Electromagnetic Waves
32-9
32.37.
IDENTIFY
and
SET UP:
Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem.
EXECUTE:
Eq.(32.12):
y
z
E
B
x
t
∂
∂
=−
∂∂
Taking
t
∂
∂
of both sides of this equation gives
2
2
2
y
z
E
B
x
tt
∂
∂
∂
. Eq.(32.14) says
00
.
y
z
E
B
x
t
μ
∂
∂
−=
∂
∂
P
Taking
x
∂
∂
of
both sides of this equation gives
22
1
, so
yy
z
z
EE
B
B
x
tx
x
=
−
∂
∂
∂
∂
P
P
. But
y
y
E
E
x
x
=
∂
∂
(The order in
which the partial derivatives are taken doesn’t change the result.) So
1
z
z
B
B
−
∂
∂
P
and
,
z
z
B
B
x
t
=
P
as was to be shown.
EVALUATE:
Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that
both fields propagate with the same speed
1/
.
v
=
P
32.38.
IDENTIFY:
The average energy density in the electric field is
2
1
,av
0
av
2
()
E
uE
=
P
and the average energy density in
the magnetic field is
2
,av
av
0
1
2
B
uB
=
.
SET UP:
( )
2
1
2
av
cos (
)
kx
t
ω
.
EXECUTE:
max
(,
)
c
o
s
(
)
y
E xt E
k
x
t
.
2
11
m
a
x
cos (
)
Ey
uEE
k
x
t
==
−
PP
and
2
1
,av
0
max
4
E
=
P
.
max
)
c
o
s
(
)
z
Bxt B
k
x
t
, so
2
max
cos (
)
Bz
uBBk
x
t
μμ
−
and
2
,av
max
0
1
4
B
=
.
max
max
E
cB
=
, so
1
,av
0
max
4
E
uc
B
=
P
.
1
c
=
P
, so
2
,av
max
0
1
2
E
=
, which equals
,av
B
u
.
EVALUATE:
Our result allows us to write
2
1
av
,av
0
max
2
2
E
uu
E
P
and
2
av
,av
max
0
1
2
2
B
B
.
32.39.
IDENTIFY:
The intensity of an electromagnetic wave depends on the amplitude of the electric and magnetic
fields. Such a wave exerts a force because it carries energy.
SET UP:
The intensity of the wave is
2
1
av
0
max
2
/
IPA
c
E
P
, and the force is
av
FP
A
=
where
av
/
PI
c
=
.
EXECUTE:
(a)
I = P
av
/
A
= (25,000 W)/[4
π
(575 m)
2
] = 0.00602 W/m
2
(b)
2
1
0m
a
x
2
Ic
E
=
P
, so
max
0
2
I
E
c
=
P
=
2
12
2
2
8
2(0.00602 W/m )
(8.85 10
C /N m )(3.00 10 m/s)
−
×⋅
×
= 2.13 N/C.
B
max
= E
max
/
c =
(2.13 N/C)/(3.00
×
10
8
m/s) = 7.10
×
9
10
−
T
(c)
F =P
av
A =
(
I
/
c
)
A =
(0.00602 W/m
2
)(0.150 m)(0.400 m)/(3.00
×
10
8
m/s) = 1.20
×
12
10
−
N
EVALUATE:
The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth
worrying about!
32.40.
IDENTIFY:
cf
λ
=
.
max
max
E
cB
=
.
2
1
a
x
2
E
=
P
. For a totally absorbing surface the radiation pressure is
.
I
c
SET UP:
The wave speed in air is
8
3.00 10 m/s
c
=×
.
EXECUTE:
(a)
8
9
2
7.81 10 Hz
3.84 10 m
c
f
−
×
= ×
×
(b)
9
max
max
8
1.35 V/m
4.50 10 T
E
B
c
−
=
×
×
(c)
21
2
2
2
8
2
3
2
a
x
(8.854 10
C /N m )(3.00 10 m/s)(1.35 V/m)
2.42 10 W/m
E
−
−
×
⋅×
=
×
P
(d)
32
2
12
8
(2.42 10 W/m )(0.240 m )
(pressure)
1.94 10
N
IA
FA
c
−
−
×
=
=
×
×
EVALUATE:
The intensity depends only on the amplitudes of the electric and magnetic fields and is independent
of the wavelength of the light.