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