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Unformatted text preview: m\
In Fig 546 (PS 32) the '
. . ' ’ plane deﬁned b _ _
. I . um I Ofpcrmeabllltypl from mediumZOfpermeabinty y fo y — 1 separates
; ts on the boundary and #2 no surface current ‘ ﬁnd B; and then evaluate your result for '
, . p1 = 5112. Hint: Start ' '
{equation for the umt vector normal to the given plane. 01“ by denwng the Solution: We need to ﬁnd ﬁz. To do so, we start by ﬁnding any two vectors in the
plane x — y = 1, and to do that, we need three noncollinear points in that plane. We choose (0,—1,0),(1,0,0),and (1,0,1).
VectorA1 is from (0,1,0) to (1,0,0): 1., ectorAz is from (1,0,0) to (1,0,1):
A2 = 21. if we take the cross product A2 x A1, we end up in a direction normalito‘ the ‘
lane, from medium 2 to medium 1, , ' :, was: r ‘ «V‘
13532 Magnelic media sepllﬂed by the plane x y = 1 (Problem 5.32). _ ﬂ _ #2
411 31’: — ”1325 + 92.5). §58 and 59: Inductance and Magnetic Energy ., 5.35 Obtain an expression for the selfinductance per unit length for the
wire transmission line of Fig. 527(a) in terms of a, d, and p, where a is
of the wires, d is the axistoaxis distance between the wires, and p is the
" 'ty of the mdium in which they reside. on: Let us place the two wires in the xz plane and orient the current in one
then to be along the +zdirection and the current in the other one to be along the firdirection, as shown in Fig. P535. From Eq. (5.30), the magnetic ﬁeld at point
~P('x,0,z) due to wire 1 is Figure P535: Parallel wire transmission line Tb’erefote, the total magnetic ﬁeld in the mgion between the wines is
' l Jr==y) uId B=Bl+32=yﬂ(: +d— Md_ 1) 1 s 1 Eq (5 93) states that the ﬂux linkag i»
 ,_ = d9 Then Eq (5. 94) gives a total inductance ovc, recognizes that the wires are thin compared to the
d > a). This has been an implied condition from the
lthe ﬂux passing through the wires themselves have
‘ ' limit' in Table 21 for the two wire line. conductor is 10 cm? Solution: From Eq. (5.99)
cable is given by $ 2 x 41: ‘7 
L — 2L’ = 2: 10 In (g) = 277 x10‘9 (H). Eq. (5.104), Wm = L12/2 = 13912 (nJ), where Wm is in nanojoules when I is in
. Alternatively, we can use Eq. (5.106) to compute Wm: 1 2
: H d‘V.
Wm Z/vvm (6.87) to ﬁnd the magnetic ﬁeld: 1 . 2 x 102 a
_. __ 1‘an
1cop ¢ R sme 8R (e )
2 ’2
1t 2 x 10 sin 9e42,“ instantaneous value, this is I ,OEI) = 6 % sinGcos (61: x 108: — 21tR) (uA/m). . e of the loop, 6 becomes —i and r becomes y.
, the loop is along —2, and the magnitude of the ﬂux is =5; 5cm —;  —i(30cm xdy) j   ofwire wrapped around n square framc
0! 19° :3: origin with each of its Sldcs parallel to
, ‘9‘ Led.   ‘s the opencircuited ends of the coil if the maﬁa),
frsinZy cos 103: (1').
me coil is not moving or changing shape, 3,, = 0V 'and
.. @466), E c d d 0.125 0.125B (‘dxd )
‘ =’ _ _ . = _N___ . z ,
9‘” th/SB ds dt 0425 —o.125 y surface normal was chosen to be in the +2 direction. d 0.125 0.125 5 1
(10cos103t/ / cosxdxdy) =62.3sin1o3:. (15y),
“0.125 y=O.125 cg: L For B = ilOcosxsin 2ycos 1031 (T), ‘ d 0.125 0.125
' = lOO— (lOcos 103t/ / cosxsin2ydxdy). =
u53:'2. .dt x=0.125 y=—0.125  ~ 7 :5. V 777777 W *7 Problem 6.5 A circularloop TV antenna with 0 01 m2 area ' '
unif   ' Is in th
armamplitude BOOMHz Signal. When oriented for maximum resmex: 10f a
, 00p develops an emf with a peak value of 20 (mV) What is the 
' Peak magm
the incident wave? tude °f B of Solution: TV loop antennas have one turn. At maximum orientation, Eq. (5 5)
evaluates to (D = f B  ds = iBA for a loop of area A and a uniform magnetic ﬁeld
with magnitude B = IBI. Since we know the frequency of the ﬁeld is f = 300m
wecan expressB as B = Bocos(o)t+oco) withm: 21tx300x105radlsandaoan
arbitrary reference phase. From Eq. (6.6), d .
Vang: —N%:2 = —A2;[Bocos(0)t+0lo)] =ABotosm(mt+ao). V.“ is maximum when sin(03t + a0) = 1. Hence, 20 x 10"3 =ABoaJ = 102 xBo x 6nx10‘, .. a. . dfthecnnentdmxwouldﬁowm sh
gap. mloophnsaninmal mistanceof CHAPIER6 223 is (p B d 15cm( IJOI ‘
—/S s—/5m —x2ny)[—xm(cmndy _ [.101 X 101 15
_ 21: 5
41: x 10‘7 x 2.5 cos(21t x 10"?) x 10‘1
21:
= 0.55 x 107 cos(21t x 104:) (Wb). X1.1 vemf = 59 = 0.55 x 21: x 104 sin(21t x10“t)><10‘7 dt
= 3.45 x 103 sin(21c x 104:) (V). (b) Vemf _ 3.45 x 103
4+1 " At t = 0, B is a maximum, it points in —idirection, and since it V b
cos(21t x 1042‘), it is decreasing. Hence, the induced current has to be CCW w en
looking down on the loop, as shown in the ﬁgure. ____________________——— sin(21t x 104:) = 0.69 sin(21t x 104:) (mA). [ind = arid: as Igl/ew a ,r Problem 6.9 A 50cmlong metal rod rotates about the zaxis at 180 revolutions per
minute, with end 1 ﬁxed at the origin as shown in Fig. 621 (P69). Determine the
induced emf V12 ifB = i3 ><10‘4 T.
Solution: Since B is constant, Vemf =
is given by u = ore), where
21: rad/ cle x 180 c cles/min
Cy ( y ) = 6n rad/s. (0=.___.__—————————' (60 s/min) Vemmf, The velocity u for any point on the bar Flgurc P69: Rotating rod of Problem 6.9. From Eq. (6.24), V m 1 ° .
12=Vm=A (“XB)dl=/ 5(¢61trx23x10‘4)f'dr 7:0 = 181cx10""'/0 rdr
r=0.5
0
= 91cx 104r2\
0.5 ~' ~'
= —91cx10‘4 x025: Problem 6.15 A coaxial capacitor of length l _ 6 cm use: an insulating d1 material withs,=9. Theradiiofthe
cylindiical cond ctors
the voltage applied across the capacitor is u m 0 5 cm m V(t) = 100sin(120m) (V),
what is the displacement current? Figure P6.15: Solution: To ﬁnd the displacement current, we need to know E m the dielectric space
between the cylindtical conductors From Eqs. (4 114) and (4.115), __ . Q
E' l.21iatzrl’ V=a,.% (2'?) 20m 44. 3
E=_f__‘_,__=_f1925.i‘l(_l.__)_ —rl———sin(120xt)w‘
nae) rm RF .VguOmmoﬁ—mﬁuovﬁwagmagﬁ”
3,8538 Ba a bsemﬁam 932 uo €93 0383:. was a, _
_ a GEE 30 .mm 5 58% §§§ 03335; 2:. 2.» 5.: .283 conﬁne 03 8328 25m 0.5 mm 5033, $5 s
. oo . H moo X "I H
25 Sans” m3 RES: 9: .5332” >30 N
minus—30:8
.Awsuu
men an A2 39 3 55% mm 8:893 158 a mo 85mg? 05. .8058
nus"? 06 8 @8858 «9:3 05 5:85 wagon 3058 nouosuuoo 05 S 3.50 on
8 an: «55o “58083”? 813332 ﬂ 5:8? 3x58 o5 35m iguana“? .93 353833 u
383303 x 52 x 1: x n: u 995$. Q2835 k V may 1 . a 5
$2 x24 .32: u m ’Fi‘gure P6.16: Parallel late ' ' '  . .
6. 16). p Capamtor comammg a [0583’ dielectric main.31 (Pmbl. (b) Obtain an expression for 14, the displacement current ﬂowing inside the
capacitor. (c) Based on your expression for parts (a) and (b), give an equivalentdram
representation for the capacitor. (d) Evaluate the values of the circuit elements forA = 2 cm2, d = 0.5 cm, a, = 4,
o = 2.5 (S/m), and V(t) = lOcos(31t x 10%) (V). Solution:
0:) CHAPTER6 +
8.0“ R 6W!) Equivalent Circuit 1 Figure P6.l6: (a) Equivalent circuit.
4 x 8.85 x 1012 x 2 x 104 0 5x104 = 1.42x 1012 1=. ...
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 Spring '08
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