the
source to find its decay constant. We can then plot a graph of the counting rate as a function of time. The decay constant is related to the
A
half-life of the source:
=
In 2 In 2 I 0.0693s -1 I tl/2 lOs =
=
The activity of the source is given by: The following graph of R program.
=
6400Bq)e-0.0( 6 93S-I)t
(
was plotted using a spreadsheet
7000 6000 5000 4000 E9.. <>:: 3000 0'
;r..;
2000 1000 0
*61
··
0
10
20
30
t (5)
40
50
60
Show that the10 9 Ag nucleus is stable against alpha decay,
1��Ag��He+I��Rh + Q. The mass of the10 9 Ag nucleus is 108.904 756 u, and the
products of the decay are 4.002 603 u and 104.905 250 u, respectively.
Nuclear Physics 441 Picture the Problem We can show that 109Ag is stable against alpha decay by demonstrating that its Q value is negative.
The Q value for this reaction is:
Q=
-[(mRh +mJ -mAJ c2 (93l.5 M_ euV_ _Ic_2 J o
Substitute numerical values and evaluate Q:
Q= =
-[ (4.002603 u +104.905250 u) -108.904756 u](931.5MeV /u) 1 -2.88 MeV I
Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus,
Q
=
Ka
+
K D Kinetic energy cannot be negative; hence, alpha decay cannot occur •
unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, mp > ma + mD • Alpha decay cannot take place unless the total rest mass decreases.
*66
··
(a) Determine the closest distance of approach of an 8-MeV
a
particle in a
lO head-on collision with a nucleus of 197Au and a nucleus of B, neglecting the recoil of the
struck nuclei. (b) Repeat the calculation taking into account the recoil of the struck
nuclei. Picture the Problem
We can solve this problem in the center of mass reference frame for
the general case of an
a
particle in a head-on collision with a nucleus of atomic mass M u
lO and then substitute data for a nucleus of 197Au and a nucleus of B. In the eM frame, the kinetic energy IS:
KeM
m 1+� _ kqlq2 - k(2e)(Ze) - ke22Z 2 ke (1.44MeV ·fm)(2Z) =
Klab
M
At the point of closest approach:
KeM
R min
or, because
KeM Solve for Rmin to obtain:
_
_
-
=
R Olin
=
1.44MeV·fm,
. Rmm
. (1.44MeV .fm)(2Z)
R rrun
=
R min
KeM
(1)
442 Chapter 40 (a) Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab. Evaluate equation (1) for 197Au: lO Evaluate equation (1) for B:
(b) Find KCM for the 197Au nucleus:
Substitute numerical values in equation (1) and evaluate Rmin:
(1.44MeV .fm)(2x 79) 8MeV =I 28.4fm I .fm)(2x 5) = (1.44MeV 8MeV =11.80fm I 8MeV =7.841MeV KCM = 4u 1+-197u .fm)(2x79) = (1.44MeV 7.84lMeV =129.0fm I .
R 01111
=
.
R mm
Rnun .
Note that this result is about 2% greater that Rmin calculated ignoring recoil. lO Find KCM for the B nucleus:
Substitute numerical values in equation (1) and evaluate Rmin :
KCM
.
R mill
= 8MeV 4u =5.7l4MeV 1+lOu .fm)(2x5) = (1.44MeV 5.714MeV =12.52fm I
Note that this result is about 40% greater that Rmin calculated ignoring recoil. The total energy consumed in the United States in 1 y is about 7.Ox 1019 J. How many kilograms of23SU would be needed to provide this amount of energy if we *70
··
assume that 200MeV of energy is released by each fissioning uranium nucleus, that all of the uranium atoms undergo fission, and that all of the energy-conversion mechanisms used are 100 percent efficient?
Picture the Problem
The mass of235U required is given by m235
=
N NA
M235, where
M23S is the molecular mass of235U and N is the number of fissions required to produce 7.0xl019 J.
Nuclear Physics 443 Relate the mass of 235U required to the number of fissions N required:
N 1n 35 = N M 35 2 2 A
(1)
where M235 is the molecular mass of mU. Determine N:
N=
Substitute numerical values and evaluate N:
N= =
Eonnua! Eperfission
7.0x1019J 19J 200MeVx 1.60x10eV 2.18x1030
Substitute numerical values in equation(1) and evaluate mm:
1n235 *73
•••
=
2.1823x1030 . (235 glmol)=I 8.51xl05 kg I 6.02x10 nucleI/mol
Assume that a neutron decays into a proton plus an electron without the
emission of a neutrino. The energy shared by the proton and electron is then O.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the
proton must be equal and opposite the momentum of the electron. This determines the
relative energies of the two particles, but because the electron is relativistic, the exact calculation of these relative energies is somewhat difficult. (a) Assume that the kinetic energy of the electron is O.782MeV and calculate the momentum p of the electron in units of MeVic. (Hin t: Use Equation 39-28.) (b) Using your result from Part (a) , calculate the kinetic energy p2/2mp of the proton. (c) Since the total energy of the electron plus the proton is 0.782 MeV, the calculation in Part (b) gives a correction to the assumption that the energy of the electron is 0.782 MeV. What percentage of 0.782MeV is this correction? Picture the Problem
E 2 = p2C2
+ E�
The momentum of the electron is related to its total energy through
and its total relativistic energy E is the sum of its kinetic and rest
energies.
(a) Relate the total energy of the
(1)
electron to its momentum and rest energy: The total relativistic energy E of the electron is the sum of its kinetic
E=K + Eo
444 Chapte 40 r
energy and its rest energy:
- c-
(K + Eo )? = p 2? + Eo2
Substitute for E in equation (1) to obtain: Solve for p:
Substitute numerical values and evaluate
p
=
p:
�(0.782MeV)(0.782MeV 2x 0.511MeV) = !1.188MeV/ c ! c
(b)Because Pp
+
=
-Pe:
Kp
Substitute numerical values (see Table
7-1
Kp
for the rest energy of a
proton) and evaluate Kp:
Kp
(c) The percent correction is:
K
*77
•••
2
2m = 1.188MeV/ cY = ! 752eV I 2938.28MeV/ c2 . . =� p
=
752eV =! 0.0962% I 0.782MeV
Frequently, the daughter of a radioactive parent is itself radioactive. Suppose
the parent, designated by A, has a decay constant AA; while the daughter, designatedB, has a decay constant
AB.
differential equation
dNBldt
= AANA
The number of nuclei ofB are then given by the solution to the
- AaNB
(a) Justify this differential equation.
(b) Show that the solution for this equation is
where NAO is the number of A nuclei present at t (c) Show that NBCt)
>0
a function of time when
whether
TB
AA > AB
= 3TA.
or AB
Picture the Problem We can differentiate
=
0
when there are noB nuclei.
> AA.
(d) Make a plot of NA(t) and NB(t) as
( )=
NA AA O -AA
NB t
As
to t to show that it is the solution to the differential equation
(e-A.At - e-A.st) with respect
Nuclear Physics 445
(a) The rate of change of
NB is the rate of generation ofB nuclei minus the rate of decay
ofB nuclei. The generation rate is equal to the decay rate of
AANA. The decay rate ofB nuclei is AsHB.
A
nuclei, which equals
dNB =AA NA _:l B N dt
(b) We're given that:
(1)
/"8
(3)
Differentiate equation
(2)
with respect to
Substitute this derivative in equation
Multiply both sides by
As - AA AsAA
(1)
t to obtain:
to get:
and simplify to obtain:
= NAO [- AAe-AA1 Ase-As1] As +
which is an identity and confirms that equation
is the solution to equation
(1).
AA As the denominator and the expression in the parentheses are both negative for t O. If AA As the denominator and the expression in the parentheses are both positive for t O. If
(c)
(2)
>
>
<
>
446 Chapter 40 (d)
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20
Chapter 41 Elementary Particles and the Beginning of the Universe Conceptual Problems *3
•
How can you tell whether a decay proceeds via the strong interaction or the
weak interaction? Determine the Concept A decay process involving the strong interaction has a very short 23 lifetime (�10- s), whereas decay processes that proceed via the weak interaction have 10 lifetimes of order 10- s. •
*9
7t, interacts with an antiproton, 15,
Based on the assumption that a pion,
is it
possible that a proton, p, could be produced by such an interaction?
Determine the Concept No. Such a reaction is impossible.
A proton requires three
quarks. Three quarks are not available because a pion is made of a quark and an anti quark and the antiproton consists of three antiquarks.
Spin and Antiparticles *12 Y
·
+ y.
y-ray.
Two pions at rest annihilate according to the reaction
1l +
1[
-
--+
(a) Why must the energies of the two y-rays be equal? (b) Find the energy of each (c) Find the wavelength of each y-ray.
Picture the Problem We can use both conservation of energy and momentum to explain
why the energies of the two y-rays must be equal. We can find the energy of each y-ray in Table 41-1 and find their wavelengths using It
=
he!E.
The initial momentum is zero; therefore, the final momentum must be
(a)
zero. The momentum of a photon is E/c. To conserve both momentum and energy the two photons must have the same momentum ma gnitude. Hence, they have the same energy.
(b) From Table 41-1:
(c) The wavelength of each y
Ey ray is
/L
given by:
Substitute numerical values and
/L
evaluate A:
447
=
=
=
I 139.6MeV I
he
E
=
1240MeV· fin E
fin 1 8.88fin I 139.6MeV
1240MeV
.
=
448
Ch apte r 41
The Conservation Laws *20
··
Test the following decays for violation of the conservation of energy, electric
charge, baryon number, and lepton number: (a) n � 7[+ +7[- +p+ +p-;(b) 7[0 � e+ + e - +y. Assume that linear momentum and angular momentum are conserved. State which conservation laws(if any) are violated in each decay. Picture the Problem
A decay process is allowed if energy, charge, baryon number, and
lepton number are conserved.
(a) Energy conservation:
Because mn
>
2m" + 2mJi' energy
conservation is not violated. Charge conservation:
Because the total charge is 0 before
0�+1 +(-1)+ 0 + 0 =0
and after the decay, charge is conserved.
Baryon number:
Because baryon number changes
1-+ 0 + 0 + 0 +0=0
from + 1 to 0, conservation of baryon number is violated.
Lepton number: 0� 0 +0+ 1 +(-1) = 0
Because LJi = 0 before and after the decay, the lepton number for muons is conserved.
The process is not allowed because it violates conservation of baryon number.
(b) Energy conservation:
Because m"
>
2me, energy
conservation is not violated. Charge conservation:
Because the total charge is 0 before
0�+1 +(-1) + 0 + 0 = 0
and after the decay, charge is conserved.
Baryon number: O�O+O+O+O=O
Because B = 0 before and after the decay, the baryon number is conserved.
Elementary Part icles and the Beginning of the Un iverse Because Le
Lepton number: O�O+O+O+O=O
=
449
0 before and after the
decay, the lepton number is conserved.
The decay satisfies all conservation laws and is allowed.
Quarks *27
••
and (c) �.
Find a possible quark combination for the following particles: (a) A0, (b) p -,
Picture the Problem
Because A 0, P -, and � are baryons, they are made up of three
quarks. We can use Tables 41- 1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles. (a) For A0 we need:
Q=O B= +l
s= -1 The quark combination that satisfies these conditions is
!
uds.
!
uud.
!
dds.
!
=
(b) For p -we need:
Q -l B=-l S= +l
The quark combination that satisfies these conditions is (c) For � we need:
!
Q=-1 B=+l S=-l
The quark combination that satisfies these conditions is
!
The Evolution of the Universe *31 · A galaxy is receding from the earth at 2.5 percent the speed of light. Estimate the distance from the earth to this galaxy. Picture the Problem
We can use Hubble's law to find the distance from the earth to this
galaxy. The recessional velocity of galaxy is
v = Hr
450
Chapter 41
related to its distance by Hubble's law:
r=-Hv l05kmIs) .025)(3x 0 ( r (230.025)c kmIs 23km/s 106c·y 106c·y I 3.26x108 c·y I
Solve for r:
Substitute numerical values and evaluate r:
=
=
=
*34
The red line in the spectrum of atomic hydrogen is frequently referred to as the Ha line, and it has a wavelength of 656.3 nm. Using Hubble's law and the relativistic Doppler equation from Problem 33, determine the wavelength of the Ha line in the spectrum emitted from galaxies at distances of(a) 5 10 c·y, (b) 50 X 10 coy, (c) 500 c·y, and (d) 5x109 c·y from the earth. 00
Hrl c = 1+ l-Hrl c
Picture the Problem
as
11 /L
x6
x106
6
Using Hubble's law, we can rewrite the equation from Problem 3 1
1 /Lo
From Problem 33 we have:
A' =Ao
v =Hr
Use Hubble's law to relate v to r: Substitute for v to obtain:
A' =Ao
l+vlc I -vic l+Hrl c l-Hrl c
(a) For r= 5x106 c·y:
A' =
656.3nm
=
(b) For r = 50 x 106 c·y:
A' =
!656.6nm I
3km1S J 50xl06c·y l+e106c· 105km/s !658.8nm ! 656.3nm ( 23kmls 50xl06c·y 11 - l06e·y J 3xlO5km1s Y
3X
=
Elementary Particles and the Beginning of the Universe 451
(c) For r= 500 xl06 c·y:
A'
=
656.3 nm
(d) For r = 5x 109 c-y:
A'
=
656.3nm -�----'-----f--'.;r--"!-
=
I 983.0 nm I
General Problems *37
••
In Problem 36, one of the reactions is nO � y + y . (a) In terms of the quark
model, show how this reaction can take place. (b) Why is it that the number of photons produced must be at least two? The nO particle is composed of two quarks, uTI. Hence, the reaction n° � y + y is equivalent to uTI � y + y . Picture the Problem
(a) (b)
I The u and u annihilate resulting in the photons. I Two or more photons are required to conserve linear momentum.
*39
Using Figure 41-2 and the laws of conservation of charge number, baryon number, strangeness, and spin, identify the unknown particle in each of the following strong reactions: (a) p + n- � 2:0 + ?, (b) P + P � n+ + n + K+ + ?, and (c) p+ K-
� �.!::.
+ ?. .
We can systematically determine Q, B, S, and s for each reaction and then use these values to identify the unlmown particles.
Picture the Problem
(a) For the strong reaction: Charge number:
+ 1 - 1=0 + Q=>Q=0
Baryon number:
+ 1 + 0 = + 1 +B=>B=O
Strangeness:
0 + 0=-1 +S=>S=+l
452
Chapter 41
Spin:
+.1+O=+.1+s =>S=O
2
2
These properties indicate that the particle is the kaon, K ° . (b) For the strong reaction: Charge number:
+1 + 1 =+1 +0+ 1 +Q=>Q=O
Baryon number:
+1 + 1 = 0 + 1 +O+B=>B=+l
Strangeness:
0 + 0 =0+0+ 1 +S=>S=-l
Spin: These properties indicate that the particle is either the LO or the AD baryon. (c) For the strong reaction: Charge number:
+l-l =-l +Q=>Q=+l
Baryon number:
+1 +0 =+1 +B=>B=O
Strangeness:
0 - 1 =-2+S=>S=-1
Spin:
+.12 + 0 = +.12 + S => S = 0 These properties indicate that the particle is the kaon, K + .
*43
•••
A ,,£0 particle at rest decays into a A ° plus a photon. (a) What is the total
energy ofthe decay products?(b) Assuming that the kinetic energy of the A ° is negligible
compared with the energy of the photon, calculate the approximate momentum of the photon. (c) Use your result from Part(b) to calculate the kinetic energy of the A 0. (d) Use
your result from Part(c) to obtain a better estimate of the momentum and the energy of the photon.
Picture the Problem
The total kinetic energy of the decay products is the rest energy of
the EO particle. We can find the momentum of the photon from its energy and use the conservation of momentum to calculate the kinetic energy of the A 0.
(a) The total kinetic energy of the decay products is given by: Substitute numerical values(see Table 41-1) and evaluate Ktat:
Elementary Particles and the Beginning of the Universe
453
(b) The momentum of the photon is given by:
py:
Substitute numerical values and evaluate
(c) The kinetic energy of the A 0 is given by:
1193MeV -(1116T)C2 Py = c =177.0� 1 KI\. - 2 2 PI\. = Py' 2 KI\. -- J!L 2 ( 77.0 MecV)2 =!2.66MeV! KI\. = 2(1116 ��V) y I\. c2-KI\. -� ml\.
or, because
ml\.
Substitute numerical values and evaluate K/\:
(d) A better estimate of the energy of the photon is:
E = E-m
Ey: =1193MeV -(1116 ��V}2 -2.66MeV =!74.3MeV! V = 174.3MeV 1 y c = 74.3Me c c
Substitute numerical values and evaluate
Ey
The improved estimate of the momentum of the photon is:
p
=
E).
