Electro Magnet

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ELECTROMAGNETISM

Principles and Applications

Paul Lorrain

University of Montreal

Dale R. Corson

Cornell University

W . H . F r e e m a n a n d C o m p a n y

S a n F r a n c i s c o



Cover: Jet-ink printer. See page 108.

Sponsoring Editor: Peter Renz; Project Editor: Pearl C. Vapnek; Manuscript

Editor: Rober t Mann; Designer: Gary A. Head; Illustration Coordinator:

Batyah J anows i ; Illustrator: George V. Kelvin Science Graphics ; Compositor:

Syntax Internat ional ; Printer and Binder: The Maple -Va i l Book Manufac tur

ing Group .

Library of Congress Cataloging in Publication Data

Lorra in , Pau l .

Electromagnet ism: principles and appl icat ions .

Bib l iography: p .

Includes index.

1. Elect ro mag net ism . I . Cor son, Dale R. , joint author . I I . Ti t le .

Q C 7 6 0 . L 6 5 3 7 7 8 - 1 9 1 1

ISBN 0-7167-0064-6

The present volume is based on Electromagnetic Fields and Waves, Second

Edi t ion, by Paul Lorrain and Dale R. Corso n. Copy right © 1962, 1970 by

W. H . F reeman and Company.

Electromag netism: Principles and Applications

Copy right © 1978 , 1979 by Paul Lorrain and Da le R. Corson .

No part of this book may be reproduced by any mechanical , photographic , or e lect ronic

process , or in the form of a phonographic recording, nor may i t be s tored in a re t r ieval

sys tem, t ransmit ted, or otherwise copied for publ ic or pr ivate use , wi thout wri t ten per

miss ion from the publ isher .

Printed in the United States of America

1 2 3 4 5 6 7 8 9



CONTENTS

PREFACE ixLIST OF SYMB OLS xiii

CHAPTER 1 VECTORS 1

CHAPTER 2 FIELDS OF STATIONARY

ELECTRIC CHARGES: I 40

Coulomb 's Law, Electric Field Intensity E , Electric

Potential V


ELECTRIC CHARGES: II 65

Gauss's Law, Poisson's and Laplace's Equations,

Uniqueness Theorem


ELECTRIC CHARGES: III 89

Capacitance, Energy, and Forces

CHAPTER 5 DIRECT CURRENTS

IN ELECTRIC CIRCUITS 110

CHAPTER 6 DIELECTRICS: I 156

Electric Polarization P , Bound Charges, Gauss's Law,

Electric Displacement D



Contents

CHAPTER 7 DIELECTRICS: II 179

Continuity C onditions at an Interface, Energy Density

and Forces, Displacement Current

CHAPTER 8 MAGNETIC FIELDS: I 201

The Magnetic Induction B and the Vector Potential A

CHAPTER 9 MAGNETIC FIELDS: II 220

Ampere's Circuital Law

CHAPTER 10 MAGNETIC FIELDS: III 234

Transformation of Electric and Magnetic Fields

CHAPTER 11 MAGNETIC FIELDS: IV 260

The Faraday Induction Law

CHAPTER 12 MAGNETIC FIELDS: V 279

Mutual Inductance M and Self-Inductance L

CHAPTER 13 MAGNETIC FIELDS: VI 311

Magnetic Forces

CHAPTER 14 MAGNETIC FIELDS: VII 333

Magnetic Materials

CHAPTER 15 MAGNETIC FIELDS: VIII 357

Magnetic Circuits

CHAPTER 16 ALTERNATING CURRENTS: I 371

Comp lex Num bers and Phasors

CHAPTER 17 ALTERNATING CURRENTS: II 398

Impedance, Kirchoff' s Laws, Transformations



Contents

CHAPTER 18

CHAPTER 19

CHAPTER 20

APPENDIX A

APPENDIX B

APPENDIX C

APPENDIX D

APPENDIX E

APPENDIX F

INDEX 499

vi i

VECTOR DEFINITIONS, IDENTITIES,

AND THEOREMS 492

SI UNITS AND THEIR

SYMBOLS 494

SI PREFIXES AND THEIR

SYMBOLS 495

CONVERSION TABLE 496

PHYSICAL CONSTANTS 497

GREEK ALPHABET 498

ALTERNATING CURRENTS: III 428

Power Transfer and Transformers

MAXWELL 'S EQUATIONS 453

ELECTROMAGNETIC WAVES 468



PREFACE

Like Electromagnetic Fields and Waves by the same authors, th is book aims to

give the reader a working knowledge of e lect romagnet ism. Those who wil l use

it should refer to a collection of essays by Alfred North Whitehead entit led Th e

Aims of Education,^ particularly the f irst essay, which carries the same ti t le. At

the outse t , White head e xplains his point: "T he w hole book is a protest against

dead kn ow led ge , that is to say, against inert ideas . . . ideas that are merely

received into the mind without being uti l ized, or tested, or thrown into fresh

combinat ions." I t i s in th is spi r i t that we have given more than 90 examples

and set 332 problems, 27 of which are solved in detail .

The book is intended for courses at the freshman or sophomore level,

either as an introduction to the subject for physicists and engineers, or as a last

course in elect romagnet ism for s tudents in other discipl ines.

I t is assum ed that the reader has had a one-term cou rse on differential and

integral calculus. No previous knowledge of vectors , mul t iple in tegrals , dif-

ferent ia l equat ions, or complex numbers i s assumed.

The features of our previous book that have been so appreciated will be

found here as wel l : the examples ment ioned above, problems drawn f rom the

current l i terature, a summary at the end of each chapter, and three-dimensional

figures, this t ime drawn in true perspective.

The book is divided into twenty short chapters suitable for self-paced in

struction. Following an introductory chapter on vectors, the discussion of elec

t romagnet ism ranges f rom Coulomb's law through plane waves in dielect r ics .

Electric circuits are discussed at some length in three chapters: 5, 17, and 18.

Chapter 16 is ent i re ly devoted to complex n umb ers and phasors. The re is no

way of dealing properly with alternating currents without using complex num

bers, and, contrary to what i s of ten assumed, complex numbers are not much of

tF ree P res s , New York , 1976 .



X Preface

an obstacle at this level. Our discussion of relativity occupies only part of one

chapter (Chapter 10)—just enough to explain briefly the Lorentz transformation

and the transformation of electr ic and magnetic f ields. This naturally leaves a

host of questions unanswered, but should be sufficient in the present context.

THE PROBLEMS

The longer problems concern devices and methods of measurement descr ibed in

the physics and electr ical engineering l i terature of the past few years. They are

"programmed" in the sense that they progress by smal l s teps and that the inter

mediate results are usually given.

These longer problems are meant to give the reader the opportunity tolearn how to make approximat ions and to bui ld models amenable to quant i ta

t ive analysis. I t is of course hoped that they can teach the heuristic process in

volved in solving problems in the f ield of electromagnetism. In fact, after

working through many such problems, s tudents end up surpr ised to see that

they can deal with real si tuations.

These problems contain a large amount of peripheral information and

should provide some interesting reading. They should also incite the reader to

apply his newly acquired knowledge to other f ields and should stimulate crea

t iv i ty . Many of them can serve for open-ended exper iments.

The easier problems are marked E, from C hapte r 2 on. They can be solvedin just a few lines, but they nonetheless often require quite a bit of thought.

Some problems are marked D. Those are quite diff icult .

A number of problems require curve plot t ing. This i s because curves are

so much more meaningful than formulas. I t is assumed that the student has

access to a programmable calculator , or possibly to a computer . Otherwise the

calculat ions would be tedious.

On the average, about two problems are solved in detail at the end of each

chapter .

UNITS AND SYMBOLS

The units and symbols used are those of the Systeme International d'UniteSj

designated SI in all languages.f The system originated with the proposal made

t S e e ASTMIIEEE Standard Metric Practice, publ ished by the Ins t i tute of Electr ical and

Elec tron ics Engine e rs , Inc . , 345 East 47 th S t ree t, New Yo rk , N . Y . 10017 .



Preface xi

by the I talian engineer Giovanni Giorgi in 1901 that electr ical units be based on

the meter -ki logram-second system. The Giorgi system grew with the years and

came to be called, f irst the MKS system, and later the MKSA system (A forampere). I ts development was fostered mostly by the International Bureau of

Weights and Measures, but also by several other international bodies such as

the International Council of Scientif ic Unions, the International Electrotechnical

Commission, and the Internat ional Union of Pure and Appl ied Physics.

Appendix D provides a conversion table for passing f rom cgs to SI uni ts ,

and inverse ly. "F urth er use of the cgs units of electr icity and ma gne tism is

depreca ted , " t

SUPPLEMENTARY READING

The fol lowing books are recommended for supplementary reading:

Electromagnetic Fields and Waves: Second Edition, by the same authors and the same

publ isher , publ ished in 1970. There are several references to this book in the foot

notes .

Standard Handbook for Electrical Engineers, McGraw-Hi l l , New York , 1968 .

Electronics Engineers' Handbook, McGraw-Hi l l , New York , 1975 .

Reference Data for Radio E ngineers, Howard Sams and Company, Inc . , Ind ianapol i s ,

1970.

Electrostatics and Its Applications, A. D . Moore , Edi to r , John Wi ley , New York , 1973 .

These books can be found in most physics and engineer ing l ibrar ies . The

reader will f ind the second and third references to be inexhaustible sources of

informat ion on pract ical appl icat ions.

ACKNOWLEDGMENTS

I am indebted first of all to my students, who have taught me so much in the

course of innumerable discussions. I am also indebted to the persons who

assisted me in writ ing this book: to Francois Lorrain, who took part in the

preparation of the manuscript and who prepared sketches of the three-dimen

sional objects that appear in the f igures in true perspective; to Ronald Liboiron,

who also worked on the preparation of the f igures; to Jean-Guy Desmarais and

flbid., p . 11 .



xi i Preface

Guy Belanger, both of whom revised the complete text; and particularly to

Lucie Lecomte, Nancy Renz, and Al ice Chenard, who typed or re typed over a

thousand pages of manuscr ipt .

I am also grateful to Robert Mann, who took such great care in editing

both this book and the previous one; and to Pearl C. Vapnek, of W. H.

Freeman and Company, for her met iculous work on the proofs.

Finally, I wish to thank collectively all those persons who wrote to me in

connection with Electromagnetic Fields and Waves, even though they all re

ceived a promp t answ er and were duly thanked individual ly . Their feedback was

invaluable .

Dale Corson's duties as President, and later as Chancellor, of Cornell Uni

versity have unfortunately prevented him from working on this offshoot of

Electromagnetic Fields and Waves, of which he is co-author.

P A U L LO RRAI N

Montreal, 1978



LIST OF SYMBOLS

SPACE, TIME, MECHANICS

Element of length dl, ds, dr

Total length / , L, s, r

Element of area da

Total area S

Element of volume dr

Total volume T

Solid angle O

Normal to a surface n

Waveleng th A

Radian length K=\/2TT

Time /

Period T=l/f

F r eq u e n cy / = l/T

Angular f requency,

angu lar velocity w = 277/

Velocity v

Mass m

Mass densi ty p

M o m e n t u m p

Mo me nt of inertia /

Force

Torque

Pressure

Energy

P o w e r

FTP

I I '

P

ELECTRICITY AND MAGNETISM

Quantity of electr ic

i ty Q

Volume charge den

sity p

Surface charge den

sity cr

Linear charge den

sity A

Electric potential V

Induced elect romo-

tance V

Electric field inten

sity E

Electr ic displace

m e n t I)

Permitt ivity of vac

uum e 0

xiv List of Symbo ls

Relative permitt ivity

Permittivity

Elect r ic dipole moment

Electric polariza

tion, electr ic di

pole moment per

uni t volume

Electric suscepti

bility

Mobi l i ty

Electric current

Volume cur r en t

densi ty

Surface current den

sity

Vector potent ia l

Magnet ic induct ion

Magnetic f ield in

tensity

Magnetic f luxMagnetic f lux l ink

age

Permeabili ty of vac

u um

Relat ive permeabi l

ity

Permeabi l i ty

Magnet ic dipole

moment per uni t

vo lume

Magnet ic suscept i

bil i ty

Magnet ic dipole

m o m e n t

Resistance

Capaci tance

e = e r € n =DIE

P

Xe

At

I

a

A

B

H

<T>

A

M O

Mr

M ;

M

Self - inductance L

Mutual inductance M

Impedance Z

Admi t t ance Y

Resistivity P

Conduct ivi ty a

Poynt ing vector

MATHEMATICAL SYMBOLS

Approx imate lyequa l to ~

Propor t ional to

Exponent ial of x

Average

Real part of z

Modulus o f z

Decadic log of x

Natural log of x

Arc tangent x

Complex conjugate

of z

Vector

Gradient

= BlH D i v e r

ge n c e

Curl

m

R

C

Laplacian

Unit vectors in Car

tesian coordinates

Unit vector along r

Field point

Source point

e 7

, e x p x

B

Re z

I"

lo g x

In x

arc t anx

Z *

E

V

V -

V x

V -

U i, K

r T

x, y, z

x', y', z'



ELECTROMAGNETISM



CHAPTER 1

VECTORS

1.1 VECTORS

1.2 SCALAR PRODUC T

1.2.1 Example: W ork Done by a Force

1.3 VECTOR PRODUC T1.3.1 Examples: Torque, Area of a Parallelogram

1.4 THE TIME DERIVA TIVE

1.4.1 Examples: Position, Velocity, and Acceleration

1.5 THE GRADIENT

1.5.1 Examples: Topographic Map, Electric Field Intensity

1.6 THE SURFACE INTEGRAL

1.6.1 Example: Area of a Circle

1.6.2 Example: Electrically Charged Disk

1.7 THE VOLUME INTEGRAL

1.7.1 Example: Volume of a Sphere

1.8 FLUX

1.8.1 Example: Fluid Flow

1.9 DIVERGENCE

1.10 THE DIVERGENCE THEOREM

1.10.1 Examples: Incompressible Fluid, Explosion

1.11 THE LINE INTEGRA L

1.11.1 Example: Work Done by a Force

1.12 THE CURL

1.12.1 Example: Water Velocity in a Stream

1.13 STOKES'S THEOREM

1.13.1 Example: Conservative Vector Fields

1.14 THE LAP LAC I AN

1.14.1 Example: The Laplacian of the Electric Potential

1.15 SUMMARY

PROBLEMS



Electric and magnetic phenomena are described in terms of the fields of

electric charges and currents. For example, one expresses the force between

two electric charge s as the prod uct of the mag nit ud e of one of the charges

and the field of the other.

Th e object of this first chap ter is to describe the math ema tic al me tho ds

used to deal with fields. All the mat eri al in this cha pt er is essen tial for a

prop er unders tand ing of what follows.

Figure 1-1 Pres sure and wind-veloci ty f ie lds over th e N o r t h A t l a n t i c on N o v e m b e r

1, 1967, at 6 h o u r s , G r e e n w i c h M e a n T i m e . The curved l ines are isobars, or l ines

of equa l p res sure . P res sures are given in ki lopasca l s . H igh-pres sure a reas are

d e n o t e d by H, and l ow-pres sure a reas by L. In this case an a r row ind ica te s th e

di rec t ion and veloci ty of t he w ind at its t i p ; a r ro w l engths are p r o p o r t i o n a l to

wind veloci t ies (the l onges t a r row in this f igure represents a wind veloci ty of

25 me te rs pe r second) . Wind vec tors are given only at a few poin t s , where ac tua l

m e a s u r e m e n t s w e r e m a d e .



Figure 1-2 A vec tor A and the th ree vec tors A xi, A yj, A.k, which , when p lacedend- to-end , a re equ iva len t t o A .

Ma them at i ca l ly , a f ie ld is a func t ion tha t desc r ibes a phys ica l q uan t i ty

at a l l points in space. In scalar fields, th is qu an t i ty i s speci f ied b y a s ingle

nu mb er for each po in t . P ressure , t em per a tu re , and e l ec t r i c po te n t i a l a r e

exam ples o f sca la r quan t i t i e s tha t can vary f rom one po in t to ano the r in

s p a c e . F o r vector fields, bo th a nu m be r and a d i r ec t ion a r e requ i r ed . W ind

veloci ty , gravi ta t ional force, and elect r ic f ie ld in tensi ty are examples of

such vector quant i t ies . Both types of f ie ld are i l lust rated in Fig . 1-1.

Vec to r quan t i t i e s wi l l be ind ica ted by boldface t y p e ; italic type wi l l

ind ica te e i the r a sca la r quan t i ty o r the magn i tude o f a vec to r quan t i ty .*

We shal l fol low the usual custom of using right-hand Cartesian

coordinate systems, as in Fig . 1-2; the posi t ive z di rect ion is the di rect ion

of adv anc e o f a r igh t -ha nd sc rew ro ta t ed in the sense tha t t u rns the pos i t ive

x-ax i s in to the pos i t ive v -ax i s th rough the 90° ang le .

1.1 VECTORS

A vector can be specified by its components a long any th ree mu tua l ly per

pend icu la r axes . I n the Car t es i an coord ina te sys t em of F ig . 1 -2 , f o r example ,

t h e c o m p o n e n t s o f t h e v e c t o r A a r e A x, A y, A z.

f In a ha nd wr i t te n text , it is con ven ient t o ident ify a vector by mean s of an arro w over

the symbol , fo r example , A .



4 Vectors

SCALAR PRODUCT

T h e scalar, o r dot product, is t he sca la r qu an t i ty ob ta i ned on mu l t ip ly ing

the mag n i tu de o f the fi rs t vec to r by the ma gn i tu de o f the second a nd by the

cosin e of the ang le betw een th e tw o vector s . In Fig . 1-3, for exa mp le,

A • B = AB c o s (q> — 6). (1-5)

I t f o l lows f rom th i s de f in i t ion tha t t he usua l commuta t ive and d i s

t r ibu t ive ru les o f o rd ina ry a r i thm et i c mul t ip l i c a t ion a pp ly to the sca la r

p r o d u c t :

A • B = B - A, (1-6)

and, for any three vectors ,

A - ( B + C ) = A - B + A - C . (1-7 )

The vec to r A can be un ique ly expressed in t e rms o f i t s componen t s

th rough the use o f unit vectors i , j , k , which are def ined as vec tors of un i t

magn i tude in the pos i t ive x , y, z d i r ec t ion s , r espe c t ive ly :

A = A xi + A,} + A :k. (1-1)

Th e vec to r A is thus the sum of th r ee vec to r s o f m ag n i t ud e A x, A y, A z,

paral le l to the x-, y~, z-axes, respect ively .

T h e magnitude of A is

A = (A2

X + A) + A2

)1

'2

. (1-2)

T h e sum of two vectors is o b t a i n e d b y a d d i n g t h e i r c o m p o n e n t s :

A + B = (A x + B x)\ + (A y + B y)\ + (A z + B J k . ( 1 - 3 )

Subtraction i s s imply ad d i t io n wi th one o f the vec to r s chan ged in s ign :

A - B = A + ( - B ) = (A x - B x)i + (A, - B y)\ + [A z - B :)k . (1-4)

The lat ter proper ty wi l l be ver i f ied graphical ly in Prob. 1-3. I t a lso fol lows

t h a t

i - i = 1, j j = l , k k = 1. (1-8)

j • k = 0, k • i = 0, i • j = 0. (1-9)

T h e n ,

A • B = (AJ + A y\ + Azk) • (B x\ + B y\ + B.k), (1-10)

= AXB X + AyB y + A-B-. (1-11)

I t is easy to chec k th at th is resul t i s cor re ct for two v ecto rs in a plane, as

in Fig. 1-3:

A • B = AB c os ( sin Q, (1-12)

= A XB X + A yB y. (1-13)

6 Vectors

1.2.1 EXAMPLE: WORK DONE BY A FORCE

A s imple phys ical example of the scalar product i s the work clone by a force F act ing

through a d i sp lacement s : W = F • s, as in Fig . 1-4.

Figure 1-4 T he wo rk d on e by a force F whose po int of app l icat i on mo ves by a

dis tance s i s (F co s <p)s, o r F • s.

1.3 VECTOR PRODU CT

T h e vector, o r cross product, of two vectors i s a vector whose di rect ion is

perpend icu la r to the p lane con ta in ing the two in i t i a l vec to r s and whose

ma gni tud e is t he p rod uc t o f the mag n i tu des o f those vec to r s and the s ine

of the ang le be twe en them . W e ind ica te the vec to r p rodu c t thu s :

A x B = C. (1-14)

Th e ma gn i tud e o f C is

C = \AB sin (cp - 0) | , (1-15)

w i t h <p a n d 0 defined as in Fig. 1-3. T he d ire cti on of C is given by th e r i gh t-

hand sc rew ru le : i t i s t he d i r ec t ion o f advance o f a r igh t -hand sc rew whose

axis , he ld perp end icu la r to the p lane o f A an d B, i s r o t a t ed in the sense tha t



1.3 Vector Product 7

ro ta t es the f i r s t -named vec to r (A) in to the second -nam ed (B) th ro ugh the

smal l e r ang le .

T h e c o m m u t a t i v e r u l e i s no t fol lowed for the vector product , s inceinver t ing the o rder o f A and B inver t s the d i r ec t ion o f C:

A x B = - ( B x A) .

The d i s t r ibu t ive ru le , however , i s f o l lowed fo r any th r ee vec to r s :

A x (B + C) = (A x B) + (A x C).

This wi l l be shown in Prob. 1-7.

F r om the def in i tion o f the vec to r p ro du c t i t f o llows th a t

(1-16)

(1-17)

(1-18)x i = 0 , j x j = 0 , k x k = 0 ,

and , fo r the usua l r igh t -handed coord ina te sys t ems, such as tha t o f F ig . 1 -2 ,

i x j = k, j x k = i, k x i = j , j x i = — k , a nd so o n . (1-19)

W r i t ing ou t the vec to r p rod uc t o f A an d B in t e rm s o f the co mp on en t s ,

A x B = (A xi + A y] + A±) x (B x\ + B y\ + B zk) , (1-20)

= (A yB z - A zB y)\ + (A._B X - AXBZ)\ + (A xB y - A yB x)k , (1-21)

i j k

AxA y A :

B x B y B :

(1-22)

We can check this resul t for the two vectors of Fig . 1-3 by expanding

sin ((p — 9) an d no t ing tha t t he vec to r p ro duc t is i n the pos i t ive z d i r ec t ion .

1.3.1 EXAMPLES: TORQUE, AREA OF A PARALLELOGRAM

A good phys ica l exa mple of the vec tor p r odu c t i s t he torque T produced by a fo rce

F ac t ing wi th a moment a rm r about a po in t O, as in Fig. 1-5, where T = r x F .

A second example i s t he area of a parallelogram, as in Fig. 1-6, where the area

S = A x B. Th e area is thu s rep resen ted by a vector perp end icul ar to the surface.



Figure 1-6 Th e area of the para l le lo gra m is A x B = S. Th e vector S is no rm al to

the pa ra l l e logram.

THE TIME DERIVATIVE

We sha l l o f t en be concerned wi th the r a t es o f change o f sca la r and vec to r

quan t i t i e s wi th bo th t ime and space coord ina tes , and thus wi th the t ime

and space der iva t ives .



9

Figure 1-7 A vec tor A , i ts i nc rem ent AA, and the i r co mp one nt s .

Th e t ime der iva t ive o f a vec to r quan t i ty i s s t r a igh t fo rward . In a

t ime Af , a vector A, as in Fig . 1-7, may change by AA, which in general

r epresen t s a change bo th in magn i tude and in d i r ec t ion . On d iv id ing AA by

At and taking the l imi t in the usual way, we ar r ive at the def ini t ion of the

t ime der ivat ive dA/dt:

dA = H m Ajt + AO - A(r)

dt A,^O At

= Hm ^ i + A ^ j + A ^ ( 1 2 4 )

A f ^ O Af

dt dt tit

Th e t ime der iv at ive of a vec tor i s thu s equa l to the vec tor sum of the t im ed e r i v a t i v e s o f i t s c o m p o n e n t s .

1.4.1 EXAMPLES: POSITION. VELOCITY. AND ACCELERATION

The t im e de r iva t ive of the position r of a point is i ts velocity v, and the t ime derivat ive

of v is the acceleration a. See Prob. 1-10.

10 Vectors

THE GRADIENT

Let us cons ider a sca la r quan t i ty tha t is a co n t in uo us an d d i f f er en t iab lefunct ion of the coor din ate s and ha s the value / ' a t a cer ta in poin t in spac e,

as in Fig . 1-8. Fo r exa mp le, / cou ld be the elec t ros tat ic pot en t ia l V. W e

wish to kn ow ho w / chan ges over the d i s t anc e dl measured f rom tha t po in t .

N o w

df = %- dx + 4- dy = A • dl . (1-26)cx oy

w h e r e

A = %- i + % j , a n d dl = dx i + dy j . (1-27)< .\- cy

Th e vec to r A, wh ose com po ne n t s a r e the r a t es o f cha nge o f / wi th

d i s t ance a long the coord ina te axes , i s ca l l ed the gradient of the scalar

qua n t i ty / . Th e op era t ion o n the sca la r / de f ined by the t e rm grad ie n t

is indic ated by the sym bo l V, cal led "del" . T hu s

A = \f. (1-28)

Figure 1-8 The quant i ty / , a funct ion of pos i t ion, changes from / to / ' + df over

the d i s t ance dl.



1.5 The Gradient 11

I n t h r e e d i m e n s i o n s ,

. . . d .3, C

V = + j — + k — .C.Y dy dz

d - 2 9 )

The par t i a l d i f f e r en t i a t ions ind ica ted a r e ca r r i ed ou t on wha tever sca la r

qua n t i ty s t and s to the r igh t o f the V sym bol . T hu s

ox dy dz(1-30)

a n d

I V I + +df 1/2

d - 3 1 )

T h u s ,

df = V / • dl = |V/'| |dl| cos 9, (1-32)

w h e r e 9 i s t he ang le be tween the vec to r s V/ and dl .

We now ask wha t d i r ec t ion one shou ld choose fo r dl i n o rder tha t

df be m ax im um . Th e answe r i s : t he d i r ec t ion in which 9 = 0, that i s , the

d i r ec t ion o f Vf. Th e g rad ien t o f / i s t hus a vec to r whose ma gn i tud e a nd

d i r ec t ion a r e those o f the max imum space r a t e o f change o f / .

The gradient of a scalar funct ion at a given point i s a vector having

the fo l lowing p rop er t i e s :

1) I ts com po ne n t s a r e the r a t es o f chan ge o f the func tion a long the

d i r ec t ions o f the coord ina te axes .

2 ) I t s m ag n i t ud e i s t he m ax im um ra te o f cha nge o f the func t ion wi thd i s t a n c e .

3) I t s d i rect ion is that of the maximum rate of change of the funct ion.

4 ) I t po in t s toward larger values of the funct ion.

Th e g rad ien t i s a vec to r po in t - func t ion der ived f rom a sca la r po in t -

func t ion .



12 Vectors

1.5.1 EXAMP LES: TOPOGR APHIC MAP. ELECTRIC FIELD INTENSITY

Fig . 1-9 s h o ws th e grad ien t of the elevat ion on a topographic map.

In th e nex t chap ter we shall see tha t , in an electrostat ic f ield , th e electric field

intensity E is e q u a l to m i n u s th e grad ien t of the e lec t r ic po ten t ia l V: E = — W.

Figure 1-9 T o p o g r a p h i c ma p of a hill. The numbers shown g ive th e e leva t ion E in

mete rs . The grad ien t of E is the slope of the hill at the po in t cons idered , and it

p o i n t s t o wa rd an mcrease in e leva t ion : V£ = (dE/dx)i + (dE/dy)\. The a r ro w s h o ws

\E at one po in t where th e e leva t ion is 400 m e t e r s .

1.6 THE SURFACE INTEGRAL

Consider the x-y plane. To locate a point in this plane, one requires two

coordinates, x and y. N ow c ons ide r a given sph eric al surface, as in Fig. 1-10.



13

z

y

x

Figure 1-10 Spherical surface of radius r. T h e t w o c o o r d i n a t e s 8 a n d q> suffice to

de te rmine the pos i t ion of a po in t P on the sur face . How do these coord ina te s

com par e wi th the l a t i t ude and the longi tud e a t t he sur face of the ea r th ?

To f ix a po in t we aga in need two coord ina tes , 0 a n d q> . Of cour se we cou ld

u s e t h e t h r e e c o o r d i n a t e s x, y, z. B u t t h o s e t h r e e c o o r d i n a t e s a r e r e d u n d a n t ,

on the given sphere, s ince x 2 + y 2 + z 2 = r 2. So we on ly need two co

o r d i n a t e s , 0 a n d (p .Th is is a genera l ru l e : one a lw ays needs to spec ify tw o coord ina te s to

ident i fy a point on a given surface.

I f we require the area of the given surface, we must in tegrate the

element of area, say dx dy, o v e r both var i ab les . We then have a surface

integral.

More genera l ly , an in t egra l eva lua ted over two var i ab les ( tha t a r e

no t necessa r i ly spac e coord ina te s ) i s ca l l ed a double integral.

Suppose the su r f ace ca r r i es a charge . Then , to f ind the to t a l charge ,

we must in t eg ra te the su r f ace charge dens i ty , mul t ip l i ed by the e l ement o f

a r ea , over both v a r i a b l e s .

So , a su rf ace in t egra l i s a do ub le in t egra l eva lua ted o ver the tw o

coord ina tes tha t a r e r equ i r ed to f ix a po in t on a g iven su r f ace .

I f the sur face is s i tuated in the x-y plane, the sur face integral i s of

the form



14 Vectors

w h e r e f{x,y) i s a funct ion of x and y , and where the l imi ts a a n d b o n x

m ay be funct ion s of y . Th e l imi ts c a n d d are cons tan t s . We f i r s t eva lua te

the integral between the braces. This gives ei ther a constant or a funct ionof y . The n th i s cons ta n t , o r func t ion , is i n t eg ra ted over y . The d oub le in t egra l

is therefore a n integral wi thin an integ ral .

As we shal l see in the next example, one ar r ives at the same resul t

by inver t ing the o rder o f in t eg ra t ion and wr i t ing the doub le in t egra l i n

the form

where the l imi ts of in tegrat ion are di f ferent : the l imi ts c ' and a" c a n n o wbe funct ions of x , whi le a' a n d b' a r e c o n s t a n t s .

We have shown braces in the above express ions so as to ind ica te

the m an ne r in which a dou b le in t egra l i s eva lu a ted , bu t these b races a r e

never used in pract ice .

EXAMPLE: AREA OF A CIRCLE

I t is a lways advi s ab le to t e s t one ' s sk i ll a t und e rs t an ding and us ing new c oncep t s

by applying them to extremely s imple cases . So, le t us calcula te the area of a circle

by mea ns of a surface integr al .

So as to s impl i fy the calculat ion, we shal l f ind the area of the top r ight-hand

sector of the c i rc le of Fig. 1-11, and then mul t iply by 4.

The e l ement o f a rea dA is dx dy. We proceed as follows. We first slide dA from

left to right, to generate the shaded strip in the figure. Then we slide the strip from

y = 0 to y = R, while adjus t ing i ts length correct ly. This generates the 90-degree

sec tor .

Let us first find an integral for the strip. The strip has a height dy . It starts at

x = 0 and ends a t the edge of the c i rc le , where

x2 + v

2

= R 1. (1-33)

T h u s , a t the r igh t-h and e nd of the s t r ip.

X = (R* - v 2 ) (1-34)

and the s t r ip has an area



15

Figure 1-11 (a ) Th e infini tesimal e lemen t of area tlx dy sweeps from the v-axis to

the periphery of the c i rc le to generate the shaded s t r ip, (b ) The s t r ip now sweeps

from the .v-axis to the top of the c i rc le to generate the top-r ight quadrant .

The to t a l a rea A of the circle is given by

A4 crfr-""*}*

Se t t ing y/R = sin 8, t h e n

co s 0. dy = R cos 0 dO.

Also , when y = 0, 0 = 0, and when y = R, 8 = n/2. T h e n

— = R R4 J "

/I = TtR 2,

c o s 2 8d8 = - R 2,4

(1-35)

(1-36)

(1-37)

(1-38)

(1-39)

as expec ted .

If we used a vert ical s t r ip, it wo uld extend from

y = 0 to y = (R 2 - x2

)112

. (1-40)



16 Vectors

(1-41)

T h e n we would have tha t

A = , , . P ( / V ( / A ,

4 Jv = o J, = o '

= \ l (R2

- x 2 ) 1 ' 2

dx . (1-42)

This integral is s imi l a r to t h a t of E q . 1-36 and the a rea of the circle is aga in nR2

.

EXAMPLE: ELECTRICALLY CHARGED DISK

If we have an electrically ch arged disk with a surface charge dens i ty a given by, say,

a = Ky2

. (1-43)

then , us ing a vert ical s t r ip and i n t egra t ing ov e r the comp le te c i rcl e , th e t o t a l cha rge is

J+ R i " v

3

^+ (R 2-* 2)" 2")K —

1 _ 3_

Using a t ab le of i n t e g r a l s / we find that

(1-44)

(1-45)

(1-46)

Q = - KR4

.v

4(1-47)

1.7 THE VOLUME INTEGRAL

In a volume integral, there are three variables, say x, y, z in Carte sian co

ordinates, or r, 0, (p in polar coordinates (Fig. 1-10). In the more general

case where the three variables can be of any nature, say velocities, or voltages,

etc . , one has a triple integral.

Here again, one integrates with respect to each variable in succession,

starting with the innermost integral.

* See, for example , Dwight , Table of Integrals and Other Mathematical Data, No . 350 .03 .



1.7 The Volume Integral 17

1.7.1 EXAMPLE: VOLUME OF A SPHERE

To ca lcu la t e the volume of a sphere, we fi rs t f ind the e lement of volume. Referr ingto Fig. 1-12, this is

(rdO)(r si n 9 dq>){dr).

T h e n

Let us integrate , f i rs t wi th respect to cp , then with respect to 0, and finally with

respect to r. This m eans th a t we f ir st ro t a t e the e lement o f vo lu me ab out the ve r t i ca l

axis to generate a r ing. Then we s l ide the r ing from 9 = 0 to 8 = n, whi le keep ing

the rad ius r a n d dr cons tant . This generates a spherical shel l of radius r and th i ck

ness dr . Then we va ry r from 0 to R t o gene ra t e the sphe re . So

V = r p = " r= "2nr 2 sin 0 dO dr , (1-49)

J r = 0 J e = 0 ' '

(1-50)

(1-51)

Figure 1-12 Elemen t o f vo lu me in sphe r i ca l co ord in a te s .

18 Vectors

This is a part icular ly s imple case because the l imits of integrat ion are a l l con

s tants . Indeed, the t r iple integral of Eq. 1-48 is real ly the product of three integrals .

In P ro b . 1 -19 we sha l l r epea t t he ca l cu la t ion in Car t e s i an coor d ina te s . The

l imits of integrat ion wil l then be funct ions , ins tead of being cons tants .

1.8 FLUX

I t i s of ten necessary to calculate the f lux of a vector through a sur face. For

example , one migh t wi sh to ca lcu la t e the magne t i c f lux th rough the i ron

co re of an elec t rom agn et . The flux of a vecto r A th ro ug h an inf ini tesimal

surface da is

c/O = A • da . (1-52)

where the vec to r da i s normal to the sur face. The f lux d<t> is t h e c o m p o n e n t

o f the vec to r A norm al to the su r f ace , mul t ip l i ed b y da . For a f inite surface S

we f ind the to tal f lux by int egr at in g A • da over the en t i r e su r f ace :

J >da . (1-53)

For a c losed surface the vector da i s t ake n to po in t outward.

1.8.1 EXAMPLE: FLUID FLOW

Let us cons ider fluid flow. We define a vector pv, p being the f luid dens i ty and v

the fluid velocity at a point. The flux of pv thr ou gh any closed surface is the net

ra t e a t which mass l eaves the vo lume bounded by the sur face . In an incompres s ib l e

fluid this flux is zero.

1.9 DIVERGENCE

Th e ou tw ard f lux O of a vec to r A th r ou gh a c losed su r f ace can be ca lcu

l a t ed e i the r f rom the above equa t ion o r as fo l lows . Le t us cons ider an

in f in i t es imal vo lume dx, dy, dz and the vector A, as in Fig . 1-13, whose



19

Figure 1-13 E lem en t of vo lu m e dx, dy, dz a r o u n d a p o i n t P, where the vec tor A has

the va lue i l l us t ra t ed by the a r row.

c o m p o n e n t s A x, A A, are func t ions o f the coord i na te s x, y, z. W e c o n s i d e r

an in f in i tes imal vo lum e and f i r s t -o rder va r i a t io ns o f A.

The va lue o f A x at the cen ter of the r igh t-h an d face can be tak en to

be the average va lue over tha t f ace . Through the r igh t -hand f ace o f the

vo lume e lement , t he ou tgo ing f lux i s

dHfx

= U c + ^ ^ \ d y d z , (1-54)

s ince the no rm al co mp on en t o f A a t t he r igh t -ha nd f ace i s t he x -c om po ne n t

of A at that face.

At the lef t -hand face,

c^ L = - ^ - ^ ^ j d y d z . (1-55)

Th e minu s s ign befo re the paren th es i s i s necessa ry here because , A xi b e i n g

inward at th is face and da be ing ou tw ard , t he cos ine of the ang le b e tween



20 Vectors

t he two vec to r s i s — 1. Th e net ou tw ar d f lux thr ou gh the tw o faces is the n

PA PAd$> R + m, = —- dx dy dz = —^ dx, (1-56)OX ox

w h e r e dx is t he vo lum e o f the in f in it es imal e l em ent .

I f we calcu late the net f lux thro ug h th e oth er pa i rs of faces in the sam e

manner , we f ind the to t a l ou tward f lux fo r the e l ement o f vo lume dx to be

\ ox Py dz J

Suppose now tha t we have two ad jo in ing in f in i t es imal vo lume e le

m en t s and tha t we add the flux th rou gh th e boun d in g su r f ace o f the fi rs t

vo lu me to the flux th rou gh th e bou nd ing su r f ace of the second . At the

co m m on f ace the fluxes a r e equ a l in m ag n i t ud e bu t op pos i t e in s ign , and

they cancel as in Fig. 1-14. The flux from the f irst volume plus that from the

secon d is the f lux th rou gh the bou nd ing su r f ace o f the comb ine d v o lum es .

To ex tend th i s ca l cu la t ion to a f in i t e vo lume, we sum the ind iv idua l

f luxes for each of the inf ini tesimal vo lum e elem en ts in the f inite volu me , an d

the to t a l ou tward f lux i s

At any g iven po in t in the vo lume, the quan t i ty

PA X PA V PA,— - H -Iox dy Pz

i s thus the outgoing f lux per uni t volume. We cal l th is the divergence of the

vec to r A a t t he po in t .

Th e d ivergence o f a vec to r po in t - fun c t ion i s a sca la r po in t - fun c t ion .

Accord ing to the ru les fo r the sca la r p roduc t ,

v . A = 8A , + 8A l + dA lt ( 1 . 5 9 )

o x dy dz

wh ere the o pe ra to r V is def ined as in Eq. 1-29. The o pe ra to r V is no t a

vec to r , of cour se , bu t i t i s conv en ien t to use the no ta t ion o f the sca la r p ro du c t

to ind ica te the oper a t io n tha t i s ca r r i ed ou t .



21

Figure 1-14 At the common face, the flux of A out oidz l i s equa l i n magn i tude , bu t

opposi te in s ign, to the f lux of A ou t of dz 2 •

1.10 THE DIVERGENCE THEOREM

Th e total o ut w ar d f lux of Eq . 1-58 is a lso equ al to the sur face integra l of

t h e n o r m a l o u t w a r d c o m p o n e n t o f A . T h u s

This i s the divergence theorem. Note that the integral on the lef t involves

only the values of A on the sur face 5 , whereas the two integrals on the r ight

invo lve the va lues o f A th roughou t the vo lume T enc losed b y S . Th e d i

vergence the ore m i s a gene ra l i za t io n o f the fundam enta l t he ore m of the

ca lcu lus ( see P rob . 1-20).

We can now redef ine the divergence of the vector A as fol lows. I f

t h e v o l u m e T i s a l low ed t o sh r ink suf fic ient ly , so th at V • A doe s not var y

apprec iab ly over i t , t hen

Js

A • da = (V • A) , ( t - > 0 ) , ( 1 -61)

a n d

V • A = lim - f A • da . (1-62)

r ^ O T JS

The d iverg ence i s t hu s the ou tw ar d f lux per un i t vo lum e, as the vo lum e t

a p p r o a c h e s z e r o .



22 Vectors

1.10.1 EXAMPLES: INCOMPRESSIBLE FLUID, EXPLOSION

In an incompressible fluid, V • (pv) is every wh ere eq ual to zero, s ince the ou tw ard

mass f lux per uni t volume is zero.

Wi th in an explosion, V • (pv) is positive.

1.11 THE LINE INTEGRAL

The in teg ra l

/ > • „ , .

eva lua ted f rom the po in t a to the po in t b on some specified curve, is a line

integral. Each e lemen t o f l eng th dl on the curve is mult ip l ied by the local

va lue o f A acco rd ing to the ru le fo r the sca la r p roduc t . These p roduc ts

a re then summed to ob ta in the va lue o f the in teg ra l .

A vector field A is said to be conservative if the line integral of A • dl

a round any c lo sed cu rve i s ze ro :

A • dl = 0. (1-63)

The c irc le on the in tegra l s ign indica tes tha t the pa th of in tegra t ion is

c losed . We shal l f ind in Sec . 2 .5 tha t an e lec tros ta t ic f ie ld is conservat ive .

/ ./ /. / EXAMPLE: WORK DONE BY A FORCE

T h e work done by a force F act ing from a t o b along some specif ied path is

W - £ F • dl, (1-64)

w h e r e b o t h F a n d dl mu st be kn ow n fun ct ions of the coo rdi na tes i f the integral i s

to be eva lua ted ana ly t i ca l ly .

Let us calcula te the work done by a force F that is in the y di rec t ion and has a

m a g n i t u d e p r o p o r t i o n a l t o y, as i t moves around the c i rcular path from a to b in

Fig. 1-15. Since

F = ky j a n d dl = dx i + dy j , (1-65)

W = S! F > d l = 5?ydy = kr 2/2 . (1-66)



23

1.12 THE CURL

Let us calculate the value of the l ine integral of A • dl i n the mor e genera l case

where i t i s not zero.

Fo r an in fin i t esimal e l e men t o f pa t h dl i n the xy-p lane , and f rom the

def in it ion o f the sca la r p rod uc t ,

A • dl = A x dx + A y dy . (1-67)

T h u s , for any closed path in the xy-plane and for any A,

A • dl = j>A x dx + &A y dy . (1-68)

Now consider the inf ini tesimal path in Fig . 1-16. There are two con

t r ibu t ion s to the f ir st i n t eg ra l on the r igh t -h and s ide o f the abov e e qua t ion ,

one a t y — {dy/2) a n d o n e a t y + (dy/2):



24

y

Figure 1-16 Closed , rec t angula r pa th in the xr -p lane . cen te red on the po in t

P(x, y, 0) , where the vec tor A has the com po nen t s Ax, A

y. The in t egra t ion

a ro un d the pa th is pe r formed in the d i rec tion of the a r row s .

Th ere i s a min us s ign befo re the second t e rm b ecaus e the pa th e l em ent a t

y + (cly/2) i s i n the nega t ive x d i r ec t ion . There fo re

A xdx = -~dydx. (1-70)dy

Simi la r ly ,

a n d

j>Aydy = -^-dxdy, (1-71)

A d l = [^-?£*)dxdyox dy

for the infinitesimal path of Fig. 1-16.



1.12 The Curl 25

If we set

dA y dAx

g3 = ^ - d-7 3)

ox dy

then

A'A\ = g3da, (1-74)

where da = dx dy is the area enclosed on the xy-plane by the infinitesimal

pat h. No te t hat the above equa ti on is correct only if the line integral is

evaluated in the positive direction in the xy-plane, that is, in the direction

in which one would have to turn a right-hand screw to make it advance in

the positive direction of the z-axis. This is known as the right-hand screw

rule.

Let us now consider g3 and the other two symmetrical quantities as

the components of a vector

/dA, dA.

\dy dzI +

dA x cA\ . ^ /dAy dAx ^

dz dx I dx dv

which may be written as

V x A (1-75)

We shall call this vector the curl of A. The quantity g3 is then its z com pon ent .

If we consider the element of area as a vector da pointing in the direc

tion of advance of a right-hand screw turned in the direction chosen for

the line integral, then da = da k and

A d l = (V x A)-da. (1-76)

This means that the line integral of A • dl around the edge of an

element of area da is equal to the scalar product of the curl of A by this



26 Vectors

element of area, as long as we observe the above sign convention. This is a

general result that applies to any element of area da, whatever i ts orientation.

Thus

(V x A)„ = l i m ^ c f A - d l : (1-77)

s-o S J

the co mp on en t of the curl of a vector n or ma l to a surface S at a given point

is equa l to the line integral of the vector ar ou nd the bo un da ry of the surface,

divided by the area of the surface when this area ap pr oa ch es zero a ro un d

the point.

EXAMPLE: WATER VELOCITY IN A STREAM

Let us cons ider a stream in which th e veloci ty v is p ro p o r t i o n a l to the d is tance

from the b o t t o m . We set the z-axis parallel to the d i rec t ion of flow, and the x-axis

p e rp e n d i c u l a r to the s t r e a m b o t t o m , as in Fig . 1-17. T h e n

i\- = 0, r v = 0, (1-78)

We shal l calculate th e curl from Eq. 1-77. For (V x v) v we c h o o s e a path paral lel

to th e yz-p lane . In evalua t ing

v d l

around such a p a t h we no te tha t the c o n t r i b u t i o n s are equal and oppos i te on the

par t s para l le l to the z-axis, hence (V x v) v = 0. Likewise, (V x v)2 = 0.

F o r the j ' - c o m p o n e n t we c h o o s e a path paral lel to the x z -p l a n e and evalua te

the in teg ra l a round it in the sense tha t wou ld advance a r igh t -hand screw in the

posi t ive y d i rec t ion . On the p a r t s of the path paral lel to the x-ax is , v • dl = 0 since

v and dl are p e rp e n d i c u l a r . On the b o t t o m p a r t of the path , at a d is tance x from

the yz-p lane ,

v • dl = cx Az,

wh e re a s at (x + Ax)

F o r th e whole pa th ,

£ v • dl = — c(x + Ax) Az.

dl -c Ax Az

(1-79)

(1-80)

(1-81)



2~

Figure 1-17 Th e veloci ty v in a viscous fluid is assu me d to be in the di rect ion of the

r -ax i s and pro por t ion a l t o the d i s t ance .x f rom the bo t to m. Th en V x v = — c'y

and the y-com pone nt o f t he cur l is

(V x v)j, = lim§ v • dl : A x A ;

s~o S Ax Az

Ca lcul at ing V x v di rect ly from E q. 1-75,

V x v

i j k

d f 0

dx r.v fz

0 0 cx

(1-82)

(1-83)

which is the sam e resul t as abo ve.

1.13 STOKE S'S THEO REM

Eq ua t io n 1-76 is t rue only for a pa th so smal l tha t V x A can be co nsid ered

cons tan t over the su r f ace da bo un de d by the pa t h . Wh at if t he pa th C is

so l a rge tha t t h i s cond i t ion i s no t met? The equa t ion can be ex tended r ead i ly

to a rb i t r a ry pa ths . We d iv ide the su r f ace— a n y su r f ace bounded by the



28

Figure 1-18 An a rb i t ra ry sur face bou nd ed by the curve C. Th e sum of the l i ne

integrals arou nd th e curvi l in ear squ ares show n is equ al to the l ine integral

a r o u n d C.

pa th o f in t eg ra t ion in qu es t ion — into e l em ent s o f a r ea d a ^ d a 2 , a n d s o

for th , as in Fig . 1-18. For any one of these smal l areas,

A d I , = (V x A ) - d a , , (1-84)

We add the lef t -hand sides of these equat ions for a l l the da ' s , and

the n we ad d al l the r ig ht- ha nd sides . Th e sum of the lef t -hand s ides is the

l ine in t egra l a round the ex te rna l boundary , s ince the re a r e a lways two equa l

a n d o p p o s i t e c o n t r i b u t i o n s t o t h e s u m a l o n g e v e r y c o m m o n s i d e b e t w e e n

ad jacen t da ' s . The sum of the r igh t -hand s ide i s mere ly the in t egra l o f

(V x A) • da over the f ini te sur face. Thus, for an y surface S b o u n d e d b y a

c losed curve C,

A • dl = J s (V x A) • da. (1-85)

This i s Stokes's theorem.

Figure 1 -18 i l l us t r a t es the s ign conven t ion .



1.14 The Laplacian 29

EXAMPLE: CONSERVATIVE VECTOR FIELDS

Stokes ' s theo rem can h e l p us to u n d e r s t a n d conservative vector fields. U n d e r w h a t

c o n d i t i o n is a vector f ield conservat ive? In o ther words , under what cond i t ion is

the l ine in tegral of A • dl a r o u n d an arb i t ra ry pa th equal to zero? From Stokes ' s

t h e o re m , th e l ine in tegral of A • dl a r o u n d an arb i t ra ry c losed pa th is zero if an d

only if V x A = 0 everywhere . Th is cond i t ion is me t if A = V/ . T h e n

£1dx '

df 81dz

(1-86)

a n d

8A ZdA,

(V x A). = — i - —1

dy dz

d 2f d 2f

dy cz cz dy

= 0, (1-87)

a n d so on for the o t h e r c o m p o n e n t s of the curl .

The field of A is therefo re conserva t ive , if A can be expressed as the g ra d i e n t

of some scalar funct ion / .

1.14 THE LAPLACIAN

The d ivergence of the gradient is of great imp orta nce in electromag netism.

Since

then

\f = f i + f i + fk, (1-88)

dx dy dz

dx cy dz* (1-89)

Th e divergence of the grad ient is the sum of the secon d derivatives with

respect to the rectangul ar coord inate s. The quantit y V • V/ is abbre viated to

\2

f, and is called the Laplacian of/ . The operato r V2

is called the Laplace

operator.

30 Vectors

1.14.1 EXAMPLE: THE LAPLACIAN OF THE ELECTRIC POTENTIAL

We shall see in Sec. 3.4 that the Luplacian of the electric potential i s p r o p o r t i o n a l

to the space charge dens i ty.

1.15 SUMMARY

I n C a r t e s i a n c o o r d i n a t e s a vector qu an t i ty i s wr i t ten in the form

A = A x i + A, j + A z k. ( / - / )

where i , j . k are unit vectors d i r ec ted a long the x, y, z axes respect ively .T h e magnitude of the ve ctor A is the scala r

A = (A 2 + A]. + A2

)112

. (1-2)

V e c t o r s c a n b e a d d e d a n d s u b t r a c t e d :

A + B = (A x + B x)i + (A y + B y)j + (A z + B :)k. (1-3)

A - B = (A x - B x)\ + (A, - B y)\ + (A z - B ._)k. (1-4)

The scalar product (Fig . 1-3) i s

A • B = AB cos (<p — 01 (1-5)

= AXBX + A yB, + A ZB._. (1-11)

I t i s c o m m u t a t i v e ,

A • B = B • A, (1-6)

and d i s t r ibu t ive ,

A • (B + C) = A • B + A • C. (1-7)

T h e vector product (Fig. 1-3) is

A x B = C (1-14)

1.15 Summ ary 31

with

C = |/4 B si n (<p - 0)\.

An oth er equ iva len t de f in i tion o f the vec to r p rodu c t i s

(1-15)

A x B

i j k

B v B t. B.

The vec to r p roduc t fo l lows the d i s t r ibu t ive ru le

A x (B + C) = (A x B) + (A x C ),

b u t no t t h e c o m m u t a t i v e r u l e :

A x B = - ( B x A).

T h e time derivative of A is

dA dA r dA, dA. ,i + — j + — k .

dt dt dt dt

T h e del operator i s def ined as fol low s:

V0 . 0 . C

ox dy cz

a n d t h e gradient of a sc ala r func tion / is

c x o z

(/-22)

( / - / 7 )

(1-25)

(1-29)

(1-30)

Th e g rad ien t g ives the max im um ra te o f chan ge o f / wi th d i s t ance a t t he

po in t cons idered , and i t po in t s toward l a rger va lues off.

T h e surface integral i s a co m m on type o f double integral. It is used for

in t egra t ing over two coord ina tes , say x and y. I t i s co m po se d of an integra l

wi th in an integra l . Th e inner in tegral i s eva lua ted f i rs t .



32 Vectors

T h e volume integral is one form of triple integral. It is used for in

t eg ra t in g func t ions o f th r ee space coo rd ina tes , say x, y, z. In this case, we

ha ve an integral tha t is wi th in an integral tha t is wi th in an integral . T he

inn erm ost in t eg ra l i s eva lua ted f i rs t, leav ing a doub le in t egra l , and so on .

T h e flux O of a vec to r A th rough a su r f ace S is

j :A • da . (1-53)

For a c losed surface the vector da p o i n t s o u t w a r d .

T h e divergence of A,

V • A

dA, dA,. dA ,

dx + dy + dz (1-59)

i s t he ou t wa rd f lux o f A per un i t vo lum e a t t he po in t cons id ered .

T h e divergence theorem s t a t es tha t

j * V • A dx = J*s A da, (1-60)

w h e r e x i s t he vo lume bounded by the su r f ace S.

T h e line integral

J>-ove r a speci f ied c urv e is the su m of the te rm s A • dl fo r each e l ement dl

of the cu rve be tween the po in t s a a n d b.

Fo r a c losed curve C t ha t bounds a su r f ace S, w e h a v e Stokes's theorem:

A • dl = J s (V x A) • da.

w h e r e

(1-85)

V x A

• J k

d d d

dx dy cz

/ L A r A,

(1-75)

is the curl of the vec tor funct ion A.



Problems 33

T h e Laplacian i s t he d ivergence o f the g ra d ie n t :

d 2f ' d 2f c 2fV • V/ ' = \2f = - L + ^ 4 + U . ( 7 - 5 9 )

o x o y dz

PROBLEMS

1-1 Sho w that the two vecto rs A = 9i + j — 6k and B = 4i — 6j + 5k are perp end icul ar

to each o th e r .

1-2 Sho w that the angle betwe en the two vectors A = 2i + 3j + k an d B = i — 6j + k

is 130.5°.

1-3 The vec tors A , B, C a re coplana r . Show graphica l ly tha t

A • (B + C) = A • B + A • C.

1-4 If A and B are adjac ent s ides of a par al le log ram , C = A + B a n d D = A — B are the

d i a g o n a l s , a n d 0 i s the angle between A and B, show tha t (C 2 + D 2) = 2(A 2 + B 2)

a n d t h a t ( C 2 - D 2) = 4AB co s 9.

1-5 Cons ide r two un i t vec tors a and b, as shown in Fig. 1-19.

Find the t r igonometr ic re la t ions for the s ine and cos ine of the sum and difference

of two angles from the values of a • b and a x b.

Solution: Fi rs t ,

a = co s a i + sin a j , (1)

b = cos fix + si n [S j . (2)

Now we can wr i t e the produc t a • b in two different ways . Fro m Eq. 1-11.

a • b = cos a cos [1 + sin a sin /?, (3)

and, from Eq. 1-5,

a • b = co s (a - /?). (4)

T h u s

cos (a — ji) = cos a cos j) + sin a sin /?. (5)

If we set \V = -/?,

co s (a + /?') = co s a co s ft' — sin a sin fi'. (6)

34

Figure 1-19 Two unit vectors, a an d b,

situated in the xv-plane. See Prob. 1-5.

Similarly, from Eq. 1-22,

a x b cos a sin a 0

cos p sin p 0

= (cos a sin p — cos p sin a)k

and, from Eqs. 1-14 and 1-15,

a x b = —sin (a — p)k.

Thus

sin (ot — P) = sin a cos p — cos a sin p.

Setting again p' = — p,

sin (x + P') = sin a cos p' + cos a sin P'.

(7)

(S)

(9)

(10)

( I D

1-6 Show that t he mag ni tude of (A x B) • C is the volume of parallelep iped whose edges

are A, B, C, and show tha t (A x B) • C = A • (B x C) .

1-7 Sh ow t ha t A x ( B + C) = A x B + A x C .

1-8 Show that a x (b x c) = b( a • c) - c( a • b).

1-9 If r • (dr/<-/>) = 0, show that r = constant.

1-10 A gun fires a bullet at a velocity of 500 meters per second and at an angle of 30" with

the horizontal.

Find the position vector r, the velocity vector v, and the acceleration vector a

of the bullet, t seconds after the gun is fired. Sketch the trajectory and show the three

vectors for some time t.

35

V

Figure 1-20 Tw o p o i n t s P' and P in

space, wi th the vec tor r and the uni t

vec tor rt. We shall often use such

p a i r s of poin t s , w i th P' at the source ,

a n d P at the poin t where the field is

ca lcu la t ed . For e x a m p l e , one m i g h t

ca lcu la t e th e electric field intensity at

P with th e e l e m e n t of c h a r g e at P'.

S e e P r o b . 1-13.

/ - / / Let r be the rad ius vec tor f rom th e origin of c o o r d i n a t e s to any p o i n t , and let A be a

cons tan t vec tor . Show tha t V(A • r) = A.

1-12 S h o w t h a t (A • V)r = A.

1-13 The vec tor r is directed from P'(x',y',z') to P(.v.y.r), as in Fig. 1-20.

a) If t he po in t P is fixed and the p o i n t P' is a l lowe d to move , show tha t th e grad ien t

of (\/r) under these condi t ions is

w h e r e rj is the uni t vec tor a long r. Show tha t t h i s is the m a x i m u m r a t e of c h a n g e of

b) Show s imilar ly that , if P' is fixed and P is a l lowed to m o v e ,

1-14 The c o m p o n e n t s of a vec tor A are

-I . - V

r y— y

< x

w h e r e /'is a funct ion of x, y, :. S h o w t h a t

A = r x Y/ ; A • r = 0. a n d A • Y / ' = O.

1-15 a) S h o w t h a t V • r = 3.

b ) W h a t is the flux of r t h r o u g h a spherical surface of r a d i u s al



36

II

Figure 1-21 Trunca ted cy l inde r . See

P r o b . 1-18.

1-16 S h o w t h a t

V • ( / A ) = / V A + A • v./;

wh ere / i s a scalar funct ion an d A is a vector funct ion.

1-17 Th e vec tor A = 3.vi + y j + 2zk , a n d / = x 2 + y 2 + :2.

a) Show that V • ( / A ) a t the po int (2,2,2) i s 120, by f irs t calcula t ing / A an d then

ca lcu la t ing i t s d ive rgence .

b) Ca lcu la t e V • (/A ) by first finding V/ 'an d V • A. an d then u sing the iden tity

of Prob. 1-16 above.

c) If x, y, : a re mea sured in me te rs , wha t a re the un i t s o f V • ( /A) ?

1-18 Calcu la t e the vo lume of the por t ion of cy l inde r shown in F ig . 1 -21 .

Set the xy-plane on the base of the cyl inder and the origin on the axis . Wri te

the e lement of volume of height / ; as h dx dy, wi th

1-19 Calcu la t e the vo lume of a sphe re of rad ius R i n C a r t e s i a n c o o r d i n a t e s .

1-20 Cons ide r a func t ion f(x). Th e fund ame nta l t heore m of the ca l cu lus s t a te s tha t

Le t A = / (x ) i , and cons ide r a cy l indr i ca l vo lume i pa ra l l e l t o the x-ax i s , ex tending

from x = a to x = b, as in Fig. 1 -22.

Show tha t t he d ive rgence theorem, appl i ed to the f i e ld A and the vo lume T,

yields the fundamental theorem of the calculus in the form s ta ted above.



37

Figure 1-22 Cylindrical volume parallel to the x-axis, extending from x = a to

x = b. See Prob. 1-20.

1-21 The gravitational forces exerted by the sun on the planets are always directed toward

the sun and depend only on the distance r. This type of field is called a central force field.

Find the potential energy at a distance r from a center of attraction when the

force varies as 1 /r2

. Set the pot enti al energy equal to zero at infinity.

1-22 Show that the gravita tion al field is conser vative. Disrega rd the curvat ure of the earth

and assume that g is a constant.

Wh at if one takes int o accoun t the curv atu re of the earth and the fact that gdecreases with altitude? Is the gravitational field still conservative?

1-23 Let P an d Q be any two paths in space that have the same end points a and b.

Sho w that , if A is a conser vat ive field,

J > - d l = J > - d l .

over P over Q

1-24 Let A be a conservative field and P0 a fixed point in space. Let

f(x,y,z) = ^ A • dl,

where the line integral is calculated along any path joining P 0 to (x,y,z).

Show that A = V/.

1-25 The azimuthal force exerted on an electron in a certain betatron (Prob. 11-9) is

proportional to J- 0' 4.

Show that the force is non-conservative.



38 Vectors

1-26 A vector field is defined by A = f(r)r.

S h o w t h a t V x A is e q u a l to zero .

1-27

1-28

1-29

1-30

1-31

1-32

Show tha t V x (/A) =

S h o w t h a t V • (A x D;

vecto rs .

S h o w t h a t V • V x A =

(Yf) x A + / (V x A), wh e re / is sca la r and A is a vecto r .

= D • (V x A) — A • (V x D), wh e re A and D are any two

0.

O n e of the fou r Maxwel l equat ions s ta tes tha t V x E = —dB/dt, wh e re E and B

are respect ively th e electric field intensity in volts pe r m e t e r and the m a g n e t i c in

d u c t i o n in tes las at a p o i n t .

Show tha t th e l ine in tegral of E • dl is 20.0 microvo l t s over a s q u a re 100 mill i

meters on the side when B is 2.00 x 10 "3

f tesla in the d i r e c t i o n n o rm a l to the s q u a re ,

wh e re t is the t ime in seconds .

In Sec. 1-14 we defined the L a p l a c i a n of a sca la r po in t - func t ion / . It is also useful

to define th e L a p l a c i a n of a vecto r po in t - func t ion A:

V2

A = V2

Axi + \2

Ay) + \2

Azk.

Show tha t V2

( V / ) = V (V2

/ ) .

Show tha t

V x V x A = V(V • A) — V2

A.

We shal l need this resul t in C h a p t e r 20.

(1)

Solution: Let us find th e .v -componen ts of these th ree quan t i t i es . Then we can

find the y- and z -c o m p o n e n t s by s y m m e t ry .

Since

V x V x A

i

8

dx

J

8

dy

K

c

dz

8A Z dAy dAx dAz

dy dz dz dx

8A,

dx

dA,

8y

12]

then its .v -componen t is

d_ fdA,

dy \ dx

8A :

~8y

dA x

The .v -componen t of V(V • A) is

8 (cAx

dx V dx

dA y dA z

~dz~(3 )



Problems 39

while that of V2

A is

Then we shou ld have tha t

d 2Ax

82

AX

82

AX

t x cy cz*

d 2A t

cy fx 5y 2

f2

/ lv

3*4,— -I ~

f z2 fz fx

d2

A, f2

/ L 82

A,

f x dy fx f z f x2

82

AX

82

AX

dy2 f z

2(5 )

which is correct .

The y-c om pon ent of the given equat ion ca n be found from E q. 5 by replacing x

by y , y by z, an d z by x . This gives anot he r ide nti ty , and s imilarly for the z-co mp one nt .



CHAPTER 2

FIELDS OF ST A TIONARY

ELECTRIC CHARGES: I

Coulomb''s Law, Electric Field Intensity E ,

Electric Potential V

2.1 COULO MB'S LAW

2.2 THE ELECTRIC FIELD INTENSITY E

2.3 THE PRINCIPLE OF SUPERP OSITION

2.4 THE FIELD OF AN EXTENDED

CHARGE DISTRIBUTION

2.5 THE ELECTRIC POTENTIAL V

2.5.1 Examp le: The Electric Dipole

2.6 SUMMARY

PROBLEMS



W e beg in our s tudy of e l ec t r i c an d mag ne t i c ph en om en a by inves t iga t ing

the f ie lds of s ta t io na ry elect r ic cha rge s.We sha l l s t a r t wi th Cou lomb 's l aw and deduce f rom i t , i n th i s chap te r ,

the elect r ic f ie ld in tensi ty an d the elect r ic pote nt ia l for any dist r i but ion of

s t a t ionary e l ec t r i c charges .

COULOMB'S LAW

I t i s f ound ex per im en ta l ly t ha t t he force be tween two s t a t io nary e l ec t r i c

point c h a r g e s Q a a n d Q b (a) acts a lon g the l ine join ing th e tw o cha rges, (b) i s

p r o p o r t i o n a l t o t h e p r o d u c t QaQ

b, an d ( c) i s i nver se ly p r op or t ion a l to the

square o f the d i s t ance r s e p a r a t i n g t h e c h a r g e s .

If t he charges a r e ex tende d , the s i tua t ion is m ore co mp l ica te d in tha t

the "d i s t ance be tween the charges" has no def in i t e mean ing . Moreover , t he

presence o f Q b can modi fy the charge d i s t r ibu t ion wi th in Q a, and vice versa,

l ead ing to a compl ica ted var i a t ion o f fo rce wi th d i s t ance .

W e t h u s h a v e Coulomb's law fo r s t a t ionary po in t charges :

F

"b

" 4 ^ ^ -r i

'( 2 _ 1 )

w h e r e F ah i s the force exer ted by Q a on Q b, a n d r t i s a uni t vector point ing in

the d i r ec t ion from Q a to Q h, as in Fig . 2-1. Th e force is repu lsive if Q a a n d

Q b are of the sam e sign ; it is a t t rac t iv e i f they are of di f ferent s igns. As u sual ,

F i s m e a s u r e d in n e w t o n s , Q in coulombs, a n d r i n mete r s . Th e con s tan t e 0

42

f a *

Figure 2-1 C h a r g e s Q a a n d Q b sepa ra t ed by a d i s t ance r. The force exerted on Q h by

Q a is F ah and is in the di rect ion of ^ a lon g the l ine joi nin g the two ch arge s .

i s the permittivity of free space:

e 0 = 8.854 187 82 x 1 < T1 2

f a r a d / m e t e r .+

(2-2)

Coulomb 's l aw app l i es to a pa i r o f po in t charges in a vacuum. I t

a l so app l i es in d i e l ec t r i cs and conduc to r s i f F a b i s the di rect force between

Q a a n d Q b, i r respe ct ive of the forces ar is ing f rom o the r cha rges wi thin the

m e d i u m .

Subs t i tu t ing the va lue o f e 0 i n t o C o u l o m b ' s l a w ,

F ^ 9 x l 0 # r i , (2-3)r

wh ere the factor of 9 i s ac cu rat e to ab ou t o ne pa r t in 1000; i t sho uld real ly

be 8.987 551 79.

T h e C o u l o m b f o rc e s in n a t u r e a r e e n o r m o u s w h e n c o m p a r e d w i t h

the g rav i t a t iona l fo rces . The g rav i t a t iona l fo rce be tween two masses m a

a n d m h separa ted by a d i s t ance r i s

F = 6.672 0 x \Q - llm am blr2. (2-4)

* F ro m Eq. 2 -1 , e 0 i s expres sed in cou lom bs squa re d pe r new ton- squa re me te r . Ho w

ever , a cou lom b squared pe r newton-m ete r , o r a cou lo mb s quare d pe r jou le , is a fa rad ,

as we sha ll see in Sec. 4.3.

2.3 The Principle of Superposition 43

The g rav i t a t iona l fo rce on a p ro ton a t t he su r f ace o f the sun (mass =

2.0 x 1 0 3 0 k i lograms, r ad ius = 7 .0 x 10 8 meters) i s equal to the elect r ic

fo rce be tween a p ro ton and one microgram of e l ec t ron s , separa ted by ad i s t ance equa l to the sun ' s r ad ius .

I t i s r em ark ab le indee d tha t we sho u ld no t be cons c ious o f these

en or m ou s fo rces in everyday life . Th e pos i t ive an d nega t ive charg es ca r r i ed

respect iv ely by the pr ot on an d the ele ct ro n are very near ly , if no t exact ly ,

the sam e. Exp er i m en ts ha ve sho wn t ha t they dif fer, if a t a ll , by at mo st on e

par t i n 10 2 2 . O r d i n a r y m a t t e r i s t h u s n e u t r a l , a n d t h e e n o r m o u s C o u l o m b

forces p reven t the acc um ula t ion o f any appre c iab le qu an t i t y o f charg e of

ei ther s ign.


W e th ink o f the fo rce be twe en the po in t charg es Q a a n d Q b i n C o u l o m b ' s

l aw as the p roduc t o f Q a a n d t h e field of Q b, or vice versa. We define the

electric field intensity E to be the force per uni t charge exer ted on a test

charge in the f ie ld . Thus the elect r ic f ie ld in tensi ty due to the point charge

Q a is

E

"=

7 r = 7 %r

> - <2

"5

>Qb 4 7 r e 0 r

Th e elect r ic field in ten si ty i s m ea su red in vol ts per m eter .

The elect r ic f ie ld in tensi ty due to the point charge Q a i s t he same,

whether the t es t charge Q h is in the field or not, even if Q b i s l a rg e c o m p a r e d

to Q„.

2.3 THE PRINCIPLE O F SUPERPOSITION

If t he e l ec t ri c f ie ld i s p ro du ce d by m or e tha n on e charge , each one p rod uce s

i ts ow n f ie ld , an d the res ul ta nt E is s imp ly the vector su m of the indiv idua l

E 's . This i s the principle of superposition.



44 Fields of Stationary Electric Char ges: I

THE FIELD OF AN EXTENDED

CHARGE DISTRIBUTION

F o r a n e x t e n d e d c h a r g e d i s t r i b u t i o n ,

E - J L f . / * * ' . (2-6)

Th e mea n in g o f the var io us t e rm s und er the in t egra l s ign is i l l us t r a t ed in

Fig. 2-2: p i s t he e l ec t r i c charge dens i ty a t t he source po in t (x',y',z')\ rj is

a uni t vector point ing f rom the source point to the f ie ld point (x ,y ,z) where

E i s ca l cu la t ed ; r i s t he d i s t ance be tween these two po in t s ; dr ' i s the e lem ent

o f v o l u m e dx' dy' dz'. I f the re exist sur face dis t r ib ut io ns of cha rge, the n we

must add a su r f ace in t egra l , wi th the vo lume charge dens i ty p r ep laced by

the su r f ace charge dens i ty a a n d t h e v o l u m e x' replaced by the sur face S' .

Th e e l ec t r i c cha rge dens i ty ins ide ma cro sco p ic bod ies i s o f cour se

no t a sm oo th funct ion of x ' , y', z'. However , nuc le i and e l ec t rons a r e so

smal l and so closely packed, that one can safely use an average elect r ic

charge dens i ty p, as above , to ca l cu la t e macroscop ic f i e lds .

When the e l ec t r i c f i e ld i s p roduced by a charge d i s t r ibu t ion tha t i s

dis tur be d by the int ro du ct i on of a f in ite test ch arg e Q', E is the force per unit

charge as the magn i tude o f the t es t charge Q' t e n d s t o z e r o :

E = lim (2-7)



2.5 The Electric Potential V 45

THE ELECTRIC POTENTIAL V

Consider a t e s t po in t charge Q' t ha t can- be mov ed a bo u t in an e l ec t r ic

f ie ld . The work W r equ i r ed to move i t a t a cons tan t speed f rom a po in t P t

t o a po in t P 2 along a given path is

The nega t ive s ign i s r equ i r ed to ob ta in the work done against the field.

H e r e a g a i n , w e a s s u m e t h a t Q' i s so smal l t ha t t he charg e d i s t r ibu t ions a r e

no t apprec iab ly d i s tu rbed by i t s p r esence .

I f the p ath is c losed , the tota l wo rk d on e is

Let us evaluate th is in tegral . To simpl i fy mat ters , we f i r s t consider the

elect r ic f ie ld produced by a s ingle point charge Q. T h e n

Figure 2 -3 shows tha t t he t e rm under the in t egra l on the r igh t i s s imply

dr/r2 o r — d(l/r). Th e sum of the inc rem en ts of (1/ r ) over a c losed pat h is

zero, s ince r has the same va lue a t t he beg inn ing and a t t he end o f the pa th .

Th en the l ine in t egra l i s ze ro , an d the ne t wo rk d on e in mo ving a p o in t

c h a r g e Q' ar ou nd a ny closed p ath in the field of a point cha rge Q, which

is fixed, is zero.

I f the elect r ic f ie ld is produced, not by a s ingle point charge Q, b u t

by some f ixed charge d i s t r ibu t ion , t hen the l ine in t egra l s co r r espond ing

to each ind iv idua l charge o f the d i s t r ibu t ion a r e a l l ze ro . Thus , f o r any

d i s t r ib u t ion o f f ixed charg es ,

(2-8)

(2-9)

(2-10)

(2-11)

An electrostatic field is therefore conservative (Sec. 1.11). I t can be

shown tha t t h i s impor tan t p roper ty fo l lows f rom the so le f ac t t ha t t he

Coulomb fo rce i s a cen t r a l f o r ce ( see P rob . 1-21).



46

Then , f rom S tokes ' s t heorem (Sec . 1.13), at a l l points in space,

V x E = 0. (2-12)

and we can wr i t e tha t

E = -W , (2-13)

w h e r e V i s a scalar poi nt funct ion, s ince V x \V = 0.

We can thus desc r ibe an e l ec t ros t a t i c f i e ld comple te ly by means o f

the funct ion V(x,y,z), which is cal led the electric potential. T h e n e g a t i v e

sign is req uire d in ord er tha t the elect r ic field in tens i ty E poi nt to w ar d a

decrease i n po ten t i a l , accord ing to the usua l conven t ion .

I t i s impor tan t to no te tha t V i s no t un ique ly def ined ; we can ad d to

i t any qua n t i ty th a t is i nde pen den t o f the coor d in a tes w i tho u t a f fec ting E

in any way .

As sh ow n in Pro b. 1-23, the work d on e in mov ing a test ch arg e at acons tan t speed f rom a po in t P { t o a po in t P 2 i s i nde pen den t o f the pa th .

W e must r em em ber th a t we a r e dea l ing here wi th e l ec t rosf a f /cs . I f

the re were moving charges p resen t , V x E would no t necessa r i ly be ze ro ,

a n d W wo uld th en des cr ibe only pa r t of the elect r ic f ie ld in ten si ty E. W e

sha l l i nves t iga te these more compl ica ted phenomena l a t e r on , i n Chap te r 11 .



47

According to Eq. 2-13,

E d l = - V K - d l = —cIV. (2-14)

Then, for any two points Px

and P2

as in Fig. 2-4,

V2

- V, = - J ^ E d l = £ E d l . (2 -15)

Note that the electric field intensity E(x,y,z) determines only the

difference betw een the pote nti als at tw o different poin ts. When we wish to

speak of the electric pote nti al at a given point, we mus t therefore arbi trar ily

define the potential in a given region of space to be zero. It is usually con

venient to choose the potential at infinity to be zero. Then the potential V

at the point 2 is

V = f" E d l . (2-16)

The work W required to bring a charge Q' from a point at which the

pote ntial is defined to be zero to the point co nside red is VQ'. Thus V is

WjQ' and can be defined to be the work per unit charge. The potential V

is expressed in jou les per co ul omb , or in volts.



48

Figure 2-5 Equipotent ia l surface and l ines of force near a point charge.

When the f i e ld i s p roduced by a s ing le po in t charge Q, t h e p o t e n t i a l

a t a d i s t ance r is

I t wi l l be observed that the sign of the potent ia l V i s the same as that of Q.

The p r inc ip le o f superpos i t ion app l i es to the e l ec t r i c po ten t i a l V as

wel l as to the elect r ic f ie ld in tensi ty E. The potent ia l V a t a po in t P due to

a charge d i s t r ibu t ion o f dens i ty p i s therefore

w h e r e r i s t he d i s t ance be tween the po in t P and the e l ement o f charge p dx'.

The points in space that are a t a given potent ia l def ine an equipotential

surface. For example , an equ ipo ten t i a l su r f ace abou t a po in t charge i s a

con ce ntr i c sph ere as in Fig . 2-5. W e can see from Eq. 2-13 tha t E is every

where normal to the equipotent ia l sur faces (Sec. 1 .5) .

(2-17)

(2-18)

2.5 The Electric Potential V 49

If we join end -to- end infinitesimal vectors rep resen ting E, we get a

curve in space—called a line of force—that is every whe re no rm al to the

equ ipot enti al surfaces, as in Fig. 2-5. The vecto r E is everyw here ta ngen t to

a line of force.

2.5.1 EXAMPLE: THE ELECTRIC DIPOLE

T h e electric ciipole s h o w n in Fig. 2-6 is one t ype of cha rge d i s t r ibu t ion tha t is en

counte red f requent ly . We sha l l r e tu rn to it in C h a p t e r 6.

The e l ec t r i c d ipo le cons i s t s of two cha rges , one pos i t ive and the othe r nega t ive ,

of th e s a m e m a g n i t u d e , s e p a r a t e d by a di s t ance s. We shal l calcula te V and E at a

di s t ance r t h a t is l a r g e c o m p a r e d to s.

A t P,

+ Q<

s

-Q

(2-19)

Figure 2-6 Th e two c h a r g e s + Q and — Q form a dipole . The e lec t r ic po ten t i a l at P

is the sum of t he po ten t i a l s due to the i nd iv idua l cha rges . The vec tor s points from

— Q to +Q, an d r, is a uni t vec tor po in t ing f rom the origin to the p o i n t of

o b s e r v a t i o n P.

50 Fields of Stationary Electric Charges: I

w h e r e

Similar ly,

a n d

rz = r + - + rs cos 0

, s y s1 + ( — + - cos 0~lr r

r I 1 + - cos 0r

1 + - cos 0r

1 - — cos 0.*2r

r

rb

V =

1 + — cos 9,2r

4n e 0r2

cos 0.

(2-20)

(2-21)

(2-22)

(2-23)

(2-24)

(2-25)

(2-26)

This express ion is val id for r 3 » s 3 .

I t i s interes t ing to note that the potent ia l due to a dipole fa l ls off as 1/ r 2, where

as the potent ia l f rom a s ingle point charge varies only as 1/ r . This comes from the

* R e m e m b e r t h a t

n(n - 1) , n(n - l)(n - 2) ,(1 + df = 1 + na + —r.—- a 1 + - £ a 3 +

2 ' 3!

If n i s a pos i t ive integ er , the ser ies s tops wh en t he coeff ic ient i s zero . Try, for ex am ple ,

n = 1 and n = 2. If n i s not a pos i t ive integer , the ser ies converges for a 2 < 1.

This series is very often used when a « 1. Then

(1 + a)" * 1 + na .

F o r e x a m p l e .

(1 + a) 2 * 1 + 2a . (1 + aC " 2 « 1 - - , e tc .



51

Figure 2-7 Lines of force (arrows) and equipotent ia l l ines for the dipole of Fig. 2-6.

The dipole is vert ical , a t the center of the f igure , wi th the pos i t ive charge c lose to

and above the nega t ive cha rge . In the cen t ra l r eg ion the l i nes come too c lose

toge the r to be shown.

fact that the charges of a dipole appear c lose together for an observer some dis

tance away, and that thei r f ie lds cancel more and more as the dis tance r increases .

We define the dipole moment p = Qs as a vec tor whose magni tude i s Qs a n dthat i s di rected from t he nega t ive to t he pos i t ive cha rge . Then

V = - -2. (2-27)

Figure 2-7 shows equipotent ia l l ines for an e lect r ic dipole . Equipotent ia l sur

faces a re gene ra t ed by ro t a t ing equipo ten t i a l l i nes a round the ve r t i ca l ax i s .

Let us now calculate E a t the point P(x.O.r) in the plane y = 0 as in Fig. 2-6.

We have tha t

p cos 0 p :

K

= 4 ?, 2

"2 8 )

and. a t P( . \ ,0 , r) ,

cV 3p xz 3r> sin 0 co s 0

£ = = — ? - — = —! , ( 2- 29 )

cx 4ne 0 r 5 4n e 0 r3

- 9 V 3 P y : = 0. ,2-30)dy 4ne 0 r

since y = 0, and

8V _ p (1 3 -2

<3z 4 7 i e 0 \

p 3 c o s 2 f) — 1

E r = - — = -y— I - 3 j - , ( 2- 31 )

dz 4ne0

\r r 5

(2-32)4n e 0 r 3

Figure 2-7 shows l ines offeree for the e lect r ic dipole .

SUMMARY

I t i s found empir ical ly that the force exer ted by a po in t charge Q a on a

po in t charge Q b is

4 n e 0 r 2

whe re /• i s t he d i s t an ce b e tween the charg es an d r, i s a un i t vec to r po in t ing

from Q a t o Q b. This i s Coulomb's law. We cons ider the fo rce ¥ a b as be ing thep r o d u c t o f Q b by the electric field intensity due to Q a,

E ^ T ^ S r , , (2-5)

or vice versa.



Problems 53

According to the principle of superposition, two or more electric

field intensities acting at a given point add vectorially. For an extended

charge distribution,

E =1

cm )4ne 0 J*' r

2

The electrostatic field is conservative,

V x E = 0, (2-72)

hence

E = -\V, (2-13)

where

1 cp dx'

4 7 T € A Jz

rne0

is the electric potential; p dx' is the element of cha rge cont ain ed within the

element of volume dx', and r is the distan ce betwee n this elemen t and t he

point where V is calculated.

PROBLEMS

2-JE 1 COULOMB'S LAW

a ) Ca lcu la t e th e elect r ic f ie ld intens i ty that would be just sufficient to b a l a n c e

the gravi ta t ional force of t he ea r th on an e lec t ron .

b) If this e lect r ic f ie ld were produced by a second e l ec t ron loca ted be low th e

f i rs t one, what would be the di s t ance be tween the two e lec t rons?

T h e c h a r g e on an e lec t ron is —1.6 x 1 0 "

1 9

c o u l o m b and its m a s s is 9.1 x1 0 " 3 1 k i l o g r a m .

2-2E SEPARATION OF PHOSPHATE FROM QUARTZ

C r u s h e d F l o r i d a p h o s p h a t e ore cons i s t s of part ic les of q u a r t z m i x e d w i t h

part ic les of p h o s p h a t e r o c k . If the m i x t u r e is vibra t ed , the q u a r t z b e c o m e s c h a r g e d

+

The le t ter E i nd ica t e s tha t th e p r o b l e m is re la t ively easy.



54

Figure 2-8 Elec t ros t a t i c s epa ra t ion of pho spha te f rom qu ar t z in c rushed pho sph a te

ore . The vibrat ing feeder VF cha rges the phospha te pa r t i c l e s pos i t ive ly , and the

quartz part ic les negat ively. See Prob. 2-2.

nega t ively and the pho sph a te pos i t ively . The ph osp ha te can then be s epa ra t ed ou t

as in Fig. 2-8.

Ov er what m ini mu m d is tanc e must the part ic les fa ll if they must be sep arat edby at leas t 100 mil l imeters?

Set £ = 5 x 1 0 5 vol ts per meter and the specif ic charge equal to 10" 5 c o u l o m b

per k i logram.

2-3E ELECTRIC FIELD INTENSITY

A charge + Q is situ ate d at x = a, y = 0, an d a cha rge — Q at x = — a, y = 0.

Ca lcul ate the e lect r ic field intens i ty a t the po int x = 0, y = a.



Problems 55

2-4 ELECTRIC FIELD INTENSITY

A c i rcu la r d i sk of cha rge has a rad ius R and carr ies a surface charge dens i ty a.

Find the e lect r ic f ie ld intens i ty £ a t a point P on the axis a t a dis tance a.

What i s t he va lue of E w h e n a « Rl

You can solve this problem by calculat ing the f ie ld a t P due to a r ing of charge

of radiu s /• an d width dr , and th en integ rat in g from r = 0 to r = R.

CATHODE-RAY TUBE

Th e elect ron be am in the cat ho de- ray t ub e of an osci l loscope is deflected vert ical ly

and horizontal ly by two pairs of paral le l deflect ing pla tes that are mainta ined a t ap

propr i a t e vo l t ages . As the e l ec t rons pas s be tween one pa i r , t hey a re acce le ra t ed by the

elect r ic f ie ld, and their kinet ic energy increases . Thus , under s teady-s ta te condi t ions ,

we achieve an increase in the kinet ic energ y of the e lect ron s wi thou t any exp end i ture

of power in the deflect ing pla tes , as long as the beam does not touch the pla tes .

Co uld th i s ph eno me no n be used a s the bas i s o f a pe rpe tua l mo t ion ma chin e?

2-6E CATHODE-RAY TUBE

In a certa in cathode-ray tube, the e lect rons are accelerated under a di fference

of pote nt ia l of 5 ki lov ol ts . After being accelera ted, they t ravel horizo ntal ly ov er a

dis tance of 200 mil l imeters .

Ca lcu la t e the downward de f l ec t ion ove r th i s d i s t ance caused by the grav i t a

t ional force .

An e l ec t ron ca r r i e s a cha rge of —1.6 x 1 0 "1 9

coulomb and has a mass o f

9 .1 x 1 0 " 3 1 k i l o g r a m .

2-7E MACROSCOPIC PARTICLE GUN

I t i s poss ible to obtain a beam of f ine part ic les in the fol lowing manner. Figure 2-9shows a paral le l -pla te capaci tor wi th a hole in the upper pla te . I f a part ic le of dus t ,

say, i s int roduced in the space between the pla tes , i t sooner or la ter comes into contact

with one of the pla tes and acquires a charge of the same polari ty. I t then f l ies over to

the oppos i t e p l a t e , and the proces s repea t s itself. The part ic le osci l la tes back and forth,

unt i l i t i s los t e i ther a t the edges or thro ug h th e centra l hole . O ne can th us obta in a

beam emerg ing f rom the ho le by admi t t ing pa r t i c l e s s t ead i ly in to the capac i to r . In

order to achieve high veloci t ies , the gun must , of course , operate in a vacuum.

Beams of macroscopic pa r t i c l e s a re used for s tudying the impac t o f mic ro-

m e t e o r i t e s .

Now i t has been found that a spherical part ic le of radius R lying on a charged

pla t e acqui re s a cha rge

Q = 1.65 x 4n ere

0R 2E

0 c o u l o m b ,

where £ 0 i s the e lect r ic f ie ld intens i ty in the absence of the part ic le , and e r i s the re la t ive

perm it t ivi ty of the p art ic le (Sec. 6.7) . W e assu me that t he pa rt ic le is a t leas t s l ight ly

c o n d u c t i n g .

Ass um ing that the pla tes are 10 mil l im eters apar t and tha t a vol tag e di fference

of 15 ki lovo l ts i s app l ied betw een t hem , f ind the veloci ty of a spherical part ic le 1



56

Figure 2-9 Device for p roduc ing abeam of f ine part ic les . See Prob. 2-7

mic romete r in d i amete r a s i t emerges f rom the ho le in the upper p l a t e . Assume tha t

the part ic le has the dens i ty of water and that e r = 2 .

2-8E ELECTROSTATIC SPRAYING

Wh en pa in t ing i s don e wi th an ord in a ry spra y gun , pa r t o f t he pa in t e s capes

depos i t ion. The fract ion los t depends on the shape of the surface, on drafts , e tc . , and

can be as high as 80%. Th e use of the ordi nar y spr ay gun in large-scale indu s tr ia l pr o

cesses would therefore resul t in intolerable was te and pol lut ion.

Th e efficiency of spray p aint in g can be increas ed to nearly 100%, and th e po l lu

t ion reduced by a large factor , by charging the droplets of paint , e lect r ical ly, and

applying a vol tage di fference between the gun and the object to be coated.

I t i s found that , in such devices , the droplets carry a specif ic charge of roughly

o n e c o u l o m b p e r k i lo g r a m .

Assuming that the e lect r ic f ie ld intens i ty in the region between the gun and the

par t i s a t leas t 10 ki lov ol ts per mete r , wha t i s the m in im um rat i o of the e lect r ic force

to the grav i t a t ion a l fo rce?

2-9 THE RUTHERFORD EXPERIMENT

In 1906, i n the course of a h i s to r i c expe r im ent th a t de mo ns t ra t ed the sma l l

s ize of the a to mi c nucleus , Rut herfo rd obse rved that an a lph a part ic le (Q = 2 x

1.6 x 1 0 " 1 9 coulomb) wi th a kinet ic energy of 7.68 x 10 6 e lec t ron-vol t s making a

head-on col l is ion with a gold nucleus (Q = 79 x 1.6 x 1 0 " 1 9 coulomb) i s repe l l edb a c k w a r d s .

T h e electron-volt i s t he k ine t i c ene rgy acqui red by a pa r t i c l e ca r ry ing one

elect r onic charg e when i t i s accele rated th rou gh a di fference of pot ent i a l of on e vo l t :

1 e l ec t ron-vol t = 1.6 x 1 0 " 1 9 j o u l e .

a ) W ha t i s t he d i s t ance of c losest appr oac h a t which the e l ec t ros t a t i c po ten t i a l

energy is equa l to the ini t ia l kinet ic energy ? Expres s yo ur resul t in fem tom eters

( 1 0 "1 5 mete r ) .

5 7

Figure 2-10 Cyl indr i ca l e l ec t ros t a t i c ana lyse r . A pa r t i c l e P can reach the col lector

C only i f i t i s pos i t ive an d if i t sa t is f ies the equ at io n given in Pro b. 2-11 . Oth erw ise ,

i t s tops on one of the cyl indrical e lect rodes . For a given charge-to-mass ra t io Q/m,

one can select di fferent veloci t ies by changing V.

A circle with the letter I represents an amm eter, while a circle with a letter V

represents a voltmeter.

b) W ha t i s t he ma xim um force of repu l s io n?

c) W hat is the max im um accele rat ion in </"s?

Th e mass of the a lpha part ic le is ab ou t 4 t imes that of a prot on , or 4 x 1.7 x

1 0 "2 7 k i l o g r a m .

ELECTROSTATIC SEED SORTING

No rm al s eeds can be s epa ra t ed f rom d i s co lored on es and f rom fore ign ob jec t s

by means of an e l ec t ros t a t i c s eed-sor t ing machine tha t ope ra t e s a s fo l lows . The s eeds

are observed by a pair of photocel ls as they fa l l one by one ins ide a tube. I f the color

is not r ight , vol ta ge is app l ied to a needle tha t dep os i ts a charge on th e seed. Th e seeds

then fa ll between a pair of e lect r ical ly cha rge d pla te s tha t deflect the undes ired one s

in to a s epa ra t e b in .

On e such ma chin e can sor t d ry peas a t t he ra t e o f 100 pe r s econd , o r a bou t

2 tons pe r 24-hour day .

a) If the seeds are dr op pe d at the ra te of 100 per second, over wha t d is tanc e

mu st they fal l if they m ust be sepa rate d by 20 mil l imeters whe n they pass between the

photoce l l s ? Neglec t a i r r e s i s t ance .

b) Assuming that the seeds acquire a charge of 1.5 x 10" 9 c o u l o m b , t h a t t h e

deflect ing pla tes are paral le l an d 50 mil l im eters apar t , and tha t the pote nt ia l di fference




be tween them i s 25 k i lovol t s , how fa r should the p l a t e s ex tend be low the cha rg ing

needle i f the charg ed seeds mu st be deflected by 40 mil l im eters on leaving th e p la tes ?

Assume that the charging needle and the top of the deflect ing pla tes are c loseto the photoce l l .

2-1 IE CYLINDRICAL ELECTROSTATIC ANALYSER

Figure 2-10 shows a cyl indrical e lect ros ta t ic analyser or veloci ty se lector . I t con

s is ts of a pair of cyl indrical conductors separated by a radia l dis tance of a few mil l imeters .

Sho w that , if the radia l dis tance betw een the cyl indrica l surfaces of avera ge

r a d i u s R is a, an d if the vol tag e betwe en the m is V, then the part ic les col lected a t /

have a veloci ty

(Q VRV12

\m a J

PARALLEL-PLATE ANALYSER

Figure 2-11 shows a pa ra l l e l -p l a t e ana lyse r . The ins t rument s e rves to measure the

energy d is t r ib ut io n of the cha rged part ic les emit te d by a sou rce. I t i s of ten u sed

for sources of e lect rons . I t has been used, for example , to inves t igate the energy

dis t r ibu t ion of e lec t ron s in the pla sm a formed at the focus of a lens i l lum inated by

a powerful laser .

If the source emits part ic les of various energies , then, for a given value of V

and for a given geometry, the col lector receives only those part ic les that have the

Figure 2-11 Para l l e l -p l a t e ana lyse r fo r measur ing the ene rgy d i s t r ibu t ion of

pa r t i c l e s emerg ing f rom a source S. The part ic les fol low a bal l is t ic t ra jectory in the

uniform elect r ic f ie ld between the two pla tes . Part ic les of the correct energy and, of

cou rse , of the correc t pola r i ty are col lected a t C. Th e othe r part ic les hi t e i ther o ne

of the two pa ra l l e l p l a t e s . The po la r i t i e s shown apply to e l ec t rons or to nega t ive

ions . See Prob. 2-12.



Problems 59

corresponding energy. O ne therefore observes th e energy spectrum of the particles

by measuring th e collector current as a function of V. As we shall see below, we

assume that th e particles all carry th e same charge.In th e figure, particles of charge Q have a kinetic energy QVQ.

This type of analyser is relatively simple to build. It can also be, at th e same time,

compact and accurate. For example, in one case, th e electrodes measured 40 mm x

115 mm, with b = 15 mm and a = 50 mm. With slit widths of 0.25 mm, the de

tector current as a function of V for mono-energetic electrons gave a sharp peak as

in Fig. 2-12 having a width at half maximum of only 1%.

a) Find a for a particle of mass m and charge Q.

Solution: We can treat the particle as a projectile of mass m, charge Q, and

initial velocity vQ. with

1\ , (1)

V2

, = QV0,

r0J

2

^) . (2,in

The particle has a vertical acceleration, downwards, equal to QE/m, or

Q(V/b)

Then th e time of flight, from slit to slit, is

Do sin 8 2v r.mb7 = 2 — = — — s i n f l , (3)

Q(V/b)/m QV

an d th e distance a is given by

2vhnba = ( r n co s 8)T = sin 8 co s 0, (4)

0

QV

AbV 0 2bV 0

= sin 6 co s 8 = sin 28. (5)V V

Note that, for a given value of V0, a is independent of both m and Q. It is propor

tional to V0. If all the particles carry th e same charge Q, a is proportional to their

energy QV 0.

b) The optimum value of 0 may be defined as that for which th e horizontal

distance a traveled by the particle is maximum. One can then tolerate a slightly

divergent beam at the entrance slit, with an average 8 of 0 o p I .

Find ()„„,.



60

Figure 2-12 The c u r r e n t / at the

col l ec tor C of Fig. 2 - 1 1 , as a funct ion

of V.

-AV

2±

Solution: Since th e di s t ance a m u s t be m a x i m u m , da/dO m u s t be ze ro . N ow

si n 20 has a m a x i m u m at 0 = 45 , so 6op, = 45°.

Note that , s ince a is m a x i m u m at 0 = 45 this is a l so the c o n d i t i o n for m a x i m u m

reso lu t ion .

c) Now find V0 as a funct ion of V, for given values of a and b, at 0opl.

Solution: F r o m Eq. 5, wi th 0 = 45°,

V0

2bV. (6)

d) F ind da/dV0 for 0 = 0opl.

Solution: At 0 = fl„ n,.

da

'IK,

2b

~V'

(7)

e) Find th e m i n i m u m v a l u e of V as a funct ion of V0 at fl opl.

Solution: In all the a b o v e e q u a t i o n s . V is a lways d iv ided by b. To find V we

use th e fact that V m u s t be sufficient to reduce th e vert ical veloci ty to z e r o :

QV > - m(v0 sin 4 5c

)2

, (8)

V > m 2QV0

1 V0

2Q m 2(9)

C o m p a r i n g now wi th Eq. 6, we see t h a t alb m u s t be smal l e r t han 4.

This ana lyse r is prac t i ca l for ene rg ies QK 0less than, say, 10 ki loe lec t ron-vol t s .

I t would be i nconvenien t to use vol t ages V l a rge r than a few ki lovol t s . For highe r

ene rg ies one w o u l d use the ana lyse r of P r o b . 2-11.



Problems 61

2-13 CYLINDRICAL AND PARALLEL-PLATE ANALYSERS COMPARED

Show tha t bo th the cy l indr i ca l ana lyse r o f P rob . 2 -11 and the pa ra l l e l -p l a t e

ana lyse r o f P rob . 2 -12 measure the ra t io

w h e r e m i s the mass of the part ic le , v i t s veloci ty, and Q i t s charge.

2-14 ION THRUSTER

Synchronous satellites desc r ibe c i rcu la r o rb i t s above the equa tor a t t he a l t i t ude

where thei r angular veloci ty is equal to that of the earth. Since these sate l l i tes are

ex t remely cos t ly to bu i ld and to l aunch , t hey mus t rema in ope ra t iona l fo r s eve ra l

years . However, i t i s imposs ible to adjus t thei r ini t ia l pos i t ion and veloci ty wi th suff i

c ient accuracy to mainta in them fixed with respect to the earth for long periods oft ime . Moreover , t hey a re pe r tu rbed by the moon, and the i r an tennas mus t rema in

cons tan t ly d i rec t ed towards the ea r th .

Synchronous s a t e l l i t e s a re the re fore equipped wi th thrusters whose func t ion

is to exert the smal l forces necessary to keep them properly oriented and on their

presc r ibed orb i t s . Even then , t hey keep wander ing about the i r r e fe rence pos i t ion . For

example , t he i r a l t i t ude can va ry by a s much as 15 k i lomete rs .

The th rus t i s m'v, w h e r e m i s the mass of propel lant e jected per uni t t ime and v

i s the exhaus t veloci ty wi th respect to the sa te l l i te . Clearly, m should be sma l l . Then

v should be large. However, as we shal l see , too large a r could lead to an excess ive power

consumpt ion . So the choice of ve loc i ty depends on the maximum a l lowable va lue for m'

and on the ava i l ab le power .

One way of achieving large values of v i s to e ject a beam of charged part ic les ,

as in Fig. 2-13. Th e pro pel l ant i s ionized in an ion sou rce and e jected as a p os i t ive

ion beam.

By itself, the beam cannot exert a thrus t , for the fol lowing reason. Ini t ia l ly, the

net charge on the sate l l i te i s zero. When i t e jects pos i t ive part ic les , i t acquires a negat ive

charge. The part ic les are thus held back by the e lect ros ta t ic force , and, once the

satelli te is sufficiently ch arg ed , they fall ba ck o n the satelli te.

To achieve a thrus t , the ions must be neutra l ized on leaving the sate l l i te . This

is achiev ed by m ea ns of a heated f ilament, as in Fig. 2-13. Th e f i lament emits e lect ro ns

tha t a re a t t rac t ed by the pos i t ive ly cha rged beam.

a) Th e curre nt / of pos i t ive ions in the be am is carr ie d by part ic les of ma ss m

a n d c h a r g e ne , w h e r e e i s t he e l ec t ron ic cha rge .

(1 2)mv 2

Q

Show tha t t he th rus t i s

b) W ha t i s t he va lue o f f fo r a 0 .1 ampe re beam of p ro ton s with V = 5 0 k i l o v o l t s ?



62

S

Figure 2-13 Sch em at ic diagr am of an ion thru s ter . The prop el la nt i s adm it ted a t P

and ionized in S: A is a beam-s hapin g e l ec t rode and B i s t he acce le ra t ing e l ec t rode ,

ma in ta ined a t a vo l t age V with respect to S; t he hea ted f i l ament F injects e lect rons

in to the ion beam to make i t neu t ra l . E lec t rode B i s part of the outer skin of the

sate l l i te . See Prob. 2-14.

c) If we call P t h e p o w e r IV spent in acce le ra t ing the beam, show tha t

Note tha t , s ince F = (2m'P)1 2

, for given values of m and P, the thrus t i s in

dependent o f t he cha rge - to-mass ra t io o f t he ions .

Also, s ince F is 2P/v, the thrus t i s inversely p r o p o r t i o n a l t o v, for a given pow er

e x p e n d i t u r e P.

Final ly, the las t express ion shows that , again for a given P, i t is preferable to

use heavy io ns carryin g a s ingle char ge (» = 1) an d to use a low accelera t ing vo l tage V.

d) If the f i lament is lef t unh eate d a nd if the be am curr ent i s 0.1 amp ere , ho w

long does i t take the body of the sa te l l i te to a t ta in a vol tage of 50 ki lovol ts?

Assu me tha t the sa te l l i te is spherica l an d tha t i t has a rad ius of 1 m e t e r .

COLLOID THRUSTER

Before reading this solved problem you should f i rs t read Prob. 2-14.

F igure 2-14 shows a so-ca l l ed co l lo id th rus t e r . A conduc t ing f lu id i s pumped

into a hol low needle mainta ined a t a high pos i t ive vol tage wi th respect to the

sa te l li t e 's ou te r sk in . Mic ro scopic d ro p le t s fo rm a ro un d the edge of the op ening ,

where the e lect r ic f ie ld intens i ty and the surface charge dens i ty are extremely high.



63

Figure 2-14 Col lo id th rus t e r . The

hol low needle TV ejects a beam of

pos i t ively charged, high-veloci ty

drople t s . E lec t rode B and f i l ament F

are as in Fig. 2-13. See Prob. 2-15.

These d rop le t s , ca r ry ing spec i fi c cha rges Q/m as large as 4 x 10 4 coulombs pe r k i lo

gra m, are acceler ated in the e lect r ic field and form a high-en ergy je t . As a rule , several

need les a re g rouped toge the r so a s to ob ta in a l a rge r th rus t .

a) Calculate the thrus t exerted by a s ingle needle for the specif ic charge given

above , when V = 10 k i lovol t s and / = 5 mic r oam peres .

Use the resul ts of Prob. 2-14.

Solution:

'2m

QV.

1/2

P. I D

1/2

1 0 4 x 5 x 1 0 " 6 , (2)V4 x 10* x 1 0 4

= 3 .5 mic rone wto ns . (3)

b) C alcu late the ma ss of f luid e jected per seco nd.

Solution: T h e m a s s in ejected per second is m/Q t imes the charge e jected per

second , which is 5 x 1 0 " 6 c o u l o m b :

tri = 5 x 10 6 ' 4 x 1 0 4 = 1.3 x 1 0 " 1 0 ki logram/second . (4 )

c) The thrus t i s so smal l that i t i s not pract ical to measure i t di rect ly. The

me tho d used is i l lus t ra ted in Fig. 2-15. Th e thru s ter i s f irst put i nto op era t ion .

Then swi t ch S i s open ed an d the vo l t age IR ac ros s R i s obse rved on an osci l losc ope,

as a funct ion of the t ime. Th e curve has the shap e show n in Fig. 2-16a.

Calculate the thrus t as a funct ion of V, / , D, T.

Solution: The th rus t i s

) = 2IV , (5)v

T V= 21V- = 2 — IT .

D D(6)



Figure 2-15 Method used for measur ing the th rus t o f t he co l lo id th rus t e r o f F ig .2-14 . The cur ren t / p rod uce d by the cha rged d rop le t s co l l ec ted on the hem ispher i ca l

e lect rode gives a vol tage IR that can be observed as a funct ion of the t ime on an

osc i l loscope .

d) If there are several types of droplets , the curve of J as a funct ion of t h a s

several pla teaus as in Fig. 2-16b.

How can one ca l cu la t e the th rus t i n th i s case?

Solution: In the las t express ion for F in Eq. 6, we see that F is 2V/D t imes the

area under the curve of Fig. 2-16a. Thus , in the case of Fig. 2-16b,

F = 2-[m

Idt. (7)D J °

T t t

(a) (b)

Figure 2-16 (a ) Current / as a funct ion of the t ime in the se t -up of Fig. 2-15, s tar t ing

at the ins tant when switch S i s opened, when al l the droplets are of the same s ize

and carry equal charges , (b) If there are several types of droplets of di fferent s izes

or charges , the curve of / as a funct ion of t has one p la t eau for each type .



CHAPTER 3

FIELDS OF STATIONARY

ELECTRIC CHARGES: II

Gauss's Law, Poissons and Laplace's Equations,

Uniqueness Theorem

3.1 SOLID ANGLES

3.1.1 The Angle Subtended by a Curve at a Point

3.1.2 The Solid Angle Subtended by a Closed Surface at a Point

3.2 GAUSS'S LAW

3.2.1 Examp le: Infinite Sheet of Charge

3.2.2 Examp le: Spherical Charge

3.3 CONDUCTORS

3.3.1 Examp le: Hollow Conductor

3.3.2 Examp le: Isolated Charged Conducting Plate

3.3.3 Examp le: Pair of Parallel Conducting Plates Carrying

Charges of Equal Magnitudes and Opposite Signs

3.4 POISSON'S EQUATION

3.4.1 Examp le: Flat Ion Beam

3.5 LAPLACE'S EQUATION

3.6 THE UNIQUEN ESS THEORE M

3.7 IMAGES

3.7.1 Examp le: Point Charge Near an Infinite Grou nded

Conducting Plate

3.8 SUMMARY

PROBLEMS



No w th a t we have d i scussed the two fun dam enta l conc ep t s o f e l ec t r i c fie ld

in tens i ty E and e l ec t r i c po ten t i a l V, we sha l l s tudy Gauss ' s l aw, Po i sson ' sequa t ion , and the un iqueness theorem. Th i s wi l l g ive us th r ee power fu l

methods fo r ca l cu la t ing e l ec t ros t a t i c f i e lds .

But f i r s t we must s tudy sol id angles.

3.1 SOLID ANG LES

3.1.1 THE ANGLE SUBTEND ED BY A CURVE AT A POINT

Co nsid er the cu rve C an d the po in t P o f F ig . 3 -1 . The y a r e s i tua ted in a

plane. We wish to f ind the angle oc sub tended by C a t P . Th i s i s

The in t egra l s a r e eva lua ted over the cu rve C: fo r the second one , we mul t ip ly

each element of length dl by sinW and div ide b y r; t hen we sum the r esu l t s .

Th e a r c c sub tend s the same ang le a a t P .

I f no w th e curv e is c lose d, as in Fig . 3-2a, wi th P insid e C, the anglesubtended at P by the smal l c i rcle of radius b is 27rb/b , or 2n r a d i a n s . T h e

ang le sub tended by C a t an y inside poin t P is there fore also 2n r a d i a n s .

I f P is outside C, as in Fig . 3-2b, then the segments C x a n d C 2 s u b t e n d

ang les o f the sam e ma gn i tud e bu t o f opp os i t e s igns , s ince s in 0 i s posi t ive

o n C t and nega t ive on C 2 . Thus the ang le sub tended by C a t an y o u t s i d e

po in t i s ze ro .

(3-1)



67

Figure 3-1 A c u r v e C and a p o i n t P s i tua t ed in a plane . The a n g l e s u b t e n d e d by C

at P is a. Th e e lement dl' is the c o m p o n e n t of dl in the di rec t ion pe rpendicu la r to

t he rad ius vec tor .

Figure 3-2 [a ) Closed curve C and a p o i n t P s i tuated ins ide . The a n g l e s u b t e n d e d by

C at P is the s a m e as t h a t s u b t e n d e d by the smal l c i rc le , namely 2n r a d i a n s .

(/)) When th e p o i n t P is o u t s i d e C. th e angle sub tended by C at P is ze ro because

segment s Cj on the r ight and C2 on the left of t he po in t s where th e t a n g e n t s to

t he curve go t h r o u g h P subtend equa l and oppos i t e ang les .

3.1.2 THE SOLID ANGLE SUBTENDED BY A CLOSED

SURFACE AT A POINT

Fig ure 3-3 sh ow s a closed surface S an d a poi nt P situate d inside. Th e small

cone intersects an element of area da on S. This small cone defines the solid

angle subtended by da at P.



68

Figure 3-3 A closed surface S and a po in t P s i tuated ins ide . Th e cone defines the

so l id angle sub tended by the e l ement o f a rea d a at P. The so l id angle sub tended a t

P by the complete surface S i s the same as that subtended by the smal l sphere of

r a d i u s b, n a m e l y 4JI s t e r a d i a n s .

By def ini t ion, th is sol id ang le is the a rea da , pro jec ted on a p l ane

perpend icu la r to the r ad ius vec to r , and d iv ided by r 2:

dQ = da cos 9 da (3-2)

w h e r e 0 i s t he ang le be tween the r ad ius vec to r r a n d t h e v e c t o r da , n o r m a l

to the su r f ace and po in t ing ou tward .

Sol id angles are expressed in steradians.

The to t a l so l id ang le sub tended by S a t P is

Qda cos 0

(3-3)

Th i s so l id ang le is a l so the so l id ang le sub te nde d a t P by the smal l sphere

s of radius a. Thus the so l id ang le sub tended by S a t an y inside point i s

Q = Ana 2/a 2 = An s t e r a d i a n s . (3-4)

N ote th a t a so l id ang le i s d imen s ion less , l i ke an o rd ina ry ang le .



I f the point P i s s i tuated outside the sur face as in Fig . 3-4, then r j • da

i s posi t ive over S Y a n d n e g a t i v e o v e r S 2, and the to t a l so l id ang le sub tended

a t an y ou t s ide po in t i s ze ro .

The s i tua t ion r em ain s unch an ge d if t he su rf ace is con vo lu ted in such

a way that a l ine drawn f rom P cu t s the su r f ace a t more than one po in t . The

total sol id angle subtended by a c losed surface is s t i l l 4n at an inside point ,

and ze ro a t an ou t s ide po in t .

GAUSS'S LAW

G au ss 's law relates the flux of E th ro ug h a closed surface to the total c har ge

enc losed wi th in the su r f ace .

By using this law one can f ind the elect r ic f ie ld of s imple charge

d i s t r i bu t io ns wi th a m in i m um of e f fo r t. Le t us suppos e tha t a po in t charge Q

i s located at P in Fig . 3-3. W e can ca lcula te the f lux of E thr ou gh the closed

surface as fol lows. By def ini t ion, the f lux of E through the element of area

da is

(3-5)



70 Fields of Stationary Electric Char ges: II

w h e r e T1 • da i s the project ion of da on a p lane normal to r t. T h e n

E • da = dO , (3-6)Ane 0

w h e r e da i s t he e l ement o f so l id ang le sub tended by da a t the po in t P.

To find the total f lux of E, we in tegra te over the whole su r f ace S. Since

the po in t P i s inside the sur face, by hypothesis , the in tegral of da is An a n d

J*s E • da = Q/e 0. (3-7)

I f m ore th an o ne po in t ch arge r es ides wi th in S, the f luxes add alge

braical ly , and the total f lux of E l eav ing the vo lu me i s equ a l to the to t a l

enc losed charge d iv ided by e 0 .

I f the charge enclosed by the sur face S' is di str ib ut ed ov er a f inite

vo lume, the to t a l enc losed charge i s

Q = J*, P dz\ (3-8)

w h e r e p i s t he charge dens i ty and x i s the volume enclosed by the sur face S' .

T h e n

f, E • da' = - f, p dr'. (3-9)

This is Gauss's law stated in integral form.

Applying the divergence theorem (Sec. 1 .10) to the lef t -hand side,

J \ V • E dr' = — £ p dr'. (3-10)

Since this eq ua t io n is val id for any closed surface, the in t eg ran ds m ust b eequal and, a t any point in space,

V - E = p / e 0 . (3-11)

This is Gauss's law stated in differential form. I t concerns the derivatives of

E wi th r espec t to the coord ina tes , and no t E itself.



3.2 Gauss's Law 71

3.2.1 EXAMPLE: INFINITE SHEET OF CHARGE

Figure 3-5 shows an infinite sheet of charge, of surface charge density a. On each

side, E = a/2e0.

Figure 3-5 Portion of an infinite sheet

of charge of density a coulombs per

square meter. The imaginary box of

cross-section A encloses a charge a A.

The flux of E leaving the box is 2EA.

Therefore, E = c/2e 0 .

EXAMPLE: SPHERICAL CHARGE

Let us calculate E, both inside and outside a spherical charge of radius R and uni

form density p, as in Fig. 3-6. The electric field intensity E is a function of the distance

r from the center 0 of the sphere to the point considered.

We shall use the subscript o to indicate that we are dealing with the field out

side the charge distrib ution, and the subscrip t / inside. The total charge Q is ( 4 /3 ) 7 iK3

p .

Consider an imaginary sphere of radius r > R, concentric with the charged

sphere. We know that E 0 must be radial. Then, according to the integral form of

Gauss's law.

4nr2

E0

= Q/e0, (3-12)

Figure 3-6 Uniform spherical charge. We can calculate the electric field at a point

P outside the sphere, as well as at a point P' inside the sphere, by using Gauss's

law.



0 R

Figure 3-7 In the case of a uniform spherical charge dis t r ibut ion, the e lect r ic f ie ld

intens i ty E rises linearly from the center to the surface of the sphere and falls off as

the inve rse squa re of the d i s t ance ou t s ide the sphe re . On the o the r hand , t he

elect r ic potent ia l fa l ls f rom a maximum at the center in parabol ic fashion ins ide

the sphere , and as the inverse f i rs t power of the dis tance outs ide .

a n d

E „ = — % (3-13)

as i f the tota l charge Q were s i tuated a t the center of the sphere .

I f t he cha rge were no t d i s t r ibu te d wi th sphe r i ca l sym met ry , E 0 would no t be

uni form over the imagina ry sphe re . Gauss ' s l aw would then on ly g ive the ave rage

va lue of the normal component o f E„ ove r the sphe re .

To ca lcu la t e E , a t an in t e rna l po in t , we draw an imagina ry sphe re of rad ius r

t h r o u g h t h e p o i n t P. Symmet ry requi re s aga in tha t E , be rad ia l : t hus

47rc2

£, . = ( 4 / 3 ) j t r 3 p / e 0 , (3-14)

a n d

E, = p r / 3 e 0 . (3-15)

F igure 3-7 shows E a n d V as funct ions of the radia l dis tance r . Note that E t =

E„ at ;• = R. This i s r equ i red by Gauss ' s l aw . s ince the cha rge conta ined wi th in a



3.3 Conductors 73

spherical shel l of zero thickness a t r = R i s zero. Note a lso that Vt

= V0 at r = R.

A di scont inu i ty in V would require an inf ini te e lect r ic f ie ld intens i ty, s ince E is

-dV/dr.

CONDUCTORS

A conduc to r can be denned as a mate r i a l i ns ide which charges can f low

freely.

S ince we a r e dea l ing wi th e l ec t ros t a t i cs , we assume tha t t he charges

hav e r eache d the i r equ i l ib r ium pos i t ions an d a r e fixed in space . Th en , ins ide

a con du cto r , there is zero elect r ic f ie ld , an d al l poi nts are a t the sam e p ote nt ia l .

I f a con du ct or i s p lac ed in an elect r ic f ie ld , cha rge s f low tem po rar i ly

within i t so as to produce a second f ie ld that cancels the f i r s t one at every

p o i n t i n s id e t h e c o n d u c t o r .

C o u l o m b ' s l a w a p p l i e s w i t h i n c o n d u c t o r s , e v e n t h o u g h t h e ne t field

is zero. Ga us s 's law, V • E = p/e 0, i s a lso val id . Then, s ince E = 0 , the

c h a r g e d e n s i t y p must be ze ro .

As a co ro l l a ry , any ne t s t a t i c cha rge on a co nd uc to r m us t r es ide on

its surface.

At the sur face of a conductor , E must be normal , for i f there were a

t ange n t i a l co m po ne n t o f E , charg es wou ld flow a lon g the su rf ace , which

w o u l d b e c o n t r a r y t o o u r h y p o t h e s i s . T h e n , a c c o r d i n g t o G a u s s ' s l a w , j u s t

ou t s ide the su r f ace , E = c / e 0 , as in Fig . 3-8, a be ing the su r f ace charge

dens i ty .

I t i s paradoxical that one should be able to express the elect r ic f ie ld

intens i ty at the sur face of a co nd uc to r in ter m s of the local sur face cha rge

dens i ty a alo ne , des pi te th e fact tha t the f ie ld is of cou rse d ue to al l t he charges ,

whe ther they a r e on the conduc to r o r e l sewhere .

E A

Figure 3-8 Por t ion of a cha rged

conduc tor ca r ry ing a sur face cha rge

dens i ty a. The cha rge enc losed by the

imagina ry box i s a da. There is zero

fie ld ins ide the conductor . Then, from

Gauss ' s l aw , E = ff/e 0.



74 Fields of Stationary Electric Charg es: II

EXAMPLE: HOLLOW CONDUCTOR

A hollow conductor, as in Fig. 3-9, has a charge on its inner surface that is equal in

magni tude and oppos i t e in s ign to any cha rge tha t may be enc losed wi th in the

hol low. Thi s is r ead i ly dem ons t ra t ed by cons ide r ing a Gauss i an sur face tha t l i es

wi th in the cond uc to r and tha t enc loses the ho l low. S ince E is eve rywh ere ze ro o n

this surface, the tota l enclosed charge must be zero.

Figure 3-9 Cross -s ec t ion of a ho l low conduc tor . A c h a r g e Q i n the ho l low induces

a tota l charge - Q on the inner surface. I f the conductor has a zero net charge, a

cha rge + Q i s induced on the outer surface.

3.3.2 EXAMPLE: ISOLATED CHARGED CONDU CTING PLATE

A n isolated charged conducting plate of inf ini te extent mu st carry equa l cha rge

dens i t i e s a on i ts two faces , as in Fig. 3-10a. Outs ide , £ = 2(a/2e0) = a/e0, while ,

ins ide the pla te , the f ie lds of the two surface dis t r ibut ions cancel and £ = 0.

EXAMPLE: PAIR OF PARALLEL CO NDUCTING PLATES CARRYING

CHARGES OF EQUAL MAGNITUDES AND OPPOSITE SIGNS

If on e has a pair of parallel conducting plates, as in Fig. 3-10b, carrying charges of egual

magnitude and opposite signs, the charges pos i t ion themselves as in the f igure and

£ = 0 every whe re excep t in the region betwee n the pla tes .

75

3.4.1

a

• :

I TO

+ <J —o

*

£ = 0

• • "T"

•

E = 0 +a

£ = — " /•: = ()

-i l l ©

£ = 0

(a ) (b)

Figure 3-10 (a) Charged conduc t ing p la t e ca r ry ing equa l sur face cha rge dens i t i e s a

on each s ide . The field inside the pla t e is ze ro , {b) P a i r of paral le l pla tes carrying

surface charge dens i t ies +a and —<j. T he electric field inside th e pla t e s and

o u t s i d e is ze ro .

3 A POISSON 'S EQUAT ION

Ac cord ing to G au ss' s law, Eq . 3-11, V • E = p/e0. Now, since E = -W,

from Eq. 2-13,

„,„ 82

V d2

V d2

V pS

2

V = — + - 3 + — y = - A (3-16)dx dy dz e 0

This is Poisson's equation.

Poisson's equation often serves as the starting point for calculating

electrostatic fields. It states that, within a constant factor, the charge density

p is equ al to the sum of the secon d de rivati ves of V with respect to x, y, z.

EXAMPLE: FLAT ION BEAM

F i g u r e 3-11 s h o w s a flat ion beam s i tua t ed be tween tw o g r o u n d e d p a r a l l e l c o n

duc t ing p la t e s . A c o m p l e t e s t u d y of the elect r ic and magnet ic f ie lds of the b e a m

w o u l d be q u i t e e l a b o r a t e , but we sha l l d i s rega rd the m o t i o n of the i ons and l imi t

ourse lves to the electric field, in the s imple case where p is uniform ins ide th e b e a m

76

p = 0 -

P

p=0-

Figure 3-11 Flat ion beam of uniform space charge densi ty p, b e t we e n t wo g ro u n d e d

conduct ing p la tes .

and where edge effects are negligible. We shal l use the subscript i on E and V inside

the beam, and the subscr ip t o ou ts ide .

Ins ide the beam,

d 2V,

~d xT

tf

e0'(3-17)

Et = - dV , = px

dx 6A.4. (3-18)

wh e re A is a cons tant of in tegrat ion. By sym me try, E,- = 0 at x = 0. ThereforeA = 0; inside the beam,

E, = px/e 0. (3-19)

F = —~x2

+ B, (3-20)

wh e re B i s ano the r cons tan t o f in teg ra t ion .

Now consider a cyl indrical volume whose upper and lower faces are outs ide

the beam, at equal d is tances from the plane x = 0, as in Fig . 3-11. Th e m ag nitu de

of E„ is the same at both faces. Let S be the area of one face. According to Gauss 'slaw.

2£„S = 2aSp/e0, (3-21)

£„ = ap/e0 x > a, (3-22)

£

„ = - a p / e 0 x < -a. (3-23)

3.4 Poisson's Equation 77

Therefo re , ou t s ide the beam.

apx

V„ = — — + C x > a. (3-24)

apxV. = — + D x < -a, (3-25)

where C and D are aga in cons tan t s o f in teg ra t ion . S ince V = 0 both at x = b

an d at x = — b,

C = D = apb/e0(3-26)

a n d

apV„ = —(b- x) x > a. (3-27)

x < —a. (3-28)

Now there can be no d i scon t inu i ty o f V at the surface of the beam, for o therwise

E = — \V would be infini te. Then Eqs. 3-20 and 3-27 must agree at x = a:

-^a> + B

2 e 0

ap(b - a). (3-29)

ap I, aB = —\b

2(3-30)

E q u a t i n g t h e V's from Eqs. 3-20 and 3-28 at x = -a gives the same resul t .

F ina l ly , in s ide the beam,

p r x2 + a h - ^e 0

2V: = —

£, = px/e 0.

(3-31)

(3-32)

Above, for x > a.

K = °Ab-x),

E 0 = ap/e0,

(3-33)

(3-34)

78

Figure 3-12 The electric field intensity E and po tent ia l F a s funct ions of .v, betw eenthe tw o pla tes in Fig. 3-11.

and below, for x < — a,

V 0 = ^(b + x), (3-35)

E 0 = -ap/e 0. (3-36)

The curves for V a n d E as funct ions of x are shown in Fig. 3-12.

3.5 LA PL A CE 'S EQ UA TION

W h e n p = 0 , P o i s s o n ' s e q u a t i o n b e c o m e s

\2V = 0. (3-37)

This i s Laplace s equation. This equa t ion app l i es no t on ly to e l ec t ros t a t i cs

bu t a l so to hea t conduc t ion , hydro- and ae rodynamics , e l as t i c i ty , and so on .

3.6 THE UNIQUENESS THEOREM

Accord ing to the uniqueness theorem, P o i s s o n ' s e q u a t i o n h a s o n l y o n e

s o l u t i o n V(x,y,z), for a given charge densi ty p(x,y,z) and fo r g iven boundary

c o n d i t i o n s . 1

f For a p roof o f t h i s t heorem, s ee our Electromagnetic Fields and Waves, p. 142.



3.7 Images 79

This is an i mp or tan t the ore m. If, so meh ow, by intui tion or by analogy,

o n e finds a function V that satisfies both Poisson's equation and the given

boundary cond ition s, then it is the correc t V.

3.7 IMAGES

T h e me tho d of images involves the conver sionof an electric field into another

equivalent electric field that is simpler to calculate.

With this method, one modifies the field in such a way that, for the

region of interest, bot h the charge distrib ution and the boun dar y condit ions

are conserved. The method is a remarkable application of the uniqueness

t h e o r e m . It is best explaine d by mea ns of an examp le.

EXAMPLE: POINT CHARGE NEAR AN INFINITE GROUNDED

CONDUCTING PLATE

Consider a point charge Q at a distance D from an infinite conducting plate connected

to ground, as in Fig. 3-13a. This plate may be taken to be at zero potential. It is

clear that if we remove the ground ed co ndu cto r and replace it by a charge — Q at a

distance D beh ind the pla ne, the n every point of the plane will be equidis tant from

Q and from — Q, and will thus be at zero potential. The field to the left of the con

duct ing sheet is theref ore unaffected.

The charge — Q is said to be the image of the charge Q in the plane.

The potential V at a point P(x,y) as in Fig. 3-13b, is given by

Q(3-38)

where

r = (x2

+ y2

)112

an d f = [(2D - x)2

+ y2

]1

'2

.

The components of the electric field intensity at P are those ofW:

dV Qx Q(2D - x)47ie nA ,. = — 47re n —— = —w - H -, ,

(3-39)

(3-40)

BV Qy Qy

- * t e 0 — = - 3 73-

dv r r

(3-41)

The lines of force and the equipotentials are shown in Fig. 3-14.



80

(a) (b)

Figure 3-13 (a ) Poin t cha rge Q nea r a g rounded conduc t ing p la t e , (b ) T h e

conduc t ing p la t e has been rep laced by the image cha rge — Q to calculate the field

at P .

At the surface of the conduct ing pla te , x = D, r = /•',

E x = 2QD/4ne0r\ E, = 0, (3-421

and the induced cha rge dens i ty i s - e0E

x. In this part icular case , the surface charge

dens i ty is negat ive , and the e lect r ic f ie ld intens i ty points to the r ight a t the surface

of the conduc t ing p la t e .

Figure 3-14 Lines of force ( ident i f ied by arrows) and equipotent ia ls for a point

cha rge nea r a g rounded conduc t ing p la t e . Equipo ten t i a l s and l ines o f fo rce nea r

the cha rge cannot be shown because they ge t t oo c lose toge the r . Equipo ten t i a l

surfaces are generated by rota t ing the f igure about the axis des ignated by the

curve d arro w. The im age f ie ld to the r ight of the cond uct ing pl a te is indica ted by

broken l ines .



3.8 Summa ry 81

I t wi l l be observed that image charges are located outside t he reg ion where

we calculate the field. In this case we require the field in the region to the left of the

plate , whereas the image is to the r ight .N ow wha t is t he e l ec t ros t a t i c force of a t t rac t ion be tween th e cha rge Q and the

gro und ed p la t e? I t i s obvio us ly the s ame as be tween two cha rge s Q a n d — Q

sepa ra t ed by a d i s t ance 2D , s ince Q cannot te l l whether i t i s in the presence of a

po in t cha rge — Q or of a grounded pla te . The force on Q i s thus Q 2/4ne0{2D)2. T h i s

is a lways t rue . The force be tween a cha rge an d a con du c tor i s a lways g iven cor rec t ly

by the Coulomb force be tween the cha rge and i t s image cha rges .

SUMMARY

T h e solid angle sub tended by a c losed su r f ace S at a point P is 4n s t e r a d i a n s

if P i s s i tuated inside S, and zero i f P i s s i tua ted ou t s id e .

Gauss's law fo l lows f rom Coulomb 's l aw. I t can be s t a t ed e i the r in

integra l or in di f ferentia l form :

(3-9)

w h e r e S' i s the sur face that encloses r ' , or

V • E = fi/e 0. (3-11)

Poisson's equation,

\2V = -p/co, (3-16)

then follows from Eq. 3-11 and from E = — W.

W he n the char ge dens i ty i s ze ro we have Laplace's equation,

\2

V = 0. (3-37)

A c c o r d i n g t o t h e uniqueness theorem, for a given p and for a given set

o f bo un da ry c ond i t ions , t he re i s on ly one poss ib le po ten t i a l sa t is fy ing

P o i s s o n ' s e q u a t i o n .

T h e m e t h o d o f images is o n e r e m a r k a b l e a p p l i c a t i o n o f th e u n i q u e n e s s

theorem. For example , t he f i e ld o f a po in t charge Q in front of an infinite



82 Fields of Stationary Electric Charges: II

gro und ed con duc ting plate is the same as if the plate were replaced by a

charge — Q at the position of the image of Q. The method of images gives

the correct field only outside the region where the images are situated.

PROBLEMS

3-1E ANGLE SUBTENDED BY A LINE AT A POINT

W h a t is the angle 0 s u b t e n d e d at a poin t P by a l ine of length a s i tua t ed at a

di s t ance b as in Fig. 3-15?

Figure 3-1 5 Line of l eng th a at a

di s t ance b from a p o i n t P. See P r o b .

3-1.

3-2 SOLID ANGLE SUBTENDED BY A DISK AT A POINT

W h a t is the so l id angle sub tended at a p o i n t P by a circular area of r a d i u s a

at a di s t ance b as in Fig. 3-16?

Hint: Consider f i rs t th e so l id angle sub tended at P by a r ing of r a d i u s /• and

width dr. and t hen in t egra t e .

Figure 3 -1 6 Circu la r a rea of r a d i u s a

s i tua t ed at a di s t ance b from th e p o i n t

P. See P r o b . 3-2.

3-3E GAUSS'S LAW

C o u l d one use G a u s s ' s law to ca lcu la t e th e elect r ic f ie ld intens i ty near a d i p o l e ?

3-4E SURFACE DENSITY OF ELECTRONS ON A CHARGED BODY

The maximum elect r ic f ie ld intens i ty that can be m a i n t a i n e d in air is 3 x 1 0 6 vol t s

pe r me te r .

a ) Ca lcu la t e th e surface charge dens i ty that wi l l produce this f ie ld.



Problems 83

b) Co ns ide r a me ta l w i th an in t e ra tom ic spac ing of 0 .3 nanom ete r . W ha t i s

t h e a p p r o x i m a t e n u m b e r o f a t o m s p er s q u a r e m e t e r ?

c ) How many ex t ra e l ec t rons a re requi red , pe r a tom, to g ive a sur face cha rgedens i ty o f t he abo ve m agn i tud e?

3-5 THE ELECTRIC FIELD IN A NUCLEUS

a) Calculate the e lect r ic f ie ld intens i ty in vol ts per meter a t the surface of an

iod ine nuc leus (53 pro ton s and 74 neu t rons ) .

b) Fin d the e lect r ic pote nt ia l a t the center of the nucleu s .

Assu me tha t the cha rge dens i ty is uniform and th at the radiu s of the nucle us is

1.25 x 1 0- 1 5

j 41 / 3

mete r , where A i s the tota l number of part ic les in the nucleus .

3-6E THE SPACE DERIVATIVES OF Ex, £

v, E

z

Lis t a l l the re la t ions between the various space derivat ives of the three com

ponents of E, us ing Gauss 's law and the fact that an e lect ros ta t ic f ie ld is conservat ive .

3-7 PHYSICALLY IMPOSSIBLE FIELDS

An elect r ic f ie ld points everywhere in the z di rec t ion .

a ) W ha t c an you con c lude a bo ut the va lues of t he pa r t i a l de r iva t ives o f Ez

with respect to x, to y. and to z , ( i ) i f the volume charge dens i ty p is zero, (ii) if the

volum e cha rge dens i ty i s no t ze ro ?

b) Sketch l ines of force for som e f ie lds that ar e poss ible and for som e that are not .

PROTON BEAM

A 1 .00-mic roampere beam of p ro ton s i s acce le ra t ed th ro ugh a di ffe rence of po

tent ia l of 10,000 vol ts .

a ) Ca lcu la t e the vo lume cha rge dens i ty in the beam, once the pro tons have

been acce le ra t ed , a s sumin g tha t t he cur re n t dens i ty i s un i form wi th in a d i am ete r o f

2.00 mil l imeters , and zero outs ide this diameter . (The current dens i ty in an ion

be am is in fact larges t in the center a nd fa lls off gradua l ly wi th increas in g radius .)

Solution: The cha rge pe r un i t l eng th k i s equ al to the curre nt I, div ided by the

veloci ty v o f t h e p r o t o n s :

;. = i/ v. i n

Also,

( l / 2 ) m i ' 2 = eV , (2)

w h e r e m i s t he pro ton mass , e i s t he pro ton cha rge , and V i s the accelerat ing vol tage.

T h u s ,

(2eVjm)m \2eVj

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84

Figure 3-17 Cylin dric al surface sur

rounding the ion beam of Prob. 3-8

for calculat ing the magnitude of E

us ing Gauss ' s l aw.

a n d

p = )JnR 2

= 2.300 x 1 0 "7

c o u l o m b / m e t e r3

, (4)

wh e re R is the radius of the beam.

b) Calculate the radial electric f ield in tensi ty , both inside and outs ide the beam.

Solution: Outside the beam, E is radial . Applying Gauss 's law to a cyl indrical

surface as in Fig. 3-17,

2nr(e0Et) = /., (5)

E„ = X/2ne0r, (6)

/ / m Y '

2

1H r'ne 0 \2eVj

= 1.299 x 10"2

/r volts /meter, (8)

= 12.99 volts / mete r at r = 1 0 "3

meter. (9)

Inside the beam, we find E by applying Gauss 's law to a cyl indrical surface of

radius r < R. T h e n

E i = nr2

p/2tie0r = pr/2e0, (10)

= 1.299 x 104

r vol ts /m eter, (11)

= 12.99 volts /m eter at r = 1 0 "3

meter. (12)

c) Draw a graph of the radial electric f ield in tensi ty for values of r ranging from

zero to 10 mil l imeters .

Solution: See Fig. 3-18.

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85

Figure 3-18 P o t e n t i a l V and electric field intensity E as funct ions of the dis tance r

from the axis for the pr oto n be am of Pro b. 3-8. The be am h as a radiu s of

1 mi l l ime te r .

d) Now le t the beam be s i tuated on the axis of a grounded cyl indrical conduct ing tube with an ins ide radius of 10.0 mil l imeters .

Are the above va lues o f E s t i l l val id?

Solution: The y a re s t il l val id: £„ de pe nd s sole ly on " /. and r , and £, o n p a n d r.

It is the value of V that i s affected by th e presen ce of the tu be.

e ) D raw a graph of V ins ide the tube.

Solution: O u ts id e th e b ea m , K = 0 a t r = 1 0 " 2 m e t e r a n d b e y o n d . T h e n

V„ = Jr' ° " E„ dr = j ' " " 1.299 x 1 0 " 2 dr/r, (13)

= 1.299 x 1 0 " 2 [ l n ; - ]r

1 0

"2

, (14)

= 1.299 x 1 0 " 2 [ l n 1 0 " 2 - In r] , (15)

= - (5 .9 82 + 1 .299 In r ) 1 0" 2 volt, (16)

= 2.991 x 1 0 " 2 vol t a t r = 1 0 " 3 meter . (17)



86 Fields of Stationary Electric Char ges: II

Ins ide the beam,

11= 2.991 x 1CT

2

+ (18)

2 .991 x 1(T 2 + 6.495 x 10 3 ( 1 ( T 6 - r 2 ) vol t , (19)

= 36 .41 x 10" 3 volt at r = 0. (20)

F igure 3-18 shows V as a funct ion of r.

3-9 ION BEAM

An ion beam of un i form dens i ty p passes between a pair of paral le l pla tes , one

at x = 0, a t zero pote nt ia l , an d on e a t x = a, a t t he po ten t i a l V 0. Neglec t t he mot ion of

the ions , as wel l as edge effects . The beam complete ly f i l l s the space between the pla tes .

3-10 A UNIFORM AND A NON-UNIFORM FIELD

Con s ide r a pa i r o f pa ra l l e l condu c t ing p la t e s , one a t x = 0 tha t i s g rou nde d ,

and o ne a t x = 0.1 me ter tha t is ma int a in ed a t a po tent ia l of 100 vol ts .

a ) Expres s V as a funct ion of x when p = 0 . Show the equipo ten t i a l s a t V =

0, 10, 2 0 , . . . 100 vol ts . Draw a curve of £ as a funct ion of x.

b) Do the same for the case where p/e 0 i s un i form and equa l t o 10 4 .

3-11 VACUUM DIODE

In a vacuum d iode , e l ec t rons a re emi t t ed by a hea ted ca thode and co l l ec t edby an anode. In the s imples t case , the e lect rodes are plane, paral le l , and c lose together ,

so edge effects can be neglected.

Since the e lect rons move at a f ini te veloci ty, the space charge dens i ty is not

ze ro . It is given by

w h e r e V0 i s the vol tage a t the anode ( the vol tage a t the cathode is zero); s i s the dis tance

between the e lect rodes ; and x is the dis tance of a point to the cathode. So, a t x = 0,

V = 0, while, at x = s, V = V 0.

This pecu l i a r cha rge d i s t r ibu t ion comes f rom the fac t t ha t t he e l ec t rons a reaccelerated in the e lect r ic f ie ld, which i tse l f depends on the charge dens i ty, and hence

on the e lect ron veloci ty. We assume that the current i s l imi ted by the space charge,

and no t by e l ec t ron emis s ion a t t he hea ted ca thode .

a ) Use Poi s son ' s equa t ion to f ind V as a funct ion of x.

b) At any point , the current dens i ty J is pv , w h e r e v i s the veloci ty of the e lect ro ns

at that point .

Use the value of pv at x = s to find J.

F i n d V and the t ransverse £ as funct ions of x.

P = ~



Problems 87

3-12 IMAGES

Tw o cha rges + Q a n d — Q a re s epa ra t ed by a d i s t ance a and are a t a dis tance b

from a large conduct ing sheet as in Fig. 3-19.F ind the x and y com pon ent s o f t he force exe rt ed on the r igh t -h and cha rge .

+ Q

Figure 3-19 Pair of charges near a

conduc t ing sur face . See P rob . 3 -12 .

3-13 IMAGES

A c h a r g e Q i s s i tuate d ne ar a cond uct ing pla te form ing a r ight angle as in Fig. 3-20.

Find the e lect r ic f ie ld intens i ty a t P .

The re a re th ree images of equa l mag ni tud es .

Figure 3-20 Po int c harg e nea r a

conduc tor fo rming a r igh t ang le . See

Prob . 3 -13 .

3-14 IMAGES

A poin t cha rge Q i s s i tuated a t a dis tance D from an infini te conduct ing pla te

connec ted to g round , a s in F ig . 3 -13 .

a) F ind th e surface charg e dens i ty on the co nd uc tor as a funct ion of the radiu s .

b ) Show tha t t he to t a l cha rge on the conduc tor i s — Q.

3-15 ANTENNA IMAGES

An tenna s a re o f ten mo un ted n ea r condu c t ing sur faces . On e example is t he wh ip

antenna shown in F ig . 18-4 .

F igure 3-21 shows s chemat i ca l ly th ree an tennas , o r por t ions of an tennas , nea r

conduc t ing sur faces . The a r rows show the d i rec t ion of cur ren t f low a t a g iven

ins t an t .

Figure 3-21 Th ree an te nna s s i tua t ed abov e conduc t ing p lanes , and the i r images .

The a r rows show the d i rec t ion of the cur ren t s in the an tennas , a t a g iven ins t an t .

See Prob. 3-15.

D raw figures show ing the di rect io n of cu rren t flow in the ima ges .

Solution: (a) Con s ide r a pos i t ive cha rge movin g dow nw ard in the ve r t i ca l

an ten na . I t s image i s a nega t ive cha rge tha t mov es upw ard . No w a nega t ive cha rge

mo ving up war d g ives a down wa rd cu r ren t . So the cur ren t i n the image is a s show n

in Fig. 3-22a.

b) Similar ly, a pos i t ive charge moving to the r ight has a negat ive image a lso

moving to the r ight . The current in the image is in the opposi te di rect ion, as shown

in Fig. 3-22b.

c ) A pos i t ive cha rge moving to the r igh t and downward has a nega t ive image

mo ving to the r igh t and u pwa rd . The image cur ren t i s i n the d i rec t ion show n in

Fig. 3-22c.

(a) (b) ^ (<)

: : _ "

Figure 3-22 R elat io n betwee n the curr ent di rec t ion s in the ante nn as and in thei r

images .



CHAPTER 4

FIELDS OF STATIONARY

ELECTRIC CHARGES: III

Capacitance, Energy, and Forces

4.1 CAPACITANCE OF AN ISOLATED CONDUCTOR

4.1.1 Exam ple: Isolated Spherical Conductor

4.2 CAPACITANCE BETWEEN TWO CONDUCTORS

4.2.1 Exam ple: Parallel-Plate Capacitor

4.2.2 Capacitors Connected in Parallel

4.2.3 Capacitors Connected in Series

4.3 POTENTIAL ENERGY OF A CHARGE DISTRIBUTION

4.4 ENERG Y DENSITY IN AN ELECTRIC FIELD

4.4.1 Exam ple: Isolated Spherical Conductor

4.5 FORCES ON CONDUCTORS

4.5.1 Exam ple: Electric Force for E = 3 x 106 volts per meter

4.5.2 Parallel-Plate Capacitor

4.6 SUMMARY

PROBLEMS



In th is chapter we shal l complete our s tudy of the elect r ic f ie lds of s ta t ionary

e lec t r i c charges in a vacuum. We sha l l s t a r t wi th the concep t o f capac i t ance .

This wi l l lead us to the energy stored in an elect r ic f ie ld , and this energy,

in turn, wi l l g ive us the force exer ted on a conductor s i tuated in an elect r ic

field.

CAPACITANCE OF AN ISOLATED CONDUCTO R

Ima g ine a co nd uc to r s i tua te d a long d i s t anc e awa y f rom any o the r bod y .

As ch arg e is ad de d to i t , i t s po ten t ia l r i ses, the m ag ni tu de of th e ch an ge in

p o t e n t i a l b e in g p r o p o r t i o n a l t o th e a m o u n t o f c h a r g e a d d e d a n d d e p e n d i n g

on the geom et r i ca l conf igura t ion o f the con du c to r .

T h e r a t i o

C = Q/V ( 4 - 1 )

i s cal led the capacitance of the i so la t ed conduc to r . A capac i t ance i s pos i t ive ,

by def ini t ion.

Th e un i t of cap ac i t a nce i s t he cou lom b per vo l t o r farad.

EXAMPLE: ISOLATED SPHERICAL CONDU CTOR

A n isolated spherical conductor of rad ius R carr ies a tota l surface charge Q. T h e n

i ts capaci tance is



91

Figure 4-1 Tw o cond uc to rs ca r ry ing equa l and oppos i t e cha rges . Th e in t egra l ofE • dl f rom one conduc tor to the o the r , a long any pa th , g ives the po ten t i a l

difference V. The capac i t ance be tween the conduc tors i s Q/V.

4.2 CAPACITANCE BETWEEN TWO CONDUCTOR S

W e hav e jus t de f ined the cap ac i t a nce o f an i so la t ed cond uc to r . W e can

proceed in a s imi l a r way to def ine the capac i t ance be tween two conduc to r s ,

as in Fig . 4- 1.

In i t i a l ly , bo th conduc to r s a r e uncharged . Le t us assume tha t one o f

the cond uc to r s i s g ro und ed , so tha t i t s po ten t i a l i s kep t equa l to ze ro . W e

then g radua l ly add smal l charges to the o ther un t i l i t ca r r i es a charge Q.

In the p rocess , charges o f opp os i t e s ign a r e a t t r a c t ed to the g ro un ded con

duc to r un t i l i t ca r r i es a charge —Q. A t t h a t p o i n t t h e g r o u n d e d c o n d u c t o r

i s s ti ll a t ze ro po ten t i a l , wh i l e the o the r co nd uc t o r i s a t a po ten t i a l V.

Th e capac i t anc e be twe en the two con du c to r s is aga in def ined as in

E q . 4 - 1 :

C = Q/V. (4-3)

Th e cap ac i t a nce be tw een tw o cond uc to r s i s a l so pos i tive , by def in i tion .

In oth er wo rds , if we establ ish a po tent ia l d i f ference V b e t w e e n t w o

c o n d u c t o r s b y g i v i n g t h e m c h a r g e s +Q a n d — Q, t h e n t h e c a p a c i t a n c e

b e t w e e n t h e c o n d u c t o r s i s Q/V.

4.2 .1 EXAMPLE: PARALLEL-PLATE CAPACITOR

Figure 4-2 shows a parallel-plate capacitor with the spacing s gros s ly exaggera t ed .

W e neglect edge effects. We hav e a unifo rm pos i t ive charg e dens i ty er on the to p



92

Figure 4- 2 Para l le l -p la te capac i to r . The lower end of the small cyl indrical f igure is

si tuated inside th e lower p la te where E = 0.

surface of the lower plate and a uniform negative charge densi ty — a on the b o t t o m

surface of the upper p la te . The electric field intensity E p o i n t s u p wa rd .

App ly ing Gauss ' s law to a cylindrical volume as in the figure, th e outward flux

of E is Eu. where a is the cross-sect ion of the cylinder. Since the enclosed charge

is (jo. then

Ea = aa/e0, E = a'e0, (4-4)

so tha tV = Es = as/e0. (4-5)

N o w Q = a A, wh e re A is the a re a of one p la te , and

C = Q/V = e0A/s. (4-6)

T h i s is the capac i tance be tween two para l le l conduct ing p la tes of a re a A s e p a ra t e d

by a d is tance s.

4.2.2 CAPACITORS CONNECTED IN PARALLEL

Figure 4-3a shows two capacitors C, and C2

connected in parallel and

carrying charges Ql

and Q2. We wish to find the value of C in Fig. 4-3b tha t

will give the same V for the same charge Qx

+ Q2.This is simple:

Qi + Q2= c,v + c

2v = cv. (4-7)



93

VQ

(a) (b)

Figure 4 -3 The s ing le capac i to r C has the s ame capac i t ance a s the two capac i to rs

C, and C 2 connec ted in pa ra l l e l .

T h e n

C = C, + C 2 . ( 4 - 8 )

S o , t he capac i t ance o f two capac i to r s connec ted in para l l e l i s t he sum

of the two capac i t ances . I f we have severa l capac i to r s in para l l e l , t he to t a l

capa c i t an ce i s t he sum of the ind iv idua l ca pac i t an ces .

4.2.3 CAPAC ITORS CONNE CTED IN SERIES

F i g u r e 4 - 4 a s h o w s t w o c a p a c i t o r s C , a n d C 2 con nec ted in series . W e f i rs t

c o n n e c t p o i n t s A a n d B t o g r o u n d , t e m p o r a r i l y . T h i s m a k e s t h e v o l t a g e s

V l a n d V 2 z e r o . T h e c a p a c i t o r p l a t e s a re u n c h a r g e d . W e n o w d i s c o n n e c t

A a n d B f r o m g r o u n d .

We then app ly a charge Q t o A. W e a s s u m e t h a t t h e v o l t m e t e r s d r a w

zero curre nt . N o cha rge can flow int o or ou t of B. A c h a r g e — Q leaves theupper p l a t e o f C 2 and goes to the lower p la t e o f Cy. This l eaves a charge + Q

on the top plate of C 2 . Then the lower p la t e o f C , t akes on a charge — Q,

as in the f igure.

T h e n

Vi = QIC,, V 2 = Q/C 2. ( 4 - 9 )



94

VQV Q

(a ) (b )

Figure 4-4 T h e c a p a c i t o r C has the s ame capac i t ance a s the two capac i to rs C, a n d

C 2 connected in ser ies .

Now what should be the value of C in Fig . 4-4b to give a vol tage V

equal to K, + V 2 wi th the same charge Q'l W e m u s t h a v e

Q Q QV = - = — + — , (4-10)

c c x c 2

o r

+ ( 4 - l Dc c , c 2

C = (4-12)

I f we have three capaci tors in ser ies ,

I * . j . J_ + _L ( 4 - , 3 )

= C ^ C j

C 2 C 3 + C 3 C , + c , c2

N ow t he inverse of the capa ci ta nc e, 1 /C, i s cal led the elastance. S o ,

if cap ac i t o r s a r e co nne c ted in se r i es , t he to t a l e l as t ance i s t he sum of the

ind iv idua l e l as t ances .



4.3 Potential Energy of a Charge Distribution 95

Note tha t , when capac i to r s a r e connec ted in para l l e l , t he vo l t ages

are the same . W he n cap ac i to r s a r e con nec ted in se r ies , t he charges a r e the

s a m e .

POTENTIAL ENERGY OF A

CHARGE DISTRIBUTION

W e can find a gen eral exp ress ion for the pote nt ia l energy of a cha rge dis

t r ibu t ion by cons ider ing the energy expended in charg ing a pa i r o f con

ductors , as in Sec. 4 .2 .

When we add a smal l charge dQ t o the l e f t -hand conduc to r , we mustexpend an energy

dW = V dQ = | dQ, (4-15)

w h e r e Q i s t he charg e a l r eady on the co nd uc to r . Then , to charg e the con du c to r

to a po ten t i a l V = QIC, we mu st supp ly an energy

W = FC %dQ = %- = 1 QV = - CV2. (4-16)

J o

c •2C 2 2

Th e reaso n for the factor of one -half shou ld be clear f rom the rea son ing

that we have used to ar r ive at i t . I t i s that , on the average, the potent ia l V

at the t ime of ar r ival of a charge dQ i s just o ne-ha lf the f inal pote nt ia l .

M ore genera l ly , if we hav e a nu m be r o f charg ed con duc t ing bod ies ,

the s to red energ y i s t he sum of s imi l a r t e rm s fo r each on e :

W = WQiV, (4-17)

^ i

F o r a c o n t i n u o u s c h a r g e d i s t r i b u t i o n o f d e n s i t y p(x',y',z') o c c u p y i n g a

v o l u m e t ' , we can r ep lace Q : by p clr' a n d t h e s u m m a t i o n b y a n i n t e g r a t i o n :

W = - { v p d f , (4-18)

w h e r e V a n d p are bo th func t ions o f the coord ina tes in the genera l case .



96 Fields of Stationary Electric Charges: III

4.4 ENERGY DENSITY IN AN

ELECTRIC FIELD

The energy stored in a parallel-plate capacitor is therefore

\ c r ' ' - H O ' l 4 • l 9 ,

= Q e0£

2

) As, (4-20)

where As is the volume occupied by the field. Th e sto red energy may therefore

be calculated by associating with each point an energy density [e0/2)E2

joules per cubic meter.

This is a general result. The energy associated with a charge distribu

tion, that is, the energy requ ired to assemble it, sta rti ng with the cha rge

spread over an infinite volume, is

W = ^ [ E2 dr, (4-21)2 J t

where the volume t extends to all the region where the field under consid

eration differs from zero.

The energy stored in an electric field can therefore be expressed, either

as in Eq. 4-18, or as in Eq. 4-21. No te th at the terms under the integral signs

in these two equations are unrelated. For example, at a point where p = 0,

Vp is zero but E2 is usually not zero.

4.4.1 EXAMPLE: ISOLATED SPHERICAL CONDUCTOR

F o r the isolated spherical conductor of radius R and carrying a charge Q,

W = f £ 2

dr.

Us i n g a spherical shell of radius r and thickness dr as element of volume dr.

W = f2 r9 JR2 J* V / W 2

Q2

47tr2 dr.

(4-22)

(4-23)

(4-24)

T h i s energy is jQV, orjCV2, where Ci s the capaci tance of the sphere , as in Sec. 4.1.1.

97

a

Figure 4-5 The local e lect r ic charge dens i ty a t the surface of a conductor gives r ise

to opposi te ly di rected e lect r ic f ie ld intens i t ies <r/2e 0 as shown by the two a r rows on

the lef t ; the other charges on the conductor give r ise to the f ie ld o/2e0 shown by

the arrow on the r ight . The net resul t i s <r/e 0 out s ide , and ze ro ins ide . Th e v ec tor

n i s a un i t vec tor no rm al to the con duc tor sur face , and i t po in t s o u tw ard .

An e lement o f charge a da on the su r face o f a co nd uc to r exp er i ences the

elect r ic f ie ld of a l l the other charges in the system and is therefore subject

to an elect r ic force. In a s ta t ic f ie ld , th is force must be perpendicular to the

sur f ace o f the conduc to r , f o r o therwise the charges would move a long the

surface. Since the charge a da i s bo un d to the co nd uc to r by in t e rna l fo rces ,

the force act ing on a da is t r a n s m i t t e d t o t h e c o n d u c t o r itself.

T o ca lcu la t e the ma gn i tud e o f th i s fo rce , we cons ider a con du c to r

wi th a su r f ace charge dens i ty a and a f ie ld E at the sur face. From Gauss 's

law, the elect r ic f ie ld in tensi ty just outside the conductor i s <r /e 0. This field

i s pe rp end icu la r to the su r f ace. No w th e fo rce on the e l emen t o f cha rge a dais not Eo da, since the f ie ld that acts on a da is the f ield due only to the other

charges in the sys t em.

Let us f i r s t calculate the elect r ic f ie ld in tensi ty produced by a da itself.

We can do th i s by us ing Gauss ' s l aw. The to t a l f lux emerg ing f rom a da m u s t

b e a da/e0, half of i t inw ard an d hal f out w ar d, as in Fig . 4-5. Th en th e elect r ic

f ie ld in tensi ty due to the local su r f ace charge dens i ty must be f r /2e 0 i n w a r d

a n d o/2e0 o u t w a r d .

FORCES ON CONDUCTORS



98 Fields of Stationary Electric Charges: III

The element of charge a da itself therefore produces exactly half the

total field at a point outside, arbitrarily close to the surface. This is reasonable,

for the nearb y char ge is mo re effective th an th e rest.

Now if a da produces half the field, then all the other charges must

produce the other half, and the electric field intensity acting on a da must be

ff/2e0.

The force dF on the element of area da of the conductor is therefore

given by the product of its charge a da multiplied by the field of all the other

charges:

dF = —o da, (4-25)2e

0

and the force per unit area is

dF a2

1— = — = - e 0E

2

. (4-26)da 2e 0 2

No te tha t this force is just equa l to the ener gy den sity of Sec. 4.4 at

the surface of the conductor. We can show that this is correct by using the

method of virtual work. In this method we assume a small change in the

system and apply the princip le of cons erv ati on of energy. Let us imaginethat the conducting bodies in the field are disconnected from their power

supplies. Then, if the electric forces are allowed t o perform mech ani cal wor k,

they must do so at the expense of the electric energy stored in the field.

Imagine that a small area a of a conductor is allowed to be pulled into the

field by a small distance x. The mechanical work performed is ax times the

the force per unit area. It is also equal to the energy lost by the field, which

is ax times the energy density. Thus the force per unit area is equal to the

energy density.

If we use the same type of argument for a gas, we find that its energy

density is equal to its pressure.Note that the electric force on a conductor always tends to pull the

conductor into the field. In other words, an electric field exerts a negative

pressure on a conductor.

In order to visualize electric forces, it is useful to imagine that electric

lines of force are real and under tension—also, that they arrange themselves

in space as if they repelled each other.

99

F o r e x a m p l e , t h e l i n e s o f f o r c e b e t w e e n t h e p l a t e s o f a c a p a c i t o r t e n d

t o p u l l t h e p l a t e s t o g e t h e r a n d b u l g e o u t a t t h e e d g e s o f t h e p l a t e s , a s i n

F i g . 4 - 6 .

4.5.1 EXAMPLE: ELECTRIC FORCE FOR E = 3 x 10 6 VOLTS PER METER

Th e m ax im um elect r ic f ie ld intens i ty th at can be sus ta ined in a i r a t no rm al tem

pera ture and pres sure i s abo ut 3 x 1 0 6 vol ts per meter . The force is then 40 new ton s

pe r squa re me te r .

EXAMPLE: PARALLEL-PLATE CAPACITOR

Let us calcula te th e forces on the p la tes of a parallel-plate capacitor of area S ca r ry inga charge dens i ty + n on one p la t e and — <r on the other , as in Fig. 4-7. We assume

tha t the dis ta nce betw een the pla tes is smal l com pa red to thei r l inear exten t , in orde r

to make edge effects negl igible .

Th e force per uni t are a on th e pla tes is equ al to the energy d ens i ty in the f ield

e 0 £ 2 / 2 , and th e force of a t t r act ion betw een the pla tes is therefore e 0E2S/2, o r

(< x 2/2 e 0)S .

We shal l calcula te this force in two other ways in Probs . 4-13 and 4-14.



10 0

Figure 4-7 A cha rged pa ra l l e l -p l a t e

capaci tor wi th i t s pla tes insula ted.

The p la t e s a re he ld in equ i l ib r ium by

mechanical forces F m that act on each

plate in a di rect ion tending to increase

the sepa rat io n s , an d by e lect r ic forces

Fe t end ing to dec rease the s epa ra t ion .

s

SUMMARY

If Q is t he ne t charge ca r r i ed by an i so la t ed c ond uc t o r , an d if V i s i t s potent ia l ,

t he r a t io Q/V is called i ts capacitance.

T h e capacitance between two isolated conductors is Q/V, w h e r e Q is

the charge t r ansfe r r ed f rom one to the o ther , and V i s t he r esu l t ing po ten t i a l

difference.

T h e c a p a c i t a n c e o f t w o o r m o r e c a p a c i t o r s c o n n e c t e d i n parallel is

the sum of the capac i t ances .

The inver se o f a capac i t ance i s an elastance.

The e l as t ance o f two o r more capac i to r s connec ted in series is the

sum of the e l as t ances .

T h e potential energy assoc ia t ed wi th a cha rge d i s t r ibu t io n can be

wr i t t en e i the r as

In the f i r s t in tegral , T ' mu s t be chosen to inc lu de a ll t he cha rge d i s t r ib u t ion ,

and in the second , m us t in c lude a l l r eg ions o f space where E i s n o n v a n i s h i n g .

The ass ignment o f an energy density e 0Ez/2 to every point in space leads

to the co r r ec t po ten t i a l energy fo r the whole charge d i s t r ibu t ion .

F ina l ly , t he force per unit area on a cha rged co nd uc to r i s equa l to

c r 2 / 2 e 0 , or to the energy density in the electr ic f ield, e 0 £ 2 / 2 .

(4-18)

or as

(4-21)



10 1

Figure 4-8 Para l l e l -p l a t e capac i to r

with seven pla tes . See Prob. 4-3.

PROBLEMS

4-1E Show tha t e 0 i s expresse d in farad s per m eter .

4-2E THE EARTH'S ELECTRIC FIELDa) The earth has a radius of 6.4 x 1 0 6 mete rs . Wha t i s i t s capac i t ance?

b) The earth carr ies a negat ive charge that gives a f ie ld of about 100 vol ts per

meter a t the surface.

Ca lcu la t e the to t a l cha rge .

c) Calculate the potent ia l a t the surface of the earth.

4-3E PARALLEL-PLATE CAPACITOR

Show th at the cap aci t anc e of a paral le l -p la te cap aci t or hav ing TV plates , as in

Fig. 4-8, is

C = 8.85(N - l)A/t picofa rads ,

w h e r e N i s t he number of p l a t e s , A i s the area of one s ide of one p la te , an d t is the

d i s t ance be tween the p l a t e s . One p icofa rad i s 10" 1 2 farad.

4-4E PARALLEL-PLATE CAPACITOR

Can y ou sugges t reaso nable d im ens ion s for a 1 p i cofa rad ( 1 0 "1 2

farad) air-

insul ated p aral le l -p la te capa ci to r tha t is to ope rat e a t a pote nt ia l di fference of 500 vol ts?

4-5 PARALLEL-PLATE CAPACITOR

A para l l e l -p l a t e capac i to r has p l a t e s o f a rea S sepa ra t ed by a d i s t ance s .

Ho w is the cap aci ta nce -affected by the int r od uc t ion of an ins ulated sheet of

metal of thickness s ' , paral le l to the pla tes?

4-6 CYLINDRICAL CAPACITOR

Calcu la t e the capac i t ance pe r un i t l eng th C between two coaxial cyl inders of radi i

R\ a n d R 2 as in Fig. 4-9.

C a l c u l a t e C w h e n R 2 = 2R l.



102

Figure 4-9 Cylindrical capacitor. See

Prob. 4-6.

Solution: Let us assume that the inner conductor is positive and that it carries

a charge of k coulombs per meter of length. Then the capacitance per unit length

is k divided by the voltage V of the inner conductor with respect to the outer one,

where

V = [Rl

Ech= r ^ d r . (1)

k R

2ne0 P ,(2)

Thus

For R2 = 2Rlt

_ _2ne^

l n ( P 2 / R 1 )

C = 27T€ 0/l n 2 = 8.026 x 10"1 1

farad/meter. (4)

DROPLET GENERATOR

There are a number of devices that utilize charged droplets of liquid. Electrostatic

spray guns (Prob. 2-8), colloid thrusters for spacecraft (Prob. 2-15). and ink-jet

printers (Prob. 4-17) are examples. As we shall see, drop size is controlled by the

specific charge (coulombs per kilogram) carried by the fluid.

a) When a charged droplet splits into two parts, electrostatic repulsion drives

the droplets apart and the electrostatic energy decreases.

Calculate the loss of electrostatic energy when a droplet of radius R carrying a

charge Q splits into two equal-sized droplets of charge Q/2 and radius R'.

Assume that the droplets are repelled to a distance that is large compared to

R'. Neglect evaporation.



Problems 103

Solution: Since

then

4 42 x -nR'

3

= ~ T I R \ (1)

R' = R/2 113. (2)

Initially, the charge Q is at the potential of the surface of the drop of radius R

and the electrostatic energy is

QV = 1 Q2

2 2 4ne0

After splitting, the electrostatic energy is

2( Q / 2 )

2

2 "3

8rce 0«

and the electrostatic energy gained is

(3)

(4)

Q 1 \ Q2

AW = —^— KTT - 1 = -0 .3 70 0 (5)

b) The energy associated with surface tension is equal to a constant that is a

characteristic of the liquid, multiplied by the surface area. The constant is called

the surface tension of the liquid. The surface tension of water is 7.275 x 10~2

joule

per square meter.

When a droplet splits, the surface area increases, and the energy associated with

the surface tension increases proportionately.

Calculate the gain in this energy when a droplet of water of radius R splits into

two equal-sized droplets.

Solution: The increase in surface energy is

AW = (2 x 4nR'2

- 4nR2

)T = - \ ]4nR2

T, (6)V 2

2

'3

= 0.2599 x 4 ; i R2

r = 0.2376K2

, (7)

where T is the surface tensi on of water.

c) A dro ple t of wate r has a radiu s of one micro mete r and carri es a specific charge

of one coulomb per kilogram.

Will the droplet split?



104 Fields of Stationary Electric Cha rges: III

Solution: Th e dro ple t will split if the tot al AM 7 i s nega t ive . The cha rge Q is nu

merical ly equal to the mass , in this case:

4 ,

Q = 1000 x - nR coulombs . (8 )

T h u s

[(4000rc/3)J?3

]2

AW = - 0 . 3 7 0 0 - — — — + 0 .2 37 6K2

, (9)Sn e 0R

= - 2 . 9 1 9 x 1016

RS + 0 .2376K 2 . (10)

F o r R = 1 0 " 6 ,

AW = - 2 . 9 1 9 x 1 0 "1 4

+ 0.2376 x 10"1 2

joule . (11)

The droplet wi l l not spl i t .

d) W ha t is the radiu s of the larges t dro plet of wate r that i s s table wi th this specif ic

c h a r g e ?

Solution: For a spec i f i c cha rge of one coulomb pe r k i logram, a d rop le t o f wa te r

will not spl it spo nta neo us ly if i ts radius is such that

- 2 . 9 1 9 x 1 0 1 6 R 5 + 0 .2376K 2 > 0.

or for radi i smal ler than

0.2376

2.919 x 10 1 6

1 /3

= 2 .012 x 10" 6 meter . (12)

4-8E ELECTROSTATIC ENERGY

a) Two capaci tors are connected in ser ies as in Fig. 4-10a. Calculate the ra t io

of the s tored energies .

b) Calculate this ra t io for capaci tors connected in paral le l as in Fig. 4-10b.

4-9E ELECTROSTATIC ENERGY

Tw o capa c i to rs o f capac i t an ces C, and C 2 h a v e c h a r g e s Q x a n d Q 2, respect ively.

a) Ca lcula te the am ou nt of energy diss ipated wh en they are con nec ted in paral le l .

b ) Ho w i s t h i s ene rgy d i s s ipa ted ?

4-10 PROTON BOMB

a) Calculate the e lect r ic potent ia l energy of a sphere of radius R ca r ry ing a to t a l

c h a r g e Q uni formly d i s t r ibu ted th r oug hou t it s vo lum e .



10 5

+ --O V o -

c ,

Figure 4-10 T w o c a p a c i t o r s : (a ) connected in ser ies , and (b ) connec ted in pa ra l l e l .

See P rob . 4 - 8 .

b) N ow ca lcu la t e the grav i t a t iona l po ten t i a l ene rgy of a sphe re of rad ius R' of

to t a l mass M.

c) The moon has a mass of 7.33 x 1 0 2 2 ki lograms and a radius of 1.74 x 10°

mete rs . Ca lcu la t e i ts g rav i t a t io na l po ten t i a l ene rgy .

d) Imagine tha t you can a s semble a sphe re of p ro tons wi th a dens i ty equa l t o

tha t of wate r . Wh at wo uld b e the rad ius of this sphe re if i t s e lect r ic pote nt ia l energy

were equa l t o the grav i t a t iona l po ten t i a l ene rgy of the moon?

4-11 ELECTROSTATIC MOTOR

Can you sugges t a rough des ign for an e l ec t ros t a t i c motor?

Draw a sketch, explain i t s operat ion, specify vol tages and currents , and make a

roug h es t ima te o f wha t i t s pow er would be .

4-12 ELECTROSTATIC PRESSURE

A l ight spherica l bal lo on is m ad e of con du ct in g mate ria l . I t i s sugges ted th at i t

cou ld be kep t sphe r i ca l s imply by connec t ing i t t o a h igh-vol t age source . The ba l loon

has a d i am ete r o f 100 mi l l ime te rs , and the max imu m brea kdo wn vol t age in a i r is 3 x 1 0 6

vol ts per meter .

a) W ha t m ust be the vo l tage of the sou rce i f the e lect r ic force is to be as large

as pos s ib l e?

b) Wha t gas p res sure ins ide the ba l loon would produce the s ame e f fec t ?


Calcu la t e the force of a t t rac t ion F betw een the pla tes of a paral le l -pla te cap aci to r .

Assum e tha t t he capac i to r i s conne c ted to a ba t t e ry supply ing a cons tan t vo l t age V.

Use the method of vi r tual work, assuming a smal l increase ds in the spacing s

between the pla tes , and set

dWB

+ dWm

= dWe,



106 Fields of Stationary Electric Char ges: III

w h e r e dW B i s t he work done by t he ba t t e ry , dW m = Fds i s t he mech anica l wo rk don e

on the sys tem, and dWe is the increase in the energy stored in the electric field.

You should f ind that

I K 2 1 ,F = - e„ — S =- e0E

2

S.2 . r 2

No te tha t on e ha lf of t he ene rgy suppl i ed by the ba t t e ry app ea rs a s me chanica l

work, and one half as an increase in the energy s tored in the e lect r ic f ie ld. This is a

general rule .


Ca lcul ate the force of a t t ra ct io n betw een the pla tes of a paral le l -pla te capa ci tor ,

a s in the preced ing prob lem, a s sumin g now tha t t he capa c i to r i s cha rg ed and d i s

connec ted f rom the ba t t e ry .In this case ,

dW m = dW e.

You should f ind the same resul t .

4-15 OSCILLATING PARALLEL-PLATE CAPACITOR

Figure 4-11 shows a pa ra l l e l -p l a t e capac i to r whose upper p l a t e o f mass m is

suppor t ed by a spr ing insu la t ed f rom ground . A vo l t age V i s app l i ed to the upp er

plate . Ther e is zero ten s ion in the sprin g whe n the leng th of the a i r gap is x 0 .

Set m = 0.1 k i log ram , x 0 = 0.01 meter , k = 150 newtons pe r me te r , V =

100 vol ts , and A = 0.01 square meter . Assume that the mass of the spring is negl igible ,

and neglect edge effects.

a) Set the var ious po tent ia l energies of the com ple te sys tem equal to zero w hen

the tens ion in the spring is zero, a t x = x 0 . Show that the potent ia l energy is then

W = mg(x - x 0) + I k(x - x 0 )2 - \ e 0A V

2

(- - —2 2 \ x x 0

Draw a curve of W as a function of x b e t w e e n x = 1 0 ~ 6 a n d x = 1 0 ~ 2 mete r .

Use a logari thmic scale for the x-axis .

b ) Sh ow tha t , a t equ i l ib r ium ,

mg + k(x - x0) + i~e

0AV 2/x 1 = 0.

You can arr ive a t this resul t in two different ways .

The re i s s t ab le equ i l ib r ium a t x eq = 3.46 x 1 0 " 3 mete r and uns tab le equ i l ib r ium

at x = 2.93 x 1 0 " 5 m e t e r .

c) If the sys tem is br ou gh t to the poin t of s table equi l ib r ium , and then dis tu rbed

s l ight ly, i t osci l la tes . For smal l osci l la t ions , the res toring force F i s p ro por t ion a l t o the



10 7

Figure 4-11 Para l l e l -p l a t e capac i to r , w i th the

lower pla te f ixed in pos i t ion and the upper

p la t e suspended by a spr ing insu la t ed f rom

g r o u n d . S e e P r o b . 4 - 1 5 .

d i s p l a ce m e n t a n d w e h a v e a s i m p l e h a r m o n i c m o t i o n :

(d 2x\ (dW\

where the de r iva t ives mus t be eva lua ted in the ne ighborhood of .x = x eq.

The ci rcular frequency is o j = (K/m)111, w h e r e

S h o w t h a t

'k - (e 0AV 2/x 3

eq)OJ = —

m

This gives a frequency of 6.16 hertz .

4-16 HIGH-VOLTAGE GENERATOR

Imagine the fo l lowing s imple -minded h igh-vol t age gene ra tor . A pa ra l l e l -p l a t e

capac i to r has one f ixed p la t e tha t i s pe rm anen t ly connec ted to g round , and one p la t e

that i s movable . When the pla tes are c lose together a t the dis tance s , the capaci tor i scha rged by a ba t t e ry to a vo l t age V. The n the mova ble p l a t e is d i s connec ted f rom

the ba t t e ry and moved ou t t o a d i s t ance ns . The vol tage on this pla te then increases

to nV , if we neglect edg e effects. O nc e the vo l tage h as been ra ised to nV , the pla te is

d i s cha rged th rough a load re s i s t ance .

a) Veri fy tha t there is con serv at io n of energ y.

b) C an yo u sugges t a rou gh des ign for such a high -vo l tage ge ner ator wi th a

m o r e c o n v e n i e n t g e o m e t r y ?



10 8

Q K

mm N

Figure 4-12 Ink-je t pr inter . A nozzle N forms a f ine je t that breaks up into droplets

ins ide e lect ro de £*. A va riab le vol ta ge V appl i ed to the e l ec t rode induces a known

charge on each drop le t i n succes s ion . The drop le t s a re de f l ec t ed in the cons tan t

elect r ic f ie ld between the deflect ing pla tes and impinge on the paper form P. U n

charged drop le t s a re co l l ec t ed a t C. See P rob . 4 -17 .

4-17 INK-JET PRINTER

Th e ink-je t pr inte r i s one solut io n to the pro ble m of prin t ing da ta a t high sp eed.I t operates more or less l ike an osci l loscope, except that i t ut i l izes microscopic droplets

of ink, ins tead of e lect rons . See Prob. 4-7.

I ts pr inciple of operat ion is shown in Fig. 4-12. A nozzle produces a f ine je t of

conduc t ing ink tha t s epa ra t e s in to drop le t s i ns ide a cy l indr i ca l e l ec t rode . The po ten t i a l

on the e lect r ode is e i ther p os i t ive or zero . At a given ins tant , the charge car r ied b y a

drop le t t ha t b re aks off f rom the j e t depend s on the cha rge induc ed on the j e t an d

hence on V.

The droplets are then deflected as in an osci l loscope, except that here the de

f lecting f ie ld is con s tan t , each drop let being deflected acco rdin g to i ts charg e. U n

cha rged drop le t s a re co l l ec t ed in an ink sump. The j e t ope ra t e s con t inuous ly , an d

only a smal l f ract ion of the dro ple ts is actua l ly u sed.

Th e deflect ion is vert ical and the je t move s ho rizo nta l ly.Th e pres sure in the nozz le is mo du la t ed a t f requenc ies up to 0 .5 m egah er t z ,

p ro duc ing up to 5 x 10 ' d rop le t s pe r s econd , w i th rad i i of t he orde r o f 10 mic ro mete r s .

a) L et the rad ius of the je t be R, and th e ins ide radiu s of e lect ro de E b e R2-

W ha t is the charg e per uni t leng th on the je t , negle ct ing end effects?

b) The j e t b reaks u p in to l a rge r d rop le t s o f rad ius 2R X. W ha t length of je t i s

requi red to fo rm one drop le t ?

c ) Ca lcu la t e the cha rge Q per d rop le t .



Problems 109

d) Ca lcu la t e the spec if ic cha rg e in cou lom bs pe r k i logram for = 20 mic rom

eters , R 2 = 5 mi l l ime te rs , V = 100 vol ts . Set the ink dens i ty equal to that of water

(1000 k i lograms pe r cub ic me te r ) .

e) Ca lcul ate the veloci ty of the drop lets i f they are pro du ced a t the ra te of

1 0 ' per s econd and if t hey a re s epa ra t ed by a d i s t ance of 100 mic ro mete rs .

f) Calcu late the vert ical deflect ion a nd the vert ical veloci ty of a dro plet up on

leaving the deflect ing pla tes , assuming a uniform t ransverse f ie ld of 10 5 vol ts per meter

ex tending ove r a hor i zonta l d i s t ance of 40 mi l l ime te rs .



CHAPTER 5

DIRECT CURRENTS IN ELECTRIC CIRCUITS

5.1 COND UCTION OF ELECTRICITY

5.1.1 Conservation of Charge

5.1.2 Charge Density in a Hom ogeneous Conductor

5.2 OHM 'S LAW5.2.1 The Joule Effect

5.3 NON-LINEAR RESISTORS

5.3.1 Examp le: Incandescent Lamp

5.3.2 Examp le: Voltage-Dependent Resistors

5.4 RESISTORS CONNE CTED IN SERIES

5.5 RESISTORS CONNE CTED IN PARALLEL

5.6 THE PRINCIPLE OF SUPERPO SITION

5.6.1 Examp le: Simple Circuit with Two Sources

5.7 KIRCHOFF 'S LAWS

5.7.1 Examp le: The Branch Currents in a Two-M esh Circuit5.8 SUBSTITUTION THEORE M

5.8.1 Examp le: Resistor Replaced by a Battery

5.8.2 Examp le: Battery Replaced by a Resistor

5.9 MESH CURRENTS

5.9.1 Examp le: The Mesh Currents in a Two-M esh Circuit

5.10 DELTA-STAR TRANSFOR MATIONS

5.10.1 Examp le: Transformation of a Symm etrical Delta and of a

Symmetrical Star

5.11 VOLTAGE AND CURRENT SOURCES

5.12 THEVEN IN'S THEORE M5.12.1 Examp le: Thevenin's Theorem Applied to a Simple Circuit

5.12.2 Examp le and Review: The Wheatstone Bridge

5.13 SOLVING A SIMPLE DIFFEREN TIAL EQUATION

5.14 TRANSIENTS IN RC CIRCUITS

5.14.1 Examp le: Simple RC Circuit

5.15 SUMMARY

PROBLEMS



Unti l now we have l imi ted ourselves to the study of the f ie lds of s ta t ionary

charges. We shal l now consider the f low of e lect r ic current , and we shal l

d e v e l o p m e t h o d s f o r c a l c u l a t i n g t h e c u r r e n t s i n c o m p l e x a r r a n g e m e n t s o fc o n d u c t o r s .

We sha l l no t be concerned as ye t wi th magne t i c f i e lds .

CONDUCTION OF ELECTRICITY

We have seen in Sec. 3 .3 that , under s ta t ic condi t ions, E is zero inside a

con duc to r . H ow ever , if an e l ec t r i c fi eld i s ma in ta i ned by an ex te rna l sou rce —

for exam ple , when on e conn ec t s a l eng th o f cop per wi r e be tween the t e r

minals of a bat tery—then charge car r iers dr i f t in the f ie ld and there is an

e lec t r i c cu r r en t .

W i t h i n t h e c o n d u c t o r , t h e current density vec to r J po in t s in the

dire ct io n of f low of pos i t ive charg e car r i ers ; for neg at ive car r iers , J po ints

in the d i r ec t io n o ppo s i t e to th e f low, as in F ig . 5 -1 . The ma gn i tud e o f J i s

the qua n t i ty o f char ge f lowing th rou gh an in f in i t es imal su r f ace perp end icu la r

to the di rect ion of f low, per uni t area and per uni t t ime. Current densi ty i s

expressed in cou lombs per square mete r pe r second , o r in amperes per

square mete r . I n good conductors, such as copper , t he charge ca r r i e r s a r e

the condu c t io n e l ec t rons , an d we hav e the s i tua t ion sho wn in F ig . 5 - lb .

Semiconductors con ta in e i th e r o r bo th o f two types o f mo bi l e c harges ,

n a m e l y c o n d u c t i o n e l e c t r o n s a n d h o l e s . A hole is a vaca nc y left by the r eleas e

of an ele ct ron by an at om . A ho le can mi gra te in the ma ter i al as if i t were a

pos i t ive par t i c l e .

F o r o r d i n a r y c o n d u c t o r s , t h e c u r r e n t d e n s i t y J i s p r o p o r t i o n a l t o

the elect r ic f ie ld in tensi ty E:

J = crE, ( 5 - 1 )



11 2

^ 0

0 — ^ 0 ^ - 0

^ 0

da

^ 0

0 -

da

^ 0

Figure 5-1 Current dens i ty vector J for (a ) pos i t ive and (b ) nega t ive cha rge ca r r i e r s .

Table 5- 1

Conductor

Conductivity a

(siemens per meter)

Alu min um 3 .54 x 10 7

Br ass (65.8 Cu , 34.2 Zn ) 1.59 x 107

C h r o m i u m 3 .8 x 1 0 7

Co pp er 5 .80 x 10 7

Go ld 4 .50 x 10 7

G rap hi te 1.0 x 10 5

Ma gn et ic i ron 1.0 x 107

M um eta l (75 Ni , 2 Cr, 5 Cu, 18 Fe) 0.16 x 10 7

Nic kel 1.3 x 10

7

Sea wa ter ss 5

Silver 6.15 x 10 7

T in 0.870 x 1 07

Zi nc 1.86 x 10 7



5.1 Conduction of Electricity 113

w h e r e a is the conductivity of the m a t e r i a l . C o n d u c t i v i t y is e x p r e s s e d in

a m p e r e s pe r vol t per m e t e r , or in Siemens11

per m e t e r . T a b l e 5-1 gives the

c o n d u c t i v i t i e s of a n u m b e r of c o m m o n m a t e r i a l s .

The dr i f t veloci ty v of c o n d u c t i o n e l e c t r o n s is easi ly calculated. It

i s su rp r i s ing ly low. If n is the n u m b e r of c o n d u c t i o n e l e c t r o n s per c u b i c

m e t e r and if e is the m a g n i t u d e of the e lec t ron charge , t hen

J = nev. (5-2)

In copper , t he re is one c o n d u c t i o n e l e c t r o n per a t o m , and 8.5 x 1 02 8 a t o m s

per cub ic mete r , so n = 8.5 x 1 02 8 e l e c t r o n s pe r c u b i c m e t e r . For a cur

r e n t of one a m p e r e in a wire hav ing a cross- sec t ion of one s q u a r e m i l l i m e t e r ,

J= 10

6

a m p e r e spe r

s q u a r e m e t e rand

1 06

x 7.4 x 105 m e t e r / s e c o n d , (5-3)

8.5 x 1 02 8 x 1.6 x 10

o r a b o u t 0.26 m e t e r per hour}

5.1.1 CONSERVATION OF CHARGE

N o w c o n s i d e r a v o l u m e T b o u n d e d by a surface S i n s i d e c o n d u c t i n g ma

ter ia l . The i n t e g r a l of J • da o v e r the surface is the charge f lowing out of Sin one s e c o n d , or the charge los t by the v o l u m e % in one s e c o n d . T h e n

l3

-Aa=

-Jtlpch

-(5

"4)

T h i s is the law of conservation of charge.

U s i n g the d i v e r g e n c e t h e o r e m (Sec. 1.10) on the lef t -hand side,

J V j * (5-5)

f Afte r Erns t Werne r von Siemens (1816-1892) . The word the re fore t akes a t e rmina l s

in th e s i n g u l a r : on e Siemens . The Siemens wa s formerly cal led a " m h o . "

{ T h e n how is it t h a t one can, say, t u rn on a l i gh t k i lomete rs away , ins t an taneous ly?

T h e m o m e n t on e closes th e swi t ch , a guided e l ec t romagne t i c wave is l a u n c h e d a l o n g

the wire . This wave t ravels at the veloci ty of l ight an d es t ab l i shes th e electric field in

a n d a r o u n d th e wire .

114 Direct Curren ts in Electric Circuits

Since this eq ua t io n is val id for any vo lum e,

V • J = —2-. (5-6)

dt

Th is is aga in the law of co ns erv at io n of cha rge , s ta ted in di f ferentia l form.

5.1.2 CHARGE DENSITY IN A HOMOGENEOU S CONDUCTOR

Under s t eady- s t a t e cond i t ions , t he t ime der iva t ive in Eq . 5 -6 i s ze ro . Then

V • J = V • <rE = 0. (5-7)

If n o w w e h a v e a h o m o g e n e o u s c o n d u c t o r w i t h a i nde pen den t o f the co

ord ina tes , V • E is zero . Th us , f rom Eq. 3-11, the net ch arg e den si ty inside a

homogeneous c o n d u c t o r c a r r y i n g a c u r r e n t u n d e r s t e a d y - s t a t e c o n d i t i o n s i s

z e r o .

5.2 OHM 'S LA W

Figure 5 -2 show s an e l em ent o f co nd uc to r o f l eng th / and c ross - sec t ion A.

If a difference of potential V i s main ta ined be tween i t s two ends , t hen

J = I/A = oE = aV/l, (5-8)

a n d

aVA V Vl

—r~m = R> ,5" 9)

w h e r e

R = 1/oA (5-10)

is the resistance of the e l ement . Res i s t anc es a r e express ed in vo l t s pe r am per e

or in ohms.



115

Figure 5-2 Element o f con du c tor o f

c ros s - s ec t ion A an d length / carry ing a

cur ren t / .

Figure 5 -3 The cur ren t / t h rough a

l inea r re s i s to r i s p ro por t iona l t o th e

V vol t age V across i t .

Equat ion 5 -9 shows tha t t he cu r r en t th rough a r es i s to r i s p ropor

t iona l to th e vol ta ge ac ros s i t , as in F ig . 5-3. This i s Ohms law. A res i s to r

satisfying this law is said to be ohmic, o r linear. Equat ion 5 -1 i s ano ther ,

mo re genera l fo rm of O hm 's l aw.

The cur r en t and the vo l t age a r e measured as in F ig . 5 -4 . I t i s a ssumed

tha t the cu r r en t f lowing th rough the vo l tmete r V i s neg l ig ib le co mp are d

t o t h a t t h r o u g h R. Th at is usua l ly the case .

V W V \ AR

Figure 5-4 Me asur em ent o f t he cur ren t / t h ro ug h a re s i s to r R. and of the vol tage V

across i t . A circle marked 1 represents an ammeter, and a circle marked V, a

voltmeter.

/16 Direct Currents in Electric Circuits

5.2.1 THE JOULE EFFECT

W h e n a cur r en t f lows th rough a c o n d u c t o r , the charge ca r r i e r s ga in k ine t i c

energy by m o v i n g in the elect r ic f ie ld . Their veloci ty never becomes large,

however , because they can t ravel only a shor t d i s t ance befo re co l l id ing wi th

a n a t o m or a m o l e c u l e . The k ine t i c energy ga ined by the car r i e r s thus se rves

to raise the t e m p e r a t u r e of the c o n d u c t i n g m e d i u m . T h i s p h e n o m e n o n is

called the Joule effect.

T h e e n e r g y t h a t is d i ss ipa ted as h e a t in t h i s way is eas i ly ca lcu la t ed .

C o n s i d e r a c u b i c m e t e r of c o n d u c t o r w i t h E para l l e l to one e d g e . The c u r r e n t

dens i ty J is the charge f lowing through the c u b e in one s e c o n d . The v o l t a g e

dif ference across the c u b e is E. T h e n the k ine t i c energy ga ined by the car r i e r s

and los t to the c o n d u c t o r as h e a t in one s e c o n d is JE, or <T EZ

, or J2

/o, fromEq . 5-1. So the h e a t e n e r g y p r o d u c e d per c u b i c m e t e r in one s e c o n d is oE

2

o r J2

/a.

If one has a r es i s t ance R c a r r y i n g a c u r r e n t L t h e n the vo l t age ac ross

it is IR and the k ine t i c energy ga ined by the c h a r g e c a r r i e r s in one s e c o n d

is I(IR). T h e n the p o w e r d i s s i p a t e d as h e a t is I2

R.

5.3 NON-LINEAR RESISTORS

A l t h o u g h O h m ' s law app l i es to ord inary r es i s to r s , it is by no m e a n s g e n e r a l .

EXAMPLE: INCANDESCENT LAMPS

F i g u r e 5-5 s h o w s th e c u r r e n t as a funct ion of vol t age for a 60-wa t t incandescent

lamp. The re s i s t ance V/I is a b o u t 20 o h m s at a few vol t s , an d a b o u t 250 o h m s at

100 volts. The reason is t ha t , in the i n t e rva l , the t e m p e r a t u r e of the tungsten filament

has changed from 300 to 2400 kelvins .

If th e t e m p e r a t u r e of the f i l ament were ma in ta ined cons tan t by exte rna l means ,

sa y by i m m e r s i n g it in a c o n s t a n t - t e m p e r a t u r e oil ba th , t hen it would fo l low Ohm's

law.

EXAMPLE: VOLTAGE-DEPENDENT RESISTORS

Voltage-dependent resistors ut i l i ze ce ramic s emiconduc tors for which V = CIB

,

w h e r e C and B ar e c o n s t a n t s , and B is l e s s than un i ty . Thi s equa t ion appl i e s at

r o o m t e m p e r a t u r e . F i g u r e 5-6 s h o w s th e cur ren t -vo l t age curve for one p a r t i c u l a r

type wi th C = 100 an d B = 0.2.



11 7

I I I I I I I I I i I0 50 100

V (volts)

Figure 5-5 Cur ren t -vo l t age re l a t ionsh ip for a 60-wa t t i ncandescen t l amp.

0 25 50 75 100

V (volts)

Figure 5-6 Cur ren t -vo l t age re l a t ionsh ip for a ce r t a in type of vo l t age -dependent

res is tor .

Vol t age -dependent re s i s to rs a re used on t e l ephone l ines fo r shor t -c i rcu i t ing

l ightning to ground, for suppress ing sparks , and so on, s ince thei r res is tance de

creases a t large vol tages .

We shal l be concerned henceforth only wi th l inear res is tors .



11 8

0

Figure 5-7 Three res is tors connected in ser ies .

5.4 RESISTORS CONNECTED IN SERIES

Figure 5 -7 shows th ree r es i s to r s connec ted in series. Since

V = IRi + IR 2 + IRi = HR, + R 2 + R 3) = IR S. (5-11)

the r es i s to r s have a to t a l r es i s t ance

R s = R l + R 2 + R 3. (5-12)

Th e resista nce of a set of resis tors con nec ted in ser ies is the su m of th e in

d iv idua l r es i s t ances .

5.5 RESISTORS CONNECTED IN PARALLEL

In Fig . 5-8 the resistors are connected in parrallel. T h e n

_ V V V _ ( 1 1 1 \

7 = R x

+ R~ 2

+ ~R 3 ~ V + R~ 2

+ R~J

and the r es i s t ance R p is given by

1 1 1 1

Rp Ri R 2 R 3'

R _R

lR

2R

3

- r;(5

-131

(5-14)

(5-15)

11 9

O- <Z>

- 0 -V \ A A A

W v A A

R 3

Figure 5-8 Three res is tors connected in paral le l .

I t i s useful to rem em be r that , for tw o resistance s in pa ral le l ,

R,R2

" R { + R 2

I t i s of ten co nve nien t to use conductance

(5-16)

G = \/R, (5-17)

ins t ead o f r es i s t ance . Th en

Gi + G 2 + G 3. (5-18)

Th e con du c ta nce o f a set of r es i s to r s con nec te d in para l l e l is t he sum of the

i n d i v i d u a l c o n d u c t a n c e s .

Conduc tances a r e expressed in amperes per vo l t o r in S i e me ns . W e

have a l r eady used the Siemens in Sec. 5.1.

THE PRINCIPLE O F SUPERPOSITION

Th e p r inc ip le o f superp os i t io n which w e s tud ied in Sec . 2 .3 app l i es to e l ec t r ic

c i r cu i t s compr i s ing sources and linear res isto rs: if a c i rcui t com pris es tw o

or more sources , each source ac t s independen t ly o f the o ther s and , a t any

po in t in the c i r cu i t, t he cu r r e n t i s t he a lgebra ic sum of the cu r r en t s p ro duc ed

by the ind iv idua l sources .



120

Figure 5-9 The current flowing through

th e circuit is the algebraic sum of the

c u r r e n t due to the 6-volt battery, plus

t h a t due to the 12-volt battery.

2C1

NA A A / ^

ion

12V

EXAMPLE: SIMPLE CIRCUIT WITH TWO SOURCES

F i g u re 5-9 shows a simple circuit comprising three resistors and two sources, all

c o n n e c t e d in series. The tota l resis tance is 15 ohm s. The 12-volt sour ce prod uce s a

c o u n t e r c l o c k wi s e cur ren t of 12/15 = 0.8 ampe re, while the 6-volt source pro duc es a

c lockwise curre nt of 0.4 ampere. The net curr ent is 0.4 ampere, counterc lockwis e.

T he power dissipated as heat in the 2-ohm resistor is (0.4)2

x 2 = 0.32 watt,

and the total power dissipated in the circuit is (0.4)2

x 15 = 2.4 wat ts .

5.7 KIRCHOFF'S LAWS

Let us consider a general circuit composed of resistors and voltage sources,

for example, that of Fig. 5-10.

Junctions such as A, B, C, D, E are called nodes; connections between

nodes, such as AB, BD, DC .. ., are called branches; closed circuits, such

as ABDE, or ABCDE, or BCD, are called meshes.

Kirchoff's laws provide a method for calculating the branch currents

when the values of the resistances and of the voltages supplied by the sources

are known.

Kirchoff's current law states that, at any given node, the sum of the

incoming currents is equal to the sum of the outgoing currents. This is simply

because charge does not accumulate at the nodes. For example, at A,

h = h+ h- (5-19)



12 1

Figure 5-10 Part of a complex circuit .

A c c o r d i n g t o Kirchoff's voltage law, the vol tage r ise , or the vol tage

d r o p , a ro un d any g iven mesh i s ze ro . For e xam ple , t he vo l t age r ise from A

t o B, to D, to £ , and bac k to A, is clearly zero, so

-V, + / , / ? , - I2R 2 + I5R 5 + I6R b = 0. (5-20)

To f ind the b ranch cur r en t s , one wr i t es down as many equa t ions o f

the above two types as the re a r e b ranches , and one so lves the r esu l t ing

s y s t e m of s i m u l t a n e o u s e q u a t i o n s .The d i r ec t ions chosen fo r the cu r r en t s a r e a rb i t r a ry . I f t he ca lcu la t ions

give a posi t ive value for , say, / , , then that current f lows in the di rect ion of

the ar ro w . I f the calc ula t ion s give a neg at ive value , then th e cur ren t f lows

in the oppos i t e d i r ec t ion .

EXAMPLE: THE BRANCH C URRENTS IN A TWO-MESH CIRCUIT

Fo r the c i rcu i t of F ig . 5 -11 , we have th ree un kno wn s , / , , I2. 13. Hence we need

three equa t ions . A t node A,

h = h + h- (5-21)

For the l e f t -hand mesh , s t a r t ing a t B, c lockwise .

V — lxR — I

3R = 0. (5-22)



12 2

i — A A / V f M A / — it /

3

K T

Figure 5-11 One can ca l cu la t e Ilt

I2, / 3 , given .R an d K

Similar ly, for the r ight-hand mesh, s tar t ing a t B, c lockwise ,

Z 3i? - 7 2 P - I2R = 0.

Solving, we f ind that

I , = 3F /5R , / , = 1 5K. / 3 = 2 F / 5 R .

(5-23)

(5-24)

SUBSTITUTION THEOREM

Let us return to Fig . 5-10 and to Eq. 5-20. Consider the resistance R 2. T h e

vol tage drop across i t i s I2R 2. Let us assume tha t I2 i s real ly in the di rect ion

of the a r row. Then the upper end o f R 2 i s pos i t ive . Now le t us subs t i tu t e

fo r R 2 a vol tage sou rce as in Fig . 5-12. Eq ua t io n 5-20 is unaffected. Simila r ly ,

t h e c o r r e s p o n d i n g e q u a t i o n f o r t h e m e s h BCD i s unaf fec ted . The n the c u r r en t s

11, I2, ^ 3 , and so for th , are unaffected.

Thus , i f a cu r r en t I f lows th rough a r es i s t ance R, one can r ep lace the

res i s t ance by a vo l t age source IR of the sam e pol ar i ty wi tho ut af fect ingan y of the cur ren ts in the c i rcui t .

Inversely , if we ha ve a vol ta ge so urce V pass ing a cu r r en t / , with the

current entering the source at the positive terminal, one can r ep lace the source

by a r es i s t ance V/I without af fect ing any of the currents in the ci rcui t .

This i s the substitution theorem.

5.8.1 EXAMPLE: RESISTOR REPLACED BY A BATTERY

In Fig. 5-13a,

h = h + h,

fR + I2R = V,

-I 2R + 1,R = 0,

(5-25)

(5-26)

(5-27)

R /, /,

+

1 AA/V^—*—< — 1

+ +

j V R * - K/3

<(6)

Figure 5-13 Th e curren ts / , , I2,1 3 are not affected by subs t i tut ing the bat tery F/3

for the res is tor R.




giving

7j = 2V/3R, I2

= V/3R, l3

= VJ3R. (5-28)

Us ing now the subs t i tu t ion theorem, we can rep lace the re s i s to r on the r igh t

by a source I3R = V/3, giving the c i rcui t of Fig. 5-13b. Then

h = h+ h, (5-29)

1,R + 1 2R = V, (5-30)

-I 2R + (V/3) = 0. (5-31)

Solving, we f ind the same values for J 1 , 1 2 , / 3 as before .

Inversely, one can subs t i tute the c i rcui t of Fig. 5-13a for that of Fig. 5-13b

without affect ing J j , I2, 7 3 .

EXAMPLE: BATTERY REPLACED BY A RESISTOR

In Sec. 5.6.1 we found that th e curre nt in Fig. 5-9 is cou nter cloc kw ise . Since a cur

rent of 0.4 am pe re f lows into the pos i t ive term inal of the 6-vol t sourc e, we can

rep lace th i s source by a re s i s tance of 6 /0.4 = 1 5 oh ms wi tho ut a f fec ting the cur re n t

f lowing in the c i rcui t . Let us check. We no w have a 1 2-vol t source an d a tota l res is

tanc e of 15 + 10 + 3 + 2 = 30 ohm s. Th en th e curre nt i s 12/30 = 0.4 amp ere ,

a s p rev ious ly .

5.9 MESH CURRENTS

I f we use Ki r cho f f ' s l aw s as such , the nu m be r o f un kn ow n c ur r en t s is equa l

to the number o f b r anches . I t i s poss ib le to r educe the number o f unknowns

in the fol lowing way . Ins tea d of usin g bran ch c urr en ts as in Fig . 5-10, on e

uses mesh currents as in Fig . 5-14, s ince there are fewer mesh currents than

bra nc h cur r en t s . Ki r ch of f ' s cu r r en t l aw i s then au to ma t i ca l ly sa ti s fi ed , andwe need to app ly on ly the vo l t age l aw.

Th e adva n ta ge o f us ing mesh cur r en t s com es from the f ac t t ha t t he

t ime r equ i r ed to so lve a se t o f s im ul t an eou s equa t ion s decreases r ap id ly

as the number o f equa t ions i s r educed .

In the end , one mu st find the b ra nc h cu r r en t s , bu t tha t is a s im ple

m a t t e r , o n c e t h e m e s h c u r r e n t s a r e k n o w n .



12 5

Figure 5-14 Mesh currents in the c i rcui t of Fig. 5-10.

5.9.1 EXAMPLE: THE MESH CURRENTS IN A TWO-MESH CIRCUIT

In the circuit of Fig. 5-15 we have two meshes . S ta r t ing a t B i n bo th cases and pro

ceeding in the c lockwise di rect ion,

a n d

V — IaR — (/„ - I„)R = o,

(/„ - Ib)R - I

b2R = 0.

7 = 3 1 7 5 / ? , h=V/5R.

(5-32)

(5-33)

(5-34)

Note tha t we have on ly two unknowns , / „ and Ib. ins tead of the three we had

prev ious ly .

R A R

- W W — f — v A A A / ^ — i

Figure 5-15 Mesh currents in the c i rcui t of Fig. 5-11.




5.10 DELTA-STAR TRANSFORM ATIONS

Figure 15-6 shows th ree nodes A, B, C , delta-connected in par t a a n d star-connected in par t b. T h e n o d e s A, B, C form par t of a larger c i rcui t , and

can be connected together e i ther as in Fig . 5-16a or as in Fig . 5-16b. We shal l

see tha t , if cer ta in rela t ion s are sat isfied betw een the three resis tance s o n

the lef t and the three on the r ight , one ci rcui t may be subst i tu ted for the other

wi th ou t d i s tu rb ing the mesh cur r e n t s . I n f ac t, if one had two box es , one

con ta in ing a de l t a c i r cu i t , and the o ther con ta in ing a s t a r c i r cu i t sa t i s fy ing

E q s . 5-38 to 5-40, wi th only th e term ina ls A, B, C s h o w i n g o n e a c h b o x ,

the re would be no way o f t e l l i ng which box con ta ined the de l t a and which

con ta ined the s t a r . The two c i r cu i t s a r e thus sa id to be equivalent.

Th is t ran sfo rm at io n is of ten used for s impl i fying the calc ulat ion ofm e s h c u r r e n t s .

We could f ind R A, R B, R c in terms of R a, R b, R c, and inversely , by

a s s u m i n g t h e s a m e m e s h c u r r e n t s IA, IB, Ic in the two ci rcui ts , as in the

f igure , an d then m ak ing the vo l t ages V A — VB,V B - V c, V c - V A in Fig. 5-16a

equa l to the co r r esp on d in g vo l t ages in F ig . 5 -16b . Th i s is t he ob jec t o f P ro b .

Figure 5-16 (a ) Three -node c i rcu i t hav ing the shape of a cap i t a l de l t a , (b ) F o u r - n o d e

circui t having the shape of a s tar . Nodes A, B, C are part of a more extens ive

circu it . It is sh ow n th at, if eith er Eq s. 5-38 to 5-40 or Eq s. 5-44 to 5-46 a re

sat is f ied, the above two ci rcui ts are complete ly equivalent .

A A

5.10 Delta-Star Transformations 127

5-19 . W e sha l l use here ano the r de r iva t io n th a t is som ew hat l ess conv inc ing ,

bu t shor t e r .

We disconnect the del ta and the star f rom the rest of the ci rcui t . I f thec i r cu i t s a r e equ iva len t , t hen the r es i s t ance be tween A a n d B must be the

s a m e i n b o t h :

RAB = „*<[V R b l = R A + R B- (5-35)

S imi la r ly ,

Ra+ R h + R c

N ote tha t , in the abo ve equ a t ion s , i s a ssoc ia t ed wi th RhRc. R B with i? c/?„ .

a n d R c with R 0R, , . Thus , s imply by inspec t ion , we have tha t

R,. + R h + R r

R B = R

f° , (5-39)

R c=

D- <

5

-4

° )

Now we a l so need the inver se r e l a t ionsh ip . We cou ld deduce i t f rom

the above th r ee equa t ions , bu t tha t would be r a ther long , and no t ve ryins t ruc t ive . I ns t ead , l e t us use the conduc tance be tween node A a n d t h e

n o d e s B a n d C shor t - c i r cu i t ed toge ther . Th i s g ives

G b + G e = r

Gf> + °i • (5 -41 .VIA + O il + (/(•



128 Direct Curre nts in Electric Circuits

Similar ly ,

G c + G " - G , + G R + G r ' , 5 " 4 2 )

G r ( G , + G JG a + G b=

c\ A . ,5-43)G .4 + G B + O c

Again , by inspec t ion .

GBGC

G A+ G B + G c

GCGA

G A+ G B + G c

GAG

B

(5-44)

(5-45)

G c = . (5-46)G A + G B + G c

EXAMPLE: TRANSFORMATION OF A SYMMETRICAL DELTA

AND OF A SYMMETRICAL STAR

In Fig. 5-16a, suppose R a, R b, R c are a l l 100-ohm res is tances . Then, from Eqs . 5-38

to 5-40, the equiva lent s ta r-co nne cted c i rcui t is m ad e up of thre e res is tances of

(100 x 100)/300 = 33 oh m s each .

If , ins tead, we had a s tar-connected c i rcui t wi th res is tances R A, R B, R c of 100 ohms

each , t hen the equiva len t de l t a -connec ted c i rcu i t would have th ree iden t i ca l re s i s

t ances w i th condu c tanc es g iven by Eqs . 5-44 to 5 -46 : 10" 4 / ( 3 / 1 0 0 ) = 1 0 " 2 /3 Siemens .

The res is tances in the del ta would thus each have a res is tance of 300 ohms.

5.11 VOL TA GE A ND CURRENT SO URCES

The idea l vo l t age source supp l i es a cons tan t vo l t age tha t i s i ndependen t

o f the cu r r e n t d r a wn , and hence o f the load res i s t ance . A go od e l ec t ron ica l ly

vol tag e-stab i l ized po w er sup ply is c lose to ideal , up to a speci f ied cu rren t ,

beyond which the vol tage fal ls as in Fig . 5-17.

S imi la r ly , t he idea l cu r r en t source supp l i es a cons tan t cu r r en t tha t

i s i ndependen t o f the vo l t age ac ross the load . Aga in , cu r r en t - s t ab i l i zed



12 9

O I

Figure 5-17 O ut pu t vo l tage as a funct ion of ou tpu t curre nt , for a vol tag e source .

/

0 V

Figure 5-18 O ut pu t cur ren t as a funct ion of ou tpu t vol tage, for a curr ent sourc e.

suppl ies are near ly ideal , but only up to a speci f ied vol tage beyond which

the curre nt de crea ses . Th is is sh ow n in Fig . 5-18.

5.12 THEVENIN'S THEOREM

I t i s f ound e xper im en ta l ly tha t t he vo l t age supp l i ed by a source decreases

wh en the load curre nt incre ases. Th is is t rue , even for vo l tage-s tabi l ize d

sources: the plateau in Fig . 5-17 is not real ly f la t but s lopes down sl ight ly .



13 0

• I

- Q -

0

V

(a ) (b )

Figure 5-1 9 (a ) Any real sourc e is equiv alent to an ideal sou rce V 0 in series with a

re s i s t ance Rt. The source i s shown he re connec ted to a load re s i s t ance R d r a w i n g a

cur ren t / . The vo l t age V be tween the t e rmina l s i s V 0 — IR h o r IR. (b) T h e t e r m i n a l

vo l t age V as a funct ion of the load current I. The s lope is — Rj .

Thevenin's theorem state s th at any so ur ce — a f lashlight bat tery , an

osc i l l a to r , a vo l t age - o r cu r r en t - s t ab i l i ze d pow er supp ly , e t c .—is e qu iva len t

to an idea l vo l t age source V 0 in series with a resistance R,, as in Fig. 5-19,

where R, i s cal led the output resistance, or the source resistance. An idea l

vol tage source, by def ini t ion, has a zero R, .

In Fig . 5-19, the current I is F 0 / (R , + R) , whi le the vo l tag e V is IR .

T h e v o l t a g e V i s smal l e r than V 0 by the factor R/(R; + R) , be cau se of the

vo l t age d rop on R, . Wi th R d i sconnec ted (R -> o c ) , V is eq ua l to F 0 .

The internal resistance R, i s g iven by the slope of the curve of V as a

funct ion of I. I f on e increase s / by decrea sing R, then V decreases and R,i s equa l to lAK /A/ l . For one par t i c u la r e l ec t ron ica l ly ro / f age- s t ab i l ized

power supp ly , t he ou tpu t vo l t age d rops by one mi l l ivo l t when the ou tpu t

cu rren t increas es f rom zero to 100 mil l iam pere s. Th us R, i s 0 .01 oh m . A

smal l ba t t e ry charger has an in t e rna l r es i s t ance tha t i s o f the o rder o f one

ohm: i f t he load cur r en t inc reases by one ampere , t he ou tpu t vo l t age d rops

by abou t one vo l t .

A cur r a i r - s t ab i l i zed power supp ly has a l a rge V 0 and a large R ; . I ts

ou tpu t cu r r en t i s V0/Rhas long as R « R, , Th us i ts ou tpu t vo l t age (F 0 / R , ) R

i s p ropor t iona l to the load r es i s t ance R. I f t he load r es i s to r i s d i sconnec ted ,

then R is inf ini te and the output vol tage r ises to some l imi t ing value thatcan be much smal l e r than V 0 an d tha t is f ixed in t e rna l ly . I n one par t i c u la r

cur r en t - s t ab i l i zed supp ly , V 0 i s 50 ki lovo l ts , R ; i s 100 ki lohms, and the l imi t ing

vol tage is 50 vol ts .

Thus , i f V 0 a n d R t are known, one can p red ic t t he va lues o f V a n d I

fo r a g iven load r es i s t ance R. Theven in ' s t heorem there fo re p rov ides us wi th

a s imple mode l fo r desc r ib ing the opera t ion o f a r ea l source .



5.12 Thevenin's Theorem 131

Th ev en in ' s th eo rem is a lso useful for calc ulat i ng cu rre nts f lowing

through ci rcui ts , as we shal l see in the fol lowing examples.

EXAMPLE: THEVENIN'S THEOREM APPLIED TO A SIMPLE CIRCUIT

Let us see how Thevenin 's theorem appl ies to the c i rcui t of Fig. 5-20a, where the

part of the c i rcui t to the lef t of the terminals is cons idered as a source feeding the

load re s i s t ance R. H e r e V x i s a vol tag e-s tabi l ized sour ce wi th an intern al res is tance

tha t i s neg l ig ib l e com par ed to R j .

Th e Th eve nin equiv alent of this sour ce is show n in Fig. 5-20b. The vol tage V 0

i s the vol tag e betwee n the term inal s of the c i rcui t of Fig. 5-20a when the res is tor R

i s removed. Also, the res is tance Rt i s t he re s i s t ance measured a t t he t e rmina l s w hen

R i s r emoved , and when V x i s reduced to zero. This is s imply R, a n d R 2 in paral le l .The cur ren t F 0 / (Ri + R) in Fig. 5-20b i s the same curre nt on e would f ind direct ly

from Fig. 5-20a using mesh currents as in Sec. 5.9.

figure 5-20 The circuit to the left of the terminals in (a ) gives the s ame cur ren t I in

the load res is tance R as the circuit to the left of the terminals in (h).

EXAMPLE AND REVIEW: THE WHEATSTONE BRIDGE

Figure 5-21a shows a Wheatstone bridge, s imilar to that of Pr ob . 5-8, except t ha t

we have shown the re s i s t ances R s a n d R d of the source and of the detector , respec

t ively, and that R d is not infinite here.

We wish to f ind the current in the detector , when the res is tance of the upper

lef t -hand branch is R( l + x) , as a funct ion of the curre nt I in tha t b ran ch .

W hen x = 0 , t he br idge is ba l ance d and the cur ren t t h r oug h the de tec tor i s

zero (Prob. 5-8) .

We could so lve th i s p rob lem by us ing e i the r b ranch or mesh cur ren t s . Ins t ead ,

we shal l solve i t in a more roundabout way that wi l l use most of the t r icks that we

have l ea rned in th i s chap te r .



13 2

Figure 5-21 (a ) Whea t s tone br idge c i rcu i t , showing the in t e rna l re s i s t ance of the

s o u r c e Rs and the res is tance of the detector R

d. W h e n x = 0, the bridge is balanced

and no cur ren t f lows th rough R d. (b, c, d) Success ive t rans fo rma t ions of the c i rcu i t

that are used in calcula t ing the current through the detector as a funct ion of x.

a ) Le t t he cur ren t t h rough the re s i s t ance R(\ + x) be / . Then, us ing the substitu

tion theorem, we can replace the res is tance Rx by a source IRx and we have the

circuit of Fig. 5-21b.

b) We now have two sources , V s a n d IRx. Accord ing to the principle of super

position, each one ac t s i ndependent ly o f the o the r . The cur ren t s due to V s a re now

those for the ba lanced br idge . Thus the cur ren t t h rough the de tec tor i s due on ly to

the source IRx, and we may d i s rega rd V 5. We now have c i rcui t c .

c ) Res i s t ances R s, bR, abR form a del ta . I f we use the delta-star transformation,

we have c i rcui t d. We now have a f i f th node, £.

5.12 Thevenins Theorem 133

Using the t rans forma t ion ru l e s ,

(bR)R s n (ahR)R s

(5-47)bR +.R S + abR' abR + R s + bR'

= BR, = aBR, (5-48)

w h e r e B is R2/R .

d) We can now use Thevenins theorem to f ind the current through the dectector

W e c o n s i d e r R d as the load res is tance, and a l l the res t of the c i rcui t as a source.

The open-c i rcu i t vo l t age K 0 i s that between D a n d B when the branch DB is

cu t . Th e cur ren t f lowing a ro un d the ou te r b ranc hes i s

IRx

R + R 2 + Ri + aR'

a n d

IRx

(aR + R 2). (5-49)R + R 2 + K 3 + aR

IRx

R + BR + aBR + aR(aR + aBR), (5-50)

ail + B)= IRx — , (5-51)

(1 + a)(\ + B)'

IRx. (5-52)1 + a

Note tha t t h i s vo l t age i s i ndependent o f R s, of b, and of the quant i ty B of Eq. 5-48.

We now requi re the re s i s t ance of R + R t of Sec. 5.12, which is the resistance

measured a t a cu t i n the branch DB, when the source IRx i s replaced by a short -

c i rcui t . We can f ind this res is tance from Fig. 5-2Id, bu t i t i s s imp ler to re tur n to

Fig. 5-21b. We imagine a source inserted in the branch DB, wi th bo th V s a n d IR x

rep laced by shor t -c i rcu i t s . Th e br idge i s ba l anced . Then a source in the bra nch

DB gives ze ro cur ren t i n the branch AC , so we may d i s rega rd the re s i s t ance Rs. So

(R + bR)(aR + abR)R ' + R > = R< + R

+bR + aR

+abR' ^

(1 + b)(\

i(l + b

1 + a

o(l + b)

Rd+ , R (5-55)



134 Direct Currents in Electric Circuits

and, finally,

V0

a IRx /(l + a)

h = = „ , r „ [\„ u , (5-56)d + R, R d + 0 ( 1 + fe)«/( l + a) ] '

/.V

( 1 + - ) (R d/R )a

(5-57)

If we wish to maximize the ra t io Idjl, t hen R d/R should be sma l l , b should be

smal l , and a large. These las t two condi t ions are not a t a l l cr i t ical and one can say,

a s a ru l e o f t humb, tha t bo th a a n d b should be approxima te ly equa l t o un i ty .

5.13 SOLVING A SIMPLE DIFFERENTIAL EQUATION

In the nex t sec t ion we sha l l have to so lve the fo l lowing equa t ion :

a d-X + bx = c, (5-58)at

w h e r e a, b, c a r e k n o w n c o n s t a n t s , a n d x is an un kn ow n funct ion of f.

This i s a differential equation. I ts general solution i s co mp ose d of two

p a r t s : (a) the particular solution, ob ta ined by conserv ing on ly the l as t t e rm

on the left,

x i = c/b, (5-59)

plus (b) the complementary function, whic h is the solu t ion for c = 0, or

x2 = Ae- {h,a)t, (5-60)

w h e r e A is a constant of integration.

T h u s

x = x , + x 2 = C-+ Ae-Wa)t. (5-61)

The numer ica l va lue o f the constant of integration A i s obtained, as a rule ,

f rom the known va lue o f x a t t = 0.

I t i s easy to check this solu t ion by su bst i tu t ing i t in the or igin al eq ua t io n

(Eq . 5 -58). A un iquen ess t heo rem s t a t es tha t t h i s so lu t ion is t he on ly p oss ib le

o n e .



5.14 Transients in RC Circuits 135

5.14 TRANSIENTS IN RC CIRCUITS

When a source is connected to a circuit comprising resistors and capacitors,transient currents flow until the capacitors are charged. Such currents are

large, at first, and then decay exponentially with time.

5.14.1 EXAM PLE: SIMPLE RC CIRCUIT

F i g u r e 5-22 s h o w s a c a p a c i t o r C, i n i t i a l ly uncharged , t ha t is c o n n e c t e d at t ime t = 0

to a vol t age source V t h r o u g h a res is tor R.

W e ca n find th e vol t age ac ros s th e c a p a c i t o r by us ing Ki rchof f ' s vo l t age law.

A d d i n g up the vol t age r i s e s a round th e l oop , s t a r t ing at the lowe r lef t -hand corn er ,

c lockwise ,

V — IRQ

c

Since / is dQ/dt,

RdQ

dt

1

0.

V

and, from th e prev ious s ec t ion ,

Q = CV + Ae-'IRC.

Since, by h y p o t h e s i s , Q = 0 a t t = 0, A = — C F a n d

Q = CF(1

After a t ime r » RC, Q % CV.

The vo l t age ac ros s th e c a p a c i t o r is

Vc = F( l I.

(5-62)

(5-63)

(5-64)

(5-65)

(5-66)

F igure 5-23 shows Vc/V as a funct ion of t i m e for R = 1 m e g o h m , C = 1 mic rofa rad ,

RC = 1 s e c o n d .

Figure 5-22 An RC circui t .

13 6

1 .0

0 2 3t (seconds)

4 5

Figure 5-23 The vol tage V c across the capaci tor C in Fig. 5-22 as a funct ion of the

t ime . I t is a s sum ed th a t V c = 0 a t t — 0 and tha t RC = 1.

T h e p r o d u c t RC is called the time constant of the c i rcui t . For t = RC, V c/V =

1 - 1/e as 2/ 3.

I t wi l l be shown in Prob. 5-23 that , i f C is discharged through R, V c = Ve~"RC.

T h e current density vector J points in the di rect ion of f low of posi t ive charge

and i s expressed in am pere s per squ are mete r .

For any surface S b o u n d i n g a v o l u m e t .

which is the law of conservation of charg e.

T h e n e t c h a r g e d e n s i t y i n s i d e a h o m o g e n e o u s c o n d u c t o r c a r r y i n g a

cur r en t under s t eady- s t a t e cond i t ions i s ze ro .

F o r o r d i n a r y c o n d u c t o r s , Ohm's law a p p l i e s :

5.75 SUMMARY

(5-4)

J = <rE, (5-1)

where J is t he cu r r en t dens i ty , a i s the conduct ivi ty , and E is the elect r ic f ie ld

in tens i ty . Th e e l ec t r i c char ge dens i ty ins ide a un i fo rm cur r en t - c a r ry ing

c o n d u c t o r is z e r o u n d e r s t e a d y - s t a t e c o n d i t i o n s .



5.15 Summa ry 137

Ohm's l aw can a l so be s t a t ed as

/ = V/R, (5-9)

w h e r e I i s the curre nt f lowing t hr ou gh a resistance R , acro ss wh ich a vol tage

V i s m a i n t a i n e d .

The power d i ss ipa ted as hea t by the Joule effect per cubic meter i s

J2/a, and the power d i ss ipa ted in a r es i s to r R car ry ing a cu r r en t I is I2

R.

Resi s to r s for which / is p ro po r t io na l to V are said to be linear, o r ohmic.

Other r es i s to r s a r e sa id to be non-linear.

The resistance of a set of resistors connected in series i s equ al to th e

sum of the i r i nd iv idua l r es i s t ances .

The conduc tance o f a se t o f resistors connected in parallel i s equ al to

the sum of the i r i nd iv idu a l co nd uc ta nce s .

T h e principle of superposition states that , as long as a l l the resistors

in a c i r cu i t a r e l i near , each source ac t s independen t ly o f the o ther s and

produces i t s own se t o f cu r r en t s .

Kirchoff's current law s t a t es tha t t he a lgebra ic sum of the cu r r en t s

ente r ing a no de , in an elect r ic c i rcui t , is equ al to zero . His voltage law s t a t es

tha t t he sum of the vo l t age d ro ps a ro un d a mesh i s equa l to ze ro . Cu r ren t s

in the var ious b ranches o f a c i r cu i t a r e usua l ly ca lcu la t ed by us ing mesh

cur r en t s and the vo l t age l aw.

A c c o r d i n g t o t h e substitution theorem, one can r ep lace a r es i s t ance Rin a c i rcui t by a vol tage source IR of the co r r ec t po la r i ty wi thou t a l t e r ing

the cur ren ts f lowing t hr ou gh the ci rcui t or , inversely , if a cu rre nt ente rs a

v o l t a g e s o u r c e V at i t s posi t ive terminal , one can replace the source by a

r es i s t ance V/I w i t h o u t c h a n g i n g t h e c u r r e n t s .

O n e c a n t r a n s f o r m t h e delta-connected circuit of Fig. 5-16a into the

star-connected circuit of F ig . 5 -16b by means o f the fo l lowing equa t ions :

RA =RhRr

R a + R b + R c

(5-38)

RBR , R „

R„ + R„ + R ,(5-39)

R,R„R„

R„ + Rh + R r

(5-40)




Also,

Gb

=

G , =

GB

GC

GA

+ GB

+ Gc

GCG

A

GA

+ GB

+ Gc

GAG

B

G, + GB+ Gr

(5-44)

(5-45)

(5-46)

A voltage source supplies a voltage that is nearly independent of the

current through the load; a current source supplies a current that is nearly

independent of the voltage across the load.

Thevenins theorem states that any source may be represented by an

ideal voltage source, in series with a resistance called it s output resistance.

When a source is connected to a circuit comprising capacitors, transient

currents flow temporarily in the circuit until all the capaci tors are charged.

PROBLEMS

5-1E CONDUCTION IN A NON-UNIFORM MEDIUM

Two p lane pa ra l l e l copper e l ec t rodes are s e p a r a t e d by a pla t e of t h i cknes s s

whose conduc t iv i ty a varies l inearly from a0. n e a r the pos i t ive e l ec t rode , to u 0 + a

n e a r th e negat ive e lect rode. Neglect edge effects . The cur ren t dens i ty is J.

F i n d the electric field intensity in the c o n d u c t i n g p l a t e , as a funct ion of th e dis

t a n c e x from the pos i t ive e l ec t rode .

5-2E RESISTIVE FILM

A square f i lm of N i c h r o m e (a n al loy of nickel and c h r o m i u m ) is d e p o s i t e d on a

sheet of glass , and copper e l ec t rodes are t hen depos i t ed on two oppos i t e edges .

S h o w t h a t the re s i s t ance be tween the e lec t rodes depends on ly on the t h i cknes s

of th e film an d on its conduc t iv i ty , as l ong as the film is s q u a r e .Thi s re s i s t ance is given in ohms per square. Nichrome f i lms range approxi

mately from 40 to 40 0 o h m s pe r s q u a r e .

5-3E RESISTOJET

F i g u r e 5-24 s h o w s the pr inc ip le of o p e r a t i o n of the res is toje t , which is used as

a t h r u s t e r for cor rec t ing e i the r the t r a j ec tory or the a t t i t u d e of a sate l l i te . See a l so

Probs . 2-14, 2-15, an d 10-11.



13 9

Figure 5-24 S che ma t ic diag ram of a res is toje t . Th e prop el la nt gas P is ad mi t ted on

the l e f t , hea ted to a h igh t empera ture by the re s i s to r , and exhaus ted th rough the

nozzle . See Prob. 5-3.

Assuming comple te convers ion of the e l ec t r i c ene rgy in to k ine t i c ene rgy , ca l cula te the thru s t for an input p ow er of 3 ki lo wa t ts and a f low of 0.6 gram of hy dro gen per

second .

5-4E JOULE LOSSES

A res is to r has a re s is t ance of 100 k i loh ms an d a pow er ra t ing of one -qu a r t e r wa t t .

Wh a t i s t he maxi mu m vo l t age tha t can be appl i ed ac ros s i t ?

5-5E VOLTAGE DIVIDER

Figure 5-25 shows a vo l t age d iv ide r o r a t t enua tor . A source suppl i e s a vo l t age

V, t o the input , and a h igh- res i s t ance load (no t shown) R » R 2 i s con nec ted acro ss P 2 .

The vo l t age on R 2 a n d R is V 0.S h o w t h a t

K = R 2

F Ri + R 2

R

Figure 5-25 Vol tag e divid er: V 0 is

R 2/(R 1+ R

2) t imes V

t. See P rob . 5 -5 .



14 0

unknown. When the s l id ing contac t i s

adjus ted for zero current , Vb/V

ais

Figure 5-26 P ote nt i om ete r c i rcui t . Th e

vol t age V a i s known, and V b is

equal to R 2/(R! + R 2). See Prob. 5-6.

5-6E POTENTIOMETER

Figure 5-26 shows a po ten t io mete r c i rcu it .

Show tha t , when7

= 0,

Th is c i rcui t i s extens ively used for me asu ring the vol tage sup pl ied by a sour ce

V b, wit hou t draw ing any curren t from i t . Th e c ircui t is used in s t r ip- cha rt rec ord ers .

In that case the current 7 i s ampli f ied to actuate the mo to r that displa ces the pen, and

s imul taneous ly moves the tap on the res is tance in a di rect ion to decrease 7. This is one

type of servomechanism.

5-7E SIMPLE CIRCUIT

Show that , in Fig. 5-27,

V = —R2 - R

VR2

+ R

if t he vo l tme te r d raw s a neg l ig ib l e amo un t o f cur ren t .

We shal l use this resul t in Prob. 15-5.

R, Ri

I — A W f A A A / — |

+ +

Figure 5-27 See P ro b. 5-7.r i



14 1

WHEATSTONE BRIDGE

Figure 5-28 shows a c i rcui t known as a Wheats tone bridge. As we shal l see , the

vol t age V i s zero when

R1 Rl

— = — • (1)

R$ R 4

Fhe bridge is then said to be balanced. As a rule , the bridge is used unbalanced, the

vol t age V be ing a measure of the unknown res i s t ance , s ay R{.

T he circu it is so widely used to da y th at i t is alm ost imp ossi ble to list all its

appl i ca t ions . Some of the be t t e r known a re the fo l lowing .

1. I f one of the res is tors is a temperature-sens i t ive res is tor , or thermistor, th e

bridge serves as a thermometer, V be ing a measure of the t empera ture .

2. If one of the res is tors is a temperature-sens i t ive wire , heated by the bridge

cur ren t and im mersed in a gas , V i s a me asu re of the gas veloci ty, or of i t s turbu lence .

One then has a hot-wire anemometer.

3. In the Pirani vacuum gauge the heated wire is in a part ia l vacuum and is

cooled more or less by the res idual gas , according to the pressure .

4. In certa in gas analyzers, t he hea te d wi re is exposed to the unkn ow n gas .

T h e n V i s a measure of the the rma l conduc t iv i ty o f t he gas . Such ana lyze rs a re

used , i n pa r t i cu la r , for ca rbo n d iox ide . The the rma l co nduc t iv i ty o f ca rbo n d iox ide

is roug hly one half that of a i r (16.6, again s t 26.1 mil l iw at ts per m eter-kelvin ) .



142 Direct Curre nts in Electric Circuits

5. In flammable-gas detectors the wire is he ate d suff ic iently to igni te a sam ple

of the gas con tain ed in a smal l cell . Th e heat genera ted by the com bu sio n increases

the tem pe ratu re and ch ang es the res is tance of the wire . Th e deflect ion of the vo l t me ter poin ter is a me asu re of the f lammabi li ty.

6. I f one of the res is tors is a f ine wire whose res is tance changes when elongated,

the bridge serves as a strain gauge for measur ing mic roscopic d i sp lacement s o r

d e f o r m a t i o n s .

7 . I f t he s t ra in gauge i s a t t ached to a d i aphragm, one has a pressure gauge.

a ) Sh ow tha t , when V = 0, Eq . 1 is satisfied.

Solution : Le t us supp ose tha t node C i s g ro und ed .

The vo l t age V being zero, there is zero current in the branch DB. T h e c u r r e n t

in the left-hand side is Vs/(R

t+ R 3 ) and the vol tage a t D is

V.

Similar ly, a t B,

R , + R 3

R 2+ R 4

Since V D = V B,

Ri + R 3 R 2 + R4

R 3 R 4

and , sub t rac t ing 1 on bo t h s ides ,

R i R 2

(2)

(3)

(4)

(5 )

b) Sup po se no w tha t R, increases by the factor 1 + x as in Sec. 5.12.2. We

r e q u i r e V/V s as a funct ion of x, whe re x can vary from —0.2 to + 0. 2. Th e vo l tme terV draws a neg l ig ib l e cur ren t .

Set

R 3 = aR lt R 2 = bR u R 4 = ab R u (6)

again as in Sec. 5.12.2. W ith x = 0, the bridg e is bala nce d.



14 3

0.2

x

0

a

Figure 5-29 Th e rati o V/V s for the Wheats tone bridge of Fig. 5-28, as a funct ion of

x and a.

Find V/V s as a funct ion of x, a, b.

No te tha t F /F tu rn s ou t t o be indep ende nt o f b.

Solution : W e now have tha t

FK 2 + K 4 £, (1 + x) + R

R4R 3

I RJ R 2 R 3/i?, (8)1 + ( R 4 / R 2 ) (1 + x) + (R 3 / R i ) '

(9)1 + a 1 + x + a '

ax(10)(1 + a)(l + a + x)

Figure 5-29 shows F/F, as a funct ion of both x and a.

c) For what value of a i s F/F m ax im um , for a given value of x? This is the

condi t ion for maximum sens i t iv i ty . F igure 5-29 shows tha t t h i s condi t ion i s = t 1.




0.05 "

i /

/

1 1 1 1 ^ i i i i1 1

- 0 . 2i i i

0.2

V/ // o . 3 /

/ 1

- 0 . 0 5 .

Figure 5-30 R a t i o V/Vs for the Wheats tone bridge of Fig. 5-28 as a funct ion of x,

fo r a eq ua l to 0 .3, 1, 3.

Solution: VjV s i s m a x i m u m w h e n d(V/V s)/da = 0. Th us we set

(1 + a) (l + a + x)x - axl(l + a + x) + (1 + a)]

(1 + a ) 2 ( l + a + x) 2 " 1 1 '

(1 + a)(l + a + x) = a(2 + 2a + x), (12)

1 + a + x + a + a 2 + ax = 2a + 2a 2 + ax , (13)

a = (1 + x )1

'2

. (14)

Since x is smal l , the con di t i on for ma xi m um sens i t ivi ty is a * 1. as we ha d

gues sed .

d) Draw curves of V/V s as a func tion of x, for x = —0.2 to x = + 0 .2 , an d for

a = 0.3, 1, and 3.



Problems 145

No te the s l igh t non- l inea r i ty o f t he curves . The n on- l inea r i ty dec reases w i th

inc reas ing a.


AMPLIFIER

F i g u r e 5-3la shows one e lement of an analog computer. Its function is to

ampli fy the input vol tage by the factor — R 2/R i '•

Th e t r iang ula r f igure is an operational amplifier whose gain is — A. Such ampli f iers

have very high gains , of the order of 10 4 t o 10 9 , and d raw a neg l ig ib l e am ou nt o f cur ren t

a t t he i r i nput t e rmina l s .

The accuracy of the gain is l imi ted only by the s tabi l i ty of the ra t io R 2/Ri- T h e

drift in R2/R i due to ag ing , t empera ture changes , and so for th , can be sma l l e r t han the

drift in A b y o r d e r s o f m a g n i t u d e .

a) Y ou can show in the following way th at the ab ov e equa t ion is corre ct . F i rs t ,

you need an expres s ion for the vo l t age V iA at the input to the ampli f ier . S ince the

ampli f ier draws essent ia l ly zero current a t i t s input terminals , you can use the c i rcui t

of Fig. 5-31b. Then, se t t ing V 0 = — AViA, you can show tha t

K = R 2 „ _ « 2

V, K , f [ ( « , + R jFVTJ * R t

The gain is — R 2/R l if A » 1 an d if A » R 2/R 1.

(b )

ri;I

VNAA- TViA

I

Rz 1v

1

Figure 5-31 (a) Mu l t ip ly ing c i rcu it . Th e

t r i ang le represen t s an ampl i f i e r whose

gain is — A. The ci rcui t has a gain of

-R 2/R!, as long as A » 1 and

A » R 2/R 1. (b ) Equivalent c i rcui t for

calculat ing the gain. See Prob. 5-9.




C

B Figure 5-32 See Prob. 5-11.

5-12E CUBE

Twelve equa l re s i s t ances R form the edges of a cube, as in Fig. 5-33.

a ) A ba t t e ry i s conn ec ted be tween A a n d G.

Ca n you f ind one set of three po ints that are a t the sam e p ote nt i a l?Can you f ind anothe r s e t ?

b) No w imag ine that you ha ve a sheet of co pp er con nec t ing the f i rs t se t , an d

an oth er sheet of cop per con nec t ing the second set . Th is doe s no t affect the cu rren ts

in the res is tors .

You can now show tha t

RAG= (5/6)R.

b) W ha t i s t he min im um va lue of A if R, = 1000 oh ms , R 2 = 2000 ohms , and

if V 0 mus t be equa l t o — (R2/R

l)V

iwithin one part in 1000?

5-10E G = el

Yo u a re g iven a l a rge num ber of 6 -ohm res i s to rs . H ow can they be connec ted

to give a c i rcui t whos e con du cta nc e is equa l to e Siemens , where e is the base of the

n a t u r a l l o g a r i t h m s ?

Hint: Write out the ser ies for e.

5-1 IE TETRAHEDRON

Six equal res is tors R form the edges of a te t rahedron, as in Fig. 5-32.

a ) A ba t t e ry i s conn ec ted be tw een A a n d B.

Show tha t , by symmet ry , po in t s C and D wil l be a t the same potent ia l . That

being so, there is zero current in branch CD, and we can remove i t .

b) W ha t is the res is tance betwe en A a n d B ?



14 7

F

Figure 5-33 See Prob. 5-12.

F G

H Figure 5-34 See P rob . 5 -13 .

5-BE CUBE

Twelve equa l re s i s t ances R form the edges of a cube, as in Fig. 5-33. To find

the re s i s t ance be tween A and C, we first flatten the cube as in Fig. 5-34.

a ) A ba t t e ry i s conn ec ted b e tween A and C.

Which four po in t s a re a t t he s ame po ten t i a l ?

Which two branches can be removed wi thout a f fec t ing the cur ren t s ?

b ) N o w s h o w t h a t

RAC= (3/4)/?.

5-14 CUBE

Firs t , read Probs . 5-11 to 5-13.

Now show that , in Fig. 5-33,

RAD= (1.4/2.4)R.



148

r ^ A A A - f - A A A - f -

550 V ~±r+ R

S a a a - ^ a a a >X20 km

Figure 5-35 A 550-vol t source feeds a load res is tance R T t h r o u g h a 20-k i lomete r

line. The pos i t ion of a fault RScan be f o u n d f r o m m e a s u r e m e n t s of the c u r r e n t / ,

with and w i t h o u t a shor t -c i rcu i t ac ros s R,. See P r o b . 5-15.

5-/50* LINE FAULT LOCATION

In Fig. 5-35. a 550-vol t source feeds a f ixed load res is tance RX t h r o u g h a pa i r

of copper (a = 5.8 x 107

Siemens pe r meter) wires 20 ki lomete rs long . The wires are

3 mi l l ime te rs in d i a m e t e r .

Suddenly , dur ing a s t o r m , the c u r r e n t I suppl i ed by the source inc reases and

t he vo l t age V a c r o s s R, falls. It is conc luded tha t a fault has d e v e l o p e d s o m e w h e r e

a l o n g the l ine, and t ha t t he re is a shunt re s i s t ance R S at some d i s t ance x from the source .

If the l oad re s i s t ance R T is d i s c o n n e c t e d , / is f o u n d to be 3.78 a m p e r e s . W h e n

the l ine t e rmina l s at the l oad ar e shor t -c i rcu i t ed , / is 7.20 a m p e r e s .

W h a t is the di s t ance x?

5-16E UNIFORM RESISTIVE NET

F i g u r e 5-36 s h o w s a t w o - d i m e n s i o n a l n e t w o r k c o m p o s e d of equa l re s i s t ances .

T h e o u t s i d e n o d e s are e i the r connec ted to ba t t e r i e s in s o m e a r b i t r a r y way, or left

u n c o n n e c t e d .

a) Use Kirchof f ' s cur ren t law to show tha t th e p o t e n t i a l at any i ns ide node is

e q u a l to the ave rage po ten t i a l at the four ne ighbor ing nodes . Fo r e x a m p l e .

VA+V M+V C+ VL

4

b ) W h a t is the ru l e for a s i m i la r t h r e e - d i m e n s i o n a l n e t w o r k ?

5-17 POTENTIAL DIVIDER

F i n d th e r a t i o VJV Tin Fig. 5-37.

P r o b l e m s m a r k e d D ar e relatively difficult .



14 9

Figure 5-36 Tw o-dim ens io na l ne twor k comp osed of equa l re s i s t ances R. A r b i t r a r y

vol tages are appl ied a t the periphery. I t i s shown in Prob. 5-16 that the vol tage on

any ins ide node is equ al to the ave rage valu e of the vol tages on the four ne igh bo ring

n o d e s .



15 0


5-18 SIMPLE CIRCUIT WITH TWO SOURCES

Show that , in Fig. 5-38,

/ =R,

r(R, + R 2)

5-19 DELTA-STAR TRANSFORMATION

Find the equa t ions for e i the r the de l t a - s t a r o r t he s t a r -de l t a t r ans forma t ion by

as suming mesh cur ren t s , a s in F ig . 5 -16 , and making the vo l t ages V A - V B, V B - F c ,

V c - V A in part a the same as those in part b.

Hint: Find on e equ a t ion of the form

( • )I A + ( • )IB + ( )Ic = 0.

Then, s ince i t must be val id, whatever the values of the mesh currents , the parentheses

must a l l be ident ical ly equal to zero. This should give you one of the equat ions of one

se t ; t he o the r two equa t io ns can be found by sym met ry .

4 KQ

i K n KQ 3 K n

5 K n

-o

Figure 5-39 See P rob . 5 -20 . The symbol Q s t ands for ohm.



Problems 151

5-20 DELTA-STAR TRANSFORMATION

Find the res is tance of the c i rcui t shown in Fig. 5-39.

5-2IE OUTPUT RESISTANCE OF A BRIDGE CIRCUIT

Show that the output res is tance of the bridge c i rcui t shown in Fig. 5-40, as

seen a t t he vo l tme te r V, is R.

Assume tha t t he source V s has a ze ro ou tpu t re s i s t ance .

5-22E OUTPUT RESISTANCE OF AN AUTOMO BILE BATTERY

I t i s found tha t the vol tage a t the term inal s of a defect ive au tom ob i le bat te ry

dr op s from 12.5 to 11.5 vol ts wh en the headl ig hts are turne d on .

W ha t i s t he appro xim a te va lue of the ou tpu t re s i s t ance?

5-23E DISCHARGING A CAPACITOR THROUG H A RESISTOR

A res is tance of 1 me go hm i s conne c ted ac ros s a 1 mic rofa rad capac i to r i n i t i a llycharged to 100 vol ts .

Ca lcu la t e the vo l t age ac ros s the capac i to r a s a func t ion of the t ime .

5-24E RAMP GENERATOR

A con s tan t - cur r en t source feeds a cur ren t / t o a capac i to r C. A t t = 0 the

capac i to r vo l t age i s ze ro .

F ind the vo l t age ac ros s C as a funct ion of the t ime.

5-25 CHARG ING A CAPACITOR THROUG H A RESISTOR

A source supply ing a vo l t age V cha rges a capac i to r C t h rough a re s i s t ance R.

Calculate the energy suppl ied by the source, that diss ipated by the res is tor ,

and that s tored in the capaci tor , af ter an inf ini te t ime.

Yo u sho uld f ind tha t one half of the energy is diss ipa ted in R, and tha t t he o the r

half is stored in C.

5-26 RC TRANSIENT

In the c i rcui t of Fig. 5-41, the cap aci tor i s ini t ia lly un cha rge d.

a) At 1 = 0 the switch is c losed. Find the vol tage V as a function of t.

b) After a long t ime the switch is op ene d. Fin d again V as a function of t, set t ing

t = 0 a t t he moment the swi t ch i s opened .




15 2


Figure 5-42 RC different ia t ing c i rcui t .

See Prob. 5-27.

5-27E RC DIFFERENTIATING CIRCUIT

Figure 5-42 shows an RC different ia t ing c i rcui t . Th e loa d res is tance con nec ted

at F> is large co m pa red to R.

Show tha t , if t he vo l t age dr op ac ro s s R is neg l ig ib le com pare d t o tha t ac ros s C ,

dVt

K.RC-^.

N o t e t h a t V„ « V,. See Prob. 5-29.

5-28E DIFFERENTIATING A SQUARE WAVE

A square wave as in Fig. 5-43 is appl ied to the RC different ia t ing c i rcui t of

Fig. 5-42.

Sketch a curve of the output vol tage as a funct ion of the t ime.

Figure 5-43 Ideal square wave. In

pract ice , the s t ra ight l ines are

L

por t ions of exponent i a l curves . See

Prob . 5 -28 .

5-29 DIFFERENTIATING CIRCUIT

The ci rcui t of Fig. 5-42 is s imple and inexpens ive, but V 0 « V t.

F i g u r e 5-44a shows a much supe r ior , bu t more complex , d i f fe ren t i a t ing c i rcu i t .

Fhe t r i ang le represen t s an ope ra t iona l ampl i f i e r a s in P rob . 5 -9 .

15 3

(b)

O-

t

I;

I

i

<> » A A A - o

I 1

Figure 5-44 (a ) Different ia t ing c i rcui t

us ing an ampli f ier , (b ) E q u i v a l e n tci rcui t used to calcula te V 0. See P rob .

5-29.

S h o w t h a t

dV t

K = -RC—1

,dt

as long as A » 1 an d |K 0 | /RC » \dVJdt\jA. N o t e t h a t RC can be much l a rge r than

uni ty, so that V0 need no t be much sma l l e r t han V

t.

Solution: From F ig . 5 -44b .

Vt=®+ViA, V„=-AViA, (1)

dV , = 1 dQ dV u _ 1 1 dV a

dt C dt dt C A dt'

N o w

so tha t

V i4 - V 0 \ 1 ,/ = — = 1 + — K, (3)

R R \ A 1

dV t 1 / 1 \ 1 dV 0

— = 1 + - IK , -. (4)

dt RC \ A I A dt1 1 / V dV"

RC " A \RC dt J'

* -VJRC, (6)

http://dvjdt/jA

http://dvjdt/jA




if A » 1 and i f |F | / i ? C » \dVJdt\/A. T h u s , w i th t h is a p p r o x i m a t i o n ,

dV ,V„=-RC-^. (7)dt

5-30 RC INTEGRATING CIRCUIT

Figure 5-45 shows an RC i n t egra t ing c i rcu it . Th e cur ren t t h rou gh the loa d

con necte d a t F„ is negl igible com pa re d to dQ/dt at C.

Show tha t , a s long as the vo l t age ac ros s C i s sma l l compared to tha t ac ros s R,

As in Prob. 5-27, V0 « F- I t i s assu me d tha t V

0= 0 a t t = 0.

5-31 E INTEGRATING CIRCUIT

A square wave as in Fig. 5-43 is appl ied to the RC integrat ing c i rcui t of Fig. 5-45.

Sketch a curve of the output vol tage as a funct ion of the t ime.

R

W V \ A

Figure 5-45 RC integrat ing c i rcui t . See

Pro b . 5 -30. ° "

5-32 INTEGRATING CIRCUIT

The in t egra t ing c i rcu i t shown in F ig . 5 -46a pe r forms in t egra t ions w i thout the

l imi t a t ion F 0 « V t that a ppl ies to the c i rcui t of Fig. 5-45. Th e t r iang le repres ents an

ampli f ier as in Probs . 5-9 and 5-29.

An alo g co mp ute rs m ak e extens ive use of the c i rcui ts of Figs . 5-31, 5-44, and 5-46.

S h o w t h a t

if A » 1 and if \V„\/RC « A\dVJdt\.

Use the c i rcui t of Fig. 5-46b and set



15 5

Figure 5-4 6 (a ) In t egra t ing c i rcu i t us ing

an ampli f ier , (b ) Equivalent c i rcui t for

ca lcu la t ing V„. See P rob . 5 -32 .

(a )

ri;

I

V W 9 ) | —Ti

c

1y.

I(b )

5-33 PULSE-COUNTING CIRCUIT

The ci rcui t shown in Fig. 5-47 can be used e i ther for count ing pulses or for

measur ing a capac i t ance , e i the r C\ o r C 2 , t he o the r one be ing known.

T h e p u l s e g e n e r a t o r G prod uces / pos i t ive squa re pu l ses of amp l i tude V p per

second and

Cj « c 2 , / c , « c 2 .

D i o d e s D x a n d D 2 pas s cur ren t s on ly in the d i rec t ion of the a r row.

Dur ing a pu l s e , no cur ren t f lows th rough d iode D u and capac i to rs C, and C 2

charge in ser ies .

Betw een pulses , the term inal s of G are effect ively sho rted a nd ca pac i tor C t

d i s c h a r g e s t h r o u g h d i o d e D l. D i o d e D 2 i s t h e n n o n - c o n d u c t i n g .

a) Sketch a curve of V as a function of the time, for V « Vp, wi th V = 0 at

f = 0.

b) W ha t is the value of F af ter a t ime f ?

Jl v,

4 ^ D ,

T

Figure 5-47 This c i rcui t can be used e i ther for measuring the repet i t ion frequency of

the cur ren t pu l s es o r ig ina t ing in G or for measur ing e i the r C, o r C 2 . See P rob .

5-33.



CHAPTER 6

DIELECTRICS: I

Electric Polarization P, Bound Charges,

Gauss's Law, Electric Displacement D

6.1 THE ELECTRIC POLA RIZA TION P

6.2 THE BOUND CHARG E DENSITIES pb AND a h

6.2.1 Example: Sheet of Dielectric

6.3 THE ELECTRIC FIELD OUTSIDE AND INSIDE

A DIELECTRIC

6.3.1 Example: Sheet of Dielectric

6.4 THE DIVERGENCE OF E IN DIELECTRICS:

GAUSS'S LAW

6.5 THE ELECTRIC DISPLACEM ENT D

6.6 THE ELECTRIC SUSCEPTIBILITY Xe

6.7 THE RELATIVE PERM ITTIVITY e r

6.8 RELATION BETWEEN THE FREE CHARGE DENSITIES

AND THE BOUND CHARGE DENSITIES

6.8.1 The Volume Charge Densities p f and p h

6.8.2 The Surface Charge Densities a ( and a b

6.8.3 Exam ple: Dielectric-Insulated Parallel-Plate Capacitor

6.9 POISSON'S AND LAPLACE'S EQUATION S

6.10 ELECT RETS

6.11 SUMMARY

PROBLEMS



Dielect r ics di f fer f rom conductors in that they have no f ree charges that can

move through the mater ial under the inf luence of an elect r ic f ie ld . In di e l ec t r i cs , a l l t he e l ec t rons a r e bound ; the on ly poss ib le mot ion in an e l ec t r i c

f ie ld i s a m inu te d i sp la cem ent o f pos i t ive and neg a t ive charges in o ppo s i t e

d i r e c t i o n s . T h e d i s p l a c e m e n t i s u s u a l l y s m a ll c o m p a r e d t o a t o m i c d i m e n s i o n s .

A d ie l ec t r i c in which th i s charge d i sp lacement has t aken p lace i s sa id

to be polarized, and i t s molecu les a r e sa id to possess induced dipole mom ents.

These d ipo les p roduce the i r own f i e ld , which adds to tha t o f the ex te rna l

cha rges . Th e dipole f ield an d the externa l ly app l ied elect r ic f ie ld can be

c o m p a r a b l e i n m a g n i t u d e .

In add i t ion to d i sp lac ing the pos i t ive and nega t ive charges , an app l i ed

e lec t r i c f i e ld can a l so o r i en t molecu les tha t possess permanen t d ipo lemoments . Such molecu les exper i ence a to rque tha t t ends to a l ign them wi th

the f ie ld , but col l i s ions ar is ing f rom thermal agi ta t ion of the molecules tend

to des t ro y the a l ignm ent . An equ i l ib r ium p o la r i za t ion i s t hus es t ab l i shed

in which th ere is , on the avera ge, a net a l ig nm ent .

This i s , in fact , a s impl i f ied view of dielect r ic behavior , because many

so l ids , such as sod ium ch lo r ide , f o r example , a r e no t made up o f molecu les

bu t o f ind iv idua l ions .

THE ELECTRIC POLARIZATION P

T h e electric polarization P i s t he d ipo le moment per un i t vo lume a t a g iven

point . I f p i s t he average e l ec t r i c d ipo le moment per molecu le , and i f N is

the nu m be r of mo lecu les per un i t vo lum e, then

p = yvp. (6-1)



Kigure 6- 1 Under the act ion of an e lect r ic f ie ld E, which is the resul tant of an

external f ie ld and of the f ie ld of the dipoles wi thin the die lect r ic , pos i t ive andnega t ive cha rges in the molecu les a re s epa ra t ed by an ave rage d i s t ance s. In the

process a net charge dQ = NQs • da' crosses the surface da', N be ing the number of

molecu les pe r un i t vo lume and Q t he pos i t ive cha rge in a molecu le . The vec tor

da' i s per pen dic ula r to the shad ed surface. Th e c i rc les indi cate the cent ers of ,

charge for the pos i t ive and for the negat ive charges in one molecule .

THE BOUND CHARGE DENSITIES p„ AND a b

W e sha l l now sh ow tha t the d i sp lac eme nt o f char ge wi th in the d ie l ec t r i c

gives r ise to net volume and surface charge densi t ies

p „ = - V - P a n d a b = P • n 1 ( (6-2)

whe re rij is t he norm al to the su rf ace , po in t ing ou tw ard .

Let us f i r s t consider the sur face densi ty a b. We imagine a smal l e l ement

of sur face da ' inside the dielect r ic , as in Fig . 6-1 . U nd er th e act io n of the

f i e ld , pos i t ive and nega t ive charges wi th in the molecu les separa te by an

a v e r a g e d i s t a n c e s. Posi t ive charges c ross the su r f ace by moving in the

d i r ec t ion o f the f i e ld ; nega t ive charges c ross i t by moving in the oppos i t e

d i r ec t ion . For the purpose o f our ca l cu la t ion , we cons ider the pos i t ive

charges to be in the form of point charges Q and the nega t ive charges in the

form of point charges — Q. F u r t h e r m o r e , w e c o n s i d e r t h e n e g a ti v e c h a r g e s

to be f ixed and the pos i t ive charges to move a d i s t ance s. T h e a m o u n t o f

c h a r g e dQ t ha t c rosses da ' i s t hen jus t t he to t a l amount o f positive c h a r g e

wi th in the imag in ary pa ra l l e l ep iped show n in F ig . 6 -1 . Th e vo l um e o f th i s



6.2 The Bound Charge Densities p b and c b 15 9

para l l e l ep iped i s

dx ' = s • da'. (6-3)

a n d

dQ = NQs • da', (6-4)

w h e r e N i s t he number o f molecu les per un i t vo lume and Qs i s the dipole

m o m e n t p of a molecu le . Then

dQ = P • da'. (6-5)

If da ' i s on th e sur face of the dielect r ic ma ter i al , dQ a c c u m u l a t e s t h e r e

as a su r f ace d i s t r ibu t ion o f dens i ty

(7 6 = ^ = P - n 1 , (6-6)da

w h e r e n, i s a un i t vec to r , no rmal to the su r f ace and po in t ing outward. T h e

bound su r f ace charge dens i ty a h is t hus equa l to the no rm al com po ne n t o f

P at the sur face.

We can show s imi la r ly tha t — V • P r ep resen t s a vo lume dens i ty o fcharg e . Th e ne t charge th a t f lows ou t o f a vo lum e x across an e l ement da'

of its surface is P • da', as we found above. The net charge that f lows out

o f the su r f ace S ' bound ing x i s thus

Q = J*g, P • da', (6-7)

and the ne t charge tha t r emains wi th in the vo lume x must be —Q. If p h is

the vo lum e dens i ty o f the cha rge r em ain in g wi th in th i s vo lum e, then

£, Pt dx = - Q = - J s„ P • da' = - £ , (V • P) dx'. (6-8)

S ince th i s equa t ion must be t rue fo r a l l x, t he in t egrand s must be equa l a t

every po in t , and the bound vo lume charge dens i ty i s

pb

= - V P (6-9)



160 Dielectrics: I

We refer to p b a n d a b as bound charge dens i t i e s , a s d i s t ingu i shed f rom

th e free charge dens i t i e s p f a n d a f. B o u n d c h a r g e s a r e t h o s e t h a t a c c u m u l a t e

th rough the d i sp lacement s tha t occur on a molecu la r sca le in the po la r i za t ionprocess . Th e o th er charges a r e ca ll ed fr ee charg es . Th e cond uc t ion e l ec t rons

in a conduc to r and the e l ec t rons in j ec ted in to a d i e l ec t r i c wi th a h igh-energy

e lec t ron beam are examples o f f r ee charges .

EXAMPLE: SHEET OF DIELECTRIC

If a sheet of dielectric i s pola r ized u niformly in the di rect ion no rm al to i ts surface,

then the surface dens i ty ofbound charge crfc is P. as in Fig. 6-2. Also. p h = — V • P = 0.

s ince P is i ndepen dent o f t he coo rd ina te s .

6.3 THE ELECTRIC FIELD OUTSIDE ANDINSIDE A DIELECTRIC

Figure 6 -3 show s a b lock o f po la r i zed d ie l ec t r i c ma te r i a l i n which P is a

funct ion of posi t ion. We must f ind the elect r ic f ie ld in tensi ty that the dipoles

in the d ie l ec t r i c p roduce a t some po in t A, e i the r ins ide o r ou t s ide . Once

ob t a ine d , th i s E can then be ad de d to tha t p rod uce d by o th er charges to

give the total E .



16 1

Figure 6-3 Block of d i e l ec t r i c w i th a d ipo le moment P per un i t vo lume . The

dipoles wi thin the e lement of volume shown ins ide the block give r ise to an

e lec t r ic po ten t i a l dV a t t he po in t A.

C o u l o m b ' s l a w a p p l ie s t o a n y n e t a c c u m u l a t i o n o f c h a r g e , r e g a r d l e s s

of o th er ma t t e r tha t ma y be p resen t . Thu s the f ie ld p ro du ced by the d ie lec t r i c ,bo th ins ide and ou t s ide , i s t he same as i f t he charge dens i t i e s p b a n d a b w e r e

s i tua ted in a vac uum .

6.3.1 EXAMPLE: SHEET OF DIELECTRIC

In the po lar ized sheet of die lect r ic of F ig. 6-2. E = 0 ou t s ide and E = o b/e 0 p o i n t i n g

to the left inside, exactly as if the sheet were replaced by its surface charges.

6.4 THE DIVERGENCE OF E IN DIELECTRICS:

GA USS'S LA W

W e have jus t seen that the elect r ic f ie ld pro du ce d by the dipoles of a p olar iz ed

dielect r ic can be calc ulate d f rom th e bou nd c ha rge densi t ies as if they w ere

si tu ated in a va cu um . Let us inve st igate the imp l ica t ion s of th is fact on

Ga uss ' s l aw (Sec . 3 .2 ) .



162 Dielectrics: I

Gauss 's law re la tes the f lux of E through a c losed surface to the charge

Q enclosed with in tha t surface:

Js

E • da = £ V • E dx = (2/e 0. (6-10)

For d ie lec tr ics , Q inc ludes bound as well as f ree charges:

Q = j(p f + p b)dx. (6-11)

If we subs t i tu t e th is va lu e of Q in to Eq . 6 -10 and equa te the in teg rands

of the volume in tegra ls , then

e 0

This is Gauss's law in i ts more genera l form. I t is one of Maxwell 's

four fundam en ta l equa t ion s of e lec t rom agne t i sm . We sha ll f ind the o the r

three in Sees . 8 .2 . 11 .4, an d 19 .2. Th is one fo l lows from (a) C ou lo m b' s law.

(b) the concept of electric field intensity, (c) the principle of superposition,

an d (d) the fact tha t the field of the dipol es situa ted in a po lari zed m ed iu m

can be ca lcula ted f rom pb a n d ab.

Note tha t we have implic i t ly assumed the exis tence of the spaceder ivat ives of E. These der iva t ives do not ex is t a t the in terface between two

media : in such cases one must rever t to the in tegra l of E • da over a c losed

surface , which is equal to the to ta l enclose d cha rge ov er e 0 , a cco rd ing to

Eq. 6-10.

THE ELECTRIC DISPLACEMENT D

Since pb = - V • P. f rom Eq. 6-9 , then

V E = -(p f - V P ) , (6-13)

o r

V • (e 0 E + P) = Pf. (6-14)



6.7 The Relative Permittivity e r 163

The vec to r e 0 E + P is therefore such th a t i ts d ivergen ce dep end s only on

the free charge density p f. This vector is ca l led the electric displacement a n d

is des ignated by D :

D = e 0 E + P. (6-15)

T h u s

(6-16)

In in tegra l form, Gauss 's law for D b e c o m e s

f D • da = f pj-d-c, ( 6 - 1 7 )

and the f lux of the e lec tr ic d isp lacement D through a c losed surface is equal

to the f ree charge enclosed by the surface .

No te tha t the divergence of D , as well as the surface integral of D , a re

bot h unaffec ted by the bo un d charge s .

6.6 THE ELECTRIC SUSCEPTIBILITY /e

In mos t d ie lec t r ic s the mo le cu la r cha rg e sepa ra t io n i s d i rec tly p rop o r t i ona l

to , and in the same d irec t ion as , E. Die lec tr ics tha t show th is s imple depen

dence of polarization on field are said to be linear a n d isotropic In prac t ice ,

most d ie lec tr ics are homogenous, as well as l inear and iso tropic .

The d ipo le momen t pe r un i t vo lume i s then

P = Np = x ee 0E, (6-18)

w h e r e N is again th e num be r of mol ecule s per uni t volu me , an d y e is a

d imens ion le ss cons tan t known as the electric susceptibility of the dielectric.

6.7 THE RELA TIVE PERMITTIVITY e r

F or a l inear and iso tr opic d ie lec tr ic we have , f rom E qs. 6-15 and 6-18 th a t

D = € 0 (1 + yjE = € 0 e r E = eE, (6-19)



16 4

Table 6-1 Relative Perm ittivities of

Dielectrics Near 25 C.

Frequency (hertz)

100 106

10"'

Bake l i t e 5.50 4.45 3.55

B u t y l r u b b e r 2.43 2.40 2.38

Fu sed silica 3.78 3.78 3.78

Luc i t e 3.20 2.63 2.57

N e o p r e n e 6.70 6.26 4.0

Polys tyrene 2 .56 2.56 2.54

S tea t i t e 6.55 6.53 6.51

S tyrofoam 1.03 1.03 1.03

Teflon 2.1 2.1 2.08

W a t e r 81 . 78.2 34 .

w h e r e

e, = 1 + l e = e / e 0 (6-20)

i s a d ime ns ion les s cons tan t l a rger than un i ty an d i s kn ow n as the relative

permittivity of the mate r i a l . I n a vacuum, y e = 0, e r = 1. Th e relat iv e

permi t t iv i ty was fo rmer ly ca l l ed the dielectric constant. The quan t i ty e i s

cal led the permittivity.

Th e relat ive perm it t iv i ty of dielect r ics l ies typical ly be twe en 2 an d 7 .

However , some non- l inear d i e l ec t r i cs have r e l a t ive permi t t iv i t i e s as h igh as

1 05

.

Table 6 -1 g ives the r e l a t ive permi t t iv i t i e s o f some common d ie l ec t r i cs .

I t wi l l be observed f rom the table that e r is f r equen cy-dep ende n t . W e sha ll

return to th is phenomenon br ief ly in Sec. 7 .7 .



6.8 Relation Between the Free and Bound Charge Densities 165

6.8 RE LATION BETWEEN THE FREE

CHARGE DENSITIES AND THE BOU ND

CHARGE DENSITIES

6.8.1 THE VOLUME CHARGE DENSIHES p f AND p„

In a l inea r and is ot ro pic dielect r ic , from Eqs. 6-15 and 6-19,

D = e 0 E + P = e r e 0 E . (6-21)

a n d t h u s

P = (e r - 1 ) e ( )E = ^ — - D . (6 -2 2)

Ta ki ng no w the divergen ce of bo th sides an d using Eqs. 6-9 an d 6-16,

p b = - ^ - P f (6-23)

N o t e t h a t p b a n d p f have oppos i t e s igns , so tha t t he to t a l charge

dens i ty i s smal l e r th an p s•:

P/ + Pt = Pf/er- (6-24)

Also, if p f if zero in a l inear an d is ot ro pic dielect r ic , which is nea r ly alw ays

the case, then p h i s a lso zero.

6.8.2 THE SURFACE CHARG E DENSITIES a f AND o„

At the in t e rf ace be twee n a d i e l ec t r ic and a condu c to r , t he re is a b ou nd

sur f ace charge dens i ty a h on the dielect r ic , and a f ree sur face charge densi ty

o f on the conduc to r , a s in F ig . 6 -4 . For s t eady- s t a t e cond i t ions , t he re i s

zero elect r ic f ie ld inside the conductor and. inside the dielect r ic , f rom Gauss 's

l a w .

€ 0E = Of + o h, D = e re 0E = a r. (6-25)



16 6

E ,D

Figure 6-4 Interface between a die lect r ic and a conductor , wi th a pos i t ive charge

dens i ty on the sur face of the con duc tor .

This eq ua t ion i s s imi l a r to Eq . 6 -24 .

6.8.3 EXAMPLE: DIELECTRIC-INSULATED PARALLEL-PLATE CAPACITOR

Figure 6-5 shows a d i e l ec t r i c - insu la t ed pa ra l l e l -p l a t e capac i to r w i th the spac ing s

exag gera ted for c lar i ty. We assum e that s i s smal l , co m pa red to the l inear exten t

of the pla tes , in order that we may neglect the fr inging f ie ld a t the edges .Th en P is unifo rm, V • P = 0, and p b = 0, which is cons is tent w i th the fact

t h a t p, = 0.

The sur face -cha rge dens i t i e s + a s on the lower pla te and — a s on the upper

p la t e p roduce a un i form e lec t r i c f i e ld d i rec t ed upward . The po la r i za t ion in the

dielect r ic gives a bound surface charge dens i ty — ah on the lower surface of the

dielect r ic and + a b on the upper sur face . These bound cha rges produce a un i form

electr ic f ie ld di rected downward that cancels part of the f ie ld of the charges s i tuated

on the pla tes . S ince the net f ie ld between the pla tes must remain equal to V/s, th e

free charge dens i t ies +a f a n d — a f on the pla tes must be larger than when the

dielect r ic i s abse nt . The presen ce of the die lect r ic thus has the effect of increa s ing

the charges on the pla tes for a given value of V, and hence of increas ing the

c a p a c i t a n c e .To ca lcu la t e the capac i t ance , we apply Gauss ' s l aw for the d i sp lacement D to a

cylinder as in Fig. 6-5. Then the only flux of D t h rough the Gauss i an sur face i s

t h r o u g h t h e to p , a n d D i s num erica l ly equa l to a f. T h e n . E = o f/e re 0, t he po ten t i a l

di fference between the pla tes is afs/e

re

0, a n d t h e c a p a c i t a n c e

T h u s

(TF

+ a h = a f/e r. (6-26)

C = = e re 0A/s, (6-27)



16 7

+ + + + +

U t p

+ + +

Figure 6-5 Diele ct r ic- in sulated para l le l -pla te capac i tor . The smal l rectangle is the

cross-sect io n of an ima gina ry cyl inder used for calcu la t ing D.

w h e r e A i s the area of one pla te . The capa ci ta nce is therefore increased b y a factor

of e r t h rough the presence of the d i e l ec t r i c .

The me asur eme nt o f t he capac i t ance of a su i t ab le capac i to r w i th and wi thout a

d ie l ec t r i c p rov ides a convenien t me thod for measur ing a re l a t ive pe rmi t t iv i ty er.

POISSON'S AND LAPLACE'S EQUATIONS

In homogeneous , l i near , and i so t rop ic d i e l ec t r i cs , D is p ro po r t io na l to E ,

a n d e r is i nde pen den t o f the coo rd in a tes . The n , from Eqs . 6 -12 an d 6 -24 ,

V • E = = p f/e re0. (6-28)

Also, s ince E is —VI 7,

r-V = -p

- ± ± l ± = - P / / M o . (6-29)e

o

This i s Poissoris equation for dielect r ics . The expressions involving e r a reval id only in l inear and isot ropic dielect r ics .

Laplace's equation i s aga in .

V 2 K = 0 , (6 -3 0)

w h e n b o t h p s a n d p b are ze ro .



16 8

Figure 6-6 (a ) Bar e lect re t polar ized uniformly paral le l to i t s axis , (b ) The E field of

the bar e lect re t i s the sam e as tha t of a pai r of c i rcula r pla tes carr yin g unifo rm

surface ch arg e dens i t ies of oppo s i te polar i t ies , (c) Lines of E (black) . Lines of D

are shown in whi te ins ide the e lect re t ; outs ide , they fol low the l ines of E.



6.11 Summary 169

6.10 ELECTRETS

A n electret i s t he e l ec tr i c equ iv a len t o f a pe rm an en t ma gne t . I n mo st d i

e l ec t r i cs the po la r i za t ion d i sappear s immedia te ly when the e l ec t r i c f i e ld i s

r emoved , bu t some d ie l ec t r i cs r e t a in the i r po la r i za t ion fo r a ve ry long t ime .

Som e po lym ers have ex t r ap o la t ed l i f et imes o f a few hu nd red ye ar s a t r o om

t e m p e r a t u r e .

One way of charging a dielect r ic i s to place i t in a s t rong elect r ic f ie ld

a t h igh t empera tu re . The bound charge dens i ty on the su r f aces then bu i lds

up s lowly as the molecu les o r i en t themse lves . F ree charges a r e a l so depos i t ed

on the su r f aces when spark ing occur s be tween the e l ec t rodes and the d ie l ec

t r ic . Th e fr ee and the bo un d cha rges hav e opp os i t e s igns . Th e samp le is

then coo led to room tempera tu re wi thou t r emoving the e l ec t r i c f i e ld .

The examples in Sees. 6 .2 .1 and 6 .3 .1 apply to a sheet e lect ret wi th

zero f r ee charge dens i ty .

Figure 6-6 shows the f ie ld of a uniformly polar ized bar e lect ret .

6.11 SUMMARY

W he n a dielect r ic ma ter i al i s p lac ed in an elect r ic field, posi t ive and neg at ive

charges wi th in the molecu les a r e d i sp laced , one wi th r espec t to the o ther ,a n d t h e m a t e r i a l b e c o m e s polarized. T h e i n d u c e d d i p o l e m o m e n t p e r u n i t

v o l u m e P i s cal led the electric polarization.

This p ro duc es r ea l acc um ula t ion s o f cha rge tha t we can use to ca lcu la t e

V and E bo th ins ide and ou t s ide the d ie l ec t r i c :

Pb V • P, (6-2)

w h e r e p b i s t he vo lu me den s i ty and o h i s the sur face densi ty of bound charge.

Th e uni t vec tor i s no rm al to the sur face of the dielect r ic an d point s

oufward .Gauss's law can be expre ssed in a form that i s val id for dielect r ics :

V • EPf + Pb

e 0

(6-12)

This i s one o f Ma xwe l l ' s f our funda men ta l e qu a t ion s o f e l ec t ro ma gne t i sm.



170 Dielectrics: I

In l inear and i so t rop ic d i e l ec t r i cs , t he d ipo le moment per molecu le p

i s p ro po r t ion a l to E , an d

P = e 0 Z , E , (6-18)

w h e r e % e i s a d im ens ion less con s tan t ca l l ed the electric susceptibility.

T h e electric displacement

D = e 0 E + P. (6-15)

= e 0 d + yjE = e0e

rE = eE , (6-19)

w h e r e e r is the relative permittivity and e i s the permittivity. A l s o ,

V • D = p y , (6-16)

o r

f D d a = [pfdx. (6-17)

Inside a dielect r ic ,

P f + P b = r V * r . (6

'2

^

whi le , a t t he in t e r face be tw een a d i e l ec t ri c and a con duc to r ,

o f + o h = o f/e r. (6-26)

Poisson's equation for dielectr ics is

R - V = _Pl±I> m _ p//€f60) (5.29)

e

o

a n d Laplace's equation i s again

V2

V = 0 , ( 6 - 50 )

w h e n p r a n d p h are bo th ze ro .

Problems 171

As usua l , express ions invo lv ing e r are val id only in l inear and isot ropic

m e d i a .

A n electret i s a p i ece o f d i e l ec t r i c mate r i a l t ha t i s pe rmanen t ly

p o l a r i z e d .

PROBLEMS

6-1E THE DIPOLE MOMEN T p

A sample of diamond has a dens i ty of 3.5 x 1 0 3 ki lograms pe r cub ic me te r and

a p o l a r i z a t i o n o f 1 0 " 7 c o u l o m b p e r s q u a r e m e t e r .

a ) C o m p u t e t h e a v e r a g e d i p o l e m o m e n t p e r a t o m .

b) F ind the ave rage s epa ra t ion be tw een cen te r s o f pos i t ive and nega t ive cha rge .

Ca rbo n has a nuc leus w i th a cha rg e + 6e , sur roun ded by 6 e l ec t rons .

6-2E THE VOLUME AND SURFACE BOUND CHARGE DENSITIES pb

AND ab

Co ns ide r a b lock of d i e l ec t r ic w i th bou nd ch a rge dens i t i e s p b a n d <j b.

S h o w m a t h e m a t i c a l l y t h a t

w h e r e t h e x i s the volume of the die lect r ic and S is i t s surface. In other words , the tota l

ne t bound cha rge i s ze ro .

6-3 BOUN D CHARGE DENSITY AT AN INTERFACE

Show tha t t he bound cha rge dens i ty a t t he in t e r face be tween two d ie l ec t r i c s 1

and 2 that i s crossed by an e lect r ic f ie ld is (P x — P 2 ) • n. Th e po lar iza t ion in 1 is P ,

and i s d i rec t ed into the interface; the polar izat ion in 2 is P 2 a n d p o i n t s away from the

interface. The uni t vector n is normal to the interface and points in the di rect ion from

6-4E COAXIAL LINE

Sho w th at V • E = 0 in the die lect r ic of a coaxial l ine .

Hint: App ly the diver genc e the ore m to a po rt io n of the die lect r ic .

6-5 COAXIAL LINE

The capac i t ance pe r un i t l eng th C of a die lect r ic- insulated coaxial l ine is equal

to that for an a i r- insulated l ine as in Prob. 4-6, mul t ipl ied by e r:

1 to 2.

C =

2n e re 0

In (R 2 /R , )

17 2

Table 6-2

D i a m et er

B & S Gaug e in Mi l l imeters

20 0.812

22 0.644

24 0.511

26 0.405

28 0.321

30 0.255

32 0.202

34 0.160

For a ce r t a in appl i ca t ion , one requi re s a coax ia l cab le whose ou t s ide d i amete r

mu s t no t exceed abou t 10 mi l l ime te rs . Then R2

< 5 mi l lime te rs . I t s capac i t ance pe r

uni t length must be as low as poss ible . Then R 2 mu st be 5 mil l im eters , and R, as smal l

as poss ible . The cable wi l l be operated a t up to 500 vol ts .

Th e insulat in g mat eria l wi ll be Teflon. Teflon ha s a re la t ive perm it t ivi ty of 2.1.

I t can op era te re l iably in such a cable a t e lect r ic f ie ld inten s i t ies as high as 5 x 10 6 vol t s

per meter . (The ma xi m um e lect r ic f ie ld intens i ty before bre ak do wn is cal led the di

electric strength of a dielectric. In a uniform field the dielectric strength is larger for

th inne r spec imens . A f ilm of Teflon 0.1 mil l imeter th ick ha s a die lect r ic s t reng th of

1 08 vol ts per meter . )

a ) W ha t i s t he mi n im um a l low able d i amete r for t he inne r co nd uc t or?

Remember tha t you can so lve a t ranscendenta l equa t ion by t r i a l and e r ror .

b ) Tab le 6-2 show s the d i amete r s of som e comm onl y ava i l ab le copp er w i res .

F ine r w i res would be imprac t i ca l because they would t end to b reak a t t he connec tors

f ixed to the ends of the cable .

W h a t g a u g e n u m b e r d o y o u s u g g e s t ?

c) What wi l l be the capaci tance per meter wi th the wire s ize you have chosen?

COAXIAL LINE

A coaxia l l ine is com pose d of an in t e rna l cond uc to r o f rad ius R , an d an ex te rna l

conduc tor o f rad ius R 2 , sepa rated by a die lect r ic . Th e die lect r ic should no rm al ly

Problems 173

f il l a ll t he space be tween the cond uc to rs . Ho wever , du e to an e r ror , t he ou te r rad ius

of the dielectric is R < R2, leaving an a i r space.

a) Assu min g that the die lect r ic i s wel l center ed, what i s the capa ci tan ce w hen

Rj = 1 mm . R2 = 5 m m . R = 4.5 mm, and e

r = 2.56 (polystyrene)'?

Solution: Th e ou te r sur face of the d i e l ec t r i c i s an equip o ten t i a l sur face . Thu s ,

we can imagine that i t i s covered with a thin conduct ing layer . This does not dis turb

the f ie ld. Then we may cons ider that we have two cyl indrical capaci tors , one ins ide

the o the r .

Be tween P , and R we have a capaci tance per uni t length of

2n e re 0

MR R t)

from Prob. 6-5. Between R a n d R 2. the capaci tance per uni t length is

l n ( P 2 / P ) '

These tw o capac i to rs a re in s e r ie s , and the capac i t ance pe r un i t l eng th be tw een

the inne r and the ou te r conduc tors i s g iven by

j _ = l n ( P / P . ) + l n ( P 2 / P )

C 2ne re 0 2n e 0

12n e

0

! P , ) + l n ( P , P ) .I n f P v / P ! ) + l n ( P 2 /P ) I. (2)

C = - ^ . (3)

— l n ( P / P , ) + l n ( P 2 / P )

2n x 8 .85 x 1 0 " 1 2 , ,= = 80.2 pico farad s /me ter . (4)

In4.5 + In —

2.56 0.9

b) Calculate the percent change in capaci tance due to the a i r f i lm.

Solution: Without the a i r f i lm, the capac i t ance would be

2n x 2.56 x 8.85 x 1 0 " 1 2

C = = 88.4 pico farad s /me ter . (5)

In 5

The a i r f i lm therefore decreases the capaci tance by about 10%.

174 Dielectrics: I

6-7 CHARGED W IRE EMBEDDED IN DIELECTRIC: THE FREE

AND BOUND CHARGES

A conduc t in g wi re ca r ry ing a cha rge / . pe r un i t l eng th i s em bed ded a long the

axis of a c i rcular cyl inder of die lect r ic . The radius of the wire is a ; the radius of the

cyl inder is b.

a) Show that the bound charge on the outer surface of the die lect r ic i s equal

to the bound charge on the inner surface, except for s ign.

b) Show that the net charge a long the axis i s /./e r per uni t length.


Th e space be tween the p l a t e s o f a pa ra l l e l -p l a t e capac i to r w i th a p l a t e s epa r a t ion

.s an d a surface are a A i s part ia l ly f i l led wi th a die lect r ic pla te of thickness t < s a n d

a r e a A.

a ) Show tha t

C =

b) Call C 0 t he capac i t ance for t = 0.

Draw a curve of C/C 0 as a function of t s for e r = 3 .

6-9 SPHERE OF DIELECTRIC WITH A POINT CHARGE AT ITS CENTER

A sphere of die lect r ic of radius R conta ins a po in t cha rge Q at i ts center. Set

R = 20 mi l l ime te rs , Q = 1 0 " 9 c o u l o m b , e r = 3.

a ) Dr aw graph s of D, E, V as funct ions of the dis tance r from the center , out to

r = 100 mil l imeters .

Note the d i s cont inu i ty in E at the surface.

b) Now ca lcu la t e a b at the surface from the value of the polar izat ion P j us t

ins ide the surface.

c ) Now use Gauss ' s l aw to expla in the d i s cont inu i ty in E.

6-10 CHARGED DIELECTRIC SPHERE

A dielect r ic sphere of radius R contains a uniform dens i ty of free charge p f.

Show that the potent ia l a t the center i s

(2 e r + l)p fR2

MEASURING SURFACE CHARGE DENSITIES ON DIELECTRICS

Figu re 6-7a shows a s chem at i c d i ag ram of an ins t ru men t tha t ha s been used to

measure charge dens i t ies a t the surface of die lect r ics . See a lso Prob. 17-9,

Th e prob e P i s sup por t ed a few mi l l ime te rs abov e the surface of the s am ple

and can be d i sp laced hor i zon ta l ly and ve r t i cal ly a long th ree or tho gon a l d i rec t ions .



17 5

Figure 6-7 (a ) Ins t ru me nt for me asu rin g surface charg e dens i t ies on die lect r ics . A

c o n d u c t i n g p r o b e P i s held c lose to the surface of the sample S. The probe i s

connec ted to an e l ec t romete r E t h rough a coaxia l cab le CC. (h) Equiva len t c i rcu i t :

C\ i s t he capac i t ance be tween the top sur face of the s ample and g roun d , C 2 i s thatbe tween the probe and the s ample sur face , and C 3 i s the capaci tance of the coaxial

cab le , p lus the input cap ac i t an ce of the e l ec t romete r . See P ro b . 6 -11 .

As we shal l see , the surface charge on the sample induces a vol tage on P. This vol tage

is read on the e le ct rom eter .

A n electrometer i s a vo l tme te r th a t has an ex t remely h igh re s i s t ance . For e xamp le ,

one type has a res is tance of 1 0 1 4 o h m s . F o r c o m p a r i s o n , c o m m o n d i g i t a l v o l t m e t e r s

have res is tances of the order of 10 6 or 10 7 o h m s . A g o o d a n a l o g v o l t m e t e r d r a w s

abo ut 100 mi cro am per es . I f i t i s ra ted a t 10 vol ts , it s res is tance is 10 10" 4 = 1 0 5 o h m s .

An e l ec t rom ete r d raw s es sen t i al ly zero cur ren t and i s t he re fore su i t ab le fo r m easur ing

the vol tage on a smal l cap aci to r , which is wh at we have h ere .

Figure 6-7b shows the equivalent c i rcui t : we have, effect ively, three capaci tors

connec ted in s e r i e s , and the e l ec t romete r s e rves to measure the vo l t age V.

a) Find the surface charge dens i ty a in terms of Clt C 2 , C 3 , V.

Solution: Let the area of the probe be A. Then the cha rge we a re conce rned

with, at the surface of the dielectric, is Aa . We neglect the fringing field at the edge

of the probe. Part of the f ie ld of Aa goes to the probe , and pa r t goes to g round . Le t

Aa = Q + Q\ w h e r e Q = Ae 0E 2, Q' = Ae^o^i, (D

as in Fig. 6-7b. The f igure a lso shows the charges induced on the conductors . S incethe probe and i ts lead to the e lect rometer are exceedingly wel l insula ted, the net

cha rge induced on them is ze ro .

T h u s V = Q C 3 , and we must f ind a, C 3 b e i n g k n o w n .

Now we have two expres s ions for the vo l t age V b e t w e e n t h e p r o b e a n d g r o u n d :

v=Q=9L.Q. ,2 ,

c 3 c, c 2

176 Dielectrics: I

So

Subst i tut ing in Eq. 1,

e' = e ^ + ^ . (3)

Aa = 61 1 + £ i + - i ) .C , C

Since 2 = C 3 K ,

a = C, + C , +

C , C , \ K

C , A (5)

b) Expres s <r in terms of the thickness t , of the die lect r ic , the dis tance r 2 b e t w e e n

the probe and the die lect r ic , and e r .

Solution:

and

d = e.eo/4/f!, C 2 = e0/ l / t

2

= i ^ + C3 + € ^ r 2 ) i /

r,

(6)

(7)6-12E VARIABLE CAPACITOR UTILIZING A PRINTED-CIRCUIT BOARD

A printed-ci rcui t board is a sheet of plas t ic , about one mil l imeter thick, one s ide

of which is covered w ith a thin coat i ng of cop per . P art of the copp er can be e tched away ,

l eav ing conduc t ing pa ths tha t s e rve to in t e rconnec t re s i s to rs , capac i to rs , and so for th .

The t e rmina l s o f t hese component s a re so lde red d i rec t ly to the copper .

Figure 6-8 shows a variable capaci tor in which a s l iding conduct ing pla te l ies

under a p r in t ed-c i rcu i t boa rd tha t has been e t ched to g ive a p resc r ibed va r i a t ion of

capac i t ance wi th pos i t ion z .

At the pos i t ion z . the capaci tance is

t Jo

w h e r e t i s the thickness of the plas t ic sheet of re la t ive permit t ivi ty e r .

Set e r = 3, t = 1 mil l ime ter , and f ind y(z) giving the fol lowing v alues of C, wi th

C expressed in farads and x i n me te rs :

a ) l ( T 9 z , b ) K T 8 z 2 .



17 7

QFigure 6-8 Var i ab le capac i to r made

wi th a p r in t ed-c i rcu i t boa rd PC B a n d

a conduc t ing p la t e P. T h e c a p a c i t a n c e

be tween the t e rmina l s depends on the

pos i t ion of the pla te P and on the

shape of the copper foi l F. See P rob .

•y 6-12.

In both cases , draw a curve of y(z) from z = 0 to z = 100 mi l l ime te rs , showing

the une tched reg ion on the c i rcu i t boa rd .

6-13E EQUIPOTENTIAL SURFACES

A l iquid die lect r ic is s i tua ted in an e lect r ic f ie ld. A thin c on duc t ing sheet carry ingzero net charge is int roduced into the die lect r ic so as to coincide wi th an equipotent ia l

surface.

a) Is the e lect r ic f ie ld di s tu rbe d?

b) W ha t is the value of a s on the surfaces of the sheet a t a point where the e lect r ic

d i sp lacement i s D]

6-14E NON-HOMO GENEOUS DIELECTRICS

Show tha t a non-homogeneous d i e l ec t r i c can have a vo lume dens i ty o f bound

charge in the absence of a free charge dens i ty.

6-15 FIELD OF A SHEET OF ELECTRONS TRAPPED IN LUCITE

W hen a b lock of insu la t ing ma te r i a l such a s Luc i t e is bo m bar ded wi th h igh-ene rgy e l ec t rons , t he e l ec t rons pene t ra t e in to the ma te r i a l and remain t rapped ins ide .

In one pa r t i cu la r i ns t ance , a 0 .1 mic roa mp ere b eam b om bar ded an a rea of 25 squa re

cen t ime te rs o f Luc i t e (e r = 3.2) for f second, and essent ia l ly a l l the e lect rons were

t rap ped abo ut 6 mi l l ime te rs be low the sur face in a reg ion abo ut 2 mi l l ime te rs th i ck .

Th e b lock was 12 mi l l ime te rs th i ck .

In the fol lowing calculat ion, neglect edge effects and assume a uniform dens i ty

for the t rapped elect rons . Assume also that both faces of the Luci te are in contact wi th

g r o u n d e d c o n d u c t i n g p l a t e s .

S h o w t h a t

a ) i n the reg ion where the e l ec t rons a re t rapped .

p, = —2.000 x 1 0 " 2 c o u l o m b / m e t e r 3 .

p h = 1.375 x 1 0 " 2 c o u l o m b / m e t e r3 ;

b) in the neu tra l region,

D„ = - 2 . 0 0 0 x 1 0 " 5 c o u l o m b / m e t e r 2 ,

P„ = —1.375 x 1 0 "5

c o u l o m b m e t e r2

.

E„ = - 7 . 0 6 2 x 1 0 5 vol t s /me te r ,

V„ = 7.062 x 10 5.v - 4,237 vo l ts ;

c) at the surfaces.

a b = — 1.375 x 1 0 " ? c o u l o m b / m e t e r 2 ;

17 8 Dielectrics: I

d) in the cha rged reg ion ,

dDc

~dx~ - 2 . 0 0 0 x 102

c o u l o m b / m e t e r3

.

- 2 . 0 0 0 x 1 0 " 2. v c o u l o m b / m e t e r 2 .

- 7 . 0 6 2 x 1 0 8 v o l t s m e t e r 2 ,

Ec

=

d2

Vc

die 2

7.068 x 1 0 8 x vo l t s /me te r ,

= 7.062 x 1 0 8 v o l t s / m e t e r 2 .

V c = 3.531 x 1 08

x2

- 3884 volts .

e) Draw curves of D, E, V as funct ions of the dis tance x t o the midplane , f rom

x = —6 to x = 6 mil l im eters .

f) Show that the s tored energy is 1.88 x 1 0- 4

j o u l e .

g) Is ther e any dan ger tha t the block wil l exp lod e?

6-16 SHEET ELECTRET

A sheet e lect re t i s polar ized in the di rect ion normal to i t s surface.

Draw a f igure showing the po la r i za t ion P . t he sur face cha rge dens i t i e s +a h

and -<r fc , as wel l as the e lect r ic f ie ld intens i ty E and the e lect r ic displacement D, both

ins ide and outs ide the e lect re t .

6-17 RELATION BETWEEN R AND C FOR ANY PAIR OF ELECTRODES

W he n a para l le l -p la te cap aci to r i s f il led wi th a die lect r ic wh ose re la t ive p er

mittivity is e r , i t s capaci tance is C. If the die lect r ic i s s l ight ly conduct ing, i t s res is tance

is P.

S h o w t h a t RC = e re 0/a , w h e r e o is t he condu c t iv i ty .

Th is rule appl ie s to any pair of e lect ro des sub me rge d in a me diu m, as long as

the conduc t iv i ty o f t he e l ec trodes i s much l a rge r than tha t o f t he m ediu m.



CHAPTER 7

DIELECTRICS: II

Continuity Conditions at an Interface, Energy Density

and Forces, Displacement Current

7.1 CONTINUITY CONDITIONS AT THE INTERFA CE

BETWEEN TWO MEDIA7.1.1 The Potential V

7.1.2 The Normal Compo nent of D

7.1.3 The Tangential Com ponent of E

7.1.4 Bending of Lines of Force

7.1.5 Examp les: Point Charge Near a Dielectric, Sphere of Dielectric

in an Electric Field

7.2 POTENTIAL ENERGY OF A CHARG E DISTRIBUTION,

ENERGY DENSITY

7.3 FORCES ON CONDUCTORS IN THE PRESENCE

OF DIELECTRICS

7.3.1 Examp le: Parallel-Plate Capacitor Immersed in a LiquidDielectric

7.4 ELECTRIC FORCE S ON DIELECTRICS

7.4.1 Examp le: Curve Tracer

7.5 THE POLARIZATION CURRE NT DENSITY dP/dt

7.6 THE DISPLACEM ENT CURRE NT DENSITY c D [ d t

7.6.1 Examp le: Dielectric-Insulated Parallel-Plate Capacitor

7.7 FREQUENCY AND TEMPERATURE DEPENDENCE,

ANISOTROPY

7.7.1 Examp les: Water, Sodium Chloride, Nitrobenzene, Compo unds

of Titanium

7.8 SUMMARY

PROBLEMS



This chap te r wi l l comple te our s tudy o f d i e l ec t r i cs .

We sha l l f i r s t d i scuss the impor tan t con t inu i ty cond i t ions tha t app ly

a t the in t e r f ace be tween two media . These cond i t ions concern the po ten t i a l

V, t h e t a n g e n t i a l c o m p o n e n t o f E , a n d t h e n o r m a l c o m p o n e n t o f D .

Then we must re turn to the energy stored in an elect rostat ic f ie ld .

Th i s s to r ed energy wi l l g ive us the fo rces exer t ed on conduc to r s immersed

in non-conduct ing l iquids. I t wi l l a lso give us the forces act ing wi thin the

d ie l ec t r i cs themse lves .

I f the elect r ic f ie ld is a funct ion of the t ime, then the mot ion of charge

wi th in the molecu les o f the d ie l ec t r i c g ives a po la r i za t ion cur r en t . The

d i sp lac em ent cu r r en t i s equ a l to the po la r i za t io n cur r e n t p lus ano the r

current that exists whenever the elect r ic f ie ld var ies wi th t ime, even in a

v a c u u m .

7.1 CONTINUITY CONDITIONS A T THE

INTERFACE BETWEEN TWO MEDIA

T h e q u a n t i t i e s V, E, and D must sa t i s fy ce r t a in boundary cond i t ions .

7.1.1 THE POTENTIAL V

A t t h e b o u n d a r y b e t w e e n t w o m e d i a , V must be con t inuous , f o r a d i scon

t inui ty wo uld im ply an inf inite ly large elect r ic field in tensi ty , which is

phys ica l ly imposs ib le .

7. / Continuity Co nditions at the Interface Between Two Media 181

Th e pote nt i a l i s no rm al ly set equ al to zero at inf inity if the ch arg e

d i s t r ib u t ion i s o f f in i te ex ten t . T he po te n t i a l i s con s tan t th ro ug ho u t any

conduc to r , a s long as the e l ec t r i c charges a r e a t r es t .

7.1.2 THE NORMAL COMPONENT OF D

C o n s i d e r a s h o r t G a u s s i a n c y l i n d e r d r a w n a b o u t a b o u n d a r y s u rf a ce , a s

in Fig . 7-1. Th e end faces of the cyl inde r are paral le l to the bo un da ry an d

arbi t rar i ly c lose to i t . The boundary car r ies a f ree sur face charge densi ty a f.

If the are a S i s sm al l , D a n d a s do not vary signif icant ly over i t and, according

to Gauss 's law, the f lux of D em erg ing f rom th e f la t cyl inder i s equ al to the

c h a r g e e n c l o s e d :

(D nl - D„ 2)S = OjS. (7-1)

The only f lux of D i s through the end faces, s ince the area of the cyl indr ical

su r face i s a rb i t r a r i ly smal l . T hu s

Dm - Dm = <>{• (7-2)

Figure 7-1 Gauss ian cy l inde r on the in t e r face be tween two media 1 and 2 . The

difference D nl — D n2 be tween the normal co mp on en ts of D is equ al to the surface

dens i ty of free charge a f.



182 Dielectrics: II

At the boundary be tween two d ie l ec t r i c med ia the f r ee su r f ace charge

dens i ty o f i s genera l ly ze ro and then D„ i s co n t in uo us ac ross the b ou nd ary .

On the o ther han d , if t he bo un da ry i s be twe en a co nd uc to r a nd a d i e l ec t r ic ,and i f the elect r ic f ie ld is constant , D = 0 i n t h e c o n d u c t o r a n d D„ = a f in

the dielect r ic , a f be ing the fr ee charg e dens i ty on the su r f ace o f the con du c to r .

7.1.3 THE TANGENTIAL COMPONENT OF E

Th is bou nd ar y co nd i t io n fol lows f rom th e fact that the l ine integra l of E • dl

aro un d a ny c losed pa t h i s ze ro for e l ec t ros t a t i c fi elds. Cons id er the pa th

shown in Fig . 7-2, wi th two sides paral le l to the boundary, of length L, and

arb i t r a r i ly c lose to i t. Th e o ther two s ides a r e perp end icu la r to the bo un dar y .In calculat ing this l ine in tegral , only the f i r s t two sides of the path are impor

tant , s ince the lengths of the other two approach zero. I f the path is smal l

e n o u g h , E, does not vary signif icant ly over i t and

E nL- E,2L = 0, (7-3)

o r

E n = E t2. (7-4)

T h e t a n g e n t i a l c o m p o n e n t o f E is t h e re f o r e c o n t i n u o u s a c r o s s th e b o u n d a r y .

Figure 7-2 C losed p ath of inte grat ion cro ss ing the interface between tw o media 1

and 2 . Wha teve r be the sur face cha rge dens i ty a f th e tangential c o m p o n e n t s o f E

on ei ther s ide of the interface are equal ; £ ( 1 = £ ( 2 -



7.1 Continuity Conditions at the Interface B etween Two Media 183

I f t he bou nd ar y l ie s be tw een a d i e l ec t r i c and a con duc to r , t hen E = 0

i n t h e c o n d u c t o r a n d E, = 0 in both media. With sta t ic charges, E is therefore

nor ma l to the su rf ace o f a co nd uc to r .

BENDING OF LINES OF FORCE

I t f o l lows f rom the boundary cond i t ions tha t t he D a n d E v e c t o r s c h a n g e

direc t ion at the bo un da ry be twe en tw o dielect r ics . In Fig . 7-3, using Eq. 7-2

wi th a* — 0.

D : cos (f = D 2 co s 0 2, (7-5)

o r

er l

e0

£

i cos 0 , = e r 2 e 0 £ 2 cos 0 2 (7-6)

and, f rom Eq. 7-4,

Ei si n 0 l = E 2 sin d2. (7-7)

Figure 7-3 Lin es of D or of E cro ssin g the interface betw een two me dia 1 an d 2.

The l ines change di rect ion in such a way that e r l t a n 8 2 = e r2 t an 0 , .



184 Dielectrics: II

Then, from the last two equations,

tan 0!

tan 0, In (7-8)

7.1.5

The larger angle from th e nor mal is in the mediu m with the larger relative

permittivity.

EXAMPLES: POINT CHARGE NEAR A DIELECTRIC, SPHERE OF

DIELECTRIC IN AN ELECTRIC FIELD

F i g u r e 7-4 s h o w s tw o e x a m p l e s of the b e n d i n g of l ines of force at the b o u n d a r y

b e t w e e n tw o dielect r ics . The l ines of D are b r o k e n bu t cont inuous , s ince l i nes of D

1

(a)

Figure 7-4 (a) Lines of D and e q u i p o t e n t i a l s for a p o i n t c h a r g e in air n e a r a

dielect r ic . They are not s h o w n n e a r th e cha rge because they get too close together .



18 5

Figure 7-4 (cont.) (b) Lines of D and equipo ten t i a l s nea r a d i e l ec t r i c sphe re s i tua t ed

in a i r in a uniform elect r ic f ie ld. In both cases , the l ines of D are indicated by

a r rows , and the equipo ten t i a l sur faces a re gene ra t ed by ro t a t ing the f igures a round

the hor i zonta l ax i s .

terminate only on free charges . Some l ines of E e i ther or iginate or terminate a t the

interface, according as to whether a b i s pos i t ive or negat ive .In Fig. 7-4b, the lines of D c rowd in to the sphe re , showing tha t D is larger

ins ide than ou t s ide . However , t he equipo ten t i a l s spread ou t i ns ide , and E i s weaker

ins ide than outs ide . The dens i ty of the l ines of E is lower ins ide than outs ide , s ince

some of the l ines of E coming from outs ide terminate a t the surface.

No te tha t , in the case of Fig. 7-4b, the field ins ide the sph ere is unifo rm. No te

also that the field ou ts ide is har dly dis turb ed a t dis tances larger than on e rad ius

from the surface.



186 Dielectrics: II

7.2 POTENTIAL ENERGY O F A CHARGE

DISTRIBUTION, ENERGY DENSITY

To calculate the potent ia l energy associated wi th an elect r ic f ie ld in dielect r ic

ma ter ia l , we refer to the paral le l -p late ca pa ci t or as in Sec. 4 .3 .

W e now h av e a dielect r ic- f i l led ca pa ci to r as in Fig . 6- 5. I f we a dd

c h a r g e s dQ un t i l t he top p la t e ca r r i es a charge Q an d is a t a po ten t ia l V,

we must expend an energy

-QV = - — = i2

V 2 C 2^ ere

0/F

~QV = - ^ = , Q 2 — , (7-9)

2 \A l e r e 0 AsA, (7-10)

= \DEsA, (7-11)

wh ere Z) i s the elect r ic dis pla ce me nt (Sec. 6 .5) , E is the electr ic f ield intensity,

a n d sA i s the volume of the dielect r ic .

T h e energy density i s now DE/2 or, if the dielectr ic is l inear, as is

usua l ly the case , € re 0E2

/2 . Th e po ten t i a l energy i s t hen g iven by the in t egra l

of e re 0E2

/2 eva lua ted over the vo lume occup ied by the f i e ld ,

2

7.3 FORCES ON CONDUCTORS IN THE

PRESENCE OF DIELECTRICS

Th e fo rces be tween con du c to r s in the p resenc e o f d i e l ec t r i cs can be ca l cu la t ed b y the me tho d o f v i r tua l wo rk th a t we used in Sec . 4 .5 .

W h e n t h e c o n d u c t o r s a r e i m m e r s e d i n a liquid dielect r ic , the forces

a re a lways found to be smaller than those in ai r by the factor e r if the charges

are the same in bo th cases . They a r e larger than in ai r by the factor e r if the

electric fields ( and hence the voltages) are the same.

The case of sol id dielect r ics wi l l be i l lust rated in Probs. 7-9 and 7-10.



7.4 Electric Forces on Dielectrics 187

EXAMPLE: PARALLEL-PLATE CAPACITOR IMMERSED

IN A LIQUID DIELECTRIC

We found in Sec. 4.5.2 that the force between the pla tes of an a i r- in sula ted paral le l -

p l a t e capac i to r i s e 0E2A/2, o r (a 2/2e 0)A , w h e r e a i s the surface charge den s i ty a nd

A i s the area of one pla te . We can perform a s imilar calcula t ion for a paral le l -pla te

cap aci to r imm ersed in a l iquid die lect r ic . Th e force per uni t area is equ al to the

energy den s i ty (Sec. 4.5) . Th en th e mech anic al force hold ing the pla tes ap art i s

F,„ = e,f E 2

A.(7-12)

For a given electric field s trength E, or for a given voltage difference between the

plates, the force is larger than in a i r .

Also.

D a 2

F m = A = A,2e re 0 2 e r e 0

(7-13)

1 Q 2

(7-14)e r 2e„A

a n d , for given charges +Q and — Q on the plates, the force is smaller than in a i r .

7.4 ELECTRIC FORCES ON DIELECTRICS

W he n a piece of dielect r ic i s p lace d in an elect r ic f ie ld , i ts d ipo les are subje cted

to e l ec t r i c fo rces and to rques .

I f the fie ld is unifo rm, a nd i f th e mo lecu les hav e a pe rm an en t dipo le

m o m e n t p. each d ipo le exper i ences a to rque .

T = p x E , (7-15)

that tends to align i t with the f ield, but the net force is zero.

I f the f ie ld is non-uniform, then the forces on the two charges of a

dipo le are un eq ua l , an d the re is a lso a net force.Le t us cons ider a d i e l ec t r i c whose molecu les do no t h a v e a p e r m a n e n t

dipole moment . We can f ind a general expression for the elect r ic force per

uni t volu me in a no n-u nif orm f ield by con side r ing the dielect r ic- f i lled

cyl indr ical capaci tor of Fig . 7-5.

Th e elect r ic field is rad ial , and i t s m ag ni tu de v ar ies as \jr. The ne t

elect r ic force on a dipole i s inward, because the inward force on — Q is



18 8

Figure 7-5 Dielectr ic-f i l led cyl indrical capaci tor .

l a rger than the ou tward fo rce on + Q. This net inward force is

dE

f= QE r - QE r + dr = Q ^ s ,

dE

(7-16)

(7-17)

w h e r e p = Qs is t h e d i p o l e m o m e n t .

Therefore, i f there are N d ipo les per un i t vo lume, the e l ec t r i c fo rce

per un i t vo lume i s

dE dEF

>= Np

Hr-= P

*>(7-18)

clE d(e , - = (€ r - l ) e 0 - ( E 2 / 2 ) , (7-19)

1 - £ ) ^ ( e r e 0 E 2 / 2 ) . (7-20)



7.4 Electric Forces on Dielectrics 189

Mo re generally, the electric force per unit volu me is pro po rti on al to

the gradie nt of the energy dens ity:

Th e dielectric tend s to move t o where the field is strong est. In the abo ve

expression, E is the electric field intensity inside the dielectric, of course.

7.4.1 I EXAMPLE: CURVE TRACER

A curve t race r is an e lec t ro-mechanica l dev ice tha t d raws curves ; the device is

u n d e r th e c o n t r o l of a c o m p u t e r . The curves are d r a w n on a shee t of p a p e r t h a t

must stay rigidly fixed to the pla t en . One way of h o l d i n g th e p a p e r is s h o w n in

Fig. 7-6. Para l l e l w i res are e m b e d d e d in the plas t i c p l a t en about tw o mil l ime te rs

a p a r t and c h a r g e d to plus and m i n u s 300 vol ts . This gives a h i g h l y i n h o m o g e n e o u s

field at th e surface, with lines of force converg ing toward th e cha rged wi res . Thus

V E 2 ins ide the p a p e r has a ver t i ca l component d i rec t ed in to th e pla t en .

Figure 7-6 Curve t race r . The elect r ic f ie ld produced by the cha rged wi res embedded

in the pla t en a t t rac t s the elect r ical ly neutra l sheet of p a p e r t o w a r d th e pla t en .

( 7 - 21 )

Platen



190 Dielectrics: II

7.5 THE POLARIZA TION CURRENT DENSITY dP/dt

W he n a die lectr ic is pla ced in an electr ic f ield th at is a func tion of t im e, the

mot ion o f the bound charges g ives a polarization current. We can f ind the

po la r i za t ion cur r en t dens i ty J h as fol lows.

F o r a n y v o l u m e i bounded by a su r f ace S, t he r a t e a t which bound

charg e f lows ou t th rou gh S must be equa l to the r a t e o f decreas e o f the bo un d

charge wi th in S (Sec. 5.1.1):

£ J „ d a = --jp„dx. (7-22)

Using the d ivergence theorem on the l e f t and subs t i tu t ing the va lue o f p h

on the r ight from Eq. 6-2,

" . " I D

f V • J„ dx = - f V • P dx = f V • — dx . (7-23)

Jr |( Ji Jr (!(We have pu t the c/ct under the in t egra l s ign because i t i s immater i a l whe ther

the der iv at ive w i th respec t to the t ime is cal cula ted f ir st or last . S ince x is

any vo lum e, the in t eg rand s must be equ a l and the po la r i za t ion cur r en t

densi ty i s

(7-24)

W e cou ld have added a con s tan t o f in t eg ra t ion ind epe nde n t o f the coo r

d ina tes , bu t such a cons tan t would be o f no in t e r es t . We have used the

par t ia l der ivat ive wi th respect to t b e c a u s e P can be a funct ion both of the

t ime and o f the space coord ina tes x, y. z.

7.6 THE DISPLACEMENT CURRENT DENSITY cD/ct

T h e displacement current density is

, 1 ) d 8 cP— = - ( e 0 E + P) = - e 0 E + — , (7-25)dt dt ct ct

w h e r e dP/dt i s t he po la r i za t ion cur r e n t den s i ty o f the p rev io us sec t ion .

Note the f i r s t term: i t can exist even in a vacuum.



7.7 Frequency anil Temperature Dependence, Anisotropy 191

7.6.1 EXAMPLE: DIELECTRIC-INSULATED PARALLEL-PLATE CAPACITOR

Let us ca lcu la t e the d i s p l a c e m e n t c u r r e n t in a die l ec t r i c - insu la t ed pa ra l l e l -p l a t ec a p a c i t o r .

C o n s i d e r a c u r r e n t / f lowing th rough a c a p a c i t o r as in Fig. 7-7. We a s s u m e

that pos i t ive charge f lows from lef t to r ight , that Q = 0 at t = 0, and t ha t edge

effects are negl igible .

O n th e lef t -hand s ide of t he capac i to r . Q increases and dQ/dt = I. On the r ight ,

a cha rge of e q u a l m a g n i t u d e bu t oppos i t e s ign bu i lds up and, aga in , dQ/dt = /.

N o w Q = Aa. D = a = Q A. w h e r e A is the a r e a of one plate , and a is the

surface charge dens i ty.

T h e d i s p l a c e m e n t c u r r e n t in the c a p a c i t o r is

dD

~dt

dQ

dt

I. (7-26)

T h e d i s p l a c e m e n t c u r r e n t A(dD/dt) is c o m p o s e d of two parts , f i rs t

d (a\ 1 dQA±(e

0E)

d DA — \ —

dt w dt dt(7-27)

a n d th e p o l a r i z a t i o n c u r r e n t

dP dA — =A-(D

dt dt

\\da dQ

dt '(7-28)

Q -Q

1 = dQ/dt I = dQ/dt

Figure 7-7 Curren t f lowing th rough a

capac i to r .

7.7 FREQ UENC Y AND TEMPERA TV REDEPENDENCE, ANISOTROPY

There are three basic polarization processes, (a) In induced or electronic

pola rizat ion, t he center of negativ e charg e in a molecul e is displaced , relative

to the center of positive charge, when an external field is applied, (b) In

orientational polarization, molecules with a perm anen t dipole mo men t

tend to be aligned by an externa l field, the magn itu de of the susceptibility



192 Dielectrics: II

be ing inverse ly p rop or t i on a l to the t emp era tu re , (c ) F ina l ly , ionic p o l a r i z a t i o n

occurs in ionic crystals: ions of one sign may move, wi th respect to ions of

the other s ign, when an external f ie ld is appl ied.

Fo r a g iven ma gn i tud e o f E . bo th the ma gn i tud e and the phase o f

p are funct ions of the frequency, f ir s t bec aus e of the va r iou s p ola r iz at io n

processes tha t come in to p lay as the f r equency changes , and a l so because

of the ex i s t ence o f r esonances . Thus , s ince e r i s a funct ion of the f requency .

e r i s s t r ic t ly def inable only for a pure sin uso ida l wav e. See Ta ble 6-1.

In many subs tances the r e l a t ive permi t t iv i ty decreases by a l a rge

fac to r as the t em pe ra tu re i s l owered th rou gh th e f reez ing po in t .

An i so t ro py i s an o th er dep ar tu re from the idea l d i e l ec t r i c behav io r .

Cry stal l in e sol ids co m m on ly hav e di f ferent dielect r ic pro pe r t ie s in di f ferent

c rys t a l d i r ec t ions because the charges tha t cons t i tu t e the a toms o f the c rys t a la r e ab le to move more eas i ly in some d i r ec t ions than in o ther s . The po la r i

z a t i o n P i s the n paral le l to E only whe n E is a lo ng cer tain prefer red di re ct io ns.

7.7.1 EXAMPLES: WATER, SODIUM CHLORIDE, NITROBENZENE,

COMPOU NDS OF TITANIUM

Water has a re la t ive permit t ivi ty of 81 in an e lect ros ta t ic f ie ld, and of about 1.8 a t

opt ical f requencies . Th e large s ta t ic value is a t t r ib uta ble to the ori ent a t io n of the

pe rm anen t d ipo le mom ent s , bu t t he ro t a t iona l i ne r t i a o f t he molecu les is muc h too

large for any s ignif icant response a t opt ical f requencies .

Similar ly, the re la t ive permit t ivi ty of sodium chloride is 5.6 in an electrostatic

f ie ld a nd 2.3 a t opt ical f requen cies . The larger s ta t ic value is a t t r ibu ted to ionic

motion, which again is imposs ible a t high frequencies .

In nitrobenzene, e r fa lls f rom ab ou t 35 to abo ut 3 in cha ngin g from the l iqu id

to the sol id s ta te a t 279 kelvins . In the sol id s ta te , the permanent dipoles of the

ni t robenzene molecules are f ixed r igidly in the crys ta l la t t ice and cannot rota te

un de r the inf luence of an external f ie ld.

Ceramic capac i to rs u t i l i ze va r ious compounds of titanium as die lect r ics . These

co mp ou nd s a re used because of the i r l a rge re l a tive pe rmi t t iv i t i e s rang ing up to

10.000. However, e r var i e s w i th the t empera ture , w i th the appl i ed vo l t age , and wi th

the ope r a t ing f requency .

7.8 SUMMARY

A t t h e b o u n d a r y b e t w e e n t w o m e d i a , b o t h V a n d th e t a n g e n t i a l c o m p o n e n t

o f E a r e con t inuous , bu t the d i f f e r ence be tween the normal componen t s o f

D i s equ al to the sur face de nsi ty of f ree ch arg e a f.



Problems 193

The potential energy stored in an electric field can be calculated from

th e energy density ere

0£

2

,'2. Hence the force per unit area exerted on a con

ductor submerged in a liquid dielectric is* e r e 0 £2 / 2 .

The force per unit volume on a dielectric is

1 - ^ V( ere

0£

2

/2). (7-2/)

The displacement current density at a point is

( D d

r i dP (7-25)

where dP/dt is the polarization current density.

Because of the nature of polarization processes, er

is a function of

f r equency . For most commonly used dielectrics, however, er

changes very

l i t t le, even when th e frequency chan ges by man y orde rs of magni tud e (Table

6-1). In anisotropic media. P is parallel to E only when E is along certain

preferred directions.

PROBLEMS

7-lE CONTINUITY CONDITIONS AT AN INTERFACE

Discuss the continuity conditions at the surface of the dielectric cylinder of Prob.

6-7.

7-2E CONTINUITY CONDITIONS AT AN INTERFACE

Discuss the continuity conditions at the surface of the dielectric sphere of Prob.

6-9.

7-3E ENERGY STORAGE IN CAPACITORS

A one-m icrof arad capa cito r is char ged to a poten tial of one kilovolt.

How high could you lift a one-kilogram mass with the stored energy, if you

could achieve 100°,, efficiency'?

7-4E ENERGY STORAGE IN CAPACITORS

The maximum stored energy per cubic meter in a capacitor is ere0a2

/2, where a

is the dielectric strength of the dielectric. The dielectric strength is the max imum electric

field intensity before rupture. See Prob. 6-5.

It is suggested that a small vehicle could be propelled by an electric motor fed by

charged capacitors. Comment on this suggestion, assuming that the only problem is

one of energy storage.



194 Dielectrics: II

A good die lect r ic to use would be Mylar , which has a die lect r ic s t rength of

1.5 x 10 s vol ts per meter and a re la t ive permit t ivi ty of 3.2.

Note that the use of a die lect r ic increases the capaci tance by e r and a l so pe rmi t sthe use of a higher vol tage . Th e die lect r ic s t reng th of a i r is only 3 x 10

6

vol t s pe r m e te r .

E, D. a, AND IF , FOR THE THREE CAPACITORS OF FIG. 7-8

Figure 7-8 shows three capaci tors . In a l l three the e lect rodes have an area S and

are se par ated by a dis ta nce .s. In A, the die lect r ic i s a i r . In B, the die lect r ic ha s a

re la t ive permit t ivi ty e r. In C, we have the same die lect r ic , plus a thin f i lm of a i r .

a) Find E. D, the surface charge dens i ty a on the e l ec t rodes , and the ene rgy

dens i ty W t for A and B.

Express your resul ts for B in terms of those found for A.

Solution : Fo r capa c i to r A ,

£ A = F / s , (1)

Z)A = e 0 £ A = e 0V/s, (2)

a

= DA

=eoV/s, (3)

W 1 A = \toEl = {-e 0V*/s2. (4)

Figure 7-8 Thr ee ca pa ci t or s : A. wi th out die lect r ic ; B, wi th die le ct r ic ; C, wi th

dielect r ic and an a i r f i lm. See Prob. 7-5.

http://toel/

http://toel/

Problems 195

For capacitor B.

£„ = Vs = £ A . . (5)

DK = e r e 0 £ B = ere0V/s = erDA. (6)

ffB = DB

= ere

0F x = e

ra

A. (7)

W 1 B= i e r e 0 E

2

= 1 e ,e 0 K2 /x 2

= e,W 1 A . (8)

b) Find £, D, H7

, for the dielectric in C, and <r on the left-hand electrode.

Express your results in terms of the quantities found for capacitor A.

Solution: Since the air film in capacitor C is thin, we m ay set

£„ = V/s = £A. (9)

Then D, a on the left-hand electrode, and W l are the same as for capacitor B:

Da= e r C A . (10)

< Ta

= erf f

A, (11)

Wia = e,W lA. (12)

c) Find £, D, W, for the ai r film of capacitor C. and a on the right-hand electrode.

Express again your results in terms of the corresponding quantities found for

capacitor A.

Solution: The electric displacement D in the air film is the same as in th e di

electric, since there are no free charges between th e plates. Then

DCa = M>A- (13)

ffc = Da, = e r D A= e

rff

A, (14)

Eca = DcJ*o = t,D A/e 0 = e ,£ A , (15)

WJ o = I e o £ i = J e 0 6 2 £ 2 = e r

2

rV 1 A . (16)

BOUND SURFACE CHARGE DENSITY

Find the bound surface charge densities on the dielectric of capacitor B of

P r o b . 7-5 in terms of V/s.



796 Dielectrics: II

7-7E EXAMPLE OF A LARGE ELECTRIC FORCE

O n e a u t h o r c l a i m s t h a t he can attain fields of 4 x 107 vol t s per mete r ove r a

2.5 mil l imeter gap in pur i f i ed n i t robenzene (« r = 35), an d t h a t th e resul t ing e lect r icforce pe r uni t area on the e lec t rodes is t h e n m o r e t h a n tw o a t m o s p h e r e s .

Is th e force real ly that large?

O n e a t m o s p h e r e e q u a l s a b o u t 105 pasca l s .

7-8 PERPETUAL-MOTION MACHINE

In P rob . 7-5 we s h o w e d t h a t , for the c a p a c i t o r C, the ene rgy dens i ty in the

dielect r ic an d in the air film are. respect ively,

e re 0 V2 e 2e 0 V

2

— — an d — r .2 s~ 2 s

N o w t h i s is di s turb ing because it a p p e a r s to m e a n t h a t the net force on thec a p a c i t o r is not z e r o . The forces on the capac i to r p l a t e s are

e re 0 V2

62

e 0 V2

r - 5 and =- S.2 s

2

2 s2

Is there a force on the dielect r ic? Well , the E ins ide is uni form and VE2

= 0, which

m e a n s t h a t the force dens i ty is ze ro . So it seems tha t the net force on the c a p a c i t o r is

e re 0 V2

(Cr-D-f-j-S.

W i t h M y l a r (e r = 3.2). s = 0.1 mil l ime te r . S = 1 m e t e r 2 , and V = 1 ki lovol t .

the force is a b o u t 3 x 10"1 n e w t o n s ! O ne could the re fore p rope l a large vehicle in

definitely with a set of capac i to rs l i ke th i s , fed by a smal l ba t t e ry supply ing a vol t age V

a t ze ro cur ren t ! Where have we e r r e d ?

Him: ( i )The va lues g iven above for the forces on the e lec t rodes are cor rec t ,

(ii) Inside th e dielect r ic , there is no force, (iii) Th e forces on the faces of the dielect r ic

a r e not z e r o , h o w e v e r . R e m e m b e r t h a t C o u l o m b ' s law appl i e s to any net a c c u m u l a t i o n

of charge, i r respect ive of the presence of m a t t e r .

7-9 SELF-CLAMPING CAPACITOR

A ce r t a in capac i to r is formed of two a l u m i n u m p l a t e s of a rea A, sepa ra t ed by a

sheet of dielect r ic of t h i cknes s /. and c o n n e c t e d to a s o u r c e of vol t age V.

For va r ious reasons , it is imposs ib le to a p p l y a force of m o r e t h a n a few t ens of

n e w t o n s to press the two pla t e s toge the r mechanica l ly . S ince ne i the r the pla t e s nor

the die lect r ic are perfectly flat, th e pla t e s are spaced by a di s t ance t, plus a good frac

t ion of a mil l ime te r . Thi s is highly ob jec t ionable because th e c a p a c i t a n c e m u s t be as

large as poss ib l e .

Y o u are a s k e d to i nves t iga te whe the r the elect ros ta t ic force of a t t r a c t i o n on

the pla tes might not be sufficiently large to reduce the air f i lm thickness to a negl igible

va lue .



Problems 197

Se t A = 4.38 x 1 0 " 2 squa re me te r , t = 0.762 mil l imeter . e r = 3.0. V =

60 ki lovol ts .

See a l so the nex t p rob lem.

7-10 ELECTROSTATIC CLAMPS

Electros ta t ic c lamps are somet imes used for holding work pieces whi le they

a re be ing machined . They u t i l i ze an insu la t ed conduc t ing p la t e cha rged to s eve ra l

tho usan d vo l t s and co vered w i th a th in insu la t ing shee t . The w ork p iece is p l aced on

the shee t and grou nde d .

On e pa r t i cu la r t ype ope ra t e s a t 3000 vo l ts and i s adve r t i s ed a s hav in g a ho ld ing

power of 2 x 1 0 5 pasca l s .

The insu la tor i s Myla r (e P = 3.2).

a) Sup po se f irst th at th ere is a f ilm of a i r on e i ther s ide of the M yla r .

Use the resul ts of Prob . 7-5 to show tha t the My lar f ilm is 15 mic rom eter s th ick.

b) Now calculate the electric field intensity in the air fi lm. You should find

over 6 x 1 0 s vol ts per meter , which is over 200 t imes the die lect r ic s t rength of a i r

(Prob. 7-4) . Sparking could occur in the a i r f i lm, and the die lect r ic could deter iorate .

Myla r ope ra t e s s a t i s fac tor i ly under these condi t ions , however .

c) Now assume that the a i r f i lm is replaced by a f i lm of t ransformer oi l wi th a

high die lect r ic s t reng th an d a perm it t ivi ty th at i s not mu ch different from that of the

M y l a r .

W ha t i s t he th i cknes s o f t he My la r now ?

7-11 CALCULATING AN ELECTRIC FORCE BY THE METHOD

OF VIRTUAL WORK

Figure 7-9 shows a pair of paral le l conduct ing pla tes immersed in a die lect r ic .

The lower pla te is f ixed in pos i t ion and grounded; the upper one can s l ide horizontal ly

and is m ain ta in ed a t a fixed vol tage V.

Use the me tho d of vi r tual work (Sec. 4.5) to calcula te the hori zon tal force on

the upper p l a t e when V = 1000 vol ts , s = 1 mil l ime ter . / = 100 mil l im eters . e r = 3.0.

Figure 7-9 P air of con du ct in g pla tes im me rsed in a die lect r ic . Th e lower one is

f ixed, whi le the up per o ne can s l ide horizo ntal l y. I t is poss ible to f ind the

hor i zo nta l fo rce on the upper p l a t e by the me tho d of v i r tua l work . See P r ob . 7 -11 .



198 Dielectrics: II

7-12 ELECTRIC FORCE

S h o w t h a t V £ 2 i s in the di rect ion of VE.

7-13 CALCULATING AN ELECTRIC FORCE BY THE METHODOF VIRTUAL WORK

Use the me th od of vi r tual w ork (Sec. 4.5) to calcu la te the force on th e die lect r ic

sheet in Fig. 7-10 when F = 1000 vol ts , s = 1 mil l imete r , / = 100 mil l im eters , e r = 3.0.

Figure 7-10 Sheet of die lect r ic p art ly

inse r t ed be tween a pa i r o f conduc t ing

pla t e s . The force on the die lect r ic canagain be calcula ted from the principle

of vi r tual work. See Prob. 7-13.

HIGH-VOLTAGE TRANSMISSION LINES

Losses on h igh-vol t age t ransm is s ion l ines a re h ighe r tha n no rma l dur ing foggy o r

ra iny we ather . O ne rea son for this i s that wa ter col lects on the wires an d forms

droplets . S ince a droplet carr ies a charge of the same s ign as that of the wire ,

e l ec t ros t a t i c repu l s ion forces the drop le t t o e long a te and form a sha r p po in t where

the e lect r ic f ie ld intens i ty can be h igh en ou gh to ionize th e a i r . Th e resul t i s a corona

discharge in the surrounding a i r . in which the ions are accelerated by the e lect r ic

f ie ld and col l ide wi th neutra l molecules , forming more ions and heat ing the a i r .

Corona d i s cha rges a re ob jec t ionable because they d i s s ipa te ene rgy , mos t ly by

heat ing the a i r . Over long dis tances , they can cause heavy losses . They also cause

radio and te levis ion interference.

Dr op lets col lect on the wire in dri f t ing by. Th ey are a lso a t t rac ted by the no n

uniform electric field.

You a re a sked to eva lua te the impor t ance of the non-uni form e lec t r i c f i e ld .

T o do this , assu me tha t the a i r is s tag nan t and ca lculate wi thin w hat dis tance of a

wire the e lect ros ta t ic force wi l l be larger than the gravi ta t ional force . Clearly, i f that

d i s t ance i s mu ch l a rge r than the cond uc to r rad ius , t hen e l ec t ros t a t i c a t t rac t io n i s

i m p o r t a n t .

Th e t ransmis s ion l ine under cons ide ra t ion co ns i s t s o f a pa i r o f cond uc to rs

11.7 mil l imeters in diameter , separated by a dis tance of 2 meters and operat ing a t

100 ki lovol ts (± 50 ki lovol ts ) .

You can f ind an approximate value for the f ie ld a t a dis tance r nea r one wi re

as follows. If / . is the charge per meter.

2 ; i e n



Problems 199

Th e expres sion on the right is the field near a single isolat ed wir e, whic h is a goo d

app rox im ati on as long as c is muc h less than th e dis tance to the other wire. No w

the capaci tance per meter for th is l ine is approximately ne 0/5^ a n d

/. = (ne0 5 )1 05

c o u l o m b s m e t e r .

The expression for the force per unit volume given in Sec. 7.4 involves e r a n d

the value of E inside a water droplet . For water, at low frequencies , e r = 81

(Table 6-1). To est imate the £ inside a droplet , one may use the formula for a d i

electric sphere in a uniform field . ' This is not a bad approximation, because even a

ra in d ro p is smal l , com pare d to the con duc to r rad ius , so tha t £ does no t c hange

much over one d rop d iameter . Then

e r + 2 / 2ne0r

Solution: Let us fi rs t calculate £ inside a water droplet s i tuated at a d is tance c

from the center of one of the wires:

(7re 0/5)10 5

= 361.4/r vol ts /m eter. (1)11 + 2) 2ne 0r

The electrostat ic force per cubic meter on a droplet is

1 \ dF = 1 —

81 dr

1 , , (361.4x 81 x 8.85 x 1 ( T 1 2 x

2 V r

12)

.d [i\ , 14.625 x 1CT

5

— \—A = - 9 . 2 5 0 x 1 0- 5

n e w t o n s / m e t e r3

. (3)dr \r2

y

The negative s ign indicates that the force is at t ract ive.

Now the gravi tat ional force per cubic meter is 1000 g newtons. so

F 9.250 x 10" 5

= 5 • (4

1 0 0 % 9 8 0 0 c3

This ra t io is equal to uni ty for c = 2 .113 mil l imeters , which is smaller tha n the

radius of the wire.

The electrostat ic force of at t ract ion on a droplet is therefore negligible.

+ Electromagnetic Fields and Wares, Eq. B-55.

* Electromagnetic Fields and Waves, Eq. 4-174.



200 Dielectrics: II

7-15 ELECTRIC FORCE ON A DIELECTRIC

a) Calculate the e lect r ic force per cubic meter on the die lect r ic of a coaxial

cab le whose inne r cond uc to r has a rad ius o f 1 mi l l ime te r and w hose ou te r con duc tor

has an inner radius of 5 mil l imeters . The die lect r ic has a re la t ive permit t ivi ty of 2.5.

Th e ou te r co ndu c tor i s g rou nde d , and the inne r cond uc to r is ma in ta in ed a t 25 k i lovol ts .

b) Show that the e lect r ic force near the inner conductor is about 300 t imes

larger than t he grav i ta t io nal force if the die lect r ic has the dens i ty of water , nam ely

1 0 3 ki lograms pe r cub ic me te r .

7-16 DISPLACEMENT AND POLARIZATION CURRENTS

A capaci tor wi th pla tes of area A, separated by a die lect r ic of thickness s and

rela t ive permit t ivi ty e r , i s connected to a vol tage source V t h rough a re s i s t ance R.

Calcu la t e the d i sp lacement and the po la r i za t ion cur ren t s a s func t ions of , the

t ime .

7-/7 DIRECT ENERGY CONVERSION

I t is poss ible to conve rt th erm al energy in to e lect r ical energy with the c i rcui t

of Fig. 7-11.

W i t h X connec ted to Z , t he ba t t e ry B cha rges the ba r ium t i t ana te capac i to r C.

T h e n X i s disconnected from Z and the capac i to r i s hea ted . The re l a t ive pe rmi t t iv i ty

of the t i tanate decreases and the vol tage on C i nc reases . Then X i s connec ted to Y,

a n d C recha rges the ba t t e ry B, while supplying energy to R. Th e capac i to r i s t hen coo led

and the cycle repeated.

Such devices have been bui l t to feed e lect ronic c i rcui ts on board sate l l i tes . The

capaci tors are in thermal contact wi th panels on the outer surface of the sa te l l i te that

are heated periodical ly by the sun as the sa te l l i te rota tes about i t s own axis in space.

L et V B be the po ten t i a l suppl i ed by the ba t t e ry ; Q the capac i t ance a t t he temp e r a t u r e T l: C 2 a n d V 2 t he capac i t ance and vo l t age a t T 2; \Y lh t he the rma l ene rgy

requi red to hea t t he capac i to r ; and We the electric energy fed to R.

Calculate the eff ic iency WJWlh

w h e n V B = 700 vol ts , V 2 = 3500 vol ts . e r l =

8000. e r2 = 1600, area of on e cap aci t or e lect rod e 1 squ are meter , thick ness of die lect r ic

0.2 mil l imeter , specif ic heat of barium t i tanate 2.9 x 1 0 6 j ou le s pe r cub ic me te r pe r

degree, and T 2 — T, = 30 C.

>

O

R

Figure 7-1 1 Circu i t fo r t r ans forming

thermal energy into e lect r ic energy on

board sate l l i tes . See Prob. 7-17. B



CHAPTER 8

MAGN ETIC FIELDS: I

The Magnetic Induction B and the Vector Potential A

8.1 THE MAG NETIC INDUCTION B

8.1.1 Exam ple: Long Straight Wire Carrying a Current8.1.2 Exam ple: Circular Loop, Magn etic Dipole Mom ent m

8.2 THE DIVERGENCE OF B

8.3 MAGNETIC MONO POLES

8.4 THE VECTOR POTENTIAL A

8.4.1 Exam ple: Long Straight Wire

8.5 THE VECTOR POTENTIAL A AND THE SCALAR

POTENTIAL V

8.6 THE LINE INTEGRAL OF THE VECTOR POTENTIAL A

OVER A CLOSED CURVE

8.7 SUMMARY

PROBLEMS



Th e next e ight cha pte rs wi ll deal wi th the m ag ne t ic f ie lds of e lect r ic c urr en ts

a n d o f m a g n e t i z e d m a t t e r .W e sha l l s t a r t he re by s tudy ing the ma gne t i c induc t ion B an d the

vec to r po ten t i a l A . The se two quan t i t i e s co r r esp ond , r espec tive ly , t o the

elect r ic f ie ld in tensi ty E and to the potent ia l V.

F o r t h e m o m e n t , w e c o n s i d e r o n ly c o n s t a n t c o n d u c t i o n c u r r e n t s a n d

n o n - m a g n e t i c m a t e r i a l s .

Figure 8-1 shows a ci rcui t car rying a current I. We def ine the magnetic

induction a t the po in t P as fol lows:

where the in t egra t ion i s ca r r i ed ou t over the c losed c i r cu i t . As usua l , t he

uni t vector I-

! p o i n t s from t he source to t he po in t o f observa t ion : i t po in t s

from t he e l ement d l to t he po in t P. Magnet i c induc t ions a r e expressed in

teslas.

T h e c o n s t a n t p 0 is defined as fol lows:

THE MAGN ETIC INDUCTION B

(8-1)

fi0 = 4n x 10 7 t e s l a m e t e r / a m p e r e . (8-2)

and is cal led the permeability of free space.

I f the current I i s d i s t r ibu ted in space wi th a cu r r en t den s i ty J a m p e r e s

per square mete r , t hen / becomes J da and must be pu t under the in t egra l



20 3

s i g n . T h e n J da dl c a n b e w r i t t e n a s J dx', w h e r e dx ' i s a n e l e m e n t o f v o l u m e ,

a n d

B ^ f , ^ , ( 8 - 3 )

a s i n F i g . 8 -2 . T h e i n t e g r a t i o n i s c a r r i e d o u t o v e r t h e v o l u m e x o c c u p i e d b y

t h e c u r r e n t s .

W e a s s u m e t h a t J i s n o t a f u n c t i o n o f t i m e a n d t h a t t h e r e a r e n o m a g

n e t i c m a t e r i a l s i n t h e f i e l d .

Figure 8-2 The magnet ic induct ion dB due to an e l ement J dx ' of a vo lume

di s t r ibu t ion of cur ren t .

204 Magnetic Fields: I

As in e l ec t ros t a t i cs , we can descr ibe a magne t i c f i e ld by d rawing lines

of B t ha t a r e everywhere t angen t to B.

Similar ly , i t i s convenient to use the concept of f lux, the flux of the

magnetic induction B t h rough a su r f ace S, d e f in e d a s t h e n o r m a l c o m p o n e n t

of B i n t eg ra ted over S:

O = J . B • da . (8-4)

Th e flux d> is exp res sed in webers. Th us the t es la is one webe r per sq uar e

m e t e r .

8.1.1 EXAMPLE: LONG STRAIGHT WIRE CARRYING A CURRENT

An e lement dl of a long s t ra ight wire carrying a current / , as in Fig. 8-3, produces

a magne t i c induc t ion

p0I dl si n 0d B = - — q>i,

47 i r

(8-5)

where (p t i s t he un i t vec tor po in t ing in the az imu tha l d i rec t ion . The pos i t ive d i re c t ions

for <p, and / are re la ted by the r ight-hand screw rule .

Figure 8-3 The magnet ic induct ion d B produced by an e l ement / dl of the cur ren t

/ in a long s t ra ight wire .

20 5

Figure 8-4 Lines of B in a p l a n e p e r p e n d i c u l a r to a l ong s t ra igh t w i re ca r ry ing ac u r r e n t / . The dens i ty of the l ines is i nve rse ly propor t iona l to the di s t ance to the

wire . Lines c lose to the wire are not s h o w n .

Expres s ing dl. sin 0. and r in t e r m s of i. and p.

B = - — I cos x d% ip , = <p,. (8-6)

4n p J~nl2

Inp

T h e m a g n i t u d e of B thus falls off inversely as the first power of the dis tance from

an infinitely long wire. T he l ines of B are concentr ic c i rc les lying in a plane pe rpen

d icu la r to the wire , as in Fig . 8-4.

EXAMPLE: CIRCULAR LOOP. MAGNETIC DIPOLE MOMEN T m

A c i rcu la r l oop of r a d i u s a ca r r i e s a c u r r e n t /, as in Fig. 8-5.

An e lement / dl of c u r r e n t p r o d u c e s a d B h a v i n g a c o m p o n e n t dB. on the axis , as

i nd ica t ed in the figure. By s y m m e t r y , the tota l B is a l o n g the axis , and

dB . =— -.cost). (8-7)

4n r

p 0I 2to j p0lcr

B. =r co s 0 = j - ^ . (8-8)

4n r2

There fore , on the axis , the m a g n e t i c i n d u c t i o n is m a x i m u m at the cente r of the r ing

a n d d r o p s off as r3

for z2

» a2

.

F i g u r e 8-6 shows l ines of B in a p l a n e c o n t a i n i n g th e axis of the l o o p .



20 6

Figure 8-5 The magnet ic induct ion d B produced by an e l ement / d l at a point on

the axis of c i rcular current loop of radius a. The pro jec t ion of d B on the axis is dB..

y \

Figure 8-6 Lines of B in a plane containing the axis of the loop of Fig. 8-5. The

direct ion of the current in the loop is shown by the dot and the cross .



8.2 The Divergence o / B 207

Far from th e l o o p , th e field is the s a m e as t h a t of an electric dipole (Sec. 2.5.1),

except that the factor 1 /4ne 0 is rep laced by p0/4n, an d t ha t t he e l ec t r i c d ipo le moment

p is rep laced by the m a g n e t i c d i p o l e m o m e n t m. The magnetic dipole moment m is a

v e c t o r w h o s e m a g n i t u d e is na 2I an d t h a t is n o r m a l to the p l a n e of t he loop , in the

di rec t ion g iven by the r ight-hand screw rule wi th respect to the c u r r e n t /.

THE DIVERGENCE OF B

Cons ider a current element / dl and a volume tk containing a point P as

in Fig. 8-7. Th e curren t element p rod uce s at P a magnetic induction

j U 0 / dl x i - !

4n r2

dB = ^ , • ' . (8-9)

As we saw in Sec. 8.1.1, the lines of B due to the element I dl are circles

situated in planes perpendicu lar to a line thro ugh dl and centered on it as

Figure 8-7 The cur ren t e l ement / d l p r o d u c e s a m a g n e t i c i n d u c t i o n B. The net

outward f lux of B t h r o u g h th e sur face of the e l e m e n t of v o l u m e dx is z e r o .




in Fig . 8-7. From the f igure, i t i s obvious that the net outward f lux of the

B due to I d l th rough the su r f ace o f the vo lume dx i s ze ro . N ow , any vo lu me

can be subd iv ided in to vo lume e lement s o f the same k ind as dx . Therefo re ,fo r any vo lume t bounded by a su r f ace S ,

j ;B • da = 0. (8-10)

Th is equ at i on is t rue of a l l m agn et ic f ie lds. The n, using the div ergen ce

theorem (Sec . 1.10),

V • B dx = 0. (8-11)

So

V • B = 0. (8-12)

Th e der iva t ives con ta in ed in the oper a to r V a re wi th r espec t to the fi eld

po in t where the magn e t i c indu c t io n i s B . Th i s is an o th er o f M axw el l ' s

e q u a t i o n s .

MAGNETIC MO NO POLES

Th e s t a t e me nt tha t V • B is equ a l to ze ro impl i es th a t m agn e t i c f ie lds a r e

due so le ly to e l ec t r i c cu r r en t s and tha t magne t i c "charges" do no t ex i s t .

Otherwise , one would have the magne t i c equ iva len t o f Eq . 6 -12 , and the

d i v e r g e n c e of B w o u l d b e p r o p o r t i o n a l t o t h e m a g n e t i c " c h a r g e " d e n s i ty .

E l e m e n t a r y m a g n e t i c " c h a r g e s " a r e c a l l e d magnetic monopoles. T h e i r

ex i s t ence was pos tu la t ed by Di r ac in 1931 , bu t they hav e never been obse rved

t o d a t e . T h e t h e o r e t i c a l v a l u e o f t h e m a g n e t i c m o n o p o l e is2h/e, or 8.271 17 x1 0 " 1 5 w e b e r , w h e r e h i s P lanck ' s cons tan t (6 .625 6 x 1 0 "

3 4

j o u l e s e c o n d ) ,

a n d e i s the charge of the elect ron, 1 .602 1 x 1 0 "1 9

c o u l o m b .

Th e surface in tegra l of B • da ov er a c losed surface is eq ua l to the en

c l o s e d m a g n e t i c c h a r g e .

P ro b le m s 8-9 an d 12-6 conc ern two m eth od s tha t hav e been used

for de tec t ing monopoles in bu lk mat t e r . See a l so P rob . 19-5 .



8.4 The Vector Potential A 209

TH E VECTOR POTENTIAL A

Since V • B = 0, it is reasonable to assume that there exists a vector A such

that /

B = V x A, (8-13)

because the divergence of a curl is identically equal to zero (see Prob. 1-29).

The derivatives in the del operator are evaluated at the field point.

We can find an integral for A, starting from Eq. 8-3,

B = £° f J x %th'. (8-14)

4tt J*' r2

From Probs. 1-13 and 1-27,

J xr

i f = v f - ) x J = V xJ

- - V x J . (8-15)r

2

Now, th e last term on the right is zero, because J is a function of x', y', z',

while V contains derivatives with respect to X, y, z. Then

47T JR

' r 4n JR

' r(8-16)

and

A Mo r Jf -dx'. (8-17)4n

If the current is limited to a conducting wire,

Note that we can add to this integral any vector whose curl is zero

without affecting the value of B in any way.

Note also that B depends on the space derivatives of A, and not on

A itself. The value of B at a given point ca n thus be calculated from A, only

if A is known in the region around the point considered.




EXAMPLE: LONG STRAIGHT WIRE

In Sec. 8.1.1 we calculated B for a long straight wire carrying a current /. startingfrom the definition of B. We can also calculate B starting from

HqI

An

(II(8-19)

One can see immediately that A is parallel to the wire, since the dl 's are all along

the wire.

We first calculate A for a current of finite length 2L. Referring to Fig. 8-8.

Ho i rL

' Anpi dl

Jo{p> + /2)>-2

^ i n [ z + (P

j

+ r-?2

y6.

(8-20)

(8-21)

Ikd.iUL

271 p(p

2

« L2

).

In p (8-22)

(8-23)

For L -* oo, A tends to infinity logarithmically. We can still try to calculate B,

however, because a function can be infinite and still have finite derivatives. For

exa mple , if y = x, y -» x as x -> x . but dy dx remains equal to 1.

To calculate B for p2

« L2

, we use the fundamental definition of the curl given

in Eq. 1-77, using the rectangular path shown in Fig. 8-8 for the line integral:

B = lim1

(j) A • dl.A p A r

as ApA: tends to zero. Since A is parallel to the wire,

1B = lim

A p A r[Aip) - A(p + Ap)] Ar

— hm — I In2tt Af.

In2/.

In Ap

p p + Ap

p + Ap

— hm —In Ap

In 1 +

(8-24)

(8-25)

(8-26)

(8-27)

(8-28)

Now In (1 + x) = x when x is small an d. finally, B is equal to p0I/2np, as previously.



21 1

Figure 8-8 An e lement / dl of a current / in a long s t ra ight wire produces an

e lement o f vec tor po ten t i a l d A at the point P . The sma l l rectangle is the path of

in t egra t ion used for ca l cu la t ing B .

THE VECTOR POTENTIAL A AND THE SCALARPOTENTIAL V

T h e v e c t o r p o t e n t i a l A of magnet ic f ie lds i s in many respects s imi lar to the

sca la r po ten t i a l ( o r , s imply , t he po ten t i a l ) V of electric fields (Sec. 2.5).

Le t us r e tu rn b r ie f ly to C ha p te r s 2 and 3 .

a) W ith elect r ic f ie lds, the imp or ta nt q ua nt i ty , in prac t ice , is the in

tensi ty E: i t i s E that g ives the stored energy and the forces on charged bodies

and o n po la r i zed d ie l ec t r i cs . I t is a l so E tha t cause s b rea kd ow n in d ie lec t r i cs .

b ) T h e q u a n t i t y V i s r e l a t ed to E th rough E = — VF. So one can

add to V any qua n t i ty tha t is i nde pen den t o f the coo rd in a te s wi tho u t a ff ect ingE in any way. Also, the value of E at a point can be calculated only i f V is

kno wn in the region a r o u n d t h a t p o i n t .

c) Fin al ly , the in tegra l for F( Eq . 2-18) i s s im pler to ev alu ate than tha t

for E (Eq. 2-6). So, in general, i t is easier to f ind E by first calculating V, a n d

then V F, than to ca lcu la t e E d i r ec t ly . How ever , if t he geom et ry o f the c harge

dis t r ib ut i on is par t icu lar ly s im ple, on e can use G au ss 's law (Sec. 3 .2). Th en

E is easy to calcu late di rect ly a nd th ere is no nee d to go thro ug h V.




No w le t us see how A com pa res w i th V.

a) Th e qua n t i ty tha t co r r esp ond s to E i s t he ma gne t i c indu c t ion B:

it is B tha t g ives the sto red ene rgy an d the ma gn et ic forces (Ch ap ter 13).

b ) The vec to r po ten t i a l A is r e l a t ed to B on ly th ro ug h B = V x A.

On e can add to A any qu an t i ty who se cur l is ze ro wi th ou t ch ang ing B, and

on e can ded uce f rom A the value of B at a poi nt only i f on e kn ow s the va lue

of A in the r eg ion a ro un d th a t po in t .

c) The integral for A (Eq. 8-18) i s a lso simpler to calculate than that

for B (Eq. 8-1). H ow ev er , wi th simp le curren t dis t r ib ut i on s, it i s mu ch easier

to use Am pe re 's c i rcui ta l law (Sec. 9 .1) to find B than to calcu late A f ir st .

So there is a grea t s imi lar i ty betw een V and A. Later on, in Sec. 11.5 ,

we shal l f ind an equat ion that re lates E to both V a n d A , i n t i m e - d e p e n d e n t

fields.

8.6 THE LINE INTEGRAL OF THE VECTOR

POTENTIAL A OVER A CLOSED CURVE

Using S tokes ' s t heorem (Sec . 1.13),

A • dl = f (V x A) • da , (8-29)

w h e r e S is any su r f ace bo un de d by the c losed curve C. But, from Eq. 8-13,

the cur l of A is B. Then

(j), A • dl = J s B • da = cp: (8-30)

the l ine integral of A • dl over a c losed curv e C i s eq ua l to the ma gn et i c f lux

t h r o u g h C.

8.7 SUMMARY

T h e magnetic induction B du e to a c i rcui t C car ry ing a curr en t / is def ined as

fo l lows:



Problems 213

Magnetic inductions are measured in teslas, an d u0

is denned to be exactly

An x 10"7

tesla meter per ampere.

The magnetic flux through a surface S is

(D = JsB - d a . (8-4)

Mag net ic flux is meas ure d in webers, and a tesla is one weber per square

meter.

The divereence of B is zero:

V • B = 0, (8-12)

since the magne tic flux th rou gh a closed surface is alway s identically equa l

to zero, if mag net ic mo no po le s do no t exist. Thi s is one of M axwell 's

equations.

It follows from this that

B = V x A, (8-13)

where A is the vector potential

Th e line integra l of A • dl over any closed curve C is equa l to the

magnetic flux through any surface S bounded by C:

(j). A • dl = (D. (8-30)

PROBLEMS

8-1E MAGNETIC FIELD ON THE AXIS OF A CIRCULAR LOOP

P l o t a curve of B as a function of z on the axis of a circular loop of 100 turns

hav ing a mean radius of 100 millimeters and carrying a current of one ampere.

8-2 SQUARE CURRENT LOOP

C o m p u t e the magnetic induction B at the center of a square current loop of

side a carry ing a current /.




8-3 MAGNETIC FIELD OF A CHARGED ROTATING DISK

An insu la t ing d i sk of rad ius R carr ies a uniform free surface charge dens i ty a

on one face . I t rota tes about i t s axis a t an angular veloci ty to.a) What is the magni tude of the e lect r ic f ie ld intens i ty E, close to the disk?

b) Show that the surface current dens i ty a a t the radius r i s wra.

c) Show that , a t the center of the charged surface,

1B = - n

0wRo.

d) Ca lcu la t e E a n d B for R = 0.1 meter , a = 1 0- 6

c o u l o m b p e r s q u a r e m e t e r ,

a n d OJ = 1000 rad ia ns pe r s econd .

8-4 SUN SPOTS

See Prob. 8-3.The Zeeman e f fec t

1

observed in the spectra of sunspots reveals the exis tence

of magnet ic induct ions as large as 0.4 tes la .

Let us assume that the magnet ic f ie ld is due to a disk of e lect rons 10 7 m e t e r s

in radius rota t ing a t an angular veloci ty of 3 x 1 0 "2

rad ian pe r s econd . The th i cknes s

of the disk is smal l co m pa red to i t s rad ius .

a) Sho w that the dens i ty of e lect ro ns requi red to achieve a B of 0.4 tesla is

a b o u t 1 01 9

per squa re me te r .

b ) Show tha t t he cur ren t i s abo ut 3 x 1 0 1 2 a m p e r e s .

c ) In v i ew of the enormous s i ze o f the Coulomb forces , such cha rge dens i t i e s

a re c l ea r ly imposs ib le . Then how could such cur ren t s ex i s t ?

8-5E HELMHO LTZ COILSOne of t en wishes to ob ta in a un i form magne t i c f i e ld ove r an apprec iab le vo lume .

In such cases one uses a pair of Helmhol tz coi ls as in Fig. 8-9.

Show that the f ie ld on the axis of symmetry, a t the midpoint between the two

coils is

(0 .8) 3 2p 0NI/a,

w h e r e TV i s the nu m be r of turn s in each coi l .

I f yo u have th e cour age to calc ula te the f ield as a funct ion of z, and i ts derivat ives ,

you wil l f ind that the f i rs t , second, and thi rd derivat ives are a l l zero a t z = 0!

F igure 8-10 shows B. as a func tion of z for value s of/- up to 0.16a.

Ro ugh ly spe akin g, B. i s unifo rm, w i thin 10%, ins ide a sphe re of rad ius 0.1a .

f W hen a gas is subjected to a s t ron g ma gn et ic f ie ld, i ts spectra l l ines are spli t into several

com pon ent s . The sp l i t t i ng is a me asure of the magn e t i c induc t ion .



21 5

Figure 8-9 Pair of Hel m ho l tz coi ls .

When the spacing between the coi ls i s

equa l t o one rad ius a, t he magne t i c

f ie ld near the center i s remarkably

uniform. See Prob. 8-5.

-0.2a

Figure 8-10 A xial co m po ne nt of B

nea r the cen te r o f a pa i r o f He lmhol t z

I coils as in Fig. 8-9, as a function of z

-0.2a and for var ious values of r.

HELMHOLTZ COILS

You are asked to des ign a pair of Helmhol tz coi ls (Prob. 8-5) that wi l l cancel

the earth 's ma gn et ic f ie ld, wi thin 10 perce nt , over a spherica l volum e having a radiu s

of 100 mil l imeters .

Th e magne t i c indu c t ion of the ea r th in the l abora to ry i s 5 x 1 0 "5

tes la andforms an ang le o f 70 degrees w i th the hor i zo nta l . The l a bor a tor y i s s i t ua t ed in the

No r th e rn Hem isphere . See the foo tno te to P ro b . 8 -11 .

a ) Ho w mu s t the co i ls be or i en ted , and in wha t d i rec t ion m us t t he cur ren t f low?

b) Specify the coi l dia me ter and sp acing .

c ) Wh a t i s t he to t a l nu m be r of am per e - tu rns requ i red?

d) T he l abo ra to ry has a goo d supply of N o. 18 enam eled copp er w i re on h and .




This wire has a cross-sect ion of 0.823 square mil l imeter and a res is tance of 21.7 ohms

per ki lom eter . An adjus tab le po we r supply is a lso avai lable . I t can sup ply from 0 to

50 vo l t s a t a maximum of two amperes .

How many tu rns should each co i l have?

Spec i fy the ope ra t ing cur ren t and vo l t age .

Wil l it be necessary to cool the coi ls? If so, wh at d o you sugg es t?

8-7 LINEAR DISPLACEMENT TRANSDUCER

Draw a curve of Bz as a function of z on the axis of a pair of Helmhol tz coi ls

(Prob. 8-5) wi th the currents f lowing in opposi te di rect ions and with the coi ls separated

by a d i s t ance 2a . ins tead of a. for values of z ranging from - a t o + a.

Note the l ineari ty of the curve over most of this region.

One could bu i ld a l i nea r d i sp lacement t ransduce r fo r measur ing the pos i t ion

of an object by fixing to it a Hall probe (Sec. 10.8.1) and having it move in the field

of such a pa i r o f oppos ing He lmhol t z co i l s .

8-8E THE SPACE DERIVATIVES OF B IN A STATIC FIELD

Consider the f ie ld of a c i rcular loop, as in Sec. 8.1.2.

On the r ight and somewhat above the axis , the l ines of B f lare out , so B. d e

creases wi th z a n d dBJvz i s negat ive . At the same point , B y increases with y, so dB y/dy

i s pos i t ive . S imilar ly, cBJdx is positive.

a ) W hy i s t h i s , ma th ema t i ca l ly? Ho w a re these de r iva t ives re l a t ed ?

b) Compare these derivat ive for other points on Fig. 8-6.

8-9E MAGNETIC MONO POLES

I t is p red ic t ed theore t i ca l ly th a t a magn e t i c cha rge Q* s i tua t ed in a magne t i c

f ie ld would be subjected to a force Q*B/u0.Calculate the energy acquired by a magnet ic monopole in a f ie ld of 10 tes las

ove r a d i s t ance of 0 .16 me te r . Expres s your answer in g igae lec t ron-vol t s (10 9 e lec t ron-

volts).

In one exper iment fo r de tec t ing magne t i c monopoles , va r ious s amples , such

as deep-sea sed imen ts , were subjected to such a f ie ld, as in Fig. 8-11. N on e were found.

Figure 8-11 Appara tus used in an

unsucces s fu l a t t empt to obse rve

magne t i c monopoles . The dev ice i s

inserted ins ide a coi l producing an

axial B of 10 tes las . A s lurry of deep-

sea sediment f lows into the chamber

on the r ight a t / and out a t O. I t was

h o p e d t h a t m a g n e t i c m o n o p o l e s i n

the s ample would be acce le ra t ed in

the magnet ic f ie ld and would leave

t racks in the photographic p l a t e P.

See Prob. 8-9.



Problems 217

8-10 MAGNETIC FIELD OF A CHARGED ROTATING SPHERE

A conducting sphere of radius R is charged to a potential V and spun about a

diameter at an angular velocity w as in Fig. 8-12.

a) Show that the surface charge density a is e0V/R.

b) Now show that the surface current density is

It turns out that B is uniform inside the sphere.

d) What would be the value of B for a sphere 0.1 meter in radius, charged to

10 kilovolt s, and spinning at 1 04

revolutions per minute?

e) Show that the dipole moment is

f) What is the dipo le mom ent for the above sph ere?

g) What current flowing through a loop 0.1 meter in diameter would give the

same dipole moment?

a = e0o)V sin 0 = M sin 0,

where M is e0coV, an d 0 is the polar angle.

c) Show that the magnetic induction at the center is

B = ( 2 / 3 ) 6 o / u 0 o j F = (2/3)/i„M.

m = (4/3 )7tPv3

M .

+ ++

++

r +

++ +

Figure 8-12 Charged sphere spinning

about a diameter. See Probs. 8-10 and8-11.

THE EARTH'S MAGNETIC FIELD

The origin of the earth 's magn eti c field is still largely unkn own . According to one

model, the earth would carry a surface charge, producing an azimuthal current

because of the rotation, and hence the magnetic field, as in Prob. 8-10.

218

Figure 8-13 Typical line of B for the

earth's magnetic field. The shape of

the field inside the earth is unknown

With the model discussed in Prob.

8-11 , the B inside would be uniform,

and the lines of B would be straight,

as in the figure.

This model is attractive, at first sight, because it gives a magnetic field that has

the correct conf igurat ion, o r nearly so, except that one has to disregard t he fact

that the magnetic and geographic poles do not coincide. As we shall see, this model

requires an impossibly large surface charge density a.

a) What mus t be the sign of <r?+

Solution: Figure 8-13 shows the earth with its North and South geographic

poles, a line of B, and the direction of the angular velocity a).

The current must be in the direction shown, and hence a must be negative.

b) With this current distribution, it can be shown that the B inside the earth

is uniform.

Use the results of Prob. 8-10 to deduce the value of M from the fact that the

vertical component of B at the poles is 6.2 x 1 0 "5

tesla. At the poles, the value of

B is the same, immediately above and immediately below the surface.

Solution: Since

2

B = -n0M = 6.2 x 10"5

, (1)

1.5

M = — 6.2 x 10 = 74 amp ere s/m ete r. (2)

/'o

c) Calculate the magnetic dipole moment of the earth.

The earth has a radi us of 6.4 x 106

meters.

Solution: Again from Prob. 8-10,

m = (4 /3 )7 tR 3

M , (3)

= (4/3)ti (6.4 x 106

)3

74 = 8.1 x 1 02 2

ampere meters2

. (4)Remember that the magnetic pole that is situated in the Northern Hemisphere is a

South" pole: the "North" pole of a magnetic needle points North.



Problems 219

d| Calc ulate the surface charge dens i t ) required to give the surface curren l

dens i ty ca l cu la t ed above .

If there were such a surface charge, what would be the value of the vert icale lect r ic f ie ld inten s i ty?

There does exis t a negat ive charge a t the surface of the earth. I t gives an £ of

ab ou t 100 vol ts per meter . See Prob . 4-2. Also, the m ax im um elect r ic f ie ld intens i ty

tha t can be sus t a ined in a i r a t normal t e mp era tu re and p res sure is 3 x 10 6 vol ts per

me te r .

Solution: At the equa tor . I) = 0

7. = av = M = 74. (5)

w h e r e v i s the tangen t ia l veloci ty. Th us

74= 0 . 16 c o u l o m b m e t e r 2 .=

,2;r x 6.4 x 10" )/(24 x 60 x 60)(6)

This surface charge dens i ty would give an e lect r ic f ie ld intens i ty

a 0.16= 1.8 x 1 0 1 0 vol ts meter . (7 )=

- 1 28.85 x 10

from Sec. 3.2, which is absu rdly large.



CHAPTER 9

MAGN ETIC FIELDS: II

Ampere's Circuital Law

9.1 AMPE RE'S CIRCUITAL LAW

9.1.1 Examp le: Long Cylindrical Conductor9.1.2 Example: Toroidal Coil

9.1.3 Example: Long Solenoid

9.1.4 Examp le: Refraction of Lines of H at a Current Sheet

9.1.5 Example: Short Solenoid

9.2 THE CURL OF THE MAGNETIC INDUCTION B

9.3 SUMMARY

PROBLEMS



This cha p te r i s devo ted to A m pe re ' s c i r cu i t a l l aw and to the cu r l o f B.

Am pe re ' s l aw is used to ca lcu la t e ma gne t i c induc t io ns in mu ch the sam eway as Gauss 's law is used to calculate e lect r ic f ie ld in tensi t ies . The value

of V x B wi l l f o l low immedia te ly f rom Ampere ' s l aw by us ing S tokes ' s

t h e o r e m .

We s t i l l l imi t our se lves to cons tan t conduc t ion cur r en t s and to non

m a g n e t i c m e d i a .

We have seen in Sec. 8 .1 .1 that the magnet ic induct ion vector B near a long

s t r a igh t wi r e is az im utha l an d tha t i t s ma gn i tud e i s p.0I/2np, w h e r e p is the

distance f rom the wire to the point considered. Thus, over a c i rcle of radius p

cen tered on th e wire as in Fig . 9-1,

AMPERE 'S CIRCUITAL LA W

(9-1)

Th is is a gen eral resu l t and , for any close d curv e C,

(9-2)

wh ere / is the curre nt enclo sed by C. Th e posi t ive di rec t ion for the in teg rat io n

is re lated to the posi t ive di rect ion for / by the r ight-hand screw rule as in

Fig. 9-1.



22 2

Figure 9-1 Posi t ive di rect ion for the

in tegra t ion pa th a rou nd a cur ren t / .

For a vo lume d i s t r ibu t ion o f cu r r en t ,

(f ) cB • dl = / t 0 J , J • da, (9-3)

w h e r e J is t he cu r r e n t dens i ty th r ou gh a ny su r face S bo un de d by the cu rve C

as in Fig. 9-2a. This is Ampere's circuital law.

In many cases the same cur r en t c rosses the su r f ace bounded by the

curve C several t imes. With a solenoid, for example, C could fol low the

ax i s an d r e tu rn ou t s ide the so leno id , a s in F ig . 9 -2b . Th e to t a l cu r r e n t c ross ing

the su rf ace i s t hen the cu r r e n t in each tu rn mu l t ip l i ed by the num be r o f

tu rns , o r the number o f ampere-turns.

c

(a)

Figure 9-2 (a ) Am pe re ' s c i rcui ta l law s ta tes that the l ine integral of B • dl a r o u n d C

is equal to p 0 t imes the cur ren t t h rough any sur face bounded by C. (b) Pa th of

in t egra t ion C for a solenoid. In this case the l ine integral of B • dl is &p0I.



9.1 Ampere's Circuital Law 223

The circuital law can be used to calculate B when its magnitude is the

same, all alon g the path of integ ratio n. This law is thu s somew hat similar t o

Gauss's law, which is used to compute E, when E is unifo rm over a surface.

EXAMPLE: LONG CYLINDRICAL CONDUC TOR

The long cy l indr i ca l conduc tor of Fig. 9-3 ca r r i e s a c u r r e n t / uni formly d i s t r ibu ted

over its cross-sect ion with a dens i ty

J = I/nR2

. (9-4)

O u t s i d e th e c o n d u c t o r , B is a z i m u t h a l and i n d e p e n d e n t of (p, so t h a t , a c c o r d i n g

to the circui ta l law,

B = p0I/2np.

Ins ide the c o n d u c t o r , for a c i rcu la r pa th of r a d i u s p,

PaJitp 1 p0Jp p0IpB

2n p 2n R 2

(9-5)

(9-6)

T h e m a g n e t i c i n d u c t i o n B therefore increases l inearly wi th p ins ide th e c o n d u c t o r .

O u t s i d e the c o n d u c t o r , B dec reases as 1/p. The curve of B as a funct ion of p is

s h o w n in Fig . 9-4.

Figure 9 -3 Long cy l indr i ca l conduc tor ca r ry ing a c u r r e n t / .



224 Magnetic Fields: II

'•0 2.0 3.0 4.0 5.0 6.0

p (millimeters)

Figure 9-4 The magnet ic induct ion B as a function of radius for a wire of 1 mil l ime te r

rad ius ca r ry ing a cur ren t o f 1 ampere .

EXAMPLE: TOROIDAL COIL

A close -wo und t oro ida l coi l of squ are cross-se ct ion, as in Fig. 9-5, carr ies a curren t / .

A l o n g p a t h a, the line integral of B i s zero, s ince there is no current l inking this

p a t h . T h e n t h e a z i m u t h a l B i s zero in this region. The sam e appl ies to c and to any

s imi l a r pa th ou t s ide the to ro id . Then the az imutha l B i s ze ro eve rywhere ou t s ide .

Ins ide , a long pa th b,

2npB = p 0NI , (9-7)

w h e r e N i s t he to t a l num ber of tu rns , a nd

B = u.N I 2np. (9-8)

The re ex is t non -az im utha l com pon ent s o f t he mag ne t i c induc t ion ou t s ide the

toro id . For a pa th such a s d in Fig. 9-5, the area bo un de d by the pat h is crossed

Figure 9-5 To roi da l coi l . Th e brok en l ines show p ath s of integ rat io n.

9.1 Ampere's Circuital Law 225

once by the current in the toroidal winding and, at a distance large compared to

the outer radius of the toroid, the magnetic induction is that of a single turn along

the mean radius.Although the magnetic induction B outside the toroid is essentially zero, the

vector potential A is not. This will be evident if one remembers that A is a constant

times the integral of / dl /r, where r is the distance between the element dl and the

point where A is calculated (Sec. 8.4). For example, at a point close to the winding.

A is due mostly to the nearby turns, it is parallel to the current, and it has approxi

mately the same value inside and outside.

The vector potential A can therefore exist in a region where there is no B field.

This simply means that we can have at the same time A # 0 and V x A = 0.

For example, A = ki, where k is a constant, satisfies this condition. We are already

familiar with a similar situation in electrostatics: the electric potential can have

any uniform value in a region where E = — VV = 0.

9.1.3 EXAMPLE: LONG SOLENOID

For the long solenoid of Fig. 9-6. we select a region remote from the ends, so that

end effects will be negligible; also, we assume that the pitch of the winding is small.

Let us choose cylindrical coordinates with the r-axis coinciding with the axis of

symmetry of the solenoid, as in Fig. 9-7.

1. We first note that B has the following characteristics, both inside and outside

the solenoid.

a) By symmetry, all three components Bp, B_, Bv are functions neither of cp

nor of z.

b) Moreover, Bp = 0 for the following reason. Consider an axial cylinder of

length / and radius p. either larger than the solenoid radius, or smaller, as in Fig. 9-6.

The integral of B • da over its surface is simply InplB^ since the integrals over the

two end faces cancel. But, according to Eq. 8-10, the integral of B • da over any

closed surface is zero. Then B (, = 0.

c) Consider a rectangular path a inside the solenoid as in Fig. 9-6. By Ampere's

circuital law, the line integral of B • dl around the path is zero. Since Bp = 0, as we

have just found, the line integrals on the vertical sides must cancel, which means

that, inside the solenoid, B. is also independent of p. The same applies to a similar

path entirely outside the solenoid.

2. Outside the solenoid.

a) We have shown abo ve that, ou tside th e solenoid. B. is independent of all

three coordinates p. <p , z. However. B is zero at infinity. Therefore B. = 0 outside

the solenoid.

b) A path such as b is linked once by the current and, outside the solenoid.

B v = p0l 2np. (9-9)



Figure 9 - 7 C o m p o n e n t s o f B a t P . in cy l indr i ca l co ord in a te s .



9.1 Ampere's Circuital Law27

3. Inside t he so lenoid .

a) B v = 0 ins ide because the l ine integral of B • dl over a c i rc le of radius p,

say the top edge of the smal l cyl inder shown in Fig. 9-6, i s 2-itpB^, and this must be

ze ro accord ing to Eq . 9 -2 because the re i s no cur ren t enc losed by the pa th .

b) Con s ide r in g now p a th c in F ig . 9 -6 , and rem emb er ing th a t B f, = 0 b o t h

ins ide and ou t s ide , and tha t B. = 0 outs ide , we see that B.S = p 0N'IS, w h e r e N'

is t he num be r of tu rn s pe r me te r . The re fore

B r = ix0N'I. (9-10)

The magne t i c induc t ion ins ide a long so lenoid in the reg ion remote f rom the

end s is theref ore axia l , uniform, an d equa l to p 0 t i m e s t h e n u m b e r o f a m p e r e - t u r n s

p e r m e t e r NT. I t i s mu ch l a rge r th an the az im utha l B outs ide , as long as (l/N')/2np «

1, or as long as the pi tch of the win din g 1/JV', divided by i ts c i rcumference 2np, isvery smal l .

I t i s obvious , f rom Fig. 9-10, that B. o u t s i d e a finite solenoid is not zero. This

ou t s ide B : can be l a rge r o r sma l l e r t han the ou t s ide B^ , depending on the geomet ry

of the so lenoid , bu t bo th a re much sma l l e r t han B. ins ide .

EXAMPLE: REFRACTION OF LINES OF B AT A CURRENT SHEET

Imagine a th in conduc t ing shee t ca r ry ing a cur ren t dens i ty o f a amperes pe r me te r ,

as in Fig. 9-8. We can unders tand the refract ion of l ines of B at a current sheet by

pro ceed ing a s in Sec. 7.1. Since the diverg ence of B is z e r o , t h e n o r m a l c o m p o n e n t

4B

Figure 9-8 Conduc t ing shee t ca r ry ing a cur ren t dens i ty o f a amperes pe r me te r .

Since V • B = 0 , t h e n o r m a l c o m p o n e n t o f B i s the same on both s ides of the sheet .

Accord ing to the c i rcu i t a l l aw , howev er , t he t angen t i a l com po nen t i s no t conse rved

and a line of B i s deflected in the di rect io n s how n.

of B i s conse rved:

B„i = B, l2. (9-11)

Also, if we app ly Am pe re ' s c i rcui ta l law to a path of length L tha t i s perp end icula r

to the sheet as in the figure.

fl„L-B ,2L = ,, 0s<L, (9-12)

B, 2= B„ - / ,„* . (9-13)

Th e l ine of force is therefore rot a te d in the c lockwise di re ct ion for an obse rver

looking in the di rect ion of the vector a .

We could have a r r ived a t t h i s re su l t i n ano the r way . The magne t i c induc t ion B

i s due to the current sheet i t se l f and to other currents f lowing elsewhere . Accordingto Am pe re 's c i rcui ta l law. the curre nt sheet pr odu ces , jus t b elow i tse l f in the f igure ,

a B that i s di rected to the left an d who se ma gn i tud e is p 0

a

/ 2 (Pr ob . 9-3). S imilar ly,

th e B jus t ab ov e the sheet i s di rected to the r ight and has the same m ag ni tu de . If we

add this f ie ld to that of the other currents , we see that the tangent ia l components of

B must di ffer as above.

EXAMPLE: SHORT SOLENOID

We can ca l cu la t e B on the axis of the short solenoid of Fig. 9-9 by summing the

co ntr ibu t ion s of the indiv idual turn s , us ing Eq. 8-8. If the length of the solenoid is /

and if i ts radius is a. t he magne t i c induc t ion a t t he cen te r i s

= Po f + i2 a 2NT dz_ Po r+i/2 j

2 J " " 2 (aa 1 + zz

Y

(9-14)

u 0a2N'I

a\a2 + z

2

)l!2

= HcjN'I si n 8„ (9-15)

as in Fig. 9-9. We hav e assum ed th at the soleno id is c lose w ou nd .

For a long so lenoid . 0m -* i t /2 , and B -> fi0NT, as in Sec. 9.1.3.

At one end. again on the axis ,

B = u 0NT si n 0J2. (9-16)

Th e magne t i c ind uc t ion thus dec reases a t bo th end s of the so lenoid , and th i s

is of course due to the fact that the lines of B flare out as in Fig. 9-10.

Upon c ros s ing the winding , t he rad ia l component o f B r e m a i n s u n c h a n g e d ,

bu t t he ax ia l com pon ent ch anges bo th i t s ma gni tude a nd i t s s ign . Fo r exam ple .



22 9

Figure 9-10 Lines of B for a solenoid wh ose leng th is twice i ts diam eter .



23 0

1.5 -1.0 0.5

00.5 1.0 1.5

Figure 9-11 The m a g n e t i c i n d u c t i o n B as a funct ion of z on t he axi s of a so lenoid

with a l eng th equa l to 5 t imes its d i a m e t e r . N o t e how sha rp ly th e f ie ld dr op s off at

t he ends .

t h e a x i a l c o m p o n e n t in the upper lef t -hand s ide of t he so lenoid in Fig . 9-10 c h a n g e s

from, say, —0.9p0N'I to +0.\p oN'I, s ince th e surface current dens i ty /. is N'l

(Sec. 9.1.4).

F igure 9-11 shows B as a funct ion of z for a so lenoid wi th / = 10a. See P r o b . 9-5.

THE CURL OF THE MAGNETIC INDUCTION B

Apply ing Stok es's theo rem to Eq. 9-3, we find tha t

(9-17)

for any bounded surface S. Then

V x B = HqJ. ( 9 - 1 8 )

Thi s is an oth er of Ma xwell' s equat ion s, but no t in its final form yet,

bec aus e we are still limited t o static fields, and we hav e no t yet co nsi der ed

magnetic materials.

Problems 231

9.3 SUMMARY

Ampere's circuital law states that the l i r ie in tegral of B • dl over a c losed

curve C i s equ a l to the cu r r en t f lowing th ro ug h any su r f ace S bo un de d by C:

(j) c B • dl = fi0 js J • da. (9-3)

U p o n a p p l y i n g S t o k e s ' s t h e o r e m t o th i s e q u a t i o n , o n e f in d s t h a t

V x B = /< 0J. (9-75)

PROBLEMS

9-1E DEFINITION OF u 0

Sho w tha t the perm eabi l i ty of free space fi 0 can be defined as follows: If an

infini tely long solen oid carr ie s a cur ren t dens i ty of on e amp ere per meter , then the

magnet ic induct ion in tes las ins ide the solenoid is numerical ly equal to / i 0 .

9-2 MAGNETIC FIELD OF A CURRENT-CARRYING TUBE

A length of tubing carr ies a current / in the longi tudinal di rect ion.

a) W ha t is the value of B o u t s i d e ?

b) Ho w i s A or i en ted ou t s id e?

c) W ha t is the value of B i ns ide?

d) Sho w that A is uniform ins ide.

9-3E MAGNETIC FIELD CLOSE TO A CURRENT SHEET

A conduc t ing shee t ca r r i e s a cur ren t dens i ty o f a amperes p e r m e te r .

Show tha t , ve ry c lose to the shee t , t he magne t i c induc t ion B due to the cur ren t

in the sheet is u. 0x/2 i n the d i rec t ion pe rpendicu la r t o the cur ren t and pa ra l l e l t o the

sheet .

9-4E VAN DE GRAAFF HIGH-VOLTAGE GENERATOR

In a Van de Graaff high -vol ta ge gen erat or , a cha rged in sulat in g bel t i s used tot ranspor t e l ec t r i c cha rges to the h igh-vol t age e l ec t rode .

a) Calculate the current carr ied by a bel t 0.5 meter wide driven by a pul ley

0.1 meter in diam ete r an d ro ta t in g a t 60 revolu t ions pe r second, i f the e lect r ic f ie ld

intens i ty a t the surface of the bel t i s 2 ki lovol ts per mil l imeter .

b) Calculate the magnet ic induct ion c lose to the surface of the bel t , neglect ing

edge effects. See Prob. 9-3.




9-5 FIELD ON THE AXIS OF A SHORT SOLENOID

A shor t so l enoid ca r r i e s a cur ren t I a n d h a s N' t u rns pe r me te r .

Show that , a t any point on the axis .

B = - / J 0 / N ' ( C O S + cos a 2 ) i

w h e r e a l and a 2 a re the angles sub tended a t t he po in t by a rad ius R at e i ther end of

the so lenoid .

Fo r exam ple , i f the coi l has a length 2 L, and if the po int i s s i tua ted a t a dis tan ce

x from the center .

L - x L + xcos a 2

1

[R2

+ (L — x )2

]1

'2

'2

[R2

+ (L + x )2

]1

'2

9-6 FIELD AT THE CENTER OF A COIL

A coi l has an inner radius R 1? an ou te r rad ius R 2, and a l eng th L .

a) Show that the magnet ic induct ion a t the center , when the coi l carr ies a

c u r r e n t 7, is

ti 0nIL a + (i 2 + /f)" 2

B = In r—T7T-,2 1 + (1 + P )

w h e r e

a n d n i s t he tu rn dens i ty , o r t he nu mb er of w i res pe r squa re me te r ,

b) Show that the length of the wire is

\ = Vn = 2nn(y.2 - l)/?7?3

,

w h e r e V i s the volume occupied by the wire .

9-7 CURRENT DISTRIBUTION GIVING A UNIFORM B

A long s t ra igh t conduc tor has a c i rcu la r c ros s - s ec t ion of rad ius R and ca r r i e s

a current 7. Ins ide the conductor , there is a cyl indrical hole of radius a whose axis i spa ra l l e l t o the ax i s o f t he conduc tor and a t a d i s t ance b from it .

Show tha t t he mag ne t i c induc t ion ins ide the ho le is un i form a nd equ a l t o

2n(R 2 - a 2)'

See a l so the nex t p rob lem.



Problems 233

Hint: Use a un i form c ur ren t d i s t r ibu t io n th rou gh out the ci rc le o f rad ius R,

plus a current in the opposi te di rect ion in the hole .

9-8D 1 SADDLE COILS

Figure 9-12 shows the cons t ruc t ion of a pa i r o f s addle co i l s . The windings

occupy the two ou te r reg ions of a pa i r o f ove r l apping c i rc l e s .

a) In what general di rect ion is the f ie ld oriented in the region between the

c o n d u c t o r s ?

b) I t i s sa id that the mag net i c indu ct io n is uniform in the hol low . Is this t ru e?

What is the value of B?

Assume that the length of the coi l i s inf ini te .

You ca n solve this one by s t ra ight integ rat io n i f yo u wish, bu t there is a mu ch

eas ier way. See the hint for Prob. 9-7.

Such co i l s a re used for magne tohydrodynamic gene ra tors . See P rob . 10-16 .

Figure 9-12 (a) Pair of saddle coils forproduc ing a t ransve rse magne t i c f i e ld

ins ide a tubula r enc losure . Only one

turn of each coi l i s shown, (b) C r o s s -

sect ion of the windings . See Prob. 9-8.

9-9D TOROIDAL COIL

A toroidal coi l of N t u rns has a ma jor rad ius R a n d a m i n o r r a d i u s r.

a ) Ca lcu la t e the magne t i c f lux by in t egra t ing B over a cross-sect ion.

b) A t wha t rad ius does B have i ts me an va lue?

f The le t ter D indicates that the problem is re la t ively di ff icul t .



CHAPTER 10

MAGN ETIC FIELDS: III

Transform ation of Electric and Ma gnetic Fields

10.1 THE LORENTZ FORCE

10.1.1 Example: The Crossed-Field Mass Spectrometer

10.2 EQUIVALENCE OF E AN D v x B

10.3 REFERENCE FRAMES, OBSERVERS,

AND RELATIVITY

10.4 THE GALILEAN TRANSFORMATION

10.5 THE LORENTZ TRANSFORMATION

10.6 IN VARIANCE OF THE VELOCITY OF LIGHT c

10.6.1 Example: The Velocity of Light is Unaffected by the Earth's

Orbital Velocity

10.7 INVARIANCE OF ELECTRIC CHARGE

10.7.1 Example: The Electron Charge

10.8 TRANSFORMATION OF ELECTRIC AND

MAGNETIC FIELDS

10.8.1 Example: The Hall Effect

10.9 SUMMARY

PROBLEMS



In th i s chap te r , we sha l l s t a r t wi th the long -kno wn f ac t t ha t a mo ving charged

par t ic le i s def lected by a m ag ne t ic fie ld . F o r exam ple, the elec t ron be am ina t e lev i s ion tube i s de f lec ted b o th h or i zo n ta l ly a nd ver t i ca l ly by the m agn e t i c

f ie lds of tw o sets of coi ls car r yin g rapi dly va rying cu rren ts .

Now why should a magnet ic f ie ld exer t a force on an elect r ical ly

charge d pa r t i c l e? The ex p lan a t ion i s t ha t t he par t i c l e " sees" an electric

f ie ld . In other words, the elect rons in the te levision tube "see" an electric

f ie ld when they pass between the def lect ing coi ls .

Le t us cons ider a more genera l case . Suppose one has some conf igura

t ion of e lect r ic charg es and ele ct r ic cu rre nts that p ro du ce an elect r ic f ie ld

in tens i ty E , an d a ma gne t i c indu c t ion in a l ab ora to ry . Bo th E t a n d B ,

are func t ions o f the coor d in a tes .v, v . z . N o w im agine an observ er mo vinga t some co ns ta n t ve loc i ty v wi th r espec t to the l abor a to r y . Th i s obse rver i s

equ ippe d wi th ap pro pr i a t e ins t rum ent s for me asu r ing the e l ec t r i c f ie ld

intensi ty and the magnet ic induct ion. At every point , he wi l l f ind a f ie ld

E 2 , B 2 that i s different f rom E , , B x (except if v. E,. B , are al l paral le l ) . This

chap te r concerns the equa t ions tha t a r e used to f ind E 2 . B 2 . g iven E , . B ^

and inver se ly .

W h e n e v e r o n e e x p r e s s e s E t , B j in terms of E 2 . B 2 , or inversely , one

transforms t he e l ec t romagne t i c f i e ld f rom one r e f e r ence f r ame to ano ther .

Most o f the p rob lems a t t he end o f th i s chap te r concern magne t i c

forces exer ted on particles; macroscopic man i fes t a t ions o f ma gne t i c fo rceswill be dea l t wi th in C ha pt er s 13 an d 15.

236 Magnetic Fields: 111

10.1 THE LORENTZ FORCE

It is obse rved exp erim entall y that a char ge Q moving at a velocity v in

a region whe re the magn etic ind ucti on is B, is subjected to a magnetic

force

F = Qy x B. (10-1)

Since this force is perpendicular to v, F • v = 0 and the powe r supp lied

to" the particle is zero. The Q\ x B force therefore chan ges the direction of

v without changing i ts magnitude.

M or e gen erall y, if the re is als o an electr ic field E,

F = <2(E + v x B). (10-2 )

This is the Lorentz force.

10.1.1 EXAMPLE: THE CROSSED-FIELD MASS SPECTROMETER

The crossed-field mass spectrometer is illustrated in Fig. 10-1. T he positive ions of

mass in and charge Q have a velocity v given by

- me2

= QV. (10-3)

In the region of the deflecting plates, they are submitted to an upward force QE and

to a downward force QiB. Fh e net force is zero for ions having a velocity

v = E/B, (10-4)

or a mass

in = 2QVB2

E2

. (10-5)

Fhe mass spectrum is obtained by observing the collector current I as a func

tion of E, for a constant B.

This type of mass spectrometer has a poor resolution, mostly because the ion

beam is defocused in the fringing electric an d magnetic fields. It is nonetheless a

simple and useful device if one is concerned only with the very lightest elements.



23 7

9 "

A

0

Figure 10-1 Crossed- f i e ld mass spec t romete r . Pos i t ive ions produced in a source S

mainta ined a t a po ten t i a l V are focused into an ion beam b. In pas s ing th rough

the superposed e lect r ic and magnet ic f ie lds , the beam spl i ts vert ical ly into i t s

va r ious com pon ent s . Ions of the pro pe r ve loc i ty a re undef l ec t ed and a re co l l ec ted

at C. Th e spec trom eter is enclo sed in a vacu um vessel A e v a c u a t e d b y a p u m p P.

Th e Lor en tz fo rce o f Eq . 10-2 i s i n t r igu ing . W hy shou ld v x B have the

sa me effect as the elect r ic f ie ld E? Cle ar ly , f rom E q. 10-2, the pa r t ic le ca nn ot

tel l whether i t "sees" an E or a v x B term . This i s i l lust r ated in the a bo ve

exa mp le. I t i s a lso i l lust r ated in Pr ob . 10-6, wh ere two typ es of ma ss spe c

t rom ete r a r e com pa red . Th e ion t r a j ec to ry is c i r cu la r in bo t h types , bu t

the deflecting force is QE i n one , and Q\ x B in the o ther . Thus v x B is,

somehow, an elect r ic f ie ld in tensi ty .

As we shal l see in the next sect ions, the explanat ion is provided by

the theory of re lat iv i ty . I t has to do wi th reference f rames.

By observer, w e m e a n e i t h e r a h u m a n b e i n g e q u i p p e d w i t h p r o p e r i n s t r u

m e n t s , o r s o m e d e v i c e t h a t c a n m a k e m e a s u r e m e n t s , t a k e p h o t o g r a p h s , a n d

s o o n , e i t h e r a u t o m a t i c a l l y o r u n d e r r e m o t e c o n t r o l .

10.2 EQUIVALENCE OF E AND v x B

10.3 REFEREN CE FRAME S, OBSERVERS,

AND RELA TIVITY



238 Magnetic Fields: III

An observer t akes h i s measurement s wi th r espec t to h i s reference

frame. F o r e x a m p l e , o n e n o r m a l l y m e a s u r e s a m a g n e t i c i n d u c t i o n w i t h a n

ins t rum en t tha t is a t rest wi th respec t to the ear th . Thi s par t ic ula r f rame is

usual ly cal led the laboratory reference frame.

Special relativity i s c o n c e r n e d w i t h t h e o b s e r v a t i o n s m a d e b y t w o

observer s , one o f w ho m ha s a con s tan t ve loc i ty wi th r espec t to the o th er .

10.4 THE GALILEAN TRANSFORMATION

Imag ine two r e f e r ence f r ames Si a n d S 2 as in F ig . 10-2, wi th S 2 m o v i n g a t

a veloci ty vi wi th r espec t to S , . N ow im agin e some even t , say a nu c lea r

r eac t ion , occ ur r ing a t a ce r t a in ins t a n t a t a ce r t a in p o in t in space . An obse rver

in S { no tes the coord ina tes o f the even t as x l s y u zu t. An observer in S 2

notes, l ikewise, x 2 , .v 2, z2, t, a n d

x, = x 2 + vt, (10-6)

y, = y2, (io-7)

- i = i 2 . (10-8)

Figure 10-2 Two Car t e s i an coord ina te sys t ems , one moving a t a ve loc i ty v\ wi th

respec t t o the o the r in the pos i t ive d i rec t ion of the common x-ax i s . The two

sys tems ove r l ap when the or ig ins 0 1 and 0 2 coinc ide . We shall always refer to

these two coordinate systems w henever we discuss relativistic effects.



10.5 The Lorentz Transformation 239

We have set t = 0 at the instant w hen th e two reference frames ov erlap .

These equations constitute the Galilean transformation.

The inverse transformation, giving the coordinates x2, y2, ?i m terms

of xit

yu

zu

is given by inter cha ngi ng the subscr ipts 1 and 2, and chan ging

the sign of v.

10.5 THE LORENTZ TRANSFORMA TION

Altho ugh the Galilean tran sfor matio n is adeq uat e for everyday p heno mena ,

it is only approximate, and the correct equations are those of the Lorentz

transformation:

x , = y(x2

+ vt2), (10-9)

J? I = yi, (10-10)

*i = z2, (10-11)

ti = y ( * 2 + 4 x A (10-12)

where

1

[1 - {v/cfY11

'(10-13)

and c is the velocity of light in a vacuum, 2.997 924 58 x 10s

meters per

second.

The inverse relationship, giving the coordinates in Sl in terms of those

in S2, is again obt ained by interchang ing the subscripts 1 and 2, and changing

the sign of v.

Note that, in general, t1

/ t2. In other words, the observer on Sj does

not agree with the observer on S2

as to the time of occurrence of a given

event.

The Lo rent z tran sfor matio n forms the basis of special relativity. On e

can deduc e from it equ at ion s of tra nsf orm ati on for the length of an object.




10.6 IN VARIANCE OF THE VELOCITY OF LIGHT c

Exp erim ents designed to measu re the velocity of light c in a vac uum , with

respect to reference frames moving at various velocities, always give the same

value. Thus c is said to be invariant.

10.6.1 EXAMPLE: THE VELOCITY OF LIGHT IS UNAFFECTED BY

THE EARTH'S ORBITAL VELOCITY

The orbi ta l veloci ty of the e a r t h a r o u n d the sun is 3 x 104 m e t e r s per second . Thi s

is two o r d e r s of m a g n i t u d e l a r g e r t h a n th e t angent i a l ve loc i ty due to the r o t a t i o n

of th e e a r t h a b o u t its own axis . At n o o n t i m e , th e orbi ta l veloci ty is wes tward whi l e ,

a t midnigh t , it is e a s t w a r d . If o n e m e a s u r e s c in the eas t -wes t di rect ion in the l a b o

ratory, first at n o o n t i m e and t hen at m i d n i g h t , one finds precisely the same va lue ,

wi thin the exper imenta l e r ror .

10.7 IN VA RIA NCE OF ELECTRIC CHA RGE

Electric charge is also invar iant . This is again an exper imen tal fact.

10.7.1 EXAMPLE: THE ELECTRON CHARGE

In th e Millikan oil-drop experiment, one m e a s u r e s th e veloci ty of an elect r ical ly

c h a r g e d m i c r o s c o p i c oil d r o p in an electric field. The veloci ty is of th e o r d e r of mil l i

m e t e r s pe r m i n u t e an d is a m e a s u r e of the e lec t r i c cha rge on the d r o p . The c h a r g e

ca r r i ed by a dr op i s a lways a mul t ip l e of t he e l ec t ron cha rge , 1.602 x 10_ 1 g c o u l o m b .

If now one m e a s u r e s in the l a b o r a t o r y f r a m e th e c h a r g e on an e lec t ron emerg ing

from an acce le ra tor w i th a v e lo c it y a p p r o a c h i n g th e veloci ty of l ight , by deflect ing

it in a known magnet ic f ie ld, one finds again 1.602 x 1 0- 1 9 c o u l o m b .

Electromagnetic Fields and Waves, C h a p t e r 5.

the duration of an event, a velocity, an acceleration, a force, a mass, and

so on /



10.8 Transform ation of Electric and Magnetic Fields 241

10.8 TRANSFORMA TION OF ELECTRIC

AND MAGNE TIC FIELDS

T h e e x p r e s s i o n for the Loren tz fo rce in Eq. 10-2 give s us a clue as to ho w to

t r ansfo rm a magnet ic f ie ld . In t h i s express ion , F, (9, E, v, B are all m e a s u r e d

in the same reference f rame S,, w h i c h is n o r m a l l y the l a b o r a t o r y f r a m e :

F , = Q{El

+ v x B,) (10-14)

N o w c a l l S2 the reference f rame of the par t i c l e , moving at the veloci ty

v at a cer t a in ins t an t .

W h a t is the field in S 2 ? Let us a s s u m e for the m o m e n t t h a t v2

« c2

,

w h e r e c is the veloci ty of l ight . In t ha t case , the force is the s a m e in b o t hreference f rames. In S

2. the par t ic le veloci ty is z e r o and t h e r e can be no

magnet i c fo rce . Then the force in S2 m u s t be QE

2, w h e r e

E , = E , + v x Bp (10-15)

So the magnetic field B, in f rame 1 b e c o m e s an electric field v x B, in

f rame 2.

T h e a b o v e e q u a t i o n is val id only for r2

« c2

. W h e n v a p p r o a c h e s c,

it is s h o w n in r e l a t iv i ty theory tha t the forces in the two reference f rames

are different, and the cor r ec t va lue of E2

is as f o l l o w s : 1

E21 = 7 ( E , i + v x B,). (10-16)

E 2 | I = E 1 H , (10-17)

w h e r e the subscr ip t s r e f e r to the c o m p o n e n t s t h a t are e i t h e r p e r p e n d i c u l a r

or paral le l to the veloci ty vi of f r ame S2 wi th r espec t to S,, as in Fig. 10-3;

7 is as def ined in Eq. 10-13.

The magne t i c f i e ld in S2 is g iven by

x E ^ . (10-18)

B 2 | , = B[||. (10-19)

Eq uat ion s 10-16 to 10-19 are the equations of transformation for E and B.

* Electromag netic Fields and Waves, pages 262 and 288.




Thus, given a field E 1 ( B ( in frame Su one can calculate the field E 2 ,

B 2 in frame S2. The inverse transforma tion is obtai ned, as usual, by inter

cha ngin g the subscrip ts 1 an d 2, an d chan gin g the sign of v.

When v2

« c2

, y % 1 an d

E 2 = Ei + v x B,, (10-20)

1

B 2 = B , - ^ v x E , (10-21)

c

Note that the expression for the Lorentz force given in Eq. 10-2 remains

t rue, even if v x c.

10.8.1 EXAMPLE: THE HALL EFFECT

S e m i c o n d u c t o r s c o n t a i n e i t h e r one or b o t h of two types of mobi le cha rges , namely

c o n d u c t i o n e l e c t r o n s and holes (Sec. 5.1). W h e n a cur ren t f lows th rough a ba r ofs e m i c o n d u c t o r in the p r e s e n c e of a t ransverse magnet ic f ie ld B, as in Fig. 10-4, th e

mobile charges dri f t , not only in the di rec t ion of the appl ied e lect r ic f ie ld E but also

in a d i r e c ti o n p e r p e n d i c u l a r to b o t h the appl ied e lect r ic and magnet ic f ie lds , because

of the Qy x B force. This gives rise to a vol tage di fference V be tween th e u p p e r and

l ower e l ec t rodes . The drift is s imi l a r to the m o t i o n of t he ions in the m a s s - s p e c t r o

m e t e r of Sec. 10.1.1.



243

(c) (d)

p-type material n-type material

Figure 10-4 Hall effect in semiconductors. In p-type materials conduction is due

to the drift of positive charges (holes) and the Hall voltage is as shown in (a). In

H -type materials conduction is due to the conduction electrons and the Hall

voltage has the opposite polarity as in (b). Ordinary conductors such as copper

behave as in (b). Figure (c) shows the two opposing transverse forces QEy

and

Qv x B on a positive charge drifting along the axis of the bar at a velocity v.

Figure (d) shows how these forces are reversed in «-type material.

If the voltmeter V draws a negligible current, the plates charge up until their

field Ey is sufficient t o stop the tra nsverse drift. This transve rse electric field is called

the Hall field.

We shall first calculate the magnitude and direction of the Hall field Ey by using

the Lorentz force. Then, as an exercise, we shall find the field E 2 , B2

in the reference

frame of the charge carriers, to arrive again at Ey. We assume that the material is

zi-type as in Fig. 10-4b and 10-4d: the charge carr iers are negat ive.

a) The field Ey

must be such that Eq. 10-4 is satisfied. Then its magnitude is vB.

Since v x B is in the positive direction of the y-axis (Fig. 10-4d), Ey must point in

the negative direction and

vB. (10-22)

as in Figs. 10-4b and 10-4d.




b) In the l abora to ry f rame Sj , the electric field intensity is Ex\ + E j , and the

magnet ic induct ion is Bk . We have omi t ted th e subscr ip t s 1 for simplici ty .

To find the field in the reference frame S 2 of the mo vin g charges, we use Eqs. 10-20

a nd 10-21. For n- type mater ia l , th e velocity is — ri, wh e re v is a pos i t ive quan t i ty .

T h e n

E 2 = £. vi + £ r j - ri x Bk . (10-231

= £ v i + (£,. + vB)i, (10-24)

B 2 = B k + ( l / c 2 ) r i x ( £ , i + £ v j ) , (10-25)

= [B + (vEt c2

)]k. (10-26)

The force in frame 2 is QE 2, and since it has no v c o m p o n e n t , £ v = —iB as

prev ious ly .

If there are n e lec t rons per cub ic meter , each car ry ing a charge e.

I = abJ = ab(nev).

vBb = Bl/nea.

(10-27)

(10-28)

Note tha t charge car r ie rs of ei ther s ign are swep t down by the magnetic f ield .

The Hall effect is c o m m o n l y u s e d for m e a s u r i n g m a g n e t i c i n d u c t i o n s and for

var ious o ther pu rposes , some of which ar e descr ibed in P r o b s . 8-7, 10-15, and 17-12.

10.9 SUMMARY

A charge Q moving at a velocity y in superposed electric and magnetic

fields E an d B is su bm itt ed to the Lorentz force

F = Q(E + v x B), (10-2)

wher e F, E, v, and B are all mea sur ed w ith respect to the same reference

frame S,.

To find the field perceived by the charge Q in its ow n reference frame

S2 moving at the velocity ti with respect to St, one mu st use special relativity.



10.9 Summary 245

which is bas ed o n the Lorentz transformation:

x l = y(x 2 + M 2X (10-9)

V l = y 2, (10-10)

(10-11)

h= y[t 2 + ^ x 2 ) t (11-12)

wi th

y[1 - (v/cf\

T 1 / 2 '(10-13)

an d c equ al to the veloci ty of l ight in a va cu um , 2 .997 924 58 x 10 8 m e t e r s

p e r s e c o n d .

I t i s poss ib le to deduc e f rom th i s set o f equ a t io ns o the r t r ans fo rm at ion

equat ions for a mass, a veloci ty , e tc . A given elect r ic charge Q and the veloci ty

of light c a lw ays ha ve the sam e value, wha teve r the veloci ty of the reference

f r ame wi th r espec t to which they a r e measured .

F or a given f ie ld E ^ I* ! in frame S,, the f ield in S 2, moving a t a ve loc i ty

v wi th res pect t o S l, is given by

E 2 ± = y(E1±

+ v x B,) , (10-16)

E 2 N = E j | | , (10-17)

B 2 1 = ? ( b U - ^ X E A (10-18)

Ban = Bj ||. (10-19)

If r 2 « c 2, t h e n y = 1 and

E 2 = E , + v x B , , (10-20)

B 2 = B i - - T y x E 1 . (10-21)




PROBLEMS

I0-1E THE CYCLOTRON FREQUENCYA par t i c l e o f mass in. c h a r g e Q. and veloci ty v. describes a c i rc le of radius R in a

plan e perpe nd icula r to the di rec t ion of a uniform ma gne t ic field B.

a ) Show tha t

BQt = mv 2 R.

b) Show tha t the an gu lar veloci ty

co = BQ/m.

The frequency BQ 2nm is called the cyclotron frequency be cau se it is the frequen cy at

which an ion c i rcula tes in a cyclotron.

N o t e h o w a) is independent of the velocity r of the part ic le . This is not s t r ic t ly

t rue , however , because m is i tself a function of the velocity:

m 0

m

~ [1 - ( r 2 / r 2 ) ] " 2 '

w h e r e c i s the veloci ty of l ight in a vacuum and m 0 is the rest mass. In pract ice . a> m ay

be cons idered to be independent of c . for c2

« c2

. T h e m a s s m i s 10 percen t large r th an

m 0 w h e n v i s ab ou t 0.4c .

The above three equat ions are val id a t a l l veloci t ies .

c) Calculate the cyclotron frequency for a deuteron in a f ie ld of one tes la . for

r2

« c2

.

I0-2E MOTION OF A CHARGED PARTICLE IN A UNIFORM B

A charged part ic le moves in a region where E = 0 and B i s uniform.

Show that the part ic le describes e i ther a c i rc le or a hel ix.

10-3E MAGNETIC MIRRORS

Figu re 10-5a show s a cross-sect io n of a solen oid hav ing a unifo rm t urn dens i ty,

except nea r the ends where ex t ra tu rns a re added to ob ta in a h ighe r magne t i c induc t ion

than nea r the cen te r .

Show q ual i ta t iv ely th at , if the axia l veloci ty is not to o large, a charge d part ic le

that spi ra ls around the axis in a vacuum ins ide the solenoid wi l l be ref lected back when

i t reaches the higher magnet ic f ie ld. The regions of higher magnet ic f ie ld are cal ledmagnetic mirrors.

Magne t i c mi r rors a re used for conf in ing h igh- t empera ture p l a smas . (A plasma

i s a highly ionized gas . I t s net charg e dens i ty is ap pro xim ate ly zero.)

In the Van Allen radiation hells, cha rged pa r t i c l e s a re t rapped in the ea r th ' s

ma gn et ic f ie ld, osci l la t ing no rth an d sou th a lon g the ma gn et ic f ie ld l ines . Th ey a re

re f l ec t ed nea r the Nor th and South magne t i c po les , where the l i nes c rowd toge the r ,

forming magnet ic mirrors , as in Fig. 10-5b.



24 7

Solar wind

Figure 10-5 (a) Cross -s ec t ion of a so lenoid wi th magne t i c mi r rors a t t he ends .

{b ) Gray l ines : magnet ic f ie ld of the earth. The earth acts as a magnet ic dipole .

Black l ines : magnet ic f ie ld near the earth, as observed by means of sa te l l i tes . The

so la r w ind* is com pos ed of p ro ton s and e l ec t rons evapora ted f rom the surface of

the sun. Since the interplanetary pressure is low, there are few col l is ions and the

conduc t iv i ty i s h igh . In moving th rough the magne t i c f i e ld o f t he ea r th , t he cha rge

part ic les are subjected to a v x B elect r ic f ie ld, and curre nts f low acc ord ing to

Lenz 's law. The black l ines show the net f ie ld.

10-4E HIGH-ENERGY ELECTRONS IN THE CRAB NEBULA

Som e as t rophys ic i s t s be l i eve tha t t he re ex i s ts , w i th in the Cra b Neb ula , e l ec t rons

with energies of ab ou t 2 x 1 01 4

elect ro n-v ol ts spi ra l l ing in a ma gne t ic field of 2 x 10 ~8

tes la . See Prob. 10-1.

a ) Ca lcu la t e the ene rgy W of such an e lect ron in joules .

Electromagnetic Fields and Waves, p. 502.



248 Mag netic Fields: III

b) Ca lcu late i t s ma ss from the re la t io n

W= H i e

2

.

w h e r e c is the velocity of l ight in a vacuum, 3 x 1 0 8 mete rs pe r s econd .

How does th i s mass compare wi th tha t o f a low-ve loc i ty (c 2 « c 2 ) e lec t ron ,

which is9.1 x 1 0 " 3 1 k i l o g r a m ?

c) Calculate the radius of i t s orbi t , se t t ing v = c and neglect ing the veloci ty

component pa ra l l e l t o B .

d) C a lcu la t e the numb er of days requi red to comple te one tu rn .

10-5E MAGNETIC FOCUSING

There exis t many devices that ut i l ize f ine beams of charged part ic les . The cathode-

ray tube tha t i s used in te levis ion receivers an d in osci l loscop es is the be s t -k now n

exam ple . The e l ec t ron mic ros cope i s ano the r exam ple . In these dev ices the pa r t i c l ebea m is focused an d deflected in mu ch the same way as a l ight bea m in an opt ical in

s t r u m e n t .

Beam s of char ged part ic les can be focused and deflected by prop erly shap ed

electric f ields . See, for exam ple , P rob s . 2-11 and 2-12. Th e e lect ron beam in a TV t ub e

is focused with electric fields and deflected with magnetic fields. Beams can be deflected

along a c i rcular path in a magnetic f ie ld as in P ro b. 10-1. Let us see how the y can be

focused by a magnet ic f ie ld.

Figu re 10-6 sho ws an e lect ro n gun s i tuat ed ins ide a long solen oid. The e le ct ro ns

tha t emerge f rom the ho le in the anode have a sma l l t r ansve rse ve loc i ty component

and, i f there is no current in the solenoid, they spread out as in the f igure . Let us see

wha t happens when we tu rn on the magne t i c f i e ld .

Figure 10-6 Foc usin g of an e lect ron be am in a uniform ma gn et ic f ield: F is a

heated filament, A i s the anode, and the e lect ron beam is focused a t P . The

accele rat ing vo l tage is F . A f ilament sup ply of a few vol ts is con nec ted betwe en

the top two t e rmina l s . Wh en the so lenoid is no t ene rg ized , t he beam d ive rges a s

shown by the dashed l ines . See Prob. 10-5.



Problems 249

Let the ve loc i ty component s a t t he ho le be v x a n d c v . with r 2 » v 2. T h e n

1 1

,eV = - m ( r

2

+ 17) % - mo*.

If v y = 0. the e lect ron cont inues paral le l to the axis . I f v y =F 0, i t follows a helical

path a t an ang ula r veloci ty cu as in Pro b. 10 -1. After one com plete tu rn, i t has re tu rned

to the axis. So. if B i s adjus ted correc t ly, a ll the e lect ro ns wi l l con verg e a t the same poin t

P, as in the f igure , and form an image of the hole in the anode.

a) Show that this wi l l occur i f

L

b) Ca lcu la t e the number of ampere tu rns pe r me te r IN ' in the solenoid, for an

acce le ra t ing vo l t age V of 10 ki lovo l ts and a dis tan ce L of 0.5 meter.

In actual pra ct ice , ma gn et ic focus ing is achieve d with sho rt coi ls, an d not wi th

so lenoids .

I0-6E DEMPSTER MASS-SPECTROMETER

Figure 2-10 shows an e lect ros ta t ic veloci ty analyser . I t was shown in Prob. 2-11

that a part ic le of charge Q. mass m, and veloci ty v pass ing through the f i rs t s l i t wi th the

proper o r i en ta t ion wi l l r each the de tec tor i f

v = QER/m.

Figure 10-7 shows a Dempster mass-spectrometer. Here again, the ion describes

a c i rc le , except that the centr ipeta l force is now Qy x B . ins tead of QE . T h e m a g n e t i c

f ie ld is pro vid ed by an e lect r om agn et .

a ) Show tha t

m = QR2

B2

/2V.

b) In one pa r t i cu la r expe r iment , a mass -spec t romete r o f t h i s l a t t e r t ype was

used for the hyd rogen ions H^.H^.Hi, wi th /? = 60 .0 mi l l ime te rs and V = 1000 vol ts .

Figure 10-7 Cross -s ec t ion of a mass

spec t romete r o f t he Demps te r t ype .

Ions produced a t t he sourse S a re

col lected a t C. The ions are accel

erated through a di fference of poten

tial V and describe a semicirc le of

r a d i u s R. See Prob. 10-6.




T h e W ,+ i on i s a p ro ton . H% i s com pose d of two pro t ons and one e l ec tron , and H^ is

com posed of th ree pro to ns and two e l ec t rons .

Find the values of B for these three ions .

c ) Draw a curve of B as a function of in under the above condi t ions , f rom

h y d r o g e n t o u r a n i u m .

Wo uld you use th is type of spec t ro mete r for heavy ions? W hy ?

10 -7 MASS SPECTROMETER

Figure 10-8 show s a mass spec t rom ete r tha t s epa ra t e s ions , bo th a ccord in g to

the i r ve loc ie s and accord ing to the i r cha rge - to-mass ra t ios .

Sho w that an ion of cha rge Q, m a s s in. and veloci ty v i s col lected a t the poin t

Figure 10-8 Mass spectrometer . Ions injected a t A fol low hel ical paths in the E a n d

B f ie lds that are both paral le l to the y-axis . Each square is a separate col lector .

See Prob. 10-7.

I0-8E HIGH-TEMPERATURE PLASMAS

One method of in j ec t ing and t rapp ing ions in a h igh- t empera ture p l a sma i s

i l l us t ra t ed in F ig . 10-9 . Mo lecu la r i ons of deu te r ium , D j ( two deu te ron s p lus one

e lec t ron , t he deu te ron b e ing compo sed of one pro ton a nd one neu t ron) , a re in jec ted

into a magnet ic f ie ld and are dissocia ted into pairs of D + i ons (deu te rons ) and e l ec t rons



25 1

D +

in a high -inten s i ty arc . Th e radius of the t ra jectory is redu ced an d the ions are t r ap ped .See P rob . 10-1 .

a) Cal cul ate the rad ius of cu rva tur e P for DJ ions having a kinet ic energy of

600 k i loe lec t ron-vol t s i n a B of 1.00 tesla.

b ) Ca lcu la t e R for the D + ions prod uce d in the arc . Th e D + ions have one half

the kinet ic energy of the D 2 i ons .

10-9 HIGH-TEMPERATURE PLASMAS

The type of d i s cha rg e shown in F ig . 10-10 has been used to p ro duce h igh- t em

perature plasmas . The discharge has the shape of a cyl indrical shel l and is s i tuated in

the space be tween two conduc t ing coaxia l cy l inde rs tha t ac t a s re tu rn pa ths fo r t he

c u r r e n t .

a) Is there a magnet ic f ie ld outs ide the outer cyl inder? Ins ide the inner cyl inder?W h y ?

b) Draw a f igure showing l ines of B .

c ) Cons ide r the upper pa r t o f t he d i s cha rge . Suppose a pos i t ive ion , moving

tow ard the r ight , has an upw ard c om po ne nt of veloci ty. Ho w is i t s t ra jectory affected

by the magnet ic f ie ld af ter i t has emerged from the discharge?

d) Wha t happens to a pos i t ive ion tha t t ends to l eave the upper pa r t o f t he

d i s c h a r g e b y m o v i n g d o w n w a r d ?

G p

Figure 10-10 Sec t ion th rough a dev ice

that has been ut i l ized for producing a

h i g h - t e m p e r a t u r e p l a s m a P . T h e

copper enc losure , r epresen ted by

heavy l ines , i s sepa rated from the

plasma by a pair of glass cyl inders G.

See Prob. 10-9.




10-lOE ION-BEAM DIVERGENCE

Let us calcula te the forces act ing on a part ic le in a beam of pos i t ively charged

part ic les having a veloci ty v a n d c a r r y i n g c h a r g e s Q. We as sume tha t t he re a re no

external ly appl ied e lect r ic or magnet ic f ie lds . We shal l cons ider pos i t ive part ic les , but

i t wi l l be a s imple mat ter to apply our resul ts to negat ive ones .

Firs t , there is an o utw ard E, as in Fig. 10-11. This is s imply a case of e lect ros ta t ic

repul s ion . A l so , t he beam cur ren t p roduces an az imutha l B . an d v x B poin t s inward ,

as in the figure.

Thus , there is an outward e lect r ic force QE and an inward magne t i c fo rce Q\ x

B. Does the beam converge or d ive rge? Exper imenta l ly , e l ec t ron and ion beams a lways

diverge, when left to themse lves , but the diverg ence is s l ight when th e part ic le veloci ty

r approaches the ve loc i ty o f l i gh t c.We consider a part ic le s i tuated a t the edge of a beam of radius R. T h e c u r r e n t

is / and the part ic le veloci ty is v.

C a l c u l a t e

a) the charge / . per mete r of bea m.

b | the outward e lect r ic force QE.

c) the inward magnet ic force QvB.

d) t he net force.

You shou ld f ind that the net force points outw ard a nd is pro po rt io na l to 1 —

e 0 / ( 0 r2 , or to 1 - (v/c)

2 (see Eq. 20-24). Th en the net force tend s to zero as r - » c.

If the part ic les are negat ive . E is inward ins tead of outward, but QE is again

o u t w a r d . A l s o . B poin t s in the oppos i t e d i rec t ion and Q\ x B poin t s aga in inward .

In pract ice , a vacu um is neve r perfect . Let u s say the ions are pos i t ive . I f thei r

energy is of the order of tens of e lect ron-vol ts or more, they ionize the res idual gas .

forming low-energy pos i t ive ions and low-energy e lect rons . These pos i t ive ions dri f t

Figure 10-11 Po rt io n of a pos i t ive-ion be am an d i ts e lect r ic and m agn et ic f ie lds .

See Prob. 10-10.

e ) E lec t rons move toward the l e f t . Wha t happens to them when they l eave the

d i s c h a r g e , b y m o v i n g e i t h e r u p w a r d o r d o w n w a r d ?



Problems 253

away f rom the pos i t ive beam. The low-ene rgy e l ec t rons , however , r ema in t rapped in

the beam and neu t ra l i ze pa r t o f i t s space cha rge , t he reby reduc ing E. T h e m a g n e t i c

force then t ends to p inch the beam . Thi s ph eno me non is ca l led the pinch effect. It is

a lso cal led gas focusing. With gas focus ing, some of part ic les in the beam are scat tered

away, in col l iding with the gas molecules .

10-11 ION THRUSTER

Figure 10-12 shows an ion thrus ter that ut i l izes the magnet ic force . Other types

of thru s ter ar e described in Pro bs . 2-14, 2-15. and 5-3. An arc A ionizes the gas , which

enters from the lef t , and the ions are blown into the crossed e lect r ic and magnet ic f ie lds .

I f t he cur ren t f lowing be tween the e l ec t rodes C a n d D i s / . then the thrus t F is Bis.

which is the force exerted on the gas ions . We assume that B an d s a r e u n i f o r m t h r o u g h

out the thrus t chamber, and we neglect the fr inging f ie lds .

If H I ' is the mass of gas flowing per unit t ime, and if v i s the exhaus t veloci ty wi th

respect to the vehicle , then F is m'v, and the kinetic power communica ted to the gas i s

1 , 1 1P = - H i ' r

2 = -Fv = - Blsr.2 2 2

If the efficiency is defined as

PG + P D

w h e r e P D i s the pow er diss ipate d as heat be tween C and D. show tha t

1 1 1~ 2 H I ~ ~ 2~T ~ IE'

1 + r ~ 1 + 1 + —aB-x (TBV Br

Figure 10-12 Ion thrus ter . See Prob. 10-11.



254 Mag netic Fields: III

10-12E GAMMA

For wha t va lue of v is the value of y one pe rcen t l a rge r than un i ty?

10-13E REFERENCE FRAMES

Co nsid er tw o reference frames , 1 and 2, as in Fig. 10-2, wh ere frame 2 has a

veloci ty v = c/2 with respect to 1.

An event is found to occur at .v, = \\ = r , = 1 me te r . t1 = 1 s econd .

F ind the coord ina te s .v 2, y 2 . - 2 . h-

10-14 REFERENCE FRAMES

See the preced ing prob lem.

If the event occurs a t x 2 = y 2 = r , = 1 me te r , t2 = 1 s econd .

F ind v , , V i , z u tx.

10-15 HALL EFFECT

Let us inves t igate more c losely the Hal l effect described in Sec. 10.8.1. W e a s s u m e

again th at the curre nt i s carr ie d by e lect rons of cha rge —e. Their effective mass is m* .

T h e effective mass takes into account the periodic forces exerted on the e lect rons , as

they t ravel through the crys ta l la t t ice . As a rule the effect ive mass is smaller t han the

mass of an i so la t ed e l ec t ron .

The e lect rons are subjected to a force

F = -e(E + v x B ) .

where E has two component s , t he appl i ed f i e ld E x and the Hall field E y. T h e a v e r a g e

drift velocity is

Fv =

e

wh ere . / / i s thei r mo bi l i ty. T he m obi l i t y of a part ic le in a med ium is i t s aver age drif t

velocity , divided by th e elec tric field ap plie d to it . Th e law F = m*a appl i e s on ly be tween

col l is ions wi th the crys ta l la t t ice .

a ) Show tha t

v x = -J/(E X + v yB) .

v, = -.M{E y - v xB) ,

r = = 0.

b) If i! is t he num ber of con duc t ion e l ec t ron s pe r cub ic me te r , t he cur ren t dens i ty

is

J = —nev.



Problems 255

Show tha t

E x - //E.Ji

J x = ne .M — T A r ,

E, + M E XBJ,. = nejf— i—

1 + .//'ti

ll J y = 0, as in Sec. 10.8.1,

.//E XB,

bV = -J/VJ3.

a

Note tha t t he Ha l l vo l t age V y i s p ro por t ion a l t o the product of the appl ied vol tage

V x a n d B. This fact makes the Hal l effect useful for mul t iplying one variable by another .

See Prob. 17-12.

When connected in this way, the Hal l e lement has four terminals and is cal led

a Hall generator, o r a Hall probe.

As a rule , rc- type materia ls are used because of thei r re la t ively high mobi l i ty.

However, the Hall effect exists in al l c o n d u c t i n g b o d i e s .

On e imp or t an t ad van tage of Ha ll e l ement s is t ha t t hey can be mad e sma l l. Fo r

ce r t a in appl i ca t ions they a re incorpora t ed in to in t egra t ed c i rcu i t s .

c) Calculate F v fo r b = 1 mil l im eter , a = 5 mil l ime te rs . .// = 1 mete rs squa red

per vo l t s econd ( ind ium an t imonide ) , V x = 1 volt , B = 1 0 - 4 tesla.

d ) Show tha t , i f £ , = 0 ,

AP , ,— = .//2B 2,Pn

0, an d A P is the

increase in res is tance up on a pp l icat io n of the mag net i c f ie ld.

Fhe Hall field E y i s ma de equa l t o ze ro by ma king c smal l , say a few micro

me te rs , and p la t ing conduc t ing s t r ips pa ra l l e l t o thev ax i s , a s in F ig . 10-13. The e l ement

then ha s only two term ina ls an d is cal led a magnetoresistor.

Magne tores i s to rs a re used for measur ing magne t i c induc t ions . See P robs . 15-3

and 15-5 .

Figure 10-13 Magne tores i s to r . See

Prob . 10-15 .




MAGNETOHYDRODYNAMIC GENERATOR

Figure 10-14 shows schemat ical ly the principle of operat ion of a magnetohyclrody-namic, o r MHD generator. Th e func t ion of an M H D gen era to r is t o t rans form the

kinet ic energy of a hot gas di rect ly into e lect r ic energy.

A very hot gas is injected on the left at a high velocity v. Th e gas i s ma de con

duct ing by inject ing a sa l t such as K 2 C 0 3 t ha t i on izes read i ly a t h igh t empera ture ,

forming pos i t ive ions and free e lect rons . Conduct ivi t ies of the order of 100 S i e m e n s

per me te r a re thus ach ieved , at t emp era tur es o f abo ut 3000 ke lv ins . (The con duc t iv i ty

of copper is 5.8 x 1 0 7

S i emen s per me te r ) . Of course r2

« c 2.

The ions and the e lect rons are deflected in the magnet ic field. With the B s h o w n ,

the pos i t ive ions a re de f l ec t ed downward and the e l ec t rons upward . The re su l t ing

curr ent f lows thr ou gh a load res is tance R. giving a vol tage di fference V. and hence

an electric field E be tween the p l a t e s. On e such M H D gen era to r is p l ann ed to have

a power ou tpu t i n the 500-megawat t r ange . I t would use a B of several teslas over aregion 20 me ters long and 3 me ters in dia met er .

F igure 10-14 Schemat i c d i agram of a magne tohydrodynamic gene ra tor . The

kinetic energy of a very hot gas. injected on the left at a velocity v, is transformed

direct ly into e lect r ic energy. The magnet ic f ie ld B i s that of a pai r of coi ls , outs ide

the chamb er . See P ro b . 9 -8. Th e movin g ions a re de f lec t ed e i the r up or do wn,

depending on the i r s ign . See P rob . 10-16 .

a) Let us sup po se that E, B . and the veloci ty v a re un i form ins ide the chamber .

These a re ra the r c rude a s sumpt ions . In pa r t i cu la r , v i s not uniform because (a) the

ions and e lect ro ns have a vert ical com po ne nt of veloci ty, an d (b) the vert ical veloci t ies

give Q\ x B forces tha t po in t backward , fo r bo th types of pa r t i c l e ; t hese brak ing

forces s low do wn th e horizo ntal dr i f t of the charge d part ic le s . I t i s thr ou gh this

dece lerat io n that pa rt of the ini t ia l kine t ic ener gy of the gas is t ran sfor me d into

elect r ic energy.



Problems 257

The elect rodes each have a surface area A and a re s epa ra t ed by a d i s t ance b.

S h o w t h a t

V 'I =

R + R,

where R, is the internal res is tance of the generator , and V is the value of V w h e n

R i s inf ini te . Thevenin 's theorem (Sec. 5.12) therefore appl ies here .

Note tha t , because v 2 « c 2, the curre nt den s i ty is the sam e in the reference frame

of the laboratory as i t i s in the reference frame of the moving gas .

Solution : W e hav e two reference frames , tha t of the lab ora tor y and that of the

moving gas .

In the reference frame of the gas,

E 2 = E + v x B. (1)

We omit the subsc ript 1 for the quan t i t ie s measu red in the reference frame of the

l abora tory , fo r s impl i c i ty . Thus

J 2 = ffE 2 = a(E + v x B), (2)

j = j 2 = a(vB - E) = a (vB - H , (3 )

I = JA = a A \vB - — , (4)

(5)

b

vB b V

an d

rtAR b R + R,

b a A

V = vBb. R t = b:aA. (6)

N o t e t h a t V is |v x B|/?. and that R i i s the res is tance of a conductor of conduct ivi ty

CT, length h. and c ros s - s ec t ion A.

b) Find an express ion for the eff ic iency

^ pow er diss ipa ted in R

t o t a l power d i s s ipa ted '

as a funct ion of / . that does not involve the res is tance R.

25 8

I oAiB I (JAIB

Figure 10-15 (a) Effic iency as a funct ion of load cur ren t for the m ag ne to hy dr o-

dy na mi c gene rato r of Fig. 10-14. (£>) O ut pu t v ol tage as a funct ion of load curr ent .

W ha t is the efficiency at / = 0?

W ha t is the value of the cur rent when the eff ic iency is zer o?Sketch a curve of S as a function of / .

Solution:

I2

R R

(8)r-R + r-R, R + R,

R <TAR

R + (b/aA) oAR + b

Since

oAvBb

oAR + b"

(9)

(10)

oAR _ RI

aAvBb ~ v~Bb'

V - Rjl vBb - (b/aA)I I

vBb vBb oAvB(12)

Th e effic iency is equ al to uni ty when / = 0.

Th e eff iciency is zer o wh en I = aAvB. In that case R = 0. f rom Eq. 4 . t he ou t pu tvo l t age V i s zero, an d / = V'/Ri, from Eq. 5.

Fig ure 10-15a sho ws & as a funct ion of / .

c) Find an express ion for V as a funct ion of / that does not involve R.

W ha t is the value of V when the cur ren t i s ze ro?

W ha t is the value of the curren t when th e vol tage is zero ?

Sketch a curve of V as a function of / .



Problems 259

Solution:

V = IR= vBb - 7 - vB b ( 1 — ).

W he n th e curr ent i s zero (R -> oo) , the output vol tage is vBb. The ions then

f low through the gene ra tor hor i zonta l ly , unde f l ec t ed . See Sec . 10.1.1.

When the vo l t age i s ze ro , R = 0 and / i s equ al to aAvB as above .

F igure 10-15b shows V as a funct ion of I.

10-I7E ELECTROMAGNETIC FLOWMETERS

Electromagnetic flowmeters operate as fol lows. A f luid, which must be a t leas t

s l igh t ly con duc t ing (b lood , fo r example ) , f lows in a nonco ndu c t ing tube be tween the

poles of a mag ne t . The ions in the f luid are then sub jected t o mag net i c forces Q\ x B

i n the d i rec t ion pe rp endic u la r t o th e ve loc i ty v of the f luid and to the magnet ic induct ion

B. Electrodes placed on e i ther s ide of the tube and in contact wi th the f luid thus acquire

charg es of op po s i te s ign s ; the resul t ing vo l tage di fference is a me asu re of v.

No te that , in the e lect ro ma gn et ic f lowmeter, ions of both s igns hav e the sam e

veloci ty v and the pola r i ty of the e lect ro des is a lway s the sam e for a given direct io n

of flow and for a given direction of B. Compare wi th the Hal l effect discussed in Sec.

10.8.1.

Faraday a t t empted to measure the ve loc i ty o f t he Fhames r ive r in th i s way in

1832. Th e mag ne t ic fie ld was of cour se that of the earth .

Consider an ideal ized case where the tube has a rectangular cross-sect ion af t ,

with side b paral le l to B , and w here the ve loc i ty is t he s ame th ro ugh ou t the c ros s - s ec t ion .

W ha t i s t he vo l t age be tween the e l ec t rod es?

In fact , the veloci ty is m ax im um on th e axis an d zero a t the inner surface of

the tube . Moreover , t he non-uni form v x B f ie ld causes currents to f low within the

fluid. I t tur ns out , curious ly e nou gh, that , for a tub e of c i rcula r cross-sect ion a nd rad ius

a, t he vo l t age be tween d iamet r i ca l ly oppos i t e e l ec t rodes i s 2avB, w h e r e v i s the average

veloci ty, which is jus t w ha t one wo uld exp ect if the veloci ty were unifo rm.



CHAPTER 11

MAGNETIC FIELDS: IV

The Faraday Induction Law

111 THE INTEGRAL OF E • dl

11.1.1 Example: The Expanding Loop11.2 THE FA RADA Y INDUCTION LA W

11.2.1 Example: Loop Rotating in a Magnetic Field

11.3 LENTS LAW

11.4 THE FARADA Y INDUCTION LA W IN

DIFFERENTIAL FORM

11.5 THE ELECTRIC FIELD INTENSITY E IN TERMS

OF V AND A

11.5.1 Example: The Electromotance Induced in a Loop by a Pair

of Long Parallel Wires Carrying a Variable Current

11.6 SUMMARY

PROBLEMS



In th is ch ap ter w e shal l con side r the l ine integ ral of E • d l ar ou nd a closed

c i r cu i t . Th i s wi l l l ead us to s t i l l ano ther o f Maxwel l ' s equa t ions .

In Ch ap te r 2 we saw that , in e lect ro stat i c f ie lds, the l ine integ ral of

E • d l a ro un d a c losed c i r cu i t C i s zero. This i s no t a general rule . I f the

ci rcui t C, or par t of i t, m ov es in a m ag ne t ic f ie ld , the n on e mus t take int o

a c c o u n t t h e v x B te rm of Ch ap te r 10 . Th i s wi l l g ive us the Far ad ay induc

t ion l aw, acco rd ing to which th e abov e in t egra l is equa l to mi nus the t im e

der iva t ive o f the enc losed magne t i c f lux .

What i f , in a given reference f rame, we have a changing magnet ic

field? Th en ele ctr ic f ield inte ns ity is eq ua l, no t to — W as in Chap te r 2 , bu tr a ther to — V I 7 - cA/dt, where A is the vector potent ia l of Sec. 8 .4 , and

F a r a d a y ' s i n d u c t i o n l a w a g a i n a p p l i e s .

We have seen in Sec. 2 .5 that an elect rostat ic f ie ld is conservat ive, or that

Th us the work per fo rme d b y an e l ec t ros t a t i c fi eld i s ze ro when a charg e

m o v e s a r o u n d a c l o s e d p a t h .

However , t he re a r e many cases where one has a c losed , conduc t ing

circui t , wi thout sources, wi th par t , or a l l of the ci rcui t moving in a magnet ic

f i e ld . The conduc t ion e l ec t rons in the moving conduc to r s then f ee l an e l ec t r i c

THE INTEGRAL OF E • dl

(11-1)



262 Magnetic Fields: IV

field v x B, as in Sec. 10.1. In such cases, the ab ov e integral, eva luate d over

the circuit, is not zero.

The v x B field in a conductor moving in a magnetic field is calledthe induced electric field intensity and the integral of E • dl around a closed

circuit is called th e induced electromotance.

EXAMPLE: THE EXPANDING LOOP

The expand ing loop of Fig. 11-1 has one side that ca n sl ide to the right at a veloci ty

v in a region of uniform B. Both v and B are m e a s u re d in the l abora to ry re ference

frame Sx. In the moving wi re , there is an electric field E , = v x B . The l ine in tegral

of E • dl a r o u n d th e circui t is thus eva lua ted par t ly in St and par t ly in S2- In the

coun ter -c lockwise d i rec t ion .

= 0 + f (v x B) • dl,

Jda

= vwB.

(11-2)

(11-3)

(11-4)

T h i s is the i n d u c e d e l e c t ro m o t a n c e . It acts in the coun ter -c lockwise d i rec t ion in

the figure. If, at a certain instant , the l o o p has a res i s tance R , t h e n th e c u r r e n t is

LWB/R.

We have assumed tha t R is large so as to m a k e th e magnetic f ield of the cu rren tf lowing around th e loop neg l ig ib le compared to B .

Figure 11-1 A conduct ing wi re ad sl ides at a veloci ty c on a pai r of c o n d u c t i n g

rails in a reg ion of un i fo rm magnet ic induct ion B . The magnet ic fo rce on the

e lec t rons in the wire p roduces a c u r r e n t / in the circui t .

11.2 The Faraday Induction Law 263

11.2 THE FARAD A Y INDUCTION LA W

Th e r igh t-h an d side of Eq. 11-4 is the are a swept by the wire per uni t t ime ,mu l t ipl ied by B, or d8>/dt, where O is the magnet ic f lux l inking the ci rcui t .

N o w it is the cu sto m to ch oo se th e posi t iv e di rec t ion for <t> relat ive to the

pos i t ive d i r ec t ion fo r the l ine in t egra l , accord ing to the r igh t -hand sc rew

ru le . Then , s ince the in t egra l has been eva lua ted in the coun te r - c lockwise

d i r ec t ion whi l e O po in t s in to the paper ,

This i s the Faraday induction law. t he e l ec t romotance induced in a c i r cu i t

i s equ al to min us th e rate of ch an ge of the ma gn et i c flux l inking the ci rcui t .

The path of in tegrat ion may be chosen at wi l l and need not l ie in

c o n d u c t i n g m a t e r i a l .

I f the re are no sou rce s in a c i rcui t , the curre nt i s equ al to the indu ced

elect romotance divided by the resistance of the ci rcui t , exact ly as i f the

e lec t romotance were r ep laced by a ba t t e ry o f the same vo l t age and po la r i ty .

Th e induced e l ec t ro mo tanc e i s expressed in vo lt s , and i t add s a lge bra ica l ly

to the vo l t ages o f the o ther sources tha t may be p resen t in the c i r cu i t .

If on e has a r ig id loo p and a t im e-d ep en de nt B. Eq. 11-5 st il l appl ies .

If one has a r igid coil and a t ime -dep end en t B, then one must add the

f luxes thr ou gh e ach tu rn. The su m of these f luxes is cal led the flux linkage.

We shal l use the symbol A for f lux l inkage, keeping O for the sur face integral

of B • da ove r a s im ple loo p. Fo r exam ple , if a coi l has N t u rns wi th a l l t he

tur ns close tog eth er a s in Fig . 11-2, A is N t imes the <D t h rough one o f the

t u r n s :

T h u s , fo r c i r cu i t s o ther than s imple loops , Faraday ' s induc t ion l aw

b e c o m e s

( H - 5 )

A = /Vd). (11-6)

(11-7)

F lux l inkage i s expressed in we ber - tu rn s .

26 4

Figure 11-2 Coil wi th several turns , all c lose toge the r .

Equation 11-7 should be used with care. It is not as general as it is

co mm on ly said to be. It is corr ect (a) if the circuit is rigid and the ma gn eti c

ind uc tio n B is tim e-d epe nde nt, or (b) if all or pa rt of the circuit m ove s in

such a way as to produce an induced electric field intensity v x B. See

Prob . 11-4. As we sha ll see in Sec. 11.5. the E in the first case is — cA/dt.

EXAMPLE: LOOP ROTATING IN A MAGNETIC FIELD

F i g u r e 11-3 s h o w s a l o o p r o t a t i n g in a c o n s t a n t and uniform magnet ic f ie ld at an

angula r ve loc i ty to. We shal l calcula te th e i nduced e l ec t romotance us ing th e v x B

field in th e wire , and then verify th e F a r a d a y i n d u c t i o n law. This wi l l i l lus t ra te th e

pr inc ip le of o p e r a t i o n of an electric generator.

T h e i n d u c e d e l e c t r o m o t a n c e is

w h e r e the integral is e v a l u a t e d a r o u n d th e l o o p .

A l o n g the top and bot tom s ides , v x B is p e r p e n d i c u l a r to dl and the t e rm

u n d e r th e i n t egra l is zero. Set t ing 0 = o>t. and r e m e m b e r i n g t h a t v = w(a!2). th e

e l e c t r o m o t a n c e i n d u c e d in the vert ical s ides is

T h i s e l e c t r o m o t a n c e is c o u n t e r c l o c k w i s e in the figure. It is pos i t ive because it is

in the pos i t ive d i rec t ion a round th e l oop , w i th re spec t to the di rec t ion of B, ac

c o r d i n g to the r igh t -hand s c rew ru le .

111-8)

/ = 2aj(a/2)(sin (ot)bB — ouibB sin cot. (11-9)

N o w

<> = abB cos 0 = abB cos tot (11-10)



26 5

Kigure 11 - 3 L o o p r o t a t i n g in a c o n s t a n t an d uniform magnet ic f ie ld B . The vec tor

ri! is n o r m a l to the l o o p .

an d

" ~dt' (11-11)

as predicted by the F a r a d a y i n d u c t i o n law.

T h e e l e c t r o m o t a n c e is ze ro when the p l a n e of the l o o p is p e r p e n d i c u l a r to B.

since v a l o n g th e vert ical s ides is then paral le l to B, and v x B is ze ro all a r o u n d

the loop .

N o t e t h a t th e vol t age 1 of E q. 11-9 is a s inusoidal funct ion of (. Such vo l t ages

are sa id to be alternating. C h a p t e r s 16 to 18 ar e d e v o t e d to a l t e rna t ing vo l t ages and

c u r r e n t s .

11.3 LENZ'S LA W

If t h e f l u x l i n k a g e A i n c r e a s e s , dA/dt is p o s i t i v e , a n d t h e e l e c t r o m o t a n c e is

n e g a t i v e , t h a t i s , the i n d u c e d e l e c t r o m o t a n c e is in the n e g a t i v e d i r e c t i o n .

O n t h e o t h e r h a n d , if A d e c r e a s e s . dA/dt is n e g a t i v e , a n d t h e e l e c t r o m o t a n c e

is in the p o s i t i v e d i r e c t i o n .




11.4 THE FARAD A Y INDUCTION LA W

IN DIFFERENTIAL FORM

The Fa ra da y law stat ed in Eqs. 11-5 an d 11-7 gives the ele ctr omo tan ce

ind uce d in a compl ete circu it when the flux lin kage is a function of the time.

It applies to both fixed and deformable paths in both constant and time-

dependent magnetic fields, with the restriction stated at the end of Sec. 11.2.

We shall now find an important equation that concerns the value of

the E induced at a given point by a time-dependent B.

If, in a given reference frame, we have a ti me- depe nden t B, then we

can state the Fa ra da y in duct ion law in differential form as follows. Usi ng

Stokes's theorem, Eq. 11-5 becomes

J s ( V x E , . d a = - ^ = - i j s . B . d a , ,11-12,

where S is any surface b ou nd ed by the int egra tio n p ath. If the pa th is fixed

in space, we may int erch ang e the ord er of differentiation an d inte grat ion

on the right-hand side and

J > x E ) d a = - J s ^ d a . (11-13)

We have used the partial derivative of B because we now require the rate

of change of B with time at a fixed point.

Since the above equa tio n is valid for arb itr ary surfaces, the in teg ran ds

must be equal at every point, and

Th e direct ion of the indu ced curr ent is alwa ys such that it prod uce s

a magnetic field that opposes, to a greater or lesser extent, the change in

flux, depending on the resistance in the circuit. Thus, if A increases, the

induc ed cur rent pr odu ces an oppo sing flux. If A decrea ses, the in duce d

cur rent pro duc es an aiding flux. This is Lenz' s law.

The currents induced by changing magnetic fields in conductors other

than wires are called eddy currents.



11.5 The Electric Field Intensity E in Terms of V and A 26 7

This i s a genera l l aw fo r s t a t ion ary m edia . W e have here ano t he r o f

the four Maxwel l equa t ions . We have a l r eady found two o ther s , namely

E q s . 6-12 and 8-12:

V • EPf + Pb

V • B = 0. (11-15)

Eq ua t io n 11-14 is a di f ferent ia l eq ua t io n tha t re late s the sp ace

der iv at ive s of E at a pa r t ic ula r poi nt to the t ime ra te of ch an ge of B at th e

sam e po in t . Bo th E an d B a re me asu red in the same reference f rame. The

e q u a t i o n d o e s no t give the value of E , unless i t can be integrated.


IN TERMS OF V AND A

Equat ion 11-14 wi l l g ive us an impor tant expression for E in terms of the

p o t e n t i a l s V and A. Since

B = V x A, (11-16)

V x E = —— (V x A) = —V x ~ , (11-17)

dt ct

or

E + - — ]

Th e t e rm be tw een p aren thes es must be equa l to a qua n t i ty w hose cur l is

ze ro , namely a g rad ien t . Then we can se t

3AE = -\V - — . (1 1 -1 9 )

Fo r s t eady cur r en t s , A i s a con s tan t an d th i s equ a t io n r educes to

Eq . 2 -13 . Th e qua n t i ty V i s therefore the elect r ic potent ia l of Sec. 2 .5 , which

i s a l so known as the scalar potential.

Eq ua tio n 11-19 is a general expressio n for E. It states tha t an electric

field intensity can arise bo th from acc um ula tio ns of charg e, th rou gh the

— V I 7 term, and from changing magnetic fields, through the —dA/dt term.

All three quantities, E, A, and V, are measured in the same reference frame.

Th us, if we have a field E ,, Bj in reference frame 1, the elect ric field

in frame 2, mov in g at a co ns ta nt velocity v with r espec t to 1, is given by

Eqs. 10-16 an d 10-17, or by Eq. 10-20. Ho we ve r, in an y given reference

frame, E is given by Eq. 11-19 an d B is eq ua l to t he cu rl of A as in Eq. 8-13.

11.5.1 EXAMPLE: THE ELECTROMOTANCE INDUCED IN A LOOP

BY A PAIR OF LONG PARALLEL W IRES CARRYING

A VARIABLE CURRENT

A pai r of paral lel wires , as in Fig. 11-4, carr ies equal cu rren t s / in oppos i te d i rec

t ions , and / increases at the ra te dl/dt. We shall first calculate th e induced e lec t ro

motance f rom Eq. 11-5 and then from Eq. 11-19.

a) From Sec. 9.1.1, th e c u r r e n t / in wire a p ro d u c e s a magnet ic induct ion

B„ = n0I/2npa, (11-20)

a n d a similar relat ion exis ts for wire b. The flux through th e l o o p in the d i rec t ion

s h o wn in Fig. 11-4 is t h u s

(11-21)

F r o m Eq . 11-5, the e lec t romotance induced in the c lockwise d i rec t ion is r<l> ft:

( fE

. d l = - ^ l n

J 2n dt

T h u s / ' flows in the coun terc lockwise d i rec t ion . Th is is in agreemen t wi th Lenz ' s

l a w: th e c u r r e n t / ' p ro d u c e s a magnetic f ield that opposes th e increase in d>.

b) Let us now use Eq. 11-19 to ca lcu la te th i s same e lec t ro mota nce f rom th e t ime

der iva t ive of the vecto r po ten t ia l A, V being equal to zero in th i s case . From Sec.

ra{rb + w)(11-22)

(11-23)



269

hA

A 8 A

IP,

Idt

Pb

Figure 11-4 Pair of parallel wires carrying equal currents / in opposite directions

in the plane of a closed rectangular loop of wire. When / increases, the induced

electromotance gives rise to a current T in the direction shown. The vector

potential A and the induced electric field —dA/dt are shown on the horizontal sides

of the loop. The induced current /' flows in the counterclockwise direction because

— dA/dt is larger on the lower wire than on the upper wire.

8.4.1. A is parallel to the wires and, if we choose the leftward direction as positive.

(I 1-24)

(1 1-25)

along the lower and upper sides of the loop, respectively. Thus

E (1 1-26)

(11-27)

and, in the clockwise direction.

(11-28)

as previously.

W e have disreg arded the field —dA/dt along the left-hand and right-hand sides,

because it is perpendicular to the wires .

T o find the elec trom otan ce induce d in the lo op by a changing current in a

single conductor, we set rh

-» oo, and then

(j)E-dl = ^ ~ l n ( — — V (11-29)J 2n dt \ra + w)

11.6 SUMMARY

The Faraday induction law can be stated as follows: For a circuit C linked

by magnetic flux O,

E d l<7<J>

dt '

The integral on the left is called the induced electromotance. The positive

directions chosen for <J> and for the integration ar ou nd C are related accor ding

to the rig ht- hand screw rule. This law is true if either B chang es, giving

a — cA/ct field, or if the co ndu ct or moves, giving a v x B field. If the in tegral

is evalua ted over a coil, the flux <J> must be replaced by the flux linkage A.

Lenis law states that the induced current tends to oppose the change

in flux.

In differential form, the Faraday induction law is

V x E =Hi"

(11-14)

This is ano th er of Maxwell's equa tio ns. Both E and B are measu red in the

same reference frame.

The electric field intensity E can be due to accumulations of charge,

or to changing magnetic fields, or both. In general,

E = —VF7 A

"fit"'(11-19)

where E, F , and A are all mea sur ed in the same reference frame.

Problems 271

Figure 11-5 See P rob . 11-3 .

11-4 INDUCED CURRENTS

Figure 1 l -6a sho ws a con du ct in g disk rota t i ng in front of a bar ma gne t ,

a) In wha t di rect io n doe s the cu rren t f low in the wire , c lockw ise or co unt er

c l o c k w i s e ? W h y ?

PROBLEMS

U-1E BOAT TESTING TANKA carr iage r uns on ra i ls on e i ther s ide of a long tank of water eq uip pe d for tes t ing

boa t mod e l s . Th e ra i ls a re 3 .0 me te rs apa r t and the ca r r i age has a maxim um speed of

20 me te rs pe r s econd .

a ) Ca lcu la t e the ma xim um vol t age be tween the ra i l s if t he ve r ti ca l co mp one nt

of the ear th 's mag net i c field is 2.0 x 1 0 " 5 tesla.

b ) Wh a t wou ld be the vo l t age if t he t ank were s i tua t ed a t t he magn e t i c eq ua t or?

11-2 EXPANDING LOOP

A conduct ing bar s l ides a t a cons tant veloci ty v a long conduc t ing ra i l s s epa ra t ed

by a d i s t ance s i n a reg ion of un i form mag ne t i c indu c t ion B pe rpendicu la r t o the p l ane

of the ra i ls . A res is tance R i s con nec ted betw een the ra i ls . Th e res is tance of the res t of

the circuit is negligible.

a) Calculate the current / f lowing in the c i rcui t .

b ) Ho w much po wer is requ i red to move the ba r?

c) How does this power compare wi th the power loss in the res is tance R!

11-3E INDUCED CURRENTS

A bar magnet i s pul led through a conduct ing r ing a t a cons tant veloci ty as in

Fig. 11-5.

Sketch curves of (a) the magnet ic f lux <t>. (b) the cu rren t / . (c) the pow er dis

s ipated in the r ing, as funct ions of the t ime.

Use the pos i t ive di rect ions for <> and / shown in the f igure .

Thi s phenomenon has been used to measure the ve loc i t i e s o f p ro jec t i l e s . The

project i le , wi th a t iny perm an en t m agn et inserted in to it s nose , i s ma de to pass th rou gh

tw o coi ls in success ion, sep arat ed by a dis ta nce of ab ou t 100 mil l im eters . Th e t ime delay

betwe en the pulses is a measu re of the veloci ty. Th e me tho d h as been used up to veloci t ies

of 5 k i lome te rs pe r s econd .



27 2

Figure 11-6 (a) C o n d u c t i n g d i s k

ro ta t in g nea r the No r th po le o f a ba r

mag ne t . The wi re is conn ec ted to the

d i sk th rough s l id ing contac t s .

(£>) Solen oid w ou nd on a b ar ma gn et .

One end of the w i re i s connec ted to a

s l id ing contac t t ha t can move a long

the length of the solenoid.

b) W ha t h app ens i f bo th th e po la r i ty of t he magne t a nd the d i rec t ion of ro t a t ion

a re reve rsed?c) Sho w that there is no curren t genera ted with the se t -up of Fig. 11 -6b, despi te

the fact that th e magn et ic f lux l inking the c i rcui t is not con s tan t .

USE INDUCED ELECTROMOTANCE

A loop of wire is s i tuated in a t ime-dependent magnet ic f ie ld wi th

B = 1.00 x 1 0 "2

co s (27t x 60)f

pe rpen dicu la r t o the p l ane of the loop .

Ca lcu la t e the induced e l ec t romotance in a 100- turn squa re loop 100 mi l l ime te rs

on the s ide .

11-6 ELECTROMAGNETIC PROSPECTION

In e l ec t romagne t i c p rospec t ion , a co i l ca r ry ing an a l t e rna t ing cur ren t causes

induced cur ren t s to f low in conduc t ing ore bodies , and the a l t e rna t ing magne t i c f i e ld

of these induced currents is detected by means of a second coi l placed some dis tance

away from the f i rs t one, as in Fig. i l -7.

Le t us cons ide r a s imple example of induced cur ren t s .

A thin con du ct in g disk of thicknes s /?, radiu s R, a n d c o n d u c t i v i ty a is placed in

a uniform al ternat ing magnet ic f ie ld

B = B 0 co s cot

parallel to the axis of the disk as in Fig. 11-8.

a) Fin d the induced cu rren t dens i ty J as a funct ion of the radius . Assu me that

the conduc t iv i ty a is smal l enou gh to ren der the magn et ic f ie ld of the induc ed curr ent

neg l ig ib l e , compared to the appl i ed B.

Choose the pos i t ive d i rec t ions for B and for J in the di rect ions shown in the

figure.

b) Sketch curve s of B a n d J as funct ions of the t ime. Explain why B a n d J are so

rela ted.



27 3

Figure 11-7 E l e c t r o m a g n e t i c p r o s p e c t i o n . C o i l A, connec ted to a source of a l t e r

na t ing cur ren t , p roduces an a l t e rna t ing magne t i c f i e ld . Coi l B is he ld pe rpe ndicu la r

t o A and is sens i t ive only to the ma gn et ic f ie ld du e to the cu rren ts indu ced by A

in the ore body C. The f igure is not drawn to scale ; the coi l diameters are of the order

of one meter or less .


11-7E INDUCTION HEATING

I t i s com mo n prac t i ce to hea t con du c tor s by sub jec t ing them to an a l t e rna t in g

magnet ic f ie ld. The changing f lux induces eddy currents t ha t hea t t he conduc tor by the

Joule effect (Sec. 5.2.1). The method is called induction heating. It is used extensively

for mel t ing metals and for hardening and forging s teel .

The powers used range f rom wa t t s t o megawat t s , and the f requenc ies f rom

60 he r t z to s eve ral hun dred k i lohe r t z .



Induc t ion hea t ing has the adv anta ge of conven ience and of no t con tam ina t in gthe me ta l w i th combus t ion gases . I t even pe rmi t s hea t ing a conduc tor enc losed in a

vacuum enc losure . Induc t ion hea t ing has anothe r ma jor advantage . A t h igh f requenc ies ,

currents f low near the surface of a conductor . This is the skin effect. See Probs . 16-17

and 18-14. So, by cho os in g the frequency c orrect ly, one can apply a brief heat t rea tme nt

dow n to a kno wn de pth . Thi s is pa r t i cu la r ly imp or t a n t because the re a re nu me rou s

purposes for which one requires s teel parts wi th a hard skin and a soft core: the hard

sk in re s i s t s abras ion and the sof t core reduces breakage . P lowshares a re hea t - t rea t ed

in this way.

Induction furnaces are used for mel t ing metals . They cons is t of large crucibles

wi th capac i ti e s rang ing up to 30 tons , t he rma l ly insu la t ed and su r rou nde d by cur ren t -

carry ing coi ls . Op er at i on is usual ly s tar ted wi th part of the load a l rea dy mo lten.

Let us con s ider F ig. 1 l -9a where a rod of radius a. l eng th L. and conduc t iv i ty a

i s placed ins ide a solenoid having N' t u rns pe r me te r and ca r ry ing an a l t e rna t ing cur ren t

/ = I0 cos 0)t.

As usual, we neglect end effects.

We a l so a s sum e tha t t he f requency is l ow eno ugh to avoid the c om pl i ca t ion s

du e to the skin effect. Th is assu mp tion is wel l sa t is f ied a t 60 her tz wi th a rod of gra phi te



Problems 275

(a = 1.0 x 10?

S i e m e n s per mete r ) hav ing a rad ius of 60 mi l l ime te rs . In o t h e r w o r d s , th e

m a g n e t i c i n d u c t i o n of the i n d u c e d c u r r e n t s is negl igible . We a l so a s sume tha t th e

c o n d u c t o r is n o n - m a g n e t i c . .T h i s is a d m i t t e d l y a highly s impl i f ied i l lus t ra t ion of i n d u c t i o n h e a t i n g . M o r e

ove r , th e power d i s s ipa ted in the g r a p h i t e is only a few wat t s , which is absurd ly sma l l

for such a l a rge p i ece . Thi s should none the le s s be a useful exercise on i nduced cur ren t s .

C o n s i d e r a r ing of r a d i u s /•. t h i cknes s dr. and l eng th L ins ide the c o n d u c t o r ,

a s in Fig. 11 -9b.

a ) Show tha t th e e l e c t r o m o t a n c e i n d u c e d in the r ing is

u.0nr2

coNT0sin cot.

b) Show tha t , for a cur ren t f lowing in the az imutha l d i rec t ion , th e r ing has a

re s i s t ance

R = InrioLdr.

c ) Show tha t th e ave rage power d i s s ipa ted in the r ing is

^ ( / (0o ;N7

0)

2

7T ( tL/- 3 dr.

d) Show tha t th e t o t a l ave rage power d i s s ipa ted in the cyl inde r is

1 , ,— (H 0o)NT 0)-naLa .16

e) Ca lcu la t e th e power d i s s ipa ted in the g r a p h i t e rod, set t ing L = 1 mete r ,

7 0= 20 a m p e r e s , N' = 5000 tu rns per m e t e r .

USD INDUCED ELECTROMOTANCE

A magnet ic f ie ld is desc r ibed by

. 2nyB = £ 0 sin —^- (sin wt)i.

A

In this field a s q u a r e l o o p of side A/4 lies in the i . -p l an e w i th its s ides paral le l to

th e y- and r -axes . The l o o p m o v e s at a cons tan t ve loc i ty vj.

C a l c u l a t e the e l e c t r o m o t a n c e i n d u c e d i n the l o o p as a funct ion of the t ime i f th e

t ra i l ing edge of the l o o p is at y = 0 at t = 0.

Set co = 2 m l A.

BETATRON

F i g u r e 11-10 s h o w s the pr inc ip le of o p e r a t i o n of the b e t a t r o n . Th e b e t a t r o n p r o

duces e l ec t rons hav ing ene rg ie s of the o r d e r of mil l ions of e lec t ron-vol t s . E lec t rons

are held in a c i rcu la r o rb i t in a v a c u u m c h a m b e r by a magnet ic f ie ld B . Th e e lec t rons



27 6

/J

i i m i L L J J i n j ^ ^ 11! 11

11111! 111

1LLL

Figure 11-10 Cross -s ec t ion th rou gh the cen t ra l r eg ion of a be ta t ron . Th e e l ec t rons

desc r ibe a c i rcu la r o rb i t i n a p l ane pe rpendicu la r t o the pape r a t 0—0, in the

f ie ld o f an e l ec t romag ne t and ins ide an evacua ted ce ram ic to rus T. Only the po le

p ieces P of the e l ec t romagne t a re shown. The core i s l amina ted . The po le p i eces

a re shaped so tha t t he ave rage B ins ide the orbi t i s eq ua l to twice that a t th e o rbi t .

The e l ec t rom agne t is ope ra t ed on a l t e rna t in g cur ren t , and the e l ec t rons a re

acce le ra t ed on ly dur ing tha t pa r t o f t he cyc le when bo th B a n d dB/dt have the

proper s igns . See Prob. 11-9 and Fig. 11-11.

Figure 11-11 Cir cul ar e lect ron orbi t in the be ta t ron . It i s assu me d that th e

ma gn et ic f lux is in the di rect ion sh own and th at i t increase s . Here .

e = - 1.60 x 1 0 " 1 9 c o u l o m b .

Problems 277

are accelerated by increasing the magnetic f lux l inking the orbi t . Betatrons serve

as sources of x-rays. There are very few that are s t i l l in use today.

Sho w that the average ma gne tic induc tion over the plane of the orbi t must betwice the magn etic ind uctio n at the orbi t , if the orbi t ra dius is to remain fixed when

bo th B and the electron energy increase.

Solution: The various vectors we are concerned with are shown in Fig . 11-11.

In the reference frame of the labo rat ory , an electron is subjected t o the Lo ren tz

force

e (E + v x B ).

Fhe electric field E is due to the t ime-d epen den t m agn etic f ield and is — dA/dt.

We do not kno w the value of A. but we do know , from F ara da y 's induc tion law,

t h a t

1 <M>E = - - - . (1)

2nr tit

We assume that the magnetic f lux <t> increases. This electric field E provides the

tange ntial accele rat ion tha t increases the electro n energy. Th e centripe tal force

cv x B keeps the electron on i ts circular orbi t .

So , in the az imutha l d i rec t ion .

d e d<t>(mv) = eE= - — — , (2)

dt 2nr dt

while , in the radial d irect ion,

,m-2

r = -Bev. (3)

In these two equations, al l the quanti t ies are posi t ive except e.

Equating the two values of mv obt ain ed from Eqs. 2 and 3 ,

emv = <t = -Ber, (4)

2nr

<t> = 2 tu -2

B, (5)

wh e re B i s the magnet ic induct ion a t the rad ius r of the electron orbi t . Now, by

defini t ion,

<t> = 7t r2

B, (6)

wh e re B is the average magnetic induction within the circle of radius r. T h e n , c o m

parin g Eqs . 5 and 6 , B is eq ual to 2B .




This effect wa s pred ic t ed by Maxwel l , and was o b s e r v e d for the first time in

1916 by T o l m a n and c o l l a b o r a t o r s .

The inverse effect , namely th e acce le ra t ion of a b o d y c a r r y i n g a var i ab le cur ren t ,

was a l so pred ic t ed by M a x w e l l , and was first observed by B a r n e t t and o t h e r s in 1930.

T h e U.S. Nat iona l E lec t r i ca l Code ru l e s tha t bo th conduc tors of a circui t

o p e r a t i n g on a l t e rna t ing cur ren t , if enc losed in a meta l l i c condui t , mus t be run in the

s a m e c o n d u i t .

Let us s u p p o s e t h a t we h a v e a s ing le conduc tor ca r ry ing an a l t e r n a t i n g c u r r e n t

and enc losed wi th in a c o n d u c t i n g t u b e .

Show tha t bo th A and dA/dt are l o n g i t u d i n a l in the t u b e .

So, with a s ingle wire , a l o n g i t u d i n a l c u r r e n t is i n d u c e d in the t ube . Thi s causes

a needless power loss , and may even cause spa rk ing at fau l ty jo in t s . Wi th two conduc tors

ca r ry ing equa l and oppos i t e cur ren t s , bo th A and cXIct are essent ia l ly zero in the

c o n d u i t .

11-11E ELECTRIC CONDUITS

11-12E THE POTENTIALS V AND A

We have s een tha t

E = - VI - , \ \ ft. B = V x A.

Show tha t ne i the r E no r B arc affected if t he po ten t i a l s V and A are rep laced by

V + (G dt and A - VG.

w h e r e G is any funct ion of v, r. z, / whose s econd de r iva t ives ex i s t and are c o n t i n u o u s .

11-10 THE TOLMAN AND BARNETT EFFECTS

If a c o n d u c t o r is given an acce le ra t ion a, the c o n d u c t i o n e l e c t r o n s of m a s s m

and cha rge — e are subjec ted to inertia forces —ma.Show tha t , if E' is the equivalent tota l e lect r ic f ie ld intens i ty and if B is the m a g

ne t i c induc t ion in the c o n d u c t o r ,



CHAPTER 12

MAGNETIC FIELDS: V

Mutual Inductance M and Self-Inductance L

12.1 MUTUAL INDUCTANCE M

12.1.1Example: Mutual Inductance Between Two Coaxial Solenoids

12.2 INDUCED ELECTROMOTANCE IN TERMS

OF MUTUAL INDUCTANCE

12.2.1 Example: The Vector Potential of the Inner Solenoid

12.3 SELF-INDUCTANCE L

12.3.1 Example: Long Solenoid

12.3.2 Example: Toroidal Coil

12.4 COEFFICIENT OF COUPLING k

12.5 INDUCTORS CONN ECTED IN SERIES

12.5.1 Zero Mutual Inductance

12.5.2 Non-Zero Mutual Inductance

12.5.3 Example: Coaxial Solenoids

12.6 INDUCTORS CONNECTED IN PARALLEL

12.6.1 Zero Mutual Inductance

12.6.2 Non-Zero Mutual Inductance


12.7 TRANSIENTS IN RL CIRCUITS

12.7.1 Example: Inductor L Connected to a Voltage Source V 0

12.7.2 Example: Horizontal B eam Deflection in the Cathode-Ray

Tube of a Television Receiver

12.8 SUMMARY

PROBLEMS



In th i s chap te r we sha l l see how one u t i l i zes the Faraday induc t ion l aw to

ca lcu la t e induced vo l t ages and cur r en t s in e l ec t r i c c i r cu i t s . I n Chap te r 5 weascr ibed to each b ranch o f a c i r cu i t a ce r t a in r es i s t ance R. I f the current

f lowing in a b r anch p roduces an apprec iab le magne t i c f i e ld , t hen the b ranch

also possesses a self-inductance L. I f t he ma gne t i c fi eld o f on e b ran ch p r o

duces a f lux l inkage in ano ther b r anch , then the re ex i s t s a mutual inductance

M b e t w e e n t h e t w o b r a n c h e s . T h u s , th e m a g n e t i c p r o p e r t i e s o f t h e b r a n c h e s

are charac te r i zed by the two quan t i t i e s L and M.

W e sha l l r e tu rn to mu tua l indu c tan ce l a t e r on , in C ha p t e r 18, which

dea l s wi th pow er t r ansfe r an d t r a nsfo rm er s .

12.1 MUTUAL INDUCTANCE M

To ca lcu la t e the e l ec t romotance induced in one c i r cu i t when the cu r r en t

changes in another c i rcui t , i t i s convenient to express the f lux l inkage in the

f irs t one in term s of th e cu rren t in the seco nd an d of a geo me tr ica l factor

invo lv ing bo th c i r cu i t s .

Co ns id er the tw o c i r cu i t s o f F ig . 12-1 . Th e cur r en t / „ in c i rcu i t a

p r o d u c e s i n b a f lux l inkage A ah t h a t i s p r o p o r t i o n a l t o / „ :

A f l i = M ahIa. (12-1)

Similar ly , i f there is a current Ib t h r o u g h c i r c u i t b, t he magne t i c f lux

of b l i nk ing a is

Ka = M haI„. (12-2)

28 1

Figure 12-1 Two circuits a a n d b. The f lux <t> ab shown l ink ing b and or ig ina t ing in a

i s pos i t ive . This is because i t s di rect ion is re la ted by the r ight-hand screw rule to

the d i rec t ion chosen to be pos i t ive a round b.

Since bo th c i r cu i t s can be of any shape , one na tu ra l ly d oes no t expec t

to f ind a genera l r e l a t ionsh ip be tween M ab a n d M ba. On the con t r a ry , i t can

b e s h o w n t h a t M ab = M haf Th e f ac to r o f p ro po r t ion a l i ty M = M ab = M ba

is called the mutual inductance be tween the two c i r cu i t s .

Mutua l induc tance depends so le ly on the geomet ry o f the two c losedc i r cu i t s an d on the pos i t ion a nd o r i en ta t io n o f one wi th r espec t to the o the r .

When mul t ip l i ed by the cu r r en t in one c i r cu i t , M gives the f lux l inkage in

the o ther .

A s imi la r s i tua t ion ex i s t s in r e l a t ion wi th the capac i t ance be tween

t w o c o n d u c t o r s . T h e c a p a c i t a n c e C depends so le ly on the geomet ry o f the

two con du c to r s and on the pos i t ion and o r i e n ta t ion o f one wi th r espec t to

the o ther . If one o f the con du c to r s i s g ro und ed , then the charge indu ced on

i t i s equal to CV, w h e r e V i s t he vo l t age app l i ed to the o ther .

S ince the mutua l induc tance i s t he f lux l inkage in one c i r cu i t pe r un i t

o f cu r r e n t in the o ther , i ndu c tan ce i s me asu red in we ber - tu rns per am pere ,or in henrys.

T h e m u t u a l i n d u c t a n c e b e t w e e n t w o c i r c u i ts is o n e h e n r y w h e n a

cu rren t of on e am pe re in on e of the ci rcui ts pro du ce s a f lux l inkage of on e

weber - tu rn in the o ther .

* Electromagnetic Fields ami Waves, p. 343 .

282 Magnetic Fields: V

The sign of M is chosen as follows. The mutual inductance between

two circuits a and b is positive if a current in the positive direction in a

produces in b a flux that is in the same direction as one due to a positive

current in b. Thi s is illus trate d in Fig. 12-1 . Th e positiv e directi ons for the

currents are chosen arbitrarily.

A pair of coils designed so as to possess a mutual inductance is called

a mutual inductor. Mutual inductors are also called transformers.

1.1 EXAMPLE: MUTUAL INDUCTANCE BETWEEN TWO

COAXIAL SOLENOIDS

Let us calculate the mutual inductance between two coaxial solenoids as in Fig. 12-2.We assume that both windings are long, compared to their common diameter 2R.

that they have the same number of turns per meter N', and that they are wound in

the same direction, as in the figure. We set /„ > lb.

Let us assume a current /„ in coil a. Then the flux of coil a linking each turn of

coil b is

<t>ah = Li 0nR2

NT a. (12-3)

and the mutual inductance is

M = \JIB = N

bQJI

a = ti

0nR

2

N'Nh. (12-4)

= ti0nR

2

NaN

h/l

a. (12-5)

I.

I,

Figure 12-2 Coaxial solenoids. The two radii are taken to be approximately equal.

12.2 Induced Electromotance in Terms of Mutual Inductance 283

We can a l so ca l cu la t e the mutua l i nduc tance by a s suming a cur ren t Ib in coil b.

T h e n

<D ba = u0nR2

NTb(12-6)

This f lux l inks only lhN' = N b turns of coi l a. since B fa l ls rapidly to zero beyond

the end of a long solen oid, as we saw in Sec. 9.1.5. Th en t he mu tua l in du ctan ce is

M = AJI„ = N„<S>JIb

= n0nR 2N'N b = p 0nR 2N aN b/l a,

as prev ious ly .

(12-7)

12.2 INDUCED ELECTROMO TANCE IN TERMS

OF MUTUAL INDUCTANCE

Fr om Sec . 11.2, the e lec t r om otan ce ind uced in c ircu i t b by a change in /„ in

Fig . 12-1 is

E d l =d A , , h

= -M—.dt dt

(12-8)

In this case, E is —dA/dt, wh ere A is the va lue of the vecto r poten t ia l of

the cu r ren t Ia at a poin t on c ircui t b (Sec. 11.5).

S imi la r ly , the e lec t romotance induced in c i rcu i t a by a change in Ib is

E d ldA

bl

dtM

dh

dt(12-9)

T h e s e e q u a t i o n s a r e c o n v e n i e n t f or c o m p u t i n g t h e i n d u c e d e l e c t r o m o

t a n c e , s ince they invo lve on ly the mu tua l induc tance and dl/dt, both of

wh ich can be measu red .

We now have a second de f in i t ion o f the hen ry : the mu tua l induc tance

between two c ircui ts is one henry i f a current changing a t the ra te of one

ampere pe r second in one c i rcu i t induces an e lec t romotance o f one vo l t in

the o the r .

12.2.1 EXAMPLE: THE VECTOR POTENTIAL OF THE INNER SOLENOID

I t i s pa rad oxica l t ha t a va ry ing cur re n t i n the inne r so lenoid sho uld induce an

e lec t romotance in the ou te r one , s ince we have shown (Sec . 9.1.3) t ha t t he magne t i c



2 8 4 M a g n e t i c F i e l d s : V

i nduc t ion ou t s ide a l ong so lenoid is ze ro . The e x p l a n a t i o n is t ha t th e i n d u c e d

electric field intensity at any given point is e q u a l to the nega t ive t ime de r iva t ive of

t he vec tor po ten t i a l at t ha t po in t (Eq. 11-19) and t h a t th e vec tor po ten t i a l A d o e s

not van i sh ou t s ide an inf ini te solenoid, despi te th e fact that B = V x A d o e s . See

also Sec. 9.1.2.

W e can ac tua l ly ca l cu la t e th e vec tor po ten t i a l j us t ou t s ide the inner solenoid

from th e m u t u a l i n d u c t a n c e . If we cons ide r the di rec t ion of the c u r r e n t /„ in the

i nne r so lenoid as posi t ive , then th e e l e c t r o m o t a n c e i n d u c e d in the o u t e r one is

- M i l J i t , or

d l ,- | ioJt* 2

JVW»-£a t

a n d th e induced e lect r ic f ie ld intens i ty — cA/ct is 2nRNh t imes sma l l e r :

PA HoRN' dl,= - — - . (12-10)

dt 2 dt

In t egra t ing , we h a v e th e vec tor po ten t i a l at any point c lose to the s o l e n o i d :

A = ^ L (12-11)

in th e a z i m u t h a l d i r e c t i o n .

12.3 SELF-INDUCTANCE L

A single circuit carrying a current / is of course linked by its own flux, as in

Fig. 12-3. Th e flux linkag e A is pro po rti on al t o the c urr ent :

A = LI, (12-12)

where L is called the self-inductance, or simply the inductance of the circuit.Self-inductance, like mutual inductance, depends solely on the geometry and

is mea sur ed in henr ys. It is always positive.

A circuit t hat is desi gned so as to ha ve a self-induc tance is called an

inductor.

An indu cto r has a self-inductance of one he nry if a curren t of one

ampere produces a flux linkage of one weber-turn.

2 8 5

Figure 12-3 Isola ted c i rcui t carrying a

current / and i ts f lux l inkage A.

A cha ng e in the cu r r en t f lowing th rou gh a c ir cu i t p roduc es w i th in

the c i r cu i t i tse lf an induce d e l e c t ro mo tanc e

dt dt(12-13)

T h e i n d u c e d e l e c t r o m o t a n c e t e n d s t o o p p o s e t h e change i n cu r r en t , accord ing

to Lenz ' s law, and ad ds to wha teve r o the r vo l t ages a r e p resen t . If a va ry ing

cur r en t f lows th rough an induc tance L , the vol tage across i t i s L dljdt, a s

in Fig. 12-4.

An induc to r the re fo re has a se l f - induc tance o f one henry i f t he cu r r en t

f lowing th ro ug h it cha nges a t t he r a t e o f one am pe re per second when the

vol tage di f ference between i t s terminals i s one vol t .

V

O

I L

Figure 12-4 Ideal ized inductor wi th

ze ro re s i s t ance . The vo l t age V is

-±r L dljdt.

1 2 . 3 . 1 EXAMPLE: LONG SOLENOID

I t was show n in Sec. 9.1.3 tha t the mag net i c induc t ion ins ide a long soleno id,

neglect ing end effects , i s uniform, and that

B = HoN'I,

w h e r e AT i s t he num ber of tu rn s pe r me te r . T hus

(12-14)

<P = nR\ (12-15)

28 6

5

Rll

10

Figure 12-5 Factor K for calcu la t ing the induc tanc e of a short solen oid.

w h e r e N i s the tota l n um be r of tur ns . / i s the length of the solen oid, an d R is itsr a d i u s . T h e n

A N<t> p 0N2 ,

L = — = = — — n R 1 . (12-16)I I I

Th e s e lf - induc tance of a long so lenoid is t hus pro por t iona l t o the square of the

nu mb er of tu rn s and to i t s c ros s - s ec t ion , and inve rse ly prop or t io na l t o i ts l eng th .

The induc tance of a short solenoid is smaller by a factor K, wh ich is a function

of R/l, as in Fig. 12-5.

EXAMPLE: TOROIDAL COIL

A toroidal coi l of N t u rn s is wo und on a fo rm of non -ma gne t i c ma te r i a l hav ing a

square cross-sect ion, as in Fig. 12-6.

Ac cord ing to the c i rcui ta l law, or from Eq. 9-8. the mag net i c induc t ion in the

az imutha l d i rec t ion , a t a rad ius p ins ide the toroid. i s

B = poNl/litp. (12-17)



287

Figure 12-6 Toro ida l co i l of square c ross -sec t ion and m e a n r a d i u s R.

T h u s th e flux linkage is

p0N

2

I r-R + W/2 w dpR + WI:

j R - w / 2

L =

2n

p0N

2

w .

w In2R + w

2R - w

2R + w

2R - iv

(12-1S

(12-19)

(12-20)

The se l f - inductance of a toroidal coi l is p ro p o r t i o n a l to the s q u a re of t h e n u m b e rof turns , l ike that of a long so leno id .

12.4 COEFFICIENT OF COUPLING k

In Sec. 12.1.1 we found that the mutual inductance between the two solenoids

of Fig. 12-2 is

Now, from Eq. 12-16,

M = n07iR

2

NaNb

/la

.

L „ = n0nR2N2Jla,

L h = n0nR2N2/lb,

(12-21)

(12-22)

(12-23)

a n d

M = (j) iKU)m

. (12-24)

T h e m u t u a l i n d u c t a n c e i s t h u s p r o p o r t i o n a l t o t h e g e o m e t r i c a l m e a n o f

the se l f - induc tances .

M o r e g e n e r a l l y ,

M = k(L aL b)112, (12-25)

w h e r e k is the coefficient of coupling, which can be e i the r pos i t ive o r nega

t ive . Moreover ,

|fe| < 1. (12-26 )

For example , i n the p resen t case , k = (l h/l a)in < 1 , by h ypo thes i s .

W h e n \k \ % 1, the tw o ci rcui ts are said to be tightly coupled; t hey a r e

loosely coupled w h e n \k\ « 1.

12.5 INDUCTORS CONNECTED IN SERIES

12.5.1 ZERO MUTUAL INDUCTANC E

Consider two induc to r s connec ted in se r i es as in F ig . 12-7a o r 12-7b . The i r

m u t u a l i n d u c t a n c e is a p p r o x i m a t e l y z e r o . T h e n

v=L4+L4={L

>+L

>)d

Tt'

<i2

-27)

and the ef fec tive indu c tan ce i s s imply the sum of the induc tan ces . The sam e

ru le app l i es to any number o f induc to r s connec ted in se r i es , a s long as

the mutua l induc tances a r e a l l ze ro .



289

-Tt

(a) lb)

Figure 12-7 (a) Tw o coils connected in series. They are assumed to be far from each

other, and their mutual inductance is approximately zero, (b) Two coils connected in

series, one perpendicular to the other. T h e flux linkage and the mutual inductance are

again approximately zero.

12.5.2 NON-ZERO MUTUAL INDUCTANCE

If the coupl ing coefficient bet ween the two in duc tors con nec ted in series

is not zero, as in Fig. 1 2 -8 , then the voltage on coil 1 is L-i t imes dl/dt, plus

M times dl/dt .in coil 2 , and the voltage on coil 2 is given by a similar ex

pression. Thus

V = ( L , + M)~ + (L2 + M ) ^ , ( 1 2 - 2 8 )

= ( L , + L2

+ 2 M ) — . ( 1 2 - 2 9 )

M

Figure 12-8 Tw o coils connected in

series with a non-zero mutual

inductance M .




The ef fect ive inductance is thus

L = Lj + L 2 + 2M . (12-30)

Remember tha t M can be e i the r pos i t ive o r nega t ive (Sec . 12 .1 ) .

EXAMPLE: COAXIAL SOLENOIDS

Figure 12-9a shows the coaxial solenoids of Fig. 12-2 connected in ser ies , wi th the

l e f t -hand wi res connec ted toge the r . Le t us ca l cu la t e the induc tance be tween a a n d b.

Supp ose the cur ren t f lows from te rm ina l A t o t e r m i n a l B. Then the f lux of a p o i n t s

left , while that of b points r ight . So the f lux of b part ly cancels that of a, a n d

La + L b2 M , (12-31)

w h e r e L a, L h, M are given in Eqs . 12-22 to 12-24 and M i s a pos i t ive quant i ty, here .

If the solenoids are connected as in Fig. 12-9b,

La + L„ + 2M. (12-32)

Figure 12-9 The coaxial solenoids of

Fig. 12-2, con nec ted in ser ies , (a) wi th

M nega t ive , and (b) wi th M pos i t ive .B A



12.6 Inductors Connected in Parallel 291

12.6 INDUCTORS CONNECTED IN PARALLEL

•

12.6.1 ZERO MUTUAL INDUCTANC E

W e hav e jus t seen tha t tw o indu c to r s connec ted in se ri es a r e t r ea t ed l ike

resistors in ser ies (Sec. 5 .4) when thei r mutual inductance is zero. One might

guess that inductors in paral le l are calculated l ike resistors in paral le l (Sec. 5 .5)

w h e n M i s zero. As we shal l see, th is i s cor rect .

We now have two induc to r s connec ted in para l l e l a s in F ig . 12-10 ,

wi th ze ro M. T h e n

T h e t o t a l c u r r e n t I is I t + I2. If L i s the sel f - inductance that i s equivalent

to L l a n d L 2 in paral le l ,

(12-34)

(12-35)

(12-36)

(12-37)

a n d

L =LXL 2

(12-38)L , + L 2

as we guessed a t t he beg inn ing .



29 2

I t

-T l

Figure 12-10 Two co i l s connec ted in pa ra l l e l w i th approxima te ly ze ro mutua li n d u c t a n c e .

12.6.2 NON-ZERO MUTUAL INDUCTANC E

Let us now cons ider F ig . 12-11. w h i c h s h o w s t w o c o u p l e d i n d u c t o r s ,

connected in paral le l . I f L is t he equ iva len t indu c tanc e ,

V = Ldl

Jt '(12-39)

dt dt(12-40)

Also, for the left-hand coil ,

dt dt(12-41)

and, for the r ight-hand coi l ,

dt dt(12-42)



29 3

1 A

A/ O U 2

Figure 12-11 Two coi l s connec ted in paral le l wi th a n o n - z e r o m u t u a l i n d u c t a n c e M.

H e r e , M is pos i t ive : a pos i t ive (downward) cur ren t in L, p r o d u c e s a d o w n w a r d

B in L 2 , whi le a pos i t ive (downward) cur ren t in L2 a l so produces a d o w n w a r d B

in L , .

From the last two equations,

( L { - M)d

± = ( L 2 - M )d

^ . (12-43)

The n, from Eqs. 12-40 an d 12-41,

and

L 2 - MJ dt \1

L2 - MJ dt

L = L,(L2 — M) + M( L. - M)( L 2 - M ) + (Lj - M )

L X L 2 - M2

L , + L2 - 2M'(12-46)

When M = 0, we revert to the pre viou s case.




12.6.3 EXAMP LE: COAXIAL SOLE NOIDS

F i g u r e 12-12 shows aga in th e coaxia l so l enoids of Fig. 12-2, t h i s t ime connec ted

in paral le l . With th e c o n n e c t i o n s s h o w n in Fig. 12-12a, M is posi t ive and the in

duc tance be tween t e rmina l s A and B is

L =M

2

La + L„ 2M(12-47)

with the L's and th e M on the r igh t as in Eqs. 12-22 to 12-24.

If th e c o n n e c t i o n s to one of the s o l e n o i d s are i n t e r c h a n g e d as in Fig. 12-12b,

L =LaL b - M

L a+ Lb+ 2 M '(12-48)

Here aga in , M is the pos i t ive quant i ty ca l cu la t ed as in Eq. 12-24

Figure 12-12 The p a i r of coaxia l so l enoids of Fig. 12-2, con nec ted in paral le l ,(a) with M posi t ive , (b) with M nega t ive .

12.7 TRANSIENTS IN RL CIRCUITS

Wh en a circuit com pris ing resistances and induct ance s is dist urbe d in some

way, whe ther by cha ngin g the appli ed voltages or curr ents , or by modifying



12.7 Transients in RL Circuits 295

t he c i r cu i t , t he cu r r en t s t ake some t ime to ad jus t t hemse lves to the i r new

steady- s t a t e va lues . We d i scussed a s imi l a r phenomenon in r e l a t ion wi th

RC ci rcui ts in Sec. 5 .14.

EXAMPLE: INDUCTOR L CONNECTED TO A VOLTAGE SOURCE V 0

Figure 12-13 shows an induc tor L connec ted to a vo l t age source F 0 . Th e ind uc tor

and i t s connec t ing wi res a re made of a s ing le con t inuous wi re o f conduc t iv i ty a

and c ros s - s ec t ion a.

a) The current is constant.

If the wire is in air (e r « 1), i ts surface c ha rge de nsit y is e 0 £ „ , w h e r e E„ i s the

nor ma l com pon ent o f E jus t ou t s ide the wi re . Thi s fo l lows f rom G auss ' s l aw

(Sec. 3.2) because £„ = 0 inside the wire, as we shall show below. Similarly, if the

wire is submerged in a die lect r ic , the surface charge dens i ty is e r e 0 £ „ .At any given point on the surface of the wire , the f ie ld £„ is proport ional to F 0 .

I t depends on the configurat ion of the c i rcui t , as wel l as on the pos i t ion and con

f iguration of the ne igh bo rin g bodies .

Inside the wire , the volume charge dens i ty is zero, as we saw in Sec. 5.1.2. Also,

Th us £ and |VF | , l i ke J, are the same everywhere ins ide the wire . The l ines of force

inside the wire are parallel and, if the length of the wire is / ,

E = - V F = J/ff. (12-49)

/ |VK| = / - = I— = IR = V 0.a ao

(12-50)

T h u s |VF| i n s i d e t h e w i r e i s s i m p l y 1,, /.

V

+ + + + +

Figure 12-13 Inductor £ connected to a vol tage source.



296

t (seconds)

Figure 12-14 Current through the inductor of Fig. 12-13 as a function of the time.

It is ass umed that / = 0 at r = 0 and that RjL = 1.

b) The current builds up from zero to its steady state value V0/R.

We assume that the time the field takes to propagate from the source to the

inductor is negligible. We also neglect the stray capacitance between the turns of

the inductor and between the wire and ground.

Since the wire has both a resistance R and an induct ance L, Kirchoff's voltage

law (Sec. 5.7) gives us that

dlL - + RI= V

0.

dt(12-51)

We saw in Sec 12.3 that the voltage across a pure inductance is L dljdt. From Sec.

5.13, the solution of this equation is

RKe (12-52)

where K is a constan t of inte grati on. By hypo thesi s, / = 0 at / = 0, and th us K

- V0jR. Then

(12-53)

The current 1 increases with time as in Fig. 12-14. No te the analogy with Fig. 5-23.

The current density J is again the same throughout the wire since, by hypothesis,

there is zero capacitance, and hence no accumulation of charge at the surface of

the wire. Thus

J I

E = - = —a aa

RI. (12-54)



12.7 Transients in RL Circuits 297

and £ is uniform throughout the wire. It builds up gradually, like /. Also.

RI = El. (12-55)

From Eq. 11-19,

RI = j)^-\V - ^Y«U, (12-56)

V F - d l - - ^ ( j ) A - d l , (12-57)

where the integration is evaluated along the wire, in the direction of E, clockwise

in Fig. 12-13.

Now, from Eq. 8-54, the line integral of A • dl around a simple circuit is equal

to the enclosed flux. Mo re generally , the integral is equa l to the flux linka ge:

A d l = A (12-58)

and

r dA

RI = - c i v K - d l , (12-59)

J

dt

r L dl

= - d ) V F - d l . (12-60)J at

from Eq. 12-12. Co mp ar in g with Eq. 12-51, we find that

) V F - d l . (12-61)

This inte gral is therefore a const ant . The negative sign comes from t he fact tha t

the integral is the voltage of the left-hand terminal with respect to the right-hand

one, or — F 0 .

Inside the wires that lead from the source to the coil, the magnetic field is weak

and |V F| is app rox ima tel y eq ual to the £ of Eq. 12-54, with / a function of the tim eas in Eq. 12-53.

Inside the coiled part of the wire, A is not negligible. The — VF field points in

the same direction as the current, while —dA/dt points in the opposite direction.

At t = 0, when the so urce is con necte d, A = 0 but I^A/f'fj is large , / = 0, and the

two terms on the right in Eqs. 12-56, 12-57, 12-59, 12-60 cancel. At the beginning,

most of the voltage drop appears across the coil while, under steady-state conditions,

the potential drop is uniformly distributed along the wire.




EXAMPLE: HORIZONTAL BEAM DEFLECTION IN THE

CATHODE-RAY TUBE OF A TELEVISION RECEIVER

The e lec t ron beam sweeps ho r izon ta l l ines across the screen , one be low the o ther ,

unt i l the whole screen has been covered. The process then repeats i tself . The image

is ob ta ined by modu la t ing the beam in tens i ty .

T h e b e a m1

is deflected by the magnetic f ields of two pairs of coi ls , one for the

hor iz on ta l m ot ion , and one for the ver ti ca l . The bea m sweeps ho r izon ta l ly a t a

cons tan t ve loc i ty , and re tu rns a t a much h igher ve loc i ty .

To sweep hor izon ta l ly a t a cons tan t ve loc i ty , the cu rren t th rough the pa i r o f

horizontal deflect ion coi ls must increase l inearly with t ime. This is achieved by

app ly ing a cons tan t vo l tage V to the coils, for then

dlL —

dt

VI « — t.

L

(12-62)

12.8 SUMMARY

T h e mutual inductance M betw een tw o c i rcu i t s is equ al to the mag net i c f lux

l ink ing one c i rcu i t per un i t curren t f lowing in the o ther:

9

Kb = M ahIa. (12-1)

Th e mu tu a l i n d u c t an ce d e p en d s so le ly o n t h e g eo m e t ry a n d o n th e re l a ti v e

pos i t ions an d or ie n ta t ion s of the two c i rcu i t s . I t has the sam e value , which ever

c i rcu i t p roduces the magnet ic f lux l ink ing the o ther . Mutual inductance i s

m e a s u r e d i n henrys.

M u tu a l i n d u c t an ce can b e e i t h e r p o s i t i v e o r n eg a t i v e . Th e mu tu a l

i n d u c t an ce b e tw een tw o c i rcu i t s a a n d b is pos i t ive if a cu rr en t in the pos i t ive

di rec t ion in a p ro d u ces i n b a f lux tha t i s in the same d i rec t ion as tha t p ro

duced by a posi t ive curren t in b.

Th e e l ec t r o m o tan c e i n d u ced i n c i rcu i t b by a change in curren t in

c i rcu i t a is

E d l = - M ^ . (12-8 )dt

T h e self-inductance L of a circuit is i ts flux l ink age p er unit cu rre nt .



Problems 299

Thus

E d l = - L ~ . (12-13)dt

The voltage across an inductance L carrying a current I is L dl/dt.

The coefficient of coupling k between two circuits is defined by

M = k(LaL„)

1

'2

, (12-25)

and can have values rangi ng from —1 to + 1 . It is zero if no ne of the flux

of one circuit links the other.

The effective i ndu cta nce of tw o inductors connected in series is

L = Ll

+ L2

+ 2M, (12-30)

where M is the mut ual ind uct ance , which can be either positive or nega tive.

The effective inductance of two inductors connected in parallel is

L = J ^ U L ^ .{l2-46)

Ll + L

2- 2M

Wh en a volt age is appl ied to an inductiv e circuit, the curren ts t ake

some time to adjust to their steady-state values.

PROBLEMS

12-1E Starti ng from E q. 12-5, show th at p0 is expressed in henry s per meter.

12-2E MUTUAL INDUCTANCE

A long straight wire is situated on the axis of a toroidal coil of N turns as in

Fig. 12-15.

S h o w that

u0bN* ( b\M = — In 1 + - .

2n \ a



30 0

Figure 12-15 See Prob. 12-2,

12-3E MUTUAL INDUCTANCE

a ) Show tha t t he mutu a l i n duc tance be tw een two coaxia l co il s , s epa ra t ed by a

dis tance z as in Fig. 12-16, one of radius a a n d N a t u rns , and the o the r o f rad ius b « a

a n d N h turns is

nu 0N aN ba2b 2

2(a 2 + z 2 ) 3 ' 2 '

We shal l use this resul t in Prob. 14-7.

You can calculate this in one of two ways . You can e i ther assume a current in

coil a and calculate the f lux through coi l b, or a s sume a cur ren t i n b and ca lcu la t e the

f lux th rough a. From Sec. 12.1, M is the same, e i ther way. Now, i f the current i s in a, you

know how to calculate the f ie ld a t b, from Sec. 8.1.2, since b « a. If the current is in b,

then the calculat ion is vas t ly more di ff icul t because you have to calcula te B away from

the axis . So you sh ould calcu late the magn et ic f lux thro ug h coi l b due to a cur ren t

th rough co i l a.

b) Ho w does the mu tua l i n duc tan ce va ry when the sma l l co i l i s ro t a t ed a ro un d

the vert ical axis?

Figure 12-16 See Prob. 12-3. N a turns



Problems 301

c) What if one ro ta ted th e l a rge co i l a round a ver t i ca l ax i s? Would the m u t u a l

inductance vary in the s a m e wa y ?

•

I2-4E A OUTSIDE A SOLENOID

S h o w t h a t the vecto r po ten t ia l at a d is tance r from the axis outside an infinite

so leno id of r a d i u s R and N' t u r n s per meter car ry ing a c u r r e n t / is

A = n0N'R

2

II2r.

The vec to r po ten t ia l is a z i m u t h a l .

I2-5E A INSIDE A SOLENOID

S h o w t h a t the vecto r po ten t ia l at a d is tance r from th e axis , inside an infinite

so leno id of r a d i u s R and N' t u r n s per meter car ry ing a c u r r e n t / is

A = n0Nir/2.

Here a l so , the vecto r po ten t ia l is a z i m u t h a l .

12-6 MAGNETIC MONOPOLES

M a g n e t i c m o n o p o l e s are the magnet ic equ iva len t of electric charges. See Sec. 8.3.

Figure 12-17 shows schemat ica l ly one device that ha s been used in the search

fo r magnet ic monopo les in bu lk mat te r . At the beg inn ing of the exper imen t . SW is

closed and t h e r e is zero cu rren t th rough the iV-turn superconducting coi l C. A sample

of ma tte r SA, with a m a s s of the o rder of one k i log ram, is m a d e to follow a closed path

P t ravers ing C.

Imag ine tha t th e s a m p l e c o n t a i n s one m o n o p o l e of charge e* weber wi th rad ia ll ines of B. As it a p p r o a c h e s a given turn of C . th e flux through that turn is e*/2 we b e r

i i

O , « . p

SA

Figure 12-17 M a g n e t i c m o n o p o l e d e t e c t o r . The sample SA is p laced on a conveyor

and fol lows path P with th e switch SW closed. After hundreds of passes , the switch

is opened and the vo l tage pu lse across R is observed on an osci l loscope. The coil C

i s superconduct ing . See P r o b . 12-6.




to the r ight , and, a moment la ter , i t i s e*/2 weber to the lef t . Thus , on emerging a t the

r igh t - hand end of C, t he flux l inkage th rou gh C due to the mo no po le has inc reased by

Ne*. Ho we ver, the res is tanc e of the coi l i s zero an d the indu ced cu rren t keep s the f luxl inkage ze ro .

Ca lcu la t e the cur ren t pe r mo no po le a f t er 100 pas ses th roug h a co i l o f 1200 tu rns

with a se l f- inductance of 75 mil l ihenrys .

The cur ren t i s de t ec t ed by opening the swi t ch SW and obse rv ing the vo l t age

pulse on R.

Al l the measurement s to da te have g iven nu l l r e su l t s .

ROGOWSKI COIL

Figure 12-18 shows a Rogowski co i l sur rounding a cur ren t -ca r ry ing wi re . The

coi l has N t u rns , and r » r'. Rog owsk i co il s a re used for measu r ing l a rge f luc tua t ingcur ren t s , pa r t i cu la r ly l a rge cur ren t pu l s es .

a) If the value of R i s chosen l a rge enough , V i s unaffected b y that p art of the

circui t that i s to the r ight of the vert ical dot ted l ine . Show that

Nr'2

dl

2Vdl'(1 )

Show that , i f C i s now chosen l a rge eno ugh to mak e V « V, t hen

Nr'2

I

V=

Po - 2r RC12)

Figure 12-18 Rogowski coi l for measuring a heavy current pulse in the axia l wire .

Wi th the RC c i rcu i t shown, the vo l t age V i s p r o p o r t i o n a l t o I. See Prob. 12-7.



Problems 303

See Prob. 5-30. The product PC must be large compared to l/f, where / is the re

petitio n frequency, dl/dt being of the order of If. The current I need not be sinusoidal.

Solution: Under those conditions, V is the induced electromotance:

dcp d ( , u0I\V = N— = N — T t r '

2

— , 3

dt dt V 2nr)

_ ^ r ' 2 d l

2r dt'

Thus the mutual inductance is n0Nr'2/2r.

Al so ,

dQV = PR + V % PR = — R, (5)

dt

dV

* P C — . (6)

Comparing now Eqs. 4 and 6,

F « u 0 . 7)' ° 2r PC

We have set V equal to zero when / is zero. For example, the current / could

be in the form of short pulses and, between pulses, both / and V would be zero.

b) Show that, if the capacitor is shorted and if R is made small so that V is de

creased by a factor of, say, 20 or more, then F * I/N.

The self-inductance of the coil is

L « p0N2

r'2

/2r. (8)

Solution: The induced electromotance is equal to the sum of the voltage drops

across the induct ance L of the coil and the resistance P :

u0

Nr'

2

dl dP™- - r = L — + RI'. (9)2r dt dt

If P were very large, then RI' would be nearly equal to the induced electro

motance. On the other hand, if R is small, we may set

L' E v ^ l

d

l u o ,dt 2r dt




Substituting now the value of L,

Li 0N2

r'2

dl' n0Nr'2

dl

— x— . (11)2r dt 2r dt

1

'

I' x I/N. (12)


An azimuthal curren t is induced in a conduc ting tube of length /, average radius

a, and thickness b, with b « a.

a, Show that its resistance and induct ance are given by

R = 2na/(jbl, L = [i0na2j\.

b) Calculate L, R, L/R for a copper tube (a = 5.8 x 107

S i emen s per meter),

one meter long, 10 millimeters in diameter, and with a wall thickness of 1 millimeter.

12-9 COAXIAL LINES

Coaxial lines, as in Fig. 12-19, are widely used for the interconnection of elec

tronic equipment and for long-distance telephony. They can be used with either direct^

or alternating currents, up to very high frequencies, where the wavelength c/f is of

the same order of magnitude as the diameter of the line. At these frequencies, of the

order of 10 1 0 hertz, the field inside the line becomes much more complicated than that

shown in the figure and quite unmanageable.

There is zero electric field outside the line. Since the outer conductor carries the

same current as the inner one, there is also zero magnetic field, from Ampere's circuital

law (Sec. 9.1).

A coaxial line acts as a cylindrical capacitor. We found its capacitance per

meter in Prob. 6-5:

_ 2nere0

~ In (Rt/Rj

where Rx and R2 are the radii of the inner and outer conductors, respectively.

/

Figure 12-19 Coaxial line. See Prob. 12-9.



Problems 305

A coaxial l ine is a lso indu ct ive , s ince there is an a zim uth al m agn et ic f ield in

the annula r space be tween the two conduc tors .

Show that the se l f- inductance per meter i s .

12-10 COAXIAL LINES

I t i s kno wn th a t h igh- f requency cu r ren t s do no t pene t ra t e in to a con duc tor a s

do low -frequen cy cur ren ts . Th is is cal led the skin effect. See Pro b. 11-7.

W ou ld you exp ect the se l f- inductance of a coaxial l ine to increase or to decre ase

wi th inc reas ing f requen cy?

12-11 LONG SOLENOID W ITH CENTER TAP

Cons ide r a long so lenoid wi th connec t ions a t bo th ends , A and C, and w ith a

cen te r t ap B. The two halves are in ser ies , and thus

w h e r e M i s t he i r mu tua l i ndu c tance .

N ow . if on e uses Sec. 12.3.1 to calcula te th e three L 's , one f inds tha t M = 0.

Th is is ab sur d be cause th e coeff ic ient of cou pl in g is surely not ze ro.

Can you expla in th i s s t range re su l t ?

12-12 TRANSIENT IN RLC CIRCUIT

The switch in the c i rcui t of Fig. 12-20 is ini t ia l ly open, and then c losed a t t = 0.

LAc = L AB + L BC + 2 M ,

C


Draw a curve ofQ as a funct ion off for V = 100 vol ts , L = 1 henry , C = 1 m i c r o

farad, R = 8000 ohms , w i th t varying from 0 to 20 mil l iseconds .




Solution : F ro m Ki rchof f ' s vo l t age l aw .

dl QRI + L h —

dt C1.

or, since / = dQjdt,

d 2Q dQ QL ~ + R-±- + ~ = V.

dt 1 dt C

The pa r t i cu la r so lu t ion of th i s equa t ion i s

Q = CV = 1C T 4,

and the comp lem enta r y func t ion i s t he so lu t ion of

d 2Q dQ QL ~ + R — + —

dt2

dt C

Let us set

Q = Ae"'

Then, subs t i tu t ing in Eq . 4 and d iv id ing by Q,

0.

1Ln 2 + Rn + -

C0 .

R

2L

R2 1

4L2 ~ LC

Note tha t t he re a re two va lues o f n, say n, and n 2. T h e n

Q = Ae"" + Be" 2'.

(1)

(2)

(3)

(4)

15)

(6)

(7)

(S )

w h e r e A a n d B a r e u n k n o w n c o n s t a n t s o f i n t e g r a t i o n .

H e r e R/2L = 400 0. the br ac ke t of Eq . 7 is 1.5 x 1 0 7 , and the general solut ion is

Q = e - 4 0 0 0 1 1 ^ 3 8 7 3 !+ Be" + 10"

= Ae~ 121' + Be- lsli' + 10"

S ince (2 = 0 a t r = 0 . A + B = - 1 0 " 4 . Also,

dQ

dt\21Ae' 7873Be"

(9)

(10)

( I D

http://21ae%27/

http://21ae%27/



30 7

1 . 5 f x 1 0 " 4

0 10 20 msf

Figure 12-21 The cha rge on the capac i to r o f F i g . 12-20, as a function of the time

for various values of R. Th e t ime is expre ssed in mil l i secon ds . W he n the res is tan ce

i s sma l l , t he cha rge ove rshoot s and osc i l l a t e s about i t s a sympto t i c va lue .

a n d I = 0 at r = 0, so th at

127/4 + 78 73 5 = 0, (12)

A = - 1 . 0 1 6 x 1 0 " 4 . B = 1.640 x 1 0 " 6 . (13)

Q = - 1 . 0 1 6 x 1 0 "4

e "1 2 7

' + 1.640 x l O "6

? "7 8 7 3

' + 1 0 "4

. (14)

As a check, we can show that L dljdt = 100 at f = 0. At tha t ins ta nt , 7 = 0

and the vo l t age drop on R i s zero.

F igure 12-21 shows Q as a funct ion of t for R = 500, 2000 , and 8000 oh ms . W i th

R = 2000 ohms , the bracke t i s ze ro . Th e r igh t -h and s ide of Eq . 8 then becom es

(A + Bt) e x p nt . W i t h R = 500 ohm s , the bracke t is nega t ive and we aga in ha ve

anothe r type of so lu t ion .

12-13 VOLTAGE SURGES ON INDUCTORS

I f t he cur ren t t h roug h a l a rge indu c tor i s i n t e r rup ted sudd enly , dljdt is large

and t rans i en t vo l t ages appea r ac ros s the swi t ch an d ac ros s the co il . The vo l t ages can



30 8

Figure 12-22 Ind uc to r L and i t s

re s i s t ance R fed by a source V

t h rough a swi t ch represen ted he re by

a res is tor R s. See P rob . 12-13.

Ia L

9

be l a rge enough to damage bo th . Care must therefore be taken to reduce the current

slowly. This is pa r t i cu la r ly im por t a n t when us ing l arge e l ec t rom agne t s . No prob lem

arises , however, on connect ing the source to the inductor . See Sec. 12.7.1.

Figure 12-22 shows the induc tor L and i ts res is tance R fed by a source V. T h e

res is tance R s plays the role of the switch. At first . R s is zero and / = VjR. T h e n R s

i s quick ly increased to some value muc h larger than R.

Set L = 10 hen rys . R = 1 0 o h m s . V = 100 volts , dljdt = - 1 0 3 a m p e r e s p e r

second .

a) Show that the vol tages across the switch and across the inductor both r iseto about 10 k i lovol t s .

Of course the insulat ion of the switch and the insulat ion between turns on the

ind uct or break d ow n mu ch before the vol ta ge can r ise to such a high valu e.

b) I t i s poss ible to suppress the t rans ient by us ing a diode. A diode i s a two-

terminal device that has e i ther a very low or a very high res is tance, depending on the

polari ty of the appl ied vol tage, as in Fig. 12-23. In other words , a diode wil l pass a

curr ent in only one dir ect ion . Of course , if the inv erse vol tage is suffic ient ly high,

current wi l l pass in the forbidden direct ion and the diode wil l be damaged.

Figure 12-23 The symbol fo r a d iode

is a t r iangle an d a ba r . In (a), th ediode has a low res is tance and can

usual ly be cons idered as a short j

cir cu it. In (/>), the re sis tan ce is hig h ' = 0

an d is usual ly con s idere d to be • . I

infinite. The curve of / as a function + * ^1 *_

* ^1

of the appl ied vol tage is non-l inear ,

even for forward vol tages . UT) (/?)



Problems 309

A given type of d iod e can be used up to a ra ted m axim um cur ren t , and up to a

ra t ed maximum inve rse vo l t age .

You are given a diode that i s ra ted a t 20 amperes and 200 vol ts peak inverse

vol t age . How should i t be used?

W ha t is the cur ren t thr ou gh the diod e as a funct ion of the t ime if the switch is

ope ned at f = 0?

W ha t hap pe ns to the energy s tor ed in the ma gn et ic f ie ld of the indu cto r af ter

the swi t ch i s opened?

12-14D TRA NSIENT IN RLC CIRCUIT

One wishes to cha rge a capac i to r a s qu ick ly a s pos s ib l e . The source suppl i e s

an open-c i rcu i t vo l t age V s and has an internal res is tance R (Sec. 5.12), as in Fig. 12-24a.

I t i s sugge s ted that the ra te of cha rge could be increased by addin g an i nd uct or

with L = R 2C/4 in ser ies wi th the capaci tor as in Fig. 12-24b.

Dr aw cu rves of the vo l t age V across the capaci tor as a funct ion of t ime for F s =100 vol ts , R = 100 ohms , C = 1 mic rofa ra d , w i th and wi tho ut the indu c tor .

Is the sugges t ion val id ?

O - i O - i

c V S - ±

3

(b)

Figure 12-24 Capaci tor C charged by a source F s having an in t e rna l re s i s t ance R.

T h e i n d u c t o r L in {b) has been added in the hope tha t t he capac i to r w i l l cha rge

more rap id ly . See P rob . 12-14 .

I2-I5D TRANSIENT IN RLC CIRCUIT

In the c i rcui t of Fig. 12-25 the switch is ini t ia l ly open. The capaci tor i s charged

to the vo l t age V and no cur ren t i s d raw n f rom the source .

Se t V = 100 volts , L = 1 henry , R = 10 ohm s , C = 1 mic rofa rad ,

a) T he switch is c losed. Show that

/ = 10(1 - e'



31 0


L R

b) After a t ime t » L!R, the switch is ope ned . Show that , f rom tha t t ime on,

Qx - 1 0 ~ V 5 ' cos 10 3 r + 1( T 4 .

I * O . l e " 5 ' sin 10 3f.



CHAPTER 13

MAGNETIC FIELDS: VI

Magnetic Forces

13.1 FORCE ON A WIRE CARRYING A CURRENT IN A

MAGNETIC FIELD

13.1.1 Example: The Hodoscope

13.2 MAGNETIC PRESSURE

13.3 MAGNETIC ENERGY DENSITY

13.4 MAGNETIC ENERGY

13.4.1 Example: The Long Solenoid

13.5 MAGNETIC ENERGY IN TERMS O F THE CURRENT I

AND OF THE INDUCTANCE L


13.6 MAGNETIC FORCE BETWEEN TWO ELECTRIC

CURRENTS13.6.1 Example: The Force Between Two Infinitely Lon g Parallel Wires

13.7 MAGNETIC FORCES WITHIN AN ISOLATED CIRCUIT

13.8 SUMMARY

PROBLEMS



In Ch ap te rs 4 an d 7 , we studie d the energ y sto red in an elect r ic f ie ld , an d

the r esu l t ing e l ec t r ic fo rces . The n , in Ch ap te r 10 , we s tud ied the ma gne t i c

fo rces on ind iv idua l charge d par t i c l es . N ow w e mu st s tudy mag ne t i c energy

and macroscop ic magne t i c fo rces . Magne t i c fo rces a r e , i n p rac t i ce , so much

la rger than e l ec t r i c fo rces tha t t hey l end themse lves to many more app l i ca

t ions . For example , e l ec t r i c motor s a lmost invar i ab ly use magne t i c fo rces!*

We con t inue to assume tha t t he mate r i a l s in the f i e ld a r e non

magne t i c . We sha l l dea l wi th the magne t i c fo rces exer t ed by e l ec t romagne t s

in Chap te r 15 .

13.1 FORCE ON A WIRE CARRYING A CURRENT

IN A MAG NETIC FIELD

Consider a wi r e o f c ross - sec t iona l a r ea a, carrying a current / as in Fig . 13-1.

T h e r e a r e n con du c t ion e l ec t rons per un i t vo lum e, each one ca r ry in g a charge

— e at a velocity v.

I f the wire i s s i tua ted in a m ag ne t ic f ie ld B , due to cu rre nts f lowing

e l sewhere , t he magne t i c fo rce on one e l ec t ron i s — ev x B (Sec. 10.1) and,

f Al thou gh e l ec t ros t a t i c moto rs a re exceeding ly ra re , t hey have been s tud ied ex tensive ly .

See, for example , A . D . Moore , Edi to r , Electrostatics and Its Applications, John Wi ley ,

New York , 1973 .



31 3

Figure 13-1 Wire of cross-section a ca r ry ing a cur ren t / . The cha rges — e m o v e

at a veloci ty v. Th e m agn et ic f ield B i s du e to curre nts f lowing elsewh ere .

for a length dl c o n t a i n i n g na dl c o n d u c t i o n e l e c t r o n s ,

d F = nadl(-ev x B). (13-1)

N ow th e charg e pass ing th rou gh a g iven c ross - sec t ion per second i s

the charge on the ca r r i e r s con ta ined in a l eng th v of the wire and

I = — nae\. ( 1 3 - 2 )

T h e n

d F = Idl x B = / d l x B, (13-3)

w h e r e dl i s in the di rect ion of the current .

Fo r a s t ra i ght leng th / of wire car ryi ng a cur ren t / in a uniform ma gn et ic

field B.

F = II x B (13-4)

F igu re 13-2 shows the l ines o f B nea r a cu r r en t - ca r ry ing wi r e , pe r pe nd icu lar to a uniform ma gn et ic f ield. The force on a length / i s the n simply

IIB, i n the d i r ec t ion shown.

As for electr ic f ields, i t is useful to imagine that lines ofB really exist ,

a r e under t ens ion , and r epe l each o ther l a t e r a l ly . Th i s s imple mode l g ives

the correct d i rect ion for magnet ic forces, as can be seen f rom Fig. 13-2.



31 4

(a) (b)

Figure 13-2 Magnet ic f ie ld near a current-carrying wire s i tuated in a uniform

magnet ic f ie ld, (a) Lines of B for a cur ren t pe rpendicu la r t o the pape r , (b) Lines of

B for a unif orm field par allel to th e pape r, (c) Su pe rp os itio n of fields in (a) an d (b)

and the resul t ing mag net ic force F . Th e wire carr ies a cur ren t of 10 am per es , the

uniform fie ld has a B of 2 x 1 0 ~4

tes la , and the point where the l ines of B are

bro ken is a t 10 mil l im eters from th e center of the wire .

EXAMPLE: THE HODOSCOPE

T h e hodoscope i s a device that s imu lates the t ra jec tory of a charge d p art ic le in a

magnet ic f ie ld. The principle involved is s imple: i f the charged part ic le , of mass m,

c h a r g e Q, and veloci ty v, is replaced by a light wire fixed at the two ends of the tra

j ec tory and ca r ry ing a cur ren t I, the wire will follow the trajectory if

mv/Q = T/L (13-5)



315

Figure 13-3 (a) C h a r g e Q moving at a veloci ty v in a magnetic f ield B. The rad ius

of curvature of the t rajectory is R t. (b) Light wire carrying a current / in theopposi te direct ion in the same magnetic f ield . The tension in the wire is T, and

its radius of curvature is R w. It is shown that the wire has the same radius of

curvature as the t rajectory i f mv/Q is equal to T/I. The ang le dQ is infinitesimally

small .

where T is the tension in the wire. This s tatement is by no means obvious, but we

shal l demonstrate i ts val id i ty .

The advan tage o f the so -ca l led floating wire lies in the fact that it is much easier

to experiment with a wire than with an ion beam.

Let us cons ider a region where the ion beam is perp end icula r to B as in Fig. 13-3a.

Then the charge Q is subjected to a force QvB. T h u s

QvB = mv 2/R„ (13-6)

wh e re R, is the radius of curvature of the t rajectory, and

R, = mv/BQ. (13-7)

In Fig . 13-3b we have replaced the charged part icles by a l ight wire carrying a

current / f lowing in the opposite direct ion. If the element dl is in equil ibrium, the

ou tward fo rce BI dl i s com pens a ted by the inward com pon en t o f the t ens ion fo rces T:

Bl dl = IT si n (dd/2) = IT dO/2 = T dl/R w. (13-8)

Fhe radius of curvature of the wire is thus

Rw

= T/BI, (13-9)

and the two radi i of curvature wil l be the same if

mv/Q = T/l. (13-10)



316 Magnetic Fields: VI

Fo r exam ple , if we have a p ro to n (in = 1.7 x 1 0 " 2 7 ki logram) beam wi th a

kinet ic energy of 10 6 e lec t ron-vol t s (10 6 x 1.6 x 1 0 " 1 9 j ou le ) , t hen T/I i s about

0.15 and. if / = 1 am per e , T i s 0.15 new ton.

Note that the part ic le wi l l be deflected downward i f the magnet ic force is down

ward, but the wire wi l l curve downward i f the force is upward. The magne t i c fo rces

mus t t he re fore be d i rec t ed in oppos i t e d i rec t ions .

If the magnet ic f ie ld is not uniform, and i f the beam is not perpendicular to B,

then the wire does not a lways fol low the t ra jectory. For example , magnet ic f ie lds

are often used both to deflect and focus an ion beam. In such cases the focus ing

forces on the bea m can b eco me defocu s ing forces on the wire , which is then deflected

away from the t ra jectory.

13.2 MAG NETIC PRESSURE

In Ch ap te r 10 we cons idered the mag ne t i c fo rces ac t ing on ind iv idua l

particles, and we have jus t seen how on e can ca lcu la t e the fo rce on a c u r r en t -

c a r r y i n g wire. I f , now, we have a current sheet, it is ap pro pr i a t e to th ink in

t e r m s o f m a g n e t i c pressure.

Let us imagine a f la t cur rent sheet car rying a a m p e r e s p e r m e t e r a n d

si tu ated in a uniform tan gen t ial mag ne t ic field B/2 du e to cu rre nts f lowing

elsewhere, as in Fig . 13-4a, wi th a normal to B.

W e a d j u s t a unt i l i t produces an aiding f ie ld B/2 on one side and anop po sin g field B/2 on the othe r s ide, as in Fig . 13-4b. Th en , f rom P ro b. 9-3,

a = B'p 0 and the force per uni t area, or the pressure, i s ot. t imes the f ield B 2

due to the currents f lowing elsewhere, or B 2/2p 0.

S o, whe neve r one has a cu r r e n t flowing th rou gh a con duc t ing shee t ,

w i t h a m a g n e t i c i n d u c t i o n B on one s ide and ze ro magne t i c induc t ion on

the other s ide, the sheet i s subjected to a pressure B 2/2p 0. This p ressure i s

exer ted in the same direct ion as i f the magnet ic f ie ld were replaced by a

c o m p r e s s e d g a s .

We have ar r ived at th is resul t for a f la t cur rent sheet , but the same

appl ies to any current sheet when B = 0 on only one side.Now we saw in Sec . 13 .1 tha t , qua l i t a t ive ly , one can exp la in magne t i c

forces by as sum ing th at the l ines of B (a , repel l each ot he r la teral ly , a nd

(b) a r e und er t ens ion . Th e ma gne t i c p r essu re we have here i s c l ea r ly asc r iba b le

to the l a t e r a l r epu l s ion , s ince the l ines a r e rough ly para l l e l t o the cu r r en t

shee t .



317

Figure 13-4 (a) A current sheet carries a current density of a amperes per meter of

its width and is situated in a uniform magnetic field B/2 due to currents flowing

e l sewhere , (b) The total magnetic field is that of (a), plus that of the current sheet.

By adjusting a so that it just cancels the magnetic field on the left-hand side of the

sheet , we have a total field B on the right.

Since the lines are under tension, there should also be a force of attrac

tion in the direction of B . This is correct, as we shall see later on in Sec. 15.3. In

fact, along the lines of B, the force per unit area is again B2

/2p0.

13.3 MAGNETIC ENERGY DENSITY

In Sec. 4.5 we found that, in an electrostatic field, the force per unit area on

a conductor and the energy density in the field are both equal to e 0E2

/2.

No w we hav e just found tha t the mag net ic pressur e is B2

/2fi0. One might

therefore expect the energy density in a magnetic field to be equal to B2

/2[i0.

This turns out to be correct/

Figure 13-5 shows the magnetic pressure, or the magnetic energy

density, B2

/2ti0, as a function of B.

+

Electromagnetic Fields and Waves, p. 367.

31 8

1 01 0

10 8

p(pascals) 10

6

'T O 4 -1 l l./"T 1 1 1

102

1 1 1 1 1 1 1

10"* 1 0 "3

1 0"2

10 "' 1 10 102

103

B (teslas)

Figure 13-5 M a g n e t i c p r e s s u r e B2/2fi0 as a funct ion of B. The magne t i c p res sure

is equal to the magne t i c ene rgy dens i ty . A tmospher i c p res sure at sea level is a b o u t

1 05 pasca l s .

13.4 MAGN ETIC ENERGY

Since the magnetic energy density is B2

/2u 0, the energy stored in a magnetic

field is

2Ho

whe re the integral is evalu ated over all space.

$yB2

ch, (13-11)

13.4.1 EXAMPLE: THE LONG SOLENOID

The magne t i c ene rgy s to red in the field of a l ong so lenoid of Af t u rns , r ad ius R, and

length /, ca r ry ing a c u r r e n t / is

2/i<A / /inR

2

l). (13-12)

lipN 2nR 2

21(13-13)

T h e m a g n e t i c p r e s s u r e is

1 [ti0NlV p 0N2

(13-14)



13.5 Magnetic Energy in Terms of Current 1 and Inductance L 319

13.5 MAGNETIC ENERGY IN TERMS OF THE

CURRENT I AND OF THE INDUCTANCE L

I f one compares Eq. 13-13 for the magnet ic energy stored in the f ie ld of a

sole noid w i th Eq. 12-16 for the indu cta nc e of a sole noid , one f inds th at

M ^ W i L / 2 (13-15)

This i s a general resul t : the magnet ic energy stored in the f ie ld of a c i rcui t

o f induc tance L ca r ry ing a cu r r en t I is \LI2

.

Yo u wil l recal l f rom Sec. 4 .3 tha t the energy s tore d in th e elect r ic

f ie ld of a capaci tor of capaci tance C charged to a vo l t age V is \CV2

.If one ha s two c i r cu i t s wi th induc tanc es L a a n d L fc , a n d h a v i n g a m u t u a l

i n d u c t a n c e M , t h e n t h e m a g n e t i c e n e r g y i s

W m = ~ LJ2

a + ~ LbI2 + MlJh. (13-16)

Th is is sho wn in Pr ob . 13-16.

13.5.1 EXAMPLE: COAXIAL SOLENOIDS

We can eas i ly veri fy the above equat ion for coaxial solenoids .

We f i rs t evaluate the r ight-hand s ide . From Eqs . 12-16 and 12-7,

- LJ 2 + - L„I2

b + MIJ„

1 fi 0N2

nR2

1 n 0N2

bnR2

H 0N aN bnR2

ti 0nR2

fN2

I2

N2

I2 2N aN bIaf

+ —\ r2 \ l„ • l„ ' la

(13-17)

(13-1J

To ca lcu la t e W m, we use the fact tha t the energy dens i ty is B 2/2[i0. Ov er the l eng th

lb we have a tota l B of

>A 'B




giv ing a magne t i c ene rgy

1 2 (N J a N„Ib\2

2p -o \ L >b JOver the length /„ — lb, we have only the f ie ld of coi l a and a magnet ic energy

2 v / . '

Then the magnet ic energy s tored in the f ie ld of the coaxial solenoids is

II,"ii 07iR

2

(N2

aI2

lb Njll N.NJJt Nil2

N1

!2

!,

2 V \l lb I. /. W. (13 m

which reduces to the express ion on the r ight-hand s ide of Eq. 13-1S

13.6 MAGNETIC FORCE BETWEEN TWO

ELECTRIC CURRENTS

I t i s w e l l k n o w n t h a t c i r c u i t s c a r r y i n g e l e c t r i c c u r r e n t s e x e r t f o r ce s o n e a c h

o t h e r . I n F i g . 1 3 - 6 , t h e f o r c e e x e r t e d by c i r c u i t a on c i r c u i t b is o b t a i n e d b y

i n t e g r a t i n g E q . 1 3- 3 o v e r c i r c u i t b:

Kb = Ib(f bd\h x Ba

(13-20)

w h e r e B i s d e f i n e d a s i n E q . 8 - 1 .



13.7 Magnetic Forces Within an Isolated Circuit 321

EXAMPLE: THE FORCE BETWEEN TWO INFINITELY

LONG PARALLEL WIRES

The force between two inf ini te ly long paral le l wires carrying currents /„ and Ib,

s e p a r a t e d by a di s t ance p, as in Fig. 13-7, is ca lcu la t ed as follows. The force act ing

on an e lement lh d l b is

d F = / „ ( d l t x BJ,

dF = Ib_d lbii 0lJ2np,

(13-21)

(13-22)

a n d th e force per uni t length is

dF— = p 0IJ b/2np.dU

(13-23)

The force is a t t rac t ive if the c u r r e n t s are in the same d i rec t ion ; and it is repuls ive if

t hey are in oppos i t e d i rec t ions .

Figure 13-7 Two l ong pa ra l l e l w i res ca r ry ing cur ren t s in the same d i rec t ion . The

e lement of force d F ac t ing on e l e m e n t dl fc is in the di rec t ion shown.

13.7 MAGNETIC FORCES WITHIN

AN ISOLATED CIRCUIT

In an isolated circuit the current flows in its own magnetic field and is there

fore subj ected to a force. Th is is illu stra ted in Fig. 13-8 : the int era cti on of



32 2

x

R S

v W— f —

o — b — f ti)

•i i

4Figure 13-8 Schemat ic diagram of a ra i l gun. The bat tery A c h a r g e s , t h r o u g h a

res i s t ance R, t he capac i to r C, which can be made to d i s cha rge th rou gh th e li ne by

clos ing switch S. The role of the capaci tor i s to s tore e lect r ic charge and to supply

a very large current to the loop for a very short t ime. If s ide D i s a l lowed to move,

i t moves to the r ight under the act ion of the magnet ic force F.

t he cu r r en t / wi th it s ow n magn e t i c induc t ion B p rodu ces a fo rce tha t t ends

to extend the ci rcui t to the r ight .

Magne t i c fo rces wi th in an i so la t ed c i r cu i t a r e usua l ly ca lcu la t ed by

pos tu la t ing a smal l d i sp lacement and us ing the p r inc ip le o f conserva t ion

of energy. See Prob. 13-25.

Th e force on an eleme nt of wire of length dl c a r r y i n g a c u r r e n t I in a region

where the magne t i c induc t ion i s B , i s

If a cond uc t i ng shee t s i tua ted in a magn e t i c f ie ld ca r r i es a cu r r e n t

tha t cance ls the mag ne t ic f ie ld on one side, then i t i s subje cted to a magnetic

pressure B

2

/2p 0, w h e r e B i s t he magn e t i c ind uc t ion on the o the r s ide . Th i sexpression also gives the energy density in the magnet ic f ie ld .

T h e magnetic energy sto red in a m ag ne t ic f ie ld is ob tai ne d by inte

g ra t ing the energy dens i ty :

13.8 SUMMARY

dF = / dl x B. U3-3)

(13-11)



Problems 323

If the field is du e to an i nd uc tan ce L , the n

Wm = (1 2)LI2 (13-15)

or, if we hav e two in duc tanc es La and Lh with a mutu al in ductance M. then

Wm = (l/2)L„/ a

2

+ ll 21/ . , , / , ; + M / a / 0 . (13-16)

The magnetic force exerted by a circuit a on a circuit b is

Ff l f t

= h§hA\

hx B„. (75-20)

M agnetic forces also exist within an isolated circuit. The y are calculate d

by postulating a small displacement and using the principle of the con

servation of energy.

PROBLEMS

13-1E MAGNETIC FORCE

C a l c u l a t e th e magne t i c fo rce on an arc 50 mil l ime te rs long ca r ry ing a c u r r e n t

of 400 a m p e r e s in a d i r e c t i o n p e r p e n d i c u l a r to a uni form B of 5 x 1 0 "2 tesla.

H i g h - c u r r e n t circuit breakers often comprise coi ls that generate a magnet ic f ie ldto b low out the arc t ha t fo rms when the c o n t a c t s o p e n .


a ) F ind the cur ren t dens i ty neces sary to float a copper w i re in t he ea r th ' s magne t i c

field at th e e q u a t o r .

A s s u m e a field of 10~4 tes la . The dens i ty of c o p p e r is 8.9 x 10

3 k i l o g r a m s pe r

cubic me te r .

b) Will th e w i re b e c o m e h o t ?

T h e c o n d u c t i v i ty of c o p p e r is 5.80 x 107

S i emen s per mete r .

c) In w h a t d i r e c t i o n m u s t the cur ren t f low? See P r o b . 8 - 1 1.

d ) W h a t w o u l d h a p p e n if the exper iment were pe r formed at on e of the m a g n e t i c

poles?


C a l c u l a t e the force due to the earth 's magnet ic f ie ld on a hor i zonta l w i re

100 meters long carrying a c u r r e n t of 5 0 a m p e r e s due n o r t h .

Set B = 5 x 1 0 "5 t e s l a , p o i n t i n g d o w n w a r d at an angle of 70 degrees wi th th e

h o r i z o n t a l .

See P rob . 8 -11 .





Show that the tota l force on a c losed c i rcui t carrying a current / in a uniform

magnet ic f ie ld is zero.

13-5E ELECTROMAGNETIC PUMPS

The conduc t ion cur ren t dens i ty in a l i qu id me ta l i s

J f = ff(E + v x B ) ,

w h e r e a i s t he condu c t iv i ty , E is the electric field intensity, v is the velocity of the fluid,

a n d B i s the magnet ic induct ion, a l l quant i t ies measured in the frame of reference of

t h e l a b o r a t o r y .

Show that the magnet ic force per uni t volume of f luid is

ff(E + v x B ) x B .

For example , in crossed e lect r ic and magnet ic f ie lds , a conduct ing f luid is pushed

in the di rect ion of E x B. The field v x B t h e n o p p o s e s E. This is the principle of

o p e r a t i o n o f electromagnetic pumps.

13-6E HOMOPOLAR GENERATOR AND HOMOPOLAR MOTOR

Figure 13-9 shows a homopolar generator. I t cons i st s in a cond uc t in g d i sk ro t a t ing

in an axial magnetic field.

In one pa r t i cu la r case , B = 1 tesla, R = 0.5 meter , and the angular veloci ty is

3000 revolu t ions pe r minute .

Ca lcu la t e the ou tpu t vo l t age .

Homopola r gene ra tors a re inhe ren t ly h igh-cur ren t , l ow-vol t age dev ices .Ho mo po la r gen e ra tors us ing supe rcon duc t ing co il s an d wi th power ou tp u t s of s eve ra l

me gaw at ts are used for the puri f icat ion of me tals by e lect rolys is .

I f a vo l t age is app l i ed to the ou tpu t t e rm ina l s o f a hom op ola r gen e ra tor , i t

b e c o m e s a homopolar motor. H o m o p o l a r m o t o r s c a n p r o v i d e l a r g e t o r q u e s . T h e y a r e

su i t ab le , i n pa r t i cu la r , fo r sh ip propul s ion .

F igure 13-9 Ho m op ola r g ene ra tor .

See Prob. 13-6.



Problems 325

13 -7 HOMOPOLAR MOTOR

See Pr ob . 13-6. N ow con s ider the device i l lus t ra ted in Fig. 13-10. Wil l the wheel

turn when the switch is c losed? If so, in which direct ion and why?*

Figure 13-10 Spiral coi l . A vol tage is

appl i ed be tween th e ax i s and the

pe r iphe ry . Contac t t o the pe r iphe ry

i s ma de wi th a ca rbon brush l ike

tha t used on motors . Does the d i sk

tur n? See P r ob . 13-7. Th e d i sk c ould

be ma de wi th a p r in t ed-c i rcu i t boa rd .

13-8 MAGNETIC PRESSURE

So as to c lar i fy the concept of magnet ic pressure , le t us cons ider , f i rs t a s ingle

soleno id, and then a pair of coax ial solen oids .

a ) F igure 13- l l a shows a longi tud ina l s ec t ion th rough a so lenoid . The sma l l

e lemen t ident i f ied by vert ical bar s (1) pro du ces i t s own m agn et ic f ie ld, bo th outs id e

an d ins ide the solenoid, a nd (2) is s i tua ted in the f ield prod uc ed b y the res t of the wind ing.

fTYTYTVTYTYn TYTYTYri rTYTYlYTYTYTYT)U A 1A L AJ A IA L J J A I A I A J J U A l A J A l A i A i A L '

0 0 © 0 0 © 0 © © 0 © 0 © 0 © 0 0

(a)

(MXKDQXDQXKD0(Tj(IXJXD(D(]XD

fTYTYTYTYTYT)U A I A l A i A JA I A l A i A JT U A l A i A i A l A l A i A i /

Figure 13-11 (a) Lon gi tud ina l s ec t ion th r oug h a long so lenoid . In P ro b . 13-8 , we

consider the f ie lds c lose to the smal l e lement ident i f ied by vert ical bars , (b) Sect ion

through a pa i r o f coax ia l so l enoids .

f See the American Journal of Physics, Vo lum e 38, 1970, p. 1273.

Sho w the f irst f ie ld by me ans of sol id arrow s, and th e second o ne by mea ns of dash ed

a r r o w s .

We are of course thinking here of the fields infinitely close to an infinitely thin

winding .

Now can you expla in why the magne t i c p res sure i s B 2 2/< 0? ^ —

b) Fig ure 13-1 lb show s a pair of coaxia l solen oid s carryin g equal cu rren ts in

opposi te di rect ions . In this example we have three f ie lds .

Show, on e i ther s ide of the e lement , (a) sol id arrows for the f ie ld of the e lement ,

(b) dash ed arr ow s for the field of the rest of the inne r soleno id, (c) wavy a rro ws for the

fie ld of the outer solenoid.

W ha t is t he magne t i c p res sure on the inne r so leno id?

13-9E MAGNETIC PRESSURE

a ) Sh ow tha t t he mag ne t i c p res sure is abo ut 4 B 2 a t m o s p h e r e s . O n e a t m o s p h e r e

i s abo ut 10

5

pasca l s .b) Draw a log-log plot of the e lect r ic force per uni t area je 0E

2. in pascals, as a

funct ion of E. The maximum elect r ic f ie ld intens i ty that can be mainta ined in a i r a t

norm al t emp era tu re and pres sure is 3 x 1 06

vol ts per meter .

c) Discuss the s imilar i t ies and differences between the magnet ic pressure and

the e lect r ic force per uni t area .

13-10E MAGNETIC PRESSURE

I t i s poss ible to a t ta in very high pressures by discharging a large capaci tor

th rough a ho l low wi re .

a) Show that , for a thin tube of radius R ca r ry ing a cur ren t / . t he inward

magne t i c p res sure i s

p,„ = tt0I

2 Sn2

R2

.

b) Ca lcu late the pressure for a cur rent of 30 ki lo am per es in a tube on e mil l im eter

in d i amete r .

O n e a t m o s p h e r e i s a p p r o x i m a t e l y 1 0 5 pasca l s .

13-1 IE MAGNETIC PRESSURE

Magne t i c f i e lds a re used for pe r forming va r ious mechanica l t a sks tha t requ i re a

high power level for a very short t ime.

Fo r exam ple , if a light a lum inu m tube is inserted axia l ly in a solenoid, an d i f the

so lenoid , is suddenly con nec ted to a cha rged ca pac i to r , the induced e l ec t ro mo tance

</<t>/</f prod uc es a large curr en t in the tube, which c olla pse s un de r the m ag ne tic pre ssu re.The tube can act as a shut ter to turn off a beam of l ight or of soft X-rays .

Le t us ca l cu la t e the pres sure exe r t ed on a conduc t ing tube p laced ins ide a long

solenoid. I f the current / in the solenoid is increased gradual ly from zero to some

arbi t ra r i ly large value, the curr ent ind uced in the tube is smal l and the ma gn et ic pres sure

is negl igible . Let us assume that dl/dt i s so large that th e induced c urr ent in the tub e

ma inta ins zero ma gn et ic f ield ins ide i t. Th en th ere is a ma gn et ic f ield only in the an nu lar

reg ion be tween the so lenoid and the conduc t ing tube .

Problems 327

a) Calculate the pressure on the tube for B = 1 tes la .

bl W ha t would th e pressure be if the cond uct ing tu be were paral le l to the axis ,

bu t some d i s t ance away?

'3-12E ENERG Y STORAGE

C om pa re the energies per uni t vo lum e in (a) a ma gne t ic f ie ld of 1.0 tesla an d

(b) an elec tro stat ic field of 106

vol ts per meter .

'3-13E MAGNETIC PRESSURE

Imagine tha t t he cur ren t i n a long so lenoid i s ma in ta ined cons ta n t , whi l e the

magne t i c p res sure inc reases the rad ius f rom R to R + dR. Then the magne t i c ene rgy

mcreases by

a ) Show tha t t he m echanica l wor k d on e by the magn e t i c p res sure p„ , i s

2nRlp„, dR .

b) Sh ow tha t t he mcrease in magn e t i c ene rgy is equa l t o the mechanica l wor k

d o n e .

c ) Dur ing the expans ion , t he magne t i c induc t ion ins ide the so lenoid remains

c o n s t a n t b e c a u s e B depends on ly on the number of tu rns pe r me te r and on the cur ren t .

So, as the radius increases , the f lux l inkage a lso increases .

Show tha t t he ex t ra ene rgy suppl i ed by the source dur in g the expan s ion , l(N (!<£>),

i s j us t tw ice the mecha nica l work done .

We conclude that one half of the energy suppl ied by the source serves to perform

mechanica l work , and the o the r ha l f s e rves to inc rease the magne t i c ene rgy . See P rob .4-13.

13-14 FLUX COMPRESSION

Flux com pres s ion is one me tho d of ob ta in ing a mag ne t i c f ie ld w i th a ve ry

large B. For example , imagine a l i gh t conduc t ing tube s i tua t ed ins ide a so lenoid pro

ducing a s teady B 0 . Th e ann ul ar space be tween the tube and th e solenoid is fi lled

with an explos ive, and t he soleno id is re inforced ex ternal ly . If the tube is now imp lode d,

i t wi l l be crushed, an azimuthal current wi l l f low, and the internal magnet ic pressure

B2

/2u 0will bui ld up unt i l it is c lose to the extern al gas pr essure .

where 6 i s t he magne t i c induc t ion when the rad ius has been reduced to P .

b) Show that the surface current dens i ty in the tube must be about 10 9 a m p e r e s

per meter to achieve a field of 10 3 tes las .

c) If the initial B is 10 teslas. and if the tu be h as initially a diam ete r of 100 milli

me te rs , wha t should be the va lue of B when th e tube i s com pres sed to a d i am ete r o f

abou t 10 mi l l ime te rs ?

a) Sho w that , if the rad ius of the tube sh rink s very rap idly.

B * B0(R0!'Rf.




d) Ca lcu la t e the resul t ing increase in magne t i c ene rgy . Assume tha t th e cyl in

d e r is 200 mil l ime te rs long an d t h a t the c u r r e n t in the so lenoid is cons tan t . Neglec t

end effects.

F l u x c o m p r e s s i o n ca n also be achieved by m e a n s of a c o n d u c t i n g p i s t o n s h o t

axial ly into a so lenoid . y

If the ra diu s of t he so lenoid is R0. and if t he rad ius of the pis ton is K, the m a g n e t i c

i n d u c t i o n in the annula r reg ion be tween th e pi s ton and the so lenoid , becomes

S 0

B %(R/R0r

13-15E PULSED MAGNETIC FIELDS

Ext remely h igh magne t i c f i e lds can be o b t a i n e d by d i s c h a r g i n g a c a p a c i t o r

t h r o u g h a l ow- induc tance co i l . Th e capac i to r l eads mus t of course have a low i n d u c

t ance . Such fa s t capac i to rs cos t approxima te ly tw o d o l l a r s per j o u l e of s tored-ene rgycapac i ty .

a ) Es t ima te th e cos t of a capac i to r t ha t cou ld s to re an ene rgy equa l to t h a t of a

100-tes la magnet ic f ie ld occupying a v o l u m e of one l i ter.

b ) Es t ima te th e cos t of the elect r ic i ty required to c h a r g e th e c a p a c i t o r s .

c ) Ca lcu la t e the m a g n e t i c p r e s s u r e in a t m o s p h e r e s , at 100 tes las . One a t m o

s p h e r e is a b o u t 105

pasca l s .

13-16E MAGNETIC ENERGY

It was s t a t ed in Sec. 13.5 t h a t , if we h a v e tw o i n d u c t a n c e s La

and Lb, ca r ry ing

c u r r e n t s Iaan d Ib,

Wm =l

-LJ2

a +\Lbll + MIJb.

Y o u can prove this qui te eas i ly.

Let coil a pro du ce flux l inkages

Aaain coi l a an d Aab

in coil b.

Similar ly, let coil h produce f lux l inkages

a n d Abb in coil b.

a ) F ind Wm in t e r m s of the c u r r e n t s and of the f lux l inkages .

b) Verify th e a b o v e e q u a t i o n .

13-17E ENERGY STORAGE

a ) Show tha t th e magne t i c ene rgy s to red in an i sola ted c i rcui t carrying a

c u r r e n t / and with a f lux l inkage A is

Wm

= ( l / 2 ) /A .

b) Ca lcu la t e Wmfor a l ong so lenoid us ing th i s fo rmula .



Problems 329

13-18 ENERGY STORAGE

a ) A cons tan t -vo l t age source s e t a t a vo l t age V i s conn ec ted to an ind uc tance

L of negl igible res is tance a t t = 0.Fin d, as funct ions of the t ime , ( i) the cu rren t / . ( ii ) the energy suppl ied by t he

sourc e, and ( i ii ) the energ y s tor ed in the ma gn et ic f ie ld.

b ) A con s tan t -c ur ren t source s e t a t a cur ren t / is conn ec ted to a capac i to r C

at r = 0.

Find , as funct ions of the t ime, (i) th e vol tag e V on the capaci tor , ( i i ) the energy

supp l ied by the sour ce, and ( ii i) the ene rgy s tored in the e lect r ic f ie ld.

13-19 ENERGY STORAGE

Electr ical ut i l i t ies must be able to meet peak power demands . I t i s therefore

des i rable to s tore energy during periods of low demand and feed i t back into the grid

when needed . On e way of s to r ing ene rgy i s t o pu m p wa te r in to an e l eva ted re se rvo i r .

I t has been sugge s ted that large am ou nt s of energy cou ld a lso be s tore d in themagnet ic f ie lds of superconduct ing coi ls .

a) C om pa re the energy d ens i t ies in (i ) an e lect r ic f ie ld of 10 8 vol ts per meter in

Mylar , for which e r = 3.2. and (ii) a ma gn eti c field of 8 teslas. Fhe se fields are ab ou t a s

high as present technology wil l permit . Fhe magnet ic energy dens i ty is larger by two

o r d e r s o f m a g n i t u d e .

b) I t i s sugges ted that 10 gig aw at t - ho urs be s tored in a solen oid wi th a length -

to -d iam ete r ra t io o f 20 . Th e so lenoid wo uld be ma de in four pa r t s , j o ined en d to end

to form a quas i - toroidal coi l located in a tunnel excavated in sol id rock. Disregard the

toro ida l shape in your ca l cu la t ion . The m agne t i c indu c t ion w ould be 8 t e s la s . ( i) Ca lcu

l a t e the to t a l l eng th and the d i amete r , ( ii) Ca lcu la t e the magn e t i c p res sure in a tmo spher es ,

( ii i) Cal cula te the surface area tha t wo uld req uire cryogen ic ther ma l in sula t ion .

13-20 SUPERCONDUCTING POWER TRANSMISSION LINE

A supe rcon duc t ing po wer t ransmis s ion l ine has been pro pose d tha t wo uld

hav e the fol lowing chara cter is t ics . I t wo uld carry 1 0 " wat ts a t 200 ki lovo l ts DC o v e r

a dis tance of 10 3 ki lome te rs . Th e cond uc to rs would hav e a c i rcu la r c ros s- s ec t ion of

5 squa re cen t ime te rs and would be he ld 5 cen t ime te rs apa r t , cen te r t o cen te r .

a ) Ca lcu la t e the magne t i c fo rce pe r me te r .

I t so happens that the force of a t t ract ion is the same as i f the conductors were

replace d by f ine wires , 5 cen t im eters apa rt .

b ) Ca lcu la t e B 2 / 2 p 0 m i d - w a y b e t w e e n t h e t w o c o n d u c t o r s .

c) Ca lcu late the s tore d energy and i ts cos t a t 0.5 cent per ki low at t - ho ur.

Th e sel f- inductanc e of such a line is 0.66 mic roh en ry p er me ter .

13-21 ELECTRIC MOTORS AND MOVING-COIL METERS

Elec t r i c motors , a s we l l a s moving-co i l vo l tme te rs and ammete rs , u t i l i ze the

torque exerted on a current-carrying coi l s i tuated in a magnet ic f ie ld.

Cons ide r an N- turn rec t angula r co i l s i t ua t ed in a un i form magne t i c f i e ld a s

in Fig. 13-12.

a) C alc ul ate th e forces on sides 1, 2, 3, 4.

b ) Sho w tha t t he to rque

T = NIBab sin 0.



33 0

Figure 13-12 Rec tangula r co i l whose

n o r m a l n forms an angle 0 with a

uniform magnet ic f ie ld. See Prob.13-21.

In an e lect r ic motor , we have essent ia l ly a number of such coi ls , offset by, sa ;

15 degrees , aro un d th e axis, an d con nec ted t o a se t of co nta cts cal led a commutator

f ixed to the axis . Connect ion to the coi ls i s made by carbon brushes that r ide on th

c o m m u t a t o r . T h e c o il s a n d t h e i r c o m m u t a t o r f or m t h e armature of the motor . Th

magnet ic f ie ld is suppl ied by the stator coils. Th e mag net ic f lux is carr ied by lamina tci

i ron , bo th in the a rma ture and in the s t a to r yoke.

Moving coi l vol tmeters and ammeters use a uniform radial magnet ic f ie ld as i i

F ig . 13-13. The to rqu e i s t hen s imply NIBab and i s i ndepe nden t o f 0. A light spiraspr ing exe r t s a re s to r ing to rque tha t i s p ropor t iona l t o 0. giving a deflection 0 p r o p o r

t ional to / .

Figure 13-13 Moving co i l and mag

net ic f ie ld in a vol tmeter or ammeter .

Th e mag net ic f ie ld is supp l ied by a

p e r m a n e n t m a g n e t . T h e p a r t s m a r k e d

/ are made of i ron.



Problems 331

13-22 MAGNETIC TORQUE

Consider a square s ingle- turn coi l of s ide a, ca r ry ing a cur ren t / .

a) Show that i t tends to orient i t se l f in a magnet ic f ie ld in such a way that thetota l magnet ic f lux l inking the coi l i s maximum.

b) Show that the torque exerted on the coi l i s m x B, where m is the magnet ic

m om en t of the coi l (Sec. 8.1.2) , an d B is the ma gne t ic ind uct io n w hen th e curren t in

the coil is zero.

This resul t i s independent of the shape of the coi l .

13-23 ATTITUDE CONTROL FOR SATELLITES

See the prev ious prob lem.

Many sa t e l l i t e s requ i re a t t i t ude cont ro l t o keep them proper ly or i en ted . For

exam ple , com mu nica t ion s a t e ll i t es mus t keep the i r an te nn a sys t ems on t a rge t . A t t i t ude

cont ro l requ i re s a me thod for exe r ting appr op r i a t e to r ques a s they a re requi red .

At t i t ude cont ro l can be ach ieved to a ce r t a in ex ten t by means of co i l s whosema gne t ic f ie lds interact wi th that of the ear th.

a) Show that the torque exerted by such a coi l i s

NIBA sin 0.

w h e r e N i s the nu m be r of turn s , / is the cur ren t . B is t he magne t i c in duc t ion due to

the ea r th , A i s the area of the coi l , and 0 i s the angle between the earth 's magnet ic f ie ld

and the normal to the co i l .

b ) Ca lcu la t e the nu m be r of am per e - tu rns requi red for a co il wo un d a rou nd

the outs id e surface of a sa te l l ite whose dia me ter i s 1.14 me ters . Th e torq ue requ ired

at 0 = 5 degrees i s 1 ( F 3 newton-mete r . and the magne t i c induc t ion a t an a l t i t ude of

700 ki lome ters ov er the equa tor , wher e the orbi t of the sa te l l i te wi ll be s i tuated, i s

4.0 x 1 0 " 5 tes la .

13-24 M ECHANICAL FORCES ON AN ISOLATED CIRCUIT

Sho w tha t , if the geo me try of an isola ted ac t ive c i rcui t is a l tered, the e nergy

supp l ied by the sourc e is equ al to twice the increase in ma gne t ic energy. Th us the m e

chanica l work pe r form ed is equa l t o the inc rease in magn e t i c ene rgy . Assum e tha t t he

cur ren t i s kep t cons ta n t .

I t fol lows that , on this ass um pti on , the force on an e lem ent of an act ive c i rcui t

i s equa l t o the ra t e o f change of magne t i c ene rgy .

See Pr ob s . 13-13 and 13-25.

RAIL GUN. OR PLASMA GUN

Calculate the force on the movable l ink D in the circuit of Fig. 13-8.

Ins tead of being a metal l ic rod. the l ink can a lso be an e lect r ic arc . The device

then accelerates blobs of plasma and is cal led a plasma gun. State l l i te thrus ters can

be made in this way. See Probs . 2-14. 2-15. 5-3, and 10-11.




w h e r e F i s the driving force and v i s the veloci ty of the l ink. Note that both L and

/ are funct ions of the t ime.

Now what is the value of VI T h e v o l t a g e V (a) increases the flux linkage in the

induc tance and (b) g ives a vo l t age drop IR ' on the res is tor R':

(1)

(LI) + IR\ (2 )

w h e r e LI is the flux linkage.

C o m b i n i n g E q s . 1 a n d 2 .

(3)

2(4)

I-1(5)

LI 2. (6)2

Solution: We cannot so lve th i s p rob lem by in t egra t ing the magne t i c fo rce

J x B over the vo lum e of the l ink, s ince nei th er J n or B are kno wn , or even eas i ly

calculated. Ins tead, we shal l f ind the force by inves t igat ing the magnet ic and mechanical energies involved.

If we set x = 0 a t the ini t ia l pos i t ion of the l ink, then th e indu ctan ce L i n t h l T '

circuit is L 0 + L'x, w h e r e L'x i s the indu ctan ce asso cia ted w ith the ma gne t ic f lux

l inking the ra i ls . L' being the sel f- inductance per meter .

Dur ing the d i s cha rge of the capac i to r C we can neg lec t t he ba t t e ry and the

res i s t ance R. s ince the current f lowing in that part of the c i rcui t i s negl igible . Let

the res is tance on the r ight-hand s ide of the c i rcui t be only that of the arc . R' .

At any ins tant , the power suppl ied by the capaci tor serves to (a) increase the

magne t i c ene rgy \LI2. (b) increase th e kinet ic energ y, and (c) diss ipate en ergy in

the res is tance R' . T h u s



CHAPTER 14

MAGNETIC FIELDS: VII

Magnetic Materials

14.1 THE MAGNETIZATION M

14.2 THE EQUIVALENT SURFACE CURRENT DENSITY ete

14.3 THE EQUIVALENT VOLUM E CURRENT DENSITY J (,

14.4 CALCULATION O F MAGNETIC FIELDS ORIGINATING

IN MAGNETIZED MATERIAL

14.4.1 Example: Uniformly Magnetized Bar Magnet

14.5 THE MAGNETIC FIELD INTENSITY H

14.6 AMPERE'S CIRCUITAL LAW

14.6.1 Example: Straight Wire Carrying a Current I and Embed ded

in Magnetic Material

14.7 MAGNETIC SUSCEPTIBILITY /„„ PERMEABILITY p.

AND RELATIVE PERMEABILITY p r

14.8 THE MAGNETIZATION CURVE

14.9 HYSTERESIS

14.9.1 Example: Transformer Iron

14.10 BOUNDARY CONDITIONS

14.11 SUMMARY

PROBLEMS



Thus far we have studied only those magnetic fields that are attributable

to the motion of free charges. Now, on the atomic scale, all bodies containspinning electrons that move around in orbits, and these electrons also

produce magnetic fields.

Ou r pu rpo se in this chap ter is to express the magne tic fields of these

atomic currents in macroscopic terms.

Magnetic materials are similar to dielectrics in that individual charges

or systems of charges can possess magnetic moments (Sec. 8.1.2), and these

moments, when properly oriented, produce a resultant magnetic moment

in a macroscopic body. Such a body is then said to be magnetized.

In most atoms the magnetic moments associated with the orbital and

spinning m oti ons of the electrons cancel. If the cancellat ion is not complete,the material is said to be paramagnetic. When such a substance is placed

in a magnetic field, its atoms are subjected to a torque that tends to align

them with the field, but thermal agitation tends to destroy the alignment.

This phe nom eno n is anal ogo us to the alignme nt of polar molecules in

dielectrics.

In diamagnetic materials, the element ary mome nts are not per mane nt

but are induced according to the Faraday induction law (Sec. 11.2). All mate

rials are diamagnetic, but orientational magnetization may predominate.

Magnetic devices use ferromagnetic materials, such as iron, in which

the magnetization can be orders of magnitude larger than that of eitherpara- or diamagnetic substances. This large magnetization results from

electron spin and is associated with gro up ph en ome na in which all the

elementary moments in a small region, known as a domain, are aligned. The

magnet izatio n of one dom ain may be oriented at ran do m with respect to

that of a neighb oring doma in.

The re is one impo rt an t difference betwee n dielectric and mag net ic

materials. In most dielectrics D is propor tional to E and the medium is said



14.2 The Equivalent Surface C urrent Density a,, 33 5

to be linear. Fer romagne t i c mate r i a l s a r e no t on ly h igh ly non- l inear , bu t

the i r behav io r a l so depends on the i r p r ev ious h i s to ry . The ca lcu la t ion o f

the fi elds assoc ia t ed wi th m agn e t i c ma te r i a l s is t he re fo re l a rge ly emp i r i ca l .This wi l l be the subject of the next chapter .

14.1 THE MAGNE TIZA TION M

T h e magnetization M i s t he mag ne t i c mom en t per un i t vo lum e a t a g iven

point . I f m i s t he average mag ne t i c d ipo le mo me nt p er a tom , and i f N is the

n u m b e r o f a t o m s p e r u n i t v o l u m e ,

M = A/m, (14-1)

if t he ind iv idua l d ip o le mo m en t s in the e l ement o f vo lu me cons id ered a r e

a l l a l igned in the same d i r ec t ion .

Th e magn e t i za t ion M i s me asu red in am pere s per mete r , and it co r

r e s p o n d s t o t h e p o l a r i z a t i o n P in dielectr ics (Sec. 6.1).

14.2 THE EQUIVALENT SURFACE CURRE NT DENSITY ae

Fig ure 14-1 show s a cyl in der of ma ter i al d ivid ed in to squa re cells of cros s-

sec t iona l a r ea a2

, w h e r e a -» 0 . Imagine that each cel l car r ies a c lockwise

sur f ace cur r en t o f M amperes per mete r , a s in the f igure . Then each one-

me te r l eng th o f cel l has a ma gne t i c mo m en t o f a 2M a m p e r e m e t e r s q u a r e d

(Sec. 8.1.2) a n d t h e m a g n e t i c m o m e n t p e r c u b i c m e t e r is M a m p e r e s p e r

meter in the di rect ion shown in the f igure. Now the current in one cel l i s

cance led by the cu rre nts in the adjoi ning cel ls , excep t a t the pe r iphe ry of the

m a t e r i a l . T h u s , if t h e m a t e r i a l h a s a m a g n e t i z a t i o n M, t h e e q u i v a l e n t c u r r e n t

dens i ty ct e at the surface is M.

More genera l ly , t he equivalent, or Amperian, surface current density is

a , = M x n, , (14-2)

w h e r e n, i s a un i t vec to r no rm al to the su r f ace and po in t ing outward.

These equ iva len t cu r r en t s do no t d i ss ipa te energy because they do no t

invo lve e l ec t ron d r i f t and sca t t e r ing p rocesses l ike those assoc ia t ed wi th

c o n d u c t i o n c u r r e n t s .



336

Figure 14-1 A m p e r e ' s m o d e l for the equ iva len t cu rren t in a cy l inder of m a g n e t i z e d

mater ia l .

14.3 THE EQUIVALENT VOLUME CURRENT DENSITY Je

Let us now supp ose that the magnet iza tio n M is a function of the x-c oor din ate

as in Fig. 14-2, with

where p is a positive con stan t. Since M is expressed in amper es per meter,

p is expressed in amp ere s per squ are meter .

At the surface of the material, the equivalent current density «e

is

again given by Eq. 14-2, with M equal to the local value of the magnetization

Thus ae

is a function of x, as in th e figure.

Inside the material, currents parallel to the x-axis cancel.

Currents parallel to the y-axis do not cancel. Along any of the vertical

planes inside, the left-hand side of a given cell carries an upward current of

M = px, (14-3)

while the right-hand side carries a downward current of



33 7

y

X

Figure 14 -2 Mag ne t i zed cy l inde r in which M i s a funct ion of the x-c oor d in a te :

M increases l inearly wi th x as in Eq. 1 4 - 3 .

Along any ver t i ca l p l ane , we the re fo re have a ne t downward cur r en t o f

pa a m p e r e s .

N o t e t h a t t h is d o w n w a r d c u r r e n t i s i n d e p e n d e n t o f x. This i s s imply

b e c a u s e w e h a v e a s s u m e d t h a t M i s d i r ec t ly p rop or t i on a l to x .

So, a lon g every ver t i ca l d iv i s ion be tween ce l l s, we have a ne t d ow nw ard

cur r en t o f pa a m p e r e s . T h e n e t d o w n w a r d c u r r e n t d e n s i t y is t h u s p, since

w e h a v e pa am pere s for every in t e rva l a. In o th er words , we have an equ iva len t

cur r en t dens i ty

W e c a n n o w s h o w t h a t J e i s eq ua l to V x M in the fol lowing wa y.

Along the cu rve C of Fig. 14 -2, M • dl i s zer o over the lef t -hand e nd, b eca use

M i s pe rp end icu la r to dl. The same app l i es to the r igh t -hand end . Thus

J e = -pi (14-4)

(14-5)

= -pa. (14-6)

wh ere the line integral i s eva lua ted in the di re ct ion sho wn in the f igure.



338 Magnetic Fields: VII

Since the area of the loop is a x 1, t he y -componen t o f the cu r l o f

M is — p, from Sec. 1.12, a n d

je

= (v x rv iy. (14-7^

More genera l ly , t he equ iva len t vo lume cur r en t dens i ty i s

J e = V x M. (14-8)

14.4 CALCULA TION OF MAGNE TIC FIELDS

ORIGINATING IN MAGNE TIZED MATERIAL

Let us f ir s t recal l the reas on ing we fol lowed in Sec. 6 .3 to f ind t he elect r ic

f i e ld due to po la r i zed d ie l ec t r i c mate r i a l . We found tha t t he charge d i sp lace

me nt tha t occu r s on po la r i z a t ion r esu l t s i n the acc um ula t ion o f ne t dens i t i e s

o f b o u n d c h a r g e , a b on the sur faces and p h i n s ide . We then a rgued tha t an y

net charge density gives r ise to an electr ic f ield, exactly as if i t were in a

vacuum. As a consequence , E can be ca lcu la t ed a t any po in t in space f rom

a b a n d p b.

We now have an en t i r e ly ana logous s i tua t ion in the magne t i c f i e lds

o r i g i n a t i n g i n m a g n e t i z e d m a t e r i a l s . T h e m a g n e t i z a t i o n g i v e s e q u i v a l e n t

cur r en t s , wi th dens i t i e s aL, on the su r f aces and J e i n s ide . Thes e cur r en t s

p roduce magne t i c f i e lds , bo th ins ide and ou t s ide the mate r i a l , exac t ly as i f

they were s i tua ted in a vacuum.

In pr incip le , on e can calc ulat e B at any p oint in spac e f rom the equ iv

a len t cu r r en t dens i t i e s a e a n d J e. In pract ice , the vector M is a n u n k n o w n

func t ion o f the coord ina tes ins ide the mate r i a l , and one can do l i t t l e more

th an gues s it s valu e. W e shal l re t ur n to th is subject in the next ch ap ter .

EXAMPLE: UNIFORMLY M AGNETIZED BAR MAGNET

Th e uniformly m agn et ize d b ar ma gne t of Fig. 14-1 is a go od ex am ple to use a t this

po in t .

W e have shown in Sec. 14.2 tha t the cyl indrical surface carr ies an eq uivale nt

cur ren t dens i ty x e that i s equ al to M. Ins ide the magne t , V x M = 0 , and the vo lum e

current dens i ty J f is zero. Ove r the end faces. M is paral le l to the norm al uni t vecto r

n , , a n d a, i s zero.

14.5 The Magn etic Field Intensity H 339

The B field of a cyl inder uniformly magnetized in the direct ion of i ts axis of

symmetry is therefore identical to that of a solenoid of the same dimensions with

JV' tu rns per meter and car ry ing a cu rren t / such tha t IN' = M, as in Fig. 9-10.

In a rea l bar magnet the magnet ic moments o f the ind iv idua l a toms tend to

al ign themselves with the B field , so that the magnetizat ion M is weaker near the

ends . The end faces a l so car ry Amper ian cu rren ts , s ince M x n is zero only on the

axis . Th e net resul t is that th ere are tw o "poles ," one at each end of the ma gne t ,

from which l ines of B radiate in al l d irect ions outs ide the magnet . The poles are

mos t c onsp ic uous if the bar ma gnet i s long and th in .

14.5 THE MAG NETIC FIELD INTENSITY H

W e fou nd in Sec. 9.2 tha t , for a stea dy cu rre nt density of free ch ar ge an dfor n o n -m ag n e t i c m a te r i a l s ,

V x B = /< 0 J/- (14-9)

No w we have jus t seen tha t magn et ize d ma ter ia l can be rep laced by i t s

equiv alen t cu rre n ts for ca lcu la t ing B . Co nse que nt ly , i f we have ma gne t ized

mater ia l as wel l as s teady curren ts ,

V x B = nQ{i f + j e ) , (1 4 -1 0 ,

w h e r e J e i s the equ ivalen t vo lum e curre n t dens i ty . Th is equ at io n i s o f course

val id on ly a t po in ts wher e the deriva t ives of B wi th respect to the coo rd i nat es

x, y, z exis t . I t i s therefore no t ap pl ica b le a t the surface of m agn et ic ma ter ia l s

an d w e d i s reg a rd a (,.

Su b s t i t u t i n g V x M for J t > ,

V x B = p.0(J f + \ x M), (1 4 -1 1 ,

V x ^— — mJ = J f. (14-12 ,

Th e vector be tween pare n the ses , whos e curl i s equa l to the free c urre n t

density at the point , is the magnetic field intensity

H = — - M , (14-13 ,I h

and i s express ed in am pe res pe r m eter . N ot e tha t H and M are expressed

in the same uni t s .



340 Magne tic Fields: VII

T h u s

B = n0(H + M). (14-14)

This equat ion is to be compared with Eq. 6-15 , which appl ies to d ie lec tr icsrf

E = — (D - P). (14-15)

e 0

14.6 AMPERE'S CIRCUITAL LA W

Rew rit ing no w Eq. 14-12, we hav e tha t , for s teady curr ents , e i ther ins ide or

ou ts ide magne t ic ma te r ia l ,

V x H = J f. (14-16)

In tegra t ing over a surface S,

J" s(V x H ) d a = JsJ / - d a (14-17)

or , us ing Stoke 's theorem on the lef t -hand s ide ,

( f c H •«« = / „ (14-18)

w h e r e C is the curve bounding the surface S, a n d I r is the current of free

cha rges l ink ing the cu rve C. N o t e t h a t I f d o e s not inc lude the equ iva len t

currents . The term on the lef t is ca l led the magnetomotance. A m a g n e t o m o -

tance i s exp ressed in amp eres , o r in a mp ere - t u rns .

This is a more genera l form of Ampere's circuital law (Sec. 9.1), in

tha t i t can be used to ca lcula te H even in the presence of magnet ic mater ia ls .

I t is r igorously val id , however , only for s teady currents ; we shal l dea l with

va r iab le cu r ren t s in Chap te r 19 .

14.6.1 EXAMP LE: STRAIGHT WIRE CARRYING A CURREN T I AND

EMBEDDED IN MAGNETIC MATERIAL

Fi gu re 14-3 shows a s t r a i gh t wi re car ry i ng a cu r r en t / and emb edde d i n m agne t i c

mat er i a l . Th e ma gne t om ot an ce a ro und t he pa t h shown c i r c li ng t he cu r r en t I is

equal to / , and H = I/2nr, exact ly as if the wire were s i tuated in a v ac uum .



34 1

Figure 14-3 St ra igh t w i re ca r ry ing a cur r en t / and emb edd ed in ma gne t i c m a te r i a l .

I f the ma teria l i s i sot ro pic , and if i t has the shap e of a c i rcular cyl ind er wi th /

a lon g i ts axis , M is azi mu tha l , l ike H .

14.7 MAGNETIC SUSCEPTIBILITY xm, PERMEABILITY n,

AND RELATIVE PERMEAB ILITY \i r

As for dielectr ics (Sec. 6.6), i t is convenient to define a magnetic susceptibility

~/ m s u c h t h a t

N o w , s i n c e

t h e n

w h e r e

M = Z m H .

B = p 0 ( H + M),

B = jU 0(l + z J H = p 0p rH = pU,

P-R = 1 + Xm , / ' = M O / V

(14-19)

( 1 4 - 2 0 )

(14-21)

( 1 4 - 2 2 )

T h e q u a n t i t y p i s t he permeab i l i ty and p r is the relative permeability. B o t h

X m a n d p r a r e d i m e n s i o n l e s s q u a n t i t i e s .




Eq ua t io n 14-20 is gene ral , bu t Eqs. 14-21 an d 14-22 are base d on th e

ass um pt io n tha t t he mate r i a l is bo th i so t rop ic and l inear , i n o the r word s ,

tha t M i s p ro po r t io na l to H and in the same d i r ec t ion . Th i s ass um pt io n i su n f o r t u n a t e l y no t val id in fer romagnet ic mater ials , as we shal l see in

nex t sec t ion .

The magne t i c suscep t ib i l i t y o f paramagnetic substances is sma l l e r th an

un i ty by severa l o rder s o f magn i tude and i s p ropor t iona l to the inver se o f

t h e a b s o l u t e t e m p e r a t u r e .

In diamagnetic materials, t he mag ne t i za t i on i s i n the d i r ec t ion opposite

to the external f ie ld; the relat ive permeabi l i ty i s less t han un i ty and i s i nde

p e n d e n t of t h e t e m p e r a t u r e . If o r i e n t a t i o n a l m a g n e t i z a t i o n p r e d o m i n a t e s ,

the r esu l t an t r e l a t ive perm eab i l i ty i s g r ea te r than un i ty .

14.8 THE MAGN ETIZA TION CUR VE

O n e c a n m e a s u r e B for var ious values of H with a Rowland ring w h o s e

minor r ad ius i s much smal l e r than i t s majo r r ad ius , a s in F ig . 14-4 . The

func t ion o f win d ing a, which has N a t u rns and ca r r i es a cu r r en t / „ , i s t o

produce a known magne t i c f i e ld in t ens i ty

H = NJJlnr (14-23)

Figure 14-4 Rowland r ing for the determinat ion of B as a funct ion of H in a

fe r romagne t i c subs tance .



34 3

H (ampere-turns meter)

Figure 14-5 Magne t i za t ion curves for va r ious magne t i c ma te r i a l s : a, P e r m a l l o y ; b,

a n n e a l e d p u r e i r o n : c, duct i le cas t i ron; d, Alnico 5 . Th e num bers sh own a re the

re l a tive p e rmeabi l i t i e s B//i0H.

i n the sample . The magne t i c induc t ion i s

B = cp/S, (14-24 )

w h e r e S i s the cross-sect ional area of the core, and O is the magnet ic f lux.

O ne can mea sur e change s in 0 , and hence chan ges in B, by chan g ing / and

i n t e g r a t i n g t h e e l e c t r o m o t a n c e i n d u c e d i n w i n d i n g b as in Prob. 14-16. Both

B a n d H are the re fo re eas i ly m eas urab le .

If we s t a r t wi th an unm agn e t i z ed s am ple and inc rease the cu r r en t in

coi l a, t he magne t i c induc t ion a l so inc reases , bu t i r r egu la r ly , a s in F ig . 14-5 .

Al l f e r romagne t i c subs tances have such S- shaped magnetization curves. N o t e

the l a rge var i a t ions in u, for a g iven mate r i a l . W hen ever o ne uses p r in relat ion

wi th f e r romagne t i c mate r i a l s , one r e f e r s to the r a t io B/u0H, for specific

values of B or of H.

A chara c te r i s t i c f ea tu re o f the mag ne t i za t io n curve i s t he saturation

i nduc t ion beyond which M increases no fu r ther . Maximum a l ignment o f

the dom ain s is t hen ach ieved , a ft e r which dB = fi0 dH . T h e s a t u r a t i o n

ind uc t ion l ies in the ran ge f rom 1 to 2 teslas , dep en din g on the m ate r ial .



34 4

Figure 14-6 M a g n e t i z a t i o n c u r v e ab

and hys teres is loop bcdefgb.

14.9 HYSTERESIS

Co nsid er now F ig . 14-6. Cu rve ab is t h e m a g n e t i z a t i o n c u r v e . O n c e p o i n t b

has been r eached , i f t he cu r r en t in wind ing a of Fig . 14-4 is reduced to zero,

B d e c r e a s e s a l o n g be . T h e m a g n i t u d e o f t h e m a g n e t i c i n d u c t i o n a t c is the

remanence or the retentivity fo r the par t i cu la r sample o f mate r i a l . I f t he

cur r en t i s t hen r ever sed in d i r ec t ion a nd inc rease d . B r eaches a po in t d w h e r e

it is r educ ed to ze ro . The m agn i tud e o f H a t th i s po in t i s kn ow n as the coercive

force. On fu r ther inc reas ing the cu r r en t in the same d i r ec t ion a po in t e,

s y m m e t r i c a l t o p o i n t b, i s r eached . I f t he cu r r e n t i s no w reduc ed , r ever sed ,and inc reased , the po in t b i s aga in r eached . The c losed curve bcdefgb is

k n o w n a s a hysteresis loop. If, a t any po int , the cur ren t i s var ied in a sma l ler

cycle , a smal l hysteresis lo op is descr ibe d.

Energy i s r equ i r ed to desc r ibe a hys te r es i s cyc le . Th i s can be shown

by cons ider in g F ig . 14-7. W hen the cu r r en t in win d ing a of Fig. 14-4 is

inc reas ing , t he e l ec t romotance induced in the wind ing opposes the inc rease



345

Figure 14-7 The shaded area gives the

energy per unit volume required to go

from « t o b on th e hysteresis loop.

T h e energy per unit vo lume required

to describe a complete hysteresis loop

is equal to the area enclosed by the

loop .

in current, according to Lenz's law (Sec. 11.3), and the extra power spent by

the source is

where S is the cross-sectional area of the ring, N is the number of turns in

winding a. and B is the average magnetic induction in the core. Also,

where / is the me an circumf erence of the ring, T = SI is its volume, and

IN/I = H, as in Eq. 14-23. Thus

W, = T [b

HdB (14-27)




i s the energy suppl ied by the source in going f rom the point g t o the po in t b

in Fig . 14-7. Th is in tegral cor r es po nd s to the shad ed are a in the f igure a nd

i s equ a l to the energy sup p l i ed per un i t vo lum e o f the magn e t i c co re .

W hen the cu r r en t i s i n the sam e d i r ec t ion b u t i s decreas ing , t he polarivf

of the induced e l e c t ro mo tanc e i s r ever sed , acc ord in g to Lenz ' s l aw, wi th the

resu l t t ha t t he energy

W 2 = t P H dB (14-28)

i s r e tu rne d to the source .

F ina l ly , t he energy supp l i ed by the source dur ing one cyc le i s

W=tj)HdB, (14-29)

wh ere the in t egra l i s eva lua ted a ro un d the hys te r es i s loop . The a r e a o f the

hys te r es i s loop in t es l a - ampere tu rns per mete r o r in weber - ampere tu rns

per cub ic mete r i s t he re fo re the num be r o f jou le s d i ss ipa ted pe r cub ic mete r

and per cycle in the core.

Hys te r es i s losses can be min imized by se l ec t ing a mate r i a l wi th a

n a r r o w h y s t e r e s i s l o o p .

14.9.1 EXAMPLE: TRANSFORMER IRON

Fo r the t ransfor me r i ro n of Fig. 14-7. the area enclosed b y the hys teres is loo p is

appro xim a te ly 150 web er -am pere tu rn s pe r cub ic me te r o r 150 jou les pe r cub ic

me ter and per cycle , or 1.1 wa t ts per ki log ram at 60 hertz . This i ron is sui tab le for

use in a power t ransformer, but other a l loys are avai lable wi th lower hys teres is

losses .

14.10 BO UNDA R Y CONDITIONS

Let us exam ine the con t inu i ty co nd i t ions tha t B and H mu st obey a t t he

inter face betw een t wo m edia . W e shal l pro cee d as in Sec. 7 .1 .F igure 14-8a shows a shor t cy l indr i ca l vo lume whose top and bo t tom

faces are para l le l an d inf ini te ly close to the inter face. Since th ere is ze ro ne t

f lux through the cyl indr ical sur face (Sec. 8 .2) , the f lux through the top face

m u s t e q u a l t h a t t h r o u g h t h e b o t t o m , a n d

B„i = B n2. (14-30)

Th e no rm al com po ne n t o f B i s the re fo re con t i nu ou s ac ross the in te r f ace .



34 7

(b)

Figure 14-8 (a) Gauss ian surface a t the interface between two media , (b) C l o s e d

path cross ing the interface.

Co nsi der n ow F ig . 14-8b . Th e c losed pa th has two s ides para l l e l t o

the inter face and close to i t . From the ci rcui ta l law (Sec. 14.6) ,

H d l = 7, (14-31)

whe re / is t he con duc t ion cu r r en t l i nk ing the pa th .

If the tw o sides par alle l to the inte rface a re infinitely close to i t , 7 is

equa l to ze ro ,

H n = H t2, (14-32)

a n d t h e t a n g e n t i a l c o m p o n e n t o f H i s con t inu ou s ac ross an in t e rf ace .

If we can set B equal to uH in both media, the relative permeabilities

being those that correspond to the actual values of H. then

uriu

l}H

l cos 0l

= p,r2fj,

0H

2 cos 92, (14-33/f^

from the continuity of the normal component of B, and

Hx

sin 0j = H2

sin 02, (14-34)

from the continuity of the tangential component of H. Then

tan 6j nri

tan 02

Uri(14-35)

Th e lines of B, or of H. are farther awa y from th e nor mal in the med iu m

having the larger permeability.

14.11 SUMMARY

The magnetization M is the magnetic mom ent per unit volum e.

One can calculate the magnetic field of a piece of magnetic material

by assuming an equivalent surface current density

a, = M x n, . (14-2)

where n, is a unit vector nor mal to the surface and poi nti ng out war d, a nd

an equivalent volume current density

Je = V x M. (14-8)

The magnetic field intensity is

H = — - M, (14-13)

l<0

an d

V x H = JF

. (14-16)



Problems 349

This i s Ampere's circuital law in terms of H. In in tegral form,

i cH • dl = I (14-18)

w h e r e I f i s the current of f ree charges l inking the curve C.

I f a magn e t i c ma te r i a l i s l i near and i so t rop ic , t he n

15 = /'o(l + Zm)H = Mort-H = pH . (14-21)

14-1E

14-2E

14-3E

14-4

w h e r e ym is the magnetic susceptibility, p r is the relative permeability, a n d p

is the permeability.

F e r r o m a g n e t i c m a t e r i a l s a r e h i g h l y n o n - l i n e a r , a n d o f t e n a n i s o t r o p i c .

Th e r e l a t ion be twe en B an d H de pe nd s on the p rev io us h i s to ry o f the f ie ld ,

and i s r ep resen ted by the hysteresis loop. The a r ea o f a hys te r es i s loop g ives

the energy d i ss ipa ted in the mate r i a l pe r cub ic mete r and per cyc le .

A t t h e b o u n d a r y b e t w e e n t w o m a t e r i a l s , t h e t a n g e n t i a l c o m p o n e n t

o f H a n d t h e n o r m a l c o m p o n e n t o f B a r e c o n t i n u o u s .

PROBLEMS

MAGNETIC FIELD OF THE EARTH

A sphere of rad ius R i s uniformly magnet ized in the di rect ion paral le l to a

diameter . The external f ie ld of such a sphere is c losely s imilar to that of the earth. See

Prob . 8 -11 .

Show that the equivalent surface current dens i ty is the same as i f the sphere

carr ied a uniform surface charge dens i ty a and ro ta t ed a t an angula r ve loc i ty co such

EQUIVALENT CURRENTS

An i ron to ru s whose ma jor rad ius i s mu ch l a rge r than i ts minor rad ius is ma g

net ized in the azimuthal di rect ion with M uni form.

What can you s ay about at,'?

EQUIVALENT CURRENTS

A long tube is uniformly magnet ized in the di rect ion paral le l to i t s axis . What

is the value of B ins ide the tube?

DIELECTRICS AND MAGNETIC MATERIALS COMPARED

a ) Imagin e a pa ra l l e l -p l a t e capa c i to r whose p l a t e s a re cha rge d and insu la t ed .

Ho w are E and D affected by the int ro du ct io n of a die lect r ic betwe en the pl a tes?




b) Show a polar molecule of the die lect r ic or iented in the f ie ld.

c ) I s i ts ene rgy maxim um o r min im um ?

d) Now imagine a long so lenoid ca r ry ing a f ixed cur ren t .Ho w are B an d H affected by the int ro du ct io n of a cyl inde r of fe rrom agne t ic

ma te r i a l i ns ide the so leno id? /

e ) Show a sma l l cur ren t l oop represen t ing the magne t i c d ipo le moment o f an

a tom, and show the d i rec t ion of the cur ren t on the loop .

f ) I s i ts ene rgy ma xim um or m in im um ?

14-5 MAGNETIC TORQUE

Show tha t t he to rq ue exe r ted on a pe r ma nen t magne t o f d ipo le mom ent m

situ ated in a ma gn etic field B is m x B.

See Prob. 13-22.

I4-6E MEASUREMENT OF M

The fo l lowing me thod has been used to measure the magne t i za t ion M of a small

sph ere of mat eria l , induce d by an appl ied uniform ma gne t ic f ie ld B 0 as in Fig. 14-9.

In the figure. HG is a Hall gen er ato r (Sec. 10.8.1) with its cu rre nt flow ing par allel

to B 0 . S ince Hal l generators are sens i t ive only to magnet ic f ie lds perpendicular to thei r

current f low, this one measures only the dipole f ie ld originat ing in the smal l sphere .

Figure 14-9 Se t -up for measur ing the

m a g n e t i z a t i o n M induced in a smal lspherical sample of materia l by a

uniform field B 0 . Th e Ha l l gene ra to r

HG i s or iented so as to be sens i t ive

to the v-component of the f ie ld

or ig ina t ing in the sphe re , and no t t o

B„. See Prob. 14-6.



Problems 351

If the magnetization is Af, the magnetic moment m of the sphere is ( 4 / 3 ) 7 r P3

M ,

and it turns out that the x-component of the dipole field of the sample at HG is

3p0m sin 0 cos 0

An r3

Thus, at a given ang le 0, thi s field decreases as 1/r3

.

At what angle 0 will the measured field be largest?

14 -7 MICROMETEORITE DETECTOR

Figur e 14-10 shows an instrum ent t hat has been devised to detect ferromagnetic

micrometeorites falling to the ground. The instrument detects particles as small as

10 micr omet ers in diam eter and me asure s their magnetic mome nt, which is a measu re

of their size and composition.

A particle of radius b is magnetized in the field of the solenoid and acquires a

magnetic moment in. It is then equivalent to a small coil having a magnetic moment m.

The mutual inductance between the particle and the coil C of mean radius a

shown in the figure is then the M of Prob. 12-3. with nb2

NhI

h= m.

I

Figure 14-10 Section through a micro-

meteorite detector. Air is sucked

through a funnel F. and a particle p

is magnetized in the field B of a

solenoid S. In passing through the

tube, the particle induces a voltage

in coil C. See Prob. 14-7.




a) Show that the vol tage induced in the coi l i s

3Li 0N aa2

where r i s the veloci ty of the part ic le .

b) Draw a curve of z(a 2 + r 2 ) - 5 2 as a function of z for a = 10 mil l ime ters . Th e

c o o r d i n a t e z should va ry from —50 to + 5 0 mi l l ime te rs .

Since the veloci ty is cons tant , z i s p ropor t iona l t o the t ime t.

14-8E MECHANICAL DISPLACEMENT TRANSDUCER

One of t en wishes to ob ta in a vo l t age tha t i s p ropor t iona l t o the d i sp lacement

of an object from a kn ow n reference p os i t ion .

If there are no magnet ic materia ls nearby, one can f ix to the object a smal l

permanent magnet and measure i t s f ie ld wi th a Hal l generator , as in Fig. 14-11. T h e

Hal l e lement is sens i t ive only to the v- com po nen t of the ma gn et ic f ie ld and. from

P r o b . 14-6.

3 / / 0 ' " xz

* = ~4V(x2 + _ -

2

)5 2

'

a ) Draw a curve of xz / (x 2 + r 2 ) ' 2 as a function of z for v = 100 mil l im eters

and for z = —50 to + 5 0 mi l l ime te rs .

b ) Over wha t range does B x dev iate by less than 5% from the tang ent to th e

curve a t z = 0 ?

The l inear region is longer for larger values of x. but B x decreases rapidly wi th x.

x

Figure 14-11 H a l l g e n e r a t o r HG used

as a mechanica l d i sp lacement t rans

duce r . The moving pa r t P s l ides a long

the z-axis . The Hal l generator is

or i en ted so a s to moni tor the x-

component o f t he magne t i c f i e ld o f

t h e s m a l l p e r m a n e n t m a g n e t M fixedto P . See Prob. 14-8.

HG

14-9 MAGNETIZED DISK

A thin disk of i ron of radius a and th i cknes s t i s ma gne t ized in the di rect io n

parallel to its axis.

C ' a lcula te 11 and B on the axis , both ins ide and o uts ide the i ron.



Problems 353

I4-I0E TOROIDAL COIL WITH MAGNETIC CORE

C om pa re the fields inside tw o s imilar toro idal coi ls , bot h carry ing a curre nt / ,

one wi th a non-magnet ic co re and the o ther wi th a magnet ic co re .How are the equ iva len t cu rren t s on the magnet ic co re o r ien ted wi th respec t to

those in the coi l?

EQUIVALENT CURRENTS

A long wire of radius a carries a current / and is s i tuated on the axis of a long

hollow iron cyl inder of inner radius b and ou ter rad ius c .

a) Compute the flux of B inside a sect ion of the cyl inder / me ters long.

b) Find the equivalent current densi ty on the inner and outer i ron surfaces ,

and find the direct ion of these equivalent currents relat ive to the current in the wire.

c) Find B a t d i s tances r > c from the wire.

H ow w ould th is value be affected if the i ron cyl inder were re mo ve d?

THE DIVERGENCE OF H

We have se en in Sec. 8.2 that V • B is zero . This eq uat ion is val id even in n on

l in e a r, n o n -h o m o g e n e o u s , a n d n o n - i s o t ro p i c m e d i a .

Set B = u.runM. Un de r what cond i t ions is V • H # 0?

wTHE MAGNETIZATION CURVE

Use Fig . 14-5 to find the relat ive permeabil i ty of annealed pure i ron at a magnetic

induction of one tes la.

14-14 ROWLAND RING

A ring of duct i le cast-i ron (Fig . 14-4) has a major radius of 200 mil l imeters and

a min or ra dius of 10 mil l ime ters . A 500-tu rn toro ida l coi l is wo und o ver i t.a) What is the value of B inside the ring when the current through the coi l is

2 .4 amperes? Use Fig . 14-5.

b) A 10-turn coi l is wo un d over th e first one. Calcu late the voltage ind uced in i t

i f the current in the large coi l suddenly increases by a small amount at the rate of

10 ampe res per second . A ssume tha t u r r e m a i n s c o n s t a n t .

14-15E THE WEBER AMPERE-TURN

Show tha t one weber am pere- tu rn i s one jou le .

14-11

14-12E

14-13E

14-16 ROWLAND RING

Figu re 14-12 show s how one can m eas ure B as a function of H to ob ta in the

magnet iza t ion and hys te res i s cu rves o f a r ing o f magnet ic mater ia l .

W i n d i n g a is fed by an adjustable power supply. As in Sec. 14.8.

H = NJJlnr.

Fhe vo l tage across wind ing b is in tegrated with the circui t of Prob. 5-32.

35 4

The cross-sect ion of the r ing is S and winding b h a s N b t u r n s .

Show tha t

B = RCV/N bS

i f . a t the beginning of the experiment . B = V = 0 . Th e in t egra t ing c i rcu it d raws

es sen t i a l ly ze ro cur ren t .

Solution: The vo l t age ac ros s w inding h is

</4> dB

~dt=

~dt'(1)

s ince there is zero current in b, and hence no vo l t age drop caused by the re s i s t ance

and in duc tanc e of the w inding .

The n , f rom Pro b . 5 -32 , d i s rega rd ing the nega t ive s ign .

1 p dBv =

RC J°N

"s

Tt= N

"S B/ RC | 2 )

a n d

B = RCV/NbS.

14-17 TRANSFORMER HUM

W h y d o e s a t r a n s f o r m e r h u m ?



35 5

2

-2\ I I I I- 4 0 - 2 0 0 2 0 4 0

H ( a m p e r e - t u r n s , m e t e r )

Figure 14-13 Hys te res i s l oop for the a l loy De l t amax. See P rob . 14-18 .

14-18E POWER LOSS DUE TO HYSTERESIS

Figure 14-13 shows the hys teres is loop for a nickel- i ron a l loy cal led Del tamax.

W ha t i s t he appro xim a te va lue of the powe r d i s s ipa t ion pe r cub ic me te r and

per cyc le when the ma te r i a l i s d r iven to s a tura t ion bo th way s?

14-19 THE FLUXGATE MAGNETOMETER AND THE PEAKING STRIP

A wire is wo uld a ro un d a s t r ip of Del t am ax (Fig. 14-13). form ing a solenoid. A

few turns o f w i re a re then wou nd ove r the so lenoid , and con nec ted to an osc i l loscope .

An a l t e rna t ing cu r ren t i s app l i ed to the so lenoid , d r iv ing the De l t a max to

sa tura t ion in bo th d i rec t ions .

a ) W ha t i s t he shape of the waveform obse rved on the osc i l loscope?

b) W ha t hap pe ns to the pa t ter n on the osci l loscop e if a s teady m agn et ic field isnow appl i ed pa ra l l e l t o the s t r ip?

It can be shown that , upon appl icat ion of the s teady f ie ld, the output vol tage

conta ins a com po nen t a t dou ble the f requency of the appl i ed a l t e rna t ing cur ren t and

pr op ort ion al to the m ag ni t ud e of the s teady f ie ld. Th is is the principle of op erat ion of

th e fluxgate magnetometer. Fluxg a te magn e tom ete rs a re o f ten used on boa rd s a t el l it e s

for measuring magnet ic f ie lds in outer space. They are useful from about 10 _ 1 0 to 1CT

tesla.




T h e peaking strip a l so u t i l i zes a so lenoid wound on a ma te r i a l w i th a squa re

hys teres is loop l ike Del ta ma x (Fig. 14-13), an d a sec on dar y w inding . I t i s used differently,

however. The peaking s t r ip is placed ins ide a solenoid whose current i s adjus ted so asto cance l t he ambien t f i e ld , making the pa t t e rn on the osc i l loscope symmet r i ca l . The

peaking s t r ip i s used to measure magne t i c induc t ions in the range of about 1 ( T 6 to

1 0 "2

tesla.



CHAPTER 15

MAGNETIC FIELDS: VIII

Magnetic Circuits

15.1 MAGNETIC CIRCUITS

15.2 MAGNETIC CIRCUIT WITH AN AIR GAP15.2.1 Example: Electromagnet

15.3 MAGNETIC FORCES EXERTED BY

ELECTROMAGNETS

15.4 SUMMARY

PROBLEMS



In gen eral , it i s not po ss ible to calculate m ag ne t ic f ie lds accura tely , w hen

magnet i c mate r i a l s a r e invo lved . There a r e s evera l r easons fo r th i s , ( a ) The

re la t ion be tw een H a nd B for f e r romagn et i c m ate r i a l s i s non - l inear . I t even

depends on the previous his tory of the mater ial , as we saw in Sec. 14.9. Also,

f e r roma gnet i c m ate r i a l s a r e o f ten no n- i so t r op ic . (b ) Th e i ron cores tha t a r e

used to conf ine and guide the magnet ic f lux are of ten qui te ineff icient . As we

shal l see, a large par t of the f lux can be s i tua ted o uts ide the core, (c) Per m an en t

magne t s a r e no t as s imple as one would l ike them to be : the i r magne t i za t ion

M i s non -un i fo rm a nd depen ds on the p resence o f ne ig hbo r ing ma gne t i c

m a t e r i a l s .

I t i s none the les s neces sary to be ab le to m ak e app rox im ate ca lcu la t ions .

Th e ca lcu la t ion s e rves to des ign a mo del on w hich mag ne t i c f ie lds o r m agn e t i c

forces can be measured, and which can then be modif ied to give the f inal

des ign .

15.1 MAG NETIC CIRCU ITS

Figu re 15-1 show s a f e r romagn et i c core a ro un d wh ich i s wo un d a shor t co i l

of N t u rns ca r ry ing a cur r en t / . We wish to ca lcu la t e the magne t i c f lux O

t h r o u g h t h e c o r e .

In the absen ce of fer ro ma gne t ic ma ter ial , the l ines of B are a s show n

in th e figure. At f irst s ight, on e expe cts the B insid e the co re to be m uc h

large r close to the win din g tha n on the op po s i te s ide. Th is is no t the case,

how ever , and B i s o f the s am e orde r o f ma gn i tud e a t a ll po in t s wi th in the

f e r r o m a g n e t i c m a t e r i a l .



359

Figure 15-1 Ferromagneti c toroid with concentrated winding . The lines offeree

s h o w n are in the plane of the toroid and are similar to those of Fig. 8-6. They

ap p l y only when there is no iron present.

This can be understood as follows. The magnetic induction of the

current / magnetizes the core in the region near the coil, and this magnetiza

tion gives equi valen t cur ren ts that bo th incr ease B and extend it along t he

core. This further increases and extends the magnetization, and hence B,

until the lines of B exte nd all aro un d t he core .

Of course some of the lines of B escape into the air and then return to

the core to pass again through the coil. This constitutes the leakage flux

that may, or may not, be negligible. For example, if the toroid is made up

of a long thin wire, most of the flux leaks across from one side of the ring

to the other and the flux at P is negligible compared to that near the coil.

Let us ass ume tha t the cross-sectio n of the toroi d is large enough to

render the leakage flux negligible. Then, applying Ampere's circuital law,

Eq. 14-18, to a circu lar pa th of rad ius r goin g all arou nd inside the to roi d,

(15-1)

InrB firft

oNI, (15-2)

B Hrfi

0NI/2nr. (15-3)

3 6 0 M a g n e t i c F i e l d s : VIII

T a k i n g R { to be the rad ius co r re spond ing to the ave rage va lue o f B, a n d

R2

to be the minor radius of the toroid .

<D = n rn 0nRlNII2nR v (15-4)

Th e flux th rou gh the core is therefore t he sam e as if we had a tor o id al coi l

(Sec. 9 .1 .2 ) of the same s ize and i f the number of ampere- turns were increased

by a factor of a r. In o ther words , for each ampere- turn in the coi l there are

u r — 1 am pe re- tur ns in the core . Th e amplif ica t io n c an be as h igh as 1 0 5 .

Th is eq uat ion sh ows that the ma gne t ic flux is g iven by the m ag ne to -

m o t a n c e NI mult ip l ied by the fac tor

UrUonRl/lnRu

wh ich is called th e permeance of the magnet ic c ircui t . The inverse of the

per me anc e is ca l led the reluctance. T h u s

^ /V/ NI

(1) = — = (15 -5)

St InRJp^nRV ' '

and the re luctance is

9t=InRJurHouRl.

(15-6)

Re luc tanc e i s exp ressed in hen rys "

Th e ana logy wi th O hm 's l aw is obv ious : if an e lec t ro mo tance V w e r e

induced in the core , the current would be

r(15-7)InRxlanRj

w h e r e 2nR\lanR\ is the res is tance of the core .

Thus the co r re spond ing quan t i t i e s in e lec t r ic and magne t ic c i rcu i t sare as fo l lows:

C u r r e n t /

Cur ren t dens i ty J

C o n d u c t i v i t y a

Magnetic flux (P

M a g n e t i c i n d u c t i o n B

Permeab i l i ty p = prp0



15.2 Magnetic Circuit with an Air Gap 361

E l e c t r o m o t a n c e ir

Electr ic f ie ld intens i ty E

C o n d u c t a n c eG = l/R

R e s i s t a n c e R

M a g n e t o m o t a n c e NI

Magnet ic f ie ld intens i ty H

P e r m e a n c e \f.

J

AR e l u c t a n c e 88 .

T h e r e is one impor tan t d i f f e r ence be tween e lec t r i c an d m a g n e t i c

c i r c u i t s : the magnet i c f lux cannot be m a d e to follow a magnet i c c i r cu i t in

t h e m a n n e r t h a t an elect r ic current fol lows a c o n d u c t i n g p a t h . I n d e e d , a

m a g n e t i c c i r c u i t b e h a v e s m u c h as an elect r ic ci rcui t would if it w e r e s u b

m e r g e d in ta p w a t e r : p a r t of the cur ren t would f low th rough the c o m p o n e n t s ,

a n d the r es t would f low th rough th e w a t e r .

If a m a g n e t i c c i r c u i t is not p r o p e r l y d e s i g n e d , the leakage f lux ca n

easily be an o r d e r of m a g n i t u d e larger t ha n tha t flowing a ro un d the circui t .

15.2 MAGNETIC CIRCUIT WITH AN AIR GAP

F i g u r e 15-2 s h o w s a circui t wi th an air gap whose c ros s - sec t ion is different

f rom tha t of the sof t - i ron yoke . Each wind ing p rov ides NI/2 a m p e r e - t u r n s .

We wish to c a l c u l a t e th e m a g n e t i c i n d u c t i o n in the air gap.

W e a s s u m e t h a t the leakage f lux is negl igible . As we shal l see, t h i sas sumpt ion wi l l r esu l t in q u i t e a l a rge e r ro r .

A p p l y i n g A m p e r e ' s c i r c u i t a l law to the circui t , we see t h a t

NI = Hik + H glg, (15-8)

w h e r e the s u b s c r i p t i refers to the i r o n y o k e and g to the air gap; /, and lg

a r e th e pa th l eng ths . The pa th l eng th /, in the i ron can be t a k e n to be the

l e n g t h m e a s u r e d a l o n g the c e n t e r of the cros s - sec t ion of the y o k e .

If we neglect leakage f lux, the flux of B m u s t be the same over any

cros s - sec t ion of the magnet i c c i r cu i t , so

BiAi = BgAg, (15-9)

w h e r e A t and A g are, r espec t ive ly , the cros s - sec t ions of the i r on yoke and of

t h e air gap.

36 2

Soft iron yoke

• Ii

•11

Figure 15-2 Electromagnet . The coi ls have been cut out to expose the i ron yoke.

C o m b i n i n g t h e s e t w o e q u a t i o n s .

NI, ( 1 5 - 1 0 )

and the magnet ic f lux is

NI

I;(15-11)

Th e ma gne t i c flux i s there fore equa l to the mag ne tom ot an ce d iv ided by

the sum of the reluctances of the i ron and of the ai r gap.

This is a general law: reluctances in ser ies in a magnet ic ci rcui t add in

the same way as res is tances in ser ies in an elect r ic ci rcui t ; permeances in

para l l e l add l ike co ndu c ta nc es in para l le l .

15.2 Magnetic Circuit with an Air Gap 363

Since we have neglected leakage flux, the above equation can only

pr ov id e an up pe r lim it for .

N ow let = IJi.irP.QAj be the reluctance of the iron, and &g = lglp 0A g

be the relucta nce of the air gap. N ot e that

/ / , / , = . ^ , < D. ( 1 5 - 1 2 )

H glg = ®g4>. ( 1 5 - 1 3 )

If 09 t « dig, as is usually the case, since pr » 1, Eqs. 15-8 and 15-11

become

NI % Hgl

g, ( 1 5 - 1 4 )

* * NI p0A

g/l

g. ( 1 5 - 1 5 )

EXAMPLE: ELECTROMAGNET

L et us ca lcula te B and the s tored energy in the e l e c t r o m a g n e t of Fig . 15-2. If t h e r e

is a t o t a l of 10,000 turns in the two w i n d i n g s , and se t t ing / = 1.00 a m p e r e . Ax =

10* square mi l l ime te rs , Ag = 5 x 103 square mi l l ime te rs , pr = 1,000,/, = 900 mil l i

mete rs , and /,, = 10 mil l imete rs , then

<1>1 0

4

0. 9 1015-16)

1 0 3

x 4;: x 1 0 "7

x 10~2

4TT X 1 0 "7

x 5 x 10"3

= 6.0 x 1 0 "3 weber, (15-17)

B g = 6.0 x 1 0 " 3 / 5 x 1 0 "3

= 1.2 teslas. (15-18)

The se l f - induc tance is

L = NQ/I = 104

x 6.0 x 1 0 ~ 3 / 1 . 0 0 = 60 h e n r y s . 15-191

T h e l e n g t h /f is m e a s u r e d a l o n g the m i d d l e of the cross-sect ion of the y o k e .

The s tored energy is

( 1 / 2 ) L / 2

= (1/2) x 60 x 1.00 = 30 joules . (15-20)

http://iji.irp.qaj/

http://iji.irp.qaj/

http://iji.irp.qaj/

http://iji.irp.qaj/



364 Magnetic Fields: VIII

In this part icular case the leakage f lux is 70% of the f lux in the gap. In other

wo rd s, the mag ne tic ind uct ion in the gap is no t 1.2 teslas, bu t only 1.2 1.7 = 0.71 tesla.

There exis t empirical formulae for es t imat ing leakage f lux for s imple geometries .

15.3 MA GNETIC FORCES EXER TED

BY ELECTROMAGNETS

E l e c t r o m a g n e t s a r e c o m m o n l y u s e d t o a c t u a t e v a r i o u s m e c h a n i s m s s u c h a s

switch es , valves , an d so for th . A switch act iv ated by an ele ct ro ma gn et i s

cal led a relay. These e l ec t romagnet s compr i s e a co i l and an i ron core tha t

i s m ad e in two par t s , one f ixed and one m ova ble , ca l l ed the armature, wi th

an a i r gap be tween the two . The co i l and i t s core toge ther a r e usua l ly t e rmed

a solenoid. The magne t i c fo rce i s attractive.

In the s imples t cases , one has a f la t a i r gap, perpendicular to B. T h e n

the magne t i c fo rce o f a t t r ac t ion i s B2

/2p:0 t imes the cross-sect ion of the

gap (Sec. 13.2) where B i s calc ulate d as in Sec. 15.2. I f the so leno id is energized

wi th d i r ec t cur r en t , t hen B i s app rox ima te ly p rop or t i on a l to the inver se o f

the gap l eng th , and s ince the fo rce i s p ropor t iona l to B2

, the force increases

rap id ly wi th decreas ing gap l eng th .

The ai r gap is of ten des igned so that the at t ract ive force wi l l vary wi th

the gap l eng th in som e prescr ibed w ay . Th e des ign o f the mag ne t i c c i r cu i t

is largely empir ical .

15.4 SUMMARY

The concep t o f magnetic circuit i s widely used whe nev er the m ag ne t ic f lux

i s gu ided mos t ly th r ou gh m agn e t i c mate r i a l . We then have a m agn e t i c

e q u i v a l e n t of O h m ' s l a w :

O = NI/M, (15-5)

w h e r e NI is the magnetomotance of the energizing coi l and J ? is the reluctance

of the magne t i c c i r cu i t .

Problems 365

T h e c o r r e s p o n d e n c e b e t w e en

fol lows:

e lec t r i c and magne t i c c i r cu i t s i s as

C u r r e n t /

C u r r e n t d e n s i t y J

Conduct iv i ty r r

E l e c t r o m o t a n c e f~

Electric f ield intensity E

C o n d u c t a n c e

Res i s t ance R

M ag ne t ic F lux <P

M a g n e t i c i n d u c t i o n B

Permeabi l i ty /<r/<„

M a g n e t o m o t a n c e NI

Magnet ic f ie ld intens i ty H

P e r m e a n c e

R e l u c t a n c e 'J)

The r e luc tance o f a bar o f magne t i c mate r i a l o f permeab i l i ty /u r /u 0 ,

c ros s - s ec t ion A, an d len gth / i s l/nrfi0A. Reluctances in ser ies and in paral lelare t reated l ike res is tances in ser ies and in paral lel .

Th e magn e t i c force o f a t t r ac t ion ex er t ed by an e l ec t rom agn et i s B2

/2p.0

t imes the cross-sect ion of the gap.

PROBLEMS

15-1E RELUCTANCE

Show that the energy s tored in a magnet ic c i rcui t i s

W = (1/2)<D 2 ^,

w h e r e <> i s the magnet ic f lux and J# i s the reluctan ce.

15-2E RELUCTANCE

Show tha t the induc tance Lof a co i l of N turns is given by

L = JV2

w h e r e J

A i s the reluctance of the magnet ic c i rcui t .

I5-3E CLIP-ON AMMETER

It is often useful to be ab le to m ea su re the cu rre nt flowing in a wire w ith ou t

dis tu rbin g the ci rcui t . Th is can be do ne with a c l ip-on a mm eter . As a rule , c l ip-on

am m ete rs are t ransform ers , as in Pr ob . 18-16, and can therefore be used only with

a l t e r n a t i n g c u r r e n t s .

I t i s a lso poss ible to m ak e a cl ip-on a mm eter th at will me asu re direct cu rre nts

as in Fig. 15-3. In this ins t rum ent the mag net ic flux thro ug h the yoke is me asur ed with



366

Kigure 15-3 Clip-on ammeter for direct currents. A

hinge A permits the iron yoke to be opened at C so

that i t can be cl ipped around the current-carrying

wire. A gap G in the yoke contains ei ther a Hall

genera to r o r a magnetores i s to r . The magnet ic

induction in the gap is a measure of the current / .

See Prob. 15-3.

a Hall gene rato r (Sec. 10.8.1), or with a mag neto resi s tor ( Pro b. 10-15) s i tuated in the

gap of length lg.

a) Show that , i f lg » IJfi,, the magnet ic induct ion in the gap i s fi0I/lg-

What is the advantage of using an iron core?

b) Is this magnetic induction affected by the posi t ion of the wire inside the r ing?

15 -4 MAGNETIC CIRCUIT

The ducti le cast- iron r ing of Prob. 14-14 is cut , leaving an air gap 1 mil l imeter

long. The current in the toroidal coil is again 2.4 amperes.

C a l cu l a t e B in the air gap. using the relat ive permeabil i ty curve of Fig. 15-4.

You wil l have to solve this problem by successive approximations. Star t by

assuming a reasonable value for / i r , say 500. Find the corresponding B . Fhis wil l prob-

6 0 0

4 0 0

2 0 0

0 0.5 1.0

B (tesla)

Figure 15-4 Relat ive permeabil i ty as a function of magnetic induction for ducti le

cast i ron. See Prob. 15-4.



Problems 367

ably not give the correct B on the curve . Then t ry ano ther va lue of p r, e tc . Draw a

table of p r and of the calculated B as a he lp in s e lec t ing your next approximat ion .

The ca lcula ted va lue of B need not agree to bet ter than 10% with the B on thecurve .

MAGNETO RESISTANCE MULTIPLIER

The m agne t i c c i rcu i t shown in F ig . 15-5 g ives an outp ut vo l t age K tha t i s pro por

t ional to the product IJ b. The vol tmete r draws a negl ig ib le cur rent .

The a i r gaps conta in magne tores i s t ances (P rob . 10-15) R x a n d R 2, c o n n e c t e d

as in Fig. 15-5b. Th is c i rcui t i s ident ica l to tha t of Pr ob . 5-7.

F r o m P r o b . 1 0 - 1 5 .

R 5 = R 0 [ l + ,J I2(B a + B„)2l

R2 = R 0 [ l + JI2

(B a - B b)2

l

where , It i s the mobil i ty of the charge carriers .

i/x

Figure 15-5 (a) Ma gne t i c c i rcu it wi th a pa i r of mag ne tores i s t an ces and R 2 for

obta in ing a vol t age V p r o p o r t i o n a l t o t h e p r o d u c t o f Ia a n d Ih. The s t ra ight

a r row s show the flux thro ugh the i ron core due to / „ . Th e wavy a r row s show the

flux due to Ib. In the left -hand gap, B = B a + B b, while on the r ight B = B„ - B b.

(b ) The electr ic c i rcui t . See Prob. 15-5.



368 Magnetic Fields: VIII

Show tha t

v = -ue2

Bfi bv 0,

as long as ,M 2B\ « I , M 2B\ « 1.

MAGNETIC FORCE ON THE ARMATURE OF AN ELECTROMAGNET

Figu re 15-6 show s a smal l e lectrom agn et . All dim ens ion s are in mil l im eters . Eac h

coi l has 500 turns and carries a current of 2 amperes .

a ) Ca lcula te B in the ai r gaps as a funct ion of the gap length x. Set ii r = 1000.

and neglect the leakage f lux.

Solution: T h e m a g n e t o m o t a n c e i s

NI = 2 x 2 x 500 = 2000 am per e turns . (1)

The i ron yoke has a cross-sect ion of 10 x 10 squa re mi l l ime te rs , and a m ean

length of 2 x 25 mil l imeters for the vert ical parts , plus 23 mil l imeters for the

hor izo nta l p a r t . Hence i t s re luc tance i s

St y = 73 x 1 0 ~ 3 , (1000 x An x 1 0 ~ 7 x 1 0 - 4 ) = 5.8 x 1 0 s h e n r y - 1 . (2)

Figure 15-6 Elec t romagne t . See P rob . 15-6 .



Problems 369

Th e arm at ure has a cross-sect ion of 50 squ are mil l imeters . Th e mean path length

of the f lux is about 23 mil l imeters . Hence

m a = 23 x l(r3

/ (1000 x 4n x 1 0 " 7

x 50 x 1 0 ~ 6 ) = 3.7 x 1 0 5 h e n r y - 1 . (3)

Finally , the two air gaps have a reluctance

m g = 2x/(47t x 1 0 ~7 x 1 0 ~

4

) = 1.6 x I 01 0

x h e n r y- 1

. (4)

T h u s , in the gaps.

2000B = XI if A = « = ^ 2 , (5)

(5.8 x 1 0s

+ 3.7 x 1 05

+ 1.6 x 1 01 0

x ) 1 0 ~4

200

• teslas. (6)'.5 + 1.6 x 105

.Y

b) Calcu la te the fo rce o f a t t r ac t ion exer ted on the arm atu re when x =

5 mil l imeters .

Solution: The force is equ al to the p rodu ct of the energy densi ty in the gap, by

its cross-sect ion A:

F = 2B2

A/2tt„, (7)

4 x 1 0 4 x 1 0 ~ 4

(9.3^*- 1.6 x 1 05 x 5 x 1 0

_ 3

)2

4 7 r x 10 ^ % 5 new tons . (8)

c) W ou ld it m ak e sense to use Eq. 6 to cal cul ate the force (i) at x = 0.1 mil lime ter,

(ii) at x = 20 millim ete rs?

Solution: At x = 0.1 mil l imeter , Eq. 6 gives a magnetic induction of about

12 teslas . This is mu ch mo re tha n t he sat ura t ion f ield of one o r two teslas for i ron

(Sec. 14.8). Eq ua tio n 6 is not val id at x = 1 0- 4

.

At x = 20 mil l imeters , the reluctance of the air gap is large and the leakage

f lux is mu ch larger tha n the f lux in the gaps. The a ctua l force is mu ch sm aller t ha n

that ca lculate d f rom Eq. 7 . Eq ua tio n 6 is again no t val id .

So the magnetic induction found in Eq. 6 is val id only over a l imited range of x .

15-7 RELAY

Calculate the force of at tract ion on the armature of a relay such as the one

shown in Fig. 15-7 whose magnetic circuit has the fol lowing character is t ics:

Coi l , 10,000 turns; resis tance, 1000 ohms; rated voltage, 10 volts .

G ap leng th , 2 mi l l imeter s ; gap cross - sec t ion , 1 square cen t imeter .

For s implici ty , we assume that the gap has a uniform length, that the reluctance

of the iron is negligible when the gap is open, and that there is zero leakage f lux.



37 0

Figure 15-7. Relay. W he n the coi l is energized by clos ing the switch, the arm atu re

falls, opens the upper contac t , and c loses the lower one . Tens of contac t s can be

ac tua ted s imul taneous ly in th i s way. See P rob . 15 -7 .

15-8 MAGNETIC FLUIDS

There exis t magnet ic f luids consis t ing of f ine part ic les (*10 nanometers in

diam eter) of i ro n oxides sus pen ded in variou s f luids .*

a ) I f the mag ne t i c f lu id i s p laced in a non -ma gne t i c conta in e r on a n on-m agn e t i c

suppor t , a pe rmanent magne t p laced in i t wi l l remain suspended near the bot tom,

but wi thout touching the conta ine r , e i the r a t the bot tom or a t the s ides . Why i s th i s?b) C an you find appl ica t ion s for ma gn et ic f luids?

f Ma nufac tured by the Fer rof lu id ics Cor por a t io n , 144 Middlesex Turnp ike , Bur l ing ton ,

M a s s a c h u s e t t s 0 2 1 0 3 .



CHAPTER 16

ALTERNATING CURRENTS: I

Complex Numbers and Phasors

16.1 ALTERNATING VOLTAGES AND CURRENTS

16.2 R M S , OR EFFECTIVE VALUES

16.2.1 Example: The Output Voltage of an Oscillator

16.3 THREE- WIRE SINGLE-PHA SE AL TERN A TING

CURRENT

16.4 THREE-PHASE ALTERNATING CURRENT

16.5 ALTERNATING CURRENTS IN CAPACITORS AND

IN INDUCTORS

16.6 COMPLEX NUMBERS

16.6.1 Examples: Using Com plex Numb ers

16.7 PHASORS16.7.1 Example: The Phasors in a Three-Phase Supply

16.7.2 Addition and Subtraction of Phasors

16.7.3 Example: The Phasors in a Three-Phase Supply

16.7.4 When Not to Use Phasors

16.8 SUMMARY

PROBLEMS

M os t elect r ic an d mag ne t ic devices use al t ern at i ng or , a t leas t , f luctuating

cur ren t s . Th ere a r e man y r easo ns fo r th i s , bu t the two majo r one s have to

do wi th power t echno logy and wi th the t r ansmis s ion and p roces s ing o f

i n f o r m a t i o n .

Firs t , wi th alternating cur ren t s , t he e l ec t r i c power supp l i ed by a

source a t a g iven vo l t age can be made ava i l ab le a t a lmos t any o ther con

ven ien t vo l t age by m ean s of t r ans form ers . Th i s ma ke s e lec t r i c pow er ad ap t

able to a broad var iety of uses .

The power suppl ied by a <77rerf -current source can also be changed

from o ne vol tag e to an oth er . But this i s do ne by f irst swi tch ing the c urre nt

per iod ica l ly , t o ob ta in an a l t e rna t ing cur r en t , t hen f eed ing th i s to a t r ans

former , and then rect i fying and f i l ter ing the output . The operat ion is rela

tively costly and inefficient.

The s econd r eason fo r us ing a l t e rna t ing o r f luc tua t ing cur r en t s i s

tha t they can s e rve to t r ansmi t in fo rmat ion . For example , a microphone

t r ans forms the in format ion con ta ined in a spoken word in to a complex

f luctuating cu rren t . Th is cu rren t then serves to m od ul at e a ra dio-f re que ncy

cur ren t tha t is fed to an an ten na . The an te nn a l aunch es an e l ec t ro ma gne t i c

wave, and so for th .

A go od pa r t o f th i s cha p te r wi ll be dev o ted to the meth od u sed for

so lv ing the d i ff e ren t ia l e qu a t io ns as soc ia t ed wi th c i r cu i t s ca r ry ing a l t e rn a t ing

c u r r e n t s . T h i s m e t h o d r e q u i r e s th e u se o f c o m p l e x n u m b e r s . N o p r e v i o u s

knowledge o f the sub jec t i s r equ i r ed .

We sha l l as sume tha t the vo l t ages , t he cur r en t s , and the charges a r e

a l l cos ine func t ions o f t ime , wi th appropr ia t e phases . Th i s i s no t a lways the

case . T h e c u r r e n t t h r o u g h a m i c r o p h o n e is n o t n o r m a l l y s i n u s o i d a l . O r o n e

might have only the pos i t ive par t of the cos ine funct ion, or a cos ine funct ion



16.2 R M S , or Effective Values 373

whose per iod i s a func t ion o f the t ime . However , any periodic funct ion can

be analyzed into an inf ini te ser ies of s ine and cos ine terms, forming what i s

cal led a Fourier series.

16.1 ALT ERISA TING VO L TAGES AND CURRENTS

Voltages are of ten of the form

V = V 0 cos cot, (16-1)

w h e r e V 0 is the peak voltage, to = 2n f is the circular frequency, expres sedin r ad ians per s econd , and / i s the frequency, expressed in hertz. Such vo l t ages

are said to be alternating (Sec. 11.2.1) . We have arbi t rar i ly set the phase cot

of V equa l to ze ro a t t = 0.

If such a vo l t age i s app l i ed to a c i r cu i t, t hen the b ra nc h cu r ren t s a r e

of the form

I = I0

co s (cot + tp), (16-2)

where 7 0 i s the peak current, cot + tp i s the phase, and tp i s the phase of the

branch cur ren t / wi th r espec t to the source vo l t age V.I t i s unfor tuna te ly the cus tom to use the expres s ions AC current a n d

AC voltage, w h e r e AC s t a n d s f o r A l t e r n a t i n g C u r r e n t .

16.2 R M S , OR EFFECTIVE VALUES

I f an a l t e rna t in g vo l t age V 0 cos to t i s appl ied across a res is tor R as in Fig.

16- l a ,

F 0 c os cot . , „/ = = I

0cos ojt, (16-3)

R

as in Fig. 16- lb . In that case the current i s in phase wi th the vol tage, and

the peak cur r en t 7 0 is V 0/R .



The ins t an taneous power d i s s ipa ted in the r es i s to r i s

V 2

P i n s t = VI =0 c o s 2 cof = IQR COS

2 rat, (16-4)

Over one cyc le , t he average power d i s s ipa t ion i s

Wl 1 ,

2 ^ = 2 / a x ( i6 -5)

as in Fig. 16-2.



37 5

Tz /a> 2n/a> 3n/a> Anjio

Figure 16-2 Ins t a n ta n e ou s a nd a ve ra ge powe r d i ss ipa t ion in a re s is to r .

Un less specified othe rwi se, on e alw ays avoid s the factor of \ by us ing

th e rms, o r effective v o l t a g e a n d c u r r e n t ,

Vvm s = V 2

1

'2

, 7 r m s = Zo/21

'2

, (16-6)

ins t ead o f V 0 a n d I0. H e r e " r m s " s t a n d s f o r root mean square, or the square

roo t o f the mean va lue o f the square o f the func t ion . See Prob . 16 -2 . T h e n

P a v = V2

rmJR = I2

msR. (16-7)

T h e p e a k v a l u e s V 0 and / 0 a re 2 1 / 2 t imes larger than the rms values , as in

Fig. 16-3, as long as V a n d I are s inusoidal funct ions of the t ime.

0.707 V 0

0

V = V 0 cos cur

\ 1 / 1 \ 1 / 1 \

' 27t/co \ 3n/o) / 47t/co \

-Figure 16-3 R e la t ion be twe e n pe a k a nd rm s valu es for a cosin e functio n.



376 Alternating Currents: I

I t i s useful to remember that

2

1 / 2

= 1.414 a nd 1/2

1

'

2

= 2

1 / 2

/ 2 = 0.707. (16-8)

Unless stated otherwise, num erical values for voltages, currents, and/

charges are always rms values.

16.2.1 EXAMPLE: THE OUTPUT VOLTAGE OF AN OSCILLATOR

An osc i l la tor supp l ies 10 vol ts a t a pair of termin als , one of which is grou nd ed .

Then the potent ia l a t the other terminal is 1.414 x 10 cos ait. The potent i a l a t tha t

t e rmina l va r i es be tween —14.14 and +1 4.1 4 vol t s wi th respec t to grou nd.

16.3 THREE-W IRE SINGLE-PHASE

ALTERNATING CURRENT

As a rule , low-power ci rcui ts , such as those used in elect ronics , ut i l ize al ter

na t in g vo l t ages and c ur r e n t s as in Sec . 16 .1 . Fo r h ighe r pow ers , such as those

used in a house , on e can proce ed dif ferent ly .

F igure 16-4a shows two a l t e rna t ing cur r en t sources connec ted to

two res is tors f t , and R 2 by mea ns o f four wi res . Th e s igns an d the a r ro wsshow the po la r i t i es and the cur r en t d i r ec t ions a t a par t i cu la r ins t an t .

The two center wires of Fig. 16-4a can of course be replaced by a

s ingle wire as in Fig. 16-4b, where the center wire car r ies only the difference

be tween the load cur r en t s . One then has three-wire single-phase current.

Thus th ree wi res do the work o f four and , moreover , t he I2R loss in the

cen te r wi re i s low, whenever the loads a r e r easonab ly wel l ba lanced .

In N or th A me r ica , e lec t r i c u t i l it i e s ma in ta in 120 vo l t s be twe en A a n d

B, 120 vo l t s be twe en C a n d B, t hus 240 vo l t s be tween A a n d C. T h e p o t e n t i a l

o f po in t D wi th r espec t to g r ou nd i s no t qu i t e ze ro bec ause o f the cur r en t

f lowing through the center wire .I t i s t he cus to m to o per a te low -pow er dev ices , such as ligh t bu lb s ,

a t 120 vo l t s , bu t h igh -po we r dev ices , such as ba seb oa rd he a te r s , a r e op era ted

a t 240 vo l t s . For a g iven power consumpt ion , a h igh-power dev ice thus d raws

only half the curr en t i t wo uld d ra w at 120 vol ts . Th is perm its the use of l ighter

wire, and reduces the cos t of the wir ing.



377

(a) (h)

Figure 16-4 (a) Two similar sources with oppo site phases connected to two resistors

•Rj and R2. (£>) Three-wire single -phase supply. If the two resist ances are equa l, the

co n n ec t i o n BD is unnecessary.

16.4 THREE-PHASE ALTERNATING CURRENT

Figu re 16-5a shows three sources of alt ern ati ng curren t feeding r esistors

R 2, ^ 3 - Th e sources are orien ted 120° apa rt, on the figure, so as to indicat e

their relative phases. For example, if the phase at A is zero at t = 0, then

VA

= V0

co s cot, (16-9)

VB = F

0co s (cot + 2n/3), (16-10)

Vc = V0 cos (cot + 4 tt /3) . (16-11)

If the three resistances are equal, the total current flowing in the three

grounded wires is

(F0/R)[cos cot + cos (cot + 2TI/3) + cos (cot + 4TT /3)].



37 8

A

B(b)

Figure 16-5 (a) Thr ee s imi la r sources of a l t e rna t in g cur re nt , wi th phase d i ffe rences

of 120 degrees , conn ecte d to three res is tors R x, R 2. R^. (b) Three-phase supply . I f

the res i s t ances a re equa l , one may omi t the connec t ion DE .

I t wi l l be shown in Prob. 16-6 that the bracket i s a lways equal to zero. (At

f = 0, i t is eq ua l to 1 - 0.5 - 0.5.) Th en th e so urc es can be co nn ec te d t othe res is tances as in Fig. 16-5b and the wire DE can be d i spensed wi th .

I f the r es i s t ances a r e unequa l , t he cur r en t s do no t comple te ly cance l

in DE. We now have four wires doing the work of s ix , wi th low I2R losses

in the wire DE.

A se t o f th ree sources , s t a r - connec ted and phased as in F ig . 16-5b ,

supp l i es three-phase alternating current. T h e m a i n a d v a n t a g e o f t h r e e - p h a s e



16.5 Alternating Currents in Capacitors and Inductors 379

current is that it can be used to produce the revolving magnetic fields that

are required for large electric motors. See Prob. 16-7.

Electric power stations generate three-phase currents. This can be

seen from the fact that high-voltage transmission lines have either three or

six wires, plus one or two light wires.

Except for a few pro bl ems at th e end of this chapt er, an d except for

the example s in Sees. 16.7.1 and 16.7.3, we shall be concerned henc efor th

solely with two-wi re single- phase cu rre nts , as in Fig. 1 -4.

(o

16.5 ALTERNA TING CURRENTS IN

CAPACITORS AND INDUCTORS

Figure 16-6a shows a source V connected to a capacitor C. Then

Q/C = V = V0

cos cot, (16-12)

I = •%•= C% = -toCV0 sin mt, (16-13)dt at

= coCV0

cos (cot + TI/2). (16-14)

Figure 16-6 (a) Capacitor C connected to a source of a l t e r n a t i n g voltage V.

(b) Voltage V and c urre nt / as functions of the time for a capac itor , with

l/a>C = 2 o h m s .



38 0

A

- \

Figure 16-7 Power t ransfer between a source and a capaci tor , as in Fig. \6-6a, fo r

C = to = V 0 = i. A: appl i ed vol t age F 0 co s co t as a funct ion oh. B: — CwVl si n cot •

cosojf, t he power absorbed by the capac i tor . C.(j) CV 1,co s2

cot, t he energy s tored

in the capac i tor .

The cur ren t is p ro po r t io na l to the f r equency an d to C. I t leads t he vo l t age

by 7t /2 as in Fig. 16-6b. The cu rre nt i s said to be in quadrature wi th the vo l t age .

Thi s expres s ion i s used whenever the phase d i f f e r ence be tween two quan t i t i es

is n/2, l ead ing o r l agg ing .

T h e i n s t a n t a n e o u s p o w e r a b s o r b e d b y t h e c a p a c i t o r i s

Th i s quan t i ty i s shown as curve B in Fig. 16-7. I t wi l l be observed that power

f lows a l t e rna te ly in to and ou t o f the capac i to r . Curve C of the same f igure

shows the energy jCV2 s tored in the capaci tor as a funct ion of the t ime. The

s to red energy osc i l l a t es be tween zero and \CV\.

Figure 16-8a shows an idea l induc tor L wi th zero r es i s t ance connec ted

VI = -mCVl cos co t si n cot. (16-15)

http://6-6a/

http://6-6a/



381

to a source V. From Sec. 12.3,

LiU

- = V = V0

cosojf, (16-16)dr

I = — sin ojf = — cos (cot - TI/2). (16-17)coL OJL

We have neglected the constant of integration because we are only interested

in the steady-state sine and cosine terms. The curre nt is inversely pro por tio nal

to the frequency and to L. Figure 16-8b shows V and I as functions of f. The

current is again in quadrature with V. It lags the voltage by JT/2 radians.

As in the case of the capacitor, the average energy dissipation in the

inductor is zero:

V2

VI = — cos cot sin cot, (16-18)coL

as in curve B of Fig. 16-9. The sto red energy oscillates bet ween zero and

iUyjcoL)2

, or ^LI2

0(Sec. 13.5).



Figure 16-9 Power t ransfer between a source and an inductor as in Fig. 16-8o, for

L = to = F 0 = 1. A: appl i ed vol t age F 0 co s tot as a funct ion of t. B: (Vl/wL) sin cof

coscof, t he power absorbed by the induc tor . C:(j) L(V 0/coL)2 s in 2 cor , the energy

s tored in the induc tor .

16.6 COMPLEX NUMB ERS

Complex numbers a r e ex tens ive ly used fo r the ana lys i s o f e l ec t r i c c i r cu i t s .

Ord inar i ly , num be r s a r e used to expres s quan t i ty , s ize , e t c . Fo r

example , one s ays tha t there a r e 20 cha i r s in a room, o r tha t a bu i ld ing i s

30 .5 me te r s h igh . Such nu m be rs a r e s a id to be real numbers.

Imaginary numbers are r ea l number s mul t ip l i ed by the square roo t

of minus one .

There a r e two symbols used fo r ( — 1 )1 / 2

. In genera l , one uses the

s y m b o l i. H o w e v e r , w h e n e v e r o n e d e a l s w i t h e l e c t r i c a n d m a g n e t i c p h e

no m ena , the cus to m i s to use j ins tead of i because , in the pas t , t he symbol

for current was i. Th e SI sym bol fo r cur r en t i s no w I. T hu s 3i, or 3/, is an

i m a g i n a r y n u m b e r .

A complex number i s the sum of a r ea l num be r an d an ima gina ry

nu m be r , like 2 + 3/ , for exa mp le. It i s useful to rep rese nt com plex nu m be rs



383

in the complex plane, as in Fig. 16-10, wi th th e real par t (2) plo t ted ho r i

zon ta l ly , and the imaginary par t (3j) ver t ical ly .

I n C a r t e s i a n c o o r d i n a t e s , t h e c o m p l e x n u m b e r z i s thus wr i t ten

z = x + jy. (16-19)

In po la r coord ina tes , f rom F ig . 16-10 ,

z = r c o s 0 + jr si n 0 = r ( cos 6 + j sin 0) , (16-20)

w h e r e r is the modulus of z, a n d 0 is its argument. H e r e

r = (x 2 + y 2 ) 1 ' 2 , (16-21)

a n d 9 i s the angle between the radius vector and the x-axis , as in the f igure.

If x is positive, z is e i the r in the f i rs t or in the four th qu ad ra nt an d

9 = a r c t a n (y/x), (16-22)

If x is negative, z i s in the s econd or in the th i rd quadran t and

0 = a r c t a n (yfx) + it, (16-23)

where the f i rs t term on the r ight i s assumed to be ei ther in the f i rs t or in the

f o u r t h q u a d r a n t .

N o w

e ie = co s 0 + j si n 6, (16-24)

if 6 i s expressed in radians. This r a ther su rpr i s ing r e l a t ion , known as Eulers

formula, can be checked by wri t ing down the ser ies for e je, for cos 6, and for

si n 0. This will be d o n e in Pr ob . 16-12. The refore , f rom Eq. 16-20,

z = x+jy = re J0, (16-25)

w h e r e x, r, r, 0 are r e l a t ed as in E q s . 16-21, 16-22, and 16-23.

C o m p l e x n u m b e r s in Car tes i an fo rm may be a d d e d or s u b t r a c t e d

b y a d d i n g or s u b t r a c t i n g the real and i m a g i n a r y p a r t s s e p a r a t e l y .

T o a d d or s u b t r a c t c o m p l e x n u m b e r s in polar form, one f i rs t t ransforms

t h e n u m b e r s i n t o C a r t e s i a n f o rm .

C o m p l e x n u m b e r s in Car tes i an fo rm may be m u l t i p l i e d in the u s u a l

w a y , r e m e m b e r i n g t h a t j 2

= - 1:

{a + bj)(c + dj) = (tie - bd) + j(ad + be). (16-26)

In polar form.

(r 1ei9l)(r 2e

ie2) = r i r 2 e m + , h ) . (16-27)

N o t e t h a t, in th e p r o d u c t , the m o d u l u s is e q u a l to the p r o d u c t of the m o d u l i ,

whi l e the a r g u m e n t is the sum of t h e a r g u m e n t s .

To d iv ide one c o m p l e x n u m b e r by a n o t h e r , in C a r t e s i a n c o o r d i n a t e s ,

o n e p r o c e e d s as fo l lows :

(16-28)a + bj _ (a + bj)(c - dj) _ (ac + bd) + j(bc - ad)

e + dj ~ (c + dj)(c - dj) ~ c2

+ d2

Divi s ion in p o l a r c o o r d i n a t e s is s i m p l e :

- = 7 L t v < o . - » 2 i (16-29)

r 2e> e> r

In the q u o t i e n t , the m o d u l u s is e q u a l to the q u o t i e n t of the m o d u l i , and

t h e a r g u m e n t is t h a t of the n u m e r a t o r , m i n u s t h a t of the d e n o m i n a t o r .

16.6.1 EXAMPLES: USING COMPLEX NUMBERS

a) From Eq. 16-24,

^ 1 2 = j_ej« = _ i e-Jxli = _j (16-30)



16.7 Phasors 385

b) Le t us expres s the complex nu mb er 2 + 3 / in pola r form:

2 + 3/ = (2 2 + 3 2 ) " 2 exp ./ (^arc tan ^ = 3.61 ex p (0.983/). (16-31)

R e m e m b e r t h a t t h e a n g l e 0 must be expressed in radians.

c) Th e squa re of 2 + 3/ is

(2 + 3/) 2 = (2 + 3/)(2 + 3/) = 4 - 9 + 12/ = - 5 + 12/. (16-32)

In polar form,

(2 + 3/) 2 = 3 . 6 1 2 e x p ( 2 x 0.983/) = 13 exp (1.966/). (16-33)

d) Finally, let us calculate (2 -(- 3/)/(l -(- j), f i rs t in Cartes ian and then in polarc o o r d i n a t e s .

2 + 3/ = (2 + 3/) (l - ,/) = (2 + 3) + (3 - 2) j

l + j (1 + ; )(! - j ) ' 2

= 2.5 + 0.5/. (16-35)

o r

2 + 3j 3.61 exp (0.983/)= — - = 2.55 ex p 0.197/ . 16-36

l + j 1.4 14 e x p [ (7 r / 4 ) / ]

76.7 PHASORS

W he n dea l ing wi th a l t e rna t ing cur r e n t s an d vo l t ages , one mus t d i f fe r en t ia t e

s ine and cos ine func t ions wi th r espec t to t ime over and over aga in . Now

the usua l p roce dur e for d i f f e r en ti a t ing these func t ions i s inconv en ien t . Fo r

example, to di f ferent iate cos cot, the cos ine funct ion is ch an ge d to a s ine,

the resul t i s mul t ipl ied by OJ, and t he s ign is cha nge d. To f ind the se con dder iva t ive, the s ine is cha ng ed bac k to a cos ine, and th e resul t is aga in m ul t i

pl ied by co , but th i s t ime wi thou t chang ing s ign , and so on .

I t i s poss ible to s impl i fy these o pe rat ion s in the fol lowing w ay.

F i r s t , r e m e m b e r t h a t

eJ"" = cos cot + j si n cot, ( 1 6 - 3 7 )



38 6

Figure 16-11 The phasors V, /o;V, and — orV, and the vol t age V i n the complex p lane .

No te tha t the pha sor ro ta t es abou t the or ig in a t the angu la r ve loc i ty co. W e hav e

m a d e co equal to 1/2 . T h e q u a n t i t i e s dV/dt a n d d2

V / d t2 are the pro jec t ions of j w V

a n d — ( o2

\ l on the real axis .

f rom Eq. 16-24. Then

V = V 0 c o s co t = R e (V 0ej""l (16-38)

where the opera to r Re means " r ea l par t o f " what fo l lows .

Re me mb er a l so tha t the t ime der iva t ive o f the exp one n t i a l func t ion

r e d u c e s t o a s i m p l e m u l t i p l i c a t i o n :

j^V^'-") = )co(V QejM

). (16-39)

C o n s i d e r n o w t h e q u a n t i t y

V = V0eJu

" = V 0 cos cot + jV 0 si n cot. (16-40)

Thi s quan t i ty , which i s ca l l ed a phasor. is a vec to r tha t ro ta t es a t t he a ng u la r

veloci ty co in the complex plane, and i ts project ion on the real axis i s V, as

in Fig. 16-11. In d i f f e ren t i a t ing V, the r ea l and the im agina ry pa r t s r em ain

s e p a r a t e :

c/V— = — coK 0 sin cof + jcoV0 c o s cot, (16-41)dt

co2 V 0 c o s cot - jco

2

V 0 sin cot. (16-42)d

2

M

' d t 2



16.7 Phasors 387

The same applies to all the derivatives. Thus

dV dy d2

V d2M— = Re — , — = Re -dt dt dt

1

dtR e - — = R e ( 1 6 - 4 3 )

One therefore replaces the real voltage

V = V0

cos cot (16-44)

by the phasor

V = V0eh

°'. (16-45)

This then adds to V the parasitic term jV0

sin at. From then on, the operator

d/dt can be replaced by the factor jw. In the end, one recovers the real part.

In this way, differential equations become simple algebraic equations.

Inversely, an inte gra tio n with respect to time is replace d by a division

by jca.

Fi gu re 16-11 also shows y'coV an d — OJ2

V . Note that multiplying a

phasor by jw increases its argument by 71/2 rad ian s a nd multipl ies its

modulus by OJ.

Ph as or s are used for F, 7, and Q. They are also widely used to represent

all kinds of sinusoidally varying quantities.

We use bold-faced sans serif letters for phasors (for example, V), but

it is mor e com mo n t o use or di nar y light-faced italic letters. For e xample ,

the letter V usually means either V0

cos ojf or F0

ex p jot, depending on the

context. Moreover V often stands for F0/ 2

1 / 2

! Surprisingly enough, this

seldom leads to confusion.1

16.7.1 EXAMPLE: THE PHASORS IN A THREE-PHASE SUPPLY

In Fig. 16-5, the voltages at A. B , C can be written either as in Eqs. 16-9 to 16-11,

o r as fol lows.

V.4 = F0

exp./cor, (16-46)

V„ = V0exp ./(cor + 2n/3), (16-47)

+

In a handwritten text, one can identify a phasor by means of a wavy line under the

s y m b o l , for e x a m p l e , ^ .

388

jy

4

Figure 16- 12 T h e p h a s o r s V 4 . V B , V ( for the voltages at A, B, C in Fig. 16-5, at

tut = nil.

V< = V 0 exp j(an + 47 i/3).

These three phasors are shown in Fig. 16-12 at o>t = nil.

(16-48)

16.7.2 ADDITION AND SUBTRACTION OF PHASORS

The r ea l quan t i ty V. o r I, o r Q. tha t i s rep rese nte d by a ph as or is given by

the p ro jec t ion o f the phasor on the x -ax i s . Thus , two vo l t ages , o r two cur

ren t s , or two charges , can be adde d by fi rs t add in g the i r pha sor s vec to r i a l ly ,

and then t ak ing the p ro jec t ion o f the sum on th e x -ax i s. Th i s is becau se the

pro jec t ion o f the sum of tw o vec to r s i s equa l to the sum of the p ro jec t ion s .

Similar ly , the di f ference between two vol tages , or two currents , or two

charge s , can be ob ta in ed f rom the d i ff e rence be twee n the c or r e spo nd ing

p h a s o r s .

16.7.3 EXAMPLE : THE PHASORS IN A THREE-PHASE SUPPLY

Figur e 16- 13 show s the th r ee phasor s V 4 . V B . V ( - of Fig. 16-12 at r = 0. W e wish to

f ind the voltage of A with respect to B, o r V A — V„. as a function of r, kno win g tha t

VAis V 0

co s cot.



389

Figure 16-13 Phasor diagram for calculating V 4 - V B for the star-connected sources

of Fig. 16-5.

We could, of course, find VA — VB by trigonometry, since

VA - VB = V0 cos cot - F 0 co s (cot + 2TC/3). (16-49)

Is it easier and more instructive to use phasors. On e can see immediately, from Fig.

16-13, that the phasor V,, — V f l has a modulus of 2(3 ' 2 2)K0 and an argument of

— 30 degrees, or — jr/6 radian at t = 0. Thus

VA

- VB

= 31 2

F 0 cos (cot - TC/6). (16-50)

If the rms voltage at A, B , C is 120 volts, the rms voltage between A and B is 31 , 2

x

120, or 208 volts.

Figures such as 16-13 that serve to perform calculations on phasors are known

as phasor diagrams.

16.7.4 WHEN NOT TO USE PHASORS

a) One can use phasors only if the time dep end enc e is strictly of the

form cos (cot + cp). where cp is a cons tant . Phas ors must not be used if one

has, say, a squar e wave for F , or even a da mp ed sine wave. They must not

be used if the com po ne nt s are non-lin ear, for exam ple, if one has n on- line ar

resis tors as in Sec. 5.3 .

b) Wh ene ver p ro duc t s of pha sor s , o r p ro du c t s o f e j'0' t e rms a re en

counte red , one r ever t s to the s ine o r cos ine func t ions .

For example , suppose one wishes to ca lcu la t e the power d i s s ipa tedin a r es i s to r . The ins t an taneous power i s VI, wi th

V = V 0 cos tot, I = 7 0 c os cot. (16-51)

I f one uses the phasor s

V = V0eic

", I = I0eJm

', (16-52)

one is t em pted to conc lude tha t

VI = V 0I0e2J

<"' a n d t h a t VI = V 0I0 co s 2cot. (16-53)

This is wrong, as one can see from Sec. 16.2.

The er ror , here, comes f rom the fact that

R e (ej<

°') R e (eiu

") # R e (e2jm

'). (16-54)

O n e c a n d i v i d e a p h a s o r b y a n o t h e r p h a s o r o r b y a c o m p l e x n u m b e r ,

however , as we shal l see in the next chapter . In those cases one has s t raight

f o r w a r d o p e r a t i o n s w i t h c o m p l e x n u m b e r s .

16.8 SUMMARY

Al te rna t ing vo l t ages and cur r en t s a r e o f the fo rm

V = V 0 cos cot, (16-1)

I = I0 cos (cot + cp). (16-2)

We have as sumed here tha t the phase o f V is ze ro at f = 0. In gen era l, the

c u r r e n t I i s no t in phase wi th the ap pl ie d vol tag e. Unles s specified oth erw ise,

one does no t use V 0, b u t r a t h e r

(16-6)



16.8 Summary 391

as a measure o f the magni tude o f a vo l t age . The s ame app l i es to a cur r en t :

one uses

7 r m s = IJ2il2

(16-6)

t o speci fy the ma gn i tud e o f a c ur r en t .

In three-wire single-phase current, one uses two sources , back to back ,

wi th a common cen te r wi re . Wi th three-phase current, one has th ree sources

with ph ase di f ferences of 120 degre es , co nne cte d to thei r loa ds wi th th ree

wi res , p lus a common cen te r wi re .

In a capac i to r C, t he cur r e n t is p ro po r t io na l to toC a n d leads t h e

app l i ed vo l t age by n/2 r ad ians . In an induc tor L , the current i s inversely

p r o p o r t i o n a l t o OJL a n d lags t he vo l t age by n/2 r ad ians . In bo th cases energy

f lows al te rna tely in an d out of the elem ent an d ther e is zer o ave rage e nergy

transfer .

Complex numbers are of the form a + bj , w h e r e a a n d b are real , and

j = ( — l ) 1 ' 2 . To p lo t complex numbers in the complex p lane , one wr i t es

: = x+jy, (16-19)

o r

z = r (co s 0 + j sin 0), (16-20)

w h e r e r is the modulus, a n d 0 is the argument of z, as in Fig. 16-10, and

r = (x 2 + y 2 ) 1 / 2 , (16-21)

0 = arc tan - or 0 = arc tan - + n. (16-22, 16-23)

x x

Also ,

e>° = cos 9 + j sin 6, (16-24)

so tha t

z = re j". (16-25)

C o m p l e x n u m b e r s m a y b e a d d e d , s u b t r a c t e d , m u l t i p l ie d , a n d d i v i d e d .

A phasor is a com plex nu mb er wh ose real part is a cosi ne function

of the time. For example.

V = V 0eiv

" = V0 C O S cot + jV0 sin tot. (16-40)

Th e additi on of the parasitic term jV0 sin rot to t he re al vol tag e

V0 co s cot trans form s the cosine function into an exp onen tial function.

Differentiation with respect to time can then be replaced by a multip licati on

by the factor jo. Inversely, an integration with respect to time becomes a

division by jco.

PROBLEMS

16-1E RECTIFIER CIRCUITS COMPARED

Figures 16-14aand bshow. respec t ive ly , half-wave a n d full-wave rectifier circuits.

W h a t are the waveforms ac ross th e res i s t ances?

B 1

(a ) (b)

Figure 16-14 (</) Half-wave rect i f ier c i rcui t . The t r i angula r f igure at the top

represent s a rectifier, or d i o d e . C u r r e n t ca n p a s s t h r o u g h it only in the di rec t ion of

t he a r row head , (b) Full-wave rect i f ier c i rcui t . The s igns show th e pola r i t i e s at a

given ins tant : poin t s A and B have oppos i t e phases . See P r o b . 1 6 - 1.

16-2E RMS VALUE OF A SINE WAVE

S h o w t h a t th e root mean square va lue of V0 co s on is V0/2112

.

Problems 393

BRIDGE RECTIFIER CIRCUIT

The vo l tage V A — V B

applied to the circuit of Fig. 16-15a is a symmetr ical sawtooth wave as in Fig. 16-15b. with a peak voltage of 100 volts. Fhe diodes in A C . A D .

etc.. pass current only in the direct ion of the ar row head.

Battery chargers use either this circuit or thos e of Fig. 16-14. Th e re sistan ce

shown in the f igures then represen ts the bat tery. Th e source for a bat te ry c harge r

is a t r ans former connected to A C power l ine, and V 0 is about 20 volts for a 12-volt

ba t te ry .

a) Calculate the average and rms values of V A — V B in Fig. 16-15b.

Solution: The ave rage value of V A — V B is zero.

The rms value is given by

( F , - F B ) r

2

m s = i j o2 ( ^ - VBfdt. (II

The l imits of integrat ion cover one per iod. To evaluate this integral , we spli t the

t ime interval into four equal par ts of 0 .5 second, giving four integrals:

\j:5

(VA-VBfdt+

l

-^(VA-VBfdt + - - :

(a) (<)

Figure 16-15 (a) Bridge rect if ier circuit , (b) Saw-tooth voltage applied between

poin ts A a n d B. (c) Vol tage be tween po in ts C and D. See Prob. 16-3.




Now these in tegra l s a re a l l equa l and

(V A - VB)rm = 100 /3 1 2 = 57.7 vol ts. (4)

b) Draw the curve of V c — V D as a funct ion of the t ime.

Solution: Con s ider the in te rva l 0- 1 second in F ig . 16-15b. where A i s posi t ive

with respect to B. During this t ime the current f lows along the path ACDB a n d

V c — VDis positive as in Fig. 16-15c.Dur ing the in te rva l 1-2 seconds , A i s negat ive with respect to B and the cur rent

f lows along BCDA. T h u s V c — V D is again positive, as in Fig. 16-15c, and so on.

c) What are the average and rms values of V c — V D1

Solution: The average value is 50 vol ts .

The rms value is the same as for V A — V B, or 57.7 vol ts .

16-4 R M S VALUE

Find the rms values for the waveforms shown in Fig. 16-16.

16-5E THREE-WIRE SINGLE-PHASE CURRENT

W hat i s the max im um va lue of the vol t age d if ference b e tween poin t s A a n d C

in Fig. 16-4b, in the 1 20/240 vol t sys tem com mo nly u sed for inter ior w iring ?

16-6E THREE-PHASE CURRENT

Show that , for any t ime r ,

cos cot + cos (cot + 2n/3) + cos (cor + 4n/3) = 0.

76-7 ROTATING MAGNETIC FIELD

Th ree iden t ical coi ls, orie nted as in Fig. 16-17, pr od uc e ma gn et ic f ie lds as

fol lows:

B a = B 0 co s cot, B b = B 0 cos (cor + 2n/3), B L = B 0 cos (cor + 4TC /3).

Show that the resul t ing f ie ld has a magni tude of 1.5B 0 and ro ta tes a t an angula r

veloc ity co.

This is the principle ut i l ized to obtain rotat ing magnet ic f ie lds in large electr ic

m o t o r s .



39 5

V

2V 0\-

Figurc 16-17 (a) Three coi l s , fed wi th a l t e rna t ing cur rent s of the same ampl i tude

and sam e f requency, prod uce m agne t i c f ie lds of equa l mag ni tud es and or i en ted as

in (ft). Th e two a rro ws in («) sho w wh ere th e coi ls tou ch. If the cur ren ts are pha sed

as in Pr ob . 16-7. the direct io n of the resul t ing m agn et ic f ield rotate s a t the

frequency / ' .




16-8E DIRECT-CURRENT HIGH-VOLTAGE TRANSMISSION LINES

From many points of view, alternating current is much preferable to direct

current for power distribution. However, as we shall see, line losses are lower withdirect current.

Consid er a long high-voltage overhea d transmission line. Its maxi mum operat ing

voltage depends on several factors, such as corona losses (Prob. 7-14), the size of the

insulators, etc. So, for a given line, the instantaneous voltage between one conductor

and ground must never exceed a certain value, say F 0 . Otherwise, the down time and

the cost of maintenance become excessive. The cost of a line increases rapidly with its

voltage rating.

Th e curren t in the line can be ma de nearly as large as one likes with out dam agi ng

it, since the conductors are well cooled by the ambient air. However, the power loss

increases as the square of the current. So. the lower the current the better.

With direct current, one has two conductors operating at + V0and — F 0 with

respect to ground. Fhe power delivered to the load is 2V0IDC.

a) With single phase alternating current, we have two wires at + F 0 cos tot an d

— F 0 cos ojt.

Show that, for the same power at the load, the rms current /s p

is 2l/2

IDC.

The I2

R losses in the line are twice as large as those with direct current.

b) With three-phase alternating current, we have three wires at F 0 cos cot,

V0

cos (cot + 2TI/3), F0

COS (cot + 4TI/3). We assume that the three load resistances

connected between these wires and ground are equal. Then the current in the ground

wire is zero.

Show that, for the same total power delivered to the three load resistances, the

rms currents / l p a re ( 2 /3 ) 21 / 2

/ D C * 7 D C .

With three-phase alternating current, the rms currents are about the same as

with direct current, but we have three current-carrying wires instead of two, so that

the losses are 50% larger th an with direct curre nt.

16-9 ELECTROMAGNET OPERATING ON ALTERNATING CURRENT

An electromagnet with a variable gap length is opera ted on alter nating current.

Show that the rms value of the magnetic flux is independent of the gap length,

for a given applied alternating voltage, and neglecting leakage flux.

16-10E COMPLEX NUMBERS

a) Express the compl ex n umb ers 1 + 2/', — 1 + 2/', — 1 — 2/, 1 — 2/ in pol ar

form.

b) Simplify the following expressions, leaving them in Cartesian coordinates:

(1 + 2/)(l - 2j). (1 + 2/ )2

,

1/(1 + 2y)2

, (1 + 2/) /(l - 2j).


Com plex num be rs in pol ar form are often writt en as rZ_0, where /• is the

modulus and 0 is the argument.

Expre ss 1 + 2/ in this way.

Problems 397


Use the series

e* =1 + x + | . \- 2!) + (x 3 /3! ) + • • •

sin x = x - (x 3 3!) + (x s / 5 ! ) - ( x 7 7!) + • • •

cos x = 1 - (x 2 2!) + (x 4 4!) - (x* 6!) + • • •

to show tha t

e>* = cos v + j sin x.

I6-I3E COMPLEX NUMBERS

Show tha t exp jn = — 1.


W hat hap pen s to a comp lex nu mb er in the complex p lane when i t i s (a) mul t ip l i ed

by j. (b | mul t ip l i ed by j2, (c) divided by /'?

16-15 THREE-PHASE ALTERNATING CURRENT

Show, by t r igonomet ry , tha t the expres s ions on the r ight -hand s ide of Eqs . 16-49

and 16-50 a re equ a l .

16-16 CALCULATING AN AVERAGE POWER WITH PHASORS

In Sec. 16.7.4 we saw that , wi th al ternat ing currents , one must no t calculate a

power by us ing the product VI.

a) Show that , for the general case where

V = V 0 co s mt, I = 70

cos (wt + </>)

in a load, the average power is

P , v = ( l / 2 ) 7 0 / 0 cos q> = F r m 5 / r m 5 co s q> .

The t e rm cos q> is called the power factor of the load (Sec. 18.2).

b) Show tha t

P„ = ( l /2)ReVI*,

where I* is the complex conjugate of I, ob tain ed by ch ang ing th e s ign of i ts imag inary

par t .



CHAPTER 17

ALTERNATING CURRENTS: II

Impedan ce, Kirchoff's Laws, Transform ations

17.1 IMPEDANCE

17.1.1 Example: The Impedance and the Admittance of a Coil

17.2 THE EQUATION 1 = V / Z IN THE COMPLEX PLANE

17.3 KIRCHOFF'S LA WS

17.3.1 Example: The Series LRC Circuit

17.4 ST A R-DEL TA TRANSFO RM A TIONS

17.4.1 Example: Transforming a Delta into a Star

17.5 TRANSFORMATION OF A MUTUAL INDUCTANCE

17.5.1 Example: Two Superposed Coils

17.6 SUMMARY

PROBLEMS



This cha p te r rough ly para l l e l s Cha p te r 5 , mo s t o f which dea l t wi th d i r ec t

cur r e n t s f lowing th ro ug h r es is t ive c i r cu i ts . I ns t ead o f d i r ec t cur r en t s , we no whav e a l t e rna t i ng cur r en t s and , ins t ead o f r es i s t ances , we have im peda nces .

As we sha l l s ee , bo th Ki rchof f ' s l aw s and the s t a r -de l t a t r ans form at ion s

remain va l id . We sha l l a l so l ea rn how to t r ans form a mutua l induc tance

in to a s t a r ; t h i s t r an s form at ion doe s no t have a d i r ec t - cur r e n t equ iva len t .

Let us retu rn t o the s im ple ci rcu i ts of Figs . 16-1, 16-6, an d 16-8, wh ere ana l t e rn a t ing vo l t age i s app l i e d succes s ive ly to a r es i s to r , t o an induc tor , and

to a capac i to r .

I f an a l te rna t in g vol t age V is app l ied to a re s is tor ,

17.1 IMPEDANCE

I = MIR. (17-1)

I f V is app l ied t o an ind uc tor , then, rew ri t ing E q. 16-16,

V = jcoLl (17-2)

a n d

I = V / j c o L . (17-3)

T h e q u a n t i t y jtvL plays the s ame ro le as R and is cal led the impedance of



400 Alternating Currents: II

17.1.1

t he induc tor . The symbol fo r impedance i s Z. I m p e d a n c e s a r e m e a s u r e d

in ohms, l ike res is tances .

W h e n V i s app l i ed to a capa c i to r ,

I = Qa>V = , (17-4)1/jcoC

f rom Eq. 16-13. H e r e 1/jcoC = —j/ojC i s the imp eda nce o f the capa c i to r .

M or e genera l ly , an imp eda nce i s com plex an d is wr i t t en as

Z = R+jX = \Z\ej", (17-5)

w h e r e R is the resistance, X is th e reactance, |Z | is the modulus or the magnitudeof the impedance , and 9 is its phase angle. A pos i t ive reac tan ce is said to be

inductive, while a negat ive reactance is said to be capacitive.

T h u s , i n genera l , t he vo l t age V a c r o s s a n i m p e d a n c e Z c a r r y i n g a

cur ren t I i s

V = Zl , (17-6)

a n d

I = V/Z , Z = V/l. (17-7)

Th e inver se o f an im ped anc e i s an admittance:

Y = 1 /Z . (17-8)

Impedances in s e r i es and in para l l e l a r e ca lcu la t ed in the s ame way

as resistances in series and in parallel (Sees. 5.4 and 5.5).

EXAMPLE: THE IMPEDANCE AND THE ADMITTANCE OF A COIL

A coi l has a res is tance of 1000 ohms and an inductance of 3.000 henrys . I ts impedance

at a frequency of 100.0 hertz is

R + jcoL = 1000 + 2n x 10 0 x 3/ = 1000 + 18 85 /oh ms , (17-9)



40 1

Figure 17-1 (a) The impedance of the coi l of Sec. 17.1.1 in the complex plane.

(h) V an d I = V Z for th is coil at t = 0. Th e mo du li of V and I are n ot to scale .

o r

1 8 8 5 \Z = (1000 2 + 1 8 8 5 2 ) 1 2 e x p ( / a r c t a n — — 1 .

= 2134 exp 1 .083/ohm s . (17-10)

Figure 17-1 shows Z in the complex plane.

If one appl ies across this coi l a vol tage V 0 cos co t with an rm s value of 10 vol ts ,

the current is

2134 ex p 1.083/ 2134exp./(ojt - 1.083). (17-11)

2 1 2 x 10/ = „ „ , co s (cut - 1.083)

2134(17-12)

The n the cur ren t / ha s an rms va lue of 10/2134. or 4 .686 mi l l iamperes . The cur ren t

lags the vol tage by 1.083 radians .




T h e a d m i t t a n c e o f th e c o i l is

y _ I _ __ J 1000 - 1885/ _ 1000 - 1885/

~ Z ~ 1 0 0 0 + 1 8 85 / ~ 1 0 0 0 2 + 1 8 8 5 2 ~ (2134) 2 '

= (2 .196 - 4 .1 39 / I1 0" 4

Siemens.

Y = e x p ( - 1 . 0 8 3 / ) ,

= 4.686 x 1 0 _ 4 e x p ( - 1 . 0 8 3 / ) S i e m e n s .

(17-13)

(17-14)

(17-15)

(17-16)

17.2 THE EQUA TION I = V / Z IN THE COMPLEX PLANE

I t i s ins t ruct ive to cons ider the equat ion I = V/Z i n the complex p lane .

W r i t i n g

\Z\eJe, (17-17)

then

\z \e* i z r(17-18)

Th us the mo du lus o f the phas or I i s the m od ulu s o f V div ided by |Z | , whi le

i t s a rgument i s the a rgument o f V minus tha t o f Z. as in Fig. 17-1.

17.3 KIRCHOFF'S LAWS

Kirchoff 's laws of Sec. 5 .7 apply to al ternat ing currents jus t as wel l as to

d i r ec t cur r en t s .

The cur r en t l aw s t a t es tha t the sum of the cur r en t s en te r ing a node

is zero. As we saw in Sec. 5 .9 , one normal ly uses mesh currents , ins tead of

branch currents , so as to avoid us ing this law expl ici t ly .



40 3

L RC

+ Q

V

Figure 17-2 Inductor L, res is tor R, a n d c a p a c i t o r C, connected in series across a

source V.

The vo l t age l aw s t a t es tha t the sum of the vo l t age d rops a round a

me sh is zer o. T o use this law, one show s plus an d m inu s s igns for the so urce sand for the charges , and ar rows for the currents , as in Fig. 17-2.

The s igns and the d i r ec t ions o f the a r rows a re a rb i t a ry , s ave fo r one

excep t ion . If a cur r en t I f lows in to a capa c i to r e l ec t rode ca r ry ing a charg e

Q , and if on e wishes to wr i te th at I = dQ/dt, t hen one mus t have the cur r en t

f lowing int o the pos i t ive Q , as in Fig. 17-2. W ith th e ar ro w s in the op po s i te

d i r ec t ion , one would have to wr i t e I = —dQ/dt.

Note tha t these s igns and a r rows a re no t meant to ind ica te the po la r i t i es

at a pa r t ic ula r ins ta nt , as they we re in Fig. 16-4. Th ey are s imply used as a n

a id in wr i t ing down the mesh equa t ions cor r ec t ly .

17.3.1 EXAMP LE: THE SERIES LRC CIRCUIT

Figu re 17-2 sho ws a ci rcui t wi th an ind uc tor L, a res is tor R, and a capac i tor C .

connected in series across a source V.

We shal l (a) wri te down the different ia l equat ion for the mesh, (b) rewri te i t in

ph aso r no tat ion to f ind Q and I . (c) wri te dow n the valu e of I direct ly from the relat ion

I = V /Z , (d) ded uce / as a funct ion of the t im e, and (e) discuss the value of Z as a

funct ion of the frequency.

a ) F r om Ki rchof f ' s vol t age l aw. s t a r t ing a t the lower l e ft -hand c orner .

dl u

-L RI - -- + V = 0.t C(17-19)

or

dQ Q+ R — + - = V.

dt C(17-20)




b) In pha sor nota t io n , th i s equ a t ion b ecom es

(jwfL + jmR + CQ = V,

so tha t

{jtofL + jcoR + (1/C)

Mul t ip ly ing both s ides by jco,

jwQ = IR + jmL + (1/jwC)

(17-21)

(17-22)

(17-23)

c) Since we have impedances in series , the total impedance is the sum of thei m p e d a n c e s :

Z = R + jtoL

T h e n

1

jcoC

Z R + jtoL + (1/jcoC)

(17-24)

(17-25)

as above .

d) To find I as a funct ion of the t ime, we f i rs t express the impedance in the polar

form |Z | exp (j9):

Z = R + j[coL - (1/coC)],

= \R2

+ [coL - ( l / c » C ) ] 2 } 1 / 2 e x p j arc t ancoL - (1/coC)

T h u s

| Z | = [R 2 + [coL - ( 1 / c o C ) ] 2 } 1 ' 2 ,

a n d

0 = arc t an

_ K0e"

0

'

1

" W '

Vo

IZ l

wL - (1/coC)

R

(17-26)

(17-27)

(17-28)

(17-29)

(17-30)

(17-31)



405

150 r

f (hertz)

Figure 17-3 Reactance as a function of the frequency for the circuit of Fig. 17-2, for

L = 1 millihenry and C = 1 microfara d. F he reactan ce X is (oL — 1/coC.

The current I is the real part of I:

/ =Z

cos (cot — 0). (17-32)

e) Note that

Since

| Z | an d 0 are both functions of the frequency.

Z = R+jX = R + jcoL 11

o) 2LC(17-33)

the reactance X is zero when co2LC = 1. At tha t frequency the voltage dro p on L

exactly cancels that on C and the circuit is said to be resonant. With series resonance,the voltages across L and across C can be larger than the applied voltage.

Fig ure 17-3 shows the reac tance as a function of the frequency for L = 1 milli

henry and C = 1 microfar ad.

This is a case of series resonance because the capacitor and the inductor are

connect ed in series with the source. Proble m 17-15 concerns parallel resonance.

In the case of parallel resonance, the currents through the capacitor and through

the inductor partly cancel, and each one can be larger than the total current.




17.4 STAR-DELTA TRANSFO RMAT IONS

In Sec. 5.10 we saw how one can t r a n s f o r m a res is t ive s tar c i rcui t in to an

equ iva len t de l t a , and vice versa.

If one has com plex im ped ance s , as in Fig. 17-4a, ins tead of r es i s t ances ,

t h e t r a n s f o r m a t i o n r u l es a r e s imi la r . Th i s comes f rom th e fact that th e r e a

s o n i n g we used in Sec. 5.10 is in n o way l imited to r e a l i m p e d a n c e s . T h u s

ZA= ^

Z

f ' , „ , (17-34)z

hz

c

za

+ zh

+ zc

Z 7

z a + z„ + z e

z„zb

za + z

b+ z

c

YBYC

YA+ YB+ YC

YCYA

YA + YB+ YC

YAYB

(17-35)

(17-36)

(17-37)

(17-38)

F. = ^ J * . (17-39)

' YA+YB+YC

[ U

" >

i A

Figure 17-4 F h r e e n o d e s A, B, C. {a) D e l t a - c o n n e c t e d , (b) S t a r - c o n n e c t e d . T h e w a v y

l ines represent impe dance s .



17.4 Star-Delta Transforma tions 407

17.4.1 EXAMPLE: TRANSFORMING A DELTA INTO A STAR

As an example, le t us t ransform the del ta of Fig. t7-5a into a s tar a t a frequency of

one ki lohertz . In this case,

Z a = Z h= ifjcoC = - If)-'/ : ; : oh ms , (17-40)

Z f = 100 oh m s. (17-41)

Fro m Eqs . 17-34 to 17-36. and rem emb er ing tha t Z = 1/7 ,

Z A = Z B = 45.51 - 14 .30 /oh ms , (17-42)

Z c = - 2 2 . 7 5 - 7 2 . 4 3 / o h m s . ( 17 -4 3 )

T h u s Z A a n d Z B are each composed of a res is tance of 45.51 ohms, in series with

a capac i tor whose impedance i s —J/OJC, w h e r e

o>C = 1/14.30. (17-44)

C = 1/(14.30 x 2TT x 1000) = 11.13 mic rofa rads . (17-45)

Similarly, Z c i s composed of a res is tance of —22.75 ohms, in series with a capaci tor

of 2.197 microfarads .

Th e equivale nt s tar is sho wn in Fig. 17-5b.

Note that the res is tances of the equivalent c i rcui ts can have negat ive values .

A real c i rcui t can therefore be equiva lent to a f ic ti tious on e that is imp ossible tobui ld .




Note a l so tha t the res i s t ances and capac i t ances in the s t a r depend on the

op era t ing frequency. Fo r exam ple, in Pr ob . 17-16, it i s sho wn th at , a t one me gah ertz ,

the del ta of Fig. 17-5a is equ ivale nt to a com pletely different s tar .

17.5 TRANSFORMA TION OF A MUTUAL INDUCTAISCE

Figure 17-6a shows a mutua l induc tance M wi th i t s tw o co i ls co nnec ted

a t C. We can t r ans form th i s mutua l induc tance in to the s t a r o f F ig . 17-6b

in the fol lowing way.

T h e m e s h c u r r e n t s p, q, r mus t be the s ame in the two c i r cu i t s . For

s impl ici ty , le t us assume that C i s a t gr ou nd po ten t ial . This wi ll not res t r ic t

the genera l i ty o f the t r ans f orm at ion .In Fig. 17-6a,

V A = (p - q ) Z a + jcoM (r - q). (17-46)

Th e sec ond te rm on the r ight is the vol tag e induc ed on the lef t -hand s ide,

o r M t imes the rate of change of the current on the r ight (Sec. 12.1) . Here,

M i s pos i t iv e if the tw o coi ls are w ou nd in such a way t ha t a cu rren t in th e

pos i t ive di rect ion in one coi l produces in the other coi l a f lux l inkage that

is in the sa m e direc t ion as that d ue t o a pos i t ive cu rren t in that co i l (Sec. 12.1) .

O t h e r w i s e , M is nega t ive . W e have as sume d here tha t c lockwise cur r e n t s

are pos i t ive.

In 17-6b.

V 4 = (p - q)Z A + (p - r ) Z c . (17-47)

Similarly, in 17-6a,

V B = (q - r)Z b -jcoM(p - q), (17-48)

while, in 17-6b.

V B = (q - r)Z B + (p - r ) Z c . (17-49)

E q u a t i n g n o w t h e t w o v a l u e s o f V^, and then the two va lues o f V B

and s impl i fying.

p(Z„ - Z A - Z c) + q(Z A - Z a - jcoM) + r ( Z c + jcoM) = 0, (17-50)

p ( Z c + jcoM) + q(Z B - Z„ - jcoM) + r(Z„ - Z B - Z c ) = 0. (17-51)

40 9

Since these two e q u a t i o n s are va l id , wha tever the va lues of p, q, r. the six

coef f i c i en t s be tween paren theses mus t be zero , and

ZA

= Z„ + jcoM, (17-52)

ZB

= Zb

+ jcoM, (17-53)

Zc

= -jcoM. (17-54)

N o t e t h a t , if M is pos i t ive , the r e a c t a n c e —jcoM is c a p a c i t i v e . On the

o t h e r h a n d , if M is nega t ive , then — joM is the r e a c t a n c e of a pure i n d u c t a n c e

\M\. A g a i n , we h a v e a c i r cu i t t ha t is m a t h e m a t i c a l l y e q u i v a l e n t to one t h a t

c a n n o t be real ized in pract ice, s ince any i nduc tor , un les s it is s u p e r c o n d u c t i n g ,

pos ses ses a r es i s t ance .

17.5.1 EXAMPLE: TWO SUPERPOSED COILS

F i g u r e 17-7a s h o w s two superposed coils w o u n d in the same d i rec t ion , on e over the

other. Each coi l has a res i s t ance R and an i n d u c t a n c e L . Since the two coi ls are

s u p e r p o s e d . \ M\ = L .

In this example, with th e cur rent d i rec t ions shown in the figure. M is nega t ive an d

the equivalent c i rcui t is s h o w n in Fig. 17-7c; the i m p e d a n c e is R/2 A- JOJL. If we

c h a n g e d th e di rec t ion of one of the c u r r e n t s , we would s t i l l arr ive at the same resul t .

T h e i m p e d a n c e R/2 + j<oL is eas i ly expla ined: in effect, we h a v e one coi l made

u p of larger wire , so t h a t th e i n d u c t a n c e is L. and the res i s t ance is R/2.



41 0

(c)

W V \ AR

V \ A AR

rrr\21.

1

L

2L R

\ A A AR

Figure 17-7 Tw o superposed coi l s , (a) w o u n d in the same d i rec t ion , and (h) w o u n d

in oppos i t e d i rec t ions : (c ) and (d), equiva lent c i rcu i t s for (a) and (h). respect ively.

F igure 17-7b shows a s imi la r pa i r of coi ls , except that now the coi ls are w o u n d

in oppos i t e d i rec t ions . For t he cur rent d i rec t ions shown. M is posi t ive, th e equiva lent

circui t is s h o w n in 17-7d. and the t o t a l i m p e d a n c e is R/2.

T h i s is correct s ince, in t h i s example , the net magnet ic f ie ld is zero , and the two

res i s t ances are in paral le l .

Th e curre nt in a circuit elem ent is given by the volta ge acro ss it, divided by

th e impedance of the element:

The impedance of a resistor is R, th at of an in du ct or is /VoL, an d t ha t of a

capaci tor \/jcoC. An impe danc e is expressed as a complex nu mber , either

in Cartesian or in polar form:

77.6 SUMMARY

I = V/Z. (17-1)

Z = R+ jX (17-5)



Problems 411

where R is its resistance, X is its reactance, \Z\is the modulus or magnitude

of the impedance, and 0 is its phase angle.

The inverse of an impedance is an admittance:

Y = 1/Z. (17-8)

Series and parallel combinations of impedances are calculated in

the same way as with resistances.

Kirchoff's laws apply to alternating currents as well as to direct

currents.

The star-delta transformations of Cha pte r 5 apply to i mpedances.

A mutual inductance can be transformed into a star circuit as in

Fig. 1 7-6 and Eqs. 17-52 to 17-54 .

PROBLEMS

17-1 IMPEDANCE

a) Calcu la te the i m p e d a n c e Z of the circuit shown in Fig. 17-8.

W h a t is th e value of Z (i) for / = 0, (ii) for /' -» oo?

b ) Wh a t is the m a g n i t u d e and the phase ang le of Z at 1 k i l o h e r t z?

c ) Wh a t is the m a g n i t u d e and the phase ang le of Y = 1/Z at that f requency?

d) Calcu la te the power d i ss ipa t ion when th e cu r r en t is 100 mil l iamperes , againat 1 ki loher tz .

e) Can the real par t of the impedance become negat ive?

f) For what f requency range is the circuit equivalent to (i) a resis tor in ser ies

with an i n d u c t o r , (ii) a resis tor in series with a cap ac i t o r ?

g) At what f requency is the circuit equivalent to a pure r es i s tance?

i o n

io n

o — A A A

5 LIF

Figure 17-8 See P r o b . 17-1.

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17-2 REAL INDUCTORS

A real inductor has not only an inductance but also a resistance and a capaci

tance. Its equivalent circuit is shown in Fig. 17-9.Show that the impedance between the terminals is

_ R + jco[L(l - OJ2

LC) - R2

C]

(1 - to2

LC)2

+ R2

to2

C2

I — v W A

Figure 17-9 Equivalent circuit of a real

inductor. Fhe stray capacitance is C.

See Prob. 17-2.

L

- I t -

17-3E COMPENSATED VOLTAGE DIVIDER

Th e voltage divide r of Prob . 5-5 canno t be used as such at high frequencies for

the following reas on. There are stray cap aci tanc es due to the wiring, etc.. in parallel

with P, and R2. If the frequency is high enou gh, these stray capac itan ces car ry an

appreciable current and V,,/Vj is a funct ion of the frequency.

Show that, at any frequency

2*1 * 2

V, P , + R2

if P , C , = R2C2 in Fig. 17-10. Capac itance s C, a nd C , are made large comp ared to

the stray capacitances. The voltage divider is then said to be compensated.

F'igure 17-10 Compensated voltage

divider. See Prob. 17-3.

17-4 RC FILTER

Four impedances are connected as in Fig. 17-11.

413

A C

o —n n n

— • -r r r

^ - m o

Figure 17-11 See Prob. 17- 4. O • • O

a) Show that

V„ _ BD

Vf ~ AB + (A + B)(C + D)'

b) Draw the curve of |V„/Vi| as a function of RwC for the circuit of Fig. 17-12for RmC = 0.1 to 10. Use a log scale for RcoC.

R C

o - W V A - f 1(—f o

V, c ^ < /< v„

Figure 17-12 See Prob. 17- 4. O • • O

17-5 MEASURING AN IMPEDANCE WITH A PHASE-SENSITIVE VOLTMETER

An unk now n imp edance Z is connecte d in series with a known resistance R'

as in Fig. 1 7 - 1 3 .

Figure 17-13 See Prob . 17-5.




Show tha t the imped ance Z can be found by me asur in g the vol tage ra t io

r = a + bj = V 2 / V ! ,

in o the r words , by measur in g the in-phase and the qua dra tur e com pon ent s of V 2 , V ,

b e i n g k n o w n .

17-6E IMPEDANCE BRIDGES. THE WIEN BRIDGE

Figure 17-14 shows an impe danc e br idge . I t i s the a l t e rna t ing-cu r rent equiva lent

of the Wheats tone bridge of Prob. 5-8. In this case, V = 0 when

Z 1/Z 2 = Zi/ZA.

In a l t e rna t ing-cur rent br idges the component s mus t s a t i s fy tw o i n d e p e n d e n t

equa t ions to s a t i s fy both the rea l and the imaginary pa r t s of th i s equa t ion .The re ex i st man y types of impe danc e br idge .* One co m mo n type is the W ien

bridg e sho wn in Fig. 17-15. As a rule , on e sets

Ri = R-i — R 3 = R*, C 3 = C 4 .

I

Figure 17-14 Impedance br idge . See

P r o b . 17-6.

Figure 17-15 Wien br idge . See P rob .

17-6.

* See. for ex am ple. Reference Data for Radio Eng ineers. Fi fth Edi t ion . Ch apt e r 11

(Howard W. Sams and Co. . Inc . . Indianapol i s . 1968) .

Problems 415

Show tha t , under these condi t ions , P 3 < a C 3 = 1.

The Wien bridge is used in tuned amplif iers and in osci l la tors , as wel l as for mea

sur ing or moni tor ing a f requency. To measure a f requency, one changes P 3 a n d P 4

s imul taneous ly unt i l R3coC } i s equ al to uni ty.

See also Prob. 17-18.

17-7E LOW-PASS RC FILTER

The circui t of Fig. 17-16 is commonly used to reduce the A C c o m p o n e n t a t t h e

output of a power supply .

As a rule , the term power supply appl ies to a c i rcui t that t ransforms 120-vol t

60-her t z a l t e rna t ing cur rent in to d i rec t cur rent . One s imple type i s the ba t t e ry charger

of P rob . 16-3 .

A power supply usua l ly compr i ses three pa r t s : a t rans former , a rec t i fy ing

circui t , an d a f i lter . The o utp ut of the rect i fying circui t is a pulsat in g direct cur ren t

tha t can be cons ide red to be a s t eady d i rec t cur rent , p lus an a l t e rna t ing cur rent . Thefunct ion of the f i lter is to a t te nu ate the al tern at in g part . Fi l ters are usual ly d ispense d

wi th in ba t t e ry chargers .

The f i l ter is often fol lowed, or even replaced, by an electronic ci rcui t that main

tains ei ther the output vol tage or the output current a t a set value (Sec. 5.11).

The circui t shown is a s imple form of low-pass f i l ter . The load res is tance R L

i s a s sum ed to be l a rge comp ared to 1 /mC,

a) Sh ow tha t the a l t e rna t ing com po nen t i s a t t en ua te d by the fac tor

|V 0/V,. * IJRcoC

if RcoC » 1. Th e vol t age dr op ac ross the res is t ance P should be smal l com pare d to

tha t ac ros s R L, so

RL » P » 1/mC.

b) Plot a curve of |V 0 / V , | as a fun ction of frequ ency from 0 to 150 he rtz for P =

1 04 ohm s and C = 100 mic ro fa rads .

c ) P lo t the cor responding curve of the a t t enua t ion in decibels as a function of

frequency, where the number of decibels is 20 log |V 0 /V f | .

R

Figure 17-16 Simp le low-p ass P C filterfor a t t enua t ing the a l t e rna t ing

com pon ent a t the outp ut of a power

supply, with P L » R » \fcoC. See

P r o b . 17-7.

17-8 PHASE SHIFTER

I t i s often necessary to shif t the phase of a s ignal . Figure 17-17 shows a s imple

circui t for do ing this . Th e res is tances are adjus table , but are kept eq ual .

http://fcoc/

http://fcoc/



41 6

Figure 17-17 Phase shif ter . See Prob.

17-8.

Use the polari t ie s show n. These me an th at V, is the vol tag e of the top term inal

wi th respect to the bo t to m one , and V„ i s the vol t age of the r igh t -hand t e rmina l wi threspect to the left -hand one.

a ) Show tha t

V„/V, = exp [2/ arc tan (1/RcoC)].

b) Draw a graph of the phase of V„ with respe ct to V, in the range RO JC = 0.1

to 10. Use a log scale for RcoC.

17-9 MEASURING SURFACE POTENTIALS AND SURFACE CHARGE

DENSITIES ON DIELECTRICS WITH A GENERATING

VOLTMETER, OR FIELD MILLIn P r ob . 6-11 , we saw how one can mea sure the sur face potent i a l V a n d t h e

surface charge densi ty a on a d ie l ec t r i c , wi th a probe connec ted to an e l ec t romete r .

Here is a s imilar method, shown in Fig. 17-18a, in which the charge induced on the

prob e i s mo dula ted to produ ce an a l t e rna t ing cur rent . The a l t e rna t ing cur ren t pas ses

thr ou gh a res is tance, giving an al ter nat i ng vo l tage that is then am plif ied to give an

o u t p u t v o l t a g e p r o p o r t i o n a l t o V, or to a. Ins tru me nts tha t mea sur e electr ic f ie ld in

tensities in this way are called generating voltmeters, o r field mills.

One such ins t rument can measure potent i a l s up to 3000 vol t s and has a f i e ld

of view only about 2 mil l imeters in diameter . I t serves to tes t the photoconduct ive

mate r i a l s used for xe rography.

F igure 17-19 shows another one tha t measures the t e rmina l vol t age on a Van

de Graaf f h igh-vol t age gen era tor .a) Use Fig. 17-18b to find E a, as a funct ion of a, d a, dD, e r, wi thout the vane of

F ig . 17-18a . The cur rent through R i s then zero.

b) No w wh at is the value of c ,?

No te tha t cr, depe nds on bot h dD and on e r.

c) What is the value of V in terms of cr. d a, dD, efl

N o t e t h a t V d e p e n d s o n d a, and is therefo re affected by the presen ce of the

i n s t r u m e n t .



41'

p A / V S

V X A aI -

T V h i

P

J_

JZZZZZ- VA

R 0D

£

Figure 17-18 (a) Ap pa ra tu s for me asu rin g the pot ent ia l a t the surface of a dielectr ic

s lab D. The pro be P is grou nde d th rou gh the res i st ance R. a n d t h e g r o u n d e d

v a n e VA i s made to pas s be tween D a n d P in such a way that P i s a l ternately

sc reened and exposed to D. An a l t e rna t in g cur rent Hows to P a n d p r o d u c e s a

vol t age drop on R. The vol t age ac ross R i s amplif ied (Prob. 5-9). giving a vol tageV„ propor t iona l to the sur face charge dens i ty . The vol t age ac ross R i s orders of

magni tude smal l e r than the vol t age a t the sur face of D. (b) Probe P near the

dielectr ic carrying a surface charge densi ty a.

Figure 17-19 Genera t ing vol tmete r for measur ing a vol t age of the order of mi l l ions

of vol ts , from a dis tance. A motor M rotates a shield S in front of a probe P in

view of the h igh-vol t age t e rmina l T enclosed in a pressure tank PT . Sec tors a re

cut out of the shield. An al ter nat ing c urr ent f lows to P throu gh P. Th e vol tage V

is propor t iona l to the vol t age on the t e rmina l . See P rob . 17-9 .

P T




d) No w ca lcula te the v ol t age ac ross R w h e n R = 1 0 m e g o h m s a n d V —

1000 vol t s , and when the capac i t ance C be tween P and the dielectr ic surface varies

s inusoidal ly between 0 and 0.1 picofarad. The vane osci l la tes a t 100 hertz and screensP once per cycle .

17-10E REFERENCE TEMPERATURES NEAR ABSOLUTE ZERO

Th e circui t sho wn in Fig. 17-20 is used to define f ixed poin ts on the tem pe rat ur e

scale near absolute zero. I t ut i l izes the fact that several e lements exhibi t sharp and

reproduc ib le t rans i t ions to and f rom the superconduc t ing s t a t e a t de f in i t e t empera tures .

A s lug of one of these elements , S , is inserted in the mutual inductance M 2.

T h e n R L/R 2 i s adjus ted to m ak e V = 0 . Whe n S becom es supe rcon duc t in g , l a rge eddy

currents f low through i t . cancel ing part of the magnet ic f ie ld, M 2 i s decreased, and

V # 0.

Show tha t , when V = 0,

M , / M 2 = RJ R 2.


17-11 REMOTE-READING MERCURY THERMOMETER

A good mercu ry th e rm om ete r i s a simple , inexpens ive device for makin g accu ra tem e a s u r e m e n t s o f t e m p e r a t u r e . T h e r e a r e m a n y c i r c u m s t a n c e s , h o w e v e r, w h e r e o r d i n a r y

mercury the rmomete rs a re imprac t i ca l . For example , one may wish to moni tor the

tempera ture a t many poin t s nea r the bot tom of a r ive r , every minute , 24 hours a day .

Or one may wish to use a t empera ture reading to cont ro l a manufac tur ing process

a u t o m a t i c a l l y .

I t is poss ib le to read a me rcury the r mo me te r e l ec t ronica l ly , and hen ce au to

matical ly and remotely, with the ci rcui t shown in Fig. 17-21.

41 9

Figure 17-21 (<v) Mer cury the r mo me te r a dap ted for rem ote reading . An osc i l l a tor

appl i es abo ut 10 vol t s a t 10 meg aher tz to an e l ec t rode sur rou ndin g the bulb an d

f o rm i n g a c a p a c i t a n c e C j . T h e c a p a c i t a n c e C 2 be tween the mercury column and a

coaxia l tube va r i es l inea r ly wi th t empera ture . The vol t age on R is amplified

(Prob. 5-9) and measured at V. (r>) Equiva lent e l ec t r i c c i rcu i t . See P rob . 17-11.

a) F ind the vol t age V as a function of to, t he source vol t age V s, C 2 , and P .

» c 2 .

b) Under wha t condi t ion wi l l V vary l inea r ly wi th t empera ture?

c) Est imate the value of C2. and suggest a value for P.

d) F hen wha t is the order of ma gni tude of V if V„ is 10 volts?

17-12 WATTMETER

Figure 17-22 shows a Hal l generator (Sec. 10.8.1) used as a wat tmeter to measure

the power suppl ied by a source to a load Z.

The coi l of a smal l e lectromagnet £ has an inductance £, with toL « |Z|. It

suppl ies the magnet ic f ie ld for the Hal l generator . The res is tances Pj and P 2 are chosen

so that P, + P 2 » |Z| . In this way, the powe r supp l ied by the sour ce is appro xim atel y

equal to that diss ipated in the load.

The vol t age and the cur rent a t the source a re

V = F 0 co s cot,

I = I0 co s {cot + cp).

Uti l ize the resul t of Sec. 10.8.1 to show that the average value of the Hal l

g e n e r a t o r o u t p u t v o l t a g e V i s pro por t ion a l to F m s ' r m s c o s V-



420

Figure 17-22 Hal l - type wa t tmete r formeasur ing the power suppl i ed to a

load imped ance Z . Th e cur rent to the

load passes through the coi l of an

e l e c t r o m a g n e t E that appl ies a

magnet ic f ie ld to the Hal l generator

HG. A small fract ion of the vol tag e

across the load is appl ied to HG a n d

the vol t age V i s pro por t ion a l to the

power d i s s ipa ted in Z . See P rob .

17-12.

17-13 TRANSIENT SUPPRESSOR FOR INDUCTOR

In P ro b . 12-13 we saw tha t , if one d i sconn ec t s a l a rge cur ren t -ca r ry in g in duc tor f rom

a d i rec t -cur rent source , h igh t rans ient vol t ages deve lop and the induc tor can be

des t royed. We a l so saw tha t the overvol t ages can be e l imina ted by connec t ing a

diode in pa ra l l e l wi th the induc tor .

I f the indu c tor i s fed by an a l t e rna t ingcur ren t so urce , one can use the p ro tec t ive

c i rcui t shown in F igure 17-23. H e r e , R i s the res is tance and L is the inductance of

the induc tor , and C = L/R2

.

Show th a t the reac tanc e be tween A and B is ze ro . Th en th e c i rcu i t be tween A

and B mu s t ac t l ike a pure res i s tance , and the vol t age ac ross the induc tor m us t becom e

zero when the switch is opened.

—. AI O ^ C ^ - f .



Problems 421

Solution:

(R + jcoDVR + ( l / j o C ) ]

2R + jwL + (1/jtoC) '

R2 + (L /C) + j[toLR - (R/coC)~]

2R + j[wL - (1/coC)] '

{R2 + (L/C) + j[toLR - (P/coC )]j {2P - j[_a>L - (1/ toC)]}

4 P2 + [wL - (1 /coC )] 2

Th e num era tor of th i s expres s ion is ze ro , Hence the reac tance be tw een A and B is

zero at a l l frequencies!

17-14E SERIES RESONA NCE

W ha t i s the locus of the impe danc e Z in the complex p lane when the f requency

app l ied to the series-re son ant c i rcu i t of Fig. 17-2 chan ges from zer o to infini ty?

17-15 PARALLEL RESONANCE

An induc tor of induc tance L and res i s t ance R i s connected in paral le l wi th acapac i tor C .

Se t R = 10 oh ms , L = 1.0 mil l ihe nry, and C = 1.0 m icrofa rad.

a ) P lo t curves of the rea l pa r t , t he imaginary pa r t , t he magni tude , and the phase

of the impedance as funct ions of the frequency, from zero to 10 ki lohertz .

N o t e t h a t , w h e n Z i s m a x i m u m , w2

LC w 1.

b) A t wh a t f requency i s the reac tance z e ro?

c) F ind the impedance a t 8 k i loher t z .

d) At that frequency, the ci rcui t has the same impedance as a capaci tor C a n d

a res is tor P ' in series . W ha t are the values of P ' and of C ?

This is a case of parallel resonance, L and C being in paral le l .

T h e bandwidth i s defined as the difference between the frequencies where the

magni tude of the impedance i s 1/21

'2

, or 0.707 t imes i ts maximum value.T h e Q (for Qu al i ty ) of the ci rcui t is defined to be the reson anc e frequency, divided

b y t h e b a n d w i d t h . T h e h i g h e r t h e Q, t he sharper the resonance peak . I t can be shown

The reac tan ce is the imagina ry pa r t of Z :

X =[ P

2 + (L/ C)] [ca L - (1/coC)] + 2 P [ O J L P - (R/toC)]

4R 2

+ [ c o L - ( 1 / w C ) ] 2

(4)

t h a t

T h e Q of the above circui t is 3.2.




Para l l e l re sonance has innumerable appl i ca t ions . Here i s one example . For

high-speed , h igh-prec i s ion coi l winding , i t i s imp or ta nt to mo ni to r the wi re d iam ete r .

For s evera l reasons , mechanica l devices a re unacceptable . The moni tor ing can be done

in the fol lowing way: a smal l coi l of a few turns is wound around a quartz tube a few

mil l im eters in diam eter thro ug h which the wire passes . Th e coi l of cou rse has an in

du ctan ce. I t a lso has a capa ci ta nc e; as in any coi l , there is a vol tag e difference, and

hence a capac i t ance , be tween turns . Thi s stray capacitance i s increased by the presence

of the wire . So the coi l acts as a paral le l -resonant c i rcui t , whose resonance frequency

depends on the diameter of the wire . If the wire diameter increases , the resonant

frequency decreases and vice versa. Th e curve of Z as a funct ion of frequency is l ike that

ca lcula ted under pa r t a above , the peak moving to the right when the wi re d iamete r

decreases , and to the left when the d iamete r inc reases .

Th e coil is des ign ed so that i ts reso nan ce frequen cy is a lwa ys s l ight ly abo ve 300

mega her tz , and an osc i l l a tor opera t in g a t 300 meg aher tz i s conn ec ted to the coi l thro ugh

a res is tance. If the wire diameter increases , the resonance peak moves to the left , Z

increases , and the vol tage on the coi l increases . Measurements are accurate within 0.5" , , .

I7-16E STAR-DELTA TRANSFO RMATION

Transform the del ta of Fig. 17-5a into a s tar a t one megahertz .

I7-17E STAR-DELTA TRANSFO RMATION

Transform the s tar of Fig. 17-24 into a del ta a t one ki lohertz .




Problems 423

17-18 T-CIRCUITS

Figu res 17-25a an d b show two ty pes of T-circui ts . the twin (or paral le l ) T and the

br idged T . Th e com po nen t s a re s e lec ted in such a way tha t the outp ut vol t age V 0

becomes zero at a specif ied frequency. Circui ts that have this characteris t ic are

called notch filters. T-circui ts are used in osci l la tors and in tuned amplif iers . Some

are a l so used for measur ing impedances , in much the same way as the impedance

br idges of P r ob . 17-6 . No te tha t the input an d the ou tpu t have a com m on te rmina l ,

which i s grou nde d. Th i s is a grea t adv antag e , s ince mos t osc i l l a tors , vol tmete rs , and

osc i l loscopes have one t e rmina l grounded.

_ n n n _

.crrx. -o o »

_ p n n .

JTCX . _ n n n _

4 - 4(a )

-O O(b)

o-m—*

_ n n r i _

Yi_ n m _

Yi

J T Y l _ n m _

Yi >4

' 2 ^

- O O -

Figure 17-25 (a) Tw in, o r pa ralle l, T, an d {b) shunted T c i rcu i t s ; (c) a n d ill).

equivalent c i rcui ts . See Prob. 17-18.




To find the conditions under which the output voltage of the twin T is zero, we

transform the star circuits into deltas, giving the circuit of Fig. 17-25c, and we set

F, + Y2

= 0.

For the bridged T. we transform the delta formed by the three impedances at

the top into a star. The bridged T then becomes a simpleT as in Fig. 17-25d. and the

out put volt age is zero when Z , + Z 2 0.

Figure 17-26 shows one common type of twin T.

4

0 * - V \ A AR

V: 2C

Figure 17-26 See Prob. 17-18. O

R

R/2 K

4 *

a) Fin d the F, and Y2 of Fig. 17-25c for the twin T of Fig. 17-26, setting RcoC = x,and show that V„ = 0 when x = 1.

Solution: In the circuit of Fig. 17-25c, let us call F, the admit tan ce ob tai ned by

transforming the R-R-2C star of Fig. 17-26. Fhen

(1/PH1/R) 1_

(1/P) + (1/P) + 2jwC ~ 2R + 2jwCR2

Similarly, for the C-C-R/2 star.

_ J'oCjcoC _ jviCjcoCR

2

~ jcoC + jo)C + (2jR) ~ 2jtoCR + 2'

Fh en V„ = 0 when

1 icoCjx 1 — x2

Fi + Y2

= + — — = = 0. (3)2P(1 + ./x) 2(./'x + 1) 2P (1 +jx)



Problems 425

or when

1 - x1

= 0, (4)

x = 1. (5)

b) Find V„/V, as a function of x.

Assume that the load impedance connected to the right-hand terminals is in

finite.

Solution: Considering first the star C-C-R/2, K3 of Fig. 17-25c is given by

= jcoC(2/R)=

jcoC

3

jioC + (2/R) + jcoC \+jx

For the star R-R-2C.

_ (l/R)2jo)C = jcoC

4

(1/R) + 2iwC + (\/R) 1 + ; x'

Now

1

v„ y3

+ r4

n + Ji

v, i _ j _ y, + y2

+ y3

y> + y2

y3

+ y4

where

2i?(l

Thus

(1 - .x

2

)/(2iv) 1 - A'

2

V, [(1 - A -2

) /2 R ] 4- 2jcoC 1 — x + 4jx

(8)

(1 - A2

) . (9)

y 3 + n = no)1 + /A

(I D

c) Draw a curve of |V 0 / V , | as a function of A between A = 0.1 and x = 10. Use

a logarithmic scale for x.




42 6

IORC


17-19 BRIDGED T

Show tha t , for the br idged T of Fig. 17-28, V 0 = 0 when

co2

C2

rR = 1, o)2

LC = 2.

w w — T r a f f i c




Problems 427

R

Li o ! M g ) L 2

3 3


17-20 MUTUAL INDUCTANCE

Find the rms va lues of the cur rent s F and l 2 in the inductances of the ci rcui t of

Fig. 17-29 when

V = 10 vol ts , R = 5 o h m s , R 1 = R 2 = 10 ohms ,

L j = L 2 = 1 henry , M = +0 .9 henry , / = 1 k i loher t z .



CHAPTER 18

ALTERNATING CURRENTS: III

Power Transfer and Transformers

18.1 POWER DISSIPATION

18.1.1 Examples: A Light Bulb, A Radio-Frequency Antenna,

an Electric Motor

18.2 THE POWER FACTOR

18.2.1 Examples: Resistors, Capacitors, and Inductors

18.3 POWER TRANSFER FROM SOURCE TO LOAD

18.3.1 Example: Whip Antenna

18.4 TRANSFORMERS

18.5 MAGNETIC-CORE TRANSFORMERS

18.5.1 The Ideal Transformer

18.5.2 The Ratio V , / V 2

18.5.3 The Ratio LJL 2

18.5.4 The Input Impedance Z i n

18.5.5 The Ratio \J\2


THROUGH A TRANSFORMER

18.7 SUMMARY

PROBLEMS



In th i s l as t chap te r on a l t e rna t ing cur r en t s we sha l l be concerned wi th thet ransfer of power f rom a source to a load, f i r s t d i rect ly , and then through

a t r ans former .

Tra nsm is s io n l ines tha t ca r ry l a rge amo un t s o f pow er over l a rge d is

tan ces m ust o pe rat e , for rea son s of ef ficiency, a t vol ta ges of a few hu nd re d

tho usa nd v o l t s ; som e even ope ra te a t c lose to one mi l l ion vo l t s . Bu t insu la t ion

p r o b l e m s p r e v e n t g e n e r a t o r s i n p o w e r p l a n t s f r o m b e i n g o p e r a t e d a b o v e

about 20 ,000 vo l t s . At the o ther end o f the l ines , consumers r equ i r e e l ec t r i c

power at vol tages ranging f rom a f ract ion of a vol t for solder ing guns to 6000

vol t s fo r e l ec t r i c motor s o f a f ew hundred k i lowat t s .

I t i s the funct ion of t ran sfo rm ers t o chan ge the rat io V/I with li t t lelos s o f power . In th i s way , e l ec t r i c power can be p roduced , t r ansmi t t ed , and

u t i l ized a t the mo s t conven ien t vo l t age s , wi th in l imi t s. T ran s form ers a r e

also used for impedance matching, as we shal l see in Sec. 18.6.

Trans formers can ac t on ly on a l t e rna t ing cur r en t s .

18.1 POW ER DISSIPATION

Let us cons ider the p roces ses whereby the e l ec t r i c and magne t i c energy

f lowing in a ci rcui t is diss ip ated . Unt i l now , we have cons id ere d only the

I2

R loss in a res is tan ce R car ry in g a cur r e n t / . Of cour se , ma ny o th er p roces ses

are pos s ib le . For example , e l ec t r i c motor s p roduce mechan ica l energy ,

s to rage ba t t e r i es can p roduce chemica l energy , l i gh t bu lbs and an tennas

p r o d u c e e l e c t r o m a g n e t i c w a v e s , l o u d s p e a k e r s p r o d u c e a c o u s t i c w a v e s , a n d

so for th .



430 Alternating Currents: III

Any on e of these devices, when connected to a source, draws a current,

an d the ratio V/l is its impedance. How is this impedance related to the

power that is withdrawn from the circuit?

Let us think of a current I flowing through an impedance R + jX.

If the applied voltage V is V0

ex p jcot,

V

R + jX R2

+ X2 s

"J

"' |ZT= 7 7 — 7 7 ; = 7 7 2 — w ( * - J*) =wr2

(R- m (18-1)

V0

exp jcot

= j ^ p lR

+x e x

P (-W2)] (18-2)

and, since I is th e real part of I,

I = [R cos cot + X cos [cot - 7t/2)] . (18-3)

The instantaneous power absorbed by the impedance is the product

of this quantity by V0

cos cot:

V2

^ins t = T^k [R cos 2 cot + X co s [cot - 7t /2) cos cot]. (18-4)

Now the average of co s 2 cot over one cycle is \, while the average value of

the second term between the brackets is zero. Thus the average power

absorbed by the impedance R + jX is

\_K _ vL2\Z\-

av - ^ 7^12 & - T^iTR

- lfmsR

- (18-5)

Note that P a v is / r

2

m s R, as for direct currents, but that it is V2

ms/R

only if X = 0.

If the impedance is that of some device that draws power from a

circuit, then I2

m!iR must account for all the electric power that is absorbed.

EXAMPLES: A LIGHT BULB.

AN ELECTRIC MOTOR

A RADIO-FREQUENCY ANTENNA,

In a light bulb, I2

m!iR is equal to the heat energy that is lost by conduction and

convection, plus th e electromagnetic energy that is radiated in one second.

18.2 The Power Factor 431

A radio-frequency antenna is usually tuned to have a purely resistive impedance

P, which is called it s radiation resistance. If the current fed to th e antenna is /, then

I2

miR is the radiated power.In an alternating-current electric motor, I

2

msR includes: Joule losses (Sec. 5.2.1)

in th e copper conductors; hysteresis losses (Sec. 14.9) in the iron; eddy-current

losses (Sec. 11.3) both in the copper and in the iron; friction losses in the bearings;

windage losses in the air; and, finally, useful mechanical power. The resistance R

of an electric motor in operation is thus the su m of many terms. It is much larger

than the resistance measured with an ohmmeter when the motor is stopped. See

Prob. 18-1.

18.2 THE POWER FACTOR

The rat io R/\Z\ of an im ped anc e is called its power factor and is usually

expressed as a perce ntage . If the impe dan ce Z is repre sente d in the comp lex

plane, as in Fig. 18 - 1 , the po wer factor is cos cp, tp being the phase angle

(Sec. 1 7 . 1 ) .

Whe neve r large am ou nt s of power must be trans mitte d from a source

to a load ov er a line of app rec iab le resistance , care must be taken to adjust

the powe r factor close to unity. Th e reason for this is tha t the c om po ne nt

of I that is in qua dr at ur e with V pro du ces no useful po wer in the load, but

nonetheless gives rise to a power loss and to a voltage drop in the line

connecting the source to the load.

/.v

R

Figure 18-1 Impedance Z represented in the complex plane. F h e cosine of the

phase angle <l> is called the power factor.

432 Alternating Currents: HI

Elec t r i c motor s a r e induc t ive loads , and . in l a rge ins t a l l a t ions , the

pow er f ac to r i s cor r ec te d by con nec t in g ca pac i to r s ac ro s s the line , as c lose

as poss ible to the motors . See Probs . 18-2 and 18-3.

EXAMPLES: RESISTORS, CAPACITORS, AND INDUCTORS

Th e po wer factor of a res is tor is 100%, and tha t of a capa ci to r is 0%.

Real inductors have both res is tance and inductance.* If R = coL, t h e n <p

45 degrees , and the power factor is 70.7%.


In Fig. 18-2. we have re place d th e sourc e by an ideal vol ta ge so urc e V, in

ser i es wi th an impedance R s + jX s. This is the general form of Thevenin 's

theorem (Sec. 5 .12) .

Source Load

Figure 18-2 Source feeding an impedance Z L = R L + jX,.

* If the frequency is qu i te high th ey be com e capac i t ive. See Pro b. 17-2.

18.3 Powe r Transfer from Source to Load 433

The average power dissipated in the load is

V2

Pr = ILSP-L = 5 Ri- (18-6)

Under what conditions is PL maximum, for a given U r m s ? First,

XL+ X

s should be zero:

XL

= -Xs. (18-7)

In other words, the reactance of the load should be of the same magnitude

as the reactance of the source, but of the opposite sign.

If tha t co ndi tio n is satisfied, or if the rea ctanc es are bo th zero,

P =R

*<V

™ ,18-8)

(RL + Rs)2

We can now find the optimum value of RL

by setting

dPL

(RL + Ks)

2

~ RL x 2(RL

+ Rs)

2

dR~r Krms = a (18"9)

Then

RL

= Rs. (18-10)

Thus, for maximum power transfer to the load,

XL

= -Xs, R

L= R

s. (18-11)

The load impedance is then said to be matched to that of the source.

The maximum power that can be dissipated in the load, for a given

open-circuit voltage K r m s is

p

- - - ( M * • - % »8

-1 2 1

Also, from Eqs . 18-8 an d 18-12.

P L RLV2

ms 4RS MRJRs)

PL**, (RL + Rs)2 v

2

msl(R,JRs) + i ]

2

(18-13)



43 4

J i ' i i i i i j i i i i i i i

0.1 10

R, R,

Figure 18-3 A: PJP Lmd% as a function of RJR S. The power d i s s ipa ted in the loadis m ax im um when t he load res is tan ce is equa l to the res is tan ce of the sou rce.

B: Efficiency as a function of R/R x. The eff ic iency tends to uni ty as R -> x . It is

only 50% a t ma xim um pow er t rans fe r .

Figure 18-3 shows P L ! P L M ^ as a function of R L/R S. I t wi l l be observed that

t h e c o n d i t i o n R L = R^ for max im um pow er t rans fe r i s no t c ri t i ca l . For

example , wi th R L = 2 K S , t he power P L diss ipate d in the loa d is s t i ll 89%

of /% m a x -

T h e efficiency may be denned as the r a t io PL/(PL + P S), w h e r e P S

is the pow er d i s s ipa ted in the so urc e :

PL ILSRL RL ( 1 8 . 1 4 )

PL + PS IU PL + RS) RL + R,'

= RJ R S

(RJRS) + 1 '

(18-15)



18.3 Pow er Transfer from Source to Load 435

Figure 18-3 also shows how the efficiency varies with the ratio RL/RS.

Th e efficiency is equ al to unit y whe n RL/RS -» oo, but the n the pow er

dissipated in the load tends to zero. For RL = Rs, the power dissipated in

the load is ma xi mu m a nd t he efficiency is only 50 %.

18.3.1 EXAMPLE: WHIP ANTENNA

A radio- f requency t ra nsm i t t e r has an output res i s t ance of 50 o h m s . It is to be c o n

nec ted to the w h i p - a n t e n n a of Fig. 18-4, which has a rad ia t ion res i s t ance of 3 7 o h m s .

What wi l l be the efficiency of power t rans fe r?

In this case,

Rs = 50, Xs = 0, RL = 37, XL = 0. (18-16)

Fhe efficiency will be 37/ (50 + 37), or a b o u t 4 3 % . T h i s is sa t i s fac tory , compared

with th e o p t i m u m v a l u e of 5 0 % . f

Figure 18-4 Q u a r t e r - w a v e w h i p a n t e n n a A m o u n t e d on a m e t a l car t o p . The w h i p

is one -qu ar te r wave length long an d is s u p p o r t e d by a feed- through insula tor F. It

is fed by a t ransmi t t e r s i tua ted at the b a c k of th e ca r t h r o u g h a coaxial l ine C. The

w h i p , t oge ther wi th its i m a g e / ( P r o b . 3-15) forms a ha l f -wave antenna . Charges

flowing in the antenna form l ines of force that detach themselves and escape in to

space at the veloci ty of l ight . This type of a n t e n n a is used at frequencies ranging

from 150 to 450 m e g a h e r t z .

1

O ne could achieve 50% effic iency by us ing a matching c i rcu i t . See, for e x a m p l e , The

Radio Amateur s Handbook, Am erican Ra dio Re lay League , New ingto n , Conn . , U .S .A.

(publ ished every other year) .




18.4 TRANSFORMERS

A t r ans fo rme r i s a mutu a l indu c tor th a t i s used to chan ge the vo l t age l eve l

of an elect r ic cu rren t , as exp laine d brief ly in the int ro du ct i on t o this ch apt er .

T h e m u t u a l i n d u c t o r c a n h a v e e i t h e r a m a g n e t i c o r a n o n - m a g n e t i c

co re . Th e fo rmer type i s co mm on ly r e fe rr ed to as a magnetic-core transformer,

and the la t ter as an air-core transformer. Air -core t r ans formers a r e used a t

h igh f r equenc ies where eddy-cur ren t and hys te res i s los ses in a magne t i c

co re wo uld be excessive (Sees. 11.3 an d 14.9).

The equivalent s tar c i rcui t (Sec. 17.5) wi l l g ive us some general resul ts

tha t a r e exac t fo r a i r co res , and on ly approx imate fo r magne t i c cores .

F igure 18-5a shows a t r ans former inser t ed be tween a source supp ly ing

a vo l t age V, , and a load impedance Z L . T h e i m p e d a n c e R x + jcoLx of the

pr imary wind ing i s Z u a n d t h e i m p e d a n c e R 2 + jcoL 2 of the s econdary

wind ing p lus Z L , is Z 2 .

T h e m u t u a l i n d u c t a n c e i s

M = fc(L1L 2)1/2. (18-17)

Since the ci rcui t of Fig. 18-5b is equ iva len t to th at of Fig. 18-5a,

f rom Sec. 17.5, the im pe da nc e "see n" by the sourc e, or the input impedance

Z i n of the t r ans former i s

v v j.i AA i + JcoM)(-ja>M) M f i i mZ i n = Z , + JcoM + , — — , (18-18)Z 2 + JCOM — JCOM

co2

M2

= Z 1 + ^ — . (18-19)

Th e inpu t impe dan ce i s there fore Z u plus the t e rm co 2M 2jZ 2, which is

cal led the reflected impedance of the s e cond ary .

T h e c u r r e n t 11 in the pr imary is of course

l i = V , / Z i n . (18-20)

Th e cur re n t in the s econd ary can be found by app ly ing Ki rchof f ' s

vol t age law to th e ci rcui t of Fig. 18-5b, us ing th e value of I given in Eqs . 18-19

and 18-20 :

(18-21)



43 7

Figure 18-5 [a) Trans former fed by a source and feeding a load impedance Z L.

(b ) Equiva lent c i rcu i t .

18.5 MAGNETIC-CORE TRANSFORMERS

Ma gne t i c - co re t r ans form ers as in F ig . 18-6 a re used a t fr equenc ies r ang in g

f rom a f ew her t z to abou t one megaher t z .

A magne t i c - core t r ans former has two es sen t i a l f ea tu res . F i r s t , f o r a

g iven core c ros s - sec t ion , the magne t i c f lux per ampere- tu rn in the p r imary

i s l a rger than wi th an a i r core by s evera l o rder s o f magni tude . Second , the

ma gne t i c flux th ro ug h the s eco nda ry i s near ly equa l to tha t th ro ug h the

p r i m a r y .

T h u s , t he magn e t i c core per mi t s the des ign of t r ans fo rme rs tha t a r e

smal l e r and have f ewer tu rns , bo th in the p r imary and in the s econdary .



43 8

S

(a) (b)

Figure 18-6 («) Co m m on type of magn e t i c -core t rans fo rmer . Th e core has the

shape of a f igure e ight . The pr imary and secondary windings P a n d S a r e w o u n d

on the center leg. The core is made of two types of laminat ion, one having the

shape of an £, and the other the shape of an / , placed one next to the other. The

lamina t ions a re as sembled a f t e r the co i l has been wound, (b) S c h e m a t i c d i a g r a m

of a mag ne t i c -core t rans former .

Also , with a mag ne t ic core, there is very l it t le s t ray m ag ne t ic f ie ld . F or

example , one can opera te two such t r ans formers s ide by s ide wi th neg l ig ib le

i n t e r a c t i o n .Cores made o f i ron a l loys a r e usua l ly fo rmed of para l l e l i nsu la t ed

sheets cal led laminations. This r educes the los ses due to the eddy cur ren t s

induced by the chang ing magne t i c f i e ld . The l amina t ions have a th i cknes s

of a f ract ion of a mil l im eter . At aud io f requenc ies th e al loy is som et im es in

the form of a fine pow der m old ed in a bon d in g agen t . O ne the n has a powdered

iron core .

Ferrites are used a t aud io f r equenc ies and above . These a re ce ramic-

type mate r i a l s tha t a r e molded f rom ox ides o f i ron and var ious o ther meta l s .

The i r main a dv an t ag e i s tha t the i r cond uc t iv i t i es a r e low, o f the o rde r o f

1 S iemens per mete r , which i s ab ou t 1 0 ~

6

t imes tha t o f t r ans form er i ron .The eddy current losses in fer r i tes are thus low.

Th e ana lys i s o f ma gne t i c - core t r an s form ers is r end ered d i f fi cu lt by

severa l f ac to rs . F i r s t, t he r e l a t ion sh ip be twe en B and H in f e r rom agne t i c

m ate r ials i s no t l ine ar (Sec. 14.9). F or exa mp le, if the f lux l ink age in a ci rcui t

is A when i t car r ies a current I, t hen the s e l f - induc tance L is A// , as in

Sec. 12.3. If the re are no m ag ne t ic m ate r ial s in the f ie ld . A is pr op or t io na l

to / and A/ I depends so le ly on the geomet ry o f the c i r cu i t . However , i n

18.5 Magnetic-Core Transformers 439

t he p resence o f ma gne t i c mate r i a l s , A is no t p rop or t i on a l t o / and the self-

i n d u c t a n c e L r equ i r ed fo r the ca lcu la t ions can on ly be approx imate .

M o r e o v e r , t h e l o s s e s i n a m a g n e t i c - c o r e t r a n s f o r m e r a r e c o m p l e x :the re are ed dy -cu rre nt losses (Sec. 11.3) bo th in the i ron core an d in the c op pe r

windings , hys teres is losses in the i ron core (Sec. 14.9) , and Joule losses

resul t ing f rom the currents f lowing in the windings . Al l these losses can be

expres sed as an I2

R loss in the pr imary, but , for a given t ransformer , R d e

pends on the vo l t age app l i ed to the p r imary , on the cur r en t d rawn f rom the

secondary , and on the f r equency .

18.5.1 THE IDEAL TRANSFORMER

W e sha ll mak e the fo llowing th ree as sum pt io ns , ( a) Th ere a r e no Jou le o r

eddy-current losses , (b) The hys teres is loop for the core is a s t raight l ine

th ro ug h the o r ig in . Th en B i s p ro po r t io na l to H and there a r e no hys te res i s

losses ei ther , (c) Th ere is zer o leak age f lux. The n th e coeff icient of co upl ing

k is equ a l to un i ty , an d the f lux th rou gh the p r im ary is equ a l to tha t th r ou gh

t h e s e c o n d a r y .

As a consequence o f the f i r s t two as sumpt ions , the t r ans former i s

lossless, and its efficiency is 100%.

A f ict i t ious t ransf orm er h avi ng these cha racte r is t ics i s cal led an ideal

transformer.

T h e a s s u m p t i o n t h a t B is p ro po r t io na l to H, and henc e to / , is mos t

undes i rable , but di f f icul t to avoid.

As to the eff iciency, i t is c lose to 100% for large ind us tr i al t ran sform ers ,

bu t on ly ab ou t 70 o r 80% for a smal l pow er t r ans form er supp ly ing , s ay ,

10 wat t s .

F ina l ly , we sha l l cons ider the usua l s i tua t ion , in which the load

imp eda nce i s r ea l an d mu ch sm al l e r than the r eac ta nce o f the s e cond ary

w i n d i n g :

Z L = R L « o)L 2. (18-22)

18.5.2 TH E RATIO \IJ\I2

With these as sumpt ions , the vo l t age app l i ed to the p r imary i s

d, (18-23)

and the vo l t age acros s the s econdary i s

(10V 2 = N 2-- = N 2jc»& = \2R L, (18-24)

at

w h e r e TV, a n d N 2 are the num be rs o f tu r ns in the p r im ary a nd in the

secondary wind ings , r espec t ive ly , and where 4> i s the magnet ic f lux in the

c o r e :

0 = 0 , + $ 2 , (18-25)

= ^ + ^ (18-2 6,

w h e r e 31 i s the reluctance of the core, f rom Sec. 15.1.

T h u s

V , / V 2 = NJN 2 (18-27)

and . for an ideal t ran sfor m er and for a given V 2 is inde pen den t o f the

load cur r en t . Al so , V 2 i s e i th er in ph ase w i th V u o r n r ad ians ou t o f phase .

Of cour se , the phase o f V 2 can be changed by n radians at wi l l by inter

chan g ing the wi res com ing ou t o f the s econd ary wind ing .

N o t e t h a t V t i s equal to iV, jt»4>. Thus , fo r a g iven app l i ed vo l t age ,

the total f lux 4>, + <J> 2 is a p p r o x i m a t e l y i n d e p e n d e n t o f t h e c u r r e n t d r a w n

f rom the s econdary .

Also , O is B t imes the c ros s - sec t ion o f the core . S ince the maximum

poss ible value for B depends on the core mate r i a l , Eq . 18-23 shows tha t an

increase in f requency permits the use of a core having a smal ler cross-sect ion.

18.5.3 THE RATIO LU'L 2

N o t e t h a t

L , = . 1 . I, = Nj fp j / | j = Nj/.M. (18-28)

S imi la r ly ,

l 2 = Ni/a, (18-29)



18.5 Magnetic-Core Transformers 441

co L^ L 2 . / jo)L= > L , ( 1 - „ ; , ) . ( 18 -3 3)

(18-34)

RL + jtoL 2 R L + jcoL2

RJoLi

R L + JOJL 2

'1

R L = ( ^ A2 R L - [R L«COL 2) (18-35)

L2 \N

18.5.5 THE RATIO l , / l 2

From Sec. 18.4,

1 , = ^ = ^ , , (18-36)2m R LJo)L {

l 2 =jC

°M

2 2 V 1 = ^ . V „ (18-37)

j w M

V ,, (18-38)R LjojL i

so that

LJL 2 = Nl/Nl (18-30)

18.5.4 THE INPUT IMPEDANCE Z i n

Since the coupl ing coeff icient k is eq ua l to unity , the n, from Sec. 12.4,

M = ( L , L 2 ) 1 / 2 . (18-31)

Also ,

Z{ =jcoLu Z 2 = R, + jwL 2, (18-32)

w h e r e R L is the load r es i s t ance con nec ted be twe en the t e rmin a l s of the

s e c o n d a r y .

T h u s , f rom Eq. 18-19,




1 2 jcoM

Disregarding the sign,

Ij/lj = L2IM = ( L

2/ L i )

1 / 2

= N2/Nl. (RL

« wL2)

18.6 POWER TRANSFER FROM SOURCE TO

LOAD THROUGH A TRANSFORMER

As we saw in Sec. 18.3, the po wer di ssipat ed in a loa d is ma xi mu m when its

resistance R, is equa l to the inter nal re sistance of the sour ce Rv, and when

XL = — Xs. As a rule, the reactances are zero. If it is impossible to vary

either RL or R

s, then one can still achieve optimum power transfer by in

serting a transformer between source and load as in Fig. 18-7a. Let us see

how this comes about.

Let us assume that the transformer has a magnetic core, that the ap

proximations used in Sec. 18.5.1 are satisfactory, and that Xs = X, = 0.

The n, from Eq. 18-35, the tra nsfo rmer has an inpu t imp eda nce of (TV i/N 2)2

RL

ohms. In other words, the current and the power supplied by the source

are precisely the sam e as if the transf orme r and its load r esistanc e RL

were

replaced by the resistance (Ar

1/A

7

2)

2

i7

\L, as in Fig. 18-7b. Remember that,

with our approximation, the transformer has an efficiency of 100%.

(18-39)

(18-40)

Figure 18-7 (a) Source feeding a load resistance RLthrough a magnetic-core

t r a n s f o r m e r . (/>) Equivalent circuit.



18.7 Summary 443

Then, with a transformer as in Fig. 18-7a, the power transfer is optimum

when

N1

.

RS

= ^ RL

, (18-41)

or when the turns ratio is

NJN2

= (RJRL)

12

. (18-42)

Wh en thi s con dit ion is satisfied, the power dissip ation in the loa d is max i

mu m, bu t the efficiency is aga in on ly 5 0 ° o . The trans for mer is then said

to be used for impedance matching.

18.7 SUMMARY

Any device that transforms the energy flowing in an electric circuit into

some other form has a resistance R such that I2

R accounts for all the energy

withdrawn from the circuit.

The power factor of the load is the ratio R/\Z\, or the cosine of the

phase of I with respect to V. and is usually expressed as a percentage.

For maximum power dissipation in a load, the following two con

ditions should be satisfied:

XL

= -Xs, R

L= R

s. (18-11)

where the subscripts L and s refer, respectively, to the load and to the source.

The impedances are then said to be matched, and the efficiency is

5 0 % . The efficiency is defined as the pow er delivered to the load, divided by

the power produced by the source.

The input impedance of a transformer is equal to the impedanc e Z ,of the primary winding, plus the reflected impedance of the second ary circuit,

which is w2

M2

/Z2, where Z

2is the imp edan ce of the second ary windin g,

plus the load impedance ZL.

For an ideal magnetic-core transformer,

V2/ V, = l , / l

2= N

2/N

l

(18-27, 18-40)




and the input impedance is

An = ^ R L = ~ R L . (18-35)

L2i \

2

A trans form er can serve to match the impe dan ce of a load to tha t of a source,

by making

NJN2 = (RS/RL)12

. (18-42)

The efficiency und er these cond itio ns is again 50%.

PROBLEMS

18-1E DIRECT-CURRENT MOTORS

Let us see how a d i r e c t - c u r r e n t m o t o r o p e r a t e s .

F igure 18-8 shows a highly schemat ic d iagram of a series motor. The t e rm "se r i es "

refers to the fact that th e f ie ld winding and the rotat ing coi l , cal led th e armature, are

i A Xi

Figure 18-8 S c h e m a t i c d i a g r a m of a di rec t -cur rent s e r i es motor . The a r m a t u r e A

r o t a t e s a b o u t th e vert ical axis AX. It is c o n n e c t e d in series with th e coi l s produc ing

the magnet ic f ie ld B t h r o u g h a c o m m u t a t o r C. The c o m m u t a t o r is a spl i t copper

r ing . Contac t to the r ing is made wi th ca rbon brushes . The di rec t ion of t he cur rent

t h r o u g h th e a r m a t u r e is inverted twice every turn. The m a g n e t i c c o r e is not

s h o w n . See Prob. 18-1 .



44 5

Figure 18-9 Equiva lent c i rcu i t V,

R of the electr ic motor of Fig. 18-8.

in series. In a shunt motor they are in paral le l . Compound motors obtain part of theirf ie ld with a series winding, and part wi th a shunt winding.

W hen the m oto r i s s topp ed, i t ac t s a s a resi s t ance . Fh e res i s t ance of the a rm atu re

is mu ch sm aller tha n that of the f ie ld wind ings .

When the motor i s in opera t ion , the mot ion of the a rmature in the magne t i c

fie ld generates a counterelectrom otive force V that opposes the f low of current as in

Fig. 18-9. Here R i s the res i s t ance as soc ia ted wi th the va r ious los ses enu mer a ted in

Sec. 18.1.1.

a) Use the subs t i tu t ion theorem to show tha t the mechanica l power i s equa l

t o IV .

b) How would you define the eff ic iency?

c) If a direct-cu rren t series m ot or is con necte d to a sou rce of po we r an d run

wi thout a load , i t wi l l ga in more and more speed and, ve ry probably , the a rmature wi l lburs t . A d i rec t -cur rent series motor that is not connected direct ly to i ts load is therefore

dangerous.

Why is this?

R e m e m b e r t h a t V i s propor t iona l to the speed of the motor , mul t ip l i ed by the

m a g n e t i c i n d u c t i o n a t t h e a r m a t u r e w i n d i n g .

d) What happens when the mechanica l power i s inc reased?

18-2 POWER-FACTOR CORRECTION

A load has a power factor of 65° „ and draws a current of 100 amperes at 600 vol ts .

a ) Ca lcula te the magni tude of Z , i t s phase angle , and i t s rea l and imaginary

pa r t s .

b ) C a l c u l a t e th e in - p h a s e a n d q u a d r a t u r e c o m p o n e n t s of t h e c u r r e n t .c) What s ize capaci tor should be placed in paral le l wi th the load to cancel the

reac t ive cur rent a t 60 he r t z?

18-3 POWER-FACTOR CORRECTION W ITH FLUORESCENT LAMPS

A fluorescent lamp cons i s t s of an evacua ted tub e conta in ing merc ury vap or an d

coa ted on the ins ide with a f luorescent pow der. A disch arge is es tabl ish ed ins ide the

t ube , between electrodes s i tuated at each end. The discharge emits most of i ts energy




a t 253.7 nanomete rs , in the u l t rav io le t . The f luorescent coa t ing absorbs th i s rad ia t ion

and re-emits vis ible l ight .

The discharge wil l operate correct ly only when connected in series with an

impedance . A res i s tor could be used , but would d i s s ipa te an excess ive amount of

energy, so i t i s the custom to use a series inductor. There exis t many types of c i rcui t .

A par t icu lar f luorescent f ixture op era t ing at 120 vol ts has a pow er d iss ipa t ion

of 80 wat ts . I ts power factor is 50%.

a) F ind the reac t ive cur rent .

b) Wh at is the s ize of the cap aci t or con nec ted in paral le l wi th the disch arge

tube and i ts inductor that wil l make the power factor equal to 100", ,?

18-4 ENERGY TRANSFER TO A LOAD

Figu re 18-10 sho ws a ci rcui t that ha s been used to me asu re the energy abs orb ed

by a load Z when the source V produces a pul se of cur rent . The cur rent s / ' and I" are

negl ig ib le compared to / . and the vol t age drop ac ross r i s negl ig ib le compared to V.As we shal l see, this energy is proport ional to the area enclosed by the curve on the osci l

loscope screen. The method is val id is val id even i f Z i s non- l inea r .

We sha l l demons t ra te th i s for one cyc le of an a l t e rna t ing cur rent , bu t the same

appl ies to an individu al curre nt pulse of any sh ape .

Q i

v

Q

RR

O r

O A

C

Figure 18-10 Circui t for measuring the energy t ransfer to a load Z d u r i n g a c u r r e n t

p u l s e ; r « Z , r « R, R » 1/coC. T h e c h a r g e Q f lows out of the top terminal of the

source and in to the lower one . The t e rmina l s x and y are the input s to the

ho rizo nta l and vert ical amplif iers of the osci l loscope. See Pr ob . 18-4.



Problems 447

On the sc reen ,

Q' « 2x x — , v x —

1 V.

C • P , + R 2

Set V = V 0 cos cot.

a) Show that , i f R » 1/coC,

r

x x — Q.RC

b) Show th a t the energy t rans fe r red pe r cyc le is pro por t ion a l to the a rea enc losed

by the curve on the osci l loscope screen.

c) W ha t is the sha pe of the curve wh en th e load is ( i) a res is tance, ( ii ) a ca pac i tanc e,

( ii i) a pur e ind ucta nce , (iv) a res is tance in series with an ind ucta nce , with R = w L ?d) What is the shape of the curve i f the load draws current only when V is close

to i ts ma xim um va lue?

IH-5E ENERGY TRANSFER TO A LOAD

Figure 18-11 shows another c i rcu i t for measur ing the energy d i s s ipa ted when

a pulse of current originat ing in G passes through a load Z. Fhe res is tor R' serves to

di scharge the capac i tor C s lowly between pu lses : a negl igible am ou nt of cha rge f lows

through i t during a pulse. Also, C is large and the vol tage drop across i t i s negl igible

compared to tha t ac ros s Z .

Show tha t the a rea under the curve obse rved on the osc i l loscope i s propor t iona l

to the energy diss ipated in Z, whether Z is l inear or not .

Figure 18-11 See Prob. 18-5 .

18-6E REFLECTED IMPEDANCE

Show tha t a pos i t ive ( induc t ive ) reac tance in the secondary of a t rans former

is equiv alent t o a neg at ive (capaci t ive) reacta nce in the prim ary , and vice versa.




18-7E MEASUREMENT OF THE COEFFICIENT OF COUPLING k

A t rans forme r has a pr imary induc tance L {. a second ary indu c tance L 2 , and a

mutua l induc tance M. The winding res i s t ances a re negl ig ib le .Show tha t

Z0/Z

x, = 1 - k\

w h e r e Z 0 a n d Z x are the impedances measured a t the t e rmina l s of the pr imary , when

the secondary i s shor t -c i rcu i t ed and when it i s open-c i rcui t ed .

18-8E REFLECTED IMPEDANCE

How does the impedance a t the t e rmina l s of the pr imary of a t rans former va ry

with the res is tance R 2 i n the secondary c i rcu i t ?

Cons ider a s imple case where a = L x = L 2 = k = 1, R } = 0 .

a ) Show tha t

R 1R 2

R\ + 1 R 2

2 + 1

(Note tha t , d imens iona l ly . th i s equ a t ion i s absu rd . The rea son is tha t we have as s igned

num er ica l va lues to som e of the pa ram ete rs . For exam ple , the 1 in the den om ina t ors

is really io 2Lj.)

b) Draw curves of R, X, \Z\ as funct ions of R 2, for values of R 2 ranging f rom

0.1 to 10. Use a lo g scale for R 2.

I8-9E MEASURING THE AREA UNDER A CURVE

You a re asked to measure the a rea under a curve drawn on a shee t of paper .

You a re given a pair of He lm ho ltz coi ls , as in Pr ob . 8-5, tha t give a uniform ma g

net ic f ie ld ov er a suffic ient ly large area, a lso a sour ce of a l ter nat i ng c urre nt , an electro nic

vol tmete r , conduc t ing ink , e t c .

How wi l l you proceed?

I8-I0E SOLDERING GUNS

A solder ing gun cons i s t s in a s t ep-down t rans former tha t pas ses a l a rge cur rent

through a l ength of copper wi re .

O ne sold ering gu n diss ipa tes 100 wat ts in a piece of cop per w ire (a = 5.8 x 10 7

S i emen s per meter) havi ng a cross-se ct ion of 4 squ are m il l imete rs and a length of 100

mi l l ime te rs .a) Find V and / in the secondary .

b) Fin d the cu rre nt in the pr im ar y if i t is fed at 120 volts.

18-1 IE CURRENT TRANSFORMER

Figure 18-12 shows a s ide - looking cur rent t rans former tha t i s used for measur ing

large curren t pulses . I t i s m or e conven ient th an the R ogo wsk i coil of Pr ob . 12-7 in that

the cur rent -ca r ry ing wi re need not be threaded through the measur ing coi l .



449

I

2a

2a

Figure 18-12 Side-looking transformer.

See Prob. 18-11.bV

Show that, for a single-turn coil.


A thin-walled conducting tube has a length /. a radius a. a wall thickness £> an d

a conductivity a. When placed in a uniform axial magnetic field of the form

an induced current flows in the azimuthal direction.

Alternating currents tend to flow near the outer surface of a conductor. This

phen omeno n is called the skin effect. See Prob. 11-7. The .sAm depth 6 is (2 prp

uc>rt\

l

We are concerned here with a brass tube having a wall thickness of 0.5 millimeter

and w ith a frequency of 60 hertz. Since the skin depth is 16.3 millimete rs under those

conditions, we may disregard the skin effect and assume that the induced current is

uniformly distributed throughout the thickness of the tube.

Show that, for currents flowing in the azimuthal direction, the resistance and

self-inductance of the tube are given by

B = B 0 cos tor,

R L, =Ho™ ,

so that

coL,/P, = p0awcih/2.

* Electromagnetic Fields and Waves. Sec. 11.5.




Fo r a bra ss tub e with / = 200 mil l imete rs , a = 5 mil l im eters , h = 0.5 mil l imeter ,

a n d (7 = 1.60 x 1 0 7

S i emen s p er meter, this ra t io is only a bo ut 1°„ at 60 hertz .

18-13 EDDY-CURRENT LOSSES

Why do eddy-cur rent los ses inc rease as the square of the f requency?

18-14 EDDY-CURRENT LOSSES

Eddy-cur rent los ses in magne t i c cores a re minimized by as sembl ing them f rom

l a m i n a t i o n s .

Co nsid er a cor e of rect ang ular cro ss-sect ion as in Fig. 18-13. T h e e d d y - c u r r e n t

los s i s propor t iona l to V2

/R , w h e r e V i s the e l ec t ro mo tance induced a ro un d a typica l

cur ren t pa th such as the one show n by a das hed curve. The res is tan ce is a lso diff icul t

to define, but i t i s of the order of twice the res is tance of the upper half, o r

o(b/2)L

Sho w th at, if th e core is split int o n laminat ions , the loss wil l be reduced by a

factor of n 2.

18-15 HYSTERESIS LOSSES

Hys te res i s los ses a re prop or t io na l to the ope ra t ing f requency / , whi l e eddy-

cur ren t los ses inc rease a s / 2 (Prob. 18-13).

You a re g iven a number of t rans former l amina t ions . Can you devi se an exper i

me nt tha t wil l perm it you to eva luat e the relat ive impo rta nc e of the two types of loss?

18-16 CLIP-ON AMMETERFigure 18-14 shows a type of c l ip-on amm ete r th a t i s mo re com mo n than tha t

desc r ibed in P rob . 15-3 . The on e shown here is s imply an i ron-core t rans fo rmer whose

prim ary is the curren t-ca rry ing wire . I t is s imilar to the Rog ow ski coi l of Pr ob . 12-7,

except for the i ron core, and except for the fact that the secondary does not surround the

wire. This type of t ransformer is often cal led a current transformer.

Ins t ruments of th i s type a re ava i l ab le for measur ing a l t e rna t ing or pul sed

cur rent s ranging f rom mi l l i amperes to hundreds of amperes .



45 1

Figure 18-14 C l i p - o n a m m e t e r . T h e r i n g - s h a p e d m a g n e t i c

yoke i s made in two ha lves tha t can ro ta te about the

hinge A. Th e yoke can be c l ipped a r ou nd the wi re by

open ing the j aws a t C . Th e vol t age V i nduced in winding

S is pro por t ion a l to the cur ren t f lowing th rou gh the

wire. See Prob. 18-16.

a ) Ca lcula te V when the measu red c ur rent i s one amp ere a t 60 he r t z . The c ore

has a mean diameter of 30 mil l imeters , a cross-sect ion of 64 square mil l imeters , and a

perm eabi l i ty of 10,000. Th e seco nda ry has 1000 turn s .

b) How could you inc rease the vol t age V for a given current / and for a given

i n s t r u m e n t ?

18-17 AUTO-TRANSFORMERS

Figure 18-15a shows a s chem at ic d iagram of an auto- t ran s form er . I t s s inglewinding i s t app ed an d se rves both as a pr imary and as a s econd ary . A uto- t r ans

formers are widely used, part icularly for obtaining an adjus table vol tage at 60 hertz .

Th e left -hand te rm inal s are conne cted to the 120-vol t l ine, an d the r ight- han d

Figure 18-15 (a) Auto-transformer. ( / )) Equivalent c i rcui t . See Prob. 18-17.




terminals are connected to the load. The tap can slide along the winding. The power

ratings of commercial models vary from a few tens of watts to several kilowatts.

Calculate the ratio of output voltage to input voltage VJVi. The self-inductances

L t and L 2 are those of the winding, on either side of the tap. Set

Lj = AN\, L2

= AN\, M = ( L , L 2 ) "2

= ANXN2, (1)

where M is the mutua l induc tance between Lj and L 2 , and neglect the resistance of

the winding. These assumptions are reasonable, in practice.

You will find that V„ is directly pro por tio nal to the numbe r of tur ns in L 2 ,

and is independent of Z, under the above assumptions.

Solution: First let us transform the circuit of Fig. 18-15a into that of Fig. 18-15b.

Then

jcoM

Similarly,

(Z - jo)M)jw(L2

+ M)

V _ Z - jwM + jco(L2 + M)

V;~ (Z -jcoM)jco(L2

+ M) '(

'

' jco(Li + M)Z - jo.)M + jo)(L2

+ M)

Then

V

° _V

»V

' _ Zjw(L2

+ M)

Vi ~ ~ Z[jw(L2

+ M)+jco{Li + M)] + m2

M(L2

+ M) - w2

L 2 ( L l + M)

Substituting the values of L,, L 2 , M,

V; = ZjcoA(N\ + NXN

2)

V,- ZjcoA(N\ + N\ + 2 / V , N 2 ) + m2

A2

(N\N2

2 - / V f / V2

) '

N7

JV, + N2

161



CHAPTER 19

MAXWELL'S EQUATIONS

19.1 THE TOTAL CURRENT DENSITY J ,

19.2 THE CURL OF B

19.2.1 Example: Dielectric-Filled Parallel-Plate Capacitor

19.3 MAXWELL'S EQUATIONS

19.4 MAXWELLS EQUATIONS IN INTEGRAL FORM

19.5 SUMMARY

PROBLEMS



At th i s s t age we have found on ly th ree o f Maxwel l ' s four equa t ions . These

were Eqs . 6-12, 8-12, an d 11-14. O u r first ob jec tive in thi s ch ap te r is todedu ce the four th one , namely the eq ua t ion for the cur l of B . The n we sha l l

r e - examine a l l f our equa t ions as a g roup .

19.1 THE TOTAL CURRENT DENSITY J,

There are several types of elect r ic current .

a) Th ere are f i rs t the usu al conduction currents through good con

ductors such as copper , which a re as soc ia t ed wi th the d r i f t o f conduc t ionelectrons (Sec. 5.1).

b ) There a r e a l so conduction currents in semiconductors in which bo t h

conduct ion elect rons and holes (Sec. 5 .1) can dr i f t .

c) Electrolytic currents are due to the mot ion o f ions th rough l iqu ids .

d ) The m ot i on o f ions o r e l ec t rons in a vac uum — for exam ple , in an

i o n b e a m o r i n a v a c u u m d i o d e — g i v e s convection currents.

e) The motion of macroscopic charged bodies a l so p roduces e l ec t r i c

cur ren t s . A good exam ple i s the be l t o f a Van de Graaf f h igh-vo l t age

g e n e r a t o r .

Al l these cur r en t s a r e as soc ia t ed wi th the mot ion o f f r ee charges . Weshal l use the symbol J f for the current dens i ty associated wi th f ree charges .

f ) The displacement current. T h e d i s p l a c e m e n t c u r r e n t d e n s i t y dD/dt

h a s t w o c o m p o n e n t s , t h e polarization current dens i ty dP/dt, a n d e 08E/dt

(Sec. 7.6).

g ) The equivalent currents i n magne t i c mate r i a l s , wi th a vo lume

dens i ty V x M and a surface dens i ty M x n, (Sees. 14.2 a n d 14.3).

19.2 The Curl of B 455

T h u s , in the general case, the total volume current density is

j , = J / + ^ D - + V x M , ( 19 -1 )

dt

J r + e 0 ? + ? + V x M , ( 19 - 2)

' dt ct

= J,„ + C o ^ . (19-3)ct

where J„, i s the volume current density in matter, wi th

r D r E r P

" 6 0 J i+

at'

J„ , = J | + ^ + V x M . (19-5)

( 1 9 - 4 )

79.2 THE CURL OF B

In Sec. 9.2 we foun d t ha t, for sta tic f ields.

V x B = n0(J

f+ V x M ) . ( 1 9 - 6 )

For t ime-dependen t f i e lds , we mus t r ep lace the paren thes i s by the to t a l

cur r en t dens i ty . Then

V x B = /<„ (j, + e 0 ™ + ~ + V x ( 1 9 - 7 )

o r

V x B - wo ™ = /< 0 ( j / + ^ + V x M ) . (19-8)

456 Maxwell's Equations

This is one of Maxwell's equations. In more concise form.

rEV x B - €

0/-<o -^ r = MoJ

(19-9)

Since

B = u0(H + M), (19-10)

we also have that

V x H = J,- + ~di' (19-11)

19 .2 .1 EXAMPLE: DIELECTRIC-FILLED PARALLEL-PLATE CAPACITOR

Figure 19-1 shows a parallel-plate capacitor connected across a source of alter

nating voltage V. The cap acitor co ntains a slightly conductin g dielectric that has a

permittivity e r e 0

a r

>d a conductivity rj. We neglect edge effects. Then the field inside

the capacitor is uniform and both Jf

and cD/dt are independent of r.

We can find the magnetic induction inside the dielectric by transforming Eq. 19-8

into its integral form, setting M = 0. Integrating over an area S,

J s V x B d a da.

and, using Stokes's theorem on the left.

H i l l(D

da.

(19-12)

(19-13)

where C is the curve bounding the surface S. By symmetry, the vector B is azimuthal.

Not e that B, J / 5 and D are at the same time vectors and phasors, here.

If C is now a circle of radius r inside the capacitor and parallel to the plates,

as in the figure, we have the following phasor equation:

IKI-B

3D

77

(19-14)

(19-15)

45 7

lT =====V ' * ^ „

•

3Dct/3Dct

Figure 19-1 Para l l e l -p la t e capac i tor connec ted to a source of a l t e rna t ing vol t age .

T h e c u r r e n t 3 f + c D/ct gives an azimuthal magnet ic f ie ld B.

N o w ,

J f = FFE = <TV/S,

5 E

I Te r e 0 — = ; w e r e 0 - ,

dt s

(19-16)

(19-17)

a n d

B = *^(o + jcoe^r.2s

(19-18)

Thus the vec tor B i s az im utha l and i t s ma gni tu de i s pro por t ion a l to r. Figure 19-2

shows l ines of B in a plane paral le l to the plates .

Figure 19-2 Lines of B in a plane

paral le l to the plates ins ide the

capaci tor of Fig. 19-1.




The magn e t ic ind uc t ion i s the sum of tw o te r ms . The one tha t is p r o por t ion a l

to a i s a s soc ia ted w i th the cond uc t io n cur r en t . Th a t B is in phas e with V. like the

cond uc t io n c ur r en t cE . The o the r t e r m, w hich i s p r o por t ion a l to e r e 0 , is associated

w i th 80 /dt and is thu s pro po r t io nal to the t ime rate of cha nge of V. I t is in q ua d

rature with V.

19.3 MAXW ELL'S EQUATIONS

Let us g roup toge ther the four Maxwel l equa t ions , which we found suc

cessively as Eqs. 6-12, 8-12, 11-14, 19-9:

V - E = p , / e 0 , (19-19)

V - B = 0 . (19-20)

dB

V x E + — = 0, (19-21)

V x B - € 0fi 0

cE

~d t(19-22)

As usua l .

E is the elec tric f ield inten sity in volts pe r m et er ;

Pi — Pf + Pb i s i n e t o t a l e l ec t r i c charge dens i ty in cou lombs per

c u b i c m e t e r ;

Pf i s the f ree charg e d en s i ty ;

pb i s the bou nd charg e dens i ty — V • P, w h e r e P is the electric

p o l a r i z a t i o n i n c o u l o m b s p e r s q u a r e m e t e r ;

B i s the ma gne t i c indu c t ion in t es l as ;

J„, = if + dP/dt + V x M is the cu rre nt den s i ty resul t in g f rom

the f low of charges in mat t e r , i n amperes per square mete r ;

3f

i s the current dens i ty of f ree charges ;dP/dt i s the po la r i za t io n cur r en t den s i ty ;

V x M is the equ iv a len t cur r en t dens i ty in ma gne t i zed ma t t e r ;

M i s the ma gne t i za t io n in am pere s per m ete r ;

e 0 is the permittivity of free space, 8.854 187 82 x 1 0 " 1 2 farad per

m e t e r ;

p.0

i s the permeabi l i ty of f ree space, 4n x 1 0 " 7 henry per mete r .



19.4 Maxwell's Equations in Integral Form 459

In conduc tor s obey ing Ohm's l aw (Sec . 5 .2 ) ,

J f = c t E , (19-23)

where r j i s the conduct ivi ty in Siemens per meter .

Maxwel l ' s equa t ions a r e par t i a l d i f f e r en t i a l equa t ions invo lv ing space

and t ime der ivat ives of the f ie ld vectors E an d B , the to t a l cha rge dens i ty p„

and the cur r en t dens i ty J m . They do no t yield the values of E and B direct ly ,

bu t on ly a f t e r in t egra t ion and a f t e r t ak ing in to account the p roper boundary

c o n d i t i o n s .

T h e s e a r e t h e f o ur f u n d a m e n t a l e q u a t i o n s o f e l e c t r o m a g n e t i s m . T h e y

apply to a l l e l ec t romagnet i c phenomena in media tha t a r e a t r es t wi th

respec t to the coord ina te sys tem used fo r the de l opera to r s .

19.4 MAXW ELL'S EQUA TIONS IN INTEGRAL FORM

We ha ve s t a t ed M axw el l ' s equa t io ns in d i ff e ren t ia l fo rm; l e t us now s ta t e

the m in in t egra l fo rm, in o rde r to a r r ive a t a be t t e r und er s t an d in g o f the i r

p h y s i c a l m e a n i n g .

The in tegra l fo rm i s r ea l ly more in te res t ing and genera l . I t i s more

in te res t ing because one can v i sua l i ze in t egra l s more eas i ly than der iva t ives .

I t i s a l so more genera l because i t i s app l i cab le even when the der iva t ivesdo no t ex i s t , f o r example a t the in t e r f ace be tween two media .

In Sec. 6 .4 , we dedu ce d E q. 19-19 f rom G au ss ' s law in integra l form,

whe re S i s the su rf ace bo un din g the vo lum e p t i s the to t a l charge dens i ty

pf

+ p b, a n d Q, i s the total net charge Q f + Q b c o n t a i n e d w i t h i n x. T h e

outward f lux o f E t h rough any c losed sur f ace S i s the re fore equa l to l / e 0

t imes the total net charge ins ide, as i l lus t rated in Fig. 19-3.

We proceeded s imilar ly in Sec. 8 .2 , where we deduced Eq. 19-20 f rom

the fact that

w h e r e S i s any closed surface. Th us th e net ou tgo ing f lux of B thr ou gh any

closed surface S i s zero. This i s shown in Fig. 19-4.

(19-24)

(19-25)



Figure 19-4 Lines of B through a closed surface S. The net outward f lux of B is

equa l to ze ro .

46 1

Figure 19-5 Th e d i rec t ion of the e l ec t ro mo tance induced a rou nd C i s indicated by

an a r row for the case where the magne t i c induc t ion B i s in the direct ion shown

and inc reases . Th e e lec t ro mo tance is in the same d i rec t ion if B i s upw ard and

decreases .

I f we now integrate Eq. 19-21 over a surface S b o u n d e d b y a c u r v e C,

Or , if we use S tokes ' s the ore m on th e le ft an d inver t the ope ra t io ns on the

r ight , the surface S being f ixed in space,

T h e n t h e e l e c t r o m o t a n c e i n d u c e d a r o u n d a c l o s e d c u r v e C is equ a l to m inus

th e ra te of cha ng e of the m ag ne tic f lux link ing C, as in Fig. 19-5. T he

pos i t ive d i r ec t ion s fo r B an d a ro un d C a re r e l a t ed accord ing to the r igh t -h and

screw rule (Sec. 1.12). I f t he curve C has more than one tu rn , then the su r f ace

S beco me s comp l ica te d a nd the su r f ace in tegra l i s ca l l ed the f lux l inkage .

Fina l ly , if we also integ rate Eq . 19-22 ove r an area S b o u n d e d b y a

curve C ,

(19-26)

(19-27)

(19-28)

462

Figure 19-6 The direction of the magnetomotance around C is indicated by an

a r r o w for the case where the current Jf

+ (<~D dt) is in the direction shown. The

d i s p l acem en t curr ent is down war d if D is do wnw ard and increases, or if it is

u p w a r d and decreases.

If we perform the same ope rat ion on the cor re spo nd ing e qu ati on for H, as

we did in Sec. 19.2,

(19-29)

The mag ne to mo ta nc e ar ou nd C is equal to the sum of the free cur rent plus

the displacement current linking C. This is illustrated in Fig. 19-6. The

positive directions are again related by the right-hand screw rule.

19.5 SUMMARY

The total volume current density is the sum of four terms:

r E r P

J, = Jf + e0

— + — + V x M. (19-2)ct ct

where the first term is the free current density, the second and third together

form the displacement current density BD /dt, and the fourth term is the

19.5 Summary 463

equivalent volume current density i n m a g n e t i c m a t e r i a l s . T h u s

dBBE

. f3P

Also ,

ct dt

J,„ = if + ? + V x M (79-5)

is the current density in matter.

The fo l lowing four equa t ions a r e known as Maxwell's equations:

V - E = p , / e 0 , (19-19)

V • B = 0. (19-20)

V x E + ^ = 0, (19-21)ct

V X B - e oi U o ^ = Mm- (/9

"22

>f' f

I n c o n d u c t o r s o b e y i n g Ohms law,

J f = ffE.(19-23)

I n i n t e g r a l f o r m , M a x w e l l ' s e q u a t i o n s b e c o m e

js E • da = Q , /€ 0 , (/9-2^)

J* B • da = 0. (79-25)

d) E d l = - i f B - d a = . (19-27)

J f d t J s dt

(j), B • dl = ^ J ( J m + e 0 ^ j • d a = /<„/,. (79-28)




PROBLEMS

19-1 PARALLEL-PLATE CAPACITOR SUBMERGED IN

AN INFINITE MEDIUM

Consider a paral le l -plate capaci tor s i tuated in an infini te medium as in Fig. 19-7.

Fo r s implici ty, we assu me th at the plates are ci rcular . At r = 0, the plates carry

c h a r g e s +Q a n d —Q. The capac i t ance be tween the p la tes i s C . Fhe medium i s

s l ight ly conduct ing and the res is tance between the plates is R.

Fr om Sec. 5.14, the vol tag e difference F betw een the plate s decreases ex

ponent i a l ly wi th t:

V = V0e-"RC

. (1)

Fherefore, a t any point in the dielectr ic , the electr ic f ie ld intensi ty decreases in the

s a m e m a n n e r :

E = E 0e-,IRC

, (2)

where £ 0 var ies f rom one poin t to another .

a) Use the resul t of Prob. 6-17 to show that the total current densi ty J f + dD/dt

in the dielectr ic is zero. The dielectr ic has a conduct ivi ty o and a re l at ive pe rm i t t iv i ty

Solution: At any point , the total current densi ty is

3D

Jf + y t = Iff + e re 0-IE, (3)

P C

since P C = e re 0/a, from Prob. 6-17.

b) What does this resul t te l l you about the magnet ic f ie ld ' :

Solution: According to Eq. 19-13, for any closed curve C,

B • dl = 0. (5)

Now, by symmetry, i f there is a magnet ic f ie ld, i t must be azimuthal . So consider

a ci rcle C as in Fig. 19-7. By symmetry again, s ince the plates are ci rcular by hypo

thes is , B mus t be the same a l l a round the c i rc l e . Because of the above equa t ion , B

is zero everywhere on the ci rcle . But the ci rcle can be s i tuated anywhere, as long

as i t i s paral le l to the plates and centered on the axis of symmetry, so B is zero every

w h e r e .

19-2E MAXWELLS EQUATIONS

Show that Maxwell's equations can also be written under the form

V • D = pf,

V - B = 0,

3E

V x E + — = 0,

<3f

cDV x H = Jf.

ct

I9-3E MAXWELL'S EQUATIONS

Show that, for linear and isotropic media, and for fields that are sinusoidal

functions of the time.

V • e ,e0E = Pf

V • u ru 0H = 0,

V x E + jwfirfi0H = o,

V x H - jmurri0E =h

Here again, we have quantities that are both vectors and phasors.




19-4E MAXWELLS EQUATIONS

T h e eq u a t i o n for the conservat ion of charge of Sec. 5.1.1 is buil t into Maxwell 's

eq u a t i o n s .

Show tha t it follows from Eq. 19-22.

T h e J used in C h a p t e r 5 is the cur ren t dens i ty of free charges Jj.

19-5E MAGNETIC MONOPOLES AND MAXWELL'S EQUATIONS

W h e n it is as s u m ed t h a t m ag n e t i c m o n o p o l e s do exist, it is the cu s t o m to write

Maxwel l ' s equat ions in the fol lowing form:

dBV • E = p,/e0, V x E = - — - J*.

ct

( dE

V • B = p*, V x B = / t o k - + J ,

w h e r e p* is the volume dens i ty of magnet ic charge in w eb er s per cub ic meter , and J*

is the magnet ic cur ren t dens i ty , expressed in weber s per second per square meter ,

a) Show that

dp*V • J* = — — .

dt

The modi f ied Maxwel l equat ions therefo re pos tu la te the co n s e r v a t i o n of m ag n e t i c

c h a r g e .

By analogy wi th Coulomb 's law, n ea r a magnet ic charge .

Q*

4m - 2 1

In a v a c u u m ,

Q*

4np 0r

an d th e force between tw o magnet ic charges is

0*0*F

f» = e « * H

6= f ^ r

1.

4np0r

b) Show that , for any closed path C, and if cB/ct is zero ,

E d l = - / * .



Problems 467

w h e r e / * is the m a g n e t i c c u r r e n t . Note th e n e g a t i v e s i g n ; th e i n d u c e d e l e c t r o m o t a n c e

in a c o i l is e q u a l t o m i n u s th e m a g n e t i c c u r r e n t linking it . See Prob. 12-7.

19-6 Express the f o l l o w i n g q u a n t i t i e s in t e r m s of k i l o g r a m s , meters, s e c o n d s , an d a m p e r e s :

j o u l e , w a t t ,

coulomb, v o l t , ohm, S i e m e n s , f a r a d ,

weber, t e s l a , h e n r y .



CHAPTER 20

ELECTROMAGNETIC WA VES

20.1 WAVES

20.1.1 Plane Sinusoidal Waves. Propagation of a Scalar Quantity20.1.2 Plane Sinusoidal Waves. Propagation of a Vector Quantity

20.2 THE WA VE EQUATION

20.3 THE ELECTROMAG NETIC SPECTRUM

20.4 PLANE ELECTROMAG NETIC WA VES IN FREE

SPACE: VELOCITY OF PROPAGATION c

20.5 PLANE ELECTROMAG NETIC WAVES IN FREE

SPACE: THE E AN D H VECTORS

20.6 THE POYNTING VECTOR if IN FREE SPACE

20.6.1 Example: Laser Beam

20.7 PLANE ELECTROMAG NETIC WAVES IN

NON-CONDUCTORS

20.7.1 Example: Laser Beam

20.8 SUMMARY

PROBLEMS



In th i s l as t chap te r we sha l l have no more than a g l impse o f e l ec t romagnet i c

w a v e s . T h e s e w a v e s a r e e v e r y w h e r e — l i g h t w a v e s a r e e l e c t r o m a g n e t i c — a n dthe i r app l i ca t ions a r e innumerab le , some of the bes t known be ing r ad io ,

te levis ion, and radar . We shal l s tar t wi th a br ief review of plane waves , and

then discuss plane elect romagnet ic waves , f i r s t in f ree space, and then in

dielect r ics . This wi l l require al l four of Maxwel l ' s equat ions .

Unfor tuna te ly , the methods tha t a r e used fo r l aunch ing o r fo r de

t ec t ing these waves a r e bey on d the s cope o f th i s bo ok . H ow ever , i f yo u

wish to go fu r ther , you wi l l f ind many books on e l ec t romagnet i c waves . f

20.1 WA VES

The essent ial nature of waves is i l lus t rated by a s t retched s t r ing, f ixed at

one end and moved r ap id ly in a ver t i ca l p l ane in some a rb i t r a ry f ash ion a t

the oth er end , as in Fig. 20 -1.

If we call y(t) t he he igh t o f the moving end , then the he igh t y a t a

d i s t a n c e z along the s t r ing is y[t — (z/v)], assuming no losses , a perfect ly

f lexible s t r ing, and no ref lected w ave . Th e height at z i s thu s equ al to t he

he igh t o f the moving end a t a p rev ious t ime t — {z/v), t h e q u a n t i t y z/v

be ing the t ime r equ i r ed fo r the d i s tu rbance to t r ave l th rough the d i s t ance z

at the veloci ty v.

f Almost All About Waves by Jo hn R. Pierce, M IT Press , 1974, is a delightful l i t t le

book by an authori ty on the subject . See also Electromagnetic Fields and Waves,

Ch apte rs 11 to 14 .



47 0

y

Figure 20-1 Wave on a s t re t ched s t r ing .

We can genera l i ze f rom th i s s imple example and cons ider waves

propaga t ing in an ex tended r eg ion , such as acous t i c waves in a i r o r l igh t

waves in space.

The quan t i ty p ropaga ted can be e i ther a s ca la r o r a vec to r quan t i ty .

Fo r examp le , in an aco us t i c wave , we hav e the p rop ag a t i on o f a p res sure ,

which is a s ca la r . I n an e l ec t rom agn et i c wave , we have the p r op ag a t io n o f

a vector E and of a vector H .

20.1.1 PLANE SINUSOIDAL WAVES. PROPAGATIONOF A SCALAR QUANTITY

If a ce r t a in s ca la r quan t i ty a p ro pag a t in g wi th a ve loc i ty v is define d at

- = 0 by

This expres s ion descr ibes a plane wave t r ave l ing a long the z - ax i s ,

s ince a is independent of x a n d y. The wave i s a l so unattenuated, s ince the

a m p l i t u d e a 0 i s con s tan t . Th e quan t i ty be tw een b ra cke t s is the phase of the

w a v e . T h e frequency f is w/2n, a n d t h e period T is iff.

y. = 7. 0 cos ojf. (20-1)

then , fo r any po s i t ion z in the d i r ec t ion o f the p ro pa ga t ion .

(20-2)

471

Figure 20-2 The quantity i — a0c o s t y [ f

function of l.

•)] as a function of z and as a

F o r a given pos i t io n z , we ha ve a s inus oida l var iat i on of J. wi th t ime(Fig. 20-2a) ; for a given t ime f , we have a s inusoidal var iat ion of a wi th z

(Fig. 20-2b) .

The sur f aces o f cons tan t phase , ca l l ed wave-fronts, are normal to the

z-axis .

T h e q u a n t i t y v is called the phase velocity of the wave, s ince i t is the

ve loc i ty wi th which a wavef ron t i s p ropaga ted in space . For example , i f t he

bra cke t i s zero , the po s i t io n of the wa vefron t at the t ime f i s given by

z = vt . (20-3)

T h e wavelength X i s the dis tance over which coz/v changes by 2K

r ad ians , as shown in F ig . 20-2 :

coX/v = 2n,

X = 2nv/co = v'f.

(20-4)

(20-5)

Also .

A = A/27i = v/co, (20-6)

w h e r e A ( r ead " l ambda bar " ) i s the radian length. I t i s the dis ta nce o ver wh ich

the phase o f the wave cha nges by on e r ad ian ; th is is ab ou t A /6. I t turn s o ut

tha t the quan t i ty tha t appear s in near ly a l l wave ca lcu la t ions i s A, a n d n o t

A. We are al ready famil iar wi th the fact that in near ly al l osci l la t ion calcu

lat ions i t i s the ci rcular f requency <o th at is used ins tea d of the frequen cy / '.

472 Electromagnetic Waves

The intuitively simple quantity, namely the wavelength or the frequency,

is not the one that is "na tur al" from a math emat ical stand point .

In phasor notation (Sec. 16.7), a sinusoidal wave traveling in the

positive direction along the z-axis is described by

Similarly, a wave traveling in the negative direction along the z-axis

is describ ed by

20.1.2 PLANE SINUSOIDAL WAVES. PROPAGATION

OF A VECTOR QUANTITY

In the above equations, a is a scalar. Now, as we noted at the beginning of

this chap ter, the quant it y tha t is pr op aga te d can also be a vector , say the

electric field intensity E. The n, the above equa tio ns app ly to each one of

the components Ex, E

y, E.. Or, if we multip ly th e equ ati on for E

xby i, that

for Ey. by j , and that for E

:by k, and take the sum, we have an equation like

20-7 with a replaced by E.

Not e that the quanti ty E is then both a phasor and a vector. It is avector because it is oriented in space, with the usual three components, and

it is also a pha sor b ecau se its time depe nde nce is writ ten in the form

Let us calculate the second derivatives of a (Eq. 20-7) with respect to t

and to z:

y. = <x0expj[cof - (z/X)].* (20-7)

a = a0

ex p j [cot + (z/?.)]. (20-8)

exp/(cot + cp).

20.2 THE WA VE EQUA TION

7 T = - t o a,dt

2

32a

a. (20-9)

+

In the last ch ap t e r , we shall follow the usual cus tom of not us ing a specia l no ta t ion

for p h a s o r s : vecto r phasor s wi l l be writ ten l ike vec tors , an d sca lar phasor s l ike scalars .



20.3 The Electromagnetic Spectrum 473

T h u s

r 2 a(20-10)

or

1 T 2 a( 2 0

-u )

w h e r e v i s the phase veloci ty . This i s the wave equation for a plane wave

pro pag a t in g a lon g the z - ax i s. Th i s eq ua t io n i s va l id even fo r no n- s inu so ida l

w a v e s .

More genera l ly , fo r an una t t enua ted wave in space ,

1 r 2 a

V 2 a = ~ ^ . I2 0

"1 2

)

v dt

If we had a vec to r qu an t i ty ins t ead o f the s ca la r x , we would have the s am e

equ a t io n wi th a r ep laced by , s ay , E . Fo r an una t t en ua ted s inuso ida l w ave ,

2

V 2 a = (20-13)v

20.3 THE ELECTROM AGNETIC SPECTRUM

W e are now ready to s tudy the p rop ag a t io n o f e l ec t ro ma gnet i c waves in

f ree space and in dielect r ics .

Maxwel l ' s equa t ions impose no l imi t on the f r equency o f e l ec t ro

ma gne t i c waves . To da te , t he spec t ru m th a t has been inves t iga ted exp er i

men ta l ly i s sho wn in F ig . 20-3 . I t ex te nds con t inuo us ly f rom the long r ad io

waves to the very h igh energy gamma rays observed in cosmic r ad ia t ion .

In the former , the f requenc ies are ab ou t 10 her tz , and th e wa vele ngth s are

a b o u t 3 x 1 0 7 meters ; in the la t ter , the f requencies are of the order of

1 0 2 4 her t z , and the wave leng th s of the o rde r o f 3 x 1 0 " 1 6 m e t e r . T h e k n o w n

spec t rum thus cover s a r ange o f 23 o rde r s o f ma gni t ude . Rad io , l i gh t, an d

h e a t w a v e s . X - r a y s a n d g a m m a r a y s — a l l a r e e l e c t r o m a g n e t i c , a l t h o u g h t h e

sources and the de tec to r s , as wel l as the modes o f in t e r ac t ion wi th mat t e r ,

vary widely as the f requency c hang es by o rde r s of ma gni tud e . The n om en

clature of radio waves is given in Table 20-1.

~fI

I

!

-t

i

±= J

t z

_ t

O BO o

o

w £2

O. BO

- OS

C 3

Q.

^ o- oT 3 _

— -ise ~ a

"Sh >> o

-Q T < 3?

2 rr

o 1>

3 -s

.O E a

£3

?r_ OZJ •—!

o a"

3 O

-•

e

on •

MPS— >



20.3 The Electromagnetic Spectrum 475

Tabic 20-1

Band Frequency Metric

Num bers* Range Subdivi sion Abbreviation

1 3--30 Hz

2 30 -300 Hz M e g a m e t r i c w a v e s elf

3 300--3000 Hz vf

4 3- - 3 0 k H z My riam et r i c waves v lf

5 30 -300 kH z Ki lom et r i c waves If

6 3 0 0 - -3000 kH z Hec to met r i c waves mf

7 3- - 3 0 M H z D e c a m e t r i c w a v e s hf

8 30 - 3 0 0 M H z M e t r i c w a v e s vh f

9 300--3000 M H z Dec im et r i c waves uhf

10 3- - 3 0 G H z C e n t i m e t r i c w a v e s sh f

11 30 -300 G H z Mi l l imet r i c waves eh f

12 300- 3 0 0 0 G H z Dec imi l l ime t r i c waves

* B a n d n u m b e r N ex te nds f rom 0 .3 x 10v

to 3 x 10 'v

h e r t z .

Th e funda me nta l iden t i ty o f a l l t hese types o f wav es i s dem on s t r a t ed

by many exper iment s cover ing over l app ing par t s o f the spec t rum. I t i s a l so

shown by the fact that in f ree space they are al l t ransverse waves wi th a

c o m m o n v e l o c it y o f p r o p a g a t i o n . F o r e x a m p l e , s i m u l t a n e o u s r a d i o a n d

opt i ca l obse rva t io ns o f f la r e s t a r s hav e shown tha t the ve loc i ty o f p rop ag a t io n

i s the s ame, wi th in expe r ime nta l e r ro r , fo r wave leng ths d if fe r ing by m or e

than s ix o rder s o f magni tude .W e shal l fol low the cus tom of us ing H rat he r th an B in discuss in g

elect romagnet ic waves , in spi te of the fact that , unt i l now, we have used H

only in r e l a t ion wi th magne t i c mate r i a l s . The main r eason fo r us ing H in

s t ead o f B in dea l ing wi th e l ec t romagnet i c waves i s tha t E x H gives the

energy f lux dens i ty , as we shal l see.

20.4 PLANE ELECTROMAGNETIC WA VES IN FREE

SPACE: VELOCITY OF PROPAGATION c

Let us start with the relatively simple case of a plane sinusoidal wave pro

pagating in a vacuum in a region infinitely remote from matter. Then

Maxwell 's equ ati on s 19-19 to 19-22 beco me

V • E = 0, (20-14)

V • H = 0, (20-15)

V x E + ,/oj/i0H = 0. (20-16)

V x H - ;<B€ 0E = 0. (20-17)

Taking the curls of Eqs. 20-16 and 20-17,

V x V x E + jcono V x H = 0, (20-18)

V x V x H — j f coe 0 V x E = 0, (20-19)

or, usin g the identi ty of Pr ob . 1-32,

V(V • E) - V2

E + jcofio V x H = 0, (20-20)

V(V • H) - V2

H - jco€0 V x E = 0. (20-21)

Finally, substituting Maxwell's equations,

V2

E = - e0/ (

0w

2

E , (20-22)

V2

H = -e0p0co2

H. (20-23)

Comparing now with Eq. 2 0 - 1 3 , we find th at these differential equa

tions are those of a wave. Also, it follows that the field vec tor s E an d H

can propagate as waves in free space at the velocity

c = l/feo^o)1 (20-24)

20.5 Plane Electromagnetic Waves in Free Space 477

E q u a t i o n 2 0 - 2 4 l i nk s t h r e e b a s i c c o n s t a n t s o f e l e c t r o m a g n e t i s m : t h e

velocity of an electromagnetic wave c, the permit t ivi ty of f ree space e 0 , and

the permeabi l i ty of f ree space p 0. I t wi l l be rem em be red f rom Sec. 8 .1 th att h e c o n s t a n t p 0 was def ined a rb i t r a r i ly to be exactly An x 1 0 " 7 henry per

m e t e r . T h e c o n s t a n t e 0 can thus be dedu ced f rom the me asu red va lue for

the ve loc i ty o f e l ec t romagnet i c waves ,

c = 2.997 924 58 x 1 0 8 meter s / s econd: (20-25)

e 0 = l/ c

2p 0 = 8.854 187 82 x 1 0 " 1 2 f a r ad /mete r . (20-26)

We sha l l use the approx imate va lue o f c, n a m e l y 3 x 1 0 8 meter s per

second, which is accurate wi thin 7 par ts in 10,000.The permit t ivi ty of f ree space e 0 can a l so be de te rmined d i r ec t ly

f r o m m e a s u r e m e n t s i n v o l v i n g e l e c t r o s t a t i c p h e n o m e n a . T h e m e a s u r e m e n t s

l ead to the above va lue wi th in exper imenta l e r ro r , t he reby conf i rming the

t h e o r y .

20.5 PLANE ELECTROM AGNETIC WA VES IN FREE

SPACE: THE E AND H VECTORS

For a p lane e l ec t romagnet i c wave p ropaga t ing in the pos i t ive d i r ec t ion

of the z-axis , E is independent of x a n d y, a n d

V • E = <^f-£ = 0. (20-27)

CZ

The refore the z co m po ne nt of E can no t be a funct ion of z . W e shal l set

Ez = 0, (20-28)

s ince we are interes ted in waves , and not in uniform f ields .

The s ame a rgument app l i es to the H vec to r , and we can s e t

H z = 0. (20-29)

A p lane e l ec t romagnet i c wave p ropaga t ing in f r ee space i s there fore

transverse, s i n c e i t h a s n o l o n g i t u d i n a l c o m p o n e n t s .



47 8

y

Figure 20-4 Decomposi t ion of the vector E into two vectors E t a n d E , .

W e no w as sum e tha t the wave i s p l an e-po la r i zed wi th i ts E v e c t o r

a lways po in t ing in the d i r ec t ion o f the x -ax i s . Th i s does no t invo lve any

los s o f genera l i ty s ince any p lane-po la r i zed wav e can be cons idered to be

the sum of two waves tha t a r e p lane -po la r i ze d in perp end icu la r d i r ec t ion sand in phase. For example, in Fig. 20-4, the vector E can be r eso lved in to

t w o m u t u a l l y p e r p e n d i c u l a r v e c t o r s E i a n d E2.

For a p lane-po la r i zed wave hav ing i t s E vector in the di rect ion of

the x-axis (Fig. 20-5) ,

(20-30)

Let us now subs t i tu t e th i s va lue o f E in to Eq . 20-17 :

J k

0 0 (20-31)

H x Ey 0



47 9

(a)

H

E x H

E x H

• XXX

x M

(b)

Figure 20-5 Fh e E and H ve ctors for a pla ne electro ma gne t ic wave t ravel ing in the

posi t ive direct ion along the z-axis . (a) Fhe variat ion of E and H with r a t a

par t i cu la r moment . The two vec tors a re in phase , but pe rpendicula r to each o the r .(b) The cor responding l ines of force as s een when looking down on the xz -p lahe .

The l ines represent the electr ic f ie ld. The dots represent magnet ic l ines of force

coming out of the paper, and the crosses represent magnet ic l ines of force going

in to the paper . The vec tor E x H gives the d i rec t ion of propaga t ion .




On the left-hand side we have set the derivatives with respect to x and to

v equal to zero because we have a plan e wave prop agati ng a lon g the z-axis.

We have also set H2 equal to zero, from Eq. 20-29. Thus

-— Hy = jco€0E0 expjco I t - - J , (20-32)

j~zHx= 0. (20-33)

Replacing the operator d/dz by the factor —jco/c,

Hx = 0, (20-34)

H = ce0E0 expjco (t - ^ j. (20-35)

= (e0/n0)ll2Ei. (20-36)

Therefore H is perpendicular to E, and

E/H = l / e0c = u0c = (P-o/eoV2 = 377 ohms, (20-37)

E/B = c = 3.00 x 10 8 meters/second. (20-38)

The E and H vectors of the wave are perpendicular and oriented in

such a way that their vector product E x H points in the direction of

propa gat ion, as in Fig. 20-5. The E and H vectors are in phase, since E/H

is real, and they have the same relative magnitudes at all points at all times.

The electric and magnetic energy densities are equal and in phase,since

( l / 2 ) e 0 £2 = (\/2Ui

0H

2

. (20-39)

At any instant, the total energy density is therefore distributed as in Fig. 20-6.

48 1

€ 0 £ 2

Figure 20-6 The energy dens i ty e0E2 = p 0H

2 as a function of z for a plane

e lec t romagne t i c wave t rave l ing a long the r -ax i s in f ree space .

20.6 THE POYNTING VECTOR <f IN FREE SPACE

We have found tha t a p lane e l ec t romagnet i c wave in f r ee space p ropaga tes

in the di rect ion of the vector E x H. Let us calculate the divergence of this

vector for an y elect romagnet ic f ie ld in f ree space. From Prob. 1-28.

V • (E x H ) = — E • (V x H ) + H • (V x E). (20-40)

Now, us ing the Maxwel l equa t ions fo r V x H and for V x E,

V - ( E x H ) = - E - e o - ^ - H - M o ^ , (20-41)ct ct

=

" ^ ( !f

°£ 2 + 2 ' ' o / / j < 2 ( M 2 )

I n t e g r a t i n g o v e r a v o l u m e t bo un de d by a su r f ace S , and u s ing the d ivergence

theorem on the l e f t -hand s ide ,

J , (E x H ) • d a = - l- I e 0E2

+ ~ p0H2j ch . (20-43)

Th e in tegra l on the r igh t i s the sum of the e l ec t r ic and ma gne t i cenergies , acc ord ing t o Sees . 4 .4 an d 13.3. Th e r igh t-h an d s ide is thu s th e

energy los t per uni t t ime by the volume r , and, s ince there must be con

serv at io n of energ y, the lef t -hand s ide m ust be the tota l out w ar d f lux of

e n e r g y t h r o u g h t h e s u r f a c e 5 b o u n d i n g t .

T h e q u a n t i t y

! f = E x H (20-44)




EXAMPLE: LASER BEAM

A laser beam carr ies a power of 20 gigawatts and has a diameter of 2 mil l imeters .

Let us calculate the peak values of E and of B.

From Eq . 20-47 ,

21 / 2

£ _ = 21

10 3 20 x 10'

,2 .66 n x 1 0 ~6

2.2 x 10 9 vol t s /meter .

(20-49)

(20-50)

Thi s is an e no rm ou s electr ic f ield; it corres pon ds to a voltage dif ference of ab ou t a

qu ar t er of a volt over a distan ce of 1 0 " 1 0 meter , which i s abou t the d iameter o f

an a t o m .We can now find B 0 f rom Eq. 20-38:

2.2 x 10 "

3 x 1 0s

7.3 teslas. (20-51)

Thi s is abo ut ten t imes larger tha n the ma gne tic induc tion betw een the pole pieces

of a power fu l pe rma nen t magn et .

is called the Poynting vector. W he n in te gra te d over a c losed sur face , it gives

the to ta l ou tw ar d f low of energy p er un i t t ime .

I n a p l an e e l ec t r o m ag n e t i c wav e , t h e Po y n t in g v ec to r p o in t s i n th ed i r ec t io n o f p r o p ag a t io n o f th e wav e . It s i n s t an ta n eo u s v a lu e a t a g iv en

po int in spa ce is E x H an d , acco r d in g to Eq . 2 0 - 3 7 ,

Sf = — £ 2 k = c e 0 £ 2 k . (20-45)

The t ime average of ff is

•^av = (^ 0E; ms)ck = [ ( l / 2 ) e 0 E 2 ] c k = (l/2)(€0/n

0)"

2

E2

0k, (20-46)

= 2.66 x 1 0 ~ 3 £ r

2

m s k watts /meter2

. (20-47)

Th e energy can therefore be con s ide red to t ravel wi th an ave rage dens i ty

e0£

r

2

m s= ( l / 2 ) e 0 £ 2 = ( l / 2 ) e 0 £ r

2

m s + (l/2)u0H2

ms (20-48)

a t t h e v e lo c i ty o f p r o p ag a t io n ck .



20.7 Plane Electromagnetic Waves in Non-Conductors 483

20.7 PLANE ELECTROMA GNETIC WA VES

IN NON-CONDUCTORS

I n a n o n - c o n d u c t o r w i t h a p e r m i t t i v i t y e r e 0 a n d a p e r m e a b i l i t y nrpt

0,

Maxwel l ' s equa t ions a r e s imi la r to Eqs . 20-14 to 20-17 , excep t tha t e 0 a n d

p.0

are r ep laced by e r e 0 a n d fi rfJ.0- T h e w a v e e q u a t i o n s a r e t h e r e f o r e

V 2 E = - e , € 0 ^ 0 m 2 E , (20-52)

V 2 H = - e r e 0 / ( n u 0 c o 2 H , (20-53)

and al l the resul ts of Sees . 20.4 to 20.6 apply, i f one replaces e 0 b y e r e 0 a n d

Ih by n rHo •The phase ve loc i ty i s now

v = 1 ™ = C

- m . (20-54)

Th e phase ve loc i ty in no n-c on du c to r s i s the re fore l es s tha n in fr ee space ,

a n d t h e index of refraction is

n = c/v = (e rfi r)112

. (20-55)

I n a n o n - m a g n e t i c m e d i u m , p r = 1 and the inde x of ref ract ion isequa l to the square roo t o f the r e l a t ive permi t t iv i ty :

n = el12

. (20-56)

N o t e , h o w e v e r , t h a t n and e r are bo th func t ions o f the f r equency . We have

discussed br ief ly the var ia t io n of e r with f requency in Sec. 7 .7 . Since tables of

n are usua l ly compi led a t op t i ca l f r equenc ies , whereas e r i s usua l ly me asure d

a t much lower f r equenc ies , such pa i r s o f va lues cannot be expec ted to

c o r r e s p o n d .

Also , as in Sec. 20.5,

H y = (e r€ 0/p rfi 0)ll2E x. (20-57)

In non-conduc tor s , t he E and H vec to r s a r e in phase and the e l ec t r i c

a n d m a g n e t i c e n e r g y d e n s i t i e s a r e e q u a l :

( l / 2 ) e r e 0 £ 2 = ( l / 2 ) ^ / i 0 H 2 . (20-58)




The to ta l ins t an taneous energy dens i ty i s thus e r e 0 £ 2 o r p rpQH2, a n d t h e

average total energy dens i ty is e r e 0 £ r

2

m s o r p rp 0H2

mi. The average va lue o f

the Poynt ing vec to r i s

^ = \ ( ^ - T Eok = (^ f E2

mX (20-59)2 W ' o /

= ( e r e 0 £ r

2

m s ) rk . (20-60)

= 2.66 x 1 0 " 3 ( e r / i 0 1 / 2 £ r

2

m s k w a t t s / m e t e r 2 . (20-61)

Th e average va lue o f the Poy nt in g vec to r i s aga in eq ua l to the phase ve loc i ty

mul t ip l i ed by the average energy dens i ty .

20.7.1 EXAMPLE: LASER BEAM

If the 20 gigawat t laser beam of Sec. 20.6.1 is in a glass whos e in dex of refra ction is

1.6, then e r "2 is 1.6, u r i s uni ty , and £„ is smaller than in air by a factor of 1.6 1 2 o r

1.26. Then E 0 in th e glas s is 1.7 x 10 9 vol ts per m eter .

To c a lc u la t e B 0 we can use the fact tha t the P oy nt in g vec tor {j)E0H0 i s the sam e

in the glass as in the air; s ince £ 0 is sm aller in the glass by a factor of 1.26. B 0 m u s t

be larger by the sam e fac tor .

You can check tha t th is ma kes the e lec tr ic energy de ns i ty equ al to the ma gne t ic

energy dens i ty (Eq. 20-58) .

20.8 SUMMARY

For any p lane wave p ropaga t ing in the pos i t ive d i r ec t ion o f the r - ax i s a t a

veloci ty v, t he d i s tu rb anc e a t z , a t t he t ime t. is th e sa m e as th at at r = 0.

at a previous t ime f — (z/v). I t is as sum ed th a t there i s no a t t en ua t ion . T he

q u a n t i t y v is the phase velocity.

F o r a p l a n e , u n a t t e n u a t e d s i n u s o i d a l w a v e ,

x 0 cos to / (20-2)

where a 0 is the amplitude, an d the qua n t i ty be tween b rack e t s is the phase.

Sur faces o f con s tan t ph ase a r e wavefronts. The wave leng th i s

/. = V I. (20-5)



T h e wave equation is

o r , i n th ree d imens ions .

20.8 Summary 485

(20-11)dz 2 v 2 dt 2

Th e kn ow n spec t rum of e l ec t ro ma gnet i c waves ex ten ds from 10 to

1 02 4 her t z .

In f ree space, the wave equations for electromagnetic fields a re

V 2 E = -e 0p 0«)2

E, (20-22)

V 2 H = -e0p0oj2

H. (20-23)

T h e phase velocity of electromagnetic waves in free space is

c = l / ( e 0 / ( 0 ) 1 / 2 = 2.997 924 58 x 1 0 s m e t e r s / s e c o n d . (20-25)

Plane e l ec t romagnet i c waves in f r ee space a re t r ansver se . The i r E

a n dH

vec tor s a r e o r thogona l and o r i en ted so tha t Ex H

points in thed i r e c t i o n o f p r o p a g a t i o n . T h e m a g n i t u d e s o f E a n d H are related to give

equa l e l ec t r i c and magne t i c energy dens i t i es :

4 = W e 0 )1 / 2 = 377 ohm s . (20-37)

H

T h e q u a n t i t y

ff = E x H (20-44)

i s cal led the Poynting vector. I t i s exp ressed in wa t ts per squa re meter , an d,wh en i ts no rm al ou tw ard c om po ne nt i s in t egra ted ove r a c losed sur face ,

i t g ives the e l ec t romagnet i c power f lowing ou t th rough the su r f ace .

F o r a p l a n e w a v e , Sf i s eq ua l to the energy dens i ty m ul t ipl i ed by the

phase ve loc i ty .

In non-conductors, the phase veloci ty is

v = c/(erp r)l

>2 (20-54)




PROBLEMS

20-1 PLANE WAVE IN FREE SPACE

A plane e lec t romagne t i c wave of circular frequency co p r o p a g a t e s in free space

in the di rec t ion of th e z-axis .

Show tha t , f rom Maxwel l ' s equa t ions ,

k x E = n0cH,

k x H = - e 0 c E ,

w h e r e k is the uni t vec tor in the di rec t ion of th e z-axis , and c is the veloci ty of l i gh t in

a v a c u u m .

LOOP ANTENNA

A 3 0 - m e g a h e r t z p l a n e e l e c t r o m a g n e t i c w a v e p r o p a g a t e s in free space, and its

peak electric field intensity is 100 mil l ivol ts pe r m e t e r .

C a l c u l a t e the peak vol t age induced in a 1.00 square mete r , 10- turn rece iv ing

l o o p o r i e n t e d so t h a t it s p l a n e c o n t a i n s the n o r m a l to a wave front and forms an a n g l e

of 30 degrees with the e lec t ri c vec to r .

POYNTING VECTOR

a) Ca lcula te the electric field intensity of t h e r a d i a t i o n at the surface of t h e sun

from the fo l lowing da ta : power rad ia ted by the sun, 3.8 x 1 02 6 w a t t s ; r a d i u s of the

s u n , 7.0 x 10s meters .

b ) W h a t is the electric field intensity of s o l a r r a d i a t i o n at the surface of the

e a r t h ? The average d i s t ance be tween the sun an d the e a r t h is 1.5 x 1 01 1 m e t e r s .

c ) Ca lcula te the va lue of •'/'at th e surface of t he ea r th , neglec t ing absorpt ion in

t h e a t m o s p h e r e .

20-4E SOLAR ENERGY

C a l c u l a t e the area requi red for p r o d u c i n g one m e g a w a t t of electr ic po we r from

sola r energy at the surface of the ear th , a s suming tha t the average eff ic iency over one

year is 2 % . The efficiency of solar cells is present ly about 15°, , at n o r m a l i n c i d e n c e .

See P rob . 20-3 .

20-2

20-3E

a n d , in n o n - m a g n e t i c m e d i a fji r = 1 ) , t h e index of r e f r ac t ion n is r e la ted

to the r e la t ive permi t t iv i ty by the r e la t ion

n = e r

1 / 2

. (20-56)

T h e v e c t o r s E and H are in p h a s e , and the electr ic and magnet ic energy

dens i t i es are e q u a l .



Problems 48 7

20-5

20-6

20-7D

20-8E

20-9

POYNTING VECTOR

An electromagnet ic wave in ai r has an electr ic f ie ld intensi ty of 20 vol ts rms per

meter . I t i s absorbed by a sheet having a mass of 10~ 2 ki logram per square mete r and

a specif ic heat capaci ty of 400 jou les per ki log ram kelvin.

Assuming tha t no hea t i s los t , ca lcula te the ra t e a t which the t empera ture r i s es .

POYNTING VECTOR

A long superconduc t ing so lenoid ca r r i e s a cur rent tha t inc reases wi th t ime .

a ) Dr aw a ske tch show ing, on a longi tud ina l c ros s -sec t ion of the so leno id ,

A, -dA/dt, H , and E x H .

b) Can yo u expla in the ex i s t ence and the or i en ta t io n of the Poy nt ing vec tor?

POYNTING VECTOR

A c i rcula r ly pola r i zed wav e resul ts from the superp os i t ion of two waves tha t

a re (a ) of the same f requency a nd a mp l i tude , (b) p lane-po la r i zed in pe rpe ndicu la r

direct io ns , an d (c) 90 out of pha se.

Show that the average value of the Poynt ing vector for such a wave is the sum

of the average va lues of the Poynt ing vec tors for the two p lane-pola r i zed waves .

Hint: Proceed as in P rob . 16-16b.

POYNTING VECTOR

A wire of radius a has a res is tance of R' ohms per meter . I t carr ies a current I.

Sho w that the Po yn t ing v ector a t the surface is directed in wa rd and th at i t gives

the correct Joule power loss of I2R' wat t s pe r mete r .

COAXIAL LINE

A s t ructure des igned to guide a wave along a prescribed path is cal led a waveguide.

Th e s imples t type is the coaxial l ine i l lus t rated in Fig. 20-7. An electr om agn et ic

wave propaga tes in the annula r reg ion be tween the two coaxia l conduc tors , and the re

is zero f ield outs ide. Fh e me diu m of pr op ag at i on is a low-loss dielectr ic . (See Pr ob s .

6-5 and 6-6.)

We as su me tha t a wave pro pag a tes in the pos i t ive d i rec t ion of the z -axi s and

that there is no reflected wave in the opposi te direct ion. We also neglect the res is tance

o f t h e c o n d u c t o r s .

As in a cyl indrical capaci tor , E is radial and inversely proport ional to the radius .Since we have a wave.

w h e r e K is a cons tan t . We sha l l cons ide r an E poin t ing ou twa rd to be pos i t ive .

F h e r a d i a n l e n g t h X i s the same as for a plane wave in the same dielectr ic .

exp /' ojt —



488

Figure 20-7 Coa xial l ine. See Pro b. 20-9.

a ) Draw a longi tudina l s ec t ion through the l ine and show vec tors E a t a g iven

ins tant over a t l eas t one wave length , va ry ing the l engths of the a r rows according

to the s t rength of the f ie ld.


b) F ind the vol t age of the inner conduc tor wi th respec t to the oute r one .

Solution:

{ ; ; E - d r = K e x p . / U - l j j ; ; ^ ,

K i n — e x p 7 wf -

I l l

(2)

t 1 1 t t I 1 t

I t " ' " t 1 1 t t 1

Figure 20-8 Sect ion through a coaxial l ine showing the electr ic f ie ld intensi ty

(ver t i ca l a r rows) and the cur rent (hor i zonta l a r rows) over 1.5 wa vele ngth s at a

given ins tant . The wave t ravels from left to r ight . The surface charge densi ty

i s pos i t ive where the cur rent a r ro ws poin t r ight , and n ega t ive where they

point left.



Problems 489

c) The v e c t o r s E and H are re l a t ed as in a plane wave , so t h a t

The magnet ic f ie ld is a z i m u t h a l .

F i n d the c u r r e n t on the i n n e r c o n d u c t o r . An equal current f lows in the o p p o s i t e

d i rec t ion a long th e o u t e r c o n d u c t o r .

N o t e t h a t V an d / are i n t e r re la t ed .

Solution: F r o m A m p e r e ' s c i r c u i t a l law (Sec. 9.1),

/ = 2nR,H = 2n R t ^ — j ^ exp. / (at - ^Y (4)

Kef2

( z\

= exp / {cot . (5)

60 P

V * /

Figure 20-8 shows how the cur rent va r i es a long the l i ne . Note tha t £, H, and /

a re all in p h a s e .

d) Ca lcula te the t ime averaged t ransmit ted power, f i rs t from the p r o d u c t VI, and

tf^en by i n t e g r a t i n g the Poynt ing vec tor over the annula r reg ion .

Solution:

K / = K l n ^ ^ c o s

2

( c o f - ~ Y (6)R, 60 V;

-/

( ) a v =

1 2 0 " RW a t t S '

EH = -=- co s- cot — - , (8)

' \ /'o } \ *>

rR 22nrclr K2

e]2

CR 2 dr.., CR 2 Inrclr ts.-e, C-R, dr(EHV, = -K

2

ef2

— = - = ^ - I - . ( 9)(

'a l

2 J*> r2

120 JR

' r

£•21/2 n

= ^ — ^ l n — w a t t s (10)12 0

COAXIAL LINE

A coaxial l ine has an i n n e r d i a m e t e r of 5 mil l imete rs and an o u t e r d i a m e t e r of

20 mil l imeters . The o u t e r c o n d u c t o r is g r o u n d e d ; th e i n n e r c o n d u c t o r is he ld at + 2 2 0

vol ts . The c u r r e n t is 10.0 amperes .




Integrate the Poynting vector over the annular region between the conductors,

and compare with the power 220 x 10.0 watts.

20-1 ID REFLECTION AND REFRACTION. FRESNEL'S EQUATIONS.

When light or, for that matter, any electromagnetic wave, passes from one

medium to another, the original wave is separated into two parts, a reflected wave and

a transmi tted wave. There are two ex ceptions: in total reflection there is only a reflected

wave, and a wave incident upon an interface at the Brewster angle gives only a trans

mitted wave.

The relative amplitudes of the incident, reflected, and transmitted waves are

given by Fresne/'s equations.

Let us find Fresnel's equations for an incident wave polarized with its E vector

parallel t o the pla ne of incidence as in Fig. 20-9. The plane of incidence is the plane

containing the incident ray and the normal to the interface.

Of course the angle of incidence is equal to the angle of reflection and, according

to SnelTs law.

iij sin 0, = sin 0,.

Figure 20-9 Th e incident, reflected, and tran smit ted waves when the incident

wave is polarized with its E vector parallel to the plane of incidence. The

arrows for E and H indicate the positive directions for the vectors at the

interface. See Prob. 2 0 - 1 1 .



Problems 491

T h e q u a n t i t y n si n 0 i s therefo re con serve d on cross ing the interface. To tal reflect ion

occurs when this equat ion gives values of s in 0, that are larger than uni ty,

a ) F ind Fresne l ' s equa t ions

by us ing the condi t ions of cont inui ty for the t angent i a l component s of E and of H a t

the interface (Sees. 7.1 and 14.10).

b) Draw curves of E 0r/E oi a n d o f - £ 0 , / £ 0 , as funct ion s of the angle of inciden ce

for a wa ve in air incid ent o n a glas s surface w ith », = 1.5.

c) Sh ow th at there is no reflected ray w hen

The angle of incidence is then cal led Brewster ' s angle . The reflected ray disappears a t

Brewster ' s an gle only if the incid ent ray is polar ized w ith i ts E vector in the p lane of

incidence as in Figure 20-9.

d) Sh ow tha t , for l ight pr op ag at i ng in glass with an index of refract ion of 1.6,the Brewster angle is 32.0 degrees .

Bi + 0, = re/2.

APPENDIX A

VECTOR DETINITIONS, IDENTITIES,AND THEOREMS

DEFINITIONS

1) v / =dx

+ — jcy

+ 7 T k

2) V - A =6A

X

dx+ ^

dy

3) V x A =(dA

\ dy

dA

' ~ 7 z

4) V 2 / =d 2

f

dx 2 + dy 2

c . 1, i M \ . , <".-l.. <"••!

c z dx / l e x f v

5) V 2 A = V2

/4,i + V 2 / 4 y j + V2

/ lzk

IDENTITIES

6) a x (b x c) = b(a • c) — c(a • b)

7) V( /g ) = / V 0 + a V f

8) V • (./A) = (V/) • A + / (V • A),

9) V • (A x B) - B • (V x A) - A • (V x B),

10) V x ( /A) = (V/) x A + ./'(V x A),

11) V x V x A = V(V • A) - V 2 A

/ 1 \ r.

12) V I - I = - j , where the g rad ien t i s ca lcu la ted a t the source po in t (x\ y', z') and r ,

is the unit vector from the source po in t (x 1, y', z') to the field point (x, y, z).



Theorems 493

13) V j J = — " 2 -, w h e r e t h e g r a d i e n t i s c a l c u l a t e d a t t h e f ield p o i n t , w i t h t h e s a m e

u n i t v e c t o r .

THEOREMS

16 ) D i v e r g e n c e t h e o r e m : f A * da = f V • Xdx,

w h e r e S i s t h e c l o s e d s u r f a c e t h a t b o u n d s t h e v o l u m e i.

1 7) S t o k e s ' s t h e o r e m : ( £ A • dl = J* s (V x A) • d a .

w h e r e C i s t h e c l o s e d c u r v e t h a t b o u n d s t h e s u r f a c e S .

APPENDIX B

SI UNITS AND THEIR SYMBOLS

Quantity Unit Symbol Dimensions

Length meter m

Mass kilogram k g

Time second s

Temperature kelvin K

Current ampere A

Frequency hertz 11/ 1 s

Force newton \ k g - m / s 2

Pressure pascal P a N / m 2

Energy joule J N - m

Power watt W .1 s

Electric charge coulomb c A -s

Potential volt V J /C

Conductance Siemens s A /V

Resistance o h m < > V /A

Capacitance farad 1 c / v

Magnetic flux weber W b V s

Magnetic induction tesla T W b / m 2

Inductance henry 11 W b / A



APPENDIX C

SI PREFIXES AND THEIR SYMBOLS

Multiple Prefix Symbol

1 0 1 8 ex a E

10 1 5

peta P

1 0 1 2 tera T

10" giga G

10 6 mega M

10 3 kilo k

10 2 hecto h

10 d e k af da

I O - 1 deci d

1 0 " 2 centi c

1 0 " 3 milli m

1 0 " 6 micro

I O " 9 nano n

I O " 1 2 pico P

1 0 " 1 5 femto f

1 0 " 1 8 at to a

* This prefix is spelled "deca" in

French.

C a u t i o n : T h e symbol for the

prefix is written next to that fo r

the unit without a do t. Fo r ex

ample. m N stands for millinew-

ton, and m - N is a meter newton,

or a joule



APPENDIX D

CONVERSION TABLE

Examples: On e meter equals 100 centimeters. One volt equals 10s

electromagnetic units of

potential.

CGS Systems

Quantity SI esu emu

Length meter 102

centimeters 102

centimeters

Mass kilogram 103

grams 103

grams

Time second 1 second 1 second

Force newton 105

dynes 10s

dynes

Pressure pascal 10 dynes/centimeter2

10 dynes/centimeter2

107

ergsnergy joule 107

ergs

10 dynes/centimeter2

107

ergs

Power watt 107

ergs/second

3 x 10"

107

ergs/second

Charge coulomb

107

ergs/second

3 x 10" 10 "1

Electric potenti al volt 1/300 108

Electric field intensity volt /met er 1/(3 x 104

) 106

Electric displacement coul omb/ mete r2

12tt x 105

4tc x 1 0 "5

Displace ment flux cou lom b 12tt x 10Q

4n x 10 "1

Electric polarization coulomb/meter2

3 x 105

1 0"5

Electric current ampere 3 x 10° 10 "1

Conductivity siemens/meter 9 x 109

1 0 " "

Resistance ohm 1/(9 x 101 1

) 109

Conductance Siemens 9 x 1 01 1

1 0 "9

Capacitance farad 9 x 1 01 1

1 0 "9

Magnetic flux weber 1/300 108

maxwells

Magnetic induction tesla 1/(3 x 106

) 104

gausses

Magnetic field intensity ampere/meter \2n x 107

47i x 10"3

oersted

Magnetomotance ampere 12ji x 10" (47t/10) gilberts

Magnetic polarization ampere/meter 1/(3 x 101 3

) 1 0"3

Inductance henry 1/(9 x 101 1

) 109

Reluctance ampere/weber 36 ti x 10" 4n x 10~9

N o T t : We have set c = 3 x 108

meters second.



APPENDIX E

PHYSICAL CONSTANTS

Constant Symbol Value

G r a v i t a t i o n a l c o n s t a n t G 6.6720 x 1 0 " 1 1 N m 2 / k g 2

A v o g r a d o ' s c o n s t a n t N A6.022045 x 1 0 2 3 m o l " 1

Proton res t mass m p1.6726485 x 1 0 " 2 7 k g

Elec t ron res t mass1 1 1 e

9.109534 x 1 0 " 3 1 kg

Elem enta ry charge e 1.6021892 x 1 0 " 1 9 C

Permi t t iv i ty of vacuum «o 8.85418782 x 10" 1 2 F / m

Permeabi l i ty of vacuum /'n 4n x 1 0 " 7 Ft/fn

Speed of l ight in va cuu m c 2.99792458 x 10 8 m /s

APPENDIX F

GREEK-ALPHABET

Letter Lowercase Uppercas e

Alpha a A

Beta $ B

Gamma y 1

Delta 5 A

Epsilon e i:

Zeta c /

E ta '/ IITheta eIota i i

Kappa K K

Lambda "/. A

M u /' M

N u V N

Xi { —

Omicron o ( )

Pi n 11

R h o P 1'

Sigma a

T a u T TUpsilon U r

Phi0 V



C h i X \

Psi <l>Omega w



INDEX

Acce le ra t ion , 9

A d m i t t a n c e , 400Alte rna t ing cu r ren t , 265 , Ch aps . 16 -18 , 3 7 2 ,

3 7 3 , 449

A l t e rn a t in g - c u rr e n t m o t o r s , 3 2 9 , 3 9 4 , 4 3 1 , 4 3 2

Alternat ing vol tage. See Alternat ing current

A m m e t e r , 115

c l ip -on , 365, 450

A m p e r e , 111

Ampere ' s c i r cu i t a l law, 212 , 2 2 1 , 34 0

A m p e r e - t u r n , 22 2

Ampl i f i e r , ope ra t iona l , 145, 152, 154

Ampl i tude of a w a v e , 47 0

Analog compute r , 145

Analyze r :

cyl indrical e lectrostat ic , 58 , 61

g a s , 141

para l l e l -p la te , 58, 61

A n e m o m e t e r , h o t - w i r e , 141

Angle , so l id , 67

Aniso t ropy : in d ie lec t r i c s , 192

in magnetic mater ia ls , 342, 358

Antenna , 43 1

h a l f - w a v e , 435

l o o p , 486

w h i p , 435

Antenna image . 87

Area :

under a c u r v e , 44 8

as a vec to r . 7, 18

A r g u m e n t , 38 3Attenua to r . See Voltage divider

B a n d w i d t h , 42 1

Bamett effect , 27 8

Bat te ry cha rge r s , 39 3

Beta t ron , 37, 275

Boat test ing tank, 27 1

Boundary cond i t ions . See Cont inu i ty cond i

t i ons ; Uniqueness theorem

B r a n c h , 120

Br idge :i m p e d a n c e , 41 4

Wh e a t s t o n e , 131, 14 1, 151

W i e n , 414

Capac i t ance , 90, 91 , 281

stray, 422

Capac i t ive r eac tance , 400

Capac i to r :

a l ternat ing current in, 379, 400

cy l ind r ica l , 101

se l f - c lamping . 196

See also Paral le l-plate capaci tor

Capac i to r s :

in paral le l . 92

in ser ies , 93

Car te s ian coord ina te sys tem, 3

Cathode- ray tube , 55 , 248, 298

Charge :

conse rva t ion of, 1 1 3 , 466

invariance of, 240

m a g n e t i c . See Mo n o p o l e , m a g n e t i c

nea r an inf ini te grounded conducting plate .

79

sheet of, 71

sphe r ica l , 71

Charge densi ty

b o u n d , 159f, 165, 171

f ree , 160, 165, 182

in a h o m o g e n e o u s c o n d u c t o r , 114on the surface of a d ie lec t r i c , measure of,

174, 416

See also G a u s s ' s law

Charge dis tr ibut ion, potent ia l energy of, 95,

186

Ci rcu i t , e l ec t r i c , 120, 361

de l t a -connec ted , 126

different ia t ing, 152

equ iva len t , 126



500 Index

Circui t , e lectr ic (continued)

in teg ra t ing , 154

magnetic force within an isolated, 321

pu lse -coun t ing , 155Q of, 421

RC , 135, 151

rectifier, 392, 393

RL , 294

RLC, 3 0 5 , 3 0 9 , 40 3

s ta r -connec ted , 126

T - , 4 2 3 , 42 7

Circu i t , magne t i c . See Magne t ic c i r cu i t

Circui t breaker , 32 3

Circular f requency, 37 3

Clamp, e lec t ros ta t i c . 197

Coaxial l ine , 171, 172, 304 , 3 0 5 , 48 7

capac i t ance of, 171

Coefficient of coup l ing , 287, 448

Coil :

field at center of, 232

H e l m h o l t z , 215, 216, 448

induc tance of, 365

moving-co i l me te r s , 32 9

R o g o w s k i , 302 , 4 4 8 , 450

s a d d l e , 23 3

to ro ida l , 224, 286

Coi l w ind ing , 42 2

Colloid thruster , 62

Compensa ted vo l t age d iv ide r , 41 2

Complemen ta ry func t ion , 134

Complex con juga te , 39 7

C o m p l e x n u m b e r s , 38 2

add i t ion , sub t rac t ion , mul t ip l i ca t ion , d iv i s ion of. 384

Complex p lane , 38 3

C o m p u t e r , a n a l o g , 145

C o n d u c t a n c e , 119, 361

Conduc t ing p la te , 74

charge near a, 79

Conduc t ing p la te s , pa i r of, 74

Conduc t ion cu r ren t , 454

Conduc t ion e lec t rons , 111

drift velocity of. 113

Conduc t iv i ty . 113 , 360

C o n d u c t o r , 73

electric field at surface of, 73

force on , 97 , 105, 186, 196f

g o o d , Ml, 454

h o l l o w , 74

h o m o g e n e o u s , 114

magnetic field of long cyl indrical , 22 3

sem i- , 11 1, 242ff

sphe r ica l , 96

See also Interface

Con juga te , 39 7

Conse rva t ion of cha rge , 113 , 466

Conservat ive f ie ld , 22, 29, 37, 45 , 47

Cons tan t of in teg ra t ion , 134

Cons tan t s , t ab le of p h y s i c a l , 49 7

Cont inu i ty cond i t ions (at an in terface):B, 2 2 7 , 34 6

D , 1 8 1 , 185

E , 182, 185

H , 34 7

V, 180

Convec t ion cu r ren t , 454

Corona d i scha rge , 198

C o u l o m b , 41

C o u l o m b f o r c e . C o u l o m b ' s law, 4 1 , 1 6 1 , 162

Counte re lec t romot ive fo rce , 44 5

Coupling, coeff ic ient of, 287. 448

Crab nebu la , 24 7

C u r l , 23. 25

of B, 2 3 0 , 455

of E, 266

of H , 4 5 6 , 46 5

Curren t , 36 0

c o n d u c t i o n , 454

convec t ion , 454

d i s p l a c e m e n t , 190, 454

e d d y , 266 , 2 7 3 , 4 3 8 , 450

e lec t ro ly t i c , 454

equ iva len t , 454

m e s h , 124

po la r i za t ion , 190, 454

t h r e e - p h a s e , 377, 387, 388

th ree -wi re s ing le -phase , 37 6

See also Alternat ing current ; Direct current

Cur ren t dens i ty , 111 , 36 0d isp lacemen t , 190 . 454

equ iva len t , 335, 336, 338

to ta l , 454

v o l u m e , in m a t t e r . 45 5

Current sheet , magnetic f ie ld c lose to, 231

Curren t source , 128

Curren t - s t ab i l i zed power supp ly , 130

Curren t t r ans fo rmer 448 , 450

Curve t r ace r , 189

Cyc lo t ron f requency , 24 6

Del ope ra to r , lOf

Del ta -connec ted c i r cu i t , 126

Delta-star t ransformation, 126, 406

Der iva t ive , t ime , 8

Diamagne t i c ma te r i a l s , 334, 342

Die lec t r i c , 157, 334

anisotropy in, 192

electr ic force on , 187

l inear and i so t rop ic , 163 , 165, 167

l iqu id , 187

n o n - h o m o g e n e o u s , 177

polar izat ion processes in, 191f



Index 501

po la r i zed , 157

sheet of, 161

See also Interface

Die lec t r i c cons tan t , 164Die lec t r i c s t r eng th , 172, 193f

Different ia l equat ion, 134, 387

solut ions of, 134

Different ia t ing circui t , 152

D i o d e , 308f

v a c u u m , 86

Dipo le :

e lec t r i c , 49

m a g n e t i c , 20 7

Dipo le moment :

e lec t r i c , 51

induced electr ic , 157

m a g n e t i c , 20 7

Direct current , 37 2

Direc t -cu r ren t moto r , 44 4

D i s c h a r g e , c o r o n a , 198

Disp lacemen t cu r ren t dens i ty , 190 . 454

Disp lacemen t cu r ren t in paral le l-plate

capac i to r . 191

Disp lacemen t t r ansduce r , 216, 352

D i v e r g e n c e , 18, 20, 21

of B, 20 7

of D , 163

of E. See G a u s s ' s law

of H , 35 3

Divergence theorem, 21

D o m a i n , m a g n e t i c , 33 4

Drift velocity of conduc t ion e lec t rons , 113

Drop le t gene ra to r , 102D W I O H T , H. B., 16

Eddy cu r ren t , 266, 273

Eddy-cur ren t lo sses , 438 . 450

Effect ive value, 373

Efficiency of power t r ansmiss ion , 434

Elas tance , 94

Elec t re t , 169

shee t , 178

Electr ic c ircui t . See Circui t , e lectr ic

E lec t r i c condu i t s , 27 8

Elec t ri c d i sp lacem en t D , 162

d ive rgence of, 163

Electric field E, 4 3 , 44, 235, 267, 361

curl of, 266

d ive rgence of. See G a u s s ' s la w

energy densi ty in, 96 , 98 , 186, 189, 193,

194f

flux of. See G a u s s ' s law

induced , 262 . See also E l e c t r o m o t a n c e

line integral of, 2 6 1 . See also E l e c t r o m o

tance

m a x i m u m , in air, 82 , 99

t r ans fo rma t ion of, 241

Electric field intensity v x B, 23 7

Electr ic force:on conduc to r , 97 , 105, 186, 196f

on d ie lec t r i c , 187

and magne t i c fo rce , 31 2

See also Coulomb fo rce

Electr ic polar izat ion P, 157, 192

Electr ic susceptibi l i ty , . , 163

Electrolyt ic current , 454

E l e c t r o m a g n e t , 36 4

Electromagnetic f ie ld:

t ransformation of, 241

wave equa t ion for, 47 6, 483

See also Maxwel l ' s equa t ions

E lec t romagne t i c f lowmete r , 25 9

Elec t romagne t i c p rospec t ion , 27 2

Elec t romagne t i c pump, 32 4

Elec t romagne t i c spec t rum, 47 3

Elec t romagne t i c wave , 473

ene rgy dens i ty in , 4 8 1 , 4 8 2 , 4 8 3 , 48 4

energy f low in , 4 8 1 , 484

phase veloci ty of, in non-conduc to r s , 483

p l a n e , in free space, 476ff

p l a n e , in n o n - c o n d u c t o r s , 483

veloci ty of, in f re e s p a c e , 2 3 9 , 2 4 0 , 4 7 5 , 4 7 6

wave equa t ion for, 4 7 6 , 48 3

Elec t romete r , 175

E l e c t r o m o t a n c e , 3 6 1 , 46 1

i n d u c e d , 262 , 2 8 3 , 467

See also Counte re lec t romot ive fo rce

E lec t ron :c h a r g e of, 53

conduc t ion , 111

drift velocity of, 113

m a s s of, 53

Elec t ron-vo l t , 56

Elec t ros ta t i c c l amp , 197

Electrostat ic cyl indrical analyzer , 58, 61

Electrostatic field. See Electrostat ics

Electrostat ic force:

on conduc to r , 97 , 105, 186, 196f

on dielectr ic , 187

Electrostat ic paral le l-plate analyzer , 58, 61

Electrostat ic seed sort ing, 57

Electrostat ic separat ion of ore , 53

Elec t ros tat i c sp ray ing , 56

Elec t ros ta t i c s , Chaps . 2-4 and 6- 7, 45f, 73 ,

182 , 183 , 230 , 26 1

Energy.Se e Magnetic energy; Potent ia l : Poten

tial energy

Energy , so la r , 486

Energy convers ion , the rma l to e lec t r i ca l , 200

Energy dens i ty :

in electric field, 96, 98, 186, 189, 193 , 194f



502 Index

Energy densi ty (continued)

in e l e c tr o m a g n e t ic w a v e , 4 8 1 . 4 8 2 , 4 8 3 , 4 8 4

in gas, 98

in magnetic field, 31 7Energy flow in e l e c tr o m a g n e ti c w a v e , 4 8 1 , 4 8 4

Energy s to rage , 193, 328, 329

Energy s tored in magnetic field. See Magne t ic

ene rgy

Equipotent ia l surface, 48, 49

Eule r ' s fo rmula , 38 3

Farad , 90

FA R A D A Y , M., 259

Faraday induct ion law, 263, 266, 334

Fer r i t e s , 438

Fer romagne t i c ma te r i a l s , 334 , 342, 343 , 358

Fie ld . 2

central force, 36, 45

conse rva t ive vec to r , 22, 29, 37. 45 . 47

sca la r , 3

vec to r , 3

See also Electric field E; Elec t romagne t i c

field; Electrostatics; Magnetic field

Field mil l , 41 6

Filter.

low-pass RC. 41 5

n o t c h , 423

RC . 412

Flammable -gas de tec to r , 142

Floa t ing wi re , 31 5

Flow of ene rgy . See Poynting vector

F l o w m e t e r , 25 9

Fluorescen t l amp , 445F l u x , 18, 20 , 21

of E. See G a u s s ' s law

l e a k a g e , 359, 361

m a g n e t i c , 204, 212, 2 6 3 , 36 0

Flux compress ion , 32 7

Fluxga te magne tomete r , 35 5

Flux l inkage, 26 3

Focus ing :

g a s , 253

m a g n e t i c , 24 8

Force :

cen t ra l , 36. 45

coe rc ive , 34 4

C o u l o m b , 4 1, 161, 162

coun te re lec t romot ive , 445

grav i t a t iona l , 37, 42f

Loren tz , 236, 2 4 1 , 24 2

See also Electr ic force; Lines of force;

Magne t i c fo rce ; Torque

Fourier ser ies , 37 3

F r e q u e n c y , 470

c i rcu la r , 37 3

cyc lo t ron , 24 6

Fresne l ' s equa t ions , 49 0

Gal i l ean t r ans fo rma t ion , 23 8

G a m m a r a y s , 4 7 3 , 474

Gas ana lyze r , 141

Gas focus ing , 25 3Gas p ressu re , 98

G a u g e s , 141 f

G a u s s ' s law, 69, 16 1, 1 6 3 , 2 1 1 , 2 2 3

Genera t ing vo l tme te r , 416

Genera to r :

d rop le t , 102

e lec t r i c . 26 4

h igh-vo l t age , 107

h o m o p o l a r , 324

m a g n e t o h y d r o d y n a m i c , 2 3 3 , 25 6

r a m p , 151

V an de Graaff, 2 3 1 , 416

See also Hal l gene ra to r

G r a d i e n t , 10

Grav i ta t iona l fo rce , 37, 42f

Greek a lphabe t , 498

Gun:

pa r t i c l e , 55

p l a s m a , 33 1

r a i l , 322, 331

Hall effect, 242 , 254, 259

Hall field, 24 3

Hall generator (Hall probe) , 216, 255, 350,

3 5 2 , 3 6 6 , 419

Hea t ing , induc t ion , 27 3

Hea t waves , 473

Helmhol tz co i l s , 215, 216, 448

H e n r y , 2 8 1 , 2 8 3 , 285

H e r t z , 37 3

High voltage. See Genera to r ; T ransmiss ion

l i ne ; Van de Graaff generator

H o d o s c o p e , 31 4

H o l e , 111

Homopola r gene ra to r , 324

H o m o p o l a r m o t o r , 324, 325

Hys te res i s , 344

lo s ses , 346, 450

I m a g e , 79

a n t e n n a , 87

Imag ina ry numbers , 38 2

I m p e d a n c e , 39 9

inpu t , 436, 441

m a t c h e d , 4 3 3 , 443

ref lected, 436, 447, 448

Impedance b r idge , 414

I m p e d a n c e m a t c h i n g , 4 3 3 , 443

Incandescen t l amp . 116

Induced e lec t r i c d ipo le moment , 157

Induced electric field intensity, 26 2

Induced e lec t romotance , 262 , 2 8 3 , 467



Index 503

Induc tance , 284, 319

m u t u a l , 28 0

self-, 284

t r ans fo rma t ion of mutua l , in to s t a r , 408

Induct ion:

Fa raday law, 263 , 266, 334

See also Magne t ic induc t ion B

Induct ion furnace, 27 4

Induct ion heat ing, 27 3

Induc t ive r eac tance , 400

Induc to r , 284 , 295 , 412

al ternat ing current in, 379, 399

m u t u a l , 28 2

t ransient suppressor for, 308, 420

Induc to r s :

in paral le l , 29 1

in ser ies . 28 8

Ink-je t pr inter , 108

Integral :d o u b l e . 13

l i ne , 22

su r face , 12

t r ip le , 16

v o l u m e , 16

See also Line integral

Integrat ing circui t . 154

Interface, 180, 346

d ie lec t r i c -conduc to r , 165, 182, 183

d ie lec t r i c -d ie lec t r i c , 171 , 182, 183, 185

Inva r iance :

of e lectr ic charge. 24 0

of veloci ty of light in v a c u u m , 24 0

Ion beam, 75

pro ton beam, 83

Ion -beam d ive rgence , 25 2

Ion thruster , 61, 253

I ron , 36 6

p o w d e r e d , 438

t r ans fo rmer , 34 6

See also Fer romagne t i c ma te r i a l s

Joule effect , 116

Kirchof f ' s l aws , 120, 124, 402

Labora to ry re fe rence f r ame , 23 8

L a m b d a bar, 471Lamina t ions , 438 , 450

L a m p :

fluorescent, 445

incandescen t , 116

Lap lace ope ra to r , 29

L a p l a c e ' s e q u a t i o n , 78 , 167

Lap lac ian , 29

Leakage flux, 359, 361

L e n z ' s law, 265

Ligh t :

inva r iance of, 240

veloci ty of, 239, 475

Ligh t waves , 4 7 3 , 474, 475

Line faul t locat ion, 148

Line integral , 22

of A, 21 2

of B, 221

of E , 261. See also Elec t romotance

of H. See Ma g n e t o m o t a n c e

Lines of B, 204 , 216 , 313 , 316f

refraction of, 227, 348

Lines of fo rce , 49

bend ing of electr ic , 183. 185

e lec t r i c , 49, 98f

Lines of H , refraction of, 348

Loop , c i r cu la r , 205

Loop an tenna , 48 6

Loren tz fo rce , 236 , 2 4 1 , 24 2Loren tz t r ans fo rma t ion , 23 9

Losses :

eddy-cur ren t , 438, 450

hys te res i s , 346, 450

J o u l e , 116

Magne t :

ba r , 338

e lec t ro - , 364

p e r m a n e n t , 169, 358

Magne t ic cha rge . See Monopo le , magne t i c

Magne t i c c i r cu i t , 35 8

with ai r gap, 361

energy s tored in, 365

Ma g n e t i c d o m a i n , 33 4

Magne t ic ene rgy :

dens i ty , 31 7

stored in isolated circui t , 32 8

stored in long so leno id , 318, 319

stored in magne t i c c i r cu i t , 36 5

stored in magnetic field, 212, 318

Magnetic f ie ld , 23 5

Ear th ' s , 217 , 246f, 34 9

energy s tored in, 212, 318

of long cyl indrical conductor , 22 3

originat ing in magne t i zed ma te r i a l , 33 8

ro ta t ing , 394

Magnetic f ie ld intensi ty H, 339, 361curl of, 4 5 6 , 46 5

d ive rgence of, 353

line integral of. See Ma g n e t o m o t a n c e

l ines of, 348

Magnetic f luids , 37 0

Magnetic f lux, 204, 212, 263, 360

Magne t ic focus ing , 24 8

Magne t ic fo rce , 212, 312

exe r ted by an e lec t romagne t , 36 4



504 Index

Magnetic force (continued)

within an isolated circuit, 32 1

between two two electric currents, 320

on a wire, 312See also Lorentz force

Magnetic induction B, 202, 360

curl of, 230, 455

divergence of, 207

flux of. See Magnetic flux

line integral of, 221

saturation, 34 3

See also Lines of B

Magnetic materials, 334, 342, 358

anisotropy in, 342, 358

isotropic an d linear, 342

magnetic field originating in, 338

Magnetic mirrors, 24 6

Magnetic pressure, 316, 325, 326

Magnetization M, 335, 350

Magnetization curve, 342

See also Hysteresis

Magnetized body, 33 4

Magnetohydrodynamic generator, 233, 256

Magnetometer, 355

Magnetomotance, 340, 360, 361, 462

Magnetoresistance multiplier, 36 7

Magnetoresistor, 255, 366 , 367

Mass, 246

effective, 25 4

rest, 246

Mass spectrometer, 25 0

crossed-field, 236f

Dempster, 249

M A X W E L L , J . C , 278

Maxwell's equations, 162, 208 , 230, 267 ,

Chap. 19, 456 , 458 , 465, 466

in integral form, 459

Mesh, 120

Mesh current, 124

Meter:

moving-coil, 329

See also Ammeter; Electrometer; Flow

meter; Gauges; Magnetometer;

Mass spectrometer; Potentiometer;

Thermometer; Voltmeter: Wattmeter

Micrometeorite detector, 35 1

Millikan oil-drop experiment, 24 0

Mobility, 254Modulus, 38 3

Monopole, magnetic, 208 , 216 , 301, 466

M O O R E , A. D., 312

Motor (electric), 31 2

alternating-current, 329, 394, 431 , 432

compound, 445

direct-current, 444

homopolar, 324, 325

series, 444

shunt, 445

Moving-coil meter, 329

Mylar, 194, 197

/7 -Type material, 243

Nitrobenzine, 192

Node, 120

Non-linear component in a circuit, 38 9

Non-linear resistor, 116

Notch filter, 423

Observer, 237

Ohm. 114

Ohm's law , 114

Output resistance, 130

p-Type material, 243

Parallel, connection in . See Capacitors;

Inductors; Resistors

Parallel-plate analyzer, 58, 61

Parallel-plate capacitor, 91,99,105,106, 174,

194, 200

dielectric-insulated, 166 , 17 8, 191

displacement current in, 191

energy stored in, 96

in infinite medium. 464

in liquid dielectric, 187

magnetic field in circular, 456

Paramagnetic material, 334, 342

Particle gun , 55

Peak current, 373Peaking strip, 355

Peak voltage, 37 3

Period, 470

Permeability, 341, 360

of free space, 202, 231 . 477

relative, 341, 342, 366

Permeance, 360, 361

Permeances in parallel, 362

Permittivity, 16 4

of free space, 42, 477

frequency-dependence of, 164, 19 If

relative, 163

temperature-dependence of, 192

Perpetual motion machine, 55 , 196

Phase, 373, 470

Phase angle of an impedance, 400

Phase-sensitive voltmeter, 413

Phase shifter, 415

Phase velocity, 47 1

Phasors, 38 5

addition an d subtraction, 38 8

diagrams, 389

when not to use, 389



Index 505

Phosphate , separat ion from quartz , 53

P h o t o n , 474

Physical constants , table , 497

PIERCE, J .R., 469

Pinch effect , 25 3

Pi ran i vacuum gauge , 141

Planck ' s cons tan t , 208 , 474

P l a s m a , 246 , 250 , 251

Plasma gun. 331

Poisson ' s equa t ion , 75, 7 8, 167

Pola r i za t ion , e l ec t r i c , P, 157, 192

Polarizat ion current , 190, 454

Polarizat ion processes in d ie lec t r i c s , 19If

Polarized dielectr ic , 157

Polarized wave:

c i rcu la r ly , 487

p lane - , 478

Poles:

m a g n e t i c , 33 9Nor th an d S o u t h , 21 8

See also Mo n o p o l e , m a g n e t i c

Potent ia l :

e lectr ic (scalar) , V, 4 5 , 2 1 1 , 26 7

line integral of vec to r , 212

vec to r , A, 209, 2 1 1 , 267, 301

Potent ia l divider . See Voltage divider

Potent ia l energy (of a charge dis tr ibu

t i on) , 95 , 186

Poten t iomete r , 140

Powdered i ron , 43 8

P o w e r , 39 7

dissipated in a r e s i s to r , 116, 390

wat tme te r , 419

Power d i s s ipa t ion , 42 9

Power factor , 397, 431

cor rec t ion , 445

Power supp ly , 130 , 415

Power t ransfer :

from source to load , 432

th rough t r ans fo rmer , 44 2

Power t r ansmiss ion :

efficiency of, 434

See also Transmiss ion l ine

Poyn t ing vec to r , 481 , 484, 487

Pressure:

g a s , 98

m a g n e t i c , 316, 325 , 326

Pressu re gauge , 142Printed-circui t board, 176

Printer , ink-je t , 108

Product of two vec to r s :

scalar , or dot, 4

vec to r , or c r o s s , 6

Prospec t ion , e l ec t romagne t i c , 27 2

Pro ton beam, 8 3 . See also Ion beam

Pro ton bomb, 104

Pulse -coun t ing c i r cu i t , 155

P u m p , e l e c t r o m a g n e ti c , 32 4

Q of ac i rcu i t ,

42 1

Q u a d r a t u r e , 38 0

Q u a n t u m of r ad ia t ion , 474

Radian l eng th , 47 1

Radia t ion :

quan tum of, 474

Van Allen bel ts , 246

See also Cathode- ray tube ; Gamma rays ;

X- rays

Rad ia t ion re s i s t ance , 43 1

R a d i o w a v e s , 47 3

Rail gun, 322 , 331

R a y s . See Cathode- ray tube ; Gamma rays ;

X-rays

Reac tance , 400capac i t ive , induc t ive , 40 0

Rea l numbers , 38 2

Rectif ier c ircui ts , 392, 393

Refe rence f r ame , 23 8

l abora to ry , 238

Reflect ion, 490

Ref rac t ion , 490

index of, 483

of lines of B, 227, 348

of lines of D and E , 183, 185

Rela t iv i ty , 237, 238, 239, 241

Relay , 364, 369

Reluc tance , 360, 361

Reluc tances in se r i e s , 36 2

R e m a n e n c e , 34 4

Remote - read ing mercury the rmomete r ,

418

Res i s t ance , 114, 122, 3 6 1 , 400

ou tpu t , 130

radiat ion, 43 1

source , 130

Resistive film, 138

Resist ive net, un i fo rm, 148

Resistojet , 138

Res i s to r , 115, 122

l inear , 115

non- l inea r , 116

o h m i c , 115

power dissipated in, 116, 390vo l tage -dependen t , 116

See also Magne to res i s to r

Resis tors:

in paral le l , 118

in series, 118

R e s o n a n c e , 405, 421

ser ies , 40 5

Reten t iv i ty , 34 4



506 Index

Right-hand screw rule . 3, 6, 25

RMS v a l u e , 37 3

Rogowsk i co i l , 302 , 4 4 8 , 450

Root mean square,37 5

Rowland r ing , 342, 353

Rutherford experiment , 56

Sadd le co i l s , 23 3

Satel l i te , a l t i tude control lor, 331

Satel l i te thruster . See Thrus te r

Saturat ion magnetic induct ion, 34 3

Scalar potent ia l . See Potent ia l

Seed sort ing. 57

Semiconduc to r , 1 11 , 242f f

Se r ie s , connec t ion in . See Capac i to r s ;

Inductors; Resis tors

Series LRC c i rcu i t , 305 , 309, 403

Series motor , 444

Ser ie s r e sonance , 405S e r v o m e c h a n i s m , 140

Surface tension, 103

Surface as a vec to r , 7, 18

Susceptibi l i ty:

e lec t r i c , 163m a g n e t i c , 3 4 1 , 342

Sys teme in te rna t iona l d ' un i t e s . See SI sys

te m of units

T-c i rcu i t s , 423 , 427

Tef lon , 172

Telev i s ion tube , 248, 298

T e s l a , 20 2

T h e r m i s t o r , 141

T h e r m o m e t e r , 141

r emote - read ing mercury , 41 8

T h e v e n i n ' s t h e o r e m , 129, 432

Three -phase cu r ren t , 37 7

Three -phase supp ly , 387, 388

Electro Magnet

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