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32.1: a)
.s28.1
sm1000.3
m1084.3
8
8
c
d
t
b) Light travel time is:
s1072.2
)hour1(
)s3600(
)day1(
)hours24(
)year1(
)days365(
)years61.8(years61.8
8
km.108.16m108.16s)10(2.72s)m100.3(
131688
ctd
32.2:
.m180)s100.6()sm100.3(
78
tcd
32.3:
jjB
ˆ
2cos
ˆ
cos()
maxmax
t
c
z
fB
ωt)kzB(z,t
jB
ˆ
s)m1000.3(
Hz)1010.6(2cos)T1080.5()(
8
144
t
z
z,t
.
ˆ
))srad1083.3()m1028.1((cos)T1080.5()(
15174
jB tz,tz
)
ˆ
()
ˆ
()(
kjE c(z,t)Bz,t
y
.
ˆ
s)rad1083.3()m1028.1((cosm)V1074.1()(
15175
iE t)zz,t
32.4: a)
.Hz1090.6
m1035.4
sm1000.3
λ
14
7
8
c
f
b)
.T1000.9
sm1000.3
mV1070.2
12
8
3
max
max
c
E
B
c) The electric field is in the
x
-direction, and the wave is propagating in the z-
direction. So the magnetic field is in the y-direction, since
.BES
Thus:
.
ˆ
))srad1034.4()m1045.1cos(()mV1070.2()(
ˆ
sm1000.3
)Hz1090.6(2cosV/m)1070.2(),(
ˆ
2cos
ˆ
(cos),(
15173
8
143
maxmax
iE
iE
iiE
tzz,t
z
t
πtz
t
c
z
πfEωt)kzEtz
And
.
ˆ
))srad1034.4()m1045.1cos(()T1000.9(
ˆ
),(
),(
151712
jjB tz
c
tzE
tz
32.5: a)
y
direction.
b)
m.1011.7
)srad1065.2(
)sm1000.3(22
λ
λ
2
2
4
12
8
ω
πcπc
πfω
c) Since the electric field is in the
z
-direction, and the wave is propagating in the
y
-direction, then the magnetic field is in the
x
-direction
).( BES
So:
iiiB
ˆ
)sin(
ˆ
)sin(
ˆ
),(
),(
00
ty
c
c
E
tky
c
E
c
tyE
ty
iB
ˆ
)srad1065.2(
)sm1000.3(
)srad1065.2(
sin
sm1000.3
mV1010.3
),(
12
8
12
8
5
tyty
.
ˆ
))srad1065.2()m1083.8sin(()T1003.1(),
1233
i(B tyty
32.6: a)
x
direction.
b)
.Hz1059.6
2
)sm100.3()mrad1038.1(
2
22
11
84
π
π
kc
f
c
fπ
λ
π
k
c) Since the magnetic field is in the
y
-direction, and the wave is propagating in the
x
-direction, then the electric field is in the
z
-direction
).( BES
So:
.
ˆ
))srad1014.4()mrad1038.1sin(()mV48.2(),(
ˆ
))srad1014.4()mrad1038.1sin(())T1025.3((),(
ˆ
)2sin(
ˆ
),(),(
124
1249
0
kE
kE
kkE
txty
txctx
fkxcBtxcBtx
32.7: a)
m.361
Hz1030.8
sm1000.3
λ
5
8
f
c
b)
1
m0174.0
m361
2
λ
2
ππ
k
c)
s.rad1021.5)Hz1030.8(22
65
πfω
.mV0145.0)T1082.4()sm1000.3(
118
maxmax
cBE
32.8:
.T1028.1
sm1000.3
mV1085.3
11
8
3
max
max
c
E
B
So
,1056.2
T105
T1028.1
7
5
11
earth
max
B
B
and thus
max
B
is much weaker than
.
earth
B
32.9:
BEBE
KK
cB
KK
BB
vBE
00
.mV779.0
)23.1()74.1(
)T1080.3()sm1000.3(
98
E
32.10: a)
.sm1091.6
)18.5()64.3(
s)m1000.3(
7
8
BE
KK
c
v
b)
.m1006.1
Hz0.65
sm1091.6
λ
6
7
f
v
c)
.T1004.1
sm1091.6
mV1020.7
10
7
3
v
E
B
d)
.mW1075.5
)18.5(2
)T1004.1)(mV1020.7(
2
28
0
103
0
B
K
EB
I
32.11: a)
.m1081.3
Hz1070.5
sm1017.2
λ
7
14
8
f
v
b)
.m1026.5
Hz1070.5
sm1000.3
λ
7
14
8
0
f
c
c)
.38.1
sm1017.2
sm1000.3
8
8
v
c
n
d)
.90.1)38.1(
22
2
2
n
v
c
K
K
c
v
E
E
32.12: a)
s.m1034.2)m15.6)(Hz1080.3(λ
87
fv
b)
.64.1
)sm1034.2(
)sm1000.3(
28
28
2
2
v
c
K
E
32.13: a)
25
max
2
max0
mW101.1som,V090.0;
2
1
IEcEI
b)
T100.3so
10
maxmaxmaxmax
cEBcBE
c)
W840)m105.2()4()mW10075.1()4(
23252
av
rIP
d) Calculation in part (c) assumes that the transmitter emits uniformly in all directions.
32.14:
The intensity of the electromagnetic wave is given by Eqn. 32.29:
.
2
rms0
2
max0
2
1
cEcEI
Thus the total energy passing through a window of area
A
during
a time
t
is
J
9.15)s0.30)(m500.0()mV0200.0()sm1000.3()mF1085.8(
228122
rms0
μAtcE
32.15:
J105.2)m100.2()4()mW100.5()4(
25210232
av
rIP
32.16:
a) The average power from the beam is
W104.2)m100.3()mW800.0(
4242
IAP
b) We have, using Eq. 32.29,
.
2
rms0
2
max0
2
1
cEcEI
Thus,
mV4.17
s)m1000.3)(mF1085.8(
mW800.0
812
2
0
rms
c
I
E
32.17:
cIp
rad
so
23
rad
mW1070.2 cpI
Then
W105.8)m0.5()4()mW1070.2()4(
52232
rIP
av
32.18: a)
Hz.1047.8
m354.0
sm1000.3
λ
8
8
c
f
b)
.T1080.1
sm1000.3
mV0540.0
10
8
max
max
c
E
B
c)
.mW1087.3
2
)T1080.1()mV0540.0(
2
26
0
10
0
EB
SI
av
32.19:
2
0
max
2
0
2
max
2
)4(
2
r
Pc
Er
c
E
ASP
av
m.V0.12
)m00.5(2
)sm1000.3()W0.60(
2
0
8
max
E
.T1000.4
sm1000.3
mV0.12
8
8
max
max
c
E
B
32.20: a) The electric field is in the
y
-direction, and the magnetic filed is in the
z
-
direction, so
.
ˆ
ˆ
)
ˆ
(
ˆˆ
ˆ
ikjBES
That is, the Poynting vector is in the
x
-
direction.
b)
)(cos
),(),(
),(
0
maxmax
0
tkx
BE
txBtxE
txS
))).(2cos(1(
2
0
maxmax
kxt
BE
But over one period, the cosine function averages to zero, so we have:
|
.
2
|
0
maxmax
BE
S
av
32.21: a) The momentum density
s.mkg107.8
s)m100.3(
mW780
215
28
2
2
c
S
dV
dp
av
b) The momentum flow rate
Pa.106.2
sm100.3
mW780
1
6
8
2
c
S
dt
dp
A
av
32.22: a) Absorbed light:
.Pa1033.8
sm100.3
mW2500
1
6
8
2
rad
c
S
dt
dp
A
p
av
.atm1023.8
atmPa10013.1
Pa1033.8
11
5
6
rad
p
b) Reflecting light:
.Pa1067.1
sm100.3
)mW2500(2
2
1
5
8
2
rad
c
S
dt
dp
A
p
av
10
5
5
rad
1065.1
atmPa101.013
Pa1067.1
p
atm. The factor of 2 arises because the
momentum vector totally reverses direction upon reflection. Thus the change in
momentum is twice the original momentum.
c) The momentum density
s.mkg1078.2
s)m100.3(
mW2500
214
28
2
2
c
S
dV
dp
av
32.23:
EBEBc
c
E
EcEES
0
0
00
0
0
0
0
2
0
0
2
00
0
1
.
2
0
0
2
0
cE
c
EEB
32.24: Recall that
:so,
BES
a)
.
ˆ
)
ˆ
(
ˆ
ˆ
kjiS
b)
.
ˆˆˆˆ
kijS
c)
.j)i()k(S
ˆ
ˆ
ˆ
ˆ
d)
.
ˆ
ˆ
ˆ
ˆ
j)k(iS
32.25:
T1033.1
8
maxmax
cEB
BE
is in the direction of propagation. For
E
in the +
x
-direction,
B
E
is in the +
z
-
direction when
B
is in the +
y
-direction.
32.26: a)
.m00.2
)Hz100.75(2
sm1000.3
22
λ
6
8
f
c
x
b) The distance between the electric and magnetic nodal planes is one-quarter of a
wavelength =
m.00.1
2
m00.2
2
4
λ
x
32.27: a) The node-antinode distance
m.1038.4
Hz)1020.1(4
sm1010.2
44
λ
3
10
8
f
v
b) The distance between the electric and magnetic antinodes is one-quarter of a
wavelength
m.1038.4
Hz)10(1.204
sm102.10
44
λ
3
10
8
f
v
c) The distance between the electric and magnetic nodes is also one-quarter of a
wavelength
m.1038.4
Hz)104(1.20
m/s102.10
44
3
10
8
f
vλ
32.28:
.cm0.20m200.0
)Hz1050.7(2
sm1000.3
22
λ
8
8
nodes
f
c
x
There must be nodes
at the planes, which are 80.0 cm apart, and there are two nodes between the planes, each
20.0 cm from a plane. It is at 20 cm, 40 cm, and 60 cm that a point charge will remain at
rest, since the electric fields there are zero.
32.29: a)
mm.7.10mm)55.3(22λ
2
λ
xx
b)
.mm55.3
BE
xx
c)
s.m1056.1)m1010.7()Hz1020.2(λ
8310
fv
32.30: a)
)sincos2()sinsin2(
),(
maxmax
2
2
2
2
ωtkxkE
x
ωtkxE
x
x
txE
y
.
),(
sinsin2sinsin2
),(
2
2
00max
2
2
max
2
2
2
t
txE
tkxE
c
tkxEk
x
txE
yy
Similarly:
)cossin2()coscos2(
),(
maxmax
2
2
2
2
tkxkB
x
tkxB
x
x
txB
z
.
),(
coscos2coscos2
),(
2
2
00max
2
2
max
2
2
2
t
txB
tkxB
c
tkxBk
x
txB
zz
b)
tkxkEtkxE
x
x
txE
y
sincos2)sinsin2(
),(
maxmax
.
),(
)coscos2(
),(
.sincos2sincos2sincos2
),(
max
max
max
max
t
txB
tkxB
t
x
txE
ωtkxBtkx
c
E
tkxE
cx
txE
z
y
y
Similarly:
tkxkBtkxB
x
x
txB
z
cossin2)coscos2(
),(
maxmax
tkxcB
c
tkxB
c
x
txB
z
cossin2cossin2
),(
max
2
max
,(
)sinsin2(cossin2
),(
00max00max00
t
t
xE
tkxE
t
tkxE
x
txB
y
z
32.31: a) Gamma rays:
nm.104.62m1062.4
Hz106.50
sm103.00
λ
514
21
8
f
c
b)
.nm522m105.22
Hz1075.5
sm1000.3
λ:lightGreen
7
14
8
f
c
32.32: a)
Hz.106.0
m5000
sm103.0
λ
4
8
c
f
b)
Hz.106.0
m5.0
sm103.0
λ
7
8
c
f
c)
Hz.106.0
m105.0
sm103.0
λ
13
6
8
c
f
d)
Hz.106.0
m
10
5.0
sm103.0
λ
16
9
8
c
f
32.33:
Using a Gaussian surface such that the front surface is ahead of the wave front (no
electric or magnetic fields) and the back face is behind the wave front (as shown at right),
we have:
.00
0
encl
xx
E
ε
Q
Ed
AAE
.00
xx
BABd AB
So the wave must be transverse, since there are no components of the electric or
magnetic field in the direction of propagation.
32.34: Assume
.with),sin(
ˆ
)sin(
ˆ
maxmax
tkxBandtkxE kBjE
Then Eq. (32.12) implies:
.0)cos()cos(
maxmax
tkxBtkxkEx
t
B
x
E
z
y
.λ
λ
/
2
2
maxmaxmaxmaxmaxmaxmax
cBBfB
f
B
k
EBkE
Similarly for Eq.(32.14)
.0)cos()cos(
max00max00
tkxEtkxkB
t
E
x
B
y
z
.
1
/
2
2
maxmax
2
max
2
max
00
maxmax00max
E
c
E
c
fλ
E
λ
c
f
E
k
BEkB
32.35: From Eq. (32.12):
2
2
),(),(
),(
t
txB
t
txB
tx
txE
t
zz
y
But also from Eq. (32.14):
t
txE
xx
txB
x
y
z
),(
),(
00
2
2
00
),(
t
txB
z
.
),(),(
2
2
00
2
2
t
txB
x
txB
zz
32.36:
)cos(
2
1
2
1
)cos(),(
2
max0
2
0max
tkxEEutkxEtxE
Ey
B
z
E
u
B
tkxBtkx
c
Ec
u
0
2
2
max
0
2
max
2
0
2
)cos(
2
1
)cos(
2
32.37: a) The energy incident on the mirror is
AtcEIAtPt
2
0
2
1
J.105.20s)(1.00)m1000.5()mV028.0()sm1000.3(
2
1
102428
0
E
b) The radiation pressure
.Pa1094.6)mV0280.0(
2
152
0
2
0rad
E
c
I
p
c) Power
2
rad
2
24 RcpRIP
W.101.34m)(3.20Pa)10(6.94s)m1000.3(2
42158
P
32.38: a)
.Hz1081.7
m0384.0
sm1000.3
λ
9
8
c
f
b)
T.1050.4
sm1000.3
mV1.35
9
8
max
max
c
E
B
c)
.mW102.42m)V(1.35)sm1000.3(
2
1
2
1
2328
0
2
max0
cEI
d)
.N1093.1
s)m1000.3(2
)m(0.240T)10(4.50m)V35.1(
2
12
8
0
29
0
c
EBA
c
IA
pAF
32.39: a) The laser intensity
.mW652
m)10(2.50
W)104(3.204
2
23
3
2
πD
P
A
P
I
But
m.V701
)sm10(3.00
)mW2(652
2
2
1
8
0
2
0
2
0
c
I
EcEI
And
T.102.34
sm103.00
mV701
6
8
c
E
B
b)
.mJ101.09m)V(701
4
1
4
1
362
0
2
max0
Euu
avav
EB
Note the extra factor
of
2
1
since we are averaging.
c) In one meter of the laser beam, the total energy is:
42)(2Vol
2
tottot
LDuALuuE
EE
J.101.07m)/4(1.00m)1050.2()mJ1009.1(2
112336
tot
E
[...]... 1 so prad I C (ii) totally reflecting W 0 so p rad 2 C These are just equations 32.32 and 32. 33 c) W 0.9, I 1.40 10 2 W/m 2 prad W 0.1, I 1.40 103 W m 2 prad (2 0.9) (1.40 103 W m 2 ) 5.13 10 6 Pa 3.00 108 m s (2 0.1)(1.40 10 2 W m 2 ) 8.87 10 6 Pa 8 m 3.00 10 s 32. 41: a) At the sun’s surface: 3.9 10 26 W P P P IA I 6.4 10 7 W m 2 2 8... initial energy differs from the proton’s by the ratio of their masses: m (9.11 10 31 kg) E e E p e (6.00 10 6 eV) 327 3 eV mp (1.67 10 27 kg) So the fraction of its energy that it radiates every second is: (dE dt )(1 s) 8 .32 10 5 eV 2.54 10 8 327 3 eV E 32. 54: For the electron in the classical hydrogen atom, its acceleration is: 2 2(13.6 eV)(1.60 10 19 J eV) v 2 1 mv 21 ... current carrying solenoid as having entered through its cylindrical walls while the current was attaining its steady-state value But also U 32. 44: a) The energy density, as a function of x, for the equations for the electrical and magnetic fields of Eqs (32. 34) and (32. 35) is given by: 2 u 0 E 2 4 0 E max sin 2 kx sin t 1 1 , cos t cos and sin ωt sin b) At t 4 4 4 2 2 ˆ ˆ.. .32. 40: a) The change in the momentum vector determines p rad If W is the fraction absorbed, P P out P in (1 W ) p ( p ) (2 W ) p Here, (1 W ) is the fraction reflected The positive direction was chosen in the direction of reflection p is the magnitude of the incoming momentum With Eq 32. 31, and taking the average, we I get prad... in the z-direction EB ρI 2 1 ρI 0 I Its magnitude is S 0 0 a 2 2a 2 2 a 3 32. 45: a) E J d) Over a length l, the rate of energy flowing in is SA I 2 lI 2 2al 2 2 a 3 a 2 l lI 2 The thermal power loss is I R I , which exactly equals the flow of A a 2 electromagnetic energy 2 32. 46: B 2 0i q q , and S E dA EA E , so the magnitude of the 2r 0 ... the capacitor Poynting vector is S 2 cBmax 32. 47: The power from the antenna is P IA 4r 2 So 2 0 Bmax 2 0 P 4r 2 c 2 0 (5.50 10 4 W ) 2.42 10 9 T 2 8 4 (2500 m) (3.00 10 m s) dB Bmax 2fBmax 2 (9.50 10 7 Hz) (2.42 10 9 T) 1.44 T s dt d dB D 2 dB (0.180 m) 2 (1.44 T s) A 0.0366 V dt dt 4 dt 4 32. 48: I P 1 2I 0 cE 2 E A 2 0c... ship with the same magnitude of momentum as you gave the flashlight 32. 50: P IA I E P 1 0 cE 2 E A 2 2P A 0 c 2Vi A 0 c 2Vi 2(5.00 105 V) (1000 A) 6.14 10 4 V m (100 m 2 )ε0 (3.00 108 m s) A 0 c And B E 6.14 10 4 V m 2.05 10 4 T 8 c 3.00 10 m s GM S m GM S 4R 3 ρ 4GM S R 3 ρ FG 2 32. 51: a) r2 r 3 3r 2 b) Assuming that the sun’s radiation is intercepted... system 32. 52: a) The momentum transfer is always greatest when reflecting surfaces are used (consider a ball colliding with a wallthe wall exerts a greater force if the ball rebounds rather than sticks) So in solar sailing one would want to use a reflecting sail b) The equation for repulsion comes from balancing the gravitational force and the force from the radiation pressure As seen in Problem 32. 51,... energy because of its acceleration is: (1.6 10 19 C) 2 (1.53 1015 m s 2 ) 2 dE q2a2 1.33 10 23 J s 3 8 3 dt 6 0 c 6 0 (3.0 10 m s) 8 .32 10 5 eV s So the fraction of its energy that it radiates every second is: (dE dt )(1 s) 8 .32 10 5 eV 1.39 10 11 E 6.00 10 6 eV c) Carrying out the same calculations as in part (b), but now for an electron at the same speed and radius... positive, which makes sense since the direction of wave propagation by definition is the direction of energy flow 32. 42: a) S ( x, t ) b) dB di d B dB di A 0 nA 0 n dt dt dt dt dt d B di di So, E dl E 2r 0 nA 0 nr 2 dt dt dt nr di E 0 2 dt 32. 43: a) B 0 ni b) The direction of the Poynting vector is radially inward, since the magnetic field is along . is
J
9.15)s0.30)(m500.0()mV0200.0()sm1000.3()mF1085.8(
228122
rms0
μAtcE
32. 15:
J105.2)m100.2()4()mW100.5()4(
25210 232
av
rIP
32. 16:
a) The average power from the beam is
W104.2)m100.3()mW800.0(
4242.
32. 35: From Eq. (32. 12):
2
2
),(),(
),(
t
txB
t
txB
tx
txE
t
zz
y
But also from Eq. (32. 14):
t
txE
xx
txB
x
y
z
),(
),(
00