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324
Chapter 32
(b)
22
r
a
d
0.0174 rad/m
361 m
k
ππ
λ
==
=
(c)
36
2
(2 )(830 10 Hz)
5.22 10 rad/s
f
ωπ
π
×
=
×
(d)
Eq.(32.18):
81
1
max
max
(2.998 10 m/s)(4.82 10
T)
0.0144 V/m
Ec
B
−
×
×
=
EVALUATE:
This wave has a very long wavelength; its frequency is in the AM radio braodcast band. The electric
and magnetic fields in the wave are very weak.
32.12.
IDENTIFY:
max
max
.
B
=
SET UP:
The magnetic field of the earth is about
4
10 T.
−
EXECUTE:
3
11
8
3.85 10 V/m
1.28 10
T.
3.00 10 m/s
E
B
c
−
−
×
= ×
×
EVALUATE:
The field is much smaller than the earth’s field.
32.13.
IDENTIFY
and
SET UP:
vf
=
relates frequency and wavelength to the speed of the wave. Use Eq.(32.22) to
calculate
n
and
K
.
EXECUTE:
(a)
8
7
14
2.17 10 m/s
3.81 10 m
5.70 10 Hz
v
f
−
×
×
(b)
8
7
14
2.998 10 m/s
5.26 10 m
c
f
−
×
×
(c)
8
8
1.38
×
=
×
c
n
v
(d)
m
so
(1.38)
1.90
nK
K
KK
n
=≈
=
EVALUATE:
In the material
and
vc
f
<
is the same, so
is less in the material than in air.
<
always, so
n
is
always greater than unity.
32.14.
IDENTIFY:
Apply Eq.(32.21).
max
max
E
cB
=
.
=
. Apply Eq.(32.29) with
m0
K
μμ
=
in place of
0
μ
.
SET UP:
3.64
K
=
.
m
5.18
K
=
EXECUTE:
(a)
8
7
m
(3.00 10 m s)
6.91 10 m s.
(3.64)(5.18)
c
v
KK
×
=
×
(b)
7
6
6.91 10 m s
1.06 10 m.
65.0 Hz
v
f
×
(c)
3
10
max
max
7
7.20 10
V m
1.04 10
T.
6.91 10 m s
E
B
v
−
−
×
=
×
×
(d)
31
0
82
max
max
0
(
7
.
2
01
0 Vm
)
(
1
.
0
41
0 T
)
5.75 10
W m .
(
5
.
1
8
)
EB
I
K
−−
−
××
=
×
EVALUATE:
The wave travels slower in this material than in air.
32.15.
IDENTIFY:
/
IPA
=
.
2
1
0m
a
x
2
Ic
E
=
P
.
max
max
E
cB
=
.
SET UP:
The surface area of a sphere of radius
r
is
2
4
Ar
=
.
12
2
2
0
8.85 10
C /N m
−
=
×⋅
P
.
EXECUTE:
(a)
2
(0.05)(75 W)
330 W/m
4(
3
.
0 m
)
P
I
A
−
=
×
.
(b)
2
max
12
2
2
8
0
(
3
3
0
W
/
m
)
500 V/m
(8.85 10
C /N m )(3.00 10 m/s)
I
E
c
−
=
×
P
.
6
max
max
1.7 10 T
1.7
T
E
B
c
−
×
=
.
EVALUATE:
At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb.
Our calculation approximates the filament as a point source that radiates uniformly in all directions.
32.16.
IDENTIFY
and
SET UP:
The direction of propagation is given by
×
!
!
E
B
.
EXECUTE:
(a)
&& &
&
()
.
=×− =−
S
i
j
k
(b)
&&
& &
.
=×=
−
S ji k
(c)
&&&
&
=− ×− =
() .
S
ki
j
(d)
&& & &
.
=×− =
S
ik
j
EVALUATE:
In each case the directions of
E
!
,
B
!
and the direction of propagation are all mutually perpendicular.
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View Full DocumentElectromagnetic Waves
325
32.17.
IDENTIFY:
max
max
E
cB
=
.
×
!!
E
B
is in the direction of propagation.
SET UP:
8
3.00 10 m/s
c
=×
.
max
4.00 V/m.
E
=
EXECUTE:
8
max
max
1.33 10
T
BE
c
−
==
×
. For
E
!
in the +
x
direction,
×
!
!
E
B
is in the +
z
direction when
B
!
is in
the +
y
direction.
EVALUATE:
E
!
,
B
!
and the direction of propagation are all mutually perpendicular.
32.18.
IDENTIFY:
The intensity of the electromagnetic wave is given by Eq.(32.29):
22
1
0m
a
x 0r
m
s
2
.
Ic
E
c
E
PP
The total
energy passing through a window of area
A
during a time
t
is
IAt
.
SET UP:
12
0
8.85 10
F/m
−
P
EXECUTE:
21
2
8
2
2
0r
m
s
Energy
(8.85 10
F m)(3.00 10 m s)(0.0200 V m) (0.500 m )(30.0 s) 15.9 J
μ
−
×
×
=
cE
At
P
EVALUATE:
The intensity is proportional to the square of the electric field amplitude.
32.19.
IDENTIFY
and
SET UP:
Use Eq.(32.29) to calculate
I
, Eq.(32.18) to calculate
max
,
B
and use
2
av
/4
IP
r
π
=
to
calculate
av
.
P
(a) EXECUTE:
25
2
1
a
x
m
a
x
2
;
0.090 V/m, so
1.1 10 W/m
IE
E
I
−
=
×
P
(b)
10
max
max
max
max
so
/
3.0 10
T
Ec
B
c
−
=
×
(c)
2
3
2
av
(4
)
(1.075 10 W/m )(4 )(2.5 10 m)
840 W
PIr
ππ
−
×
×=
(d) EVALUATE:
The calculation in part (c) assumes that the transmitter emits uniformly in all directions.
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 Spring '06
 Buchler
 Physics

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