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321
E
LECTROMAGNETIC
W
AVES
32.1.
IDENTIFY:
Since the speed is constant, distance
.
xc
t
=
SET UP:
The speed of light is
8
3.00 10 m/s
c
=×
.
7
1 yr
3.156 10 s.
EXECUTE:
(a)
8
8
3.84 10 m
1.28 s
x
t
c
×
==
=
×
(b)
87
1
6
1
3
(3.00 10 m/s)(8.61 yr)(3.156 10 s/yr)
8.15 10 m
8.15 10 km
t
×
×
=
×
=
×
EVALUATE:
The speed of light is very great. The distance between stars is very large compared to terrestrial
distances.
32.2.
IDENTIFY:
Since the speed is constant the difference in distance is
.
ct
Δ
SET UP:
The speed of electromagnetic waves in air is
8
3.00 10 m/s.
c
EXECUTE:
A total time difference of 0.60 s
μ
corresponds to a difference in distance of
86
(3.00 10 m/s)(0.60 10 s) 180 m.
−
Δ=
×
×
=
EVALUATE:
The time delay doesn’t depend on the distance from the transmitter to the receiver, it just depends on
the difference in the length of the two paths.
32.3.
IDENTIFY:
Apply
.
cf
λ
=
SET UP:
8
c
EXECUTE:
(a)
8
4
3.0 10 m s
6.0 10 Hz.
5000 m
c
f
×
= ×
(b)
8
7
3.0 10 m s
5.0 m
c
f
×
(c)
8
13
6
3.0 10 m s
5.0 10
m
c
f
−
×
×
(d)
8
16
9
3.0 10 m s
6.0 10 Hz.
5.0 10 m
c
f
−
×
×
EVALUATE:
f
increases when
decreases.
32.4.
IDENTIFY:
=
and
2
.
k
π
=
SET UP:
8
c
.
EXECUTE:
(a)
c
f
=
. UVA:
14
7.50 10 Hz
×
to
14
9.38 10 Hz
×
. UVB:
14
×
to
15
1.07 10 Hz
×
.
(b)
2
k
=
. UVA:
7
1.57 10 rad/m
×
to
7
1.96 10 rad/m
×
. UVB:
7
×
to
7
2.24 10 rad/m
×
.
EVALUATE:
Larger
corresponds to smaller
f
and
k
.
32.5.
IDENTIFY:
=
.
max
max
Ec
B
=
.
2 /
k
πλ
=
.
2.
f
ωπ
=
SET UP:
Since the wave is traveling in empty space, its wave speed is
8
c
.
EXECUTE:
(a)
8
14
9
6.94 10 Hz
432 10 m
c
f
−
×
=
×
×
(b)
max
max
(
3
.
0
01
0
m
/
s
)
(
1
.
2
51
0
T
) 3
7
5
V
/m
B
−
==×
×
=
32
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Chapter 32
(c)
7
9
22
r
a
d
1.45 10 rad/m
432 10 m
k
ππ
λ
−
==
=
×
×
.
14
15
(
2
r
a
d
)
(
6
.
9
41
0 H
z
) 4
.
3
61
0
r
a
d
/
s
ωπ
=×
=
×
.
71
5
max
cos(
)
(375 V/m)cos([1.45 10 rad/m]
[4.36 10 rad/s] )
EE
k
x t
x
t
ω
=−
=
×
−
×
67
1
5
max
cos(
)
(1.25 10 T)cos([1.45 10 rad/m]
[4.36 10
rad/s] )
BB
k
x
t
−
=
×
×
−
×
EVALUATE:
The cos(
)
kx
t
−
factor is common to both the electric and magnetic field expressions, since these
two fields are in phase.
32.6.
IDENTIFY:
cf
=
.
max
max
Ec
B
=
. Apply Eqs.(32.17) and (32.19).
SET UP:
The speed of the wave is
8
3.00 10 m/s.
c
EXECUTE:
(a)
8
14
9
3.00 10 m/s
6.90 10 Hz
435 10 m
c
f
−
×
=
×
×
(b)
3
12
max
max
8
2.70 10 V/m
9.00 10
T
E
B
c
−
−
×
=
×
×
(c)
7
2
1.44 10 rad/m
k
π
×
.
15
24
.
3
4
1
0
r
a
d
/
s
f
×
. If
max
ˆ
(,
)
c
o
s
(
)
=+
G
zt
E
k
z
t
Ei
, then
max
ˆ
)
c
o
s
(
)
+
G
B
k
z
t
Bj
, so that
×
E
B
GG
will be in the
ˆ
−
k
direction.
371
5
ˆ
( , )
(2.70 10 V/m)cos([1.44 10 rad/s)
[4.34 10 rad/s] )
−
×
+
×
G
z
t
and
12
7
15
ˆ
( , )
(9.00 10
T)cos([1.44 10 rad/s)
−
×
×
+
×
G
z
t
.
EVALUATE:
The directions of
E
G
and
B
G
and of the propagation of the wave are all mutually perpendicular. The
argument of the cosine is
kz
t
+
since the wave is traveling in the
direction
z
−
. Waves for visible light have very
high frequencies.
32.7.
IDENTIFY
and
SET UP:
The equations are of the form of Eqs.(32.17), with
x
replaced by
z
.
B
G
is along the
y
axis;
deduce the direction of
.
E
G
EXECUTE:
14
15
2
2(
6
.
1
01
z
) 3
.
8
31
r
a
d
/
s
f
×
=
×
15
7
8
2
2
3.83 10 rad/s
1.28 10 rad/m
f
k
cc
ππ ω
×
=
=
=
×
×
4
max
5.80 10 T
B
−
84
5
max
max
(3.00 10 m/s)(5.80 10 T) 1.74 10 V/m
B
−
==×
×
=
×
B
G
is along the
y
axis.
E
B
G
G
×
is in the direction of propagation (the +
z
direction). From this we can deduce the
direction of
,
E
G
as shown in Figure 32.7.
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 Spring '11
 Shaefer

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