[SSM]
(a)
(b)
(c)
(d)
True or false:
Maxwell’s equations apply only to electric and magnetic fields that are constant over time.
The electromagnetic wave equation can be derived from Maxwell’s equations.
Electromagnetic waves are transverse waves.
The electric and magnetic fields of an electromagnetic wave in free space are in phase.
(a) False. Maxwell’s equations apply to both time-independent and time-dependent
fields.
(b) True. One can use Faraday’s law and the modified version of Ampere’s law to
derive the wave equation.
(c) True. Both the electric and magnetic fields of an electromagnetic wave oscillate
at right angles to the direction of propagation of the wave.
(d) True.

 
8
•
Show that the expression E  B 0 for the Poynting vector S
(Equation 30-21) has units of watts per square meter (the SI units for
electromagnetic wave intensity).
 
Determine the Concept We can that E  B 0 has units of W/m2 by
 
substituting the SI units of E , B and  0 and simplifying the resulting
expression.
N
N
C
Nm
J
T
N
A
N

m
W
C
 C   
 s2  s2  s2 
Tm m C m
C m
m
m
m2
A
A
30 •
An electromagnetic wave has an intensity of 100 W/m2. Find its
(a) rms electric field strength, and (b) rms magnetic field strength.
Picture the Problem We can use Pr = I/c to find the radiation pressure. The
intensity of the electromagnetic wave is related to the rms values of its electric
and magnetic field strengths according to I = ErmsBrms/0, where Brms = Erms/c.
(a) Relate the intensity of the
electromagnetic wave to Erms and
Brms:
I
Erms Brms
0
or, because Brms = Erms/c,
I
Solving for Erms yields:
Erms Erms c
0

2
Erms
0c
Erms  0cI
32 •
The rms value of an electromagnetic wave’s electric field strength is
400 V/m. Find the wave’s (a) rms magnetic field strength, (b) average energy
density, and (c) intensity.
Picture the Problem Given Erms, we can find Brms using Brms = Erms/c. The
average energy density of the wave is given by uav = ErmsBrms/0c and the
intensity of the wave by I = uavc .
(a) Express Brms in terms of Erms:
Brms 
Erms
c
Substitute numerical values and
evaluate Brms:
Brms 
400 V/m
 1.334 T
2.998 108 m/s
 1.33 T
(b) The average energy density uav
is given by:
uav 
Substitute numerical values and
evaluate uav:
u av 
Erms Brms
0c
400 V/m 1.334 T 
4 10
7

N/A 2 2.998  10 8 m/s

 1.417 J/m 3  1.42 J/m 3
(c) Express the intensity as the
product of the average energy
density and the speed of light in a
vacuum:
I  uav c
Substitute numerical values and
evaluate I:
I  1.417 J/m 3 2.998 108 m/s



 425 W/m 2
37 ••
[SSM] An electromagnetic plane wave has an electric field that is
parallel to the y axis, and has a Poynting vector that is given by

S x, t   100 W/m 2 cos 2 kx  t  iˆ , where x is in meters, k = 10.0 rad/m,
 = 3.00  109 rad/s, and t is in seconds. (a) What is the direction of propagation of
the wave? (b) Find the wavelength and frequency of the wave. (c) Find the electric
and magnetic fields of the wave as functions of x and t.
Picture the Problem We can determine the direction of propagation of the
wave, its wavelength, and its frequency by examining the argument of the cosine

function. We can find E from S  E 2 0 c and B from B = E/c. Finally, we can

use the definition of the Poynting vector and the given expression for S to find


E and B .
(a) Because the argument of the cosine function is of the form kx  t , the wave
propagates in the +x direction.
2
(b) Examining the argument of the
cosine function, we note that the
wave number k of the wave is:
k
Examining the argument of the
cosine function, we note that the
angular frequency  of the wave
is:
  2f  3.00 109 s 1
Solving for f yields:
f 

(c) Express the magnitude of S in
terms of E:

 10.0 m 1    0.628 m
3.00 109 s 1
 477 MHz
2

 E2
S 
 E  0c S
0c
Substitute numerical values and evaluate E:
E
4 10
7



N/A 2 2.998 108 m/s 100 W/m 2  194.1 V/m

E x, t   194 V/m  coskx  t  ˆj
Because

S x, t   100 W/m 2 cos 2 kx   t  iˆ
 1  
EB:
and S 
0
where k = 10.0 rad/m and
 = 3.00  109 rad/s.
Use B = E/c to evaluate B:
B
194.1V/m
 647.4 nT
2.998  10 8 m/s
 1  
E  B , the direction
Because S 
0

of B must be such that the cross


product of E with B is in the
positive x direction:

Bx, t   647 nT coskx  t  kˆ
where k = 10.0 rad/m and
 = 3.00  109 rad/s.
42 •
Show by direct substitution that Equation 30-8a is satisfied by the wave
function Ey  E0 sin kx   t  E0 sin k x  ct  where c = /k.
Picture the Problem We can show that Equation 30-8a is satisfied by the wave
function Ey by showing that the ratio of 2Ey/x2 to 2Ey/t2 is 1/c2 where c =
/k.
Differentiate E y  E0 sin kx  t 
with respect to x:
E y

E0 sin( kx  t )
x x
 kE0 cos( kx  t )

Evaluate the second partial
derivative of Ey with respect to
x:
2Ey
Differentiate E y  E0 sin kx  t 
E y
with respect to t:
x
2

E0 sin( kx  t )
t
t
 E0 cos( kx  t )
2Ey
Divide equation (1) by equation (2)
to obtain:
2 Ey
•
(1)

Evaluate the second partial
derivative of Ey with respect to t:
43

kE0 cos(kx  t )
x
 k 2 E0 sin( kx  t )

t
2

 E0 cos(kx  t ) (2)
t
  2 E 0 sin( kx  t )

2
2
x 2   k E0 sin kx  t   k
 2 E y   2 E0 sin kx  t   2
t 2
or
2
2Ey k 2 2Ey
1  Ey
 2
 2
x 2
 t 2
c t 2
provided c = /k.
Use the values of 0 and  0 in SI units to compute 1 0 0 and show
that it is equal to 3.00  108 m/s.
Picture the Problem Substitute numerical values and evaluate c:
c
4 10
1
7
N/A
2
8.854 10
12
C / Nm
2

2
 3.00 108 m/s

44 ••
(a) Use Maxwell’s equations to show for a plane wave, in which E and

B y
E z B y
E z
B are independent of y and z, that
and
.

  0 0
x
t
x
t
(b) Show that Ez and By also satisfy the wave equation.
Picture the Problem We can use Figures 30-5 and 30-6 and a derivation similar
to that in the text to obtain the given results.
In Figure 30-5, replace Bz by Ez.
For x small:
E z  x2   E z  x1  

Evaluate the line integral of E
around the rectangular area xz:
 E  d  


E z
x
x
E z
xz
x
(1)
Express the magnetic flux through
the same area:
 B dA  B xz
Apply Faraday’s law to obtain:
 


E
  d     t S Bn dA    t By xz 
B
  y xz
t
B
E
 z xz   y xz
x
t
or
E z B y

x
t
Substitute in equation (1) to obtain:
In Figure 30-6, replace Ey by By and

evaluate the line integral of B
around the rectangular area xz:
S
n

y

 B  d  
0
0

S
En dA
provided there are no conduction
currents.
Evaluate these integrals to obtain:
B y
x
(b) Using the first result obtained in
(a), find the second partial
derivative of Ez with respect to x:
Use the second result obtained in
(a) to obtain:
  0 0
E z
t
  Ez    By 


 
x  x  x   t 
or
 2 Ez
  B 
  y 
2
x
 t  x 
 2 Ez  
Ez 
 2 Ez


  0 0
  0 0
x 2
t 
 t 
t2
or, because 00 = 1/c2,
 2 Ez
1  2 Ez

.
x 2
c2  t 2
Using the second result obtained in
(a), find the second partial
derivative of By with respect to x:
  By 
  Ez 

  0 0


x  x 
x   t 
or
 2 By
  E z 
  0 0


2
x
 t  x 
 By

   0 0
2
x
t2

or, because 00 = 1/c2,
2
 2 By
1  By
 2
.
x 2
c t2
46 •
An electromagnetic wave has a frequency of 100 MHz and is traveling

in a vacuum. The magnetic field is given by Bz, t   1.00 10 8 T coskz  t iˆ .
(a) Find the wavelength and the direction of propagation of this wave. (b) Find the

electric field vector E z , t  . (c) Determine the Poynting vector, and use it to find
the intensity of this wave.
Use the first result obtained in (a)
to obtain:
 2 By
  B y

  0 0
 t   t
2
Picture the Problem We can use c = f to find the wavelength. Examination of
the argument of the cosine function will reveal the direction of propagation of
the wave. We can find the magnitude, wave number, and angular frequency of
the electric vector from the given information and the result of (a) and use these

results to obtain E (z, t). Finally, we can use its definition to find the Poynting
vector.
(a) Relate the wavelength of
the wave to its frequency and
the speed of light:

c
f
Substitute numerical values
and evaluate :

2.998 108 m/s
 3.00 m
100MHz
From the sign of the argument of the cosine function and the spatial dependence
on z, we can conclude that the wave propagates in the +z direction.


(b) Express the amplitude of E :

E  cB  2.998 108 m/s 10 8 T
 3.00 V/m

  2 f  2 100 MHz   6.28  108 s 1 and
2
2
k

 2.09 m 1
 3.00 m
Find the angular frequency
and wave number of the
wave:


Because S is in the positive z direction, E must be in the negative y direction in
order to satisfy the Poynting vector expression:



E z, t    3.00 V/m  cos 2.09 m 1  z  6.28 108 s 1 t ˆj
(c) Use its definition to express and evaluate the Poynting vector:



1    3.00 V/m  10 8 T
S z , t  
EB 
cos 2 2.09 m 1 z  6.28 108 s 1 t ˆj  iˆ
7
2
0
4 10 N/A
or
 
 




S z, t   23.9 mW/m 2 cos 2 2.09 m 1  z  6.28 108 s 1 t kˆ
The intensity of the wave is the
average magnitude of the
Poynting vector. The average
value of the square of the
cosine function is 1/2:
50

••

I  S  12 23.9 mW/m 2


 11.9 mW/m 2
A 20-kW beam of electromagnetic radiation is normal to a surface that
reflects 50 percent of the radiation. What is the force exerted by the radiation on
this surface?
Picture the Problem The total force on the surface is the sum of the force due to
the reflected radiation and the force due to the absorbed radiation. From the
conservation of momentum, the force due to the 10 kW that are reflected is twice
the force due to the 10 kW that are absorbed.
Express the total force on the
surface:
Ftot  Fr  Fa
Substitute for Fr and Fa to
obtain:
Ftot 
2 12 P  12 P 3P


c
c
2c
Substitute numerical values and
evaluate Ftot:
Ftot 
320 kW 
 0.10 mN
22.998 108 m/s 
51 ••
[SSM] The electric fields of two harmonic electromagnetic waves of
angular frequency 1 and 2 are given by E1  E1,0 cosk1x  1t ˆj and
by E  E cosk x   t   ˆj . For the resultant of these two waves, find (a) the
2
2,0
2
2
instantaneous Poynting vector and (b) the time-averaged Poynting vector.
(c) Repeat Parts (a) and (b) if the direction of propagation of the second wave is
reversed so that E2  E2,0 cosk2 x   2t   ˆj
Picture the Problem We can use the definition of the Poynting vector and the


relationship between B and E to find the instantaneous Poynting vectors for
each of the resultant wave motions and the fact that the time average of the cross
product term is zero for 1  2, and ½ for the square of cosine function to find
the time-averaged Poynting vectors.
(a) Because both waves propagate
in the x direction:
Express B in terms of E1 and E2:
 

E  B  0 Siˆ  B  Bkˆ
B
1
E1  E2 
c
Substitute for E1 and E2 to obtain:

1
B x, t   E1, 0 cosk1 x  1t   E 2, 0 cosk 2 x   2 t   kˆ
c
The instantaneous Poynting vector for the resultant wave motion is given by:

1
E1,0 cosk1 x  1t   E2,0 cosk 2 x   2t    ˆj
S  x, t  
0


1
0c
1
E1,0 cosk1 x  1t   E2,0 cosk 2 x   2t   kˆ
c
2
cosk1 x  1t   E 2, 0 cosk 2 x   2 t    ˆj  kˆ
1, 0
E
c
1


E
2
1, 0

cos 2 k1 x  1t   2 E1, 0 E 2, 0 cosk1 x  1t 
0

 cosk 2 x   2 t     E 22, 0 cos 2 k 2 x   2 t    iˆ
(b) The time average of the cross
product term is zero for 1  2,
and the time average of the square
of the cosine terms is ½:

Sav 
E
2 c
1
2
1, 0

 E22,0 iˆ
0

(c) In this case B2   Bkˆ because the wave with k = k2 propagates in the
 iˆ direction. The magnetic field is then:

1
B x, t   E1, 0 cosk1 x  1t   E 2, 0 cosk 2 x   2 t   kˆ
c
The instantaneous Poynting vector for the resultant wave motion is given by:

1
E1,0 cosk1 x  1t   E2,0 cosk 2 x  2t    ˆj
S  x, t  
0


1
0c
1
E1,0 cosk1 x  1t   E2,0 cosk 2 x  2t   kˆ
c
E
2
1, 0

cos 2 k1 x  1t   E 22, 0 cos 2 k 2 x   2 t    iˆ
The time average of the square of
the cosine terms is ½:

Sav 
1
2 0 c
E
2
1, 0

 E22,0 iˆ
Bz
E
 o 0 n (Equation 30-10) follows from
x
t
 
E y
CB  d    0 0 S t dA (Equation 30-6d with I = 0) by integrating along a
suitable curve C and over a suitable surface S in a manner that parallels the
derivation of Equation 30-9.
52
••
Show that
Picture the Problem We’ll choose the
curve with sides x and z in the xy
plane shown in the diagram and apply
Equation 30-6d to show that
 Ey
 Bz
.
   0 0
x
t
Because x is very small, we
can approximate the difference
in Bz at the points x1 and x2 by:
Bz  x2   Bz  x1   B 
Then:
 B  d  


C
0
0
E y
t
Bz
x
x
xz
The flux of the electric field through
this curve is approximately:

Apply Faraday’s law to obtain:
E y
Bz
xz   0 0
xz
x
t
or
E y
Bz
   0 0
x
t
S
En dA  E y xy
54 ••
The intensity of the sunlight striking Earth’s upper atmosphere is
1.37 kW/m2. (a) Find the rms values of the magnetic and electric fields of this
light. (b) Find the average power output of the Sun. (c) Find the intensity and the
radiation pressure at the surface of the Sun.
Picture the Problem We can use I = ErmsBrms/0 and Brms = Erms/c to express
Erms in terms of I. We can then use Brms = Erms/c to find Brms. The average power
output of the Sun is given by Pav  4R 2 I where R is the Earth-Sun distance. The
intensity and the radiation pressure at the surface of the sun can be found from
the definitions of these physical quantities.
(a) The intensity of the radiation
is given by:
I
Erms Brms
Substitute numerical values and evaluate Erms :
0

2
Erms
 Erms  c0 I
c0
E rms 
2.998 10
8



m/s 4  10 7 N/A 2 1.37 kW/m 2  718.4 V/m
 718 V/m
Use Brms  Erms c to evaluate Brms :
Brms 
718.4 V/m
 2.40 T
2.998  10 8 m/s
(b) Express the average power
output of the Sun in terms of the
solar constant:
Pav  4 R 2 I
where R is the earth-sun distance.
Substitute numerical values and
evaluate Pav:
Pav  4 1.50  1011 m 1.37 kW/m 2


2
 3.874 10 26 W
 3.87  10 26 W
(c) Express the intensity at the
surface of the Sun in terms of
the sun’s average power output
and radius r:
Substitute numerical values (see
Appendix B for the radius of the
Sun) and evaluate I at the
surface of the Sun:
Pav
4 r 2
I
I
3.874  10 26 W

4 6.96 108 m

2
 6.363  10 7 W/m 2
 6.36  10 7 W/m 2
Express the radiation pressure in
terms of the intensity:
Pr 
Substitute numerical values and
evaluate Pr:
6.363 10 7 W/m 2
Pr 
 0.212 Pa
2.998 108 m/s
I
c
