Solutions Manual for Electricity and Magnetism

194 Pages·1986·4.898 MB·English

by  Morton K. Brussel, Munir H. Nayfeh

S O L U T I O N S M A N U A L
to accompany
E L E C T R I C I T Y A N D MA G N E T I S M
MUNIR H. NAYFEH
MORTON K. BRUSSEL
University of Illinois
JOHN WILEY & SONS
New York Chichester Brisbane Toronto Singapore
Copyrigh1t9@8 b6y J ohWinl ey& SonsI,n c.
Thisma. terima.ayl b er eprodfuocret eds tionrg
instructpiuornpoaslbe ysp eoplues intgh et ext.
7
ISB0N 4 718 0692
Printientd h eUn iteSdt atoefsAme rica
9 7
10 8 6 5 4 3 2 1
CHAPTER 1
+ + + +
1.1
Thec rossp roduAcxt B isp erpdeincultaorb otAh a ndB .
Sincwee w anat univte ctowre,j usdti vibdyet he
+ + +
magndie t:u -n- +A BX/ IXA Bl
+ + X y z
A X B = 2 -6 -3 l5x- lO+y 3oz
4
3 -1
IAX s1= /1s2 + 102+ 302 35
3x-
HencTIAe =, 1s x l-O+y 30; 72v + 6z
3'i
1. 2
Thep ositivoenc toorfst hesteh repeo inatrse
+ -A A + A -A
rl 2x y+ z,r 2 = 3x+ 2y z,a nd
+
r -x+ 3y+ 2z. Thep ositivoenc toofra na rbitrpaoriynits
3
+ A A A
r = xx +y y+ zz. Ifa llv ectolrasyi nt hep lanet,he tnh e
followtirnigp clreo spsr oduvcatn is.h es
whicghi velsl +x Sy+ 13z- 30 O.
1. 3
Thep oistiovne ctoorfst hespeo inatnsd a na rbitrpaoriyn t
+ A + A -,,- A
arerl Jx+ y+ 2z, r2 = X ,._y 4z, and
-
t = xx yy+ zz. Thee quatiooftn h ep lanies g overnbeytd h e
condit(ito nt )•(r-r) =w h0i cghi vetsh ee quation
2 2 1
2x +3 +y 6z+ 28 =f o0rt hep lane.
r dtnltl=/ rd/rr
V(l/r)= r d(l/r)/d=r - ;/2r
1.5C onsitdheesr u rfadceef inbeyfd where
f(x,y=, 2zx)z2 -3xy- 4x =- O7. Recatlhla Vtf i sn ormtaol
surfafc(ex ,y=, Oz,)t hen
afA afA afA
Vf - x+- y+- z
ax ay az
At( 1,,-2)1 weh aveVf: = (8+ 3 -)x 43-y +8 z
A A
7x- 3y +8 z.T heu nivte ctnoorr mtaolt hes urfaactte h si
point is:
n= Vf = 7i- 3i+ s:£ 7;_- 3i+ 8:i
"fvTT
/22 2 2 ✓122
+ 3+ 8
1.6B yd efiniwteih oanv He/ ds = IV t1 -NowV t=2xy3z; +
max
x2z3y+ 3x2yz2;,t huast X 2, y = l, z = -1w eh ave
A A A
Vt= -4x - 42yz . H+e n1ce, idstm /adxsi maulmo ng
2 2 2
(-4i- 4y+ 12�)//+4 4 + 12 direction and
2 2 2
IVtl/ 4+4 +12 =✓176
2
1.7U sinEgq .1 .57
+ 3 +3 3+ -3 + 03
V•(r/)r = V(lr/ )•r+V •r/r rA• r + = 0
4
r
1.8U sintgh ev ectiodre ntgiitvyei nnE q.( l.6w0e)h ave
=
+ + + + + + + + t+
V•(xA B =) B•(xV A ) -•A(Vx B) O, sincex =�V V x o.
+
1 .A9 i si rortatniaoli fV X wAh icghi ves
-
x(c+ 1 ) -y(a- )4 +� (b 2)= 0. Therefa=o r4,e b 2,
C= -1. HencA+=e (x +2 y 4+z )x( 2x +3 -y -z)y +
+
(4X - y 2 )+ Nzo wA z--."v�w-- � xA� + YA�+ Az . Thus
ax ay �z
ar,,
+ +
- = X 2y 4z Partiailnltye grwaittrehe spetcoxt :
ax
2
+
<I>=�+ 2xy 4xz+ f,(zy). Now�=2 x 3-y - zt,h us
2 ;)y
partiailnltye grawtiitnrhge spetcoyt gives
2 +
<I>= 2xy - 3y2 -/yz +g (xz),. No�w = 4x - y 2z, thus
az
partiailnltyea gtrinwgi trhe spte tcoz gives:
+ 2
<I>= 4xz- yz z +h (xy,).Co mparitnhge3 <!> ',s wes eet haitf
f(y,z)= -32/y2 + 2 z-yz, g(,xz)= 2x/2 +4 xz + 2,z h(,xy)
x2/2 + 2xy- 3y2/2 thewne g eta w<I>h erVe<I> A=+. Thaits
<I>= x2/2 +2 xy +4 xz- 3y2/2 + 2 z-yz + cao ntsatn.
+
+ +
1.10V •A l 1 a 2 + 0a w hicghi veas= -2.
+ + 2A
E= r/r= r/ri sc onserv(astpihveer ciocoarld inaartee s
+ + E
impeldi)i fV Ex = O.Notteh at= E;
r
3
r re rsinei
+ + a a a
VX E 2 ar TI �
r sine
0
+ +
V X E (rileE /a�- rsini8il E /38=) Q. Wec anq uicyk l
2 r r
rsin8
seet hat�s pheriyc syamlmlet(r8i,c�i ndepe)n rdaednital
fielidsc onservatTihvuVes X. ( r/=r )o .
+ A A il<I> A lil <I> A l il<I>
E =r /=r -V<I>= (- r - + e ----.+-- )�•
ar ra e rsrnae� -
Thu1s/r -il<I>/ar= -d<I>/drO.r- d<I>= dr/wrh icihn tegrates to
<I>= -tnr+ c. Appyilng<l>( r a)=0 givecs =t nat;h us
<I>= tna- tn=r -tn(r/)a.
1.12N otitchea itf w ec an finad s calfaurn ct<I>i(oxn), syuch
thaAt= V<I>a sr equiirnet dh es ecopnadr otf t hipsr oblem, then
+
itw oulbdea necesys aanrds ufficcioenndtii otnt haAt be
conservatiHvoew.e vweerw ,i lglo a heaadn de xpliyc sihtolw
+ +
thaVt Ax= 0 tos hotwh aAt i sc onservative:
A 2 2
VX A +x[il(3-xyz) /il-yil (3x- z)/ilz)
2 3
-y[il(3-xyz)/xil -il(6yx z+) /ilz]
2 3
+�[il(-3zx)a/x il-(6yx -)z/il]y
2 2
�(--1( -1)-)y (3z-(3z))+ ;(6-x( 6)x)= o.T herefore
+ +
Vx A= 0 + A isc onservatNiovwwe ef. i nd <I> such that
+
A V<I>.
4
+ 3 A 2 A 2 A
A (6x+y z )(x3 x+ z)y+ (3xz- y)z
3 2 3
Integrga at�ia/nx 6xy z+ give�s = 3xy +XZ + g(y,z),
2 - 2
Integraat�ia/ny g 3x z give�s =3xy z-y+ h(x),.z
2 3-
IntegirnagHt /az 3xz- ygive�s x= z zy+ f(x,y).
Byi nspectoifto hne set hreree altiso nwec ane asislhy ow
2 3
+
tha�t( x,y,z3)xy xz -zy+ C,
+ ax ay+ -az 3,
V•r= -+ - = Thes urfaicnet egral
ax lly az
+A
I= (! r•ndcaa nb ew rittiennt ermosf a voluimnet egral using
S
thed ivergetnhceeo rIe =m rv•dtv = fd3v= 3Vw herVe i st he
V V
voulme oft hes urf,a ce
aA aA aA
1.14a ) V•=A � + __:j_ + __z = 4 4-y+ 2z,
ax ay az
b)Fromh eroen ,i ti se asietsots wittcohc ylindrical
+
cooirndatweshe reV •A= 4 -ps4i¢n + 2z, Inc ylinrdical
coorndaitexs2, + y2 = 4i st hes amaes p = 2, From ftihgeu re
A A
theu nivte cotrsn ormtaols isp , normtaols isz ,a nd
1 2
s .!�
normatlo 3 is-_z. t Thed i vergetnhcoeer esmt ates
o 1 -rl!is;
+
f V• Adv= A• d = (4- 4psi¢n 2z)pd pd¢dz (,·.
V
:....__-
�.
3 2'1T 2
;
f dzf d¢f pdp(4 -ps4i¢n + 2z=) 84'1T
0 0 0 I I
t,/
Noww ec alcultahtese u rfaicnet egral
·,.J
+ +
1A •ad 'j
5
+ y
p icos+• y sina.n,dA 4xx - Zy2 + 2;z
2A 2A A A
(4xx - 2y+ yzz )• ( xcos+• y sni•p)d� z
(4pco2s• -z/si3n•)pd�z
-
+ + 2 3
Thigsi veAs• da= 8(co•s 2sin•) d• zdatp = 2
1
+ + 2A A
A•da(zz )•( z)pd.dwph icihse quatlo9 pdp d.a tz = 3.
2
A+ •d+a (z2z A )• (A- pzd).pd whicihsz eraot z = u.Additnhge
3
threceo ntrtiibounwseg et
1 x.aJ = 9 2 fpd p 2lf f d .+ 2f8 (lfc o2•s - 2si2•n)d3f• d z
S O 0 0 + 0
whicghi ve8s4 lft,h es amaes f V•Adv.
1.15a ) Wef irsfti ntdh e unit nTohrecm yalli.n drical
surafcei sd escrbeidb yf (x,)y x2 + 2y -16= o. We
n A A
use = ll/flVflT.h eg raidenotf f isll f= Zxx 2+y y.T he
magnitouftd hee g raedniti s lvfl Bute verywhere
/2
II
ont hes ufrcaew eh aev 2 + 2 y 4,t herefVofr =e 8. Hence
n =( x� y+y )/4.
b) + + A
SubtsitutfionrgA andn inA •nw eg etA •n(xz+ yx)4/.
Writitnhgii sn c ylindrcioocrdailn atweegs e t
+ A + A
(pzcos• Thu/s•A nda= fA• npdtdz
lf/4 5
f f (zcos+• 2sin.co=s 5.()5d+t<l1z) 2/.
72
0 0
6
X y z
1.16 V x A a/ax a/ ay a/ az
2 2
2x - y -yz -y z
+
V X A x(-2yz + 2yz) + ;(o - 0) + z(O + 1) z. Using
n = r = xA sin8cos<j, + yA sin8sin<j, + zcos8,
x = rsin8cos<j,, y = rsin8sin<j,, and z = rcos8, we have
2
(;)•(;sin8cos<j, + ysin8sin<j, + zA cos8)r sin8d8d<j,
cos8•d<j,d8sin8
at the surface of the sphere (r = 1). There fore
f V x A+ •nA da = 2Tf Tff/ 2 cos8sin8d8 = -2TT cos 2 8/2 TTI/2 'If. Now we
0 0
,, + + dt
calculate~ A•dr. In this case is along a unit circle on the
C
x,y plane and its direction is along <j,. Therefore
spherical coordinates we have
= -(2sin8cos¢ - sin8sin<j,)sin<j, - (sin8sin<j,)(cos28)cos¢
+ 2
Now, 8 = Tf/2 along C, therefore A•dt = -'?. cos<j,sin<j, + sin <j,
2Tf TT
f (-2cos<j,sin<j, T sin2.p)d<j, cos 2 <j, + 'If TT
0 ()
Thus f v x A+ •nA da
s
C
f~
1.17 a) <la=(). From problem l.l:\ we have
fVIP dv = ¢IP :i da. Let ~ be a constant function such that
7
V4' = O, and we can take 4' outside the surface integral. Thus
0 = 4if~ da or f~ da = O.
¢t
b) X n da = l) • From problem 1. Ul we have
Jv X -B+ dv = ¢~ X -8+ da. Taking -8+ = r, the radius vector, then
V X r-+ = 0 and we have 1t X nA da = o.
+ + + -+ -+
1.18 a) (Vx8dv=(nA x8da. Let A Bx C in the
-+
divergence theorem where C is a constant vector. Then
-+ -+ -+ -+
JV•( B X C) ()( B X C)•n da
-+ -+ -+ -+ -+ -+
Since V • ( B X C) C•( V X 8), and (8 X C) •n
C-+ •( n X -B+ ) then c-Jv x B-+ dv = c• tf nA x B-+ da or
8
Jv x dv
-+ -+
b) Jv4' dv Let A 4'C in the divergence theorem
-+
where C is a constant vector. Then
JV•4'C dv
-+ -+
Now we have V•(4'C) C• V4', then
-+
C• (4' n da or (V4' dv ( 4' n da
8