26477320 electromagnetic fields and waves hw4 solution iskander
TRANSCRIPT
Chapter 2
Maxwell's Equations in DifferentialForm2.1 f r .,-l
v=E,lr-lgll..ore'L \o) lin cylindncal coordinates
a ta aV =
=-a^ r---O^ T:a_dp ' pdQ' dz
dv tdw dwvlil = _ a-*-=;a^*-=-O
= oJu.,'+,',,0,1^.'
4,-i:l'.l.,," *^ * u,l,-1,u)' 1.",*.n p" 'P ol \n/l -'L \o)l--''-'
2.2 A= -ya +.ra : find V x A.v'
f: ": ':lcurlA=VxA=l* + +ll* dv *ll-v " ol
Solving determinant yields:
^.( o-+)- u(o*+). "(4*?) =an+r)=2a'\ Az) '/\ Az) '[a" A) -"-''' z
CHAPTER 2. MAXWELL'S EQUATIONS INDIFFERENTIAL FORM
.'. V x A=Zaz
t-Evaluate $l.aV around ,' + y' =1.
J.To pe*orm the integration around the specified contour, we must transform the vector A
from rectangular to cylindrical. To change, we use:
where
x = pcosQ
= psinQ
A= Aoar+ Ara,
Ao = A"cos@ + A sin@ where A, = -l
A, = -A"sin@ + A, cosp where A = x
A, = -ycos@+xsin@
A, = -(-y)sin@+xcosp
Substituting the above formulas for x and y for Ao and Ar:
Ao = -(psin@)cos@+pcos@sin@=0
A, = psin@sin@+pcos@cos@=p(sin2@+cos2 Q)= p
... A= pao
Butp=l=constant
f-frtf2o.'. 0A .dl. =Qpa^ ldp^"+ pdta^+ dza.l= | p'dQ=znpt
Jc J - \ Jrt
l-.'. SA'dV =2x(l)2 =2nI.c
over the surface bounded by x2 + )2 = I in cylindrical coordinates.Evaluate JV
t,l,.a"
IRM
, . pdw pa, =
I,=' I" r row o = [l=' z orl^" o, = 2op,l,o = 2x
P^ alFA'vlvxA=14 4 4lIdP dQ o2lr- illu P\P)',
frI VxA.4s= l2a.lr J.r
Stokes' Theorem states:
=70 *-a,(o- r.?(#-r=|trr)^, =t^.
Jv"r.as={1 av
Jv"a ar=f.a.dv=2r
.'. Stokes' Theorem is satisfied.
2.3 F=2pa, wherep=3, 0SQS2n, 0<z<2
The Divergence Theorem is
dr a, = f aiu F.auJ, J.
divF= v F=l+?t3.+4.+) -!a?p' +0+0 =4p _4\p dp ppa| az) p 0p p
+y.F=4
Jaiu r a" = [,+a" = I f" f apdzd@p
= II"^ 2pdQdp=J', 2npdp
t6r,l' r= ,. p'1"=J27t = J"div
F dv
Right-hand side:
Left-hand side:
f,n a' = I I"'r",' pd@za,= f, t"ro'o*,
= (zp'oli" a, - 2p, . 2,al: = 4xp2 . 2 - BEp2
But p=3'
CHAPTER 2. MAXWELL,S EQUATIONS IN DIFFERENTIAL FOI
I.'. 0F .ds=8np2 =8x.9 =72xJ.
Prove in Cartesian coordiaates that v(MT) = TvM + MyT.
.[+q-4)=o'\dp dQ)
T and M are scalar fields.
.'. By definition:
.'. V(rW =
2.4
r=r(x,y.z) M=M(x.y,z) v =(*-&-*)
(*. *. *)t a,,, z) u (x, y, z)l = (*. ft . !)r,rt#. #. ry =, ff *, # *, ff *, *4., #. * #'(# - #. #). *( *q. K. 9= rv M + MY r
(Vx r) of:
_ ZL,
la" all; uQ ;lla a alPp ao al
lo' | -,1
Iu.( "r," -+) - "^f+a - 4) * 1"p'\oQ az)-'ldp E)-p- ya, - x'a,
lt' u, ". Ilz a al
lax 4 azll^ rllrxz -y -x'l
Proven.
Find the curl
(a) A= p2ao
VxA =
(b) B=3xaa,
VxB =
2.5
= ".W ry)-",(ry-ry)."{+ ry)= -ar(-2x -3x) = Jyv"
(c) c =;f a,
-2pao
I a, aA aal
l.2sl,td t*to ;lVxc=14 4 4l- ldr a0 aal
l"'' o o|
I t (r-4:l*lu-fo -d"''\= ,-a.(o-o)- -a,[ dQ ) r ,\ * 1=o
(d) D= pao*pza,
la^ arlrarlID A DIt'.1VxD = l+ + +lldp dQ czll"llo p'o p'l
= !""( +-+l -"^( {-+).1" f4q-4)= -a.(2p) =p ,\dE dz ) -\dp o?.) p ,\dp dQ) E
(e) E= xzar+yzar-f'a,
la a alI' v .lVxE = Ig 4 4l
l* oY *,1lxz yz -y-l
= ^.(a( -tl-4pl -^(u(?')-4+r) *u( ar!,,-egl)t dy oz ) '(. dx dz ) "\* "y )
(f) F = Kr'a
VxF =
2.6
CHAPTER 2. MAXWELL'S EQUATTONS IN DIFFERENTIAL FORM
{.r at
S6
*ffr u.**ffr -^,0,*fi -u.*'ill,=,.
f:r . I ;!' *1,=,= ,' *l'^1,-, = |
= If ",0'
= r*.1.(a)
latiVxF=l*l*l_2l<"
(b)
Jv"r a, =
a al: ,'l ( ^ a(-y')') (^ ,(.,)') (a(-y,) &,)fr, ftl="|0--# l-""1 0-+ l+a.l -f -+ l-22a,
-t \ dz , '( dz ) '\ dx oY)-v' 0l
f fr fr pr I'.'.
J.v x F. ds =
J, J. 22a,. a dxdz =
Jo ruldr= .'ll = r
4," .. tn ,
1z=l;r=l | ;z=t;x=t | ;z=t pr=t I
l,^l ^rr",.adydzl *|"|
^2za,.adxdzl *|^I 2ru,-adydzl
rrz=0rrr=0 lr=t Jz=oJ t=o lr=, Jz=OJy=o lr=o
- l)=J.=,'^, u "*orl,=,. L,E,"u, -^,*orl,u
= l'=' ['='rr*0, = l'=' 2rd, = zrll^ = tJz=O J r=O J z=o 'u
Sl is the only one in which the dot Foduct equals I
Theorem = [v"F'ds= 6r aZJ, J,
M
I,: ou*!": o*f a-t
'
f'' ̂ , roo, * Lo ̂ ,' rsinodfa, * f,,F'' rdoa'
0 because all dot Products = 0
aol
A= **a,(0 - o) - -l= ",[. - #). i',(o - #) = o
I
0l
Jv xF'ds = fo a' =o = {Y'
a-t
VxF=
6E.d!.t
==
I a aa
lr2 sin0 rsmu
la d
la, ao
lr o
Proven.
B = p^o- zl"rstokes' Theorem = I,o t F'ds = {' O
2.8
CHAPTER 2. MAXWELL'S EQUATIONS IN DIFFERENTIAL FORM
l+ No 7lvxF=14 4 +l=
lop dQ *llo P P -zl
(a)
IV,. r.a. =J.
{1 az = f", ,o*,1,=,= f. ,rrroQl,=,= f,' ,'orl,=,= Qll" =zn
;",(w yl ",(T- o). ;",(#- o) =,".
!za,- papdpa,= f" l,zpdpdQ
= fo"
aE =zo
Jvrr,r' =
sl
pr p2r | 1t P2r
= )o J" '",' '4M'z^rl,-* J, J, 2a,' PdQdPa,
pt 12n fl
), ), zoawo = )oztaP =zn
--!-
5, is the only side for which the dot product =l
(b)
"FI Vxr ds+ lVxF.dsJs, Js,
s2
2.9 (a) A = Yzax+ xzay+ xYLz
(b) B = pa.
DivA=v.A= ry.T.ry=o
DivB=v B=;(#.#.ry)=,(c) C = ra.
Divc=v.c=, 1 [r(''i""r)*ry.+q] ]l',in! =t- r2 sing I a' ae a0 ) r'sino
(d) D= 2r'a,1,-,
Div D= v D= #u+t'!.#.PJ,=, =#1,=,=8'l'='=24
(e) E= 3xa,+(Y-3)a, +(2-z)a'
Div E=v'E= ry.N#.ry#=3+1 -r=3
2.10 F - pvoiza. ; Diver8ence Theorem $"r at = liu r a"
t_O(P' il *ag) = Llzp) +r = 2 +I = 3divF=O.r=, dp dz p
Fd PrlZ ?h Pu Pnl2
J ai' r a" =
=r{,}.j A!r:',,':os=s,+sr+sr+So+s,
S, = ds = -dpdza, Sr+ ds = -pd'pdQa" St + ds - pd@zao
so = ds = drydzao S, + ds = PdPdfa'
CHAPTER 2. MAXWELL'S EQUATIONS IN DIFFERENTIAL FORM
ft -aparur* lt -papqu..l,I 'pd@'za,* lr aoa'u' * I*t pdpdQ
Js, ,rz , ah ert2 I * o. l;,'l; zododQ\,=^
o.l:"1; -,papdQ\,="* .l. J. , odv'\o-.
o I e=lt' ' n
= o+o+ P':hl +o+z;7\ =a';h+a'-!hZ lp=d - -t1-h
aa t . 3n )l= ",r Arh+ j_A"h-
=A-n44+
Proving the Divergence Theorem'
| - ^'':r :-2V@, where P=2,z=1' 0(Q32tt2.ll 92P"lz+
r')srn e
L=2p1(z+ l)sin2 @a
,-<..(A\7
l0
$," " ==
Stokes' Theorem J
iv t n a' = {,1 aV
\+ xq ?l
lr A LlcurtA=VxA = Ia, ao &l
\t olzP'1'+ 1)sin'?Pl o l
= ! u,[ o - !7, p', r+ I ) sinz o l) - ", ro - o'.. I^,1*lz
p' rz+ I ; sin) o] - o )
P o\ dz'
= -2P2 s\nz tar+6P(z+ 1)sin2 @a'
4s= pdpdfa.
I,to x A)'ds
= -
i," fur'rr+1)sin? MpdQ= J,"u4[t'+1)sin'z@@
= 2 8(z+r) t-#\l" =rcQ+l)+ withz=1' =32n
check ={y av = Il',r',r+l)sin2 *rl,=,=I",.8.2sin,@Qz=l
= zz g - r^Ul,','" = rro=
I,o x A) .ds
1)n2.12 (a) s= pao, V.o=;fr=,
The number of flux lines in = thenumber of flux lines out.
.'. Net flux = 0 .'. Div = 0
(b) B=E60, V.B= 1aQD =Lpdp p
There is more flux coming out ofthe test region than going into it.
.'. Div * 0
(c) C=)a,, V.C= fr=O
Flux in = flux out.
.'. Net flux = 0 .'. Div = 0
12 CHAPTER 2. MAXWELL'S EQUATIONS IN DIFFERENTIAL FORM
(d) D=ra,, V'D= ##=o
t 0(r'r2sin0) l^,(e) E = ra, , V'E = ;.s*'- ar = -Z 5r- = 3
Flux in = flux out.
... Net flux = 0 .'. Div = 0
Flux in < flux out.
.'. Div * 0
(f7 F=zar, V'F=;ry=;
Flux in I flux out.
.'. Div * 0