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**Unformatted text preview: **CHAPTER 7 241 Chapter ‘7 Section 7-2: Propagation in Lossiess Mania. Prohiem 7 .1 The magnetic ﬁeld of a wave propagating through a cenain
nonmagnetic material is given by H : ﬁSOcosUOgr — 5y) (mAim). Find (a) the direction of wave propagation, (in) the phase velocity, (c) the wavelength
in the material, (C!) the relative permittivity of the material, and (e) the electric ﬁeld
phasor. ' 5' Solution:
(a) Positive y-direction.
(b)03=109 radfs,k=5radlm. new.“ ' 6) 109
MP:E:?=2XIOBWS. (6)1.2 2K/k=21t/5= 1.26111. 2 3 2
(3 3x10 3:
d = — : = -25.
(m (up) (2x108) 2 (e) From Eq. (7.3913), .. E = —T1§i X E,
,u 120:: 1201:
= -— = -— = —-—~ = 1.33
if: = 3;, and ii = ssoe-m x 10"3 (Aim). Hence, if = 451.333? x 2502—ij x 10'3 = 41257515? (Wm), and sum) = maﬁa“) = —s12.57cos(109r — 5y) (Wm). Ritchie-n: 722 Write genera} expressions for the eiectric and magnetic ﬁelds of a
I-Gi—iz sinusoidal plane wave traveling in the +y-direction in a iossless nonmagnetic
medium with relative pemﬁrtivity e, = 9. The electric ﬁeld is polaxized along the
x—direction, its peak value is 3 Vim and its intensity is 2 Vim at t = 0 and y = 2 cm. 242 CHAPTER 7 Seluﬁon: For f = 1 GHZ, pr = 1, and 8, = 9, (a) = 23f: 2115x109 radfs, k_g__2n “a _2r:x109
_x*;\0 8” c a" 3x108 E(y,r) = sscosan x 109: — ZOny-Hbg) (Wm). \/§ = 201: radfm, Atr:0andy=2¢m,E=2VImz 2 : 3cos(—201t x 2 x 10-2 +¢9) = 3 cos(—0.41t + (pg). Hence,
¢o —o.4n = cos—1 z 0.84 rad,
which gives
$0 = 2-11311 = 120.19"
and my) = 3:3 005(27: x 109: — 20w+120.19°) (Vim). Emblem 7.3 The elecn'ic ﬁeld phase: of a uniform plane wave is given by
E = 5? 1024"”-2z (me). Ifthe phase velocity of the wave is 1.5 X 103 m/s and the relative
permeability of the medium is ,ur : 2-4, ﬁnd (a) the wavelength, (b) the frequency f
of the wave, (c) the relative permittivity of the medium, and (d) the magnetic ﬁeld H(z,t).
Soiutien: H
(a) From E = ﬁlOeﬂ'z" (Vim), we deduce that k = 0.2 radlm. Hence,
2:: 21':
A. = — : — = 1017:: -4 .
k 02 31 2 m
(b)
_ £1; _ M _ 6 _
f_ l _ 3U; _4.77><10 32—43mm;
(c) From c 1 c 2 1 3 2
u: , a,=_ * =— — =1.57.
p Murat Fr (Hp) 2-4 (1-5) CHAPTER 7 243 (d) .u #r 2-4
= —:120’n’. —= 201’: —=4 . 4 Q
T' a a, 1 V1.6? 519 ( )’ .. _ 1 . .
H = $4) XE = .392) X gleam-k = £22.13eﬂ‘2‘ (mAz’m). ER, 1'] = 1222.13 (305(0)! + 0.22} (mAJm), with m = 2nf = 9.54:: x 105 radjs. Problem 7.4 The electric ﬁeld of a plane wave propagating in a nonmagnetic
material is given by E = [333 5in(2:rt x 107: — O.41tx)+ 24mm x 107: — 0.4m] (Wm). Determine (a) the wavelength, (b) er, and (e) H. Solution:
{a} Since k = 0.41:, 21': 27:
A, = — : —-'— =
k 0.41: 5m
(3))
a) 21:): 107 7
Ep E 0-4:": -— 5 X IRIS.
But
u _ i
P Hence,
2 8 2
at: : (3X10?) :36
up 5 x10
(at)
E = it x E = 1% x [3:351:1(21: x 107:: — U.41:x)+ extcosan x 107: — O.41cx)] 4
%sin(2tt x 107: — 0.4m) — fﬁcosa‘m x 107: — 0.4m) (Aim), 244 CHAPTER 7 with “0 ~ 120“ : 201:: 62.83 (9.). Problem 7.5 A wave radiated by a source in air is incident upon a soil surface,
whereupon a part of the wave is transmitted into the soil medium. If the wavelength
of the wave is 30 cm in air and 15 cm in the soil medium, what is the 5011’s relative permittivity? Assume the soil to be a very low loss medium.
Solution: From A. = 10/ Problem 7.6 The electric ﬁeld of a plane wave propagating in a lossless,
nonmagnetic, dielectric material with S; = 2.56 is given by E = §f20005(81t x 109! — k2) (Vim). Determine: f: up! A» k, and 1") 311d
(53) the magnetic ﬁeld H. Soluﬁon: (a) m: 21rf=81t><109 mus,
f=4x1o9Hz=4eHz, c 3 x103
u z —= =1.875x103m;s,
p V's—r «2.56
8
A: 32: Mzmgcm,
f 4x 109
k— 2— — ——3L =134.04Iadfm, TE
2. _ 4.69 no-2
no 377 377 = —-«-~=235.629.
11 J8? 2.56 1.6 WEE 7 245 (b)
H = —ﬁ% cos(81t X 1092‘ — kz)
— 4i cos(81t x109: 134 04 )
‘ 235.62 ‘ 3 4418.49 x 10'2cos(81c x 109: — 134.04g (Aim). 11 Section 7-3: Wave i3olarﬁzation E’mbiem 7.“? A11 RHC—polaﬂzed wave with a modulus of 2 (V 1111) is traveling in free
space in the negaﬁve z—direction. Write down the exptession for the wave’s electric
ﬁeld vector, given that the wavelength is 6 cm. J’ Figure P757: Locus of E versus time. Seﬁuﬁon: For an REC wave traveling in —i, let us try the following:
E = iacosMn‘ + k2) + @esimmt + kz).
Modulus IE] = 1/232 + a2 = m/E = 2 (Vim). Hence, 2
a=$:\/i b 246 CW 7 Next, we need to check the sign of the jacomponent relative to that of the
ﬁ-componcnt. We do this by examining the locus of E versus I at z = 0: Since
the wave is traveling along —2, when the thumb of the tight hand is along —E (into
the page), the other four ﬁngers point in the direction shown (clockwise as seen from
above). Hence, we should reverse the sign of the zit-component: E: itx/Ecos(cot+kz) —§‘\/2_sin(cor+kz) (Vim) Problem 7.8 For a wave characterized by the electric ﬁeld E(z,r) = ﬁaxeoswx — kz) + ﬁrst). cos(cot «— kz-l— 5), identify the polarization state, determine the polarization angles (ﬁx), and sketch the
locus of EU], I) for each of the following cases: (a) s:Jr = 3 Vim, a), = 4 Vim, and 5 = 0, {l3} :2: = 3 Vinny = 4Virrt, 311cm: 180°,
(c) a; = 3 Vim, a), z 3 Vim, and 5 = 45°,
(ll) ax = 3 Vlmay = 4 Vim, and 5 = —135°. Solution;
v0 = mites/ax), [Eq- (7.60)], tan2'y: (tan211fo) cosﬁ [Eq. (7-5%)],
sinzx = (SiDZWo)SiT15 [Eq (7.5%)}. “1.. 4 0 0 Linear 0 Linear
22 5" Left elliptical
Right elliptical (a) E(z,r) = §3cos(tor — k2) + §4cos(wt - la). (b) E(z,z) = it3cosfco: m kz) — §4cos(mr — kz)- (c) E[z,t) = ﬁ3cos(w~r - kz) +§3cos(cot — kz-I— 45°).
(d) Hm) = i3cos(tot — kz) +§r4cos[mr - k: — 135°). mbﬁem 7.9; The clean-it: ﬁgﬁd of a uniform plane wave propagating in free Space
is given by E : (:2 + _;'§')2€}e'1"‘z4’6 (Wm). Specify the modulus and dimction of the
616cm; ﬁeld intensity at the z x 0 plane at 1’ = 0., S and 10 as. 248 CHAPTER 7 Solution:
Hm) = meﬁefm’]
= 91¢[(i+ j§)20e—Wﬁef‘°’]
: mum ﬁeﬁﬂﬂOe—jm’léej‘m]
= 3220005(mr—1rz/6) +§20cos(cer— wax/6+ﬁ/2)
= i20cos(mt — 1112/6) - §203in(mr — aux/6) (Vim),
IE; = [534-531;2 = 20 (Wm),
1!; = tarr1 = «~(mr— 152/6)
Ex
From
a
=£=E;M=25X107m,
R 21: 2.11:
(o = 2nf= 5m 107 rad/s.
At 2 = 0,
0 at r z 0,
w = —(er = —5‘.I1: x107: : —0.251r = —45" at: = 5m, —0-5:rr = —90° at r = 10115.
Therefore, the wave is LHC polarized. Pmbiem 7.10 A linearly polarized plane wave of the form g: = iaxe'jkz can be
expressed as the sum of an RHC polarized wave with magnitude (3;: and an LI-IC
polarized wave with magnitude 2L. Prove this statement by ﬁnding expressions for
aR and aL in terms of (1;. SoEution: E = ﬁnes—M1
RHC wave: ER = aﬂi+fire—Jim)eF'ﬂ’ac = aﬁi— jﬁrk‘ﬂ“,
LHC wave: EL = are + we} 6.2-1“ = (1112“: + j§)e'i*z,
E = ER +EL,
iax = mi — i3?) +a1.(i+j§')- Bjr equating real and imaginary parts, ox = Q; +aL, O = —ag +aL, or aL : (Ix/2,
GR 2 lax/2- CHAPTER 7 249 Problem 7.11 The electric ﬁeld of an elliptically polarized plane wave is given by E(z, r) = [—i 10 sin(tor — kz — 60°) + 3’20cos(cor — kz)] (Wm). Determine (a) the polarization angles (73%) and (b) the direction of rotation. Soiution: (a) E(z,r) z [—ﬁlO sin(cor — kz — 60°) +§20cos(cor — kz)]
= [ilOcos[mr — kz + 30°) + $20 cos(oar - kz)] (Vim). Phasor fomn:
if = (sloef3°° + game-#2.
Since 5 is deﬁned as the phaSe of Ely relative to that of Ex, 5 = —30°,
20
_ —1 _ : t on
@1127: {taanxﬁcosﬁ = —l.15 or 1!: 655°,
sinzx = (sin2ipﬁsin5 = —0.40 or x = —11.';'9°. Prohiem 7.12 Compare the polarization states of each of the following pairs of
plane waves:
{a} wave 1: E1 = ﬂcosw): — kz) + ﬁZsinﬁm —kz),
wave 2: E2 : ﬁZcostr + k2) + $2 Siam)! + kz).
(b) wave 1: E1 = facosﬁm — kz) — fasinm): — kz),
wave 2: E; : i2cos(cm + Scz) — frZsinwat + k2)- Solution:
(a)
Eli : §2cos(mz — kz) + 3325mm - kz)
: i52cos{tor — kz) + § 2cos((or v» kz — It/Z),
E1 = isle—ﬂ“: + ﬁrms-W2, 250 m 7 Hence, wave 1 is RHC.
Similarly, H _ _ _
E2 = 15226sz + ﬁZe’kzehmﬂ.
Wave 2 has the same magnitude and phases as wave 1 except that its direction is along —2 instead of +2. Hence, the locus of rotation of E will match the left hand
instead of the right hand Thus, wave 2 is LHC. (b) E; = ﬁZcos(cor - kz) — 925mm: — k2),
E; = 32.2—ij +92e-i‘neiﬁf2. Wave 1 is LHC- 3732 = szei’q + yzeikzeﬁ‘f 1. Reversal of direction of propagation (relative to wave 1) makes wave 2 RHC- ﬁoblem 7.13 Plot the locus of E(0,t) for a plane wave with
E(z,r] = itsin(car + kz) + 3‘? 2cos(mr + k2). Determine the polarization state from your plot. Solution:
E = i: sin(mt+kz) + ﬁZcosw): +kz).
Wave direction is —ﬁ. At 2 = 0, E = it sinner + 37200303. Tip of E rotates in accordance with right hand (with thumb pointing along —§:).
Hence, ane state is REE. Sections 7—4: Pmpagaﬁom in a Lossy Medium Probﬁem “LE4 For each of the following combination of parametets, determine if
the maxerial is a low-loss dielectric, a quasi-conductor, or a good conductor, and then
calculate on, 1., up, and 11¢: (a) glasswithpr= 1, a. = 5, and 0': 1042 Sim at 1061-12,
(b) animal tissue withy, = 1, Sr :12, and 0' = 0.3 Sim at 100 MHZ,
(e) woodWithpr:1,Er: 3,316.5: 10‘4Shnat 1 kHz. Soﬁution: Using equations given in Table 7-1: 3.6 x 10‘13 4.5 600
ﬂow-loss dielectric quasi-conductor good conductor
8.42 x 10““ Nplm 9.75 prm 6.3 x 10-4 Nplm
468.3 radio: 12-16 Iadfm 6.3 x 19-4 mm
1.34am 51.69am 10km
1.34x108mxs ' 0.52x108sz 0.1 x 103mm
2 168.5 9; 39.54+j31.72£l 6-28(1+j) Q . -e Frebﬁem 7.15 Dry soil is characterized by 8r 2: 7.5, ,u, = 1, and 6 = 10—4 (Sim).
At each of the following frequencies, determine if dry soil may be considered a good
conductor, a. quasi-conductor, or a low-loss dielectric. and then calculate on, 5, 7L, ,up,
and Tie: (a) 60 Hz, (b) 1 kHz. (c) 1 MHz, {d) 1 GHz. Solution: 5, = 2.5, ,u, = 1, e z 10-4 Site. —' (300d conductor Quasi-conductor
1.54x10“ 6-23x10“ 1.13,>:1o-2
.
Mm) 4-08X10‘ Problem 7.16 in a medium characterized by a, :2 9, ,u, = 1, and c = 0.1 Sim, determine the phase angle by which the magnetic ﬁeld leads the electric ﬁeld at
100 MHZ. Soluﬁen: The phase angle by which the magnetic ﬁeld leads the electric ﬁeld is —611
where 81.. is the phasa angle of m. 0.1 x3611: _
211x103 X10—9x9 _ 3» 2.
038 Hence, quasi—conductor. _ 'u I Isrr)--X]2 120ﬁ(1 ' G )-I;’2
ne— 8’ ’19 J87 Image = 125.67(1~— jzrlﬂ = 71.49 + j44.18 = 84-04ﬂ1-72". CHAPTER 7 253 Therefore 9“ = 31.72.”.
Since H = {Una}; X E, H leads E by —9n, or by —31.72°. In other words, E". lags
E by 3132". Problem 7.3.7 Generate a plot for the skin depth 55 versus frequency for seawater
for the range from 1 Id-Iz to 10 SH: (use log-log scales)- The constitutive parameters
of seawater are ,u, = 1, E, = 80 and c3 = 4 Sim. Soﬁuﬁenz
2 —1{2
1 l #8! an
55 a a? “(3) ‘1 1
(0:21'cf, “5‘ “cl—creme)?
ﬂ'
ei 5 cs 4x361l: _£X109_ s! = E: (08081- : 2nfx10-9xso _ 80f
See Fig. £37.17 for plot of 55 versus frequency. Sﬁﬁn death vs. frequency for We:
10 an Skin depth [m] .a
D 1c.4 19'2 10" 1o” 10‘ 102 1o” 10" 254 CHAPTER 7 E’roblern 7.18 Ignoring reﬁmtion at the air-soil boundary, if the amplitude of a
2-GI-Iz incident wave is 10 Vim at the surface of a wet soil medium, at what depth will
it be down to 1 mVltn? Wet soil is charaeterized by it, = 1, E,— = 16, and 6 = 5 X 10—4
Sim. Soiuﬁon: Hz) = Egg—“z = 103'”, —4
E: &:2_3X10—4_
me 21tx2x109x10—9x 16 Hence, medium is a low-loss dielectric. o ,u 5 12011: 5 x 10‘4x 1201:
cc:— —:--h=———:0.024 )' ,
2 E 2 N/E; ZXV 16 (Np In) 10-3 = 103-09242 11110-4 = 4.0243,
2 = 383.76 at Probiem 7.19 The skin depth of a certain nonmagnetic conducting material is 2 ,trrn
at 5 6H2. Determine the phase velocity in the material. Soitttion: For a good conductor, or = B, and for any material 55 = 1 /0t. Hence, 2
up=g=g=2nf55=2itx5x109x2xlﬂ“6=6.28xi{)4 (refs). Froblem 7.2% Based on wave attenuation and reﬂection measurements conducted
at 1 MHz, it was determined that the intrinsic impedance of a certain medium is
28.14152 (52) and the skin depth is 5 n1. Determine (a) the conductivity of the
material, (in) the wavelength in the medium, and (c) the phase velocity. Soiution:
(a) Since the phase angle of 11,; is 45°, the material is a good conductor. Hence, - a '45” o A -
71,: = (1+3; = 28-121 = 28.103545 +128.1 511146", = 28.1 cos45° : 19.37. QIR CHAPTER 7 255 Since 0.21/55 =1/5 = 0.2 prm, 6* a — 0'2 +0015;
‘ 19.37 ‘ 19.87 ‘ ' m'
(h) Since at : B for a good conductor, and CL = 0.2, it follows that B = 0.2.
Therefore,
21: 21:
1-: F: 0—2: 101t=314m (c) u = fl=106x31.4= 3.14 x107 mis. Fromm “7.21 The electric ﬁeld of a plane wave propagating in a nonmagnetic
medium is given by E : i25e'"3°xcos(21r x 109: — 40x) (Wm). Obtain the corresponding expression for E. Solution: From the given expression for E, (0 = 21m 109 (mus),
a = 30 (Nplm),
l3 2 40 (radius). From (7.65:1) and (7.65%)), '3 7 r ‘3 1' (92 i' 0:2 — £3“ = 43118 = —€D‘.uo€o€T = —;2—s,,
:02 n 20L?) = (02,142” : gar. Using the above values for (0, 0t, and [3, we obtain the following: CHAPTER 7
E = 225e‘30‘e‘f40‘,
~ 1 .. ~ 1 . a
H = F3; x = 157 96mg? 3‘: x 2256—3Qre“’40x : ~§0.1ée*3°*e-4°*e‘f35-35 1
c .
H = mqﬁef‘ﬂ'} = —§0.16£‘30‘cos(21r X 109 ~ 40x — 36.85°) Wm). Sectian 7-5: Current Flew in Cenducters Problem 7.22 In a nonmagnetic, lossy, dielectric medium, a 300—MH2 plane wave
is characterized by the magnetic ﬁeld phasor Q = (i—j42)e*2ye_j9y (Aim). Also, we are given than f = 300 MHz 2 3 x 103 Hz. From (7.653.), a2_ﬂ2=_m2 r, —9
4—81 = —(2:r><3x 108)2x41:x10‘7><£;x 10 361': ’
whose solution gives
a; = 1.95.
Simian-1y, from (7 65b),
2043 = mzue”?
8 2 —T I! 10kg
2x2><9=(21t><3>< 10 ) x472x10 xeT x —-—,
361:
which gives
4' = 0.91. 257 0.91 "/2 77 . ..
11° (l—j—) =3f§3{0.93+j0.21)=256.93J12'5. Hence, E = 456.9 aim-6‘s x (s:— j42)e'2ye_"9v" = {ij4+i) 256.9.e—2ye'j9yej12'6n
= mm” + e) 256-9 reg-Weill?
E = {Fiﬁﬁej‘m}
= $1.03 x 103e‘zycos(mt — 9y+ 10245")
+ $256.9 e_2ycos(cot — 9y+12.6°) (Vim),
H = S‘te-[ﬁeJ-w}
= Erie-{(3% + j4ﬁ)e_2ye_j9yej‘m}
: ire—2}” cos(cm‘ - 9y) «1- 248'23’ 5111(0): — 9y) (Aim). Probﬁem 7.23 A rectangular copper block is 30 cm in height (along 2). in response
to a wave incident upon the block from above, a current is induced in ﬁle block in the
positive x—direction. Determine the ratio of the a—c resistance of the block to its d—c resistance at 1 kHz- The relevant properties of copper are given in Appendix B. Figure P723: Copper block of Problem 7.23. 258 CRAP] ER 7
Solution: 5! c es'smoeR I I - 1' = — z 1 dc GA 0.36?!) 7
I
-c s' R = —.
a re astance 39 55 0.5 cm and 1 cm, respectively. The conductors are made of copper with 8, = I,
p, = 1 and o = 5.8 X 107 Sim, and the outer conductor is 0-1 cm thick. At 10 MHz: (3) Are the conductors thick enough to be considered inﬁnitely thick so far as the
ﬂow of content through them is concerned? {53) Determine the surface resistance Rs. (e) Determine the a-c resistance per unit length of the cable.
Solution:
(a) From Eqs. (7.72) and (7.77m, 55 = [amid—“2 = [n X10T x 41tx10‘7 x 5.8 x 1071-”2 : 0.021mm. Hence, 0.1 cm ~ 50. d
55 0-021 mm W Hence, conductor is plenty thick.
(b) From Eq. (7.92a), CHAPTER 7 259 Section 7—6: EM Power Density Premiere 7.25 The magnetic ﬁeld of a plane wave traveling in air is given by
H = 5:25 sin(21t x 107: — ky) (mAJm)- Determine the average power density carried by the wave. Soiution: e = 3225511121: x 107: — Icy) (mom),
E = 4103? x E = 211025513191: x 107: — Icy) (mwm), 3 12
saw = (2 x and?) x10"6 = *%{25)2 x10—6 = 390.12 (Wlmz). ?roblem 7.26 A wave traveling in a nomagnetic medium with a, = 9 is
characterized by an electric ﬁeld given by e = [ﬁcosm x 107: + 10:) — i2cos(n x 107: +kx)] (Wm). Determine the direction of wave travel and the average power density carried by the wave- Sointion:
T30 1207C
T]: —-— = —=407|2 d8? \/§
The wave is traveling in the negative x-direction.
, 32 + 22 , 13 A ,
Ba“. : —X[—2ﬁ—] = —X2 X 4011 I (Will-ﬂ"). Problem 7.27 The electric-ﬁeld phesor of a uniform plane wave traveling
downward in water is given by E = eioe‘c‘lze'i‘J-Zz (Wm), where i is the downward direction and z = 0 is the water surface. If G = 4 Sim, (a) obtain an expression for the average power density, (b) determine the attenuation rate, and
{c} determine the depth at which the power density has been reduced by 40 dB- 260 CHAPTER 7 Solution:
(3) Since at z B = 0.2, the medium is a good conductor. (X. 0-2 . o
m = (1+1); 2 (1+3? = (1 +j)0.05 = 0.0707945 (9).
From Eq- (7.109), |Eo|2 ﬂu A 100
a — .—
°°S ‘1 z2x0.0707e (b) A = —8-680Lz = —3-68 X 0.22 = --1-74z (dB).
(e) 40 dB is equivalent to 10". Hence, ‘0'4Zcos45° = 250050“ (“Til-:13). 10—4 = 52°“ = rm, 113(10—4) = —0.4z, or z = 23.03 In. Pinblem 7.28 The amplitudes of an elliptically polarized plane wave traveling in a
lossless, nonmagnetic medium with a, = 4 are Hyo = 6 (mAhn) and Hg) : 8 (mAfm). Determine the average pewer ﬂowing through an aperture in the 59-2 plane if its area
is 20 1112. Solution: 11* no _ 1201:
x/E—r— x/Z
.1385 a. = i§[Hfo+Hfo] = x 2 [36+64] x 10-6 = 9.43 (mW/mz),
P = 3a.}; = 9.43 x 110-3 x 20 = 0.19 W. = 601: =188.5 Q, Problem “7.29 A wave traveling in a lossless, nonmagnetic medium has an electric
ﬁeld amplitude of 24.56 Vim and an average power density of 4 Win12. Determine the phase velocity of the wave. Solnﬁnn:
S {5012 I150!2
av — T u 'l
21] 2.3av
01f
(24.56)2 _
n — 2 x 4 H 75.4 $2 CHAPTER 7 261 But Hence, Problem “7.30 At microwave frequencies, the power density considered safe for
human exposure is 1 (kaniz). A radar radiates a wave with an electric ﬁeld
amplitude E that decays with distance as E(R) = (3,000/R) (Vim), where R is the
distance in meters. ‘What is the radius of the unsafe region? Solution:
2
Sav_l52(:)1 , 1(mWIm2):10—3chm3=10W!m2,
0
10_ 3x103 2X 1 _1.2x104
_ R 2x12011:_ R2 ’ Fromm “7.31 Consider the imaginary rectangular box shown in Fig. 7-19 (P731).
(.3) Determine the net power ﬂux PU) entering the box due to a plane wave in air given by E = mcosmr — ky) (Wm).
(b) Determine the net time~average power entering the box. Soiution: (a)
i: = ﬁEoCOS(W—ky}a H = —Q E c0582: -— Icy). ﬁe
52
so) = Exﬁ: ﬁr—Ocosz(mr~!gz),
710
- 53 2 2
HI) = S[Z)A|y=o-S(E)A[y=b = —-ac[cos (or -—cos (cor—kb)]. T10 262 CHAPTER 7 MWHN 4’
a, ’5
a @4- E
“ e a
fun-«SI. u u m m an.
i u g
E z E u
E 3
CE ﬁnnwnumJnanHy
z ’ 3 "
I ‘F‘
E .r E y
5’ 4?
/*v‘nﬁwuﬂmuw I Figure P731: Imaginary :ectangular box of Problems 7-31 and 732- (b) 1 T
P3v=?-/ﬂ P(I)d[. where T = Zn/m. 2 21th:)
pav=E°ac [coszmr—cosz(mr~kb)]dt} =0.
0 1'10 Net average energy entering the box is zero, which is as expected since the box is in
a lossless medium (air). Preblem 7.32 Repeat Problem 7.31 for a wave traveling in a. lossy medium in which E = i lme'wcosﬂn X 109: — 40y) (Wm)1
H = —-£0.64e_30ycos(27c x 109: — 40y — 3635") (Aim)- The box has dimensions A = 1 cm, I) = 2 cm, and c = 0.5 cm.
Soiutiun:
(a)
S(I) = E x H
: i100€‘3°yeos(2ﬁ x 109; F 40y)
x (—ﬁ0.64)e‘3uycos(27c x 109: — 40y * 3635")
= fMe'mycosQit x 1092‘ -— 40y) 605(21'5 x 109: — 40y — 3635" ). CHAPIER 7 263 Using the identity cos 8cos¢ = %[cos(9 + o) + cos(6 — (13)], 64
so) = fré‘b’kosm x 109: — 80y— 36.85") + c0536.85°], P0) = 3(tjAl}—_—o-— S(I)A[y=b
= 32ac{[cos(41r x 10% — 36.85“) + c0536.85°]
— e_60b[cos(41t x 10% — 80y — 36.85°) + cos 36.85°]}.
(b)
Pay—fl) P(I)I—E/O 0):. The average of cos( 031’ + 8) over a period .T is equal to zero, regardless of the value
of 6. Hence, PaV = 32ac(1 — fa”) cos 36-83”. Witha=1cm, b=2cm,andc=0.5cm, Pa“, = 8.95 X104 (W). This is the average power absorbed by the Eossy material in the box. Pmbiem 7.33 Given a wave with
E = 22E0c05(mz — k2), calculate:
(a) the time-average electric energy density 1 T 1 T a
(waavzfjg wedt=ﬁfﬂ 5E“dr, (b) the time-average met“: energy density 1 T 1 T 2
(Wm)av= WmdiL:2_T_/D tuH di,
and
(e) Show that {wehv = (Wm)av- Seiaﬁem:
(a)
1 T 2 2
(we)... = E if) EEO cos (a)? — was. 264 W1thT=2§, (h) (e) ...

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