Download CH 31 solutions to assigned problems

CHAPTER 31: Maxwell’s Equations and Electromagnetic Waves
Solutions to Assigned Problems
1.
The electric field between the plates is given by E 
E
2.
V
d

dE
dt

 dV
d dt

d
, where d is the distance between the plates.
V m

120 V s   1.1  105

s
 0.0011m 

1
The displacement current is shown in section 31-1 to be I D   0 A
dE
.
dt
dE
V 
2 
8
ID  0 A
 8.85  1012 C2 N  m 2  0.058 m   2.0  106
  6.0  10 A
dt
m s


8.
V

Use Eq. 31-11 with v  c.
E0
B0
 c  B0 
E0
c

0.57  104 V m
3.00  108 m s
 1.9  1013 T
11. (a) If we write the argument of the cosine function as kz + t = k(z + ct), we see that the wave is
traveling in the – z direction, or kˆ .
(b) E and B are perpendicular to each other and to the direction of propagation. At the origin, the
electric field is pointing in the positive x direction. Since E  B must point in the negative z
direction, B must point in the the – y direction, or jˆ . The magnitude of the magnetic field is
found from Eq. 31-11 as B0  E0 c .
18. The length of the pulse is d  ct. Use this to find the number of wavelengths in a pulse.
8
12
ct   3.00  10 m s  38  10 s 

N

 10734  11,000 wavelengths

1062  109 m 
If the pulse is to be only one wavelength long, then its time duration is the period of the wave, which
is the reciprocal of the wavelength.
9
1  1062  10 m 
T  
 3.54  1015 s
8
f c  3.00  10 m s 
23. The energy per unit area per unit time is given by the magnitude of the Poynting vector. Let U
represent the energy that crosses area A in a time T .
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
303
Physics for Scientists & Engineers with Modern Physics, 4th Edition
S
2
cBrms
t 
0

0 U
2
AcBrms
U
At
Instructor Solutions Manual

 4  10 T m A   335J 
m  3.00  10 m s  22.5  10 T 
7

1.00  10
4
2
9
8
2
 0.194 W m 2
 2.77  107 s  321 days
27. From Eq. 31-16b, the instantaneous energy density is u   0 E 2 . From Eq. 31-17, we see that this
instantaneous energy density is also given by S c . The time-averaged value is therefore S c .
Multiply that times the volume to get the energy.
S
1350W m2
U  uV  V 
1.00m3   4.50  106 J

8
c
3.00  10 m s
29. The radiation from the Sun has the same intensity in all directions, so the rate at which it reaches the
Earth is the rate at which it passes through a sphere centered at the Sun with a radius equal to the
Earth’s orbit radius. The 1350W m2 is the intensity, or the magnitude of the Poynting vector.
2
P
S
 P  SA  4 R2 S  4 1.496  1011 m  1350W m2   3.80  1026 W
A
32. (a) Example 31-1 refers back to Example 21-13 and
d
Figure 21-31. In that figure, and the figure included
here, the electric field between the plates is to the right.
The magnetic field is shown as counterclockwise
r0
circles. Take any point between the capacitor plates,
B
I
I
and find the direction of E  B. For instance, at the
E
top of the circle shown in Figure 31-4, E is toward the
viewer, and B is to the left. The cross product E  B
points down, directly to the line connecting the center
of the plates. Or take the rightmost point on the circle.
E is again toward the viewer, and B is upwards. The cross product E  B points to the left,
again directly to the line connecting the center of the plates. In cylindrical coordinates,
ˆ . The cross product kˆ  φˆ  rˆ.
E  E kˆ and B  Bφ
(b) We evaluate the Poynting vector, and then integrate it over the curved cylindrical surface
dE
between the capacitor plates. The magnetic field (from Example 31-1) is B  12 0 0 r0
,
dt
1
dE
evaluated at r  r0 . E and B are perpendicular to each other, so S  E  B  12  0 r0 E
,
0
dt
inward. In calculating
 S dA for energy flow into the capacitor volume, note that both S
and
dA point inward, and that S is constant over the curved surface of the cylindrical volume.
dE
dE
2
 S dA   SdA  S  dA  SA  S 2 r0d  12  0r0 E dt 2 r0d   0d r0 E dt
The amount of energy stored in the capacitor is the energy density times the volume of the
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
304
Maxwell’s Equations and Electromagnetic Waves
Chapter 31
capacitor. The energy density is given by Eq. 24-6, u  12  0 E 2 , and the energy stored is the
energy density times the volume of the capacitor. Take the derivative of the energy stored with
respect to time.
dU
dE
U  u  Volume   12  0 E 2 r02d 
  0 E r02d
dt
dt
dU
We see that  S dA 
.
dt
49. The light has the same intensity in all directions, so use a spherical geometry centered on the source
to find the value of the Poynting vector. Then use Eq. 31-19a to find the magnitude of the electric
field, and Eq. 31-11 with v  c to find the magnitude of the magnetic field.
P
P
S  0  0 2  12 c 0 E02 
A 4 r
E0 
 75W 
P0

 33.53V m
2
2
2 r c 0
2  2.00 m   3.00  108 m s 8.85  10 12 C2 n m 2 
 34 V m
B0 
 33.53V m   1.1  107 T
E0

c  3.00  108 m s 
63. (a) To show that E and B are perpendicular, calculate their dot product.
E B   E0 sin  kx  t  ˆj  E0 cos  kx  t  kˆ   B0 cos  kx  t  ˆj  B0 sin  kx  t  kˆ 
 E0 sin  kx  t  B0 cos  kx  t   E0 cos  kx  t  B0 sin  kx  t   0
Since E B  0, E and B are perpendicular to each other at all times.
(b) The wave moves in the direction of the Poynting vector, which is given by S 
1
0
E  B.
ˆi
ˆj
kˆ
S  EB 
0 E0 sin  kx  t  E0 cos  kx  t 
0
0
0 B0 cos  kx  t   B0 sin  kx  t 
1
1
1
1
ˆi   E B sin 2  kx  t   E B cos2  kx  t   ˆj  0   kˆ  0    E B ˆi
0 0

0  0 0
0 0 0
We see that the Poynting vector points in the negative x direction, and so the wave moves in the
negative x direction, which is perpendicular to both E and B .
(c) We find the magnitude of the electric field vector and the magnetic field vector.


E  E   E0 sin  kx  t    E0 cos  kx  t 
2
  E02 sin 2  kx  t   E02 cos2  kx  t 
1/ 2

 E0
B  B   B0 cos  kx  t    B0 sin  kx  t 
2
  B02 cos2  kx  t   B02 sin 2  kx  t 
1/ 2

2 1/ 2

2 1/ 2
 B0
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition
Instructor Solutions Manual
(d) At x = 0 and t = 0, E  E0kˆ and B  B0ˆj. See the figure. The x axis is
coming out of the page toward the reader. As time increases, the
component of the electric field in the z direction electric field begins to
get smaller and the component in the negative y direction begins to get
larger. At the same time, the component of the magnetic field in the y
direction begins to get smaller, and the component in the z direction
begins to get larger. The net effect is that both vectors rotate counterclockwise.
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
306