Solutions Manual for Shen and Kong's Applied Electromagnetism

Solutions Manual for

AppliedElectromagnetism

SECOND EDITION

Shen

~ Huang

,

solutions Manual

for

Shen and Kong's

APPLIED ELECTROMAGNETISM

Second Edition

by

Liang C. Shen and Frank S. C. Huang

f]~HePWS-KENT Publishing CompanyBoston

.'

PWS-KENTPublishing Company

20Puk Pl~uBolton. ~tusa,husms 02116

Copyright <01987 by PWS Publishers.

All rights reserved. No part of this book may be reproduced,stored in a retrieval system, or transcribed, in any form or byany means - electronic, mechanical, photocopying, recording orotherwise - without the prior written permission of PWS-KENTPublishing Company.

PWS-KENT Publishing Company is a division of Wadsworth, Inc.

ISBN 0-534-07621-1

Printed in the United States of America91 -- 10 9 8 7 6 5 4

CONTENTS"

Chapter 1 Complex Vectors 1Chapter 2 Maxwell's Equations 5Chapter 3 Uniform Plane Waves 8Chapter 4 Reflection and Transmission

of Waves 12Chapter 5 Waveguides and Resonators 16Chapter 6 Transmission Lines 20Chapter 7 Antennas 25Chapter 8 Topics in Waves 30Chapter 9 Electrostatic Fields 32Chapter 10 Electric Force and Energy 36Chapter 11 Solution Techniques 40Chapter 12 Direct Currents 44Chapter 13 Magnetostatic Fields 49Chapter 14 Magnetic Materials and

Magnetic Circuits 54Chapter 15 Electroquasistatic Fields 56Chapter 16 Magnetoquasistatic Fields 58

lectors

~RandTheir

wI

(1.19)

to half"B. B*.ter the

(1.20)

-domalnJoJe have-z2. and

Problems

Notice that A . B* ~ -2; and that (AIL)" B(I) -11/2)RelA . B* 1-O.We shouldalso note that A" B - 0 and that A(t) . BIt) - O. Thus, the two time-domain vectorsare always perpendicular to each other.

Example 1.20

Consider the two vectors A ~ i+ if and B - it ;y. (They are actually the samevector.) We find that A x 8 - 0 and that A " B - O. Are the two vectors parallel toeach other or perpendicular Loeach other?

SoJution: Consider instead A x B* and A . B*. We find that A x 8* - -z2; and IhaLA . B* - 2.Thus. (A(t) x B(t)} - 0, and that (A(t) • B(t)) = 1. Furthermore, Alt) x B(t) - O. andA(t) • Bft] = 1. Thus, the time-domain vector is parallel to itself all the lime.

Problems

1.1 Let 0- 8 + ;2 and h - -3 + j. Calculate (a) a t b, (b)a - h, (c) au, and (d) a/b. Givethe answer in real and imaginary parts,

1.2 Repeat [c] and (d) in Problem 1.1with the answer given in phasor form.

1.3 Find the real part, the imaginary part. and the magnitude of a''''. where wand tarereal numbers.

-1.4 Let c be a complex number, Are the following statements always trus?

(a) (c I c*) is real.(b) [c - c*)isima)(intll'Y.(e) c/c" has a magnitude equal to 1.

1.5 Consider the equation z" - 1 T ;. Find two values of 7. that satisfy this equation.

1.6 Let 0 be a real number, and lellul« 1. Show that the square root of (1 + jo) isapproximately equal to J. (1 + jo/2).

1.7 Let a be a positive real number. and let a» 1. Show that the square root of (1 + jal isapproximately equal to ;t (1 I ;)(0/2)112.

~.8 Obtain the phasor notation of the following time-harmonic functions (if possible]:

(a) VIt)-6coslwt+ 71'/41(b) f(tj- -8sinfwt)(e) AII)=3sin(wt) 2 cos 1",1)(d) c.:p)=ncos(120J!'I-lf/2)(e) Oft) ~ 1 - cos (WI)(f) lI(t) - sin Iwl + '71'/3) sin fwt + J!'/6)

1.9 Obtain e(n in terms of w from the following phasors: [u] C - 1 + j. [b] C - 4 exp ] ;0.0),and [c] C - 3 exp (j1r/2) + 4 exp r jO.Il).

.1.10 Show that, if V - r + jx and U - g I jy, then V(tll1lt) * Re{VU e''''}. Find thecorrect expression for V(i)lf(t) in tarrns of r, x, g, y, and wt.

17

18 1 Complex Vectors

1.11 L~tA - -fl)1 + ~y - 2 and R - 2x - 49' 1 32. Find (a) A T B,lb)A n, lclA· B, and(d) A x B.

-1.12 Find the angle between A and B that are given in Problem 1.11.

1.13 Show that for

V(tI- Vo cos (wt 1- "'1- RelV ejw'l

iJ V(I) _ jwVat

1.14 Find a vector C that is perpenriicula1'lo A - 0)1+ 9y - 2, has no 2 component. andhas a magnitude equal to 1.

1.15 Flnd the vector C that is parollel to A - 5x - 8y 1 22 and hal; a magnitude equalto 1.

-1.16 Find a unit vector il that points in the some direction as on arrow drawn frompoint A to point B where the rectangular coordinates of A and B are (1.0,2) and(-1,3, - 2). respectively.

1.17 Show that the definition of the dot product V· V given by (1.10a) is equivalent tothat given by (1.10b). To simplify the algebra, you may choose the coordinates sothat the x axis is along V and the z axis is perpendicular to both V and V. In otherwords, let V = oX and V =M + cy .

.;1.18 Prove n.nb) using the approach suggested in the text.

1.19 Show that the definition of the cross product V x U given by (1.12) is equivaJenttuthat given hy (1.14). To simplify the algebra, do what is suggested in Problem 1.17.

-1.20 Figure Pl.20 shows that a vector V is along the x axis and a vector U is on the x·yplane forming a 1:150 angle with the x axis. The magnitudes of these vectors are vand u, respectively. Use (1. J 4) to express V x U in terms v and u. Now, we can alsosay that the angle between V and U is 225°. Use (1.14) and explain why the vectorV x U is the same whether or not we choose the angle between them to he 1350 01'225'. Iliot: Follow the right-hand rule. In determining tho direction vector a, yourfingers must always point from V to V in the direction you measure the angle be-tween them.

1.21 Prove (1.15) using the approach suggested in the text.

_1.22 Find the phasor notations of the following time-harmonic vectors:(a) V(t) a 3 cos (wt)x + 4 sin (wt)y + t cos (wt I 'lI'/2)(b)E(t) = [3 cos (wI) I 4 sin (wt)JX + 8(cos (wt) - sin (wL)jz[c]H(I) = 0.5 cos (k:>: - wt}x

1.23 Find tho phasor notation of the following vector:

C(z.t) : (aUH) E(z. t)

where E is given in Problem 1.22(b).

-1.24 From the following complex vectors, find CIt) in terms of wI: (0) C - )1 - ;9, (b) C -j(i - jY). and (c) C - axp f - jkz)x I i exp (jkz)Y.

Problems 19

1.25 LetA-~+W+(1 + i2)2,andletB=-~-(1 j2)y I jz.Find(o)A+B.[b)A-B.(c)A . B, ami (d) A x B.

z

u

y

Flgur. Pl.20

1.26 Find A . A • am} Re(A x B"j for the values of A and B given in Prohlem 1.25.

1.27 Sketch the trace of the tip of the vector A(t). where (0) A - x - jyand where (b) A-4x + j3y.

1.28 Calculate A . B. given A - x + j2y and B - 2x I jY. AreA(t) and B(t) perpendicular011111 times?

!.:!. (4.) ~ ... b a S''''J3 OJ ~-.k c II+j (C) g,'.I!..-J~4ojz. (i) ,g/b. =-2.Z-jl.4

hJ. .!= ,.,j'2.. I. IS' u,4,of" J b a -3-#-;' =3.1'1161.S7~B:..~. 2~.()8LI7S,'I', Soil!. c ~.tl L-I4-7.~J·

!J. He I ej4IJtJ • (DSlNtoJ I,." [ejwt'J • SI~W't ,I e,iutJ =1f1:. (4) ye s (b> yes (c) yesh!.. Z 'l:. 1-+j • .;r e i (27l'fr+ ~) ~ Z I: Z ~ e=:TPiI) I "11. aD/ I

.; Z, s 2t+ eFt?, I Zc. 1~ ei(lr<# 1t'/.) • -2~ eJ r.!:.!.. (l+jA)~ = ±['+i()'a.)+-ja.1+···1 ~:t I I+j(%>} I t'f /4./«/

/.;1 a,»/, (I+J(J,.)~ ~ (Ja.)J( ::;±[J(a.)'I,. = :t(J+j)(%)~

J.:.! (o.) ~ = 6 e .;~ (b) I· j 8 (c.) .d • -j 3'" 2 Cd) £. = -j" (e) impC5~i.6/e. (f; (hll'0lSZ61e.

1:1 (a.) CC-e)- Re{(J-tj)ejwtJ eRe IVI'ej1n.ejio>t1 c yz US(wt~~)

(b) CC"t)¥/i't.{4.eJ'Q··ejwt} = .tf.~l(W~+O.e)

~C.) C Cot) I: /?t [3 eJ ~ eJwt + .je j~.8e ilVt 1 c :3 ~$( I4Jt -# ~) + 4 C_lJ$(t..Je .~. S)

!:!! s= r~Jx q V(t)= rcos~ -~ s,....we; u » 8+-j y ~ Utt:)a g.'~Awe - YS;",wtR~f 'iY. ejwt 1 :./?e f (erg -)(i'.) +j (lx+rY)J e j..,t 1_ (r; -XYJco.swt -( gX-I'r I) s,,,wtBut Vt-t) oco « ra '05~u>t + xy Si,,'UJe - 'gx + r y) S/~ kit ceswt :p Re [ ':i !l e jl4lt J

- - ;< #. A X lIr ~ A 4'" r-!J.! (~) A<t8 =-6X+SY-l2z (b) "-0 :::.-/ox·nSY- Z ct.) ,-..'8=-I"-36-3"-S5<tI) A1C8= 23X+229~/4Z

1./1 IA I '" 1- a ~ + q y - i I : JU6 , 161:: I2:- 4- 9., 3 i I : .(iiA'·a -S"S'CcSJ.: -,:: = -".84-~JAil 8 ~ 'AfiiI~ d '~ , j~!:.!1 ftWt): -rt[Voco!'(W~"'~)J = -Vow S/~(Wt.,.¢) I:: Rt. [j 'lowe J e Jl.4Jt J _; V.we

t3~ Vl-t):: V.C()5(wt +;) = Rc f~ ej~ejl.Jt J c fit! ~ejwt J ~ .!C It: Voe iiJ:. reVtt)- jwv.tj~;:jW~

" - '" " ..../,14 L,t C: 'X;"'IJY C.J.A a» C·A.,c, wAert. A=-8lf"'QY-l!

C'A--8x-+q~:;'lJ :. x=';~ Aho IA/::.I ~ ~z""\"1 =? (fJ.: ji JOt '1:r:.:t.:Js... ;K::t~- tv\4:/. C=:tr.b(9;~BY)

r/4S r 14S - .. ,.. "J.I~ Let c = ~ ; 'I 1Y T , ;. , /1A ~ A x c;:0 I wAt.re, A • 5 ~ - 8 r -I' 2 a

A'I C = ( .. e ~-1d) ; + (~X - S) ) Y +' ( S 'J .., s x) i G()

,', -21-8JDo L2X-S~50 J * ~.{~~ ~#-4)S'l~8"·o

8""t IC:I:./ ~ ')(Z.,.~2+ '}l;:/ ~ ,'1(~',4:+ ,,+);/- I ,.. " ....:. c.:: ~ fti C 5)( - 8 r +2 i)

"

. ",.~2.f ~ _V'ij

1

_ ~ A ~AB = {-/-I)x+f)-o)J + ("1.-2,)J

~ ~ "'.: -2X +JJ - 4J

..n =(-2X'" 3:J~- 4-J i/ 14+f+lb I " " ")= - (-2 J( +3 J - 4J .69y

L!..1 0. 100...) ,,'vlt V·u = It bfl. loA) J,'{tJ ii·V = A.. Iba1:~ C.oS 8

~wt' (..OJ 8-:; J, / f,'~t, 1.~ V'U = A.J ~~C\. tA1J9 ::: 4.1,

-IJ

- '" ~ "..C. = C, ~ -+ la. ~ + C) ~

A. (a t c ) ~A-. [( 13, +c,) : + ( ill,. + (', ) 1" + ( 13J+ CJ );" J

== A I B, + c,A, + Az Bt tAl CL + AJ 6) + ,A) CJ

... --- -- _.A· B + A·c -= A, 13,+ Azl3l. .,.,4j 8) + Aiel + Az r~+ AJ CJ

Thtr~f~""~ Ii· (6t C. ) :: A· iJ+ A· C-

/. I~ (/.{1.) )II)~J Vk' U .: a.c J'0. I Y. ) j,'VlS V X U ::: 4t.' 1>1.+(.- s....t) rbj, ,s,'h f) = C IJb'.,.,'"

/.20-

2

.'x

/.n..-

/. ,,}-

- t\ " 1 - '" A 8 1\ Alst A = II, x tAl J ..,. AJ J I 8:: 6, x + B.. 'j'" '.J J J A-

- " A ~ 1'\ .J.,'C. = C', X + (1 , of- -J J I T A~ ~

B~c=;( 8~CJ -C&8))+9(B]C, -C)8,)1'J(B,(~- c,BlJ

AX(B~c) = ~ [ At (8,c2. -r,B~)-A) (BJe/-CjB,)]

+ 9 [ AJ (B) (J - Cz. B.,)-A, { (3, C.,. - ~ c,J]

...; [ A I (8J C, - ~ (,) -A~ ( e.~ .;8JCl.)]O)t +k, ();kf r ~Ih''( ..

7f (A' [ ) ::.(fj, ; .,.B~ 1of' ~ S ) ( A .c, + A1,. C'z. .,. AJ cJ )r (A, B ) = (c, ~ -#- i. q~ ("I ; ) ( A I 8, + A.. 131. + III 6J )

.'. 73 (A' E ) - c f,A'"8 )

.'

Cd.) 2:: 3;-j4; ...j;(') g .. (3-J4);'" ~(I"'jJ a(c) E" 0.5' e -JAJ-;

3

1.:1:1 (~) C(t). ~'o)wi; .,.; S/~wt

,II) c (ot) c-'~ ~'',..I.J-t .. Y cc~",t((.) C ce)» 9 CbS (' eve -Ie: )) - y 5;,._(lAIt + 1l.J.)

_ _ .,. • A

!.:.Y. (0..) ~ + ~ I: (./ +J 3) Y -I cHJ 3) l_ _ "... .". ,A

(0) ~. § = 2 ~...(1-J) Y + (II J) i

(c) A o'B :. -1-(j+2)+(i-2).-S'

(J) [x e ::",.;-(J+j))Y-(/.j3?il.z6 ~.A·: [;-1 jy~(ltj2)2)( ~-jY+(':J2)iJ = I+/#-S':: 7

A x§ ..:::(;.,.j y + (l+j2)iJ x (.X-Otj2 )9- j ~] :: (-2--j+) X - (.'T jJ 9 -(I"'j)!J:. Rt [ dlt~'1= -2; - ; - i

/.27 ( .. ) g_:o: X -J 9 ~ Att) D Ccjwt f.,. ~;".wt ; AIVI... IAtt) I:: I

'Yl A{t)wt·o

1r I"~_..:+--~~~--~ ~

,~(6) A =49.j3; _A(e)c4u>Sw'i. £-3St~~Y

y,,~

- ,., ~ - #II.'" _ ...Ac )l~J2YJ §'=2.X"JY "A·g#.2-2=O

,. ,,-- "., ..,£"f:; ....t AU);: coSW/; ac - '2 S,A.wt y ta,nd., {3(*) r:r 2~.swt x - S,"I.""'" Y:. Atot) AlII/.. 8(-t) Ar~ ""(: p.rp."dicuJ-A.,.. IJ.t 4VlY -i,'me..

4-~-----.------ --

18

dte)-

r-

t-nd3.

n

)-

•C

at~

2.5 35Poynting's Theorem

Solution: E(t) - RclxEo e-th c""/- iEo cos Iwl - k7.)

HI II - Rely ~ Eu p,-Ikl e""/ - :Y ~ Eo cos (wI - hz)WJ,l WJ.l.

? kSIt) - E x H = z - Eg cos' (wI - kz]

WJJ.

1 I I k,(5) - - Re E x H" - ~ - Eo2 2wJ.l.iE~ 2

Ut• - TCOS (wI - kz]

k~E'Ufl - ~ cos·lwl. kz)

2w'J,I

(U)_fE~f. 4 0

Jc2(till) - -2- E~

4w 101

Problems2.1 Let A - 5R+ tiyzy + X3~; finn \I x A and 'ii . A.

-2.2 Let I/> ~ xyz: find \It/> and v . \11/>.

2.3 Let a - O,X I 0zY + 03Z and b - b,x + b2y + bJz. Show that equations (2.9). (2.11a}.and (2.121 Are true .

-2.4 Show that \I x (0 + b} - \I x 0+ 'il x band 'il . (0 I h) - 'ii . d + 'ii . b.

2.5 Show that \1(<1>, "'2)- <1>, \I tf>z+ <1>2\I </>\ and that 'il . ( A) == A . 'il\I • A.,.2.8 Show that \I x (AI- \I</> x A..,. </>\1 x A.

2.7 In a source-free region. H - zy + yz. Does D vary with time?

-2.8 What is the charge density in a region where D - ad?2.9 Find the magnetic field B(y.L) associated with thE!electric field E(y,t) gtven as fol-

lows:

E(Y,L) = x 0.3 cos(wt + kyJ

where wand k are constants.

-~.10 Express k in terms of the magnetic permeability and dieieotric permittivity of themedium when the electromagnetic fields arc given in Problem 2.9 in a source-freeregion.

2.11 Let £" B,. H" lind D, satisfy equations (2.1j-(2.4j with given II and Pvl' Let also E2, B~.Hz. and D2 satisfy equations (2.1)-(2.4} with given 12 lind Pv2' What are the electromag-netic fields due to a current /, and charge Pi;,where I, - 11+ 12 and Pw - (>,.\ + Pv2? Youmust show that your proposed solution satisfies Maxwell's equations. What is theappropriate name for the theorem you have just proved?

_2.12 (a) It is known that the vector a is equal to zero at one point. Does that imply that~ x a ~ 0 at that point? Give a counter-example if your answer is no.

(b) Does E = 0 on a line always imply ~ x E - 0 on that line? Give acounter-example if the answer is no.

(c) It is found that the E field is zero on a surface. Does it foJlow that aBI at - 0 onthat surface?

2.13 Show that equations (2.22c) and (2.22d) can be derived from equations (2.22aJ, (2.22bJ,and the conservation equation (2.23).

~2.14 To represent time-harmonic fields, most physics books usc the factor e- ...• instead ofe"", which most electrical engineering books use. For a time-harmonic real functionA(x, y, z, IJ - alx, y, 7.) cos [wI + .pI, find the phasor notation that corresponds to thephysicists' convention. What is the corresponding conversion rule hy which phasorscan be transformed back to the real-time expression?

2.15 Refer to Problem 2.14 about the notation a-I"'I adopted in most physics books.Write the time-harmonic Maxwell's equations using that notation.

2.t8 Whal is the range of effective perrntttivtry of the Ionosphere at AM bcoadcastlngfrequencies'? Use the following data: N - lO,a m-a and f - 500 kHz to 1 MHz.

2.17 Show that the dimension of each term of squatton (2,361 is watts pet' cubic meter.

'2.18 Indicate in watts, meters. and joules the dimansions of the following quantities:(0) E ' D. [b] H . B, and [c] S.

2.19 Let E _ (i + jy) e j2 and H - Iy - ji) R-". Find S in terms uf z and wt and find <S>.

2.20 Show thal S * Re {Ex H ei"'}.2.21 Show that S * Re {E e"" x H d"'}.2.22 Compare the energy stored in a cubic; region one meler on a side which has a

uniform E field of 10' V1m lo the energy stored in a similar region with a uniform Bfield of 10' G. (One C = 10 4 Wb/mz]. The medium is air.

2.23 Repeat Problem 2.22 for the case where the medium is water instead of air, UseF : 80 fO and JJ. : JJ.o for water.

36 2 Maxwell's Equalions

"

VLF

5

V)(e, =-;tB" -Vtli1,cJ,+ /to,) '11,5,.0" v'D;=/!,f:/x Ez "'_ ~ 81. , v)(!lz"'.Jz + liD,., v,S;. D I V·~. 111'1

C:;" E; • '1'1( ( ~ .,.Ez ) = V J(i,.,.V j( ~ ::; - Ii it - It 5" • - Ie (6, ~~ ) • -k~V)(.lii • V 'I. (ll,"'N,_)::: VII. ii, +-v If ~ = J,+ It p, + ~ +-ItPi -(j,~h) + ~ ( 5,,..li ) = r. + ~ Of.V, 8t =: Q. ( s; 040 ~2. ) r: !7,K, + r:;' e;, =: e + 0 = 0tf·D~Ir o- c P, t Dz) ~ v·5, r'V,Dz. C fr/~ &1:' fVf-.', ~ a E,,..~ . Hoi: ~ Ji, .,.Hz. , ~. ~., 01 G "'" D~. ~ r Dz. .sAt;r{y i11Ax4l1l1; ~!UA/"()'"

SCk.,..' es J~t: '5", ~ jl. ~d fvt• fv, f Iv: .TA/$ ., Me. S'jJtreos,'lt't;,." .,.hUt,,,,,-.

2.:.!.: (tl.) No. l£:.; r;""J .,.Y-S,'"",X+ ; s;....1 $0 4ft c». o, OJ I b~'t V"A: .. £CDS~'" f UJ 3 + 2A..t ce,«. D)

2./3

2. If#.

2.a V· D = V' (2 X';) = 2. :=a» t;, ., 2 C%,J- ,{vg = 0 oj ej oJ ;

V)( g :: -~ (0 ,J i -It ej fa, ) :: - j '-l :§~ = 0.3 ~ e ji.' ;

iJty,t): o,~"CM"((.Jt+4tJ);

".tH -= J ~ :t, ~"'~1f;~H"J H:: ;; E = :;~ CMJ(lJt -t4t.J) J ,J := 0... i> = E. E = 0.] £ CAJ (wt of -i)l X ,'",fo t/,.'t .!k ..l:lltt ~/',fl/'"

it: W'At.. I/'f -1(= wm .

1..'1

1../0-

A.. "'. • A ""

(0) NG). F= 2:.S,....)·Y$,,,,,X =0 .,11. y-~/!., b"'-t VICE'" yCoS}+-r"".s;t:'~o c:nt. '1_(C,) No, 'E= ~J'·"'_J::.o ~~ }=o~lo."e/b.,;t VX·Ec;e.c5}~0 (1?I.~.DIt.-..::!P ~.o

o~ ~t p'~e.V-CfJX¥.. )1:0 =;.jWV·S=O ~ V'§60

'ii' ('V'ld )~C => q.j +jWrJ·g =() ~ )w(-8-. fl·g )=0 ~ V·Q =- tvA('](,'t. ),(): a(r.",}) Cl>5(I.JI.I-'J ::: a(~~. J) U5(-wt ..t/'Jeiwt: Afx,'.J.t)-= 1f'e[a.(r,1,J)eiiJe;wt} ~ (J.(x.1,J)ej4e -i ..t : II (71'. V·l.'O. /?e [a..fx.j.D iife-ilJtJ - A(;a:.j.)) e-i ~\lttE::iwE 17)(/i.-=J-t'wD I7·B=o v,g.f..- .. 21 _ -I - J __

£ = [I' ( (- :::L) W1' = INe.' =[-ltllt.",(I."./~-t'J'. "lTrl,,"JJf I: s,'4-x ID '1lit,-,'- IYtE. q.""/~-J' ,

al: S'OOKJh : E «t. [,-/' ~.('1-1I./~'" )lJ = - :I':J 3 f' 0 '·:2IT1f.5'JI,I()1 J •

or: I MJlJ: E:: t.[/-(:::;.co'i" /] = - 80.43 E.

6

,7 ell:: %,[iij= Aj"", r. [V'(EKN)j::: v~ ::.w::.,[ II H'~]::' v-.s~, A.::: Wak- Su. r, [..1. (.J. ...'H-)' 7::: t"..ttr: 1'1"" .M :), it,. ""H ~ ",..

(c l·E1 ::._£ ,.!:L = 10...,,_ • [~(.l. [ ~ e-)17 w.:ttm~ In ...,J , " ~ ~ c» 'J: ,..,3

[j.i] e _1L,L::. ~WI J t>'I It1,

U! (4.) [ g ,j)] ::.¥; ;,.. = :~e.L ell 8] -..t!. v-s., 1)JJi.. - J~' .]'-.J._

(p) • - ..... ~ = ".,] - - ~(c) [~1':[ l s( HJ = ~ ~ = ':,~

!1 !=(;#Jr;e,jl .......E(O=;(()5(~~~)-;~(wt-J)B. = (r -j; )~-j J ~ #(~):o& 9Cb.s(c.lt-~)+; s;,.".(u-E - J)5': Ese;r:: ~ [CD$l(e.Jt'J)+ S'',..,'(4.Ji-IJ1 = ;I=l~i"=(;.,j9)~(;oI-j:)· 2Z -+ <5>c-fli'e{i,J= Z

~ t-e ]» EI. f j t, _....F· ~~j Hre;rit,[ieilAJt]::: ~CDSblt-Ezs;,...,t; Ji-Rc{!ej..,tJ ,,~(.,bS~-}lr$;A.wt

S ~g Jl i1:.{E,. )ll4.)~SI4Jt +(E;xHr )s;,...'4Jt - (E,.'rliir'" ~)( ~) s,~wt GDSUlt

BloAot g eiwt -= (NAUjwt -ilz ri"II,)~) +-j ( 17; j;~l4t +ilrGljc.Jt) AJtd.

Ref.g 7CieiraJtl = ~J(~ uswt - ~ x iI f,-"'ItJ't - E;x#A S,',..wt - E.r)( #z t.oswt :f: 5

2.1 I L~t!.:1 11f.#j fz .n.t ii· HI.. "jillff ejI<Jt = (~ u,we - £; 1,,,wt) f-j ( i;z. j,~,.)t.,. £z C6J 1Ji, )

il ei",t := (~ClJ5wt - Hr $,',..,fJ'6)'" j (~ '$;~lJt + ii.r U$wt)

Kt.[ieiwt)/.F e! ....tJ =(ff;z74~)US'Lwt +(Er )(~)j; ...2...,t - (ill)/. ii.r + lr~~) s,"-wt UJ5AJt

-(4~~)CAS"'wt- (~,«J( lJt)s; ..}t.Jt - ( 4'11.Jl,r 0# Ez ~N/l. ) S.~tlos",t• ( ~lC ~ - Ezy. #r)( e..os'wt.- s: ........wt) - ~(rp~ilz..,.'lz~iI~) s,',w.)tU$w-q

:. s I- I?t [ €.ljwtx B e jwt JHI U,.,. ~!.E.e::; X J:.,,~//)·~(lD./ =- 4.42,)f.ID-,I-(JiJAi'/,w.J)

tJ I -- ((;;;-VI, :. TJ.l.H·H ::T;c.o·(3 =

... VA/V! ~ 9xloi

Ue= -L£ s.s .;«, 8t> X/D-9"'(IOf)t..2. 2 J[,7f

UA:: 3,98)(10) (.1~~, )

7

orm Plane Waves

)f mankind lurwing made be-ture 3.11). Mod-, vary greatly individual comet

cornet's laiJ? uwhether or nothis observation16 sun that art!the sun. Other-resent oxplana-51pa rlicles andpressure of them the sun, ande ionized gasese plasma formsflow of protonst a speed up Laet's plasma and

1.Vol.199, Octobor

4, April 1964. pp.

Problems 65

Figure 3.11 Various comet shapes drawn onsilk found in China. These figures werepainted between 24610 177 B.C. Below thesefigures 01'0 Chinese names for these comets.

Problems

3.1 Estimate the power density of electromagnetic radiation from the sun received onearth in the same frequency band as that of the VHF television channel 2 (54-60MHz].

3.2 Consider the sun as an isotropic radiation source. Calculate the total power radiatedby the sun in the television channel-z frequency band (see Problem 3.1). Thedistance between the sun and the earth is approximately 1.5 x 108 km,

3.3 Assume that solar radiation is isotropic. Estimate the total power radiated by thesun. The solar power density received on the earth is 1.4.kW/m~. See Problem 3.2for other data.

3.4 Derive (3.Sb) from (3.Sa), assuming that E = Ex Jl:, and Ex is a function of z only.

3.5 The star a Centauri is approximately 4.331ight.years distant from Earth. A light-yearis a unit of length that is the distance a light wave covers in one year. How distant is aCentauri in kilometers?

3.6 An electromagnetic pulse is sent from an earth station to the moon. and the reflectedpulse is received 2.56 s later, How far is the moon from the Earth? [An electromag-netic pulse consists of a wide spectrum of electromagnetic waves at differentfrequencies. J

3.7 Find the 81 units of the following quantities associated with a uniform electromag-netic wave: (a)w. (h) k. (elf. (d) T. and (e) A.

r 3.8 A helium-neon laser emits light at a wavelength 6.328 x 10-7 m in air. Calculate itsfrequency. period, and wave number.

Two-wiretransmissionline

66 3 Uniform Plane Waves

3.9 Figure P3.9 shows a dipole antenna, II is VAry affective in receiving television sig-nals when its length is approximately equal tu one-half the signal wavelength.What are approximate antenna lengths tor receiving signals for the following: (a)Channel 2 (f = 57 !vlHz) and (b) Chunnel l J (f = ;!1:3 MHz)?

r- - >./2 ----t~ I

I Figure P3-9

. 3.10 The following set of slactromagnetic fields satisfies the time-harmonic Maxwell'sequations in free space:

E - Eo e+1kz X

andH - f-lo e Ikzy

Express Ho and k in terms of Eo and (0 and Jl-o·

3.11 Do the fields in the previous problem represent H uniform plane wave? III whatdirection does the wave travel? Find its velocity and determine the time-averagePoynting vector (S).

3.12 The Federal Communications Commission of the United States requires a minimumof 25 mV/m field intensity for AM stations covering the commercial area of a city.What is the power density associated with this minimum field? What is the intensityof the minimum magnetic field H'i

- 3.13 Study the following E field in a source-free region:

E = x Eo e-lkx

Does it satisfy Maxwell's equations? If so, find the k and the H field. U not. explainwhy.

-3.14 Show that in 13.13J, if rf>. - 1/1., - -rr/2 and 0 - b, the wave is right-band circularlypolarized .

• 3.16 Find the polarization (linear, circular, or elliptical and left-hand 01'right-band) of thefollowing fields:(a) E = (ix + y) e-11et(b) E - ((1 I il Y I (1 - ili) e-lk•(c) E - ((2 + ilx + 13- il z) e-1ky(d) E ~ (j i + j2yl el/Iet

3.16 Show that. if 0 = h and t/J. - rI>lt - 11'/4. the wave is elliptically polarized. (Refer to(3.131.)Do not try to obtain an analytical expression for the locus. [ust obtain a pair ofparametric equations similar to (3.141, calculate E. and Ey at ten points (wI = 0.10· .. . . , 90°). and sketch the locus.

ave? In what 3.22time-overage

3.23sa minimumrea of a city.the intensity 3.24

mPlane Waves

~ television sig-.al wavelength.e following: (a)

ric Maxwell's

nol, explain

id circularly

hand] of the

d. (Refer tosin a pair ofwt - U, 10°,

Problems

3.17 Show that an elliptically polarized wave can be decomposed into two circularlypolarized waves. one left-handed and the other right-handed. Hint: Let

E - (ux + lJy}~) Jk',

where a and h are, in general, complex numbers. Then, let

E - [a'x I ia'y) e I'" T (1/x - iL'y) «=and solve fur H' and 1.1' in terms of a and h.

- 3.18 Show that a linearly polarized wove can be decomposed into two circularlypolarized waves,

3.19 A dipole an tenua is in the x-y plane And makes a 45° angle to the x axis. A receiverattached to the antenna is calibrated to read directly the component of the E fieldthat is parallel to the dipnls. What are the readings when the fields are thoss given ..in (a)-(d) of Problem 3.151

- 3.20 An electromagnetic wave in vacuum has frequency rn, WAvelength >-0, wave numberkg, and velocity Vn. When it entars a dielectric medium characterized by fJ-o and E - 4f",what are the f, A, k. and v of the wave in this medium?

~3.21 Aluminum has f - (0' jJ. = fJ-o. ann" - ~.54 x '107 mho/m. If an antenna for UHFreception is made of wood coated with a IUytH' of aluminum ann if its thickness oughttu be five limes greater than the skin depth of the aluminum at that frequency.determine the thickness of the aluminum layer. Is ordinary aluminum foil thickenough for that purpose? Use r - '1 Gllz Ordinary aluminum fuil is approximately1/1000 in. thick.

Calculate the attenuation rate and skin depth of earth for 0 uniform plane wave of 10MHz. Take the following data Icr the earth: fJ- - ILl)' f ~ 4Eu, and if - '10 "mho/rn.

Find tile power density in earth where the field intensity is 1 Vim. Use the data inProblem 3.22.

Suppose that an airplane uses a radar 10 measure its altitude. l.At the frequency of theradar be 3 GHz. Suppose further that the ground is covered with a meter of hard-packed snow.

67

Airplane

<-. J:2--3-I II I, II II I

h I AI I Ail'I II II I, I Snow

W'W/'/////ff)71J#////;Mwj~ Ground

Figure P3.24

(a) What is the difference between thA Apparent altitudo measured uy the radar Annthe true altitude?

(b) How much attenuation in dB does the radar signal suffer because of the snow?Consider only the attenuation of the WHVt: ill the snow, anti neglect the effect ofsnow on the reflection at air-snow and at snow-ground interfaces. Refer to Fig-ure P3.24. Use f = 1.51.'0 and tan (j = ~ x 10 4 for hard-packed snow at 3 GHz.

- 3.25 The following data are given for a uniform plane wave in a dissipative medium:

(i) amplitude of E. at z = 0 is 1Vim.[ii] phase of E. at z = 0 is zero,(iii) k = 0.5 - j 0.5 (11m),[iv] direction of propagation is in i,(v) intrinsic impedance of the medium is 1+ j ohms

(a) Find the phasor expression for E, as a function of z.(b) Find the phasor expression for H as a function of 7..

[c] Sketch Ex at z = 0 and at z = 2 m. CIS Iunctiuns uf wI.(d) Sketch the time-domain H fields at z s 0 and z a 2 m as functions of wt.

3.26 Consider that a small space vehicle with 100 kg of mass is located in outer spacewhere the gravitational field is negligible and the fuel has been exhausted. Asearchlight of 1kW is turned on. with hopes that the vehir.1e will gain some speed.How much speed will it finaJly gain if the searchlight uan last 48 hours? Hint: Thelight wave carries radiation pressure, and there is a reaction force on the source ofthe light.

3.27 An icA particle of radius Q is r distance away from the sun. Tho gravitational forceacting on the particle is given by (3.46). The ice particle's mass can be obtainedfrom its volume and its density which is assumed to be 1gram/ern". The ice parti-cls is also subject to radiation pressure which is given in (3.47). The force actingon the ice particle due to radiation pressure is approximately aqual to the cross-sectional area of the particle times the radiation pressure. Show that, when theparticle's radius is less than a critical value, the radiation force will be greater thanthe gravitational force. and this critical radius is independent of r, the distancefrom the sun. As a result. all particles with radii smaller than this critical radiuslend to be blown out of the solar system. Find the value of this crttlcal radius.

68 3 Uniform Plane Waves

3.3-

~.I-CI/APreR 8

power dUl(;1y = 0.' tllO ell WIM 1_HI-

... power dUls,ry i.A, Sol -&'DMHJ ,.. (6D-5"4»)( 10'JiCO./' x /0-11 =-,s.' '14/0·,6 "'/"'~,

PDUll,.. dtlls/fy I~ Vi.e. .f;.guvaey rrusg,t. S"4- 6' "',,)8 "., xlf)-U WI""-NS

,'. t-«..L pOIlJ~" ~(.IIJ;-Iy .('0-'5'4) ~/D")(D.')//O-I'x4rr)((,I.St!lO")1.. 10/. g MW

a ,Ip'; (/'''- ~/01)1I -1-" tI ( t.s Jt to'") = 4- JC /0 hi

3.S

3,(,

sxH.3.9-

3.12.

),13

3.11/-

3.J1

:3"'10''/f., )( ~b Ii Ito K ~4)( J'S"I< ./..'3 ;I 4.o'l~ID"", CI 4,09)( IOJ' ~A1.

3)(loS mAu x ('1.·$'{,/l ) St(: I/:: .3. gl-)(/D 6m

(A) [wJ =.ra4/su" on«:» Vm) (C) [fJ· ,Vs«-" cd>(rJ· $«., (t){)..];m

f c 3111D ~/ ".326110'" = J/.. 7+x /014-11 7= '1J. 2.//Xlp·'S'.su.. J '* = .2o/A ~ 9. 93}(10~ (Yn.)(tJ) ;'1./0 if (2.x S'7 'II10' ) r= :2. "3 rn , eb) 3KID ,/( 21( :;'13 XID") - O. 7c<9- WI

v-. §: 'j~'/l.ii * E = JV~~ c - -:-L V-A(E" eilt' £) = _ J:._ EDe jJS:- = 1-1 eiAJ~,r It.- -r. ",.,;It- ....j<. ~ ~ 0-

:. JI~= - tAJ)'tI E. AL$t:> Vz.g.,.w;tt.t. € -0 ~ Jt '. w~" flit r. Ie::: w/~.tll 1:lN/. !/D.- (];_G.17. •

IjtS; - J cI.,r~~I/t:>,.._ i IJT":::. T :::.P:i.; <5> '= -I: Ke[ e-rs a f(_£.xJ!.)}:::. - d ~ E/jIHJ: 2S,tJl)-.y ({ c: 2S')cID·~:ZOlT c; j,I3x/tJ-$ A/'H.I 1< r~/-= -J:({; /H/'1c o.e3 ;,W/..,zNo. o« €. = v" (E. e -jl): ~ ) ::: 0 ~ jj co (?j() coH"eS~i~ H-{it-Id tAl"{{:$).

If, <P4. - t)~'IT U. a-« a#: b; &f,r Co a c~~(wt - IIJ + ~ .. ).1 E~= 4.,S;,,- (Alf. - tlJ'" ~4..)1 I Z Elf

£", .,.£" = a. tvlL .d: AJ:= ¢.c.. pf.u..e.! tVt'~ _E(t) ri4At-ho,yJ. Clrc"u..!ilrl, p'{A/'U~,!-

3./S (tt) £:(J';~Y)e·jl~ =£e-j(kJ-f).,.yeJi} ~ e(~)::I-£~iljp(w-e-A.J)+;GC5(wt-~J)

,f('jAt - Aa.,-.e{. C., r4l,./a.rt, pol.Lr'Jed..(b) Er:[(I+j)Y+(I-j)z] e-jk~ ~ E(~):::.Yl£c()5't«-*x"'¥)+;UCtJ.s(u.)'t.-Ax-!)

RIJl.,f - f,,,,,d Cir-cu.l./Lrt.:; p"t-iju,

(C) g = [_(:J+j) X +(3- j) ~J e -jk" ~ E(t)c ; r; tDS(l,oJt-Al1'~.46) + i(iD (,,()S(we -ki-O• 32.)

L~f'f; - ha"..d .e.IUpll"A!l:; po.l~r'J CJL

(d.) i.:: (jx+j2y)ejAI => E(-e)= -XS,~(I.)t."'AJ)-i~$,·t\,(.n.+~})Lit1€4rLy p" LAn'3d ,

8

)

)

a ~/a..:.b" ?a.-¢lIwl j cx(t)=d.CCS(fNt-i.J+tP(4), E,-a_CDj{tVt-~J+f/J4-;f)tXt -k J ::0 rPa... I E,k(.t:):: a.~~ ~t; ANi. &1" a. (..II $(l.o)~ - ~5D)wt 00 10" 2.0- 30" 4-fJ- So· 1>0· ?Oo eoD t:?l>D

E",' a. .geS"~ .Q4o... .fJ66a.. .7./'a. .6.f.3a.. .SA .3424., .1'144 0

Ef .7o'Ja.. .906a.. .9"'4. .9964.- . 9'16el. .'i6lJ~ .9,,6a.. . BIt/a. .7070...

EIl,;pti.Ca.llJ pDh,.';Je4. wa.ve, ~ 5VoU*'A.L ~,..,.., I g-(;&"'YiJe;J·J.J.) w)!~

~ 4/ld. J.. art. ~",~/,~ 1?u.""~erJ .

Let §:- ( ; ~I + 9j~')e oj RJ 4- ( , E. ' - 9j .!') e -j1<)::(Le/t-hand.. U·,-t;J.(./.4rt'j pDIA"'!J~)t(I&·~ltt ..~ t;i~t.JA.r1;J pl>lAriJU)

,r.e", (R ~ of- 9E ) e -.iA} ::::(; e/ +; j ~ '; e .j~$ + (; i.' ..; j,i ') e. "J*}-••• a.' -r.E_' :. E: tVta. j (E.-: k' )=1 =9 e:I. f (g,. - j R.) ~ .!'= d (g..,.j k )

}/«fJct/ d~.U>""I'l>Hd.. Wll.//u a..re, : [; i(4·;g) + Yi(j~.,. k.lj e ..j. ~ (Lt/6-AIJrv/u.)~ [ % f(B·j!) - 9 t (j~ - k. >J ( r'·9h.t -"~ed.)L;,,~rl, I'DIAlAje4. wtW€, 1~!J~t1i.rAf/~rm, gc(fA.+Yh)e"'jll}J fNl.~ a., U1A.'6 (Ire,. rtAi. "u,..,buS.Let: i. :=(fa.+Yb)e·j~~= (:'2:'+yje/)~·J·~~+(£j/-Yjl2')e·j/q,J, -Mtl.n.E;~~.! '=a.. alW( j(IJ:'-_!'):=b =* fd:,': ~(a.-jh) 1JJt~!/c-f(a~j~)... i)tc~""'I'P5d ~.s ~~ lt~(4-jb)"';/:(ja..+b)] e-jk;. (j'lt-luJ"eI.~"'-) t:u'\d_( x f( a~jb) - 9 -}6a. -b)J ( ri'll..:f. - h~,,"u{ ). .

_ "/ rr···/g·A. =tt :!.'. I g.$. I = II.~ r: 1.5'0

. I E"~/I/r 2 ::2.IZ•. - {1

9

5.%0 i=f~I 'IT =I)i.:.u.) -=.:;., ..t. ~ .: ~ (~4,) c ~O/t, ~= 1£. :2 (~) = 2.Ap

T/ ~.~-Ix/D7 1 L ~ ( 10"',.2\Y. -,~ ~(.UE::' '""10' >tJ'1f,(/o »1 r, tI ~ {~c .,,-../()4~4-"'>( 3,~X/~'./ 2 - :1,''''JiCU> ".,

S~ = 1. .JI-JlID-S''''' < 1,'5')(J()-S"~ ,', IJJf,I.~"""_f_il,~ "&·e..t.~"~4.4.flv.tr/. /~-4- 3~Tr"/l>' 1

3.;'2. /w!:: 1"" ID7 Jt 4- = 4. S'>i/~- «I ,1<1 = fff -flii. =-f~/3.D" = ID-~)(3(i1T'c A~94(}(,.) :. tit- -,;;-10,.1(,..,)

3.:'3 7 = J;( ~ = ...!... a: I .':- c .l: rFj (,- j~'()4f)-YL"" -J- IF{; :: bt>""- Yll-; 4Jl) .2 IT. I-Jcr/41t 1(7. _I T.r E, 'l. • ~ I Fe'

:. Ai J:: 0 I 1< s:>( -= 2"m- eJ - Tt;;;; = 2. '5" ,"W/m"~ (a.) Let lIoVt. c ve./ot.i..ty;"w (fUr- I .,~C" t JII 2. [~';,;:. +ryllm) J = :2 (-~=)

:. n. -I -#- (/.f = A~ ~ hW - It. • m-/ r:: o, '22S ,.., (It_",- 'Y'P4r.,.,/: Jtl./v.k)(b, E. ::::111/"0 (1.5[.) ( 1- j 9x/o-4)Ys.. - wJpp(I.Sf.) (1-j4.5'xID-f-) ~ {q.l=WI,M,t,JI.5 (4,5'IC/O· *)(~)

IItfflnt..l,6.·I-r:Dn dull -It> s,,~w I-'lU- : /q~ 2(Kt) =- """'3YJ:9{ri1<~",#.~~/D"f == O.<J6'1{N'f"')" ~KID

= ~./)'91l 8.~8"" o.66/8)3.~S' (A) E~;: e'p(-jlo,S--jo,S)J) ~ hrl-,·t} JhrC-j',,/)

I - A' ) C' ...l.).tf =:J , .. j Q)(f{-'S; .tIlf -Jo.J;): J (o.r-jo.~)tYf(-f).rJ)~tf(-j'.r)),-c) r )

~x IJ,t :I ~~r(- O.t') , Cool (lJt - tI.i)

~~) H,ll,'f:>= o.7~7~~r(-D ..rJ)c...S(&.Jf;-D ..tJ-"/'t J

-I

-2

10

)

f _-~S,,'> A = __,.....;..IO_l~_3 JL (01'

.I

tl:: _f_ =m

3.17- -IIF, :: '·'1~10

F, = r, ~

J.1 "Itl. I"'e.F. = /·Yl/o - ~~ t r·i

q J I1'l.= T1'(&1. -,0 t::.

).?n, = /. 'I f Jl 10 -1..;. , "I".e = /.J' J( , 0 ""

11

92 4 Reflection and Transmission of Waves

Flgur .... 23 Tn rp.cp.ivf) linearly polarizedelectromagnetic waves. wire grating.~ rnayre-place metal plates Ior retlector antennas.

Thus.

J. - ~(2~O)cos8e

Nole Ihal Ihecurrent flows in the y direction and that no current flows in the~ direction. In fact. if the concluding plHtH is replaced by a grating of parallelconducting wires arranged in the ~ direction. these wires also serve as areflector that is as effective as a solid conducting plate. Experiments havefound that grates are effective when the spacing of the wire in the grate ismuch smaller than the wavelength of the wave. Grates are used instead ofconducting plates to reflect linearly polarized electromagnetic wavesbecause they reduce weight, save material. and decrease resistance to wind.Based on these considerations, some reflectors use wires to replace metaldishes for transmitting and receiving electromagnetic waves. An example ofsuch a structure is shown in Figure 4.23.

(4.54)

Problems-4.1 ThAE field measured ill oil"just above a glass plate is equal to 2Vim in magnitude

and is direct at 45° away from the boundary, as shown in Figure P4.1. The magnitudeof the E field measured just below the boundary is equal to 3 V1m. Find the angle 8for the E field in the glass just below the boundary.

- 4.

4.

3Ves

54)

thelIelsalVe

~ isof/es,d..tal, of

de:Ie~e

Problems 93

Flgur. P4.1

x

/\Vz

- 4.2 The H field in air just above a perfect conductor is given byH, - 3i I 42: amperes per meter

as shown in Figure P4.2. Find the surface current J. on the surface of the perfectconductor. The conductor occupies the space y < O.

4.3 Match the following descriptions with the figures shown in Figure 1'4.3. Fields arenear the interface but on opposite sides of the boundary.(a)(b)( c)(d)(e)( t)

medium 1 and medium 2 are dielectrics with E, '» Ez

medium 1 and medium 2 are dielectrics with E, < E2

impossibleimpossiblethere is a positive surface charge on the boundary between two dielectricsmedium 2 is a perfect conductor .t

.....r

Figur. P4.3

, ,

1IE

2 .E, - 0

Iii) __ ~iii!~~

~~ I_-...,_~

Pz t,.. :> l>, e..,t-<~ ).~ I t;,

Ii)I , ),

, I Co

~:1.~ 2 ,

[Iv] (v)

94 4 Refleotion and Transmission of Waves

4.4 Calculate the critical angle 9. uf an air-glass interface similar to the interface shownin Figure 4.8. The dielectric constant of glass at optical frequencies is ~.25 times thatof air.

- 4.5 A pearl is emherlded at the middle of II cubic heavy-load glass [s, = 3.6). Is it possibleto cover a portion of the surface of the cube so that from outside the pearl will not beseen at any viewing angle? If so. find the shape and the m.inimum area of the cover[in terms of the cubic surface area}. Hint: consider conditions of total reflection, andneglect multiple internal reflections.

4.6 In the three-media configuration shown in Figure P4.fi, the wave numbers are k,. K,-,and 1>3' Find the transmission angle in medium 3 in terms of III and the wavenumbers. Assume all k's are real.

zFi,ureP4.0

k, k,

1-/ __ -=-=--_,.....-=--17 :-...---7- ( Closs rod '1~Light beam ;7

FI,ure P4.7

4.7 Solid-state lasers (ruby or glass) are often fabricated of rods with the ends bevelled atthe Brewster angle. Let ( = 2<0 for the rod. Sketch the propel' arrangement of theexternal mirrors and their angles. Indicate the bevelled angle of the glass rod. Whatis the polarization of the uutput of the laser beam? (See Figure P4.7.)

4.8 A parallel-polarized wave is incident from medium 1 on the plane boundarybetween medium 1 and medium 2. I:!othmedia are dielectrics with J.LI - !-t2 - !-til andreal permittivilies EI and Ez. We know that. when the incident angle is larger than thecriLical angle 9e- no rime-average power is transferred to medium 2. Also. when theincident angle is equal to the Brewster angle 0b. the reflected power is zero. Nowimagine a situation in which the Brewster angle is greater than the critical angle. Awave incident at the Brewster angle will not be reflected, because the incident angleis equal to Oh. nor will it he transmitted. because the incident angle is greater than 0,.Is this situation possible? Why?

4.9 Consider an electromagnetic WAveof 1 MHz impinging at 60° on the ionosphere.This case is similar to that shown in Figure 4.13. Assume the the plasma frequency ofthe ionosphere iswp = 211' X 9 X 106 rad/s, and plot] E las a function of z [like in Figure4.14). Mark the scale of z in meters. Solve only for the case of parallel polarizationwith Eo-1 Vim.

n -4.11

J

4.12

elled atI of thei.Whal

undaryPo andhan theaen theo. Nowngle. AIt anglethan IIr•

4.13sphere.ency of

4.14Figure·i2ation

of Waves

ce shownirnes that

Ipossible-ill not be.he coverlion. and

Problems 95

,.

I E I [volts per meter)McdiuIII2

10

0.5-:--7 z

[meters]-3 -2 -J o 1 2

Figure P4. 10 Flgur. P4. 11

-4.10 A perpendicularly polarized electromagnetic wave impinges from medium 1 (char-acterized by !-t, = !-to and l, - 4foJ 10 medium 2 [characterized by J.l: - !-to andE2. = <=0]' This situation is shown in Figure P4.1U. I v::::;;

'I ...~.

(a) What is the critica I angle? fJ, - ~', ., ."(b) Let the incident angle be 1$0"; find h.linu k, in terms of ko - w~,(c) Find kl• in terms of ko.(d) In the second medium. find the distance ~a at which the field strength is l/e of

that at z - 0 I.

(e) Find] R,I and the phase shifllil',I! (Rrl,A uniform plane wave in air impinges nuruially on Ii dielectric wall. The magnitudeof the total E field measured in front of the wall is shown in figure P4.11.

(a) What is the permittivity of tha dielectric wall? Assume #2 ~ JI-o.(b) What is the frequency of the wave?

A uniform plane wave in air impinges on a lossless dielectric material at a 45" angle.as shown in Figure P4.12. The transmitted wave propagates in a 30° direction withrespect to the normal. The frequency is 300 MH:l.(a) Find fz in terms of f9•

(b) Find the reflection coefficient RII•(c) Obtain the mathematica I expressions for

the incident E field, the reflected E field.and the transmitted E field.

(d) In hoth media, sketch the standing wavepattern of IEx 101"1 Ias a function of z.

x

Figure P4.12

For two isotropic media wlrh s, :;t: Ji.z and EI :;t: (3. find the Brewster angle for both theperpendicular polarization and the parallel polarization.

II a wire antenna is attached parallel to the metallic surface of a vehicle and isinsulated from the surface by a thin layer of dielectric material with a thicknessapproximately equal to 1mm, would it receive an AM signal r f = 1MHz)? Hint: Wireantenna interacts only with E field in the direction of the wire.


- 4.15 It is known that the transmitting antenna of an FM station Is located in the directionperpendicular to a metallic plate, as shown in Figure 4.15. The frequency of thesigna I is 94 MHz.(a) Where should a receiving antenna be placed to receive maximum signal? The

antenna is II dipole that interacts with the E field.(b) If the amplitude of the incident E field is 1 V1m, what is the amplitude of the

E field at this optimum position'?

- 4.16 It is found that by placing a conducting plate 0.8 m behind a dipole antenna, thereceived signal coming (rom the normal direction is twice as strong as the incidentfield. What is the frequency of the signal'? What would the strength of the total E fieldbe if tbe frequency of the wave is changed to 98 MHz while the antenna is stillplaced 0.8 m from the plate'?

-".17 What would the r; field be if the receiving system in Prohlem 4.15 were Lo 'detect awave coming in at an angle 10" off the normal? Assume that all other parametersremain thp. same.

- 4,18 Derive (4.53).

- 4.19 For a parallel-polarized uniform plane wave impinging on a perfect conductor at anAngle O. find the electric and magnetic fields E and H for the incident and for thereflected waves.

4.20 Consider a 90° "corner reflector" shown in Figure P4.20. It is made of twoconducting plates placed perpendicularly to each other. A uniform plane wave withE = "Eo expljkx cos 0 + jky sin 0) impinges on the structure at an angle O. Show thatthe total electric field is E - -24En sin(kx cos BJ sin (ky sin BJ. Hint: The field is theslim of four waves with four k-vectors shown in Figure P4.20.

4.21 Use the formula given in Problem 4.20 for the total electric field. Find the optimumposition of a dipole antenna placed in front of the 90" corner reflector. The 0 angle ofthe incident wave is 30°. The frequency is 100 MHz. Express the position in x - ycoordinates in meters. What is the "gain" of this receiving antenna? Gain is definedas follows:

Gain - 20108,0 I~: I (d8)

where E. is the E field at the antenna position and Eo is the field strength of theincident wave.

P'lgur. P4,20 Top view of a 90° corner re-flector and the four k-vectors.

wes

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Problems

x

7.

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(a)

Figure P4.22

z

)z

---~ 0.67--- - -o >

'>z (b)

Figure P4.22

97

z

z

>z

z


4.22 Match the following descriptions to the standing wave patterns shown in FigureP4.22. The inr.idenL wave in medium 1 has an amplituda equal to 1 Vim. Note:There are three patterns that do not fit tiny of the following descriptions. Cross outthese patterns.

(i) Plot (If I E, .ntul I. with medium "1 being air, medium 2 having (2 = 4€n aud J.l2 =I~o'Normal incidence.

[ii] Plot of I t::, .0.n.1 . with medium 1 being chaructertzed by €, - 4Eo and J.I, .. ~(o.and medium 2 being HiI'. Normal incidence,

(iii) Plot of II';) LOIO' I .with medium 1 being characterized hy El .. 4(0 and 11..1 - !Joo.and medium 2 being air. The incidence angle is greater than the crttical angle.

(iv) Plot of I E'loln' I .incidence angle is equal to the Brewster angle.(v) Plot of I Ez tOlal I . incidence angle is equal to the Brewster angle. {1 is greater

than Ez.(vi) Plot of I Ey lulal I . Medium "1 is air and medium 2 is perfect conductor,(vii) Plot of IIIYlo(~1 I ("d.Medium 1 is air and medium 2 is perfect conductor.

4.23 Consider the CC:lSOof normal incidence of a uniform plane wave on a perfect con-ductor as shown in Figure 4.15. rt can be seen in (4.47) that an oscillating currentis induced on the surface of the conductor. Therefore. the following expressionmay be written for the velocity of a charge on the conductor:

v = ~ d q Eo cos(wl - kz)

The above equation is exactly the same as Equatiun (3.39). Continue to work alongthis line and prove thai' the lime-average radiation pressure on the perfect conduc-tor is twu limes tbat given in (3.4!i).

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Problems

-5.1

he::utIceiceithors.ild.ildthe 5.2

-5.3

r8

:18

:al

:b)

ltl

Id.at

ra]

Ib)

I arve:heIce

me

Problems 127

Using (5.45),we obtain

VxA_;poAp_~ -ike jk.OZ P

Example 5.8

Calculate the total time-average electromagnetic power transmitted along a coaxialline when the fields are given by (5.4RJ.

Solution: The time-average Poynting vector (5) is given by

1 I {V . V } V"(8) = - Re[E x H*) = Re _! e-'·· p x cf> _! elk. = z_o2 2 P TIP 21)/

Therefore P = (11" V~/71)lln (b/a).

Show that the complete fields of the TE wave in a parallel-plate waveguide are givenas follows:

-k EoH. = --'- sin lI.x e-I~.,WIJ.

ik.Eo 'kH, = --cos k,x e /,IWJ.I

E. - 0; E, - 0; H) - 0

Find the complete fields of the TM wave in a parallel-plate waveguide.

What is the lowest frequency of an electromagnetic wave that can be propagated inthe TE mode in the earth-ionosphere waveguide? Model the latter as two perfectlyconducting parallel plates separated by RO krn.

- 5.4 Find the surface-charge density P. on the upper and the lower plates of 8 parallel-plate waveguide for (a) the TEIll mode. and (u) the TMm rnoda,

- 5.5 Find the mathematical expressions for a TEM wave in a parallel-plate waveguidethat propagates in the z direr.tion (see Figure 5.11. Sketch the parallel-platewaveguide, and indicate the directions of E, H, and T•.

-5.6 A rnicrostrip line has the dimensions a - 0.15 cm and w ~ 0.71 ern, and thepermittivity uf the substrate is e = 2.(; (u. I-' - f.Lo. (1 - O. Estimate the time-averagepower that is transmitted by the line when IE 1= 10i V1m.

128 5 Waveguidefl and Resonators

'5.7 The breakdown voltage of the dielectric substrate used in the stripline described inProblem 5.6 is 2 x 10' V1m. Use tI safety factor of 10 so that IE I is less than 2 x 10°V1m everywhere in the line. Find the maximum time-average power that thestripline can trnnsmit. Neglect the ohmic loss.

5.8 With the fields in a rectangular waveguide. find the surface current 1., on the topf y - b) of the waveguide. rr we want to cut a slot along z, where should the slot be cutin order to minimize the disturbance it will r.ause? Assume that only the TE,,, modeexists in the waveguide.

5.9 Show that. if the wavelength of an alectromagnetlc wave in an unbounded mediumcharacterized by 11 and f is greater than 20. then this WtlVAcannot propagate in therectangular waveguide (shown in Figure 5.8) with the dielectric inside the wave-guide also characterized hy Il and c.

Exhaust air duct

t4 HI meter

IFigure P5.10

5.10 An AM radio in an automobile cannot receive any signal when the car is inside atunnel. Why'( I.Atus assume that the tunnel is the Lincoln Tunnel. which was buill inHl3911nder the Hudson River in NAW York. Figure P5.10 shows a cross secticn of theLincoln Tunnel. *

5.11 find the frequency ranges for TE,n mUUI:!operation for those rectangular waveguideslisted in Table l.

5.12 Design an air-filled rectangular waveguide to be used for transmission of electro-magnetic power a12.45 CHz. This frequency should be at the middle of the operatingfrequency band. The design should also allow maximum power transfer withoutsacrificing the operating Irequency bandwidth. Find the maximum power thewaveguide can transmit. Use a safety factor of 10. Neglect ohmic loss, Thebreakdown E in air is assumed to be 2 x 106 Vim.

5.13 Repeat Problem 5.12. but assume thai a dielectric material is used to fill thewaveguide. The material is characterized by f - 2.50lo. Il = J.Lo. and u = O. Thebreakdown E fisld in the dielectric is J07 Vim.

5.14 Consider the size of a,rectangular waveguide to explain why it is not used to transmitelectromagnetic waves in the VHF range. (Take r - 100 MHz.)

*G. E. Sandstrom. Tunnels, New York: Holt. Rinehart & Winston. 1963, p. 242.

Dnators

riueuin2 l( 106

hat the

the topJt be cut,. moue

nediume in the~ wave- 5.15

inside as built in)n of the

/cguides

electro-peratingwithout-wer theJSS. The

• fill the- O. The

transmil

Problems 129

Figur. P5.16Air

o z

The electromagnetic fields associated with the TElo mode propagating in the z di-rection are given by (5.23). Find the electromagnetic fields associated with theTRIO rnude propagating in the - z direction, with maximum electric field equal toE1•

5.16 Consider a rectangular waveguide shuwn in Figure P5.1S. For the region z < 0, themedium is air and for z > 0 the medium is characterized by ~2 and 1'-2' A TEwmode with maximum E·field equal to En impinges on tha boundary from lhfl left.The result is that some power is reflected and some is transmitted. Assume thatthe retlected wave is also TE,o. with maximum E-field equal to E I' and the trans-mitted wave is TElo mode with maximum E-field equal to E2. Find the ratio EllEnin terms of Q. w, 1;0' /lU' E2' and 1'02'

5.17 The corner refter.tor studied in Problem 4.20 requires the solution

E - -z4Eo sin [kx cos OJsill (ky sin 0)

Show that although the coordinates art! different thts solution is in fact the resonatormode that we studied in Section 5.2. Placing conducting plates at x - a and y - b toform a cavity resonator as shown in Problem 4.20. what (Ire the restrictions on theincident angle 8?

5.18 (a) Find the real-time expression of the fields of the TElOl mode in the rectangularcavity shown in Figure 5.9.

(b) Find the total stored electric snergy in the cavity as a function of time. Find thecorresponding total stored magnetic energy.

(c) Show that energy is stored alternatingly in electric and in magnetic fields. thatthe maximum stored electric energy is equal to the maximum stored magneticenergy. and that the total stored electromagnetic energy in the cavity is aconstant independent of time. Note that these properties are similar to those ofthe low-frequency LC resonant circuits.

5.19 Find the lowest resonant frequency of the TE,u, mode in an air-filled rectangularcavity measuring 2 x 3 x 5 ern". Note thai there are three different choices fordesignating the z axis and that these result in three different TElol modes.

5.20 Electromagnetic waves in air with wavelengths ranging from 1 to 10 mm are calledmillimeter waves. Millimeter waves may be guided by dielectric slabs. Consider adielectric slab with f, - 10to and tz - flf• as shown in Figure 5.12. What should itsthickness be in order that only the TEomode may be excited for frequencies up to 300GII7.'~

-5.21 Use direct substitution into Maxwell's equations to show that the fields given by(5.48)are solutions of Maxwell's equations in cylindrical coordinates,

5130 Waveguides and Resonators

5.22 Use the formulas of divergence and curl in cylindrical coordinates to prove that'il . 'il x A ~ 0 for any vector A.

5.23 Find the rectangular coordinates of a point P where the cylindrical coordinates arep - 1. 4> ~ 30°. and z = 2.

5.24 Find the cylindrical coordinates of a point Q where the rectangular coordinates arex, y. and z,

5.25 Show that the differential volume in the cylindrical coordinates is pup dcp dz.

5.26 To convert a vector expressed in cylindrical components into the same in rectangularcomponents, or vice versa, it is convenient to prepare R table fur dot productsbetween unit vectors in these coordinate systems. For example . .x . p - cos f/J. asshown in the fullowing table. Complete Lhe table.

Dot Products Between Cylindrical andRectangular Unit Vectors

p Ic/J z

-it cos r/>

E5.27 Use the above table to find the rectangular components of the following vector

located at p - 2. cf> = 30°. and Z ~ 3:

A = 8p + 44> - 3 i5.28 What is the maximum time-average power 1:1 coaxial line can transmit without

causing breakdown? Assnme Lhal the coaxial line is air-filled and that the break-down E of the air is 2 x 106 Vim. lise a safety factor of 10 so that the maximum Efield anywhere in the line does not exceed 2 x 105 V1m. The dimension of the line is20 - 0.411 ern and 2b - 1.14:1 ern. Neglect ohmic: loss.

-5.29 Consider the coaxial line shown in Figure P5.29. Half uf the line (z < 0) is filledwith air, ann half of it (z > 0) is filled with tI material characterized by EI and 1-11'

The electromagnetic wave incident Irum the left bas thfl following fields:

E' • Vo e-'''-''=p-p

I • VII i~H -</>-efloP

The fields of the reflected wave may be expressed as follows:

V'E'_ p~ei~p

• - V:,H' _ t/J __ u eik.,.floP

ators

~that

!S are

!S are

gularductst/!. as

zector

ithoutireak-.wn Eline is

fillednd J.ll'

Problems 131

(a) Write down the fields of the transmitted wave in z > O. What wave number kshould he used?

(b) Find V ~ and the amplitude of the transmitted fields in terms of Yo. 711' and 710 bymatching the boundary conditions at 7. - O. Compare your result with thereflection and transmission coefficients obtained in Chapter 4 for wavesreflected from dielectric boundaries.

lE,. /11

---+-; .,.....,.·~·:I$" . >z

Figure P 5.29------~air ...•.• ..... :'

l=O

V)lgwjWEg ~ f11t§c-j"';Jl.B wi#.. 'ah,,·O9 -h51 --J""p.t:J%

'3 E " rJU-'I = -.J w,u.. ~l;c fI", - },. J.J .. :: j4AJ! f,

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18

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19

6180 Transmission Lines

We can find the reflected wave by carrying out an analysis similar to the onefor a transmission line with a capacitor. The result is as follows:

(6.58)

Figure 6.39b shows the voltage V(z) on the line during the time periodT < t < 2T.

Problems

&.1. What is the voltage in the stripline discussed in Example 6.1when the time-averagepower being transmitted is 10 kW?

&.2. Consider ths coaxial line discussed in Example 6.3. Calculate the maximumtime-average power that may be transmitted in the line, I Jss the hrea kdown E - 2 x10' Vim and a safety factor ofl0.

&.3. Two coaxial lines have equal characteristic impedances: 50 n. Both art: air-filled,ThA first line has a power capacity of 1MW. and the second line's capacity is 1kW.Find the ratios %z and bl/bz. Consider only the breakdown voltage.

-&.4. Use (6. tb] and Lheboundary condition (4.3) to obtain the surface-current density J.on the lower plate of the parallel-plate waveguide. Then calculate the total currenton the lower plate. Compare the current wilh the definition of I[z] given by (6.3bl

~.5. Find the surface-current density J. on the inner conductor of a coaxial line. Thencalculate the total current on it. Compare the total currant with I [z] defined for thecoaxial line.

~8.8. A transmission line is short-circuited (Zl. - 0)

(a) Find tha expressions for] V(z) Iand Iliz) Ias a function of kz, Zo, and V•.(b) SketchIV(z)landll(z)1(c) Find VSWR on the line.

-6.7. Repeat Problem 6.6 for a transmission line with an open circuit at the load (ZL= <10).

&.s. Repeal Problem 6.6 for a transmission line wilh a matched loaoll,. - 2.,J.

&.8. A transmission line is terminated with a normaltzed load of O.R + j1.0. Calculate (aJthe VSWR. (bl the position of the voltage minimum, anti (e) the percentage of theincident power that is reflected by the load. Sketch IViz) Ias a function of z/):

6.10. Solve the problem discussed in Example 6.6 by using the Smith chart. Find theposition of a shunt susceptance that can tune the line to have a perfect match.Determine the value (in mhos) of the shunt susceptance.

&.11. For an open-circutted 5011 transmission lin a of length P,the input impedance at theother end is j33 n. Find the length Q (in X).

8.12. Repeal Problem 6.11 when the line is short-circuited at one end.

6.13. For the first waveguide in Tahle t of Section 5.2. design an iris that will give a j1.57admittance ilt f - 8 GHz.

6.1

.6.1

&.1

&.1

8.2

...8.2:

&.2:

&.2·

&.21

8.21

-n Lines

8.14. From the Smithchart, find rL for the following ZLII:(a) 1 + jl.. (b)co, (c) 0, and (d) 0.55- jO.38.

- 8.15. Use the Smith chart 10 find ZLn from the following rL (a) 0.6 el<;'. (bJ -0.3. and(c) O.

8.18. For a load impedance of 0.4 - jO.5. find the location of the first voltage minimum andthe first voltage maximum at the load end .

.-e.17. From the Smith chart. find the admittances for the following impedances: [a] Z, =0.3 - jO.6 and (b) Zw = 5 + i3.

8.18. A transmission line is terminated with a normalized impedance ZI,n - 2 + i2. asshown in Figure 6.19a. The incident V. - 1.0. and ths characteristic impedance ofthe line is 1.0. Show that V ma> - 1mu - 1.62.1 VIOl 1- 1.55.1 V( - O.219A11- 0.78, Vmin -Imln - 0.31:1,11(0)1= 0.55, and 1If -0.219X) 1- 1.45.

8.1V. A shunt admittance of YLn - - i1.57 is added to the transmission line that isterminated by a load ZLn~ 2 + i2, as shown in Figure 6.19b. The position of the shuntis 0.219Xg from the load, so that the line is perfectly matched. Let V • ~ 1.0 and Zo - 1.0and show that Vmax - 2.08,1 VIOl 1- 2.00, 1m •• - 1.B6,1l(0) 1- 0.71, and 1mln - 0.49.

8.20. In Example 6.B, find another set of solutions of RI and Rz (in centimeters).

8.21. For the solution found in Example 6.B, how much total time-average power can befed to the array without causing breakdown in the dielectric? Use the value lBl.000V Icm as the breakdown strength of the dielectric, use a safety factor 10, and leta - 2 mm. Hint: consider the standing wave on the stub tuners as well as on thetransmission lines.

-8.22. For the circuit shown in Figure 6.25a, let Zo - 50 n, RI. - 70 n, Hg - 50 n. 11- 2 m. v _10' m/s, 6. - 10 9 s. and Vo - 1. Plot the voltage and current at z - 11./2 as a function oftime.

8.23. Calculate the percentage of energy generated by the pulse generator that is absorbedby the load in the circuit of Problem 6.22.

8.24. For a four-digit code system. design a D-A converter similar to that discussed inSection 6.5 using the transmission line shown in Figure 6.27a. Specify the value of R.the location of the sampler. and the time that a sample should be taken.

-6.25. In plotting Figure6.32. it is implicitly assumed that R,.> z.., And that H8> Zo. so thatboth rL and rs are positive numbers. Sketch a similar diagram for the case in whichRL - 0.5Zo and Rg = 0.5Zo.

8.28. Draw the voltage and the current reflection diagrams for the transmission line whichis shcrt-circuited as shown in Figure P6.26. Plot V and I as functions of time atz - 11/2.

Problems

he one

(6.58)

period

,verage

ximum~- 2 x

"-filled.51 kW.

,nsity J.::urrent5.3b).

'. Thenfor the

L =.:<:>1·

ilate [a], of the

ind thematch.

e at the ]·1

'Igur. N.26

a jl.57

181

6 Transmission Lines182

I-n

_r-iV'l = z, Jz.11--2--1

Figur. P8.27

-6.27. Draw the voltage and the current reflection diagrams for the transmission line that isperfectly matched. as shown in Figure P6.27. Plot V and I as functions of time at? - 11/2.

6.28. Refer to Figure 6.31. and let Rg = 220 and RL - O.57~" Draw the voltage reflectiondiagram for 0 <. t < 6T. and plot V at z = 3V4 Ior 0 <. t <.6T.

e.2e. Refer to Figure 6.31. and let Rg - 2 Zo and RI• - 0.5Zo. Draw the current reflectiondiagram for 0 <. t <. 6T. and plot 1 al z - 3R/4 Iur 0 <. r <.6T.

6.30. Refer to Figure 6.38a, and obtain an expression for 1 . Sketch L(t) and I(z) versus?for the time period T < t < 2T. The sketch should be similar to Figure 6.38u amiFigure 6.38c.

6.31. Derive (6.56).

-,~

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hr C&lt:lt(,'ot .li"e.., E~:sJY~I/{Q..k(,/ ..)J ~ I y,...".J .. 1.¥IXIO'Al1-IO·JJ'I.~~'" • 4j.57/' V0". #Ie ~tA.h. IY(·".CI41A)(i;;1.lyplI$,."x(·~.u7A)I. a.~S4-'Y.1••• 'Y.+'/'~""Jtl =1 Y(.8.4~7J..)1/2IY_' ~ J!ltl =IY"' .. /I~(~O.44'1A)1/2Iypl = "-5'7& x().6S'4/2e.lt;96 .y

:. p"'u = IY+I'l2ip - ( 11f? b >},(2lCS'O) = ze.» 1J.wv+:: SO~SD+S(»). Y2.} i:» V+/~O = I).Of, r.... ('10-so)/('J't>t'SO) -C.I(,", P; =0

T = .lIt! It: ~,(IO" $tG lilt) ~ (l•.I/~) f(.u.J_ 0::1/2)ItIO

V-c r... V" - o./~71t.(i):O.D8JS nLI.-- -II t: - -O.I'''1f.(O.DI)--O.CVJ/,r D t

6.22--

6.2.&' t-D

- ..(1....'_-_...:__2 :-" T.l/v

).,t

r:. :-1

t t

r

•• v;v; c v.T.:l..z-It s v-ls,» 2'1:

Iz

•(Hl~ (AtJ"'4',)

J;. ---------l' ---L-..-I- __ .J,-__ to "l)l T 'Yt :rr

23

('.2'1-~~----to T/1 T o~--:T.~lO:-~T:----t

",- (U.-i.)jCZi ....-l.)= tV. v.

V.,.C RS+i •• !••T't

$T

V.li.

v.-li.

24

annas

LJ-1_

thatfind

.49)

:l=

:.Itvaii

Problems 221

to Puerto Rico. This radio telescope system is called the Very Long Base-line Array. The angular resolution will be of the order of 10 9 radian,which is smaller than the angle spanned by a dime located in New YorkCity when it is viewed from Los Angeles.

In ordinary arrays, individual antenna elements are connected bytransmission lines. For the very long baseline array, such physical inter-connection of antennas is not practical. Instead, the signal received byeach antenna in the array is recorded on magnetic tapes which arc latertransported to a central facility where the tapes are replayed simultane-ously. The key to such processing is a very accurate time standard for allrecorded data. At present. time synchronization of the recordings isprovided by hydrogen masers which are accurate within 20 nanoseconds*.

Problems

- 7.1 Find the rectangular coordinates of a point P where the spherical coordinates are(r - 1.0 = 600• ¢ = 30°).

- 7.2 The rectangular coordinates of a point Q are (1. 2. -4). Find its sphericalcoordinates.

7.3 Show that \l . \l x A - 0 in spherical coordinates for any vector A.

- 7.4 Show thatlhe differential spherical surface element is equal to ds » r2 sin H do dr/l.Hint: ReIer to Figure P7.4.

- 7.5 To convert a vector in spherical coordinates to the same in rectangular coordinates. itis convenient to prepare a table for dot products between unit vectors in these

OK. r. Kellermann and A. R. Thompson. "The very long baseline array;' Science, Vol. 22f1.No. 4709. July 1985. pp. 123-130.

7 Antennas222coordinate systems. For example, X • f' - sin 0 cos t/J. as indicated in the fullowingtahle. Complete this table.

X sin 8 cos</>

-7.6 USA the table prepared in the preceding problem to express the following vector Alocated at (I' - 1.0 = _00°. </>= 45°) in rectangular coordinates:

A = 12i' + 88 - sJ_ 7.7 Show that the distance function Ir - r'] that appears in (7.7) and (7.8) can be

expressed in spherical coordinates as

Ir - r' IZ = 1'2 + r" 2rr' cos 'YcOS'Y= cos 8 cos (J' + sin (J sin (f cos(tP- .p')

where 'Yis the angle between the vectors rand r' and (1', 8, <1» and (1". (f, 4>') arespherical coordinates of rand t', respectively.

7.8 A vertical receiving dipole antenna at P is 15 km away from a capacitor-plateanten na that is also placed vertica lly, as shown in Figure P7.8. The receiving an tennameasures an E field equal to 10mV/ m. What is the va Ilia of E that the same receivingantenna will detect at a height 3 km above P? What must the orientation of thereceiving dipole be to obtain a maximum reading? (A maximum reading is obtainedif the dipole is parallel 10 the E field.)

~""",""-T__--- I

I -----I_---- I_--- I-~ --------------p~

Figur. P 7.8

7.0 The power lost on a cylindrical conductor that is ~z long and that carries I amperesof current is given by

Pnhm = %12R. ~zwhere Polun is the loss due to finite cond.uctivity of the wire. R. is the surfaceresistance given by 1/(ud.2?ro}, and d. is the skin depth. The efficiency of the antennais given by

Power radiated'10 - Power radiated. + PolunAssume that a short antenna of length ~7. has an efficiency of ten percent. Is the

efficiency improved. by increasing the length to 2 ~z while maintaining the samecurrent and, if so, by how much? Assume that the antenna is still a short antennaafter its length is increased to 2 ~z.

Antennas

e following

Ig vector A

'.8) can be

, 0', 1/>') are

acitor-platengantennae receivinglion of theis obtained

[ amperes

ne surfaceia antenna

cent, Is theI the samert antenna

Problems 223

y

x

..Igur. P 7.10

7.10. Consider the antenna system consisting of two short dipoles arranged perpendicu-larly to each other in space, as shown in Figure P7.10. These dipoles are driven by thesame amount of power from a common source. However, the current on thex-oritmted dipole has a -900 phase with respect to that on the }I-oriented dipolebecause of a phase shifter inserted in the transmission line that leads to the former.Find the total radiated electric field on the z axis. Verify that this antenna systemradiates a circularly polarized wave in the z direction. Is the wave left-hand orright-hand circularly polarized?

7.11. Find the expression of the total radiated electric field on the x axis that is due to theantenna system discussed in the preceding problem. What is its polarization on the xaxis?

7.12. A certain application requires that a field strength of 1V1m be maintained at a point1 km from an antenna loco ted in free space. What power must be fed to the antennaif it is (0) an isotropic antenna, [h] a short dipole, and (c) a half-wave dipole? Neglectohmic loss. An isotrupic antenna radiates an equal amount of power in alldirections.

7.13. Tho current at the center of an antenna is 100A; what is the E field 1 km away from iton the horizontal 10 - 90Q

) plane at 10 MHz if the antenna is (0) a dipole with hI -hl - 0.5 m, [b] a capacitor-plate antenna with D.z - 1 m, and [c] a half-wave dipole?

7.14. Show that if the radiation field pattern shown in Figure 7.4 for the infinitesimaldipole or the capacitor-plate antenna is plotted in x-z plans in linear scale thepattern is exactly formed by two circles.

7.15. Find the directivity uf (a) an isotropic antenna. (b) a capacitor-plata antenna, and [c] ahalf-wave dipole.

7.16. Find the radiated electric field of a linear antenna that is 3 m long r~= 3 m] and thatoperates at 100 MHz in air. plot its radiation pattern.

7.17. Consider a uniform linear array of two half-wave dipoles that are 1.5 wavelengthsapart. The currents on these two dipoles are in phase. Sketch the radiation pattern inthe horizontal (8 = 90") plane. Show clearly the number of lohas in this pattern. Also,estimate the beam width of each of the major lobes. The beam width is the anglebetween two directions in which the radiation intensity is one-half (-3 dB) themaximum value of tho boom.

Chapter 7224 Antennas

7.18. IQgure 7.23(bJ shows the array factor of a two-element array separated by 20>-.Findthe beam width (in terms of the angle between two adjacent nulls) of this arrayfactor near rP" 90° and 0 .. 30°.

(a) Use the approximate formula given by (7.49).(b) Find the exact value starting from (7.45).

7.19. Find the directivity of the two-wire transmission line shown in Figure 7.27 withradiation fields given by (7.42J.

7.20. Find the field pattern of a two-element array with d - >./4and", - O. Sketch the fieldpattern on the x-y plane.

7.21. Find the field pattern of a four-element array with d - >.!4 and ",-0. Sketch the fieldpattern on the x-y plane. (a) Use (7.37) to obtain the field-pattern formula. and [b] usethe result obtained in the preceding problem and in Figure 7.16 and the pattern-multiplication technique.

7.22. Write a computer program to plot field patterns of a ten-element phased array withd - >-/4 and varying phases.

7.23. A uniform linear array consists of 6 short dipoles. The spacing between adjacentAlements is ),,/4, as shown in Figure P7.23.

(a) What should the phase shift", be, in order to point the maximum radiation inthe 4> = 90· (that is,.9')direction?

(b) Supposo that the E-field due to the first element (the dipole at far loft) is givenas follows:

1000 -jJ.:r • 0E = -- e smBo r

Calculate I £81 of the entire array at point A(0,1000,0), point 8(1000,0,0), pointqo, -1000.0), and point O( - 1000,0,0), separately. All positions are given inrectangular coordinates in meters. Use the phase shift found in (a).

lc:) Sketch the field pattern of the array in the x-y plane.(d) Sketch the field pattern of the array ill the x-z plane.

x

Figure P7.23

-I 1-),,/4

y

8.1Raleig

CHAPTER. '7

. " e ~r~ s,,,eC.C)tP Cl>S6u,t1 -'n,<I-IC,.

'Ds8 $;,,'1y ~;,@J;"f/j '-cst/>s U>~6 - Jlhe 0

?6 Ax'" x·A =' 12 S,,.O'~Hi .,. 8 cos () ~st/> + s- sil'td> :: /:J. S;,,(-ttP)lDJO;")'" 8 ~S(-W)(DS(4$D) + ss;"{4J~)1I: <a. ((B-1)A~.. Y'A :. 12 .s/~ 8P",,¢ + £3 (.$IU;,.,4> - SU>S'f - /2. SI;' (-")Si .. (4~)+- 8ust-l.oO) $1""Ur') - St.Oj (4sO):: - 8.oS'~

A). ~,,4 :. /3us8 - 9 $,;'8 .. I~("S{.-/;DD)- tJJ, ....(-6o-) - 1.2..93

: • .4= ~(-o.~e4-5)+;(-8.oS-6)"l:I'.93

I F - r ' 1:Z r: ) r It+- I r '11.- 21fllr" COS 'Y= ""~r'1-zrr',os1

Cos1 = r. r.':: (; S,;'8,osd> + Y s/ ....es/~;. i'osB)' (£ '()5S'~S<P'

-+- y $,it8' II;' tfJ' + f ,DS 0' )= !.,;"9(.oscl S/~f)'l..Dul -to S",I} ', ..4:>$I""'S;"~' +c..os8 lOSS I

""' S/~~ s,.....8'(c..ostl /.()H/" .;. J/;' t$ $i,. t!>') +{,.DSO~S~ I

::: CO~8l..DSS' + S,'",,{) s/" g' '''S (f> -~')

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m.../(~,..(,(,.....rt..~,'''''' ;5 obi.,J."qa.. W/'VI.., oli.p~lf-"",..k t s II. 3/ 0 I~,... Vet" tl ,tJ...

(r;e,I')

II Pru."" (.IJ'j)l j PoIV.,"" (4})

o.J JV'lg-H1. 4) J I It.A. ::: 0.1 ... PrM/( Prr¥.+POIvn) ~ Po",.., - , p~at .I~II... 1.1J, 'tIL." 4 Prt>-d./( 4. Prt>J.. + J ePrAlL) .. 1=£ - Ig. 2 %

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ff,.:T. J~~:1.4J} e-JkJ f-X) =- J1~14l)e-jll~(_j)(_;)tWt. +0 X 4i,r(d~ ~f~J.....g=§,.,.~z::: .iB~VjdVe-ii) (j;-y)S/~t. I~,I:, I~'II_I tVtJ. 4'::. '10°" ~ IJ Nt.-IJ(1"tl c"rc.u!A,.l4. polA.l'iJt.d..(g, I I~~I ~

25

2:.!f_ On -Me. l'-tU''',s d.u.e.fr::, X orie,.,teA ~()lt... J €,0:60due. -1-0 Y orie,.,t..~ ~4J (, = J..'J£X;I.) e·;'/(x(_y)

- - - A jItHl~J_} _·ltx .g" £, 't fz ==(- r) 811"7( e 'J J J.,.:,,#.ArtJ f"lA"""JU2:!2 (0.) 1(~>llC/+Rt.[§)(.B"ll = 1~/§12.= ~1T

P» ul()rr~41T'~(JOOO)1. = /(.,.'7 KW

(b) f,,~t:J.. sJlort d-'p",,-, IE/='7 ~~~l :+ 1!::S&/§(r.:/xlt>oO:::./Ooo4TTIA1.dJI% 4"'(I*I~l/)1 «n »,« t 4" zP=T1 -:nr "'Tt 1 "'i1r ::s J1 (Islr) • ~1«IOOO) = 1/., KW

(C) FDr t:J.. /,o.ff-wewe.. ~po4..,I<S>I-= talorr ao:: ff;,'1~1Co ;:f~;!;t:::;.r:= 182~';~b'l:"1T=27?'78p= :r;.8'1';'Ift(zn) = z,)?·'1:::~D1f"1I2.4J7('= IO.J'''W

3"11.10''1.13 .,c"IOMI4~" Ac -;;;;:;or: 30m

( L t ":--fA. I=IOOI"=S"oA Il./la. IiI: 1R_ItJA = I.lO."..x(2"nD»)(SO~(D.s+O.S) = "3JA V/A) ~nor ~ J 1Co _ 4-rrr 4TrX '000 v. ¥- /WI

(b) ('c.pa.£,+or - pl",-~'_a.nie"n4J I =10011 11J\d.. I€ 1= 0.:114 X 2 • 0.618 VI",

f J. /I IEI"..!L ,OO.IZOTT v/.«(') hAt -wa.ve 0-4.P01G(.) - 'Z1Tr c ~n"looo'" 6 1m

7.14 (S > ::. r f [1~lJZSi,..20 = r A S/'Io 1f} ~ /I(),.,.. .. l..i.JeJ. < s> _ P. 5,'",19() 10· '*-10· 120· tao· -i40° tSo· i~O· 1:70· 1"80· t <10·

S;,,"tJ 0 o.c~ 0.17 0.1.5" o~, ().s</ o.»s 0.99 0.97 1.00)

III/

/"~'"_-:!:.!.§. (a.) D = ,

d. =1. SA, ~d = !ftl,.s-A,. It 37T J 8= 90·, ~s1''&.si"t/' j "'-2; "1/-==0I F<t/'>/ ::: J (.oS (;,J.':_SY+1/;) I"" I cO.$(·¥ S,;";)/I f(4)>1 =, A.t 4> = 0° IJr1d. <i = 41.So -JdG· 20b) (%_11) *IF(~)I = 0.7()8 =/co~(¥s""1')J ~ ~1T"s;"t/:c~ • <l',CSI'1..-'(-{;)=t.t.·

26

A:: :l o~ cos300 ::: J.3 ()'0 ·F(tJ,; ::f~·)= :J./e-oS(:20Tr$,';,8)/

4t 8= 30· • F ( lot, f o· ) :: ~ I (,,0S (10 11: ) I =.l (~,. ,U': ""w _ )

)1",(/ l'\f~" 9:: Jo· C~C~"J <t CO((:H1[J."tt~)=COJ(I071:-11C): 0

.2 0 It $,'" e, .: 10. s iT --:) B t " 31. '6;'2 0

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A -:: 9-, - 9" :::: 3. 30fl' 0

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1/( 8~ (kbr.l]

= 4- Si",l¢ cb~Z{fco5g)

'!.2.0 d.=N4J ~d. = 0/2. j Be qo~ C/J$ r ~si" ¢ j N::Z j 1/'==0/FUM/ = I CO)(I S;n~) I

4> o· sro" .1200 t;30· tilo· t 5'0· t~oO 170· teo· 11100IFftl»/ I 0.99 0.9' o.9t D.se 0.81. 0.78 0.'1~ 0·72 0.71

'f.'ZZ d-Jy4-J

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28

x

7.7.') (e:...) ~=_v/-t .(000 _j~r ,s ..n[i.,,(51·.PJ ......;-I))

(b) £6 = e J.'. , l17 r «:...[i7r s:"CP .....,_I))

a.t A. f= %, ct- "/1. I ,..~ /OfJO, /£11 = ~ ~~.t 61 f-= o~ 9=%, r~ 1"°01, 11i~/:f.il-I'I ~'"..t (, f ~_V~I e-:.~ . Y'"= loolJJ IF91= 0

~t 1>, f=-tr, 9=~. r« /DO." 1£.!=-/.I.JI(t vi"",

(c) l1 = nil. II£~ 1= loti" )(. / S"lI[f1T(~."i-1J J /r . s....( frr( s,...t;-OJ

J ~ .t' 7 If""

Ii I~IJ/

29

252 8 Topics in Waves

Figure 8.17 shows a typical arrangement of a liquid-crystal display.* 11is operated in the so-called "distortion-of-aligned-phases" or DAP mode.Figure 8.17a shows the normal state of the crystal before activation. The lightentering the crystal is polarized and then transmitted through the crystalwith no alteralinn in polarization. The second polaroid absorbs all the light,and no light is transmitted. 10 its activated state, the crystal changes thepolarization of the transmitted light. which propagates through the secondpolaroid and becomes visible.

Problems

8.1 The derivation of 18.5) only considers the electric field. Why is the magnetic fieldneglected? Hint: Cumpare the magnitude of E. with 1)H" near the sphere. or thestored electric-energy density (1/21~ IE I~with the stored magnetic-energy density(1/2jlll HI··

8.2 Wby is the rising 01' setting sun red?8,3 The smoke emitted from engines of boats contains fine particles. Against a dark

background the smoke looks blue but agains! a bright background it looks yellow.Why?

8.4 Explain the appearance of shafts uf sunlight through breaks in a cloud-covered sky.

8.5 Shuw that

II - f: e'~ dx - ,fi

Hint:

I~- f.' e' dx . t: e-I'" dy

Then. transform X-)' coordinates into cylindrical coordinates to perform the exactintegration.

8.6 Show that

t ,., .J.fir (t/)I~ - • exp( - tr«: + 4X) ux - p exp 4pz

Hint:

2 2 (q )2 q~-p x + qx = - px - - +-.2p 4J)"

Then, use the result obtained in problem 8.5 after transforming the integrationvaria hie from x to px - 4/211.

8.7 Assume that on earth a microwave beam of 10 GHz is radiated by a zo-mctcr-diamp.tAr disk antenna aimed Citthe moon. Estimate the size of the microwave heamon the moon.

"See R. W. Curtler and C. Maze. "I.iquld Crystal Displays." Il::t:(; Spectrum, November1972. p. 25.

8.8 AIf;

d,re

e,g 0hifethclte

8.10 As~Sf

::s in Waves

Iisplay. * It·AP mode.I.The light:he crystalJ the light.ranges thehe second

gnetic fieldere, or thergy density

ins! a darkoks yellow .

•vered sky.

1 Ihe exact

ntcgratlon

20-mcler-..ave haam

~ovember

Problems 253

8.8 A person leaving his home by train mails a letter home every day. Suppose that thetrain travels 200 miles per day and that the mail moves at a speed of 200 miles perday. How frequently do his letters arrive home? Try to solve this problem by simplereasoning, not by substituting numhers in some formula.

8.8 On a foggy day, the driver of an automobile stepped at a railway crossing hecause heheard a whistle from a moving train. The sound of the whistle came from his left. Afew seconds later he heard the echo, and the pitch of the first sound was lower thanthat of the echo. Assume that the echo was due to reflection from a nearby mountainclose to the track. If you were the driver, would you cross the track-that is. could youtell whether the train was approaching or leaving you? (See Figure PS.9.)

8.10 A Doppler radar sends a signal at 8.8UOGHz, and the receiver displays a frequencyspectrum of returned signals as shown in Figure P8.l0. What CClIl you say about thespeed of the targetls]?

I'Amplitude of the returned sign ..1

10 kHz 15 kHz- Flgur. P8.10

rrequenC"ya.aoo r.Hz

8 Topics in Waves254

AbsorptionaxisIII Passing

axisRandornlvpnlllri7.pclIIgltt

~~/A,\Jl

~

~Z ~/.;)--

Ctobserver]

Figur. P8.15

8.11 Fur the FM-CW Doppler radar discussed in Section 8.4. aSSUJll~ that to. the upperfrequency of the rad a I'. is 8.8 GJ 17..Suppose the radar is to measure target speedsran~in)! from 0 to 3 Mach and distances from 1 km to 10 km. Find the system'sapproximate frequAnci bandwidth and the time interval the system must be ahls toresolve.

8.12 If d - >'0/4, as shown in Figure 8.JGa. and if reflections at interfaces z - 0 and z - clare negligible. a linearly polarized wave incident from the left will become acircularly polarized wave, as discussed in the text. What is the polarization of theexiling wave if the roflections at these interfaces arc not negligihle?

8.13 II d - >'n/2 AS shown In Figure 8.16a. what is the polarization of the exiting wave if theincident wave from the IAft is circularly polarized?

8.14 For a quartz crystal. I, - 2.41 to, and ( = 2.aSEo. Find the minimum thickness of II

quartz quarter-wave plare for a li)(ht having>. - 6500 A.8.15 In FigurA PB.15 the Polaroid film at A is oriented such that it passes light polarized in

the ic direction and absorbs light pulnrizcd in the 5' direction. The film at B passesy-polarizerl light and absorbs x-polnrized light. A randomly polarized light source.such as a flashlight. sheds light from the left along z, Can an observer at C sec thelight? Explain.

8.16 Consider the orrnngemant shown ill Figure P8.'lo. This figure differs from FigureP8.15 only in the placement of a third Polaroid film at D between A and B. Theabsorption axis of the third film is 45n Irum either the x or the y axis. Nuw, can theobserver at C sap. thp. light? (If you do not believA in your answer, do an experimentwith three pairs uf polarized sunglasses and see for yourself.]

Figure PS.16l1fjO

I ;I II II

7.

(;Iouservcr)

9.1Ele

8.'1-

j

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81la. L411..,~;~ St:.A.1Ilr U ,"()r~ s-h-eJ"Jb:J ~an. rU ~kt.

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:. I;: 1"e -;r.'1eJx.:& Iff_. t " j"

1410 f·lx'.tJC) flit> -(Px-"l #1I, = e ~x. = e .e ' AX =

-- -til>

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10"/0 Z -hw2. vrw1. sr 2. ~ 8

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_', W,OIofI....: WZ/2j = IOX.3.SXIOS;/'()I.'7tc.lo4-: 3.(PJ)(/O~ co; .1"3 It",el/er~ oHlu- dLJ

-- I " I I I I I I I

.f, = 1.-.1~.".Yof ,..~ ~,r-sc JOU"'J..fa. • fUlu~l1C)' 4'" HJ~ ~"u:>

.'. 1f'6':"" ,-S n.el/,,,) ~....,ord. -tJ~ I;'T~'J~d,"" .

g.IO rhu<.. c.r~ SC.WH'''A-I. -l4"'J~lJ :

TtU·'d."t # I: if.:; ~J, .-=- c -1(/" 3./08

_ ../7A. ~ ,.,/~ r -Htc rece! eroJ I -E.. Z i.i"'IJ'''Z - v. ~ /1. I l10WAJ .,.,..",... III.

30

u,

T

AI .. L (:U.J/,) .. 8.8)(10-'111.J( q93 • .t""O ~ H

:To 31<Ic' ;;)02$"w"J

arIe - 2)t le'/ 3)( IcB:: /, . .., It. 10-' 5 c" 7 ;4($

,', 04.....Lt.J.·a_-IJ....= 5~ 1:11) ; r;-,." ".u~futf.·_ ; /,tM - ~.7,.uS .

.!..!J_ £11,p·hcQ..((, ptJfA"iJ~ wt:U/t!.

8.11

c ,.r.cu LA,. ~k."'J-'- hu.t opPOJde. 1,4" .

%. % A ' ._.' 'SDoA·d. = "c.- ~.. e &o/h Crt;. -If) C' ""(Ii,. -Ii) .+(/iH- {Lii) C ''','1(, P""

8,1S' NO.

BJft, YES.

31

286 9 Electrostatic Fields

Problems

g.1 Consider the dipole arrangement shown in Figure 9.20. Let q = '1.11X 1O-1Y C. Find <I.>

at:(a) x-O.l,y=O.l,Z-O.1(b) x-l,y=l,z-l

Use the exact formula (9.14) first. Then use the approximate formula (\1.17),and findthe accuracy of the latter, The medium Is air.

e.2 Three point charges are located on the x axis with q, - Q at x - 0, th - 2q at x = 1. andq, - -3q at x - 2. Find the position(s) on the x axis where ~ is equal to zero.

e.3 Four point charges are located on the corners of a rectangle, as shown in Figure P9.3.Find the planes on which the potential is equal to zero, Sketch these planes.

y

-2q

2m I.. ..q ----+TI

I s m

II

y

>x+11 -Q x

Figure P9.4

9.4 Two point charges arc separated by 0 meters in air. as shown in Figure P9.4.

(a) Find the potential function lx, y, z).(b) Calculate at x - 1000, Y = 100a. z - O.(c) Show that. for distances much greater than U away from these charges, the

potential is approximately given by-Q 1

,.,------~4'11'Eo(XZ + y2 + r)1I2

WhArA(x~ + y2 + Z2JI/2 >:> o. USA this approximate formula to calculate W(1000,1000,0), and compare it with the result obtained in [h).

g.5 Find the E field in air due to a point charge of 106q.(Q. - -1.6 x 10 10 C). Sketr.h adiagram similar to Figure 9.4.

9.6 For the charge distribution given in Problem 9.2, calculate E. at (aJ x - -1, [b] x - 0.5,

[c] x - 2.5, and (d) x - 3.9.7 Apply E = - 'Vto [9.14Jto find the E field located at the origin and prod need by two

charges +Qand -Q located at ro, 0, l1Jand (0,O. h], respectively. as shown in Figure9.2a. where h = 1 em. Show that E - -!(q/2'11'Eh%j.

9.8 Solve the same problem as in 9.7, but use (9.19c).

g.9 Skelch the direction of the E field located at the center of a square shown in FigureP9.9. The E field is produced by four charges at four corners of the square OABG.tbese four charges carry Q. q. -Q, and Q coulombs, respectively.

Ic Fields

:. rind 

and find

-1. and

He P9.3.

:es.lhe

), 1000.

ketch a

x= 0.5.

by twoFigure

FigureDARC,

Problems 287

y

Flgur.pe.e

x

8.10 A line charge 2h meters long is located along Inc z axis as shown in Figure 9.6a. Thecharge density is Pr coulombs per meter.

(a) Calculate the electric field at f> - O.lh. cp - 0, and z - U using ths exact formular~.21J.

(b) Calculate the electric field at the same point using the assumption that the line isinfinitely long.

(c) Find the percentage error of the value obtained in [b] as compared with the exactvalue.

D.11 For the same line-charge described in Prohlem 9.10.

(a) Calculate the electric field at P - 20h. cp - 0, z - U using the exact formula.(b) 1)0 the same using the assumption that the line is a point charge at the origin.(c) Find the percentage error of value obtained in (b).

8.12 A plane charge of p, coulombs per square meter is located on the x = 0 plane, andanother plane of -p, coulombs per square meter is located on Ihe x - 1plane. Findthe total electric-field in the region (al x '> 1.(b) 1> x » 0, and (e) x < o.

8.13 Consider the problem discussed in Example 9.11. Assuming that everything is thesame except for the Iact that ths total charge on the conducting shell is now equal tozero, calculate E everywhere. and sketch E, versus r similar 10 the sketch shown inFigure 9.15.

9.14 A charge distribution of the following Iorm is set up in air [spherical coordinates):

(

0 O<.:r<.:o

o; = 10-R U < r < h

o b< r

(a) Find the 1J field for 0 < r <. c.(b) Find D Ior a < r < 0.(c) Find 1J for b < r

9.15 A charge distribution uf the following form is spt up in air:

P" - 10 " . e " coulombs per cubic meterUse Gauss' law to find the E field everywhere. Hint: To find IhA total charge in aGaussian surface. yuu must rio the integration because the charge is not uniformlydistrihuted f lowever. symmetry still exists with respect to t/> and O.

9.16 Electric charges are distributed uniformly in the region 0.1 < x < +0.1 with densityP, - 10 6 C/m3• Elsewhere. the density is equal 10 zero. Ftnd the E field everywhere.

9288 Electrostatic Fields

Plot E. versus x. Find the potential difference V. - Vo for a point x with respect tothe origin.

9.17 Find the potential difference V A - Vo for two points A and B located at r = 0 andr - 1 in the E field obtained in Problem 9.15.

9.18 TIIIl solution for the electric field of an oscillating Hertzian dipole with angularfrequency w is given in (7.14) as-follows;

E - {; jkI ::-'hr {r [;~r I (i:r)2]2 cos 0 + 8 [1 + ;~ + (j:rIZ] sin oJ

Derive the solution (9.20) for a static dipole by selling w - O. Notice that k = w(lJtfZand I~z = apia! - jwp.

9.19 In the electric field E - 3:i + 4y - 5i. find VA - Va if A is located at (1,1,2) and 1J is atthe origin. Does the difference depend on the path of the integration?

9.20 Consider the spherical-shell problem shown in Figure 9.14. Find the potential <Il( rJ at«t) r = c: (b) b « r < c; (c) b > r » 0, and (d) r - o. Assume <Ii - 0 at infinity. Plot (rJversus r.

9.21 Repeat the preceding problem for the case in which the total charge on theconducting shell is equal to zero while all other conditions remain unchanged. Youmay want to use the result obtained in Problem 9.13.

9.22 Consider the coaxial line shown in Figure 1-'9.22. The inner conductor is a solidconducting cylinder with a radius equal to 0.1 m. The outer conductor has an in-net' radius equal to 0.4 m and an outer radius equal to 0.5 m. The medium betweentha inner and the outer conductor is air. The inner conductor carries a net chargeof - 3Eo Clm and the outer conductor carries a net charge of -lSEo elm. The sym-bol '0 used here represents 8 constant equal to 8.854 x 10-12•

(a) Find E" in the region 0.1 m 0.5 m.(d) Find atp; O.2m,knowingthat - Onlp -1m.(e) Sketch Ep as function of p for 0 ( r] atinity Plot (r)

harge on thechanged. YOIJ

:tor is a solidtor has an in-lium between; a net chargo1m. The sym-

lp.

Problems 289

9.23 Model the dome of a Van de Graaff generator as a conducting sphere. The dome ischarged to hold the maximum amount of electric charge Qm before the air sur-rounding the dome breaks down. Use the following data:

radius of tho dome - 0.11 m.breakdown E of air = 3 x 106 Vim.

(a) Calculate the maximum Qm accumulated on the dome just before the break-down.

(h) Calculate the voltage of the dome ill reference to the potential at infinity justbefore breakdown occurs.

[c] When the dome is charged with the maximum charge Qm. a person uses a con-ducting rod to discharge the electricity. Assume that the discharge takes 0.01seconds to complete. how Iltrong is the discharging current (on the average)?

~ ~ ,'.0-1 -+--""1 f· ,I I

· '.1-0.-t-.,-· ,! :

32

v;)

tfl·J·4l4

'I

q.1.Q (11) f = D.I h • .I-() c +M-'( "'/().II-.) • 84.S·-. A PL • .)_'" & ( )e p Z11Uf;.Jlt) S",(84.3 - f zre» 11.115

(b> FDr i",h"j.f./y b"J l.i.nl. J ~., 2.7rIII).,h) cp 2:i" (/0)

(C) En-or = IDq.::.s- ~ IOC"_ c o.5"03 ~

q.1I f ..204 • tf..o -= 'ku\'( 11/201-.)• Z.8'·) );:" P4 '( 8'-) ~ P, ( "It)(el t: = f z"E ('oJ.,) S'" 1. C t" """Hi 0.04.,.,

(h> po;"t Cllo,.St. App~xi.,"A~t't>"} S..~ f..p"J, t - ~h~~Jj (O.D.5')r f.7T£(201t) r _'7r~,

(C) ,.. .. ().(JS-O'~f9.f)CIO"" c a".~~,,...,. Q•.o~Q' '..6> ,.

~ (.. ) ~ eO 10,. ~>I(b> E;: 2 (Ii ~• - ; (4) f.,,.. D< X e I

(C) 'l:0 for' X <,0 t,. .\,\

"

'/.9- ,~------;~-,, / I

"'~I" I, I

, ", , I &o .' ~, -I

--t:x~.~o-+,a:-.,--"

9.14 (a.)./.,. 0< r « (l. IS. 0 (~ih(1! "'" f" 'nc./D.ed)

(b> f<,,. a e r e b , (4-rrrl)0,. ·r"fvd.~ -f". ~7r(r3_1).1)- ,. ~ (r'-eJ.·l ..,

., D z:: r 3' r t I 10);11.. fv. If)

(c) .f",. he r-, (4-rrr2.)(),. a f. fvd.r z:: fv i" (.3_a!), J ... '0 F Pv (b - a. ) .~ ~ O-it.. cT' , I I WI......, ( V .. I

9./~ App1:Jt;.w~$' ~: Er(411r'l.) -+. {"tv4v.::i f:foU'L1T(,O-'~-") rJs,"" 8"rdld~

r .. :: 10""!:!!..f. r'e-rdr-e_ •

IC-' r f{r~ -, >J,.. tr:::Wfo rt.e-rdr~ ~ri[-t-"(r'+'r+2)]: r::1:fl[2.e-,.(,.I,nr+Z)

9./ftJ tor' /'XI <o.t J 71~t(r>,,)-TT/"(-PJC) = 21T1. t(DIt.). flv "J.t.(z~).'. p)Ccfvx. E-f~x liIAI",e (v·/()-'

~,. ~ )D., J ~rrla'l.(I)~). fv 11';,'I.(fJ.l.)

.', 1),,1: 0./ Pv ... E· ;-9;(0.,) ,whue f,,:lIO-'

/0,. 'X<-,,., I 2n;,'(-l>Jt). f" 7f"I(fJ.'l)- ,. (*" I.A_~ ~ A-".'. D~ 'I: - C.I p" 9 ~. -x r.(D.I); -.JIJU"~ I V ·1.,

33

fo,.. -x e=o.t VZO·- r:te)l~x ·-l-A'-t'''Ztlx -1% - '0'';.&4''). _. 1).1

:-t{O.ODso - C.I'X' - ".~ /J. [v-(4./~.4.4DS)I' (z (1f. fir ~v Xl.,.~,. 1%1< 0", V~O·- J. kzd% =-J

l1t;XdX --T.T

lor 'k: >0.', V".: -J.7.EzJ" c- f:/~xtl't- fit:.!;;(4I)tI~~f 1 ~D-T. O.ODS'+O.'X-(J.DI --r.(o.""r-(J./z)

'1.17 ~ -Va =_fA E,d.r c _JD 1°-:[:J._~-r(r1..1.r-+ t)J at" ::~"'{!'-;4r-I'e·'dr-21'(L~~'j'rJ- ~ I • r '" 0"), Jo r r

e s,;',c. 'tA&r~ I~ A s"":JtAlkvi.~ J, f"&D I ~ ~w .. r- ,tu...~ "I 'tALA E~r~l) UtI41rJ.;dnS Yo b~ ~ lIoIiH.w ~ PUM·IIt. aN4 icc I.~ .", r' e·" sz: (, .,..e-r. e·'"cr·o) .. VA-V6.t:o~ lIs +,~,.-~e-"i1I.r-zJ,(-;:-"',. )drJ I ~J,4.,.t;l~ +7f)dr·-T

:.~~,,;'f.'f -t ..t+f-e-~. f -fe-S J .J,~ttf se=-t -e-~..f(I-li$)1D 1£:''[3e-'-2 -I +i~ Z('~~')J J = t:' t 3e-'-, j

JJ..!. E = (7I_.jAOl4Jfl e-ilar{fr..-J.: .. --LJ2'O~{;.3 r,...±+ -L] 5;"81- (T 4"'" lJRr (j~r)" (., j"r (j'r)&. of

= ft· 4~:~1.~~2e·jll,. [; (j~r-+ 'J 2cos8 ...§[(jRr)t ...jkr+'J sinS 1= ft· !,,~1::rfr[;kr+'J ZeDs8 + B[(jkr/+j*rt'lj Silt, J= ::;~: [F(j.r.'JZ<As8 .i[(jltr)I+JJtr+IJ $/,,'1 "'-0 oIlirJ [;2'OJ9+is/~8j

4-0

f/.,q VA - Va cs - f: s.a, - J02s.a, -J~:.,,13J where E.3~+4 ;-~i=-S·I-(·S)·2 -4" = :3 V

v« '€ E 0 ~ VA -V8 is ;"t:I~("tJt.."t oj 'VI' po.~ "f-Me. ""f~.Jf'tLUI}",

4JTficr)

Q.21a/.$D" g,(r>e) = -J: 1:l:,I1.,.· 4~~~

Le't 2'&0 ~ P,..JICAIt 'P.ID

fJ!il')-L.J. ..J. ..... _ ..... ~ to

34

--r

-.3 •. / <r< 0./1.~. 2Z. (,,) f, = J.lff

l~ ) 'r : a ,.~<,< ,.r(C) tf = -l-/~ -, hY f > «.s:

)'If • "r(d ) J 0. t~( •. l)=_ fl'tJ./

I

f,·r -f ' 'I ••l -Jtlf -f' 0 tAr -I Af-_ Tf ~trf, •.I •.¥

=- J_ J. (I.~ ) + .L.J.. ( •.S' )7T ~17

- 2.H· V

l~ )t fr'f

,)

1

I 0.5 '.0 ,-,-I-,)

-II-

-t

r.:~3 &W\ , ..n.- t( ) f~= E.= 3~ 10 A = D· II ) E = 1./.1'1 ~ I ()47Tl (t.~ ,

l~) :: ]30 -h VV=-r &II

35

332 10 Electric Force and Energy

This electrostatic adhesive surface is widely used in desk-top calcu-lator-driven curve tracers. A typical voltage used to charge the embeddedconductors is 300 volts. and typical spacings between them are approxi-mately 2 mm. *

Problems10.1 A point charge of Q coulombs is located at the origin (0. O. 0). and a second point

charge of Q' coulombs is at (1. O. 0). A small test-charge is placed at [3. 0, 0). and it isfound that the total force on the test charge is equal to zero. Find q' in terms of Q.

10.2 Two identical small balls are attached to weightless strings 15 COl long. Each hallcarries 10 9 C of charge. and eoch has a mass of 1 ~. They achieve an equilibriumstate under the influence of electrostalic force and gravitational force. as shown inFigure Pl0.2. Find the angle a. Hint: a is sma 11.

Flgur. P10.2

+Q

10.3 Consider a long line-charge with fl' - 10-' C/m. Find the force acting on a dustparticle carrying -10 9 C, 1 rn away from the line charge.

10.4 A line charge with p, ~ 10 fi C/m is located in air at x - 1. Y - O. A plane charge withP. = 10-6 C/m1 is located at x - O. A positive point charge of 10-9 C is at (112, 0, 0) inrectangular coordinates. What is the total force acting on this point charge?

10.5 Charge is uniformly distributed in the spherical volume r oS a with PI' - 2 X 10-6C/m

l

and PI' - 0 for r > u.

(a) Use Causs' law to find E for r :s o.(b) find the force acting on a I~sl charge uf 10 I~ C at r - (1/2.(c} Is the force obtained in [b] to be changed if the charge distribution extends to r =

20 instead of being limited to r -~ o?10.6 In a seed sui-ting machme, ulluesiroble seecis are deposited with an electrostatic

charge while they pass an automatlr. colur-sensitive or size-sensitive monitor. Thegood seeds are passed uncharged. All seeds are dropped between a high-voltageparallel-plate region to sort out the undesirable seeds. Let the charge on theundesirable seed be q. its mass be Ill, the voltage batween the parallel plates be V.and the plate separation be J. Assume thai the seeds enter the parallel-plate regionat velocity vo, and find the displacement y of the bad seed as a function of x. FigureP10.6 illustrates this situation. Consider only IhA trajectory inside the parallel-plate.

*P. Lorrain and 0 R. Corson. EIClt.'ll'omogOl!lism (San Francisco: W. II. Freeman and Co,

1978). p. 189.

d Energy Problems 333

p calcu-

IIIV (volts)

iheddedapproxi- 100

+ ~y

0

-I

+[ruilllseccnds]

+Figur. P10.e -lOU

- I T

nd point - Figur. P10.0and it is Juf 1.1,

.ach ballilibrinmhown in

--3I:m_1,- . 41110V

1r ---t-"""",=::;;;:: --Js. ~~22Z2ZZZZ:zz:z::!:Z:::ZZZZ3 0 V

Pigur. P10.11

a dust

10.7 At room temperatura (ZO°C) and standard atmosphere. what should be the size of thecorona wire if b ~ :l cm. V, - 10 kV. and the roughness factor of the wire is equal to0,8? (Refer to Figure 10.4,)

10.8 What should the lowest voltage on a Van de Craaff generator be in order to have itproduce corona on its surface? Assume that F:c - 4 x lOij V1m and that the radius ofIhA metal sphere is equal to 0,6 m.

10.0 Refer to Figure 10.1. If the voltage applied to the parallel plate is the sawtooth signalshown in Figure P10.9. find the locus of tht: electron on the fluorescent screenlocated at x - 20 cm.

10.10 For the cathode-ray tnhe shown in Figure 10,8. what should the voltages V. and V, bsin order to make the electron beam trace a circular path on the screen at 60revolutions per second? Assume that the vertical and the horizonta I deflection platesare identical.

10.11 An electron is accelerated by a difference in potential of 1 kV between the anodeand the cathode. It enters the parallel-plate region with this kinetic energy, Itsvelocity makes a 5° <Ingle with the plane of the parallel plate at the entrance end. <ISshown in Figure PIO,l1.

(a) Find Vo. v,.. and Vox at t = O.(b) Obtain two equations for the coordinates of the electron [x, z] as functions of I.

Note that x. = 0 and z = 0 tit t - 0,(c) Find the position of the electron at the exit end of the parallel plate.

gewith0,0) in

; 10 r-

ostaticr The-oltage10 thebe V,regionFigureplate.

x

m.-9.11)( 10 J1kgQ, - -1.60 x 10 I·e

334 10 Electric Force and Energy

10.12 Consider the ink-jet printer shown in Figure 10.10. Define

qd - charge on the ink dropmd - mass of the dropVo - deflection-plate voltaged - ddlt:ction-plate spacing

Vd - velocity of the ink drop at entry to the deflection plate~dP - deflection-plate lengthzp _ distance from the deflection-pistil entry to the print plane

Show that the vertical displaoernent of the ink drop is given by

flrl VO~dp ( 1)Xd- --.,-2 Zp - -~dpmrluvd 2

10.13 Find the capacitance of the spherical capacitor shown in Figure 10.13 by using (10.42)and (10.50). Start from

E - ~ r for b > t » 0411'Ef

and show that your result agrees with (10.47).

10.14 Find the capacitance of the cylindrical capacitor shown in Figure 10.14 by using(10.42) and (10.50). Start from

E _ ~ p for h e- o » u211'fP

and show that your result agrees with (10.49).

10.15 Consider the parallel-plate capacitor shown in Figure 10.12. What is the maximumcapacitance one can obtain hy using mica as the insulator? Let the area of the plate be10 ern" and the voltage rating of the capacitor be 2 kV, with a safety factor of 10. UseTable 10.1 for the value of ( for mica.

10.18 Consider the cylindrical capacitor shown in Figure 10.14. What is the maximumcapacitance one can obtain by using oil as the insulator? Ta ke a - 1 em, h - 2 em andthe voltage rating - 2 kV. with a safety factor of 5. Use Table 10.1 for the value of e foroil.

10.17 A parallel-plate capacitor is filled with two dielectric materials in a configurationshown in Figure Pl0.17. The total area of the plate is A. (a] Find the capacitance C interms of A, d, f,. and f2' (b] Suppose that the positive plate carries Q coulombs ofcharge. and find Q, and Q2 in terms of Q. where Ql and Q2 are charges on the 1eft-and on the right-hand sides of the plate, respectively. Neglect fringing fields.

10.18 Consider the capacitor shown in Figure Pl0.17. Let fl - 3to. Cz - 5f.n,J - 0.6 mrn, andA _ 20 ern", The potential between the plates is 300 V. Plnd the total stored electricenergy In this capacitor.

I"lgur.P10.17

wi;! w/2

Forceand Energy

13 hy using 110.42)

re 10.14 hy using

is the maximum'ea of the plate befactor of 10. Use

is the maximumem, h - 2 cm and. the value of E for

1a configurationcapacitance C in5 Q coulombs ofarges on the left-ing fields.J = 0.6 rnm, andal stored electric

Problems 335

10.19 Find the capacitance per unit length of II coaxial capacitor with two layers ofinsulating materials, as shown in Figura to 15c. Express CI I, in terms of 0, b, C, {I' andEz·

10.20 Find the capacitance C of 8 parallel-plata capacitor with two layers of insulatingmaterials, as shown in Figure PIO.20. Express C in terms of A (the area of the plate),dl, dz. El' and E!.

'igur. P10.20

10.21 Refer to the capacitor shown in Figure PI0.20. Lel E] - 3Eo,~z = 5Eu, d, - 0.3 mm, d2 ~

0.3 rnrn, and 1\ = 20 cm '. The voltage across the capacitor iii 300 V. Find the totalstored electric energy in this capacitor.

10.22 Derive (10.t!2j.

10.23 A parallel-plats capacitor cernes +Q on one plats and - Q on the other plate. Thearea of each plate is A and the separation between the plates is S. The medium isair.

(a) Find the total stored energy V t,; in this capacitor in terms of Q. A, S and '0'

(bJ What is the clcctrostatlc force Acting on the plates? Is it attractive or repulsive'?Hint: find the change in UE with respect to S,

CIIAPTEI/. /0

!E:J. ,-!.._ ...i, f,,_, ....( _ X f'· -/-I

!E:.!. ~t-Fe --

"'.9

at ~~4brilUft: T's';'(;)CFc· 4"l{'\~"f).~ TCIJ$(':;)="'"

..."""Ci)· "'~~71'lt:2J~t)i I

~,""~ .(~cl, fcIJ\{t>-t 1$'''(1)- -t (l-c.. :-.., ,i".ltf"'.9(-'l3)i, at It (iD")'x'."~/D' _4>•• a.) .... ,.,,(11""3 - /'11'/({o",i-/O"lIf,i -IO,2~/()

,'.1-2,1'11110" rU. -t 0(.- 4,3S 1<./o'J /'aIL c 1.4"-..!J_ - _fL_'" - - ('It)"').lff~1'''IlI()f -1'.

~ ~(2.USS' f4MJ: Dp. l1fP ,', E(fJ• Z1f'f(J f ~ F=tE= Z".", W=-I,9'tCIO ~ AI (~.J.·'It)o 'I - P. ... ,_."J.L:.._ ~~rc.e ~c fD p~c '~r9t! : F,c ZiT' Ii , ,0

!#,.,( ~..e. ~D J..,';tt,_ c.Mr:1!: ~ J: I (-:~lX ) (.i) ,,/ 7i7i)''',.1,{orce.D#'\ g: ~.~+Ii.:rt(-!I:-" ae,,"'Jl); l"ID·9,~:IC.o.s.. I ; •

.. , F = X *"(f& - t-~)-~':·!3''''J(t'oq(/c·,,-s:» x 2.()sr",,;5" /II )4.4--+--- "-~ (40) ~.r r ~a.. J I:>r(£Trrl). fv (/".,.1) .. %),.... *r

:. Ers:~={rrc 2~()"'/ll'"XID' ..r:(z.f1fJ(ID.)'" I(0) ~(t% )=/t:r'l('1S) • rIO"\ 1,471'X10·":- • r l,zrra.tI./o'" (r~pil(S;tI.)

(t.) /'10 J s'~e ~t!. ~mDu.,,-i ~I ck..-9f.s 4-nc./"su/.. ,', ~L S"d,,"& A.s ~.t ~~(b}

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368

Conductor

4>, - II

CaS!' ,

Problems

Conductor

rj>l - 0

Case II

11 Solution Techniques

Conductor

Figur. P1t.1

<1>3 = 0

Case III

11.1 Consider the three boundary-value prohlems shown in Figure Pf t.t. The solution ofCAse r is <1>" and the solution of Case II is <1>2' In Case III, the charges 4, and qa arc thesame charges that appear in cases I and II. and they appear in exactly correspondingpositions. Express <1>3 in terms of 'I>, and ....a-

11.2 Consider the three boundary-value problems shown in Figure Pl1.2. The solution ofCase T is <1>,. lind the solution of Case II is <1>2' In Case III. the charges (/, lind qz are theslime charges that appear in cases [ and Il, and they appear in exactly correspondingpositions. Note the differences in the boundary conditions for the three cases. Can <1>3he expressed in terms of <1>, and <l>2? If so, obtain the expression. If not. explain why.

11.3 The radius of the inner conductor of a coaxial line is a and that of the outer conductoris b. The potential of the inner conductor is V and that of the outer conductor is zero.There is no volume charge density between band u. Start from the Laplace equationto obtain the potential in the coaxial line.

11.4 Two concentric conducting spheres have radii u and b, respectively [h > 01.The uutersphere is at zero potential, ann the inner sphere is maintained at V volts. There is nospace charge he tween the conductors. Start from the Laplace equation to obtain thepotential {rl for b > r » o.

Couductnr

C':ilse I

Conduclnr

Casp. II

Conductor

Figur. P11.2

~~4>~ - 0

Case 111

rechniques

1.1

Problems 369

In Figure Pl1.S a conducting cone is at a potential Yo. and a small gap separates itsvertex from a conducting plane. The axis of the cone is perpendicular to theconducting plane. which is maintained at zero potential. The angle of the cone is (J,.

BecaUSAof the symmetry of this problem and the fact that the boundary conditionson the potential <fJ involve fI only. is independent of rand q; when sphericalcoordinates are used. Find the potential "'(0) in the region OJ s. IJ s. 90°. Hint: J (l/sin0) dO = In [tan 0/2). Find the surface charge density on the cone.

The upper plate of the parallel-plate capacitor discussed in Example 11.1 ismaintained at 100 V. and the lower plate hi at 80 V. All other conditions remainunchanged. Find <1>.

Model a de vacuum-tuba rectifier as two parallel plates with a space charge inbetween. as shown in Figure 11.3. Let the separation be 1 cm. Find the voltageneeded tu produce IA/m2 current.

Find the surface charge distribution on the vertical and the horizontal conductingwalls for the case discussed in Example 11.8. Plot p. for 7. :> 0 and x = y _ O.Let q =10-6 C. and a - b = 1.

Find the images of a point charge near a corner of a conductor similar to the oneshown in Figure n.ll except that rPo = 45°.

Find the electrostatic force that acts on the point charge q at (0, 0, d) and is due toinduced surface charges at z - O.as shown in Figure 11.5.

Calculate the capacitance per meter of a 12-inch (0.3048 ml-diametsr steel pipelocated 6 ft [1.R!! m Iabove and parallel to the ground.

Example 11.10 states that the maximum electric field on the surface of the conduct-ing cylinder is located at the point nearest the ground. Show the validity of thisstatement by plotting out E, on the surface as a function of (p. Use the following data:V, - 100 V. h - 2 m, and a - 1 m.

11.13 For the point charge q located d meters from a grounded conducting sphere shown inFigure 11.14. find the surface charge distribution as a function of 8.

11.14 Repeat the preceding problem for an isolated conducting sphere carrying no netcharge.

370 11 Solution Techniques

Conductor 1<1>- 0)

Flgur. P".'6 ~igur. P".i.

11.15 Equation (11.42) ).!ivl::sthe potential dus to a point charge in the presence of II

grounded conducting sphere. Equation (11.44)gives the potential due tu a point chargein the presence of an isolated sphere carrying no net charge. From these results. finelthe potential due to a point charge Q. cl meters from an isolated conducting spherecarrying a net charge of 1111'

11.16 A llna charge PI is inside a conducting tunnel of radius a, as shown in Figure P J Ll6.Notice that the linA charge is b motors off CAnter. Find the potential function in tht!tunnel. lIint: This is a complementary problem of the one shown in Figure 11.12.

11.17 Calculate the force per meter acting on the line charge in the tunnel shown in FigurePI1.16.

11.18 A point charge q is inside a spherical cavity of a conductor. as shown in FigurePl1.1S. The radius of the cavity is 0 and the cavity is filled with air.

(a) Find the potential 41in the cavity when b s O.(b) Find the surface charge density of the cavity wall when b = O.(c) Find the potential 41in the cavity when b= en,(d) Find the surface charge density of the cavity wail when b = 0/2.

11.19 Calculate the aleetrostatic force acting on the point charge in thA cavity shown inFigure Pl1.1S.

11.20 Sketch the E lines due to a point charge near the interface of two dielectric media.The situation is similar to the one shown in Figure 11.17, except that (2 - 0.5f1·

11.21 A rectangular conducting trough of width a and height h is maintained at zeropotential, as shown in Figure Pll.:.!1. The potential on the top plate. which COVArsthetrough. is known to be 4l(x, b) - 200 sin(211"x/u)volts. Find the potential 41in thetrough. There is no volume charge in the trough.

11.22 Three sides of a rectangular conducting pipe are grounded, while the Iourth side ismaintained at 100 V. as shown in Figure Pll.22. find thApotential in the pipe. TherA

is 110 volume charge in the pipe.

Flgur. P11.21 FigureP11.22

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l-u -I x

Problems 371 .

yI 100 V

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Figure P11.23 "gure P11.24

of aargsfind1ere

11.23 The boundary potentials of a rectangular conducting pipe are shown in FigurePl1.23. Find the potential in the pipe. There is no vulume charge in the pipe.

11.24 Consider the boundary value problem shown in Figure Pll.24. The upper and thelower conducting plates are maintained at :Gt!I'U potential. The plats tit the left ismaintainecl at 100 V. Two gaps insulate the side plate from the ground. There is novolume charge in the region and cI> approaches zero <18 x approaches infinity.(a) IJse the method of separation of variables tu obtain two ordinary differential

equations.(b) Solve the differential equations. (The function involving y must be a sine

function.)(c) Match the boundary conditions, and find the final solution.

11.25 /\. spherical capacitor is filled with a dielectric material of 1;1 in half of the spaceand with another material of f~ in the remaining space, as shown in Figure Pll.25.

(a) Find the potential function <l>(r) in the region a <r< b. The potential at r = a isVo and it is zero at r = h. Hint: The potential satisfies the Laplace equation(J 1.2) and it may be assumed that it is a functiou of r only.

(b) Find the electric field in the region a < r < h.(c) Find the D field in the region c c r-c b. Hint: The D Fiald in medium 1 is differ-

Ant from that in medium 2. Note: the boundary conditions on the tangential Eand on the normal D fields are sarisftsd using the suggested approach.

(d) Find the total charge on the inner conductor and the capacitance of this ca-pacitor.

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Figure P11.26

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The resistivity read hy the sonds will he influenced hy medium 2. Thus, the sondewill nol read Pr - 1U n-1I1. although it is located entirely in medium 1. To find theexpected rending. we must first calculate the potential detected at IJ. We solved thepotentia I problem in Example 12.5. In the present esse. we have (11 - U.1. x = u. y = 0,z h - 16 in. x (2.54/100) m/in. = 0.406 rn, Z f h - 32 x 2.54/100 m - 0.813 m, and

t: _ 1 (0.1 0.01) _ 0.81810.1 + 0.01

Therefore, according Lo (12.22t1),

1 [1 0.818] 1<1'0 - -- + -- = - (34.7)471' x 0.1 0.406 0.813 471'

Substituting the above value into 112.2I1J.WA obtain

2.54 1Pr - 471'x 16 x - x - x 34.7 - 14. I nom

Ino 41T

rrrsnts Problems

Solution:

Problems

387

12.1 A parA1IAIplate is filled with two materials in tI configuration shown in Figure P12.1.The total area of the plate is A. The dielectric constant and the conductivity of onematerial are f, and 1T" respectively. Those of the other material tire tz and C1Z' Find theequivalent circuit for this parallel plate, and express the circuit parameters in termsof A. d. t" 0" t2' and tJz.

12.2 A parallel plate is 1'i11I~dwith two materials in 0 configuration shown in figure P12.2.Find its equivalent circuit, and express IhAcircuit parameters in terms of A, the areaof the plate. and U1• U1• f" tz. (1" anti. (1z, which are defined in the figure.

12.3 A coaxial line has two layers of insulation. Figure P12.3 shows the geometry. Find

(a) IIIl! potential <P, for a «. p -c: b(b) the potential <t>t for h < f1 < c(c) the resistance of a section of such a line Q meters long

12.11.'t and

ldalVly. a~exactarAnt

Figure P 12. 1 Flgur. P12.3

y tool [::·:::r:~::]1dif-boun- r1"-W/2-~W/2_11 re-alive I I I

p I I I

L I ,I I I~igu,.P12.2 /1 I 1-

[: ,: ::::~n, lid'I I I ''\

\ ..... _-./ I........__ ---.-I

388 12 Direct Currents

x

Flgur. P12.6

'lgur. P12.4Xn~-- 1'1

/

Perfect conductory

Aif -,',20 em

;;; "o ~I-I0A/. ~

,~~ ,10 cm //' I,/ ',./

//'8 120cm/ I I

l10cm I

"uur. P12.7

12.4 A spherical conductor of radius a is inside a spherical conducting shell of radius c.Two materials are used to fill the space between these conductors. The dielectricconstants and the conductivities of these materials are EI• 0'1' Ez, 0'2. respectively.Figure P12.4 shows the configuration. Find the equivalent circuit of this system, andexpress the circuit parameters in terms of a. b, c. £,. Ez. 0'" and 0'2'

12.5 Two oil wells are 1km apart. The resistance between two steel pipes in these wells ismeasured at 3.4111. What is the conductivity of the )(round near these wells? Use thefollowing data: the length of both pipes - 1 km. and the diameter of hath pipes - 10cm.

12.6 A current electrode is near a perfectly conducting plate that is bent to form a 90·corner, AS shown in Figure P12.6. The output from the electrode is I amperes, and thematerial filling the space has a conductivity equal to 0'. Find the potential function4>( X'. y, 7.1.

12.7 A current electrode is near a perfectly conducting plate that is bent to form a 60·corner. as shown in Figure P12.7. The electrode produces 10 A of current, and thematerial filling the region defined by 0 < t/> < 60° is water with conductivity equal to0.Q1mho/m. Find the potential at point B shown in the figure.

12.8 A point electrode puts out I amperes of current above a conducting plane. as shownin Figure P12.8.

(8) Find 4>{x. y, z] for z > o.(b) Find the current density] ,Ix. yJ at the surface of the conductor.(c) Sketch the paths of the current flow.

-rents Problems 389

T r3m I

!! 00.15/111

12.9 For the case shown in Figure 12.9, find the pArr.AntagAof tha currant emitted fromthe electrode crosses the boundary and ental'S in medium :.1.

12.10 A source 4 meters below an interface of two conducting media emits 2 A of directcurrent. as shown in Figure P12.10.

(a) Calculate the potential at point B.(b) Calculate the potential at point C.

12.11 A well-logging resistivity tool similar to the one shown in Figure 12.12 is near aboundary between two beds, as shown in Figure P12.11. The boundary is makinga 60° angle with the well. Find the apparent resistivity measured by tins tool at theposition shown.

12.12 Refer to Example 12.6. Obtain Po (the apparent resistivity measured by the tool) asa function of tool position for Zo - ~ 160 in. to Zo - 160 in .•where Zo is the posi-tion of the center of the tool (the midpoint between electrodes A and B) relative tothe boundary. Calculate Po for at least 21 points, lind pial Po versus ~o.

12.13 Repeat Problem 12.12 for the situation shown in Figure P12.1l.

ius c.ectricively.I. and

ells is;6 the•= 10

a 90°d the.ction

H 60°d theual to

lawn

Figure P12.11

12 Direct CUrrents39012.14 A point electrode is located at (0. Yt' 0), and a perfectly conducting sphere of ra-

dius a is located at (-i, 0,0) as shown in Figure P12.14. The electrode gives I am-peres of current. The conductivity of the medium is (1. Find the potential ~ on they axis. Hint: usc (11.44).

12.15 Consider a well-logging resistivity tool similar to the one shown in Figure 12.10.Let the spacing between the current electrode A and the potential electrode Bbe6rn. The tool measures the conductivity of the earth formation as it travels in a well.Assume that the well passes near a mineral deposit modeled by a perfectly con-ducting sphere, as shown in Figure P12.15. Find the apparent resistivity measuredby the tool as a function of y. Use the. following data: (1 a 0.01 mho/m for theground; the radius of the mineral deposit '" 50 m; and the distance between thecenter of the sphere and the weU = 70 m. Plot O'epparenr versus Yfor -70 < Y< 70.Hint: use the result obtained in the preceding problem.

FIgure P12.14

x

IT = 0.01mhn/m

y

B

A

-70m-

FIgure P12.15

x

13.1Magn

CHIi PT~R 12 .as: cU Btc~st. +ch;t"f,',J_ E. - f,'~ Ie( Q, 1:1&

E,I, r' T 0-

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COrtS£j" IKtly, V= G, d = 6i~ cur, f2,er.~I

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.E.:.1. ~l

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l~ l.,~ tI., (1 C;10.",", LoJ(. c.~ CUt Ji .r.,. it-'5 +W# i"'fuf~"t cttfc.e,·hY'1 ,." serce«,

E.,,~ =r+: ;~(l(.r(,~Jf ..t'tA. ., AM ,''/( .. 1Cs: 6 At _t111i

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44

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45

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46

14

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eli; (..r /J,/ < ih; 11;",hluk.._ 2 .,,1/. B"~ ""~i.4 ... ,./ 1"= If;tr. Icr,~&.~ r 4:;'). = ~:z(r,!,.)* t. -41f.l£/r c: ~~r~

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47

13 Magnetostatic Fields422Because magnetic field is present in the coaxial line. we know that

magnetic energy is stored there. The magnetic field is given by (13.7):

{

I- b>p>a

Hoi, ~ 27rp

o elsewhere

Substituting the above expression in (1~i.~i4).we obtain

1 12. I' ]2 Jl12 (b)UH - 2 Jl 0 d¢ n pdp 47r2pl - 47r In ~

This result is the stored magnetic energy per unit length or the coaxial line.Consequently. we can calculate the inductance per unit length or the linefrom 113.41):

L = __t:_ In (~)27r 11

(13.49)

This inductance per unit length also appears in the transmission-linerepresentation of the coaxial line in (6.19) of Chapter 6.

Problems

13.1 Find the magnetic field " at tha CAnter uf a square loop carrying a current I. The sideof the square loop is b meters long.

13.2 A circular loop that has radius a and that curries a current I produces the samemagnetic-field strength at its center as thai at the center uf a square loop thai has sideb and that carries the some current I. Find the ratio of b to o.

13.3 Consider a larga conducting plate of thickness d located at -d/2 :; y :; d/2. as shownin Figure P13.3. Uniform current of density Tis flowing in the z direction. Find H inall regions.

13.4 The earth's magnetic field at the north magnetic pole is approximately 0.62 G (1C .. 10 • Wb/m2). Assume that this magnetic field is produced by a loop of currantflowing along the equator. Estimate the magnitude of this current. The radius of theearth is approximately 6,50U km.

y

FIgure P13.3

Ids Problems 423

lat

Figure P13.6

Figure P13.5

ie.ne

,91Pigur. P13.7

le

-nIII

13.5 An infinitely long tubular conductor of inner radius 0 and outer radius b carries adirect current of I amperes. as shown in Figure P13.5. Find Ihe H fiald at o; where (a)p s o. (b) 0 S p s b. and (e) IJ s: p.

13.6 All infinitely long tubular conductor has outer radius b and inner radius 0 offset by adistance c from the axis of the outer cylinder, as sbown in' Figure P13.6. ThisAccentric tubular conductor carries 0 direct curren I of 1 amperes, Find the H field atpoint A shown in the figure. Hint: Consider the tube 10 be 0 superposition of two solidcylinders that have radii b and a and thai carry uniform current density 1in oppositedirections,

13.7 An infinitely long wire is bent to form a 90° corner, as shown in Figure P13.7. Adirect current I flows in the wire. At point A find the H field due 10 this current.Follow the steps given below.

(a) Use the Biot-Savart law to express the H field at A due to a typical segment ofwire dyon the wire axis. Express the field in rectangular coordinates.

(b) Jntegrate the result obtained in [a] to find the H field due to the semi-infinitewire oe, Note: to facilitate integration, let y - a tan II, so that dy - a sec" 0 dO.

(c) Find the H field at A due 10 the semi-infinite wire BO.(d) Add the results obtained in [h] and (c) to yield the total field at A due to the

current in the wire BOC.

13.8 Follow a similar procedure 10 the one r!escribed in Problem 13,7 to find the H field atpoint 1\', as shown in Figure P13.7,

13.9 Consider a circular loop currying a current I counterclockwise. as shown in Figure13.11. Plot the magnetic field Ll, on the z axis for -0/2 < z < o/z. Find the value z, interms of a. such that, if I z 1< Zo, then H, is uniform within 10% of the value of H. atthA center of the loop.

Ie

1e-Ie

1111Ie

13 Magnetostatic Fields424

xFigur. P 13.1 0 Helmholtz coils

13.10 To improve the uniformity of the magnetic field along the axis of a circular loop {seeProblem 13.91,one may use two identical loops separated by a distance equal to theirradii. as shown in Figure P13.10. Such a pair of current-carrying loops is calledHelmholtz coils. Find Hz as a function of z on the axis of the Helmholtz coils. PlotH,for a < z < o. Find, in terms of 0, the value Zo such that. within the range I z I < Zo. H,is uniform within 10% of the magnetic field at the middle of ths two coils. Compareyour result with that obtained in Prohlem 13.9 for a single loop.

13.11 A square conductor loop 2u meters long on each side carries a direct current 1 asshown in Figure P13.11.[a] Calculate the magnetic field B at (b,O,O).Express the magnetic field in terms of

4 integrals, where each represents the contribution from the current on eachside of the square. Use the Biot-Sevart law. Du not try to integrate those inte-grals.

(b) Assume that b is much greater than o, Now, evaluate the integrals approxi-mately to obtain an approximate value of Bat (b.O,O).

Ylt-.

e

T,

I2u

[b. u. UJ x

Figure P13.11

_"-

A 8

13.12 A surface charge of p. C/m2 is uniformly distributed on a record disk. The innerradius of the disk is 0 and the outer radius is b. The record disk is turning at aconstant angular velocity w rad/s in the clockwise direction. Find the magneticfield at the center of the disk due to the surface charge on the turning disk. Ignorethe presence of the metal post on the turntable.

Fields Problems 425

"2 coils.

13.13 The earth's magnetic field at the equator is Approximately B = 10-4 Wb/m2• Calcu-late the cyclotron frequency of the electron in the ionosphere.

13.14 Aecause natural uranium contains a slight amount of Uranium 234, the electro-magnetic isotope separator can also yield 2l4U.If the radius of the circular path for2JUU particles (see Figure 13.14) is equal to 10 rn, where should one place collec-tors for 235Uand 234U particles? Express spacings in meters.

13.15 Refer to Figure 13.17. The magnetic field is changed from 5 x 10-4 to 10-3 Wb/m2•

All other parameters remain unchanged. Find the following:

(a) the position of the electron at the exit siele of the magnetic-field region(b) the exit angle (the angle between tho trajectory and the x axis after the electron

has passed through the magnetic field)

13.16 Consider an electron having initial kinetic energy IIIe v~/2 and entering a region ofuniform magnetic field, as depicted in Figure P13.16. This situation Is similar tothat shown in Figure 13.17, except that the electron in the present case is inclinedat an Q angle with respect to the x axis.

(a) Show that v, and v, of lite electron "'fter it enters the magnetic field are givenby

::>(seetheiralladot Hzz; H,ipare

.t J AS

rns ofeachinte-

V. ~ VII cos(w,t ~ (Xlv z - Vo sin[wcL+ o]

where We = 1./J3.lm. And t - 0 corresponds Lothe moment the electron enters tbemagnetic field.

(b) Find the coordinates x and z of the electron Attime t. Note that x - 0 and z - 0 att - O.

(c) Find the point where the electron leaves the magnetic field. Assume Vo = 2 X 107m/s, a-50• Wi' - 8.77 X 107 rad/s. and d - 4 em.

(d) Find the angle between the x Axis and the trajectory of the electron after it hasleft the magnetic field. Sketch the entire trajectory, and compare it with the oneshown in Figure 13.17.

-roxi-

z

-d-j------,x )( x )( I FIgure P13.16

x x : : rElectron

x x ~_:~i----__~xx .:_x_x_l

nerat aeticare

13.17 1'wo parallel wires are carrying 100 A of current in opposite directions. On eachwire find the force per unit length due to the magnetic field produced by the otherwire. Is the force repulsive or attractive? Assume that the lines are 1.5m apart.

13 Magneloslatic Fields426

13.18 Two identical circular loops of radii 0 ara separated by a distance d, where d«Q.

One of the coils carries I amperes of current clockwise, and the other carries Iamperes counterclockwise. Find the force between these coils. Hint: Becausethese coils are close together, you can approximate the magnetic field that is at onecoil and is produced hy the current on the other as HI = 1~/l2?1"d), the field due Laan infinitely long wire. Let 0 - 1 m and d = 0.05 m. How much current is neededto produce a force of 9.8 N?

13.19 A circular loop of radius 0.5 m and 100 turns is excited by fI 2 A direct current.This loop is placed in the Earth's magnetic: field, which is approximately equal to5 x 10-5 Wbfmz pointing north. How do you orient this loop to produce II maxi-mum torque? What is the value of this torque? Find that orientation uf the 1001J In

which it experiences no torque.

13.20 The square conducting loop ABCD shown in Figure P13.20 carries 2 A of directcurrent. Each side of the loop is 0.1 m long. The loop is placed in a uniform mag-netic field B. Find the force on eacb side of the loop and Lhe torque on the entireloop if:

lal B = k 0.2 Wb/m2

(b) B = - Z 0.2 Wb/m2

y

D c

~TU.1 m

1=2 A

x

Figure P13-'--

A B

.20

13.21 An infinitely long conductor of radius 0 carries a direction current I as showninFigure P13.21.

(a) Find the H field in the region O<p<o.(L) Calculate the stored magnetic energy pel' unit length in the region 0< p < a,(c) Find the inductance pel' unit length of the conductor. Consider only the mag-

netic energy in the region O<p<o.

13.22 Three infinitely long parallel wires each carry 10 A of current in the 1 direction, asshown in Figure P13.22, Find the force per unit length acting on the 113wire due tothe magnetic fields produced by the other twu wires. Give the numerical value ofthe force. its direction, and its unit.

static Fields

aers d « o.ier carries Int: Becausehat is at onefield due tort is needed

ect current.sly equal toIce a maxi-the loop in

!\ of direct.form mag-the entire

-hown in

<a.he mag-

:tion, aseduetozalue of

Problems 427

T1m

j #2

x~--lm--...j

Figure P13.22

Flgur. P13.21

13.23 The magnetic field in a coaxial line is given by

H~ = (Ol/P forO.lm < P < 0.201~ elsewhere

The medium is air. What is the total stored magnetic energy per unit length in theline? Give the numerical value and indicate its unit.

13.24 (a) Calculate the stored magnetic energy por unit length of the parallel-plate con-ductors shown in Figure 13.5.

(1:» Ifthe parallel plate is used as a capacitor to store electric energy, find the volt-age Va for which the stored electrtc energy is oqual to the stored magnetic en-ergy found in (a). Let l = lA. w '" 10 em, and a - 1 em. Express Va in volts.The medium is air.

13.25 Calculate the inductance pel' unit length of tho coaxial line shown in Figure 13.3a.

13.26 Calculate the inductance pel' unit length of the parallel-plate conductors shown inFigure 13.5.

/3./

1'"

/3.2--,.,

a:L (a.)

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49

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13

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Ij/3 1 __ IB II: 1'"ID"'I"O-+~.7, ..."..." ~1I"lIf.J/./O-jl II: 2.8 HN~

~ PS/Ils:, (mr/IH,)~ :: (IJS"/Z'8)~ = ().99J7 .. Rs- c 9.9037'"~/~8 s(ITJ4/",,)V."(U.f/H8)~ - f),flf/S'''' A'4=9.9/61r1

IS. It' R. ,.., v: ".11./6-". 2.,0 If- .,.180 - I.(,~/O·". urJ • I).II~ '"

(4) V; c ~ C#s..,~ ~ ttJ.,. - V, s,;"..,e J wJ,,,,,1. 14)- Vi/R.:. ):..: J t'l.f.,c (-e .r -i# $/"W'6/ t Co J:!i. s,~w4:. - . ~} • J/' Vi tit - ~ '~hJt I: If ': (c.jlA~ -I)

(1..~ ~ =«. ~It.) c: IJ.&J,I=-ff!,JJ;"wt. * ~t~ = ': _ ().JS-

J,U.). ;::[/I-s'J..,'. - U - ~ ({ 1-1'.1$:>' -I) - -d.~f)7~""l~) Itt Nt, I.¥i",,-., -t. -l.. ~,. .. s;,..w~•• f).'S'

lI;(i.) = V. Go,s~. c V.{1-5;,,~lJ'to ... ~ {I-( ••,r)l ; ().t;.37 Y.~ loot.) c - V;S,'" u i.c - 6,1$ C/O

,{_:- .fr,..,;'{c!j (4V1!it-4>j II: .,..,.-'(_().I%. 9.17) = - Z().S·In N.(. "'45"'~J.,·(..lteU. r4,i~1\,1 ~'.Jr~J(i)~ v.Z * y~t) .. ~CDJ(Wct+6) ANI. 1oJ(~.~$',"'(t..tt+4

~,;'c.e. 1I£{D)C: V; CDs.t..~ vJ(6)=--"!'~ ::> -v.(.t-)c:""c.-s'~t.,. e() a.-ut 1i(')=-VoSi,,(~t+,.()

<b, I" tIIf. ~J"tli' .;.'..,«.. "~ji(N1.~ ::({oi)-r tt;r(~)J-t ""... Jlof).1.41.5 ,t).t:1.10 i« tV; .... .>:. ~{ ..) - III, s.,,, ~" "''''')'''', ~ 3(1;)" ~ ClJSCIJ~T;0I0r/. + 'l

. ....!i. --~ C v;~/~u.. X(D) r= i(I:J) Ir 0 I C,:: - W, 'IA..(_ __ ~. - ~ ~J...

:. 'Xt.t-)= :: [~''''(cJc.t.tL)- f;"," J ."..c. ~(~). *[CDJ{.tJ"t+-.>- '.~.(J(C) Ai:"";' .. ~1I.jS~ .. ~ r~. a".'" %(~.). d. --rt[S/~(",,~.",A.)-Si""'J. "".J' i.'1'1llID'IX4=)/'P-Z') ~_:. SI"(W,~T.J.)C 1i;"" +SI"," I: - :lIC/I,'i "'$N\,(S· • (),2,Z.,

:. eo$ (ulct.+.() .y1-s;".'(IlJ,t.o#.,.) ={'-(D.'J''J(,)t • D.II'$'

: a.i:. #It ~)iisi~ ~(".).".()I-I'I'I tUtt:l "(l.>- ~"'D",,[I •• 'S"-U,(st;J --O.CQ~/4-m• ~ 1.'711/fJ

(d> ~ ~t. LJC.'se~ ~(+.)~ 'Vi CD$("'~~.-I;"') - P.q,S"fT. --' ~(.j.) .. -11,2.,2.' ~ey.isl: ~/t. B. fM-I(SljI~~1S(4.)J " ~-I(_~.2'-z"/ ~.9'~)_ -1$.u·

51

/3.11 d.- -{~

i-{t.-Lt><>f sl..f4Itt, 'e. pla.,uJ, A.ri)O'>-laJI, Dr

f1U"-i.L·uJ~ I;'" -Hoc. £-w 14, .."f'('" +0 obh../,..

!3.J. 0

TITIA)A= mB-= 11'(D.fi)2.xICOX.2.'K!jX./o·rlC f,SS"I(IO-3 tI-m

No +orfllt... ~Jlf""'i""'C" wise". b.p i, p/~'fA.. ·1Iu+lc..JI, 'A rJ.~ N- $ dl',.,,,,J.,·/M ..

{C\) FIt8

,:() I Fa,,: z,ilJCjj ~ ./IrO./J(,1X ~.z;1I0.fJ'f II (-j) . 'Fe.: 0I

f. - I•• I ( ") T ( )~ A ,. • J .,PA - O. D Y' N J; = 0.' 11 1J X I.1)i = It 11 '0 AI-,.. ('j )

lb) F. 1\ A '\ '"~B= O./~2.'X1I(-o.2)= I.Df.l.IJ); FO,O::: -O.I)'I-/.1:j Fe,= .o.o~)./'f

FM = D.O\j,rJ: ; .T .: 0 .

.!l.2:J. (A) 1.71' t Ii+ =- r· r-:f;\ ) _.. ~f ::rf for u<f c ~~1r~\.

lP UI( '= JJJ t _.4A Ht dv I I a. 1..111 f.4 \.''i''AA 1t-1i4 II- " I t 1. (f tiel ) -= .AI. r

tJ c> nn(C ) UH :: t L I \. L=

L till 2 .es: _ .A r:.t JI "II HI ....::- IO~ A A r- [1. 1~1r --1J.2.~ B = ~ +)' In

, .) v Jt. VilOA( A11.:_\1

• Jo7l" i

52

ft

53

14 Magnetic Materials and Magnetic Circuits452

From B(O" we find the corresponding f1101(Figure 14.17b):

1 1101 ~ 2400 Aim

The iterative method calls for substituting the above value into (14.23) to obtain the"first-order" approximation of B, which is denoted as BP':

nPI = (1000 - 2400 x O.121)J.lo_ 0.178 Wb/rn!0.005

The corresponding 11111 may be read from Figure '14.17b:

HilI _ 2100 Aim

We obtain the "second-order" approximation of B by substituling the above H(1l

value for H in (14.23);

BI'I (1000 - 2100 x 0.121},.,.0 Wt I .- - - 0.1117 () m-

0.005

This procedure can be repeated to find the n-th iterative result of BInI. When a digitalcomputer is available, the magnetization curve can be approximated by a standardpolynomial-curve filling and stored in the computer memory, A simple program maybe written to carry out the iterative procedure, which requires very little computertime [see Problem 14.7).

The problem at hand can also be solved by a graphical method. Note that (14.22)or, equivalently, (14.23) is an equation of a straight line on the B-H plane. As shownin Figure 14.17b this line intersects the B axis at 0.251 Wb/m2 and the H axis at0264 AIm. It also intersects the nonlinoar magnetization curve at B ~ 0.'19 Wb/m2

•

This result agrees fairly well with the result ubtained by the iterative method.

Problems

14.1 Refer to the magnetization curve shown in Figure 14.3. The material is a nonlinearmedium because p. depends on the magnitude of H. For magnetostatic fields, 1.1 isequal to the slope of the line joining the origin to the (H, B) point on themagnetization curve. In this way. Figure 14.3b is obtained from Figure 14.3a. Now. ifthe material is placed in a time-harmonic field, the effective p. will be different fromthe J.lfor the magnetostatic fields. Consider a field H - Hu+ HI cos (wt + 4», whereHo is the bias magnetostatic field and HI is the amplitude of the lime-harmoniccomponent of the total field. Let HI « 110: then the effective permeability of amaterial is the slope of the tangent of the magnetization curve at Ho. Sketch theeffective J.l versus Ho for the curve shown in Figure 14.3a. Compare it with thernagnetostatic /Jo shown in Figure 14.3b. and show that the !L'S in these two cases areequal to each other at 1'3'

14.2 Point out the differences between the following pairs of terms: (a) diamagnetic vs.paramagnetic, (b) remanence vs. retentivity, and (c] coercive force vs. coercivity.

14.3 What are approximate values of the retentivity and the coercivity of the ferriteshown in Figure '14.9?

:ic Circuits Problems 453

14.4 Consider the carbon steel. alnico V. and eunico materials listed in Table 14.2. Whichhas the highest permanent magnetic-field strength? Which has the most difficultyin losing its permanent magnetism once it is magnetized'?

14.5 A permanent magnet of radius 1.5 cm and thickness 0.3 em is put in a magnetic fieldthat is parallel tn the disk, as in the situation depicted in Figure 14.7. The torque onthe disk is equal to 1.2 x 10-3 N-m. and the magnetic field is equal to 10-' Wb/mz.What is the remanence of the permanent magnet?

14.6 To write "one" in the memory core X2YJ shown in Figure 14.11, how should thecurrent pulses he sent along the wires? Specify the polarity of these pulses.

14.7 Consider the magnetic-core memory sketch Ad in Figure '14.11 and the correspondinghysteresis curve for the cores shown in Figure 14.9. Now suppose that, because ofmalfunction in the circuitry, a positive pulse of amplitude I, which alone is capableof producing the switching magnetic field slrength HI, is sent down the line yz andthat simultaneously an identical pulse is sent down the line Xl' Assume that all coresarc initially in the "zero" state, which corresponds to having the magnetic flux cir-culation pointing either toward the upper left or the lower left (using the right-hand rule). What are the states of nil of the cores after these pulses have passedthrough?

14.8 Compare the hysteresis loops of two ferrites shown in Figure P14.8. The curvelabeled ttl is "thinner" than thai labeled tt2. Which ferrite core requires lessswitching current? Which ferrite has a better ability to withstand magnetic interfcr-ancss?

14.9 Consider the magnetic circuit shown in Figure P14.9. The material is steel, andFigure 14.17 shows its magnetization curvs. The flux density in the air gap is 0.5Wb/m". Find the current I needed to produce this flux.

14.10 The magnetic circuit shown in Figure P14.10 is made of a material with ~ - 600~).Find the flux densities Al and B2• and indicate their directions.

ohtain the

above Hili

n a digitalstandardgram maycomputer

hat (14.22)As shownH axis at.9 Wb/m2•

ad.

on linearslds, p. ison the

. Now. ifent Cromj, wherearmoniclity of aetch thevlth theases are

B

ill(112

Hj Figur

Jferrite

• P14.8 Hysteresis loops for twos.

_lt 0.5 em

Flgur. P14.8

tstic vs.vity.

! ferrite

lengths.PIP, - gemp,p. - P,P, - 10 emP2P,p. - 26 emP,Pu - gem

eross-seetiona I areas:

PzP, - tz cm'all other branches - 9 em'

P, Figure P14.10

454 14 Magnetic Materials and Magnetic Circuits

r-8CIlI 6cm-1

Pigure P14.111tz cm

J14.11 To produce a magnetic flux of 0.5 Wh/rn" ill the air gap of the magnetic circuit shown

in Figure P14.11. what should be the magnitude of the current in the coil? Take IJ.-200~. The cross-sectional area of all branches is equal to 4 ern",

14.12 Write a computer program 10 carry out the iteration procedure outlined in Example14.5. First, approximate the nonlinear curve in Figure 14.17b by a polynomial offifth order. Then carry out the iteration five times to obtain the fourth-order ep-proxtmation for B.

14.13 Find the approximate value of B in the magnetic circuit shown in Figure 14.178for excitation current J = 15 A instead of 10 A. All other conditions given remainunchanged. Carry out the iteration a sufficient number of times to obtain an accu-racy to tha third digit.

15.Qu,

CHAPTER 14-

oJL

o

I~. 2. f t:/.A.·d,h1~,.,t.f'G : ",~" • .w, .t..J~ p~t.d. oy' or6"'.l and..spi""';,9 d~ "1--.,,, ctIt",d ow:fCOllfpl"{f/; .

pa.r4.hlAj"H-,'c, : AlA9""J/c yOutls ~t. -h '«e.. O,.~,".H.t""4 spi"","'_' ~'f,cA-ro'" b ..,.,r:- C,IJ,.,d6....t C-"'tpiddt.

~ IT·o I H.&. -~,,;.f.u.I~ tJ/.;,(e. "ltfl'~ B.cS C4..1tuL ~".a.n""ce.'f,4t. relt14llt.",e 8 is CAlIu/. r¥';'.,~J·"d.1y IlIn",., i»-M e.nJ.s. ~I -Me I'IAJn~t'J4.J,·O"Cu,r,,~ 4I'~ ...t s~11'ofll..

[

coercive: icrce: n« vo.I..u.e t4 /I ;" -Me. "~h·"e. dt'rc.c.Ii()I't of /;'/1,,, (, 1f'f"1"t:I"JA"'~"".joe """'lily -Me. e ';;,,14..

C'oerCiv;~,: rlae.. tDc.,.,ive. ..tor<... H ,~ c"Jcd. (.ce.r'~I/'·/.y £4111"" ~,Ht ~s PI- -lite,"''''''JAC.+cj«'',P" e~v.e. t1V4. ~ SAnu-o..+i.".

1+.5 rel-bl.J.ivr!y - D.22. we/l"1a. ; e.M.rciv/+y. 100 AI,.,14.4 nA,f/Al.!,e PCN·"fJJ1t.ne m4,rle.It·, .t.'~ s,JrcnfH-... ,.,a.,~",,·It/' ;~ tt.L,ic.o V

/'hOSe a/.//f,'u.c..d +0 .L.sL perW'lt;A.,.,.."t ""~"f,Ii$h'l "-'A/p,A.l is "",,'CO

1+."- T=I.ZX/O-J =",8 =/17)(,,/0·' .. m-,.2.

m:::NrC"at) 9 NI-c ".';" - n-(~~:;S)1.e /&'9'1. s«J{r::"Z = a'r'j C ."9'/.'~_ s.« r

IY,/<4. 3"10.' iii. :KIt:>

.._ e w:,JIDH- 41f')lID·")f. 5'. '''x/()''::: 0.·" ""'/IPlJ.14.'

"'0" St.-It.: trl4Jne,I" ... c,.,e"f. PP'''-/:''J -Iokl4rw:{_ I.;JI.,rul'P~ ·.I'le .,. 4"1tI.,,. _ L../t (;:It',~&I-tD"

.~ x,~ a I I X. Y,c I j }(.J >t :::0; X.. Yl. I

')(.J Y, - I ; XJ)I." D i '1 Y.I ~'i )(,Y.. ,,0

/4-,9 #/ r~Jwre.£ 4"$ s,.,i-lulI·"J cu.,.,.."I:,#% ~ A. kHer- a./J.'/i(..y ""0 IAIitJ.s"-'w/. ,"~".w" 1"/u-/',,...,,, ....

.!!:.! Frt>"" Figu.rc. /4-.I'1~~ ;;9~reP14.9J

",iN-.. 84..- f).~WD/ItI"~ I\.~'.oAI,.., I~ ~c. .I...t. (8';"4;,.. «s». frttl~IOOCZ. !;.K(S)(JD-J)+ H,rt(~~/fr'-S'~llr')

c.Se s~,c-I .J- 4''''/0'''' -I-ZI,,>U(M"ID"-S,,'D) - 1I~3"+-

:. I = 4Z'33.4/1{)()(J = 4.25 A

/.O()pI: 5_100_ H.Jl.0.28+H,xo./ ••• -. - - - - - _. (])./.()t>I'II: (). 1./, 1C~.2""ii,~f).I .. 101,. z., 1-11. .@

~# -= ~ +~. ~141/.(9lt:Jo-#) • .J.lH,X(ISWID-'4)+Mllzx(fX /1,-')

... Ho- -tH,<l-HJ.-(..,.)(~.'.')H, -4.'H7HI.-··~s.~jJ;.j.J,',..J ® 4~ (i) ;,,-10 (1) ~

SCO- (~.26"'4.4'7+"./t(I.")Hl -> HI.. !'~- 33/H, • s. t,H, - :I. ,'It J, I • 96c. S"

:. 8, ::~I-I, c (,oo)(.f.7T_lti·x 860.5" - ().US8 wl/",' tvJ.. 8, -,u.)lI.: ""DII"''')I//)'''')(3J1 =().2.49' wJ/w,

•

14-./1- Z" -Me Air 9¥ I S-a ..- D.$' ""'/,.,.' AI\iL HA· 8"./p.._ os/u,¢"cr¢,+¢1. ~ 8t>k(~1fJO-4)::;(8,+81.)Kf-M/~~q8 •• 1J;+8~ ,u.-zo~,,/l. •81!.,--,'e, of Hte ~r~ 8,-BL ~ 130 •• 8,5'~cL B,",.. Aif" -= l3 i", ~re ~ lJo• B ..• C.$W~' +H.- ".SIIA-

S. ,... ,I 'J8,- T -T •c.1~ ~"M1. .. 14,- o.ZS,:I.LFro- ~fJ Z, ;J()CI .. H~S"'lti_'+ lIor«4./2-S'~D·')"I-f,.(D.#8+I./1+".~B)

.: 300I" ~;JCS)(/,,-J+ ~~!.Xi?"S'+ ";;,!!JtD.26 = ~X3.'37SJ(/D-3""'--------"'"

r. I - ~.I3"trJt/O·lj(3(>()~.,..".I!I.~-") c: 8.32. A~

1020304060708090100

H=OIMAGE 7A.DD,7A~D.DDD,17A~DDDDD.D,6AFOR 1;1 TO 5J=I-1B=(1000-.121*H)/3978.9H=(25815.47-(129360.27-(354706.29-(447912.425-214034.26f.B) f.B)f.B>f.B)*BPRINT USING 20;" ORDER=",J," B=",B," (L>Jb/m**2) H=",H," (AIm)"NEXT IEND

ORDER'" (I B" .251 (Wb/mf.f.2) H= 2375.6 (AIm)ORDER:: 1 B= :-179 (Wb/mf.*2) H= 2090.4 (AIm)ORDER:: 2 B== .188 (Wb/m**2) H= 2127.8 (AIm)ORDER""'- B= .187 (l-Jb/m**2) H= 2123.0 (AIm).:;.

ORDER= 4 B= .187 (Wb/m**2) H= 2123.6 (Aim)..lL »o.oos + He. J( (J./Z /./$'14/00 -/5'DO a> B _ (ISDC. ().J;" Nd )(Jl.""'. o.O()

s" = I~()O 1tC f"" 10'''/ (;I.coS .:: C.377 -9' Htll);' 3000 (/T.".. J:..),.",e 14./'7)

8(')= (lSO()-I).I"/~3000)xl"IlID·,,/".()OS == 0.28' • HM_ 2/JfJD (rr-- ... ~ig""c 14./17)

~1I:J=(lS'CC-~./2/1( ~(,.O)1(4NIO-"/ ••DDS'. D.2.98 ... ""fl), ~'()C (~1'f>i""I ~"guHl. /4.1")

r;r--.pAic...L "'L~: H -0 I B - D.~7?' J B - ().2 J H. (ISOD- ~!l(o.()()~)/".J2/. 5"820.. Sr1t. D.' wYnia

55

468 15 Electrequasistatlc Fields

Flgur. 15.0 Poynting's theorem for quasi-static fields.

p - -):[ do • (E X HI - #. do . [(\7<1» x H]~ A

= fffv dV Y' . [[\7<1>) X H] - fffv dV \7 . [Y' X (<1>Hl 4.>\7 X H]

= - fffv dV \7 . (cIl\7 X H) - - f1 da . <1>\7X H

- -.#do . <1> (1+ ~~)

Ignoring the displacement current aDlat, we find

p - - .#do • <1>1- t V n ( - fl. do • J) - t. V"1"

where In represents the currant flowing into the volume through the surfaceAm whose surface normal is pointing outward. Thus, we have establishedthat the circuit theory's concept of power input is valid only when thedisplacement current is negligible.

Problems15.1 Consider the short-circuited parallel plate shown in Figure P1S.l, with width w,

length II.and separation n. Finn Ihp, zaroth-, Iirst-, second-. and third-order electricand magnetic fields. Show that the sum of the quasistatic solution is Aqual to the fullwave solution as presented in Chapter 6 for a short-circuit transmission line. Assumethat the current at z - 0 is 1(1)- 10 COS(wl) and that all fields are functions of t and zonly.

15.2 Calculate the total zeroth-order stored electric energy in the parallel-plate regionshown in Figure 15.1. What is the zeroth-order stored magnetic energy in the sameregion'?

15.3 Calculate the total first-order stored magnetic energy in the parallel-plate regionshown in Figure 15.1. What is the first-order stored electric energy in the samevolume? Denote UW" the maximum total first-order stored magnetic energy in the

19ioniarne1 the

15.4 Calculate the total second-order stored electric energy in the parallel-plate regionshown in Figure 15.1. Compare it with the zeroth-order stored electric energy. If 2 is0.1'>' long. what can you say about the relative magnitudes of the zeroth-order storedelectric energy, thc first-order stored magnetic energy. and the second-order storedelectric energy?

15.5 Find the total zeroth-order stored electric and magnetic energies in a parallel platewith Ii short-circuit current I - 10 cos(w!) at z = 0 (refer to Problem 15.1).

15.8 Find the higher-order stored electric and magnetic energies in the parallel-plateregion shown in Figure Plri.l up to the third order. If Q - O.lX. compare the relativemagnitudes of these stored energies. Use the total zeroth-order stored magneticenergy found in Problem 15.5 Ior comparison.

15.7 A coaxial line 2 meters long is filled with a material characterized by f and CT. Theradii uf the inner and the outer conductors arc U lind b. respectively. The voltagebetween the coaxial conductors is Vo cos[w!). find the zeroth-order electroquasistaticfield E10). the current 1101, the charge QIOI. anti the first-order current 1111. Express thesein terms of the parameters Yo. c, h. Q. f. CT. and wI.

15.8 Two concentric spherical electrodes of radii a and b. respectively. are filled with amaterial characterized by e and 11. The voltage between the electrodes is VO t;OS (wI).Find the zeroth-order electroquasistatic field F:;IOI. the current J'", the charge QIDI, andthe first-order current Ill). Express these in terms of Vo. a, b. f. (J', and wI.

15.9 Show that the time needed to charge a Van de Craaff joIener'iltOI' shown in Figure9.24a with radius It to a maximum voltage of Vmo. by applying a charging current Iis equal to 41!'EoRVmA.lI. Calculate the charging time I if R. - t rn, VIIIax = 10RV and1- lO-s A.

ic Fields Problems 469

quasi-

r .~---.---/1_l_¢=========~~ w

l----/-~I/Figure Pt5. t A parallel plate with short-circuit current T - 10 (;US (wI) at z = O.

region and U~~ the maximnm total zeroth-order stored electric energy in the slimevolume. Show that

Umn ?ururu ~ ~ (k2J"F.m

:rfaceished:1 the

th w,ectricefullsumeand z

egionsame

CIIAPTe~ IS'!E.:.l I(<t) = Ib cos cJt cp Y("J = ~ !.CtI,Wt:

iv-oH.. cr4e.I"': E (0) lit 0

1,._,.,., -Me bOlM'ltl.,.y e"I'Illhl.""" J CIl #te. l~.r plAle." x _Hf.)"'J(D).'.1i(t1)_ -9!;,oswt

E"J ~ (I) ,. T.1ft arder s v~ell). -kuJr·) .. y !zK .-9 "';JJ.lIJ$/"'UJ~. ~ E C -xW.llWi ri"lAIe

lItl>.O2"J orde.r: E (S) • 0

"m 2v)(fjl"_iEE(') => ~/_!!!l.)=-£';:.Il£Wlc ClJSwt .. Hmc Ay.bJ,(£W1...1.'D~W'Cse l" a ~ JY W z.

3rt/. arder « VX Em .. - ASLRm .. y ( ~1")- y ;;JleW':,"S'-"uJt ~ ECI) c ~ ;: J4.'elIJJ /; $,"",_t;:;lJ). ()

:. E - - ';.~ wp.S;nwt (w -.;; UJ~(. + _._) II: - x-srT sillwt; [.U"'ir(~""'··J= -x!f1$/',.wt; si,,(Ai)

HZ: - y!:,~.s",t( t> ~a w~£ .... _.) • - y fr co.s~ t4$(/U)

L p/llll.$#r npr4H_;SS J &. - x l$'{ (e-jU_ejii) tMd.. g ... ?-k (e-jl!+ ej4~ ) j -!#lese

are. -IIa~ 4// WAV& 6oLu..ft;'" *' a.. sA"rt - Ciruc.'i ~Slf'liSS;t1" LlIl' .

EtC). -; ~ COlwt, R(o)~ 0

.'. U;lI'. ; f Ei,iiJi'·)dV - -iE (-~lc()5zwi((J..VJi.) == : f~No~Co521A1t anti. U:lI)_o-CI> -(I)" fV... tIE :so, H c: -'I w2- -r Smw ~

Uti) - 0 A_J I ,(1).:; J_fUH6)RC')~&I" _ .!:!:.(.lli)1",lSI;,lwt.(a.W)' (1i,Z,tli ... ~(!!l.)ilV(/S;"~tE - .-'" VII .2.' 2. 4 )." Ao.

Fro,.. +/Ie re St.llt (4 P~bl""'!!:.!.., vi:': ""-t-<~) Vo 'J..

u//,:,. (£'J!/2(~'1~Lfv.1q 'uH'::'/u;:;' _ ('~~~.'(~ )1''l.lJjc f(~J)v.~J r ~ w,!#(J.% = f(It!.)1"EO>::: X (~-:')~'~"'D$'4Jt ~ u:)= -f J£. E(~)i(')d"" :. f-(i:..i' Il.if.c.DSZwt (a.w) J_t-i"'''' i' - -i;-(:_.i )t+~"'V,z'D~lltJtut1u;') 7$ [f.-(.!i!-)J.4-J.~V/(A~/wtJ/lf(:ti!) v.zC/)~1.OJtJ = -te(hit:: U,':,..)/u,':;''f ic~./A, 1 Id.!}xc.l)" ;0.2"

U.tl'/ (II) I (ll I. I( ~ 013:. 1111'1 V.WI &: Jill' C'T e'-l71). •

()::;'/U~=t;(~l.l.,~(o'2.TTl'" 0.00'18

..15.4-

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u;::/ U'::") = ..fo( ~I.)~ cb(4.",,/. b.o'o'lB

U::';/CI~::" c: z;z_(AJ.)'. m(o.zrr)61& 0./)0009{,

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56

IS.S-

57

510 16. Magnetoquasistattc Fields

Solution: From (16.721. w~ find

2 x 47r X 10-7 X 2 x 1O~ x 2 x 10' x 1.0 $/>'eoll - - 64 x 10 N

7r x 0.5

- 6.5 metric tons

Problems1«1.1 A small circular loop of 5 mm rad ins is placed 1 rn away Irorn a 60-Hz power line.

The voltage induced on this loop is measured at 06 microvolt. What is the current onthe power line?

16.2 Assume that the current on the infinitely long line shown in Figure 16.1 is thetriangular pulse shown in Figure P 16.2. Find thp. induced voltage on the rectangularloop. Use the following data: a - 2 ern, U - 4 ern, and d - 1 cm.

16.3 Consider the network shown in Figure P16.3. The magnetic flux is increasing at arate of 0.5 Wb/s in the direction pointing into the paper. Find the readings of thevoltmeters shown.

16.4 Find the readings of the voltmeters shown in Figure PJ6.4. The magnetic flux isincreasing at a rate of 0.5 Wb/s in the direction pointing into the paper.

1 fampflrAS)

3 I [microseconds]

Flgur. P1e.2

,-----------,I 3kO I

''-'--, ¢I/ ~

4.5 kO / V$

v.r\+ I'r:! I

'- / II '-,- IL_~ J

,..-----------,I S kfl I

I'-''-) I.4.5 kfl 1/ 0'V,

v~O O~z I- \ / - I\ .> I <, I

V'/ I '-'J jL. _

,/

,../',..,\\

v~\\ ,/\/

Flgur. Pie.S Figur. Pie."

Problems 511

~----~kO --- <,

I p~ " PI

IIIIII\\\

1 kG

ne.on

:helar

t ahe

16.5

is

1 kO

Figure P16.5

Four resistors form a circuit 8S shown in Figure P16.5. The total magnetic fluxlinking the circuit is increasing at a rate of 0.5 Wb/s, in the direction painting outof the paper,

(a) Find the direction and magnitude of the induced current in the circuit.[b] Find the readings of the voltmeters Vt and V2•

16.6 Two resistors are connected by wires to form B circuit as shown in FigureP16.Ga. ThA magnetic nux linking the circuit varies with time. Figure P16.6bshows the time variation of the magnetic nux. The positive value of the fluxcorresponds to the flux directed into the paper. The magnitude of the nux isfor a single turn of a circuit loop that encircles the magnetic flux.

(a) Plul the current 'It) versus lime. Be sure to mark the scale of the current.(h) Plot the voltage VeL) versus time. Mark the scale.

7 kO

,....----------,I 1(1) I

II

'kilO VIIi

II

I IL ...J

(a)

\}I (Webe,·s)

1

Seconds

0.5

-0.5

1

(b)

Figure PUI.S

512 16 Magnetoquasistatic Fields

16.7 What is the EMF induced on 0 propeller hlade that is 1.5 rn long and is rotating at10,000 r/min in the earth's magnetic field (0.5 x 10 • Wb/m2)?

16.8 Find the voltage induced in the rectangular loop shown in Figure 16.1 if it is rotatingabout the axis parallel to tha z axis located at x - d + ~.Assllme that the angularfrequency of the rotation is wand that the infinitely long wire carries a direct currentof I amperes. Show that thA induced EMF is not a pure sinusoidal voltage. It isapproximately sinusoidal when d » a.

16.9 A magnetic coro is made of a material whose hysteresis loop is shown in FigureP16.9. Note that this hysteresls curve is not a "square loop." To read the content ofthe core, two pulses al'e applied to the wires. The currents generate an Ii equal to200 AIm. The core has an area of 3 x 10 -7 m2•

(a) What is the voltage induced in the sensing wire if the core is originally at the"zaro" state fat point C)? Assume that switching from C to A is linear with timeand that it is completed in a microsecond.

(b) What is the voltage induced in the sensing wire if the core is originally at the"one" state (at point A)? Assume that switching from A to A' is linear with timeand that it is completed in U.5 J.l.S. This voltage is the "noise" voltage because itwould ideally be zero if the hysteresis loop were 1;1 perfect square.

B (webers per square meter]

0.1

100 200 H (amperesper meter]

tSensing wire 0.220.3

-200 -tOO-0.1

Figure P16.9 Ferrite core memory and Itshysteresis loop. -0.3

16.10 Find the total expansion force acting on the surface of an air-core solenoid that has100 turns of coil and radius a - 1 cm, length ~ = lU CII1, and current I - 10 A.

16.11 Repeat Problem 16.10 for the case in which 100 turns of coil are wound over aferromagnetic core with J.I. = 10001-'0' The current is 10 mA, with a = 1 em andP = 10cm.

16.12 A copper pipe of radius a - 2 cm and thickness d ~ 0.1 em is placed in a solenoid thathas 200 turns per meter and is excited by 0 1000 Hz 10-A current. The conductivity ofthe COpPHf is 5.92 X 107 mho/m. Calculate the puwer per meter dissipated in thecopper pipe.

16.13 A transformer similar to the one shown in Figure 16.11 is made of a steel with relativepermeability equal to 1100. The effective length of the core is 40 em, and tho fluxdensity is B - 0.3 Wb/m2• N1 - lOa, Nz - 1,000, I. - 60 A.

(8) Find 12, assuming that the transformer is an ideal transformer.(b) Find 12, using (16.37).(c) Compare the two answers.

itic Fields Problems 513

(amperesper meter]

16.14 Consider a magnetic circuiL similar to the one shown in Figure 16.11. The effec-tive length of the core is 0.4 m and its permeability is 2000 1-'0'The cross-sectlonalarea of the core is 4 x 10-4 mZ

• Let I] = 10 A, I2 = 24 A, N, = 50, and Nz m 20.

(a) Calculate the B field in the core. Give both the direction and the magnttnda,(b) If I] is a-c with f= 60 Hz, what are IV 1 1 and 1V 21 ? Assume that the magnetic

flux always slays in the core without any leakage.

18.15 The primary coil of a transformer has 150 turns !U1dthe secondary coil has 450turns. The effective length of the core is 0.5 m and the flux density in the core is0.25 Wb/m2• The transformer is similar to the one shown in Figure 16.11. Assumethat 11 " 60 A ami there is no nux leakage.

(a) Find 12, assuming ideal transformer condition.(b) Find Vz, assuming ideal transformer condition and V, = cos(120wt).(c) Find I2, taking into consideration that the core material has a finite permeabil-

ity equal to 10001-'0'(d) The hysteresis loop of the core material has an area equal to 90 Wb-AlmJ.

What is the power loss due to the hysteresis in the transformer? Assume thatthe core has a cross-sectional area equal to 4 em",

16.16 Figure P16.16 shows a magnetization curve of a core used in a transformer, Noticethat the hysteresis is negligible in this case and that the curve is linear in the rangeo s 1 H 1 :S 150 AIm but saturated when H is increased beyond this range. Let usnow review Example 16.10. Because 1 VI 1 = wN1'lt, we want to use maximum -v inorder to minimize the number of coils in the transformer. Using Figure P1B.16,explain what will bappen to the shape of the -V(L) and consequently to the shapeof Vt(L) if v is too high-for example. if -v is so high as to correspond to B =1,2 Wb/m~.

16.17 Estimate the approximate power loss attributed to hysteresis in the ferrite coreshown in Figure P16.9 if the core is switched back and fortb between "zero" and"one" states 1000 times in a second, Assume that the core has an average radius ofG x 10-4 m and that its cross-sectional area is 3 x 10·' m2,

B [webers per square meier)

~otating at

is rotating.e angular-ct currenttags. It is

in Figurecontont of1 equal to

illy at thewith time

illy lit thewith time:Jecause it

are meier)

1 that has\.

1.0

0.5,d over a1 em and

200 H (amperesper meier)mold that

ictivity ofed in the

-0,5

-1.0Figure P16.16

h relative1 the flux 16.18 Show that the mechanical torque required to drive an ac generator is nnt constant

with time or. 10 he exact, that it consists of a constant term and a term that variessinusoidally with time with an angular frequency 2w, What is tbe time average of thetorque? Express the torque in terms of the area of the winding A, the current I. themagnetic flux density B. and the phase angle a between the voltage and the current.Plot T as a function of t for a - 0

514 16 Magnetequaaistatic Fields

16.19 Figure 16.15 depicts an ac generator with a single coil being rotated in a constantmagnetic field. It illustrates the operating principle uf 8 single-phase ac generator.Lei us nuw consider a three-phase ac generator. How would yon physically arrangethree sets of coils in order to generate three-phase electricity? To illustrate yourdesign, sketch a diagram similar 10 Figure 16.15.

18.20 What is the total mechanical torque needed La drive the three-phase generator thatyou have designed for Problem 16.19? Express this torque as a function of time interms of the appropriate parameters. Plot T AS a function of time, and compare itwith that obtained in Problem 16.18. Is the instantaneous mechanical torque"smoother" (does its time-average value fluctuate less) compared with that for asingle-phase generator?

U5.21 Design a coil configuration similar to the one shown in Figure 16.17. Design it in sucha way that it will produce a rotating magnetic field in the armature-stator air gap andthat the field will have an angular speed equal to w/2 when fed with the a-phasecurrent shown in Figure 16.18. Prove that your design is correct by drawinginstantaneous-current diagrams similar tu those shown in Figure 10.17.

16.22 Show Qualitatively that the torque generated by an induction motor may be varied bychanging the resistance of the windings of the rotor. Figure 1fl.21 shows torquecurves versus vJv ; with three different values of rotor conductivities. What are therelative magnitudes of crt.CT2' and "3?

16.23 Refer 10 the synchrunous motor shown in Figure 16.22. What happens when thetorque angle is negative-that ill, when the position of the 1'0101' magnetic-moment isahead of the magnetic fieW?

16.24 Consider the coils of the magneplane's track shown in Figure 16.25. For themagneplane tn travel at a speed of 250 km/h, what should be the distance 2, inmeters? Assume that the power is provided by IJ three-phase 60-Hz power line.

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61

AppendixE Answers to Odd-Numbered Problems

Chapter 1

Chapter 2

Chapter 3

Chapter 4

1.1 (a) 5 + j3 (b) 11 + j [c] - 26 + j2 (d) - 2.2 - j1.4 1.3 Cooswt; sin wt; 1

1.5 ± 21/4ei(w/8) 1.7 Proof 1.9 (a).J2 cos (wt r~) (b) 4 cos(wt + 0.8)

(e) 3 cos (wt +~) + 4 cos(wt + 0.8) 1.11 (a) -6x + 59 + 2!

(b) -10i + 13y - 4Z (e) - 55 (d) 23x + 22y + 142 1.13 Proof

1.15 ~(5X - 8, + 2i) 1.17 Proofomilled 1.19 Proof omitted'193

1.21 Proof omitted 1.23 jW[(3 - j4)x + 8(1 + j) 2] 1.25 (a) (-1 + j3) 9 + (1 + ;3)Z (b) 2 i + (1 - j)ji + (1 + j) 2 (e) - 5 (d) 4i - (1 + j3H + (-1 + j3) Z 1.27Sketch omitted

2.1 - 6yi - 3x2y,6z 2.3 Proof 2.5 Proof 2.7 No 2.9 B(y,t) = O.3(k/w) cos(wt + ky) z 2.11 £1 + Ez, H, + H2 B1 -t B2 and D] + D2, superposition theorem2.13 Proofomilled 2.15V x E = iwB, V x H = J - iwD,V' B - o and v : D =Pv 2.17Proofomitted 2.19t,t 2.21Proofomitted 2.23UII/Ue"" 1.13 x 107

3.13.6 X 10-1U W/m2 3.34 x 1026 W 3.54.1 x 10'Jkm 3.7 (a) rad/sec (b)m"(e) sec-' (d) sec (e) m 3.9 (a) 2.63 m (b) 0.704 m 3.11 Yes, -z direction,

f 1 ,- ~ j E~z 3.13 No, Maxwell's equations not satisfiedv !loJ,~ 2 J.lo

~1a) Right-hand circular polarization (b) Right-hand circuJa.r_polarization(e) Left-hand elliptical polarization (d) Linear polarization @)roofomitted 3.19 (a) 1 (b) 1 (ell.58 (d) 2.12 3.211.34 x 10-5 m, aluminum foil isabout se thtck S.232.65mW/m2 3.25 (a) E. = e-o.Gze-Ju.~z (h)H. 9(0.5 - jO.5)e-O.51 e-}U'~' (e) Sketch omitted (d) Sketch omitted 3.270.6 x 10 6 ill

4.lfi1.9° 4.3 (i) c (il) f (iii) b (Iv) a (v) d (vi) e 4.6Yes, circular shape,0.3 ff on each surface (0 is the length of each side of the cube.) 4.7bevelled angle_ 3So;mirror making 70° with z axis; R polarized 4.91ElK 1~ ~IEol11 - e-,0ll>3' ei""I.

1El.l- v'31EoII1 + e-/Z53' elk.. l. 1E2, 1- 0.51 1';u1l1- e-'2~.a·le-~""'.1 Ezzi ~ .J32' 160

IEo111 + e-1Z53'1 e- JII<l76k.z 4.11(a) 9Et, (h) 75MHz

520

ChapterS

Chapter E

Chapter'

Chapter

Chapter 5

+ j3)27

Chapter 6

em)=

: 107

I)m-t

1,

Chapter 7

)D

l foil is- jO.5)

mgleChapter 8

f360

Answers to Odd-Numbered Problems 521

4.13 cos " Jl.l(Il'lE~ - }l, f',l c 1 tl(Jl.Slz Jl.lflJZ 2 • OS

f](}l~ - }l,) J.ll(li l~J4.15 (a) 80 cm in front of the plate(b)2V/m 4.171.996IEol 4.19H'- "llop'-I."inl-/k .. ooll.Ei _ (x cosO - Z sinB)HOlle-I"'.'n'-lk ee'O,;:H" _ Ylloe-1k"''''-lkzco,B.

Et = (x cosO + z sinB)Holle ,h<int·lb"",'. where 11 - ~~

4.21 x = 0.87 m, Y - 1.5m. 12.04dJi 4.23 Proof omitted

• Eo 'kz-z-P.''1]

5.9 Proof

E5.1 Proof 5.3 1.875 kHz 5.5 E .. x Enelkz, H = - y ~ elkz• J. =

Eo '~ '1]on lower plate, Is = i - e' Z on upper plate 5.789.33 kW

1/5.11 5.26-10.52 GHz for 2.85 x 1.26:l (cm) waveguide, 21.1-42.2 GHz for 0.711 x0.355 (cm) waveguide 5.131.318 MW5.15 Ey = E1 sin ('IrX/O) elk," 5.170= tan-l(na/mb) 5.195.83 GHz

Hx = (l':lk./WIL) sin (1I'x/o) ejkt~ .Hz = (j E 11I'/WJ.lO)cos (1I'x/o) ejk.~ 5.21 Proof omitted 5.23 (0.866. 0.5. 2)where k~ = [w~ ILE - (11'/0)21'12 V

5.25 Proof omitted 5.27 A = 4.93 i + 7.469 - 3t 5.29 (a) E' = P 4-,1:]·,P

I A Vt _jk. I •H = c{> - e 1 (b) Vo = Va (111 - 1/oJ/('I11+ Ilo). V, = 2VO'1]I/(7]1 + 1/0)'1]lP

rrr:': Vu Jkz 27rVa kz6.12000V 6.3 (°1/°2) '" (b,lU2) - "1000 6.5 j.=! - e- '.1 = --e-'

1/a "I

6.7 (a) I V (z) I = 21 V + II cos kz 1.1 I (z) I = 2 ~ I sin kz IZo

(b) Sketch omitted (e) 00 6.9 (a) 2.96 (b) z = - 0.35).. (e) 24.5%6.11 0.342).. 6.13 d = 0.25 cm 6.15 (a) 1.26 + jl.61 (b) 0.54 (e) 16.17 (a) 0.61 + jl.33 (b) 0.15 jO.09 6.19 Proof omitted 6.2114.2 kW6.23 48.6% 6.25 Sketch omitted 6.27 Sketch omitted 6.29 Sketch omitted6.31 Proof omitted

7.1 (0.75,0.433,0.5) 7.3 Proof omitted 7.5 ~ . 8 = cos8 cosc/l.~ . ' t= sinOsine, y . 8 = coso sin4>,$>. ~ = 1;084>.Z . i' • cosO. i . 0 = - sinB,Z . ~ = 0

·kID. -IJoe7.7 Proof omitted 7.9 Yes, improved to 18% 7.11 E = (- y) J ZIl8 •

811'xlinear 7.13 (a) 0.314 VIm (b) 0,628 VIm (e) 6 VIm 7.15(a) 1 (b)1.5(el1.64 1.17 Six lobes; beam width = 19.2° along c/l= 0; beam width = 26.4°along e = 41.8° 7.19 D=4 7.21Sketchomitted 7.23 (a) -9Uu (b) 6; 1.414; 0;1.414 Vim (c) Sketch omitted (d) Sketch omitted

8.1 UH'UB - (krJ~«1 8.3 When background is dark, one sees light scattered bythe smoke particles. Blue light is scattered rnure strongly than red lighL Against abright background. one sees lighl passing through the smoke. The blue light getsscattered. and red and yellow lights suffer less scattering. 8.5 Proof omitted8.7360 km in radius 8.9 Train is moving toward the intersection. 8.11 Band-width = 59 kHz. 6.7 lAoS for 1 krn resolution 8.13 Circular but opposite hand8.15 No

Chapter 9

Chapter 10

Chapter 11

Chapter 12

522 Answers to Odd-Numbered Problems

9.1 Exact: (a) 5.5302 x 10-10V (b) 5.54244 x 10 12V Approximate: (a) 5.5426 x10-10y (b) 5.54256 x 1O-12V 9.3 x = 1m plane, y = 1.5m plane 9.5 (- i)1.44 X 10-3 pi• z Ylm 9.7 Proof omitted 9.9 Sketch omitted 9.11 (a) P __

r Pi 40nl1(0.0499) VIm (b) P __ (0.05) VIm (c) 0.2% 9.13 E - 0 for r <

401!'fflo and b ~ r < e, E = t _3_ for a ~ r < h and I' ~ e 9.15 10 -6

411'€l'2 fT2

l2 - e-'(r2 + 21' + 2)1 9.17 1~-1i (3e-1 1) Ilint: S dre '(1 + 1')11'2 = -e-'/r

9.19 3 V, Independent of path 9.21 (a) _q_ (b) __iL (c) ..sL(~ + 1. - 1.)4?rcc 4?rft; 4n I' e b

(d) ..sL(:! + 1. _ 1.) 9.23 (a) 4.03 x 10-0 C (b) 330 kV (c) 0.4 rnA411"~u e b

10.1 - i Q 10.31.8 x 10-5 N (attractive) 10.5 (a) 1'(2.411" x 104] VIm9

(b) a x 1.211"xlO-8N (repulsive) (e) No 10.70.36 mrn 10.9 z = ::t: 3.14 cm10.11 (a) Vo = 1.1374 X 107 mIs, vo. = 1.867 X lu7lnls, Vox = - 0.163 X lU

7mls

(b) x(t) = (8.78 x 1014)e - (0.163 x 107)1 m, z(t) a (L8n7 x 107)L III (e) x = - 3.5210 4 3 10-2x m, Z - .x m . 10 . A (€I + ~2)

10.13 Proof omitted 10.155.3 x 10- farad 10.17 (a) J -2-

(b) Qj = ~. Q2 = ~ 10.19211" / [lin (clal + 1 In (hIe)]~I 1 f2 fl + €z El E2

10.214.97 x 10-11 J 10.23 (a) Q2S12Afo (b) - Q2/2Ato (attractive)

11.1~3 = ~l + q,2 11.3~ = VIn(blp)lln(b/o) 11.5~(6) - VoIn(tan~)/

In (tan~). P. = - VOfo/[r In (Lant) SiDO] 11.712.3 V 11.9 Sketch omitted

11.1117.5p.p. F/m 11.13 _Q(dZ - o2)/[41ro(a2 + cF - 2ud COsB)3/Z)q (1 a) q(o/d) f qo11.15 - -R - ,Jo + . where Rl - (r2 + dt - 2rd eos6)1/2.

411't 1 uu~ 4wtr

(u' rei )1/2Ra ~ 1'2 I IF - 2 CfcosB 11.17 (pd/1211'E(d - blJ where d - at/h. attractive

11.19 (d/o)q2/[4wE(d - bfl, where d - o2/b, attractive 11.21200 sin (2wx/ajsinh(2'lTyl o)/sinh(21!'bl c]

I L-1{sin1nWX/a] sinhtney/c] sin(nwylbj ~inh(nWXlb)}

11.23 (400 w) _ + __;:,.__:.._,....:.--.:,....-_....:..n sinhln-rh/u] sinhf ns-u/b]

fl-udd

11.25(a) Vo (1. _ 1.) (b) Vo

(~ _~) r b (~ _ ~) ri

Cha

Cha

Ch,

12.1 C = A(Fl + Ez)/2d,C = A(u1 + (2)f2d; C in parallel with G 12.3 (a) lln [o/p)11 lin [b/o) In (e/b)]

(hulf) (b) I ln (a/b)/(2'lfoli) + IIn (blp)/(2'lfu2iJ (e) - I211'f °1 U2

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Answers to Odd-Numbered Problems 523

12.50.92 x 10 ~mho/m 12.7 10{jV

(112 ~12.9 -- x 100% 12.11 12.3 n-m 12.13 Sketch omitted

111 + 11

12.15 p = 600 f.! _ 50 + 50 )u l6 d[(y + 6 - foyld)2 + (70 - 70ro/d)2]112 d[(y + 6f + (70)211125'

where d = ll7012 + i]1/2. To = (50f/f(70f I' lll/2

13.1 Z 2 .J2 IJ( 7I"b) 13.3 H - (- ilJy for Iy I < Q, H = (-x)J(d/2) for y > en.2

H _ x ](d/2) for Y< (-d/2J 13.5 (a) 0 (b) 1(/ - o2)/[2'Jlp(h2 - 02)] (e) Ii21rp

13.7 (a) (-z)Io(dy)/l411'(a2 + y)3rl) (b) (-i)I/(4?!'o) (e) 0 (d) ( z)I/(4?!'0)13.9 zo = ! 0.27u

13.11 HAB = z.£!_ ra dx411' J -a [02 + (h _ X)2}3/2

H _. I(b - 0) fa dxBU - Z 47r J n II+ [b 0)2]3/2

IT _ z /(b + u) ru _ dxDA - 47r J -0 Ii + (b + 0)2t/2

13.13 2.8 MI-l7. 13.15 (a) x - 0.04m, Z = - 0.00725m (b) - 20.5° 13.171.33 x10- 3 N/m (repulsive) 13.19 (a) Loop should bs placed horizontally or vertically inthe east-west direction (b) 7.85 x 10-3 N-m (e) VerticaJlyin north-south direction

13.21 (a) H _..1L (b) UH

~ ~ (e) 5 x 10-8 HIm 13.232.74 x 10-8p 27r02 I6'1r

• fJ.o [1 c4 in (db) c2

(;2 + b2 J

Joule/meter 13.25 - - + In (blu)·~ 2 22 - -2--2 + 2 22?!' 4 (c - b ) c - b 4(c - b )

14.1Sketch 14.30.21 weber/rn', 100Aim 14.50.71 weber/rn! 14.7X3Y, - 1.X~Y2 - O. x,y, - I, X~Y4 = O. XIY2 - 1. X2Y2 = 1. X.Y2 - 1 14.94.233A 14.118.32A 14.130.29 weber/m'

15.1EIO) _ 0, Hlol __ y 10 cos[wtJ/w: Ell) ~ -x IowlJ.Z sin(wtJlw. HIli - 0; EIZI = O.

HIZI _ Y Iow2tp. (~) 7! cosfwt)/w: Ei31 - X /rI"lllE H) (~)t sin(wt)/w: H131 - 0

15.3uW - ~ W2E2V~ (~) sin2(wtl(~) uwR~. U~I - 0 15.5UWI - O.

!l [2UI~I - - -; cos2(wt)(waRl

2w15.7 E~ol = Vocos(wt)/[pIn (b/a»), 1(01= 2;rVu'fPcos(wt)/ln (b/o), Qlol_2?1"Vo&cos(wl)/ln [b/o), 1111 - -2;rVuEPwsin(wl)/ln (b/o] 15.911.1 sec

16.1101 A 16.3 VI ~ 0.25 V. v, = -0.15 V. Vs = -0.25 V 16.5 (8)0.125 rnA(clockwise) (b) VI = - 0.25 V; V2 = 0.625 V 16.758.9 mV16.9(8) -0.l2V (bl -0.012V 16.110.395 x 10-JN 16.13(a)6A (b)5.91A16.15 (a) 20 A (b) 330 cos (1201l't) (el 19.78 A (d) 1.08W 16.170.16 mW16.19 Sketch omitted 16.21 Sketch omitted 16.23 It becomes a generator.

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Solutions Manual for Shen and Kong's Applied Electromagnetism

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