32.33: Using a Gaussian surface such that the front surface is ahead of the wave front noelectric or magnetic fields and the back face is behind the wave front as shown at right, we have
Trang 132.1: a) 1.28s.
s m 10 00 3
m 10 84 3
8
8
c
d t
b) Light travel time is:
s 10 72 2 ) hour 1 (
) s 3600 ( ) day 1 (
) hours 24 ( ) year 1 (
) days 365 ( ) years 61 8 ( years
61
km
10 8.16 m
10 8.16 s) 10 (2.72 s) m 10 0 3
ct
d
32.2: d ct (3.0108 m s)(6.010 7 s)180m
32.3: B ) maxcos( ˆj maxcos 2 ˆj
c
z f B
ωt) kz B
s) m 10 00 3 ( Hz) 10 10 6 ( 2 cos ) T 10 80 5 ( )
B(z ,t)(5.8010 4 T)cos((1.28107m 1)z (3.831015 rad s)t ˆj
E(z,t)(B y (z,t) j)(c k)
E(z,t)(1.74105 V m)cos((1.28107 m 1)z(3.831015 rad s)t) i.ˆ
m 10 35 4
s m 10 00 3 λ
14 7
8
f
s m 10 00 3
m V 10 70
8
3 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 SEB.Thus:
.ˆ ) ) s rad 10 34 4 ( ) m 10 45 1 cos((
) m V 10 70 2 ( ) (
ˆ s m 10 00 3 ) Hz 10 90 6 ( 2 cos V/m) 10
70 2 ( ) , (
ˆ 2
cos ˆ
( cos )
,
(
15 1
7 3
8 14
3
max max
i E
i E
i i
E
t z
z,t
z t
π t
z
t c
z πf E
ωt) kz E
t
z
AndB( , ) ( , ) ˆj (9.00 10 12 T)cos((1.45 107 m 1)z (4.34 1015rad s)t ˆj
c
t z E t
Trang 232.5: a) direction.y
) s rad 10 65 2 (
) s m 10 00 3 ( 2 2 λ λ
2
12
8
ω
πc πc
πf
ω
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 (SEB) So:
B( , ) ( , )iˆ 0 sin( iˆ 0 sin( y t iˆ
c c
E t
ky c
E c
t y E t
i
) s m 10 00 3 (
) s rad 10 65 2 ( sin s m 10 00 3
m V 10 10 3 )
,
8
12 8
5
ˆ ) s rad 10 65 2 ( ) m 10 83 8 sin((
) T 10 03 1 ( )
(
32.6: a) direction.x
2
) s m 10 0 3 ( m rad 10 38 1 ( 2
2
π π
kc f c
f π 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 (SEB) So:
ˆ ) ) s rad 10 14 4 ( ) m rad 10 38 1 sin((
) m V 48 2 ( ) , (
ˆ ) s rad 10 14 4 ( ) m rad 10 38 1 sin((
)) T 10 25 3 ( ( ) , (
ˆ 2 sin(
ˆ ) , ( )
,
(
12 4
12 4
9 0
k E
k E
k k
E
t x
t y
t x
c t
x
f kx cB
t x cB
t
x
Hz 10 30 8
s m 10 00 3
8
f c
m 361
2 λ
k
c) ω2f 2π(8.30105 Hz)5.21106 rad s
(3.00 108m s)(4.82 10 11 T) 0.0145V m
max
E
s m 10 00 3
m V 10 85
8
3 max
max
c
E B
T 10 5
T 10 28
5 11 earth
B
B
and thus Bmaxis much weaker than Bearth
Trang 3B E B
cB K
K
B B
vB
0
) 23 1 ( ) 74 1 (
) T 10 80 3 ( ) s m 10 00 3
) 18 5 ( ) 64 3 (
s) m 10 00 3
B
E K K
c v
Hz 0 65
s m 10 91 6
f
v
s m 10 91 6
m V 10 20
7
3
v
E
B
) 18 5 ( 2
) T 10 04 1 )(
m V 10 20 7 ( 2
2 8
0
10 3
0
B
K
EB
I
Hz 10 70 5
s m 10 17 2
14
8
f v
Hz 10 70 5
s m 10 00 3
14
8 0
f
c
s m 10 17 2
s m 10 00 3
8
8
v
c
n
d) 2 (1.38)2 1.90
2
2
v
c K K
c
E
32.12: a) v fλ(3.80107 Hz)(6.15m)2.34108 m s
) s m 10 34 2 (
) s m 10 00 3 (
2 8
2 8 2
2
v
c
K E
max
2 max
2
max max
max
E
c) (4 2) (1.075 10 5 W m2 (4 (2.5 103 m)2 840W
P
d) Calculation in part (c) assumes that the transmitter emits uniformly in all directions
Trang 432.14: The intensity of the electromagnetic wave is given by Eqn 32.29:
2 rms 0
2
max
0
2
I Thus the total energy passing through a window of area Aduring
a time t is
J 9 15 ) s 0 30 )(
m 500 0 ( ) m V 0200 0 ( ) s m 10 00 3 ( ) m F 10 85 8
2
rms
32.15: (4 2) (5.0 103 W m2 (4 (2.0 1010m)2 2.5 1025 J
P
32.16: a) The average power from the beam is
W 10 4 2 ) m 10 0 3 ( m W 800 0
IA P
b) We have, using Eq 32.29, 2
rms 0
2 max 0 2
m V 4 17 s) m 10 00 3 )(
m F 10 85 8 (
m W 800 0
8 12
2 0
c
I E
rad 2.7010 W m
cp
I
Then P av I(4r2)(2.70103 W m2)(4)(5.0m)2 8.5105 W
m 354 0
s m 10 00 3 λ
8
8
c
f
s m 10 00 3
m V 0540
8
max max
c
E
B
2
) T 10 80 1 ( ) m V 0540 0 ( 2
2 6
0
10 0
EB S
I av
max 2
0
2 max
2 )
4 (
Pc E
r c
E A S
P av
) m 00 5 ( 2
) s m 10 00 3 ( ) W 0 60 (
2
0 8
E
s m 10 00 3
m V 0
8
max max
c E B
Trang 532.20: a) The electric field is in the -direction, and the magnetic filed is in the z y -direction, so Sˆ Eˆ Bˆ (j)kˆi.ˆ That is, the Poynting vector is in the -x
direction
b) ( , ) ( , ) ( , ) cos( )
0
max max 0
t kx B
E t
x B t x E t
x
(1 cos(2( )))
2 0
max
E
But over one period, the cosine function averages to zero, so we have:
2
|
0
max max
B E
S av
s) m 10 0 3 (
m W
2 8
2
c
S dV
b) The momentum flow rate 2.6 10 Pa
s m 10 0 3
m W 780
8
2
c
S dt
dp A
av
s m 10 0 3
m W 2500
8
2 rad
c
S dt
dp A
atm Pa 10 013 1
Pa 10 33
5
6 rad
p
s m 10 0 3
) m W 2500 ( 2 2
8
2 rad
c
S dt
dp A
5
5
atm Pa 10 1.013
Pa 10 67
momentum vector totally reverses direction upon reflection Thus the change in
momentum is twice the original momentum
s) m 10 0 3 (
m W
2 8
2
c
S dV
c
E Ec E
E S
0
0 0 0 0
0 0
0 2
0
0 2
0 0
2 0 0
2
0
cE c
E
Trang 632.24: Recall that S E B ,so:
a) Sˆ iˆ(j)kˆ
b) Sˆ ˆjiˆkˆ
c) Sˆ (kˆ)(iˆ) ˆ.j
d) Sˆ iˆ(kˆ) j.ˆ
max max
B
B
E is in the direction of propagation For E
in the + x -direction, E B
is in the + z
-direction when B is in the + y-direction
) Hz 10 0 75 ( 2
s m 10 00 3 2
2
λ
6
8
f
c x
b) The distance between the electric and magnetic nodal planes is one-quarter of a
2
m 00 2 2 4
Hz) 10 20 1 ( 4
s m 10 10 2 4
4
10
8
f v
b) The distance between the electric and magnetic antinodes is one-quarter of a
Hz) 10 (1.20 4
s m 10 2.10 4
4
10
8
f v
c) The distance between the electric and magnetic nodes is also one-quarter of a wavelength
m
10 38 4 Hz) 10 4(1.20
m/s 10 2.10
4
4
3 10
8
f
v
λ
) Hz 10 50 7 ( 2
s m 10 00 3 2
2
λ
8
8
f
c
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) λ 2 2(3.55mm) 7.10mm
2
b) x E x B 3.55mm
c) v fλ(2.201010 Hz (7.1010 3 m)1.56108 m s
Trang 732.30: a) ( , ) 2 ( 2 maxsin sin ) ( 2 maxcos sin )
2 2
2
ωt kx kE
x ωt kx E
x x
t x
E y
2 0 0 max
2
2 max
2 2
2
t
t x E t
kx E
c t kx E
k x
t x
Similarly: ( , ) 2 ( 2 maxcos cos ) ( 2 maxsin cos )
2 2
2
t kx kB
x t kx B
x x
t x
) , ( cos
cos 2
cos cos 2
) , (
2
2 0 0 max
2
2 max
2 2
2
t
t x B t
kx B
c t kx B
k x
t x
x x
t
x
E y
) 2 cos sin sin
sin 2
( )
,
(
max
) , ( )
cos cos 2
( )
,
(
sin cos 2
sin cos 2
sin cos 2
)
,
(
max
max
max max
t
t x B t
kx B
t x
t
x
E
ωt kx B
t kx c
E t
kx E
c x
t
x
E
z y
y
x x
t x
B z( , ) (2 maxcos cos )2 maxsin cos
c t kx B
c x
t x
B z( , ) 2 maxsin cos 2 2 maxsin cos
, ( )
sin sin 2
( cos
sin 2
)
,
(
0 0 max
0 0 max
0 0
t
t x E t
kx E
t t
kx E
x
t
x
Hz 10 6.50
s m 10 3.00
21
8
f c
Hz 10 75 5
s m 10 00 3 λ
: light
14
8
f c
m 5000
s m 10 3.0 λ
4
8
c
f
m 5.0
s m 10 3.0 λ
7
8
c
f
m 10 5.0
s m 10 3.0 λ
13 6
8
f
m 10 5.0
s m 10 3.0 λ
16 9
8
f
Trang 832.33: 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:
0
encl
x
ε
Q E
E
Bd AB x A0B x 0
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 E Emaxˆjsin(kxt)and BBmaxkˆsin(kxt),with
Then Eq (32.12) implies:
0 )
cos(
)
kx kE
x t
B x
λ
λ / 2
2
max max
max max
max max
k E
B
Similarly for Eq.(32.14)
0 0 maxcos( ) 0 0 maxcos( ) 0
t
E x
/ 2
2
max max
2 max 2
max 0 0 max max
0 0
c
E c
fλ E
λ c
f E
k B
E
Trang 932.35: From Eq (32.12): 2
) , ( )
, (
t
t x B t
t x B t x
t x E t
z z
y
t
t x E x
x
t x B x
y
0
0
2 0
0
) , (
t
t x
B z
2 0 0 2
2
t
t x B x
t x
2
1 2
1 )
cos(
) ,
max 0
2 0
E t
x
c
E c
0
2 2
max 0
2 max
2 0
2 ) cos(
2
1 ) cos(
32.37: a) The energy incident on the mirror is Pt IAt cE2At
0
2
1
J 10 5.20 s) (1.00 ) m 10 00 5 ( ) m V 028 0 ( s m 10 00 3 ( 2
b) The radiation pressure 2 (0.0280V m)2 6.94 10 15 Pa
0
2 0 rad
c
I p
rad
I
W
10 1.34 m)
(3.20 Pa) 10 (6.94 s) m 10 00 3 (
Trang 1032.38: a) 7.81 10 Hz.
m 0384 0
s m 10 00 3 λ
9
8
c
f
s m 10 00 3
m V
8
max max
c
E
B
c) (3.00 10 m s)(1.35V m) 2.42 10 W m
2
1 2
0
2 max 0
I
s) m 10 00 3 ( 2
) m (0.240 T) 10 (4.50 m) V 35 1 ( 2
12 8
0
2 9
0
c
EBA c
IA pA
F
m) 10 (2.50
W) 10 4(3.20
2 3
3
π D
P A
P I
) s m 10 (3.00
) m W 2(652 2
2
1
8 0
2 0
2
c
I E
cE I
s m 10 3.00
m V
8
c
E
B
4
1 4
0
2 max 0
of
2
1
since we are averaging
c) In one meter of the laser beam, the total energy is:
4 2
) ( 2
tot
J
10 1.07 m)/4 (1.00 m) 10 50 2 ( ) m J 10 09 1 (
tot
Trang 1132.40: a) The change in the momentum vector determines prad If W is the fraction absorbed, P PoutPin (1W)p(p)(2W)p.Here, (1W)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 getprad (2W)C I Be careful not to confuse p, the momentum of the incoming wave,
with prad, the radiation pressure
b) (i) totally absorbing
C
I p
W 1so rad
(ii) totally reflecting
C p
W 0so rad 2
These are just equations 32.32 and 32.33
10 00 3
) m W 10 40 1 ( 9 0 2 ( W/m
10 40 1 ,
9
s m 8
2 3 rad
2
W
Pa 10 87 8 10
00 3
) m W 10 40 1 ( 1 0 2 ( m
W 10 40 1 ,
1
s m 8
2 2 rad
2
W
32.41: a) At the sun’s surface:
Pa 21 0 s m 10 00 3
m W 10 4 6
m W 10 4 6 m) 10 96 6 ( 4
W 10 9 3 4
8
2 7
rad
2 7
2 8
26 2
c
I p
R
P A
P I IA
P
Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the radiation pressure:prad(Rsun/2)0.85Pa
At the top of the earth’s atmosphere, the measured sunlight intensity is
2
m
W
1400
,
Pa
10
5 6 which is about 100,000 times less than the values above
b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and halfway out of the sun the gas pressure is believed to be about 6 1013
times greater than the radiation, pressure Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure
Trang 1232.42: a) (1 cos2( ))ˆ ( , ) 0 cos2( ) 1,
2 ) , (
0
max
t
S
which never happens So the Poynting vector is always positive, which makes sense since the direction of wave propagation by definition is the direction of energy flow
b)
dt
di nA A
dt
dB dt
d dt
di n dt
dB ni
dt
di r n dt
di nA r
E dt
d
0 0
l
2
0
dt
di nr
E
b) The direction of the Poynting vector is radially inward, since the magnetic field is along the solenoid’s axis and the electric filed is circumferential It’s magnitude
2
2 0
di ri n
EB
2 )
( 2
2
) ( 2
2 2 2 0 2 2
2 0 0
2 0 0
a ul lA u U i n ni
B
2
2 2 0
2 2 2 0
2
ila n i
la i n i
U Li
Li
U and so the rate of energy increase due to the increasing current is given by 2 2
0
dt
di ila n dt
di Li
d) The in-flow of electromagnetic energy through a cylindrical surface located at the
2
0
2 0
dt
di ila n al
dt
di ai n πal S
S e) The values from parts (c) and (d) are identical for the flow of energy, and hence we can consider the energy stored in a current carrying solenoid as having entered through its cylindrical walls while the current was attaining its steady-state value
Trang 1332.44: 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:
t kx E
E
u 4 2 sin2 sin
max 0
2
2
1 4 sin sin
and 2
1 4 cos cos
,
ωt t
t
2
k
x
And for ,sin 0,cos 0 ˆ ˆ ˆ ˆ ˆ ˆ
2 kx kx S EB j k i
k
x k
2
1 4
3 sin sin
and 2
1 4
3 cos cos
, 4
3
t t
For 0 ,sin 0,cos 0 ˆ ˆ ˆ ˆ ˆ ˆ
k
x
k
x k
c) the plots from part (a) can be interpreted as two waves passing through each other
in opposite directions, adding constructively at certain times, and destructively at others
Trang 1432.45: a) 2
a
I A
I J E
, in the direction of the current
2
0 0
a
I B I d
l
B counterclockwise when looking into the current
c) The direction of the Poynting vector Sˆ Eˆ Bˆ kˆˆρˆ, where we have used
cylindrical coordinates, with the current in the z-direction.
2 0
2 0
1
a
ρI a
I a
ρI EB
S
d) Over a length l, the rate of energy flowing in is 2
2 3
2
2
a
lI al a
I SA
The thermal power loss is 2 ,
2 2
2
a
lI A
l I R I
which exactly equals the flow of electromagnetic energy
0 0
0
r
q E
q EA d
r
i
so the magnitude of the
2
0 3 2 0
dq r
q r
qi EB
S
Now, the rate of energy flow into the region between the plates is:
2 2
1 )
( 2
1 )
2
0
2 2 0
2
dU C
q dt
d q A
l dt
d dt
q d r
l dt
dq r
lq rl
S d
A
S
This is just rate of increase in electrostatic energy U stored in the capacitor
32.47: The power from the antenna is 4
2
2 0
2
cB IA
T 10 42 2 ) s m 10 00 3 ( ) m 2500 ( 4
) W 10 50 5 ( 2 4
8 2
4 0
2
0 max
c r
P B
max
dt
dB
V
0366 0 4
) s T 44 1 ( ) m 180 0 ( 4
2 2
dt
dB D dt
dB A dt
d
) s m 10 00 3 (
) m 36 W 10 80 2 ( 2 2
2
1
8 0
2 3
0
2
c
I E
cE A
P
I