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Unformatted text preview: 1 A coil consists of 100 turns of wine wrapped around a square frame
of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to
the x— or yaxis. Find the induced emf across the opencircuited ends of the coil if the
magnetic ﬁeld is given by ’ (a) B = i10e'2’ m.
(b) B = i lOcosx cos 103: (T).
(e) B = i lOcosx sin2y cos 103t (T). Solution: Since the coil is not moving or changing shape, 3;“ = 0 V and chf = Vc‘rmf. 0.125 0.125 d d
V =  —  t: — —— . "
emf Nd! st ds Nth/4.125 4.1sz (Rudy), where N = 100 and the surface normal was chosen to be in the +2 direction.
(a) For B = ilOe‘z' (T), chf= —ioo%(ioe2'(o.25)2) = use2’ (V) (b) For B = ilOcosxcos 1031‘ (T), 0.125 0.125
cosxdxdy) = 62.3 sin103: (kV). d 3
= _ _ O t
vcmf IlOOd:(10cosl 54.125 F4125 . (c) For B = ilOcosxsinZycos 103t (T), d 3 0.125 0.125 . 2 d d) 0
=— 00— 10 0510! / cosxsm y x y = .
vcmf 1 dt( c x=—0.l25 y=0.125 2 .
_ p O A Circularloop TV antenna with 0.01 m2 area is in the presence of a
uniformamplitude 300MHz signal. When oriented for maximum response, the loop develops an emf with a peak value of 20 (mV). What is the peak magnitude of B of
the incident wave? » Solution: TV loop antennas have one turn. At maximum orientation, .
. . (I) = f B  ds = :tBA for a loop of area A and a uniformlmagnetic ﬁeld
With magnitude 8 = lBl. Since we know the frequency of the ﬁeld is f = 300 MHz, we can express‘B as B = Bocos(o)t + do) with a) = 21: x 300 ><106 rad/s and (loan
arbitrary reference phase. I dd) (1
chr = NdT = —AE[Bocos(o1t +ozo)] = ABocosin((ot+ao). chf is maximum when sin(mt + 0(0) = 1. Hence, 20><10‘3 =ABoCD=lO_2XBoX61tX108, ' , _.
which yields Bo = 1.06 (nA/m). ,/
K3) . A SOcmlong metal rod rotates about the :axis at 180 revolutions per
minute, with end 1 ﬁxed at the origin as shown in Fig. P69. Determine the induced emf v.2 ifB = 23 ><10‘4 "1'. Solution: Since B is constant, chf = Vf‘mf. The velocity u for any point on the bar is
given by u = ﬁrm, where a) = 27': rad/cycle x (180 cycles/min)/(60 s/min) = 611: rad/s. l o
V12= ean= 1(uxB)dl= (961trxi3xlO“)r"dr
a I=0.5
0
= 181tx10‘4/ rdr
1:0.5
' 0
=91tx10‘4r2l
. ' 0.5 = —91tx 10‘4 x 0.25 ='—7o7 (w). @ The plates of a parallelplate capacitor haveﬁareas 10 cm2 each and are
separated by 1 cm. The capacitor is ﬁlled with a dielectric material with a = 480, and
the voltage across it is given by V(t) = 20cos 21: x 1061‘ (V). Find the displacement current.
Solution: Since the voltage is of the form given with V0 = 20 V and
co = 21: X 106 rad/s, the displacement current is given by '
Id =‘—::Vowsintot
4 .8 4 0"12 O 0‘4
= —X—SMI_ x 20 x 27: ><10‘5 sin(21t x 1062) 1 x 102
: —445 sin(21t x 105:) (nA). @ The conducting cylinder shown in Fig. $.11 rotates about its
axis at 1,200 revolutions per minute in a radial ﬁeld given by 8:536 (T). _. Z / Sliding contact Figure P6.1 1: Rotating cylinder in a magnetic ﬁeld I The cylinder, whose radius is 5 cm and height 10 cm, has sliding contacts at its top
and bottom connected to a voltmeter. Determine the induced voltage. Solution: The surface of the cylinder has velocity u given by 1,200
60 L .1
V12=/0 (uxB)dl=/oﬂ ($21txj6)2dz=—3.77 (V). u = (imp: $21: x x 5 xio2 = «321: (m/s), The loop showu in Fig. P6.8 moves away from a wire carrying a
current I, = 10 (A) at a constant velocity u 2 5'5 (m/s). If R = 10 K2 and the direction
of (2 is as deﬁned in the ﬁgure, ﬁnd 12 as a function of yo. the distance between the
wire and the loop. Ignore the internal resistance of the loop. 7
L. 4— CHI—O!
R
I]: 10A T “
20cm [1
u
R
YO Figure P6.8: Moving loop Solution: Assume that the wire carrying current 11 is in the same plane as the loop.
The two identical resistors are in series, so [2 = chf/ZR, where the induced voltage
is due to motion of the loop and is given by vﬂ=vgrnf= j£(uxB)d1.
C The magnetic ﬁeld B is created by the wire carrying [1. Choosing i to coincide with the direction of II , the external magnetic ﬁeld of a long wire is
e “01:
B = —— .
21v For positive values of yo in the yz plane, 3? = i“, so Holt“
21W ' uxn=yluxa=ﬁux$%"=2
r Integrating around the four sides of the loop with dl = idz and the limits of integration
chosen in accordance with the assumed direction of 12, and recognizing that only the two sides without the resistors contribute to Vc'“mf. we have
0.2 I 0 1
vgg,=/ (2“ W) (idz)+ (2“0 W) (idz)
0 21" r=y0 02 znr r=yo+0.l
_41t><10‘7><10><5><0.2(l 1 )
— 27‘ Yo Yo + 0 1
1 1
=2x10'6<—— ) (V),
Yo Yo + 0.1 and therefore ’1
Q A coaxial capacitor of length l = 6 cm uses an insulating dielectric
material with e, = 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is
V(t) = 100 sin(1201rr) (V), what is the displacement current? I
j.
23
T
1
Solution: To ﬁnd the displacement current, we need to know E in the dielectric space
between the cylindrical conductors.
= —f' Q D = 8E
21:31 ’  _ E
 Q In 2  — 21:81 a '
Hence,
. = —f
V 1 .
E = _e b = 4% = 4144 3 sin(1201tt) (V/m),
rm (E) ran r The displacement current ﬂows between the conductors through an imaginary
cylindrical surface of length l and radius r. The current ﬂowing from the outer
conductor to the inner conductor along ——x" crosses surface S where S = —i"21trl.
Hence,
—8
[d = 5% .s = 4% sin(1201tt)) (—f2an) = 1.15 x 10‘8 x1201: x 21tlcos(1201tt)
= 1.63 cos(1201tt) (11A). Alternatively, since the coaxial capacitor is lossless, its displacement current has to
be equal to the conduction current ﬂowing through the wires connected to the voltage
sources. The capacitance of a coaxial capacitor is given by (4.116) as _21tel
c_ln(). hlb‘ The current is W 211:8!
I=C—=
m mm a [1201: x 100cos(1201tt)] = 1.63cos(1201tt) (,uA), which is the same answer we obtained before. lemmew»! = —r9 x 8.85 ><10‘12 x 1443 sin(120m‘) , l /// "I C *7 I l
1 An electromagnetic wave propagating in seawater has an electric
ﬁeld with :1 time variation given by E = iEocosrm‘. If the permittivity of water is
I 8180 and its conductivity is 4 (S/m), ﬁnd the ratio of the magnitudes of the conduction canent density to displacement current density at each of the following frequencies:
(a) 1 kHz, (b) 1 MHz. (c) 1 GHz, (d) 100 GHz. Solution: the displacement current density is given by
a a
= —D = ——E
3“ a: 88:
and, the conduction current is J = 0E. Converting to phasors and 'I taking the ratio of the magnitudes,
j _ GE _ 0'
3d (0580 . (a)Atf= 1 kHz, to: 21cx 103 rad/seand 4
=——————=88 13,
21rx 103x81x8.854x1012 8* 0 l
Jd The displacement current is negligible.
(b)Atf= 1 MHz. (0: 21tx 106 rad/s, and an I
T; The displacement current is practically negligible.
(c)Atf= 1 GHz,(o= 21rx109rad/s,and 4 = _—_= 88.
21!:x10ls x 81 x 8.854 x 1012 8 4 2 21: x 109 x 81 x 8.854 x 10—12 : 0'888' i
1. Neither the displacement current nor the conduction current are negligible.
(d) Atf: lOOGHz, a): 21: x10”rad/s, and 3 4 3
~— _—_ ———————_— = . 8 10" .
1d 2nx1011x81x8854xio12 88 X
The conduction current is practically negligible.
@ At t = 0, charge density pvo was introduced into the interior of a material with a relative permittivity e, = 480. If at t = 1 us the charge density has
dissipated down to 10‘3pvo, what is the conductivity of the material? Solution: We start by ﬁnd 1.: mm = pvoe‘m', Of _6
10—3on = one—lo m,
which gives
0—6
rum3 = —1 ,
Tr
OI' ' 6
10‘ _
1,: —ln10_3 = 1.45x 10 7 (8). But 1, = 2/6 = 480/0“. Hence
450 4 x 8.854 ><10“2 62—: t 145Xm_7 =2.44x1o“ (S/m).
, . If the current density in a conducting medium is given by “LEEt) =  + izx>COS (DI, determine the corresponding charge distribution pv(x, y,z;t). Solution: .
VJ = 33%. (6)
t
The divergence OM is
A 3 . a A a . . 2 A
V J_l(x§;+ya—y+za—) (L—y3y +z2x)cosu)t
a a
= —3— 'cosmt = —6 coscut
ayo ) y Using this result in Eq. (6) and then integrating both sides with respect to t gives pV= —/(VJ)dt= —/ —6ycosa)tdt= 66ysinmtlCO, where Co isaconstant of integration. . ' '________,_‘__,_, ._ .. t . g V ___ _ ﬁe“‘éie¢:ﬁc ﬁeld of an electromagnetic wave prooagatingin air is :
given by
E(z,t) = R4cos(6 x 108: — 2z) + 5'3sin(6 x 103: — 22) (Win). Find the associated magnetic ﬁeld H(z,t). Solution: Converting to phasor form, the electric ﬁeld is given by ' Em = sate121 — meg1'21 (V/m), H(z) = _ VxE . I 2‘ 1 i y 2
_—.— a/ax a/ay a/az
‘1‘” 4w?! j3eﬁz o
1
J'(Dt1 (ﬁée'nz — §j8e'ﬂz) — 6x108 x41cx10—7
Converting back to instantaneous values, this is H(t,z) = —i8.0sin(6 x 108: — 22) + 910.6cos(6 x103t— Zz) (mA/m). ' j (:26 — way1'22 .—_ Jameﬁx + muse12" (mA/m). @ i L2 0 = O is given by The magnetic ﬁeld in a dielectric material with e = 450, u : no. and 4 H(y,t) = :25 cos(21t x 107t+ky) (A/m).
Find k and the associated electric ﬁeld E. Solution: In phasor form, the magnetic ﬁeld is given by fl = :‘(Seﬂ‘y (A/m). which, together with the original phasor expression for f1, implies that 7 4
wﬁ_2nx10ﬂ n (rad/m). have”: c f 3x103 ~30 Inserting this value in the expression for E above, ~ . 4"/30 I '4 /3o _ A '4 /3o 2
E = ‘12:: x 107 x 4 x 8.854x10’125ej “y “ “294161 U (Wm ____ The electric ﬁeld radiated by a dipole antenna is givenin“!
spherical coordinates by . . ' ' .2" 102 E(R,9;t)=9 x sine cos(61: x 108: — 21tR) (V/m). Find H(R,6; t).
Solution: Converting to‘phasor form, the electric ﬁeld is given by .2x102 17:(R,e) = éEe = a sin eeW (V/m), which can be used to ﬁnd the magnetic ﬁeld: I ~ 1 ' ..
H(R,9) = _jwuVXE= [RRsine ad, +¢Ea7é(RE9)] :¢6EX108X4EX10'7 . 3 i
=¢§Esin8e"2"k (uA/m). Converting back to instantaneous value, this is H(R,e;t) = $ 51—: sinecos(6n x108t— 21:12) (uA/m). at the internal dimension of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. l The homogeneous material inside the capacitor has the parameters 6 =.10‘11 F / m, a = 105
3/111, and cr = 10—5 S/rn. If the electric ﬁeld intensity is E =t(106/p) cos(lO5t)ap V/rn, ﬁnd: a) J: Use _
‘ J = 0E = cos(105t)ap A/In2 b) the total conduction current, Ia, through the capacitor: Have Ic = f/J  d5 = 27rle = 207rlcos(105t) = 871' cos(105t) A c) the total displacement current, Id, through the capacitor: First ﬁnd
105 1 11 6 I ' _'
——————( X 0 )(10 ) sin(105t)ap = —£ sin(10°t) A/m l
I
I
an i
P P l
l
l i ' 3 Now
I4 = 27mm = +27rlsin(105t) = '——0.81rsin(105t) A d) the ratio of the amplitude of L1 to that of IC, the quality factor 'of the capacitor: This will _ be ‘ [IA 0.8 ._ = _. = 0.1 Show that the displacement current ﬂowing between the two conducting cylinders in a loesles's 1 coaxial capacitor is exactly the same as the conduction Current ﬂowing in the external circuit if the applied voltage between conductors is VD coscth volts. '
From Chapter 7 ,_ we know that for a given applied voltage between the cylinders, the. electric ﬁeld is _ Vcoswt  'elﬁcoswt I
E=EgﬂmapV/m => D: agC/mz Then the displacement current density is " I. I _a__13 _ —we%sinwté
8t — .pln(b/a) P Over a length If, the displacement current will be ..  3D ' I 6D QtZwel/B sinwt ' . V I
= —. = —— _‘_——— = _ = I
Id / f at d5  2"” at 11n(b/a) Colt c where we recall that the capacitance is given by = 21rel/ ln(b/d).'\ I Tow/a) ' 7 C Consider the region deﬁned by ix}, lyl, and < 1. Let 6, = 5', p, = 4, and cr = 0. If .34 = 20 coe(l.5 >< 103t"— bm)ay pA/rn’"; a) ﬁnd D and E: Since Jd = BD/at. we write  20x10‘6." '3
D=/Jddt+0=msm(l.5x10 —ba:)ay = 1.33 x 1013 ein(l.5 x 10%  may C/m2 where the integration constant is set to zero (assuming no dc ﬁelds are. present). Then. D 1.33 x 1013 ' —=———— ' st—b
e (5x8.85x10'12)sm(15xm ' ﬂay = 3.0 x 103sin(1.5 x 10% — bx)a.y V/m _ E: b)’ Iuse the point form of Faraday’s law and an integration with respect to time te ﬁnd B' and H: In this case, .' _6E.‘J _ _3v "8 _ 8B
V x E — Ea, — —b(3.0 10 )cos(1.5i>< 10 t baaa; — at Solve for B by integrating over time: ——— . .I .8  = 2. I_n 1.5 Ost— '. I: 1.5x 108 s1n(15V>< 10t b:L')az ( O)b><10 sm( ><1 bm)ad T _ i _ B 2._ —11 ' ’ H = Z = 10875  bm)az = (4.0 x 10‘6)bsin(1.5 x 1032:  bm)az A/m c) use V x H = J4 + J to ﬁnd Jd: Since a = 0, there is no conduction current, so in this case i v x H = %5:iay = 4.0 x 10sz cos(1.5 x108t — b )ay A/m2 = Jul d) What is the numerical value of b? We. set the given expression for J 4 equal to. the result of part c to obtain: 20 x10‘6 = 4.0 x 1061;2 2 b: V5.0 ml _ b(3.0 x 103) . ‘ ‘\ \ Ti
J
,
a3.
:1
l
.4 +1. Let ,u. = 3>< 10‘s H/In, e = 1.2 X 10‘10 F/In, and 0' = O everywhere. IfH = 2 cos(lOl°t—,Ba')a:
A/rn, use Maxwell’s equations to obtain expressions for B, D, E, and ,3: First, B = pH = 6 X 10'5 cos(10mt — Haﬁaz T. Next we use
K 3H 53:)
=——— = s' {1 lot— a =—
V X H aw 2,8 m\ .0 ,8 )ay at from which 2 n
D = /2ﬁ sin(101°t —— 52:) dt + C' = —10—€0 cos(101°t — [390% C/m‘ where the integation constant is set to zero, since no dc ﬁelds are presumed to exist. Next, D _ I 26 10 _ 10 _ r
E = —€— _ —(1.2 X 10_10)(10m) cos(10 t —;39:)a., —— —1.67,8 cos(10 t [Sway V/m
Now
_ 3Ey _ 2 . 10 _ BB
V x E — Ea, — 1.67;? sm(10_ t —ﬁ:r)az — —at SQ _ ,
B = — f1.675?sin(101°t — ﬁ$)azdt = (1.57 x 104%2 cos(1ol°t — ﬂm)az We require this result to be consistent with the expression for B originally found. So (1.67 x 1010m2 = 6 x 105 => s = i600 rad/m The electric ﬁeld intensity in the region 0 < < 5, O < y <I1r/12, O < z < 0.06 m in free space is given by E = C sin(12y) sin(qz) cos(2‘ x 10107?) ag V/m. Beginning with the V x E
relationship, use Maxwell’s equations to ﬁnd a numerical value for a, if it is known that a is
greater than zero: In this case we ﬁnd 8E3, 8E,
VXE— 62 ay— ay a.z
= C [a sin(12y) cos(a;)a;y — 12 cos(12y) sin(az)az] cos(2 x 101°t) = %§« H=—1—/V XE n+01
#0
C'
,uo(2x101° [a sin(l2y) cos(az)ay — 12 cos(12y) sin(az)a=] 5111(2) x 101°t) A /m where the integration constant, 01 = 0, since there are'no initial conditions. Using this result, we now ﬁnd C(l44 +a2) _ aHz aHy __ . . . 10 _ V x H — [ 6y m — #00 x 1010) sm(12y) sm(az) sm(2 x 10 t) — ET
Now I I
sin(1:2'y) sin(az) cos(2 x 10101;) a¢ 2
E=':—3=/%V><Hdt+02= 0044+“)
0 p.060(2 X 1010)2
where Cg = O. This ﬁeld must be the same as the original ﬁeld as stated, and so we require IC'(144 + a2) a 1' i  w H " 102 " 1/2 '
“50(2 x 1010)2 = 1 Y USing #060= (3 X 108)2: ' a = _. 144] = 66 m—1 (3 x 105)2 . _ I . _ w
@ In a region where p, = 5., = 1 arm 0' = O, the retarded potentials are given by V = 2(2 — ct) V' and A = J:[(z/c) —‘b]az W'b/m, where c = l/V’poeo. a) S” ow that V .A = —;le(8V/3t): First,
A,
v A = = 3 = mm
Second,_ _ I
3V :1:
at ' «#1060 so We obeerve that V A = ’—/.1.geo(3V/ at) in free space, implying that the givén statement
would hold true in general media. ' : b) Find‘B, H, E, and D: Use _ V .'
’__ v '_ 314a _ 2.
B“VXA‘ 3:1: ‘(t'c‘)ayT
Then~ : B 1 z ' ', I:
'Now, I I v l  I .
' .' BA' V  ' ‘ '
E=—VV—79{=—(z—ct)am—zaz+zaz=(ct—z)a¢V/m
Then I D = 60E = 60(Ct — z)a, C/m2
c) Show that these results satisfy MaxWell’s equations if J and [5,, are zero;
i.VD=Veo(ct—z)a¢=0 ' VB=V(tz/c)'a,y=0 ‘ 5H  1 e '
Vxﬁ=§£a¢=—am= °a¢
Z #06 #0
which we require to equal BD/at:
9.12 _ 6 ca _ gnaw
,,,, _. 3t — '° — #0
iv.  ' »—~—~—.____...; ;._,__ __
V x E = 853% —ay ...
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This note was uploaded on 12/06/2010 for the course ECE 302 taught by Professor F during the Spring '10 term at Cal Poly.
 Spring '10
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