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**Unformatted text preview: **CHAPTER 6 219 Shaptet' 6 Sections 6-1 to 6-6: Faraday’s Law and its Applications Problem 6.1 The switch in the bottom 100}: of Fig. 6-17 (Pol) is closed at r = 0
and then opened at a. later time :1. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? Figure P6.1: Loops of Problem 6.1. Solution: The magnetic coupling will be strongest at the point where the wires of
the two 100ps come closest. When the switch is closed the current in the bottom 100p
will start to ﬂow clockwise, which is from left to right in the top portion of the bottom
loop. To oppose this change, a current will momentarily ﬂow in the bottom of the
top loop from right to left. Thus the current in the top loop is momentarily clockwise
when the switch is closed- Similarly, when the switch is opened, the current in the
top loop is momentarily counterciockwise Problem 6.2 The loop in Fig- 6-18 (136-2) is in the x—y plane and B = $50 sine):
with Bo positive. What is the direction of I (h or 453) at (a) r = o, (b) {or = 1t/4, and
(c) cur : 112/2? Solution: I = Vemf/R. Since the single-turn loop is not moving or changing shape
With time, Vg‘ﬂﬁ = O V and Vemf 2 ngf. Therefore, from Eq. {6.8}, if we take the surface normal to he +i, then the right hand rule gives positive
ﬂowing current to be in the +4: direction. —A 3 -AB (0 I = ———30 sine)! = 0 RBI cos cur (A), .. matin..:..:-m:. - 220 CHAS-HER 6 Figure P62: Loop of Problem 6.2. where A is the area of the loop. (a) A, to and R are positive quantiﬁes. At t = 0, cosmr = 1 so I < 0 and the
current is ﬂowing in the me direction (so as to produce an induced magnetic ﬁeld
that opposes B). (b) At tot = T's/4, cos (or = x/i/Z so I < 0 and the current is still ﬂowing in the 4?:
direction. {c} At a): = 75/2, cos {or = 0 so I = 0. There is no current ﬂowing in either direction. Problem 6.3 A coil consists of 100 turns of wire 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 y—axis. Find the induced emf across die open-circuited ends of the coil if the
magnetic ﬁeld is given by (a) B = 2 my?“ (T), (b) B = 210cm (£6th ('1‘), (c) B = a 10cosx sin2y cos 1031' (T). Soluﬁon: Since the coil is not moving or changing shape, Ve‘g‘f = 0V and
Vunf = mef. From Eq. (6.6),
d d 0.125 0.125
rim: FN— feds = —N—j[ e-(adxdy),
dr 5 dr -0125 —o.125 where N = 100 and the surface normal was chosen to be in the +3 direction-
(a) For B = 3108—2: (T), d
and: —100-&;(10e_2‘(0.25)2) = 125x” (V)- CHAPTER 6 221 (b) For n = tiOcosxcos 10% (T), d 0.125 0.:25
Vemf = — 300-— (10c051032‘f ff cosxdxdy) = 62.3 sin 103: (kV)-
dt m—mzs y=—0.125 (c) For B = 210 cosxsin2ycos 103: (T), d 0.125 0.125
Vemfz —100— (10c03103tf / cosxsinlydxdy) :- 0.
dz m—mzs y=—0.125 Problem 6.4 A stationary conducting 100p with internal resistance of 0.5 .Q is placed in a time-varying magnetic ﬁeld. When the loop is closed, a current of 2.5 A
ﬂows through it. What will the current be if the loop is opened to create a small gap
and a 242 resistor is connected across its open ends? Sointion: Vemf is independent of the resistance which is in the 100p- Therefore, when
the loop is intact and the internal resistance is only 0-5 9, Vm=2.5Ax0.5n= 1.25 v. When the Small gap is created, the total resistance in the loop is inﬁnite and. the
current ﬂow is zero. With a 2-9 resistor in the gap, 1 = rem/(znws n)=1.25 W25 n = 0.5 (A). Problem 6.5 A circular—loop TV antenna with 0.01 1112 area is in the presence of a
nnifonn—arnplitude 300-MI-Iz signal. When oriented for maximum response, the ioop develops an emf with a peak value of 20 (mV). What is the peak magnitude of B of
the incident wave? Soinﬁon: TV loop antennas have one turn. At maximum orientation, Eq. (6.5)
evaluates to CD = f B - d5 = :IzBA for a ioop of area A and a nnifonn magnetic fieid
with magnitude .8 = |B|. Since we know the frequency of the ﬁeld is f = 300 MHZ,
we can express 8 as B = Bocos(tot+txg) with o} = 211: X 300 ><106 raids and {10 an
arbitrary reference phase. From Eq. {6.6), Vemf = "—N? = —A%[B{JCOS{£DI + = Aﬂowsinﬁm + me is maximum when sin((m‘ + 01.0} = 1. Hence, 20 x10‘3 = Aing : 10-2 x an x 6:: x108, .—_-,:':ZWzM<-“""'- 222 MR 6 which yields Bo = 1-06 (nAlm). Mblem 6.6 The square loop shown in Fig. 619 (P66) is coplanar with a long,
Straight wire carrying a current £(r) = 2.503521: x 104: (A). (3) Determine the emf induced across a small gap created in the loop. (b) Detemﬁne the direction and magnimde of the cunent that would ﬂow through
a 441 resistor connected across the gap. The 100p has an internal resistance of 1 9.
Figure P6.6: LOOP coplanar with long wire (Problem 6.6).
Solution:
{3) The magnetic ﬁeld due to the wire is
_ ~ 1101 _ A #01
B _ 21w ' X 2ng where in the plane of the 100p, @ z: —§; and r : y. The flux passing ﬁmmgh the loop CHAPTER 6 223 is 15 cm
«in: [345:] (+sﬁ‘1{)[—s10(cm)]dy
5 5 2101 cm I 10‘! 5
=M—D x 1n~1— 21: 5
_ 4:: x104 x 2-5cos(2n x104r)x10'1 X1 1
s 21: .
: 0.55 x 10-7cos(2n x 104;) (Wb).
do . 4 -7
11m: —3 =0.55x21t><1045m(21:x10 z) x 10
= 3.45 x 10-3 5111(21: x 104:) (V).
(b)
V _ —-3
11m = m — w sin(21: x 104:) = O.69sin(21: x1041) (mA). 4+i_ 5 At 1‘ = 0, B is a madmum, it points in —i-direction, and since it varies as
cos(2:n: x 104:), it is decreasing. Hence, the induced current has to be CCW when
looking down on the loop, as shown in the ﬁgure. Problem 6.”? The rectangular conducting loop shown in Fig. 6-20 (P6,?) rotates at
6,000 revolutions per minute in a uniform magnetic ﬂux density given by B = 3750 (1111'). Determine the current induced in the loop if its internal resistance is 0.5 Q.
Soiuﬁoti: <12 = feds = 550 x10‘3-§z(2 x 3 x10“)cos¢(t): 3 X10'5cos¢(t),
S ¢(r) = (02’ = W? = 2001:: (mils),
at: = 3 X 10—5cos(200m) (Wb),
Vemfz 4:? z 3 x 10-5 x 2001asin(200m) = 18.85 x 10-33in(200m) CV),
and ~— Vi‘“ = 37.75in(200m) (mA). 224 CHAMBER 6 Figure 136.7: Rotating loop in a mgnetio ﬁeld (Problem 6.7). The direction of the current is CW (if looking at it along —%—directiorl) when the loop
is in the ﬁrst quadrant (0 g Q) g n/Z). The current reverses direction in the second
quadrant, and reverses again every quadrant. Problem 6.8 A rectangular conducting loop 5 cm x10 cm with a small air gap in
(me of its sides is spinning at 7200 Evolutions per minute. If the ﬁeld B is normal to
the loop axis and its magnitude is 5 X 10‘6 '1', what is the peak voltage induced across
the air gap?
Solution:
27: radicycle x 7200 cyclesfmin
60 slmin
A z 5 cm x10 era/(100 onv‘n'l)2 = 5.0 x10'3 1112. From Eqs. (6.36) or (6.38), Vemf = AcoBosinmr; it can be seen that the peak voltage is
VP“ =AcoBo = 5.0 x10‘3 x 240:: x 5 x10'6 =1s.ss (m7)- (1): = 2401': radfs, E’E'ohlem 6.9 A 50-m-long metal rod rotates about the z—axis at 180 revolutions per minute, with end 1 ﬁxed at the origin as shown in Fig. 6-21 (136.9). Determine the
induced emf V12 EB = 23 X 10—4 T. Solution: Since B is constant. Vang : :31? The velocity u for any point on the bar
is given by e = era), where
21': radlcycle x (180 cycles’min)
(60 51min) a): = 61tradls. cm}? 6 225 Figure P69: Rotating rod of Problem 6.9. From Eq. (6.24), 1 o A
Vn=vggﬁej£ (uxB)-dl=jos(¢61trxi3><l0“)-f'dr 0 = 18?: x10“‘[ rdr
n05 0
= 911: X 10—431
0.5
= —9n: x 10'4 x 0.25 = 407 (w). Problem 6.18 The loop shown in Fig. 6-22 $6.10) moves away from a wire
c3113ng :1 current I; = 10 (A) at a constant velocity at = $15 (mfs). HR = 10 ﬂ and
the direction of 12 is as deﬁned in the ﬁgure, ﬁnd I: as a function of yo, the distance
between the wire and the loop- Ignore the internal resistance of the loop. Solution: Assume that the wire carrying current I; is in the same plane as the loop.
The mo identical resistors are in series, so 12 = Vemf/ 2R, where the induced voltage
is due to motion of the loop and is given by Eq. (6.26): Vemfz Vergf= j£(ﬂXB)'£ﬂ.
C The magnetic ﬁeld E is created by the wire carrying 31 . Choosing 2 to coincide with
the direction of 11, Eq. (5.30) gives the external magnetic field of along wire to be
A1101 1 .: .' unimaginsw ' - _ ' axe-t 3%
'3‘? ".‘zJ..;'. - 226 CHAPTER 6 Figure P610: Moving 100p of Problem 6.10. For positive values of yo in the 31-: plane, 37 : i", so . A “HOII #0115!
B: uxB: ox —= . .
ax sll rs 1 tam 2.2m Integrating around the four sides of the 100p with dé = i dz 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 V3139 we have 0-2 I u 0 polls:
Vﬂ‘] = f “‘10 1 ‘ A d A
emf o (1 27C? mm (2 z)+ 02 z 2753' 475X10‘7X10X5x0.2(1 1 j
21: yo yo+0-1 1 1
=2 10‘6 — ———— ,
X (Yo yo—i-OJ) CV) - (ﬁdz)
T=Yo+0J and therefore 6.11 The conducting cylinder shown in Fig. 6-23 (P6.11) rotates about its
axis at 1,200 revolutions per minute in a radial ﬁeld given by Bzi'ﬁ (T). CHAPTER 6 227 Sliding contact V Figure P613: Rotating cylinder in a. magnetic ﬁeld (Problem 6-11). 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. Soiution: The surface of the cylinder has velocity u given by 1,200
60 u = $0):- = $27: X X 5 ><10'2 = $21: (rials), 0-1 L H
1112:] (uxB)-di= (eznxrs).edz=_3_77 (V)-
0 0 Probiem 6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an
electric bulb with a resistance of 100 Q- If the loop area is 0.1 m3 and it rotates
at 3,600 revolutions per minute in a uniform magnetic ﬂux density 39 = 0.2 T,
determine the amplitude of the current generated in the light bulb. Soiution: Front Eq. (6.38), the sinusoidal voltage generated by the a—c generator is
chf = AmBo sin(tuz + Co) = V0 sin(tut + Co). Hence, V0=AmBO=O.1x%§(—)Ex0-2=ls4 (V),
v0 7.54 =—=——=7_4 mA. I R 100 5 ( ) Probietn 6.13 The circular disk shown in Fig. 6-24 (P658) lies in the xH-y plane
and rotates with uniform angular veiocity to about the z-axis- The disk is of radius a
and is present in a uniform magnetic ﬂux density B = 2280. Obtain an expression for
the emf induced at the rim relative to the center of the disk. é ..v.‘mz" Main. Va _.. 4mmm.-...Mm.. \. Likwé..w-n.... .'..tt;,z.-.m-.stx;;,.a- .. . 223 CHAPTER 6 Figure P613: (3.) Velocity vector u. Soluﬁen: At a radial distance r, the velocity is
I! = $ cur where 4) is the angle in the x—y plane show in the ﬁgure. The induced voltage is
(I {I A
V =] (ex B)-dl :/ [(mr) x230] -i“dr.
o o é x i is along En Hence, a
V=C030f rdr: .
0 Section 6-7: Displacement Cum-em Problem 6.14- The plates of a parallel-plate capacitor have areas 10 cm2 each
and are sepaxa‘ied by 1 cm. The capacitor is ﬁlled with a dielectric maierial with CHAPTER 6 229 E = 450, and the voltage across it is given by VG) = ZUcos 21': x 106: (V). Find the
displacement current. Seluﬁon: Since the voltage is of the form given by Eq. (6.46) with V0 z 20 V and
co : 211: x 106 radls, the displacement current is given by Eq. (6.49): EA
L; = —fd—V9msin 031' 4 8‘ —12 -4
x 354 x10 x10 x10 X 20 X 2R X 105 Sine“ x 106:) 1 x 10*2
—445 sin(21tx 106:) 01A). Problem 6.15 A coaxial capacitor of length 3 = 6 cm uses an insulating dielectric
material with g = 9. The radii of the cylindrical conductors are 0-5 cm and 1 cm. If
the voltage applied across the capacitor is VU) = 1005in(120m) W); What is the displacement current? Figure P615: Solution: To ﬁnd the displacement cunem, we need to know E in the dielectric space
between the cyiindiical conductors. From Eqs. (4-114) and (4.115), _ A Q
E 3 21cm”
21:3! 5:
Hence,
' 44.
E 3: 4L :- _§-1005m(120m) : 41 3 sin(120m‘) (Vim), r1211 (5) r1112 3“ {it-11%.: . 230 cma 6 D=EE = arsoE
144.3
3" = —r9 x 8.85 x 10-12 x sin(1207tr) 1.15 10-8
=—i‘—:——sin(120m) (0:112). The displacement current ﬂows between the conductors through an imaginary
cylindrical surface of length I and radius r. The current ﬂewing from the outer conductor to the inner conductor along hf crosses surface S where
S = —i’ 21ml. Hence, rd——-S=—r-— — a: a:
= 1.15 x10"3 x 120:: x 2micos(120m‘)
= 1.63 cos( 120m) 01A)- F -3
BB a(—1'15X10 sin(120mt))-(—i'21tri) Alternatively, since the coaxial capacitor is iossless, 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 21:81
C=ln( hlﬁ' The current is dV _ 21ml,
d! W103?) [1201: x lOOcos(1201tt)] = 1-63cos(1207u) (as), which is the same answer we obtained before- E’rohiem 6.16 The parallel-plate capacitor shown in Fig. 6-25 (P6.16) is ﬁlled
with a 105531 dielectric material of relative permittivity e, and conductivity 0'. The
separation between the plates is d anci each plate is of area A. The capacitor is
connected to a time—varying voltage source V0)- (a) Obtain an expression for IC, the conductiOn current ﬂowing between the plates
inside the capacitor, in terms of the given quantities. CHAPTER 6 231 Figure P616: Parailel-plate capacitor containing a lossy dielectric material (Problem
6. 1 6). {b} Obtain an expression for lg, the displacement current ﬂowing inside the
capacitor. {c} Based on. your expression for parts (a) and (b), give an equivalent-circuit
representation for the capacitor. {d} Evaluate the values of the circuit elements forA = 2 c1113, «1‘ 2 0.5 cm, Er = 4,
0' = 2.5 (Sim), and V0) 2 10.20581: x 103:)01). Solution:
(3) d V VA
6
R 5A” ‘ R d
(b) V BE BE EABV
*x— —:——_
E d” d a: may are: (c) The conduction current is directly proportional to V, as characteristic of a
resistor, whereas the displacement current varies as 317/32, which is characteiisﬁc
of a capacitor. Hence, d EA
R = — = — .
6A and C d
{d}
0.5 x 10-2 Rzmﬂm 232 Actual Circuit Equivalent Circuit Figure P6.16: (a) Equivalent circuit. 4 X 8.85 x 10-“2 x 2 x10—4 _ _ +12
C_ O'Sxma _1.42x10 1:. ?roblem 6.17 An electromagnetic wave propagating in seawater has an electric
ﬁeld with a time variation given by E = iEocosmt. If the permittivity of water is
8180 audits conductivity is 4 (Sim), ﬁnd the ratio of the magnitudes of the conduction current density to displacement current density at each of the following frequencies:
(a) 1 kHz, (b) 1 Nil-11,63 1 GHz, (:1) 100 GHZ. Sointion: From Eq. (6.44), the displacement current density is given by a a
= "D = s—E
3“ a: a:
and, from Eq. (4.67), the conduction current is IE = (3E. Converting to phasors and taking the ratio of the magnimdes, i
lid (a)Atf=1kI-Iz,m=21tx103radls,and @— — 4 z 888 x 103
“jd ‘ 231:)(103 x 81 X 8.854. ><10"12 '
The displacement current is negligible.
(b)Atf=1MI-Iz,m=2nx105 radfs,and
3 4
~— = = gd 21EX106 X 81x 8-854 x10—12 CIMPTER 6 233 The displacement current is practically negligible.
(c) Atf = 1 GHZ, ﬂ) = 27tx109 malls, and 3 4
_ ‘ 21:: x109 >< 81 x 3.354 XIO'"12 ‘ 0338' 3d Neither the displacement current nor the conduction current are negligible.
(d) Atf: 100 GHz, to: 212x 10“ radfs, and i
is 4 3
= — = .88 10‘ .
21: X 1011 x 81 x 3-854- x 10—12”1 8 X The conduction current is practically negligible. Sections 6-9 anﬁ 6—10: Cont?an Equation anti Charge Dissipation ﬁoblem 6.18 At t = 0, charge density pvo was introduced into the interior of a
material with a. relative permittivity er :- 480. If at z = 1 ,us the charge density has
dissipated down to 10-35)“), what is the conductivity of the material? Sciatica: We start by using Eq. (6.61) to ﬁnd If: M!) = EMF”? 01'
104% = Owe—104m,
which gives
10-6
m 10—3 = — ,
tr 01' 10—6 ;=_ =Lexm4 @. 11110-3
But Ir 2 8/6: 483/6. Hence 4 _ 4 —12
43° — ﬂ = 2.44 x 10"1 (Sim). ‘52 1r _ 1.45x10—7 234- Pmbiem 6.19 If the current density in a. conducting medium is given 3(9513’523): (iz * i3?2 + 2276) COS 931‘, determine the corresponding charge disnibution pv (x, 31,2; r). Soﬂnﬁon: Eq. (6.58) is given by 39v
V- : - _
3 8:
The divergence of J is
V‘J = ($3 +3>3 +ﬁi {ﬁz—fﬁyl +22x)cosmr
8x ay 82
= *353-(yzcos (or) 2 —6ycostet. Using this result in Eq. (E4) and then integrating both sides with respect to; DV = —J€{(V-J)dr = —/ ~6ycoswtdr = gsinmI+Cog_ where C9 is a constant of integration. Fromm 6.28 In a certain medium the direction of current density J-
radial direction in cylindricai coordinates and its rangintude is indepen '
and. 2. Determine 3, given that the charge density in the medium is Pv : porcoscor (0:133). Sointﬁon: Based on the given information,
3 = i‘fr(r)- With £3, = I: 2-— 0, in cylindrical coordinates the divergence is given by V-J— liker _r5‘r From Eq. (6-54), = —% = -—§(porCOSW) = pﬂmsinw‘ CHAPTER 6 235 Hence
3— $01,) = permsiner,
r
%(Dﬂ-) I porzmsinmr,
f imam = pomsincm / rzdr,
0 3r 0
r3 :-
mug = (pomsin our) «- ,
3 o
’)
J, = powr sincor,
3
and Problem 621 If we were to characterize how good a material is as an insulator by
its resistance to dissipating charge, which of the following two materiais is the better
insulator? Dry Soil: 5, 2 2-5, G = 10-4 (Sim)
Fresh Watch a, = 80, o = 10-3 (Sim) Sokmjon: Relaxation time constant 'tr = E. G
. 2.5
For dry soﬂ, "Cr — if k 2.5 x104 5.
0
For fresh water, 1:r : 10—4; 2 8 x 104 5. Since it rakes Eonger for charge to dissipate in fresh water, it is a better insulator than dry soil. Sections 6411: Electromagnetic yo’renﬁais Pmbﬂem 6.22 The electric ﬁeld of an electromagnetic wave progragating in air is
given by E(z,r) = s4cos(6 x 108: _ 22:) +§r3sin(6 x 108.: — 2z) (Wm). é
'§
2
'5'
'I .. m... :. :. ... 236 CHAPTER 6 Find the associated magnetic ﬁeld H(z, I). Solution: Converting to phasor form, the electric ﬁeld is given by
ﬁe) = tale—ﬂ: — mas—m mm.
which can be used with Eq. (6.87) to ﬁnd the magnetic ﬁeld: 1
him
1 x z y
=_— a/Bx 3/3)! 3/33
'3‘” 49*!“22 —j3e‘J?“’- 0 1
-J'£0# Vxﬁ ﬁe.) = (rims-J?z — wage-131) j = mﬁﬁ—wsk'fk = jt8.0e—ﬁz+5r10.6e-ﬁz (menu). Converting back to instantaneous values, this is H(I,z) : —§8.0sin(6 x 10% — 22) + §10.6cos(6 x 103: — Zz) (mAJm)- Problem 6.23 The magnetic ﬁeld in. a dielectric material with E r 430, p: = no, and
5 = 0 is given by my) = 15.20501: x 107: +ky) (Aim). Find 1% and the associated electric ﬁeld E. Seluﬁon: In phase}.- form. the magnetic field is given by Q = :‘iSeﬂ‘Ev (Aim). From
Eq. (6.86). .. 1 .. _ '
E = ,—V><H = 4525.2!”
10:38 ﬁne
and, from Eq. (6.87),
H 1 ._ .. 'kl .
H = we ._ _] teem,
—J£0# ‘szﬁi-i which, together with the original phasor expression for ﬁ, implies that cox/ET. 2nx107¢1 47:
“W: c =W=§a Wm)- CW6 inserting this valee in the expression for E above, 4112/30 56j435y/30 = _29¢1ei4’9’/30 (Vim)- Ez—i 21n><10'Jr x 4 X 8.854 x104:a Problem 6.24 Given an electric ﬁeld
E = ﬁEo sinaycos(mt - kz)1 where £0, a, co, and k are constants, ﬁnd E. Solution:
E = 12.50 smaycoswx — kz),
E = ﬁEosinaye'ﬂ‘z,
e = J. w e
Jam
1 a - 3 _ _ .
= —}7m—# [fa (ED sinay e‘ﬁh) — 23 (Ea smaye ﬂail
50 ﬂ _ n _ _ .kz
:: —-— k _ J
mu [3! smay z jﬂ cos aﬂe ,
H : ﬁdﬁej‘m]
= Sic {g [fksinay + 2 acosay e‘jnn]e_jhejm}
,u
_ Ev - v A i E 1
— a); [yksmaycos(mt—~ kz)+zacosaycos (cor kz 2)
£9 [ﬁksinaycosmx — kz) + 2amsay5in(m£ — kz)]. 03;” Problem 6,25 The electric ﬁeld radiaied by a short dipole antenna is given in
Spherical coordinates by E(R=8;r) = 93%sin8 cos(61t x 103: — 215R) (Vim)- Find E(R,B;it).
Solmion: Converting to phasor form, the electric ﬁeld is given by ~ a - 1 ‘1 .
E(R,8) = 859 = 821% smﬂeﬂm (Vim), 238 CHAPTER 6 which can be used with Eq- (6.87) to ﬁnd the magnetic ﬁeld: ~_1~_1.iaz~:e-1a
BULB) — _jmpVXE—— [RRSine FEE-F i 1 A2x10-3 a Z __ b - _ —J?.1tR _jmp¢ R smB BR(e ) n 21: 2x10"2 _ ﬂaw
_¢61t>< 108 X41tx 10-7 R mag .53
=¢Esinaeﬁ2ﬂ (pA/m). Converting hack to instantaneous value, this is £02, 63) : a} €53mb5 (6n: x108: — 21:3?) (,uAfm). thlem 6.26 A Hertzian dipole is a short conducting wire carrying an
approximately constant current over its length I. If such a dipole is placed along
the z-axis with its midpoint at the origin and if the current ﬂowing through it is
£(r) = Io cos tot, ﬁnd (a) the retarded vector potential E(R,6,¢) at an observation point Q(R,B,¢} in a spherical coordinate system, and (s) the magnetic ﬁeld phasor rim, 8, ¢)_
Assume E to be sufﬁciently small so that the observation point is approximately
equidistant to all points on the dipole; that is, assume that R‘ 2 R- Solution: (3.) In phasor form, the current is given by f: 10. Explicitly writing the volume
integral in Eq. (6-84) as a double integral over the wire cross Section and a single
integral over its length, ., U2 " . —ﬁcR‘
A: if / More;
41‘: 432 s R’ where sis the wire cross section. The wire is inﬁnitesimaﬂy thin, so that R’ is not a
function of x or y and the integration over the cross section of the wire applies only to
the current density. Recognizing that J = 310 / s, and employing the relation R’ x R, ... U2 -ij‘ £32 —ij I _
Aziﬂjr e dzzﬁtﬂf e -Hfo ﬁne
47!: 4/2 R’ 4'3: -1112 R 43R CHAPTER 6 239 In spherical coordinates, i = ﬁcos 8 — 3 sin 6, and therefore .. .. A ' HIDE _ .3
= * _ J _ A (Roosﬂ 851n9)4 e (13) From Eq. (6-85), e=im=iﬁ
jLi 41': A A 9‘“ij
Vx [(RcosB—ﬁsine) R ] _Ioz.1 a _ 4R a rib?
—E¢§(3+R(—sm6e 1' )v—a—e(cose R Prob} 6.27 The magnetic ﬁeld in a given dielectric medium is given by
H = iﬁcosksina x 107: — 9.1x) (Aim), where x and z are in meters. Determine:
(a) E,
(b) the displacement current density id, and
(e) the charge density pv. Solution:
(3)
e = §6c0322sin(2 x107: — 0.1x) = $‘6coschos(2 x107: —~ 0.1x— M2),
Eleni = ﬁ6cosZze‘jo'l‘e'J-“f2 2 —§Ij6c052ze'j°'h“1 )- A Z Y
. 3/8.: 8/33! 3/32 ﬂog 0 +j6c052z e'jo'lx 0
= J'— {3 ['3(—j6cos22£'j°'”)] +2 [3(rj6coslze*j°'1‘)]}
Jim 63 3x _ ' _6
= it («-2 sinks—50'”) +2 (—126 cos2ze*-’D‘”) . From the given expression for E, to : 2 x107 (reefs), . 240 CHAPTER 6 I3 = 0.1 (radfm). Hence, up+§=2x103 (mfs), c 2 3 x108 3
up 2X108 Using the. values for (o and 2, we have: i7: = (—ﬁ305in2z + 2j1.5c052z)x life—JILLr (Vim),
E = [—1230 sinchos(2 x 107: — 0.1 x) — z 1.5coszzsin(2 x 107: _ 0.130] (lem).
(b)
ﬁ = 2% = 2,20%": = (—12% siu22+i 111.03 coszz) x lo-ﬁe-In-lx (Clmz),
3])
Ed — E a
01' 3cl = jmﬁ = (—ﬁlesinZzw £0.6coszz)e*j0.1x:
351 = = {i 12sin225in(2 X 107: — 0-1x) — i0.6coschos(2 X 107t — 0- 1 x)] (Mmz). (c) We can ﬁnd 1:)v from V - D = [3‘r
or from a
Pu
V- = — .
“E 3:
Applying Maxwell’s equation,
BE BE
m V- = v -E _—_',, x _z
9‘? D 8 ax + az )
yields
3 . 7
9V = 2,20 3 [—30sm2zcos(2 x 10 t—O-1x)] +353; [—1.5coslzsi_n(2 x107 —0.1x)]} = 2,20 [—35i112zsin(2 x107r— 0.1x) +3sin2zsin(2 x107r—0-1x)] : 0. ...

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