Electromagnetic Waves
329
32.37.
I
DENTIFY
and
S
ET
U
P
:
Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem.
E
XECUTE
:
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 t
t
∂
∂
= −
∂ ∂
∂
. Eq.(32.14) says
0
0
.
y
z
E
B
x
t
μ
∂
∂
−
=
∂
∂
P
Taking
x
∂
∂
of
both sides of this equation gives
2
2
2
2
0
0
2
2
0
0
1
, so
y
y
z
z
E
E
B
B
x
t x
t x
x
μ
μ
∂
∂
∂
∂
−
=
= −
∂
∂ ∂
∂ ∂
∂
P
P
. But
2
2
y
y
E
E
x t
t x
∂
∂
=
∂ ∂
∂ ∂
(The order in
which the partial derivatives are taken doesn’t change the result.) So
2
2
2
2
0
0
1
z
z
B
B
t
x
μ
∂
∂
−
= −
∂
∂
P
and
2
2
0
0
2
2
,
z
z
B
B
x
t
μ
∂
∂
=
∂
∂
P
as was to be shown.
E
VALUATE
:
Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that
both fields propagate with the same speed
0
0
1/
.
v
μ
=
P
32.38.
I
DENTIFY
:
The average energy density in the electric field is
2
1
,av
0
av
2
(
)
E
u
E
=
P
and the average energy density in
the magnetic field is
2
,av
av
0
1
(
)
2
B
u
B
μ
=
.
S
ET
U
P
:
(
)
2
1
2
av
cos (
)
kx
t
ω
−
=
.
E
XECUTE
:
max
( , )
cos(
)
y
E
x t
E
kx
t
ω
=
−
.
2
2
2
1
1
0
0
max
2
2
cos (
)
E
y
u
E
E
kx
t
ω
=
=
−
P
P
and
2
1
,av
0
max
4
E
u
E
=
P
.
max
( , )
cos(
)
z
B
x t
B
kx
t
ω
=
−
, so
2
2
2
max
0
0
1
1
cos (
)
2
2
B
z
u
B
B
kx
t
ω
μ
μ
=
=
−
and
2
,av
max
0
1
4
B
u
B
μ
=
.
max
max
E
cB
=
, so
2
2
1
,av
0
max
4
E
u
c B
=
P
.
0
0
1
c
μ
=
P
, so
2
,av
max
0
1
2
E
u
B
μ
=
, which equals
,av
B
u
.
E
VALUATE
:
Our result allows us to write
2
1
av
,av
0
max
2
2
E
u
u
E
=
=
P
and
2
av
,av
max
0
1
2
2
B
u
u
B
μ
=
=
.
32.39.
I
DENTIFY
:
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.
S
ET
U
P
:
The intensity of the wave is
2
1
av
0
max
2
/
I
P
A
cE
=
=
P
, and the force is
av
F
P A
=
where
av
/
P
I c
=
.
E
XECUTE
:
(a)
I = P
av
/
A
= (25,000 W)/[4
π
(575 m)
2
] = 0.00602 W/m
2
(b)
2
1
0
max
2
I
cE
=
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
E
VALUATE
:
The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth
worrying about!
32.40.
I
DENTIFY
:
c
f
λ
=
.
max
max
E
cB
=
.
2
1
0
max
2
I
cE
=
P
. For a totally absorbing surface the radiation pressure is
.
I
c
S
ET
U
P
:
The wave speed in air is
8
3.00
10
m/s
c
=
×
.
E
XECUTE
:
(a)
8
9
2
3.00
10
m/s
7.81
10
Hz
3.84
10
m
c
f
λ
−
×
=
=
=
×
×
(b)
9
max
max
8
1.35 V/m
4.50
10
T
3.00
10
m/s
E
B
c
−
=
=
=
×
×
(c)
2
12
2
2
8
2
3
2
1
1
0
max
2
2
(8.854
10
C
/N
m
)(3.00
10
m/s)(1.35 V/m)
2.42
10
W/m
I
cE
−
−
=
=
×
⋅
×
=
×
P
(d)
3
2
2
12
8
(2.42
10
W/m
)(0.240 m
)
(pressure)
1.94
10
N
3.00
10
m/s
IA
F
A
c
−
−
×
=
=
=
=
×
×
E
VALUATE
:
The intensity depends only on the amplitudes of the electric and magnetic fields and is independent
of the wavelength of the light.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3210
Chapter 32
32.41.
(a)
I
DENTIFY
and
S
ET
U
P
:
Calculate
I
and then use Eq.(32.29) to calculate
max
E
and Eq.(32.18) to calculate
max
.
B
E
XECUTE
:
The intensity is power per unit area:
3
2
3
2
3.20
10
W
652 W/m
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Energy, Magnetic Field, Emax

Click to edit the document details