Download 32_InstructorSolutionsWin

ELECTROMAGNETIC WAVES
32.1.
32
IDENTIFY: Since the speed is constant, distance x  ct.
SET UP: The speed of light is c  3.00 108 m/s . 1 yr  3.156 107 s.
x 3.84 108 m

 1.28 s
c 3.00 108 m/s
(b) x  ct  (3.00 108 m/s)(8.61 yr)(3.156 107 s/yr)  8.15 1016 m  8.15 1013 km
EVALUATE: The speed of light is very great. The distance between stars is very large compared to terrestrial
distances.
IDENTIFY: Since the speed is constant the difference in distance is ct.
SET UP: The speed of electromagnetic waves in air is c  3.00 108 m/s.
EXECUTE: A total time difference of 0.60 s corresponds to a difference in distance of
EXECUTE: (a) t 
32.2.
32.3.
ct  (3.00 108 m/s)(0.60 106 s)  180 m.
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.
IDENTIFY: Apply c  f .
c  3.00 108 m/s
c 3.0 108 m s
EXECUTE: (a) f  
 6.0 104 Hz.

5000 m
SET UP:
(b) f 
(c) f 

c


3.0 108 m s
 6.0 107 Hz.
5.0 m

3.0 108 m s
 6.0 1013 Hz.
5.0 106 m
3.0 108 m s
 6.0 1016 Hz.
 5.0 109 m
EVALUATE: f increases when  decreases.
2
IDENTIFY: c  f  and k 
.
(d) f 
32.4.
c
c


c  3.00 108 m/s .
c
EXECUTE: (a) f  . UVA: 7.50 1014 Hz to 9.38 1014 Hz . UVB: 9.38 1014 Hz to 1.07 1015 Hz .
SET UP:
(b) k 
32.5.
2


. UVA: 1.57 107 rad/m to 1.96 107 rad/m . UVB: 1.96 107 rad/m to 2.24 107 rad/m .
EVALUATE: Larger  corresponds to smaller f and k.
IDENTIFY: c  f  . Emax  cBmax . k  2 /  .   2 f .
SET UP: Since the wave is traveling in empty space, its wave speed is c  3.00 108 m/s .
c 3.00 108 m/s
 6.94 1014 Hz
EXECUTE: (a) f  
 432 109 m
(b) Emax  cBmax  (3.00 108 m/s)(1.25 106 T)  375 V/m
32-1
32-2
Chapter 32
2 rad
 1.45 107 rad/m .   (2 rad)(6.94 1014 Hz)  4.36 1015 rad/s .
432 109 m
E  Emax cos(kx  t )  (375 V/m)cos([1.45 107 rad/m]x  [4.36 1015 rad/s]t )
(c) k 
32.6.
2


B  Bmax cos(kx  t )  (1.25  106 T)cos([1.45  107 rad/m]x  [4.36  1015 rad/s]t )
EVALUATE: The cos(kx  t ) factor is common to both the electric and magnetic field expressions, since these
two fields are in phase.
IDENTIFY: c  f  . Emax  cBmax . Apply Eqs.(32.17) and (32.19).
SET UP: The speed of the wave is c  3.00 108 m/s.
c 3.00 108 m/s
EXECUTE: (a) f  
 6.90 1014 Hz
 435 109 m
E
2.70 103 V/m
(b) Bmax  max 
 9.00 1012 T
c
3.00 108 m/s
2
(c) k 
 1.44 107 rad/m .   2 f  4.34 1015 rad/s . If E( z, t )  iˆEmax cos(kz  t ) , then

B( z, t )   ˆjBmax cos(kz  t ) , so that E  B will be in the kˆ direction.
E ( z, t )  iˆ(2.70 103 V/m)cos([1.44 107 rad/s) z  [4.34 1015 rad/s]t ) and
B( z, t )   ˆj(9.00 1012 T)cos([1.44 107 rad/s) z  [4.34 1015 rad/s]t ) .
32.7.
EVALUATE: The directions of E and B 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  z -direction . Waves for visible light have very
high frequencies.
IDENTIFY and SET UP: The equations are of the form of Eqs.(32.17), with x replaced by z. B is along the y-axis;
deduce the direction of E.
EXECUTE:   2 f  2 (6.10 1014 Hz)  3.83 1015 rad/s
2 f  3.83 1015 rad/s
 
 1.28 107 rad/m

c
c
3.00 108 m/s
Bmax  5.80 104 T
k
2

Emax  cBmax  (3.00 108 m/s)(5.80 104 T)  1.74 105 V/m
B is along the y-axis. E  B is in the direction of propagation (the +z-direction). From this we can deduce the
direction of E , as shown in Figure 32.7.
E is along the x-axis.
Figure 32.7
E  Emax iˆ cos(kz  t )  (1.74 105 V/m)iˆ cos[(1.28 107 rad/m)z  (3.831015 rad/s)t ]
B  Bmax ˆj cos(kz  t )   5.80  104 T  ˆj cos[(1.28 107 rad/m)z  (3.83 1015 rad/s)t ]
32.8.
EVALUATE: E and B are perpendicular and oscillate in phase.
IDENTIFY: For an electromagnetic wave propagating in the negative x direction, E  Emax cos(kx  t ) .   2 f
and k 
2

.T
1
. Emax  cBmax .
f
SET UP: The wave specified in the problem has a different phase, so E   Emax sin(kx  t ) . Emax  375 V/m ,
k  1.99 107 rad/m and   5.97 1015 rad/s .
E
EXECUTE: (a) Bmax  max  1.25  T.
c
NElectromagnetic Waves
(b) f 
32.9.
32-3

2
1
 9.50 1014 Hz .  
 3.16 107 m  316 nm. T   1.05 1015 s . This wavelength is too short
f
2
k
to be visible.
(c) c  f   (9.50 1014 Hz)(3.16 107 m)  3.00 108 m/s . This is what the wave speed should be for an
electromagnetic wave propagating in vacuum.
   2  
EVALUATE: c  f   

  is an alternative expression for the wave speed.
 2  k  k
IDENTIFY and SET UP: Compare the E ( y, t ) given in the problem to the general form given by Eq.(32.17). Use
the direction of propagation and of E to find the direction of B.
(a) EXECUTE: The equation for the electric field contains the factor sin(ky  t ) so the wave is traveling in the
+y-direction. The equation for E ( y, t ) is in terms of sin(ky  t ) rather than cos(ky  t ); the wave is shifted in
phase by 90 relative to one with a cos(ky  t ) factor.
(b) E ( y, t )  (3.10 105 V/m)kˆ sin[ky  (2.65 1012 rad/s)t ]
Comparing to Eq.(32.17) gives   2.65 1012 rad/s
2 c
2 c 2 (2.998 108 m/s)
  2 f 
so  

 7.11104 m


(2.65 1012 rad/s)
(c)
E  B must be in the +ydirection (the direction in
which the wave is traveling).
When E is in the –z-direction
then B must be in the –xdirection, as shown in
Figure 32.9.
Figure 32.9
2

2.65 1012 rad/s
 8.84 103 rad/m

c 2.998 108 m/s
Emax  3.10 105 V/m
k


Emax 3.10 105 V/m

 1.03 103 T
c
2.998 108 m/s
Using Eq.(32.17) and the fact that B is in the iˆ direction when E is in the kˆ direction,
B  (1.03103 T)iˆ sin[(8.84 103 rad/m)y  (2.65 1012 rad/s)t ]
Then Bmax 
32.10.
EVALUATE: E and B are perpendicular and oscillate in phase.
IDENTIFY: Apply Eqs.(32.17) and (32.19). f  c /  and k  2 /  .
SET UP: The wave in this problem has a different phase, so By ( z, t )  Bmax sin(kx  t ).
EXECUTE: (a) The phase of the wave is given by kx  t , so the wave is traveling in the  x direction.
kc (1.38 104 rad m)(3.0 108 m s)
2 2 f

 6.59 1011 Hz.
(b) k 
. f 

2
2

c
(c) Since the magnetic field is in the  y -direction, and the wave is propagating in the  x -direction, then the
electric field is in the  z -direction so that E  B will be in the  x -direction.
E ( x, t )  cB( x, t )kˆ  cB sin(kx  t )kˆ.
max
E ( x, t )  (c(3.25  10
9
T))sin  (1.38 10 4 rad/m) x  (4.14 1012 rad/s)t  kˆ.
E ( x, t )  (2.48 V m)sin  (1.38 10 4 rad/m) x  (4.14 1012 rad/s)t  kˆ.
32.11.
EVALUATE: E and B have the same phase and are in perpendicular directions.
IDENTIFY and SET UP: c  f  allows calculation of  . k  2 /  and   2 f . Eq.(32.18) relates the electric
and magnetic field amplitudes.
c 2.998 108 m/s
EXECUTE: (a) c  f  so   
 361 m
f
830 103 Hz
32-4
Chapter 32
2 rad

 0.0174 rad/m
 361 m
(c)   2 f  (2 )(830 103 Hz)  5.22 106 rad/s
(b) k 
32.12.
32.13.
2
(d) Eq.(32.18): Emax  cBmax  (2.998 108 m/s)(4.82 1011 T)  0.0144 V/m
EVALUATE: This wave has a very long wavelength; its frequency is in the AM radio braodcast band. The electric
and magnetic fields in the wave are very weak.
IDENTIFY: Emax  cBmax .
SET UP: The magnetic field of the earth is about 104 T.
E 3.85 103 V/m
EXECUTE: B  
 1.28 1011 T.
c
3.00 108 m/s
EVALUATE: The field is much smaller than the earth's field.
IDENTIFY and SET UP: v  f  relates frequency and wavelength to the speed of the wave. Use Eq.(32.22) to
calculate n and K.
v 2.17 108 m/s
EXECUTE: (a)   
 3.81107 m
f 5.70 1014 Hz
c 2.998 108 m/s

 5.26 107 m
f
5.70 1014 Hz
(b)  
(c) n 
c 2.998  108 m/s

 1.38
v 2.17  108 m/s
(d) n  KKm  K so K  n2  (1.38)2  1.90
32.14.
EVALUATE: In the material v  c and f is the same, so  is less in the material than in air. v  c always, so n is
always greater than unity.
IDENTIFY: Apply Eq.(32.21). Emax  cBmax . v  f  . Apply Eq.(32.29) with   K m  0 in place of 0 .
K  3.64 . K m  5.18
SET UP:
EXECUTE: (a) v 
(b)  
c
(3.00 108 m s)

 6.91107 m s.
KK m
(3.64)(5.18)
v 6.91107 m s

 1.06 106 m.
f
65.0 Hz
(c) Bmax 
Emax 7.20 103 V m

 1.04 1010 T.
v
6.91107 m s
Emax Bmax (7.20 103 V m)(1.04 1010 T)

 5.75 108 W m2 .
2 K m 0
2(5.18)0
EVALUATE: The wave travels slower in this material than in air.
2
IDENTIFY: I  P / A . I  12 P0cEmax
. Emax  cBmax .
(d) I 
32.15.
SET UP: The surface area of a sphere of radius r is A  4 r 2 . P0  8.85  1012 C2 /N  m 2 .
EXECUTE: (a) I 
P
(0.05)(75 W)

 330 W/m 2 .
A 4 (3.0 102 m) 2
2I
2(330 W/m2 )
E

 500 V/m . Bmax  max  1.7 106 T  1.7  T .
12
c
P0c
(8.85 10 C2 /N  m2 )(3.00 108 m/s)
EVALUATE: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb.
Our calculation approximates the filament as a point source that radiates uniformly in all directions.
IDENTIFY and SET UP: The direction of propagation is given by E  B .
EXECUTE: (a) Sˆ  iˆ  ( ˆj )  kˆ.
(b) Sˆ  ˆj  iˆ  kˆ.
(b) Emax 
32.16.
(c) Sˆ  (kˆ )  (iˆ)  ˆj .
(d) Sˆ  iˆ  (kˆ )  ˆj.
EVALUATE: In each case the directions of E , B and the direction of propagation are all mutually perpendicular.
NElectromagnetic Waves
32.17.
Emax  cBmax . E  B is in the direction of propagation.
IDENTIFY:
c  3.00 108 m/s . Emax  4.00 V/m.
SET UP:
EXECUTE:
32.18.
32.19.
Bmax  Emax c  1.33  108 T . For E in the +x-direction, E  B is in the +z-direction when B is in
the +y-direction.
EVALUATE: E , B and the direction of propagation are all mutually perpendicular.
2
2
 P0cErms
. The total
IDENTIFY: The intensity of the electromagnetic wave is given by Eq.(32.29): I  12 P0cEmax
energy passing through a window of area A during a time t is IAt.
SET UP: P0  8.85  1012 F/m
2
At  (8.85  1012 F m)(3.00  108 m s)(0.0200 V m) 2 (0.500 m 2 )(30.0 s)  15.9  J
EXECUTE: Energy  P0cErms
EVALUATE: The intensity is proportional to the square of the electric field amplitude.
IDENTIFY and SET UP: Use Eq.(32.29) to calculate I, Eq.(32.18) to calculate Bmax , and use I  Pav / 4 r 2 to
calculate Pav .
(a) EXECUTE:
2
I  12 P0 Emax
; Emax  0.090 V/m, so I  1.1105 W/m 2
(b) Emax  cBmax so Bmax  Emax / c  3.0 1010 T
32.20.
(c) Pav  I (4 r 2 )  (1.075 105 W/m 2 )(4 )(2.5 103 m) 2  840 W
(d) EVALUATE: The calculation in part (c) assumes that the transmitter emits uniformly in all directions.
2
.
IDENTIFY and SET UP: I  Pav / A and I  P0cErms
EXECUTE: (a) The average power from the beam is Pav  IA  (0.800 W m 2 )(3.0  104 m 2 )  2.4  104 W .
I
0.800 W m2

17.4 V m
12
P0c
(8.85 10 F m)(3.00 108 m s)
EVALUATE: The laser emits radiation only in the direction of the beam.
IDENTIFY: I  Pav / A
SET UP: At a distance r from the star, the radiation from the star is spread over a spherical surface of area
A  4 r 2 .
EXECUTE: Pav  I (4 r 2 )  (5.0 103 W m 2 )(4 )(2.0 1010 m) 2  2.5 1025 J
(b) Erms 
32.21:
32.22.
EVALUATE: The intensity decreases with distance from the star as 1/ r 2 .
IDENTIFY and SET UP: c  f  , Emax  cBmax and I  Emax Bmax / 20
EXECUTE: (a) f 
(b) Bmax
c


3.00 108 m s
 8.47 108 Hz.
0.354 m
E
0.0540 V m
 max 
 1.80 1010 T.
c
3.00 108 m s
(c) I  Sav 
Emax Bmax (0.0540 V m)(1.80 1010 T)

 3.87 106 W m2 .
20
20
2
EVALUATE: Alternatively, I  12 P0cEmax
.
32.23.
IDENTIFY:
2
Pav  IA and I  12 P0cEmax
SET UP: The surface area of a sphere is A  4 r 2.
 E2 
P c
(60.0 W)(3.00 108 m s) 0
Pav  Sav A   max  (4 r 2 ) . Emax  av 20 
 12.0 V m.
2 r
2 (5.00 m)2
 2c  0 
E
12.0 V m
 max 
 4.00  108 T.
c
3.00  108 m s
EXECUTE:
Bmax
EVALUATE:
32.24.
32-5
Emax and Bmax are both inversely proportional to the distance from the source.
IDENTIFY: The Poynting vector is S  E  B.
SET UP: The electric field is in the +y-direction, and the magnetic field is in the +z-direction.
cos 2   12 (1  cos 2 )
EXECUTE: (a) Sˆ  Eˆ  Bˆ  ( ˆj )  kˆ  iˆ. The Poynting vector is in the –x-direction, which is the direction of
propagation of the wave.
32-6
Chapter 32
(b) S ( x, t ) 
E ( x , t ) B ( x, t )
0

Emax Bmax
0
cos 2 (kx  t ) 
Emax Bmax
1  cos(2(t  kx)) . But over one period, the
20
Emax Bmax
. This is Eq.(32.29).
2 0
EVALUATE: We can also show that these two results also apply to the wave represented by Eq.(32.17).
IDENTIFY: Use the radiation pressure to find the intensity, and then Pav  I (4 r 2 ).
cosine function averages to zero, so we have Sav 
32.25.
SET UP:
I
c
 2.70  103 W/m 2 . Then
For a perfectly absorbing surface, prad 
prad  I c so I  cprad
EXECUTE:
Pav  I (4 r )  (2.70 103 W m 2 )(4 )(5.0 m) 2  8.5 105 W.
EVALUATE: Even though the source is very intense the radiation pressure 5.0 m from the surface is very small.
IDENTIFY: The intensity and the energy density of an electromagnetic wave depends on the amplitudes of the
electric and magnetic fields.
2
SET UP: Intensity is I  Pav / A , and the average power is Pav = 2I/ c, where I  12 P0cEmax
. The energy density is
2
32.26.
u  P0 E 2 .
EXECUTE: (a) I = Pav /A =
316,000 W
2(0.00201 W/m2 )
= 0.00201 W/m2. Pav = 2I/ c =
= 1.34  1011 Pa
2
2 (5000 m)
3.00 108 m/s
2
(b) I  12 P0cEmax
gives
Emax 
2(0.00201 W/m2 )
= 1.23 N/C
(8.85  10 C2 /N  m2 )(3.00 108 m/s)
2I
=
P0c
12
Bmax = Emax /c = (1.23 N/C)/(3.00  108 m/s) = 4.10  109 T
(c) u  P0 E 2 , so uav  P0 ( Eav ) 2 and Eav =
Emax
, so
2
2
8.85  1012 C2 /N  m2  (1.23 N/C)2 = 6.69  1012 J/m3
P0 Emax

2
2
(d) As was shown in Section 32.4, the energy density is the same for the electric and magnetic fields, so each one
has 50% of the energy density.
EVALUATE: Compared to most laboratory fields, the electric and magnetic fields in ordinary radiowaves are
extremely weak and carry very little energy.
IDENTIFY and SET UP: Use Eqs.(32.30) and (32.31).
dp Sav I
EXECUTE: (a) By Eq.(32.30) the average momentum density is


dV c 2 c 2
dp
0.78 103 W/m2

 8.7 1015 kg/m2  s
dV (2.998 108 m/s) 2
uav =
32.27.
Sav I 0.78 103 W/m2
 
 2.6 106 Pa
c
c 2.998 108 m/s
EVALUATE: The radiation pressure that the sunlight would exert on an absorbing or reflecting surface is very
small.
IDENTIFY: Apply Eqs.(32.32) and (32.33). The average momentum density is given by Eq.(32.30), with S
replaced by Sav  I .
(b) By Eq.(32.31) the average momentum flow rate per unit area is
32.28.
SET UP:
1 atm  1.013105 Pa
EXECUTE: (a) Absorbed light: prad 
prad 
8.33 106 Pa
 8.23 1011 atm.
1.013 105 Pa atm
(b) Reflecting light: prad 
prad 
I 2500 W m2

 8.33 106 Pa. Then
c 3.0 108 m s
2I 2(2500 W m2 )

 1.67 105 Pa. Then
c
3.0 108 m s
1.67 105 Pa
 1.65 1010 atm.
1.013 105 Pa atm
NElectromagnetic Waves
(c) The momentum density is
32.29.
dp Sav
2500 W m2
 2 
 2.78 1014 kg m2  s.
dV c
(3.0 108 m s)2
EVALUATE: The factor of 2 in prad for the reflecting surface arises because the momentum vector totally reverses
direction upon reflection. Thus the change in momentum is twice the original momentum.
IDENTIFY: Apply Eq.(32.4) and (32.9).
SET UP: Eq.(32.26) is S  P0cE 2 .
EXECUTE:
S
P0
P
P
E
P
1
E 2  0 E 2  0 Ec  c 0 EB 
0
0 c
0
P0 0
P0 0
0
EB 
EB
0

E2
 P0cE 2
0c

3
2
c
3.00 108 m s

 0.200 m  20.0 cm. There must be nodes at the planes, which
2 2 f 2(7.50 108 Hz)
are 80.0 cm apart, and there are two nodes between the planes, each 20.0 cm from a plane. It is at 20 cm, 40 cm,
and 60 cm from one plane that a point charge will remain at rest, since the electric fields there are zero.
EVALUATE: The magnetic field amplitude at these points isn’t zero, but the magnetic field doesn’t exert a force
on a stationary charge.
IDENTIFY and SET UP: Apply Eqs.(32.36) and (32.37).
EXECUTE: (a) By Eq.(32.37) we see that the nodal planes of the B field are a distance  / 2 apart, so
 / 2  3.55 mm and   7.10 mm.
(b) By Eq.(32.36) we see that the nodal planes of the E field are also a distance  / 2  3.55 mm apart.
(c) v  f   (2.20 1010 Hz)(7.10 103 m)  1.56 108 m/s.
EXECUTE:
32.31.
P0
EVALUATE: We can also write S  P0c(cB)  P0c B . S can be written solely in terms of E or solely in terms of B.
IDENTIFY: The electric field at the nodes is zero, so there is no force on a point charge placed at a node.
SET UP: The location of the nodes is given by Eq.(32.36), where x is the distance from one of the planes.
  c/ f .
2
32.30.
32-7
xnodes 

EVALUATE: The spacing between the nodes of E is the same as the spacing between the nodes of B. Note that
v  c, as it must.
32.32.
IDENTIFY: The nodal planes of E and B are located by Eqs.(32.26) and (32.27).
c 3.00 108 m/s
SET UP:   
 4.00 m
f 75.0 106 Hz
EXECUTE: (a) x 
32.33.

 2.00 m.
2
(b) The distance between the electric and magnetic nodal planes is one-quarter of a wavelength, so is
 x 2.00 m


1.00 m.
4 2
2
EVALUATE: The nodal planes of B are separated by a distance  / 2 and are midway between the nodal planes of E.
(a) IDENTIFY and SET UP: The distance between adjacent nodal planes of B is  / 2. There is an antinodal plane
of B midway between any two adjacent nodal planes, so the distance between a nodal plane and an adjacent
antinodal plane is  / 4. Use v  f  to calculate  .
EXECUTE:


v 2.10 108 m/s

 0.0175 m
f 1.20 1010 Hz
0.0175 m
 4.38 103 m  4.38 mm
4
4
(b) IDENTIFY and SET UP: The nodal planes of E are at x = 0,  / 2,  , 3 /2, . . . , so the antinodal planes of E

are at x   / 4, 3 /4, 5 /4, . . . . The nodal planes of B are at x   / 4, 3 / 4, 5 /4, . . . , so the antinodal planes
of B are at  / 2,  , 3 /2, . . . .
EXECUTE: The distance between adjacent antinodal planes of E and antinodal planes of B is therefore
 / 4  4.38 mm.
(c) From Eqs.(32.36) and (32.37) the distance between adjacent nodal planes of E and B is  / 4  4.38 mm.
EVALUATE: The nodes of E coincide with the antinodes of B, and conversely. The nodes of B and the nodes
of E are equally spaced.
32-8
Chapter 32
32.34.
IDENTIFY: Evaluate the derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) that are given in Eqs.(32.34) and
(32.35).




sin kx  k cos kx , sin t   cost .
cos kx  k sin kx , cost   sint .
t
t
x
x
 2 E y ( x, t )  2

 2 (2 Emax sin kx sin t )  (2kEmax cos kx sin t ) and
EXECUTE: (a)
x 2
x
x
2
 2 E y ( x, t )
 2 E y ( x, t )

2

2
k
E
sin
kx
sin

t

2
E
sin
kx
sin

t

P

.
max
max
0
0
x 2
c2
t 2
 2 Bz ( x, t )  2

Similarly:
 2 (2Bmax cos kx cos t )  (2kBmax sin kx cos t ) and
x 2
x
x
 2 Bz ( x, t )
2
 2 Bz ( x, t )
 2k 2 Bmax cos kx cos t  2 2 Bmax cos kx cos t  P0 0
.
2
x
c
t 2
E y ( x, t ) 
(b)
 (2Emax sin kx sin t )  2kEmax cos kx sin t .
x
x
E y ( x, t )

E
  2Emax cos kx sin t   2 max cos kx sin t   2 Bmax cos kx sin t .
x
c
c
E y ( x, t )

Bz ( x, t )
  (2Bmax cos kx cos t )  
.
x
t
t
B ( x, t ) 
Similarly:  z
 (2Bmax cos kx cos t )  2kBmax sin kx cos t .
x
x
Bz ( x, t )



  2Bmax sin kx cos t   2 2cBmax sin kx cos t .
x
c
c
E ( x, t )
B ( x, t )

 z
 P0 0 2Emax sin kx cos t  P0 0 (2 Emax sin kx sin t )  P0 0 y
.
x
t
t
EVALUATE: The standing waves are linear superpositions of two traveling waves of the same k and  .
IDENTIFY: The nodal and antinodal planes are each spaced one-half wavelength apart.
SET UP: 2 12 wavelengths fit in the oven, so  2 12    L, and the frequency of these waves obeys the equation f = c.
SET UP:
32.35.
EXECUTE: (a) Since  2 12    L, we have L = (5/2)(12.2 cm) = 30.5 cm.
(b) Solving for the frequency gives f = c/ = (3.00  108 m/s)/(0.122 m) = 2.46  109 Hz.
(c) L = 35.5 cm in this case.  2 12    L, so  = 2L/5 = 2(35.5 cm)/5 = 14.2 cm.
32.36.
f = c/ = (3.00  108 m/s)/(0.142 m) = 2.11  109 Hz
EVALUATE: Since microwaves have a reasonably large wavelength, microwave ovens can have a convenient size
for household kitchens. Ovens using radiowaves would need to be far too large, while ovens using visible light
would have to be microscopic.
IDENTIFY: Evaluate the partial derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) .
SET UP:



sin(kx  t )  k cos(kx  t ) , sin(kx  t )   cos(kx  t ) .
cos(kx  t )  k sin(kx  t ) ,
x
t
x

cos(kx  t )   sin(kx  t )
t
EXECUTE: Assume E  Emax ˆj sin(kx  t ) and B  Bmax kˆ sin(kx  t   ), with       . Eq. (32.12) is
E y
Bz
. This gives kEmax cos(kx  t )   Bmax cos(kx  t   ) , so   0 , and kEmax   Bmax , so
x
t
E y
B

2 f
gives
Emax  Bmax 
Bmax  f Bmax  cBmax . Similarly for Eq.(32.14),  z  P0 0
x
t
k
2 / 
kBmax cos(kx  t   )  P00 Emax cos(kx  t ) , so   0 and kBmax  P0 0Emax , so

P0 0
2 f
f
1
Emax  2
Emax  2 Emax  Emax .
k
c 2 / 
c
c
EVALUATE: The E and B fields must oscillate in phase.
Bmax 
NElectromagnetic Waves
32.37.
32-9
IDENTIFY and SET UP: Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem.
E y
B
EXECUTE: Eq.(32.12):
 z
x
t
2 Ey
E
2B
B


  2 z . Eq.(32.14) says  z  P0 0 y . Taking
Taking
of both sides of this equation gives
of
xt
t
x
t
t
x
2Ey 2Ey
 2 Ey
 2 Ey
2B
1  2 Bz

both sides of this equation gives  2z  P0 0
.
But
(The order in
, so

xt t x
x
tx
tx
P0 0 x 2
which the partial derivatives are taken doesn't change the result.) So 
 2 Bz
2B
 2 Bz
1  2 Bz
and
 P0 0 2 z ,

2
2
2
x
t
t
P0 0 x
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 v  1/ P0 0 .
32.38.
IDENTIFY: The average energy density in the electric field is uE ,av  12 P0 ( E 2 )av and the average energy density in
the magnetic field is u B ,av 
SET UP:
EXECUTE:
 cos (kx  t ) 
2
av
1
( B 2 )av .
2 0
 12 .
2
2
E y ( x, t )  Emax cos(kx  t ) . uE  12 P0 Ey2  12 P0 Emax
.
cos2 (kx  t ) and uE ,av  14 P0 Emax
Bz ( x, t )  Bmax cos(kx  t ) , so uB 
2
. c
uE ,av  14 P0c2 Bmax
1 2
1 2
1 2
Bz 
Bmax cos 2 (kx   t ) and u B ,av 
Bmax . Emax  cBmax , so
20
20
4 0
1 2
1
Bmax , which equals uB ,av .
, so u E ,av 
2 0
P0 0
1 2
Bmax .
20
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.
2
SET UP: The intensity of the wave is I  Pav / A  12 P0cEmax
, and the force is F  Pav A where Pav  I / c .
2
EXECUTE: (a) I = Pav /A = (25,000 W)/[4π(575 m) ] = 0.00602 W/m2
2
EVALUATE: Our result allows us to write uav  2uE ,av  12 P0 Emax
and uav  2u B ,av 
32.39.
2
(b) I  12 P0cEmax
, so Emax 
2I
=
P0c
2(0.00602 W/m2 )
= 2.13 N/C.
(8.85  1012 C2 /N  m2 )(3.00 108 m/s)
Bmax = Emax/c = (2.13 N/C)/(3.00  108 m/s) = 7.10  109 T
32.40.
(c) F =Pav A = ( I/ c )A = (0.00602 W/m2)(0.150 m)(0.400 m)/(3.00  108 m/s) = 1.20  1012 N
EVALUATE: The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth
worrying about!
I
2
IDENTIFY: c  f  . Emax  cBmax . I  12 P0cEmax
. For a totally absorbing surface the radiation pressure is .
c
SET UP: The wave speed in air is c  3.00 108 m/s .
c 3.00 108 m/s
 7.81109 Hz
EXECUTE: (a) f  
 3.84 102 m
E
1.35 V/m
(b) Bmax  max 
 4.50 109 T
c
3.00 108 m/s
2
 12 (8.854  1012 C2 /N  m 2 )(3.00 108 m/s)(1.35 V/m) 2  2.42  103 W/m2
(c) I  12 P0cEmax
IA (2.42 103 W/m2 )(0.240 m2 )

 1.94 1012 N
c
3.00 108 m/s
EVALUATE: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent
of the wavelength of the light.
(d) F  (pressure)A 
32-10
Chapter 32
32.41.
(a) IDENTIFY and SET UP: Calculate I and then use Eq.(32.29) to calculate Emax and Eq.(32.18) to calculate Bmax .
EXECUTE: The intensity is power per unit area: I 
I
P
3.20 103 W

 652 W/m2 .
A  (1.25 103 m)2
2
Emax
, so Emax  20cI
20c
Emax  2(4 107 T  m/A)(2.998 108 m/s)(652 W/m2 )  701 V/m
Emax
701 V/m

 2.34 106 T
c
2.998 108 m/s
EVALUATE: The magnetic field amplitude is quite small.
(b) IDENTIFY and SET UP: Eqs.(24.11) and (30.10) give the energy density in terms of the electric and magnetic
field values at any time. For sinusoidal fields average over E 2 and B 2 to get the average energy densities.
EXECUTE: The energy density in the electric field is uE  12 P0 E 2 . E  Emax cos(kx  t ) and the average value of
Bmax 
cos 2 (kx   t ) is 12 . The average energy density in the electric field then is
2
uE ,av  14 P0 Emax
 14 (8.854 1012 C2 / N  m2 )(701 V/m)2  1.09 106 J/m3. The energy density in the magnetic field
B2
B2
(2.34 106 T)2
. The average value is uB,av  max 
 1.09 106 J/m3.
20
40 4(4 107 T  m/A)
EVALUATE: Our result agrees with the statement in Section 32.4 that the average energy density for the electric
field is the same as the average energy density for the magnetic field.
(c) IDENTIFY and SET UP: The total energy in this length of beam is the total energy density
uav  uE ,av  uB,av  2.18 106 J/m3 times the volume of this part of the beam.
is uB 
32.42.
EXECUTE: U  uav LA  (2.18 106 J/m3 )(1.00 m) (1.25 103 m) 2  1.07 1011 J.
EVALUATE: This quantity can also be calculated as the power output times the time it takes the light to travel L =
1.00 m
 L


11
1.00 m: U  P    (3.20 103 W) 
  1.07 10 J, which checks.
8
c
2.998

10
m/s
 


IDENTIFY: Use the gaussian surface specified in the hint.
SET UP: The wave is in free space, so in Gauss’s law for the electric field, Qencl  0 and AE  dA  0. Gauss’s law
for the magnetic field says AB  dA  0
EXECUTE: Use a gaussian surface such that the front surface is ahead of the wave front (no electric or magnetic
Q
fields) and the back face is behind the wave front, as shown in Figure 32.42. AE  dA  Ex A  encl  0 , so Ex  0.
0
AB  dA  Bx A  0 and Bx  0.
EVALUATE: The wave must be transverse, since there are no components of the electric or magnetic field in the
direction of propagation.
Figure 32.42
32.43.
I
c
SET UP: Assume the electromagnetic waves are formed at the center of the sun, so at a distance r from the center
of the sun I  Pav /(4 r 2 ).
IDENTIFY:
I  Pav / A . For an absorbing surface, the radiation pressure is prad 
NElectromagnetic Waves
EXECUTE: (a) At the sun’s surface: I 
32-11
Pav
3.9 1026 W

 6.4 107 W m2 and
2
4 R
4 (6.96 108 m) 2
I 6.4 107 W m2

 0.21 Pa.
c 3.00 108 m s
(b) Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the radiation pressure:
I  2.6 108 W/m2 and prad  0.85 Pa. At the top of the earth’s atmosphere, the measured sunlight intensity is
prad 
32.44.
1400 W m 2  5  106 Pa , which is about 100,000 times less than the values above.
EVALUATE: (b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and
halfway out of the sun the gas pressure is believed to be about 6 1013 times greater than the radiation pressure.
Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure.
P
2
IDENTIFY: I  . I  12 P0cEmax
.
A
SET UP: 3.00 108 m/s
P 2.80 103 W
EXECUTE: I  
 77.8 W/m2 .
A
36.0 m2
Emax 
32.45.
2I
2(77.8 W/m2 )

 242 N/C .
P0c
(8.854  1012 C2 /N  m2 )(3.00  108 m/s)
EVALUATE: This value of Emax is similar to the electric field amplitude in ordinary light sources.
IDENTIFY: The same intensity light falls on both reflectors, but the force on the reflecting surface will be twice as
great as the force on the absorbing surface. Therefore there will be a net torque about the rotation axis.
SET UP: For a totally absorbing surface, F  Pav A  ( I/c) A, while for a totally reflecting surface the force will be
2
. Once we have the torque, we can use the rotational form
twice as great. The intensity of the wave is I  12 P0cEmax
of Newton’s second law, net = I, to find the angular acceleration.
2
PcEmax
A 1
2
 2 P0 AEmax
c
2
For a totally reflecting surface, the force will be twice as great, which is P0cEmax
. The net torque is therefore
EXECUTE: The force on the absorbing reflector is FAbs  pav A  ( I/c) A 
1
2 0
2
 net  FRefl ( L/2)  FAbs ( L/2)  P0 AEmax
L/4
2
L/4  2m( L/2)  
Newton’s 2nd law for rotation gives  net  I . P0 AEmax
2
Solving for  gives   P0 AEmax
/(2mL) 
32.46.
8.85 10
12
C2 /N  m2  (0.0150 m)2 (1.25 N/C)2
 3.89 1013 rad/s2
(2)(0.00400 kg)(1.00 m)
EVALUATE: This is an extremely small angular acceleration. To achieve a larger value, we would have to greatly
increase the intensity of the light wave or decrease the mass of the reflectors.
IDENTIFY: For light of intensity I abs incident on a totally absorbing surface, the radiation pressure is
2I
I abs
. For light of intensity I refl incident on a totally reflecting surface, prad,refl  refl .
c
c
SET UP: The total radiation pressure is prad  prad,abs  prad,refl . I abs  wI and I refl  (1  w) I
prad,abs 
I abs 2I refl wI 2(1  w) I (2  w) I
.




c
c
c
c
c
2I
I
(b) (i) For totally absorbing w  1so prad  . (ii) For totally reflecting w  0 so prad  . These are just equations
c
c
32.32 and 32.33.
(2  0.9)(1.40 103 W m 2 )
 5.13  106 Pa. For w  0.1 and
(c) For w  0.9 and I  1.40 102 W/m2 , prad 
3.00 108 m/s
(2  0.1)(1.40 102 W m 2 )
I  1.40  103 W m 2 , prad 
 8.87  106 Pa.
3.00 108 m/s
EVALUATE: The radiation pressure is greater when a larger fraction is reflected.
IDENTIFY and SET UP: In the wire the electric field is related to the current density by Eq.(25.7). Use Ampere’s
law to calculate B. The Poynting vector is given by Eq.(32.28) and the equation that follows it relates the energy
EXECUTE: (a) prad  prad,abs  prad,refl 
32.47.
flow through a surface to S .
32-12
Chapter 32
EXECUTE: (a) The direction of E is parallel to the axis of the cylinder, in the direction of the current. From
Eq.(25.7), E   J   I /  a 2 . (E is uniform across the cross section of the conductor.)
(b) A cross-sectional view of the conductor is given in Figure 32.47a; take the current to be coming out of the page.
Apply Ampere’s law to a
circle of radius a.
AB  dl = B(2 a)
I encl  I
Figure 32.47a
AB  dl = 0 Iencl gives B(2 a)  0 I and B 
0 I
2 a
The direction of B is counterclockwise around the circle.
(c) The directions of E and B are shown in Figure 32.47b.
The direction of S =
1
0
EB
is radially inward.
1
1   I   I 
S
EB   2  0 
0
 0   a  2 a 
S
I 2
2 2 a 3
Figure 32.47b
(d) EVALUATE: Since S is constant over the surface of the conductor, the rate of energy flow P is given by S
I 2
lI 2
l
times the surface of a length l of the conductor: P  SA  S (2 al )  2 3 (2 al ) 
. But R  2 , so the
2 a
 a2
a
result from the Poynting vector is P  RI 2 . This agrees with PR  I 2 R, the rate at which electrical energy is being
32.48.
32.49.
dissipated by the resistance of the wire. Since S is radially inward at the surface of the wire and has magnitude
equal to the rate at which electrical energy is being dissipated in the wire, this energy can be thought of as entering
through the cylindrical sides of the conductor.
IDENTIFY: The intensity of the wave, not the electric field strength, obeys an inverse-square distance law.
SET UP: The intensity is inversely proportional to the distance from the source, and it depends on the amplitude
of the electric field by I = Sav = 12 P0 cEmax2.
EXECUTE: Since I = 12 P0 cEmax2, Emax  I . A point at 20.0 cm (0.200 m) from the source is 50 times closer to
the source than a point that is 10.0 m from it. Since I  1/r2 and (0.200 m)/(10.0 m) = 1/50, we have I0.20 = 502 I10.
Since Emax  I , we have E0.20 = 50E10 = (50)(1.50 N/C) = 75.0 N/C.
EVALUATE: While the intensity increases by a factor of 50 2 = 2500, the amplitude of the wave only increases by
a factor of 50. Recall that the intensity of any wave is proportional to the square of its amplitude.
dB
. To calculate
IDENTIFY and SET UP: The magnitude of the induced emf is given by Faraday’s law: E 
dt
d  B / dt we need dB / dt at the antenna. Use the total power output to calculate I and then combine Eq.(32.29) and
(32.18) to calculate Bmax . The time dependence of B is given by Eq.(32.17).
 B  B R 2 , where R = 0.0900 m is the radius of the loop. (This assumes that the magnetic field is
dB
uniform across the loop, an excellent approximation.) E   R 2
dt
dB
B  Bmax cos(kx   t ) so
 Bmax sin( kx   t )
dt
EXECUTE:
The maximum value of
dB
is Bmax , so E max   R 2 Bmax .
dt
R  0.0900 m,   2 f  2 (95.0 106 Hz)  5.97 108 rad/s
NElectromagnetic Waves
32-13
Calculate the intensity I at this distance from the source, and from that the magnetic field amplitude Bmax :
I
2
P
55.0 103 W
Emax
(cBmax )2
c 2
4
2


7.00

10
W/m
.
I



Bmax
4 r 2 4 (2.50 103 m)2
20c
2 0c
2 0
Thus Bmax 
32.50.
32.51.
2 0 I
2(4  107 T  m/A)(7.00  104 W/m 2 )

 2.42  109 T. Then
c
2.998  108 m/s
E max   R2 Bmax   (0.0900 m)2 (2.42 109 T)(5.97 108 rad/s)  0.0368 V.
EVALUATE: An induced emf of this magnitude is easily detected.
IDENTIFY: The nodal planes are one-half wavelength apart.
SET UP: The nodal planes of B are at x = /4, 3/4, 5/4, …, which are /2 apart.
EXECUTE: (a) The wavelength is  = c / f = (3.000  108 m/s)/(110.0  106 Hz) = 2.727 m. So the nodal planes
are at (2.727 m)/2 = 1.364 m apart.
(b) For the nodal planes of E, we have n = 2L/n, so L = n/2 = (8)(2.727 m)/2 = 10.91 m
EVALUATE: Because radiowaves have long wavelengths, the distances involved are easily measurable using
ordinary metersticks.
IDENTIFY and SET UP: Find the force on you due to the momentum carried off by the light. Express this force in
terms of the radiated power of the flashlight. Use this force to calculate your acceleration and use a constant
acceleration equation to find the time.
(a) EXECUTE: prad  I / c and F  prad A gives F  IA / c  Pav / c
ax  F / m  Pav /(mc)  (200 W)/[(150 kg)(3.00 108 m/s)]  4.44 109 m/s 2
Then x  x0  v0 xt  12 axt 2 gives t  2( x  x0 )/ax  2(16.0 m)/(4.44  109 m/s 2 )  8.49 104 s  23.6 h
32.52.
EVALUATE: The radiation force is very small. In the calculation we have ignored any other forces on you.
(b) You could throw the flashlight in the direction away from the ship. By conservation of linear momentum you
would move toward the ship with the same magnitude of momentum as you gave the flashlight.
2
. Emax  cBmax
IDENTIFY: Pav  IA and I  12 P0cEmax
SET UP: The power carried by the current i is P  Vi .
EXECUTE:
Bmax 
Pav 1
 P0cE 2 and Emax 
A 2
2Pav

AP0c
2Vi
2(5.00 105 V)(1000 A)

 6.14  104 V m.
AP0c
(100 m2 )P0 (3.00 108 m s)
Emax 6.14  104 V m

 2.05  104 T.
c
3.00  108 m s
EVALUATE:
32.53.
I
I  Vi / A 
(5.00 105 V)(1000 A)
 5.00 106 W/m2 . This is a very intense beam spread over a
100 m2
large area.
IDENTIFY: The orbiting satellite obeys Newton’s second law of motion. The intensity of the electromagnetic
waves it transmits obeys the inverse-square distance law, and the intensity of the waves depends on the amplitude
of the electric and magnetic fields.
SET UP: Newton’s second law applied to the satellite gives mv2/R = GmM/r2, where M is the mass of the Earth
and m is the mass of the satellite. The intensity I of the wave is I = Sav = 12 P0 cEmax2, and by definition, I = Pav /A.
EXECUTE: (a) The period of the orbit is 12 hr. Applying Newton’s 2 nd law to the satellite gives mv2/R = GmM/r2,
m  2 r / T 
GmM
. Solving for r, we get

r
r2
2
which gives
1/ 3
  6.67  1011 N  m 2 /kg 2  5.97  1024 kg  12  3600 s 2 
 GMT 2 

  2.66  107 m
r 


4 2
 4 2 


The height above the surface is h = 2.66  107 m – 6.38  106 m = 2.02  107 m. The satellite only radiates its
energy to the lower hemisphere, so the area is 1/2 that of a sphere. Thus, from the definition of intensity, the
intensity at the ground is
1/ 3
I = Pav /A = Pav /(2πh2) = (25.0 W)/[2π(2.02  107 m)2] = 9.75  1015 W/m2
(b) I = Sav = 12 P0 cEmax2, so Emax 
2I
2(9.75 1015 W/m2 )

 2.71106 N/C
P0c
(8.85 1012 C2 /N  m2 )(3.00 108 m/s)
Bmax  Emax /c  (2.71 106 N/C)/(3.00  108 m/s)  9.03  1015 T
t  d/c  (2.02  107 m)/(3.00  108 m/s)  0.0673 s
32-14
32.54.
Chapter 32
(c) Pav = I/ c = (9.75  10–15 W/m2)/(3.00  108 m/s) = 3.25  10 – 2 3 Pa
(d)  = c / f = (3.00  108 m/s)/(1575.42 106 Hz) = 0.190 m
EVALUATE: The fields and pressures due to these waves are very small compared to typical laboratory quantities.
2I
IDENTIFY: For a totally reflective surface the radiation pressure is
. Find the force due to this pressure and
c
mM sun
express the force in terms of the power output P of the sun. The gravitational force of the sun is Fg  G
.
r2
SET UP: The mass of the sum is M sun  1.99  1030 kg. G  6.67 1011 N  m 2 /kg 2 .
EXECUTE: (a) The sail should be reflective, to produce the maximum radiation pressure.
P
 2I 
(b) Frad    A , where A is the area of the sail. I 
, where r is the distance of the sail from the sun.
4 r 2
 c 
PA
PA
mM
 2 A  P 
Frad  

. Frad  Fg so
 G 2sun .

2 
2
2 r 2c
r
 c  4 r  2 r c
2 cGmM sun 2 (3.00 108 m/s)(6.67 1011 N  m 2/kg 2 )(10,000 kg)(1.99 1030 kg)
.

P
3.9 1026 W
A  6.42 106 m2  6.42 km2 .
(c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance
from the sun, so this distance divides out when we set Frad  Fg .
A
32.55.
EVALUATE: A very large sail is needed, just to overcome the gravitational pull of the sun.
IDENTIFY and SET UP: The gravitational force is given by Eq.(12.2). Express the mass of the particle in terms of
its density and volume. The radiation pressure is given by Eq.(32.32); relate the power output L of the sun to the
intensity at a distance r. The radiation force is the pressure times the cross sectional area of the particle.
mM
EXECUTE: (a) The gravitational force is Fg  G 2 . The mass of the dust particle is m  V   34  R 3 . Thus
r
4 G MR3
Fg 
.
3r 2
I
(b) For a totally absorbing surface prad  . If L is the power output of the sun, the intensity of the solar radiation
c
L
L
a distance r from the sun is I 
. The force Frad that corresponds to prad is in the
. Thus prad 
2
4 cr 2
4 r
direction of propagation of the radiation, so Frad  prad A , where A   R 2 is the component of area of the particle
LR 2
 L 
perpendicular to the radiation direction. Thus Frad  
( R 2 ) 
.
2 
4cr 2
 4 cr 
(c) Fg  Frad
4 G MR3 LR 2

3r 2
4cr 2
L
3L
 4  G M 
and R 

R 
3
4c
16c  G M


R
3(3.9 1026 W)
16(2.998 108 m/s)(3000 kg/m3 )(6.673 1011 N  m2 / kg2 ) (1.99 1030 kg)
R  1.9 107 m  0.19  m.
EVALUATE: The gravitational force and the radiation force both have a r 2 dependence on the distance from the
sun, so this distance divides out in the calculation of R.
 LR2 

F
3r 2
3L
(d) rad  
. Frad is proportional to R 2 and Fg is proportional to R3 , so this

2 
Fg  4cr  4 G mR3  16cG MR
ratio is proportional to 1/R. If R  0.20 m then Frad  Fg and the radiation force will drive the particles out of the
solar system.
32.56.
v2
.
R
C . An electron has q  e  1.60 1019 C.
IDENTIFY: The electron has acceleration a 
SET UP:
1 eV  1.60 1019
NElectromagnetic Waves
32.57.
32-15
EXECUTE: For the electron in the classical hydrogen atom, its acceleration is
v 2 1 mv2
2(13.6 eV)(1.60 1019 J eV)
a   21

 9.03 1022 m/s2 . Then using the formula for the rate of energy
R 2 mR (9.111031 kg)(5.29 1011 m)
emission given in Problem 32.57:
dE
q2a2
(1.60 1019 C)2 (9.03 1022 m s 2 ) 2
dE


 4.64 108 J s  2.89 1011 eV s. This large value of
3
dt
dt 6 P0c
6 P0 (3.00 108 m s)3
would mean that the electron would almost immediately lose all its energy!
EVALUATE: The classical physics result in Problem 32.57 must not apply to electrons in atoms.
v2
IDENTIFY: The orbiting particle has acceleration a  .
R
SET UP: K  12 mv 2 . An electron has mass me  9.111031 kg and a proton has mass mp  1.67 1027 kg .
2
 q 2a 2 
C2 ( m s ) 2
Nm J
 dE 


  W   .
EXECUTE: (a) 
3
2
2
3
6

P
c
(C
N

m
)(
m
s)
s
s
 dt 
0


(b) For a proton moving in a circle, the acceleration is
v2 1 mv2 2(6.00 106 eV) (1.6  1019 J eV)
2
a   21

 1.53  1015 m s .The rate at which it emits energy because of
R 2 mR
(1.67  1027 kg) (0.75 m)
dE
q 2a 2
(1.6 1019 C)2 (1.53 1015 m s2 )2
its acceleration is


 1.33 1023 J s  8.32 105 eV s.
dt 6 P0c3
6 P0 (3.0 108 m s)3
(dE dt )(1s) 8.32 105 eV
Therefore, the fraction of its energy that it radiates every second is

 1.39 1011.
E
6.00 106 eV
(c) Carry out the same calculations as in part (b), but now for an electron at the same speed and radius. That means
the electron’s acceleration is the same as the proton, and thus so is the rate at which it emits energy, since they also
have the same charge. However, the electron’s initial energy differs from the proton’s by the ratio of their masses:
m
(9.111031 kg)
Ee  Ep e  (6.00 106 eV)
 3273 eV. Therefore, the fraction of its energy that it radiates every
mp
(1.67 1027 kg)
second is
(dE dt )(1s) 8.32 105 eV

 2.54 108.
E
3273 eV
EVALUATE: The proton has speed v 
32.58.
2E
2(6.0 106 eV)(1.60 1019 J/eV)

 3.39 107 m/s . The electron
mp
1.67 1027 kg
has the same speed and kinetic energy 3.27 keV. The particles in the accelerator radiate at a much smaller rate than
the electron in Problem 32.56 does, because in the accelerator the orbit radius is very much larger than in the atom,
so the acceleration is much less.
IDENTIFY and SET UP: Follow the steps specified in the problem.
EXECUTE: (a) Ey ( x, t )  EmaxekC x sin(kC x  t ).
E y
 Emax (kC )e kC x sin(kC x  t )  Emax ( kC )e kC x cos(kC x  t )
x
 2 Ey
 Emax ( kC2 )e kC x sin(kC x  t )  Emax (kC2 )e  kC x cos(kC x  t )
x 2
 Emax (kC2 )e  kC x cos(kC x   t )  Emax (kC2 )e  kC x sin( kC x   t ) .
2Ey
x 2
 2 Emax kC2e kC x cos(kC x   t ) .
Setting
2kC2


 2 Ey
x
2

E y
t
 Emax e kC x  cos(kC x  t ).
E y
gives 2 Emax kC2e k x cos(kC x  t )  Emax e k x  cos(kC x  t ) . This will only be true if
t
C
C


.
, or kC 
2

(b) The energy in the wave is dissipated by the i 2 R heating of the conductor.
E
1
2
2(1.72 108  m)
(c) Ey  y 0  kC x  1, x 


 6.60 105 m.
e
kC

2 (1.0 106 Hz)0
EVALUATE: The lower the frequency of the waves, the greater is the distance they can penetrate into a conductor.
A dielectric (insulator) has a much larger resistivity and these waves can penetrate a greater distance in these
materials.