Electromagnetic Fields and Waves HW1 Solution - Iskander

CHAPTER 1. VECTOR ANALYSIS AND MAXWELL'S EQUATIONS IN INTEGRAL FORM

and OP.OQ = loPlloQlcos?

.'.

cos9 = cosdcosc[, + cosBcospr + cosTcosTr

t.2

(a) B=QP oP = QP+OQ

.'.8 = QP =

OP-OQ

|

\ t^

^ + ^za")-\zax = [-/a] = -2^,- 3a, -a.

\

+41, +Jaz/

(b) Assume the smaller angle between A and B is 9. The magnirude of projection of B on is lBlcos0. ... A B = lAllBlcoso

A'B ^ = --:--.'. 'h, |II|COSU tAl

(a, +za.

-:a.X-za, -:a" -a.) .,!I'+2' -t(-3)r

=

t'^l

t.(-2) +2.(-3) 114

-5

= -414 (c)

cosg=ry=

"lt-zl'

=

-t.336

-1.336 +(-3)'z+(-1)2

.'. 0 = 110.9"

-

-

+ (-3)(-1)

(d)

la' a.. a,'l

' 2

|

I

AxB = lt 3l lr l-2 -3 -ll = (-2 -9)a,+(6+ 1)a, + (-3 + 4)a, = -1la" +7ar+a, lA x Bl =

t

.'. unit vector = ,A

Jl l' B,

+7\ l' = 13.017 r

= -0.841a + 0.535a) + 0.0765a

lAxBl

1.3 Atx= l,y=).,7={ .4.=ar_lz *4a +12a A

unit vector

IAI l^-l

L4 Atx=2,y=3

ar,yz+4a +l2a

r._.------i "ll" +4'+12' = 0.0788arlz+03152a +0.9457a

A = 3aryz+2a +3a

B=4a +4a (a) A. B =

(b)

(3x 4) + (2x 4)+

(3

x0) = 20

AB ^=:-:---: =

= 0.7538

COSU

lAllBl

.'.

I

= 41.08"

(c) The projection of B along the direction of A is l8lcos0 lBlcosg = ^[4' + 1.5

4t .0.7538 = 4.264

If A, B, and C are perpendicular to each other, then

A.B=0

B.C=O

A.C=0

CHAPTERI.vEcToRANALYSISANDMAXWELL'SEQUATIoNSININTEGRALFORM

A'C

(su, * za, + :a.)'

=

15+29+3

+ A'B

=

(r a

"

+ cra, + ^

")

=0

cr= -9

(sa,+zrr+:a.)'(4., *

Zar+ B,a,)

58,+418,=O (4u, * 2rr+ B,a,)'(r., -

B.C

9a, + a.)

34-18*8,=Q

+ B, = B' = 1.6

(a)

BxC

=

l4'5 -25'5

u'

u'l

l'lo 2r 'l

lu'

6l

= =

unit vector =

(b) Area = lBx t.'7

-

6)a, + (-12)a, +4a

-l2ar+

4a,

ffi =-ffi

Cl = "J12' + 42

(a) If the two vectors

(6

= -g'95a'+ 0'3 16a'

= 12'65

are parallel, then A x B

= 0'

CHAPTER 1. VECTOR ANALYSIS AND MAXWELL'S EQUATIONS IN INTEGRAL FORM

(b) If A and B

are perpendicular to each other, then A

A.B = t1 x (-1)l .'.

l.l2

+ (b

3b-8c=l

x3)+

[c x

,

or

'B = 0. (-8)] = Q

l+8c J

(a) A = x2 yN, + y2 za, + x2 znz

x = y .'.

Ao = A*cos@ + AQ

= -A,sin@

A, = A, = A, = A

sin@

+ A, cos@

-

psinf

p3 cos2 @sin

p

p2zsinz P

p'zcos'q

-

QsinQ

+

p2zsinj

Qcos2 Q

+

pzzsinz QcosQ

p'cos3

-p3 sint

A,.'. A

pcosQ

Q

P2zcos'Q

Aour* Arar* A,a, (p'"or'@sin@

+ p2zsin3 O)"0+(o'

(b)

x = y z = .'. A" = A, = A, = :.A

sin'@cos2

Q

+ p2zsinz $cosQ)ao+ p'zcos' pa,

rsin0cos@

rsin0sin@ rcosO

13

sin3 g cos' P sin

@

13

sin2 9cosOsin2

@

13

sin2 gcosgcos'@

A cos0

A

sinOcos@+ Arsin0sinP +

13

sin2 g(sin2 9cos3 Psinp + sinOcosgsin3 p + cos2

gcos'p)

1.13

a0

fl1

Og

ae

a0

Oe

B=-aru9 +3a^+2a

L = 2a', + a', -3a1,

a'^

=

-ar

A,,

=

a^

a.I

^;= .'. A -

2ar-a,-3a, =-&fvot2a^-32

la. aa a"l

n*n=l-r

; -;l

= 13a rdA+5a^-a.

['3zl 1 1A l.l1

AT=

0T-0A

.'. AT=3ar +5aI +5az -

BT= 91-33

*a zl)=3ar +3a) +4a bu \ y

BT=3ax +5ay + Sai -(a +ayl)= 2ar +4at +5a \ r

z

z

l0

CHAPTER 1. VECTOR ANALYSIS AND MAXWELL'S EQUATIONS IN INTEGRAL FORM

Unit vector:

3a-+3a,.+4aAT --# = lATl ^!3, +32 +4,

a - = :---Ar

= 0'5

l45a

'

BT 2a-+4a..+5aA -Br=----+=0.298a + 42 + 5' lBTl

+'0.5

l45av-t- 0.686a

x+0.596ar +0.745a

z

z

^12,

a bn, = T;; ry'ff'an _toooldLl

= = Eu"

?

;T;rr-' 36tt

t

*^'(o'st+su'

0.409 x 10ea, + 0.409 x 10ea, + 0.545 x l0ea.

= T;;m'^,, O,

=

I i : .(o.zesu,+0.596ar+0.745a,) +tt;.-xl0-'x45

=

o.otl3!^10ea + o.rlgzxl0ea +0.149x10ea-

.'. E,o*t= En, * E"t = 0'4; x 10e a, 1.15 lFl=

(t.o x to-'n)'

4o

* -J-x 36n

+0'5145a' +0'686a')

+

0'52; 10ea, + o'ega * ton

=2.304x 10-8N

lo-e x (t\ o-'o;'

1.16

AP = 0P BP = 0P

-

-

0A = (2.5

0B = (2.5

-

- l)a, * 2a, = 1.5a, + 2a, 4)a, + 2a, = -1.5a, + 2a,

^,