ch33 - Chapter 33 1. Since , we find f is equal to 8 9 c c...

1271
Chapter 33
1. Since     , we find  f is equal to
8 9
9
2 9 2
(3.0 10 m/s)(0.0100 10 m)
7.49 10 Hz.
(632.8 10 m)
c c 
 


  
 
    
 

 
2. (a) The frequency of the radiation is
f
c
 

 
 


30 10
10 10 6 4 10
4 7 10
8
5 6
3
.
( . )( .
.
m / s
m)
Hz.
(b) The period of the radiation is
T
f
 

 

1 1
4 7 10
212 3 32
3 .
min
Hz
s s.
3. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm.
(b) Similarly, the larger wavelength is approximately 610 nm.
(c) From Fig. 33-2 the wavelength at which the eye is most sensitive is about 555 nm.
(d) Using the result in (c), we have
8
14
3.00 10 m/s
5.41 10 Hz
555 nm
c
f

   

.
(e) The period is T = 1/ f = (5.41  10
14
Hz)
–1
= 1.85  10
–15
s.
4. In air, light travels at roughly c = 3.0  10
8
m/s. Therefore, for t = 1.0 ns, we have a
distance of
d ct     

( . . 30 10 0 30
8 9
m / s)(1.0 10 s) m.
5. If f is the frequency and  is the wavelength of an electromagnetic wave, then f  = c .
The frequency is the same as the frequency of oscillation of the current in the LC circuit
of the generator. That is, f LC  1 2 /  , where C is the capacitance and L is the
inductance. Thus

CHAPTER 33 1272

 2 LC
c  .
The solution for L is
 
  
2
9
2
21
2
2 2
2 12 8
550 10 m
5.00 10 H.
4
4 17 10 F 2.998 10 m/s
L
Cc







   
 
This is exceedingly small.
6. The emitted wavelength is
 
c
f
 2  c LC  2  2.998  10 8 m/s
 
0.253  10  6 H
 
30.0  10  12 F
 
 5.19 m.
7. The intensity is the average of the Poynting vector:
I  S
avg

cB
m
2
2 
0

3.0  10
8
m/s
 
1.0  10
 4
T
 
2
2 1.26  10  6 H/m
 
2
 1.2  10
6
W/m
2
.
8. The intensity of the signal at Proxima Centauri is
I 
P
4  r 2

3.0  10 6 W
4  4.3ly
 
9.46  10 15 m/ly
 




2
 1.4  10
 28
W/m
2
.
9. If P is the power and  t is the time interval of one pulse, then the energy in a pulse is
E  P  t  100  10 12 W
 
1.0  10  9 s
 
 1.0  10 5 J.
10. The amplitude of the magnetic field in the wave is
B
m

E
m
c

1.80  10
 4
V/m
2.998  10 8 m/s
 6.00  10  13 T.
11. (a) The amplitude of the magnetic field is
9 9
8
2.0V/m
6.67 10 T 6.7 10 T.
2.998 10 m/s
m
m
E
B
c
       


1273
(b) Since the -wave E

oscillates in the z direction and travels in the x direction, we have B
x
= B
z
= 0. So, the oscillation of the magnetic field is parallel to the y axis.
(c) The direction (+ x ) of the electromagnetic wave propagation is determined by
E B 
 
. If
the electric field points in + z , then the magnetic field must point in the – y direction.
With SI units understood, we may write
 
 
15
15
8
9 15
2.0cos 10 /
cos 10
3.0 10
6.7 10 cos 10
y m
t x c
x
B B t
c
x
t
c


 
 

 
 
 
   
 
 

 
 
 
 
  
 
 
 
 
12. (a) The amplitude of the magnetic field in the wave is
B
m

E
m
c
