# Chapter 33 - Electromagnetic Waves

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```					Chapter 32 - Electromagnetic Waves

fields as they travel through space.

       Qencl
 E  dA     o
Gauss's Law for Electricity

 
   B  dA  0                                     Gauss's Law for Magnetism

                    
 E  dl   t  B  dA                           Faraday's Law

                            
 B  dl   I
o c       o o
tE  dA          Ampere's Law

If we apply these equations to fields in space -- a vacuum:

                                                  
   E  dA  0                                        B  dA  0

                                                                 
 E  dl   t  B  dA                            B  dl    t  E  dA
o o

II.   Production of Traveling Electric and Magnetic Fields. Look at an AC source (like a function
generator) connected to a simple antenna. Notice that time varying electric and magnetic fields will
be produced by the movement of the charges in the antenna.

32-1
y
A.   Look at the electric and magnetic fields after they                                             Ey
have traveled a long distance away from the
1       4
antenna. The fields will have less curvature, and
at a great enough distance they can be considered
to be plane. If the fields are traveling in the +x-                                   2        3             L
direction, then the electric field points in the y-                                  3          2
direction and the magnetic field points in the z-
direction.
x

We will look at a thin, square slab of space
Bz                       L
through which the fields are traveling, see the
consequences of Maxwell’s equations, and                                             4
x   1
examine the transport of energy by the fields.                 z

1.   Apply Ampere's Law over the top face:

                      
 B  dl    t  E  dA
o o

Bz           Ey
as x  0 :                       o o                     equation 1
x             t

2.   Apply Faraday's Law over the front face:

           
   E  dl  
t  
B  dA

Ey        Bz
as x  0 :                                          equation 2
x          t

32-2
3.   By taking the derivative of equation 1 with respect to t, and the derivative of equation 2 with
respect to x, we get:

 2 Bz             2 Ey                               2 Ey         2 Bz
  o  o                        and                             .
tx               t 2                               x 2          xt

Setting them equal to each other, we get

 2 Ey               2 Ey
  o o                                       equation 3
x 2                t 2

4.   A similar result can be obtained for the magnetic field by taking the derivative of equation 1
with respect to x, and the derivative of equation 2 with respect to t.

 2 Bz          2B
  o  o 2z ,                                   equation 4
x 2             t

6.   Both of these equations are called "wave equations," the same form as the wave equation for
waves on strings. The solution is of the form f(x - vt) or f(x + vt). For equation 3 a solution can be
written as:

E = Em sin [k(x – vt)] = Em sin (kx – kvt) = Em sin (kx - t) ,             equation 5

where k is the propagation constant,  is the angular frequency = 2f, f is the frequency, and v is
the velocity of the wave. Note that the velocity is also v = /k. A similar equation is found for
the magnetic field:

B = Bm sin (kx - t) .                                   equation 6

6.   Look at some of the characteristics:

     direction of motion

     maximum value

     frequency and angular frequency

     propagation constant and wavelength

     speed of the wave

32-3
7.   Substitute equation 5 into the wave equation 3. This will give the speed of the wave.

B.   Look at the relationship between the electric field and the magnetic field. Given E, find B using
equation 1. We will get time varying electric and magnetic fields propagating through space - these
are called electromagnetic waves or EM waves.

C.   Look at the energy that is transported by the electromagnetic wave and the rate at which it is
transferred. Refer back to the diagram showing a thin, square slab of volume where the electric and
magnetic fields are present. The energy equals the electric and magnetic field energy densities
multiplied by the volume of the slab.

Electric field energy density: uE    1
2
 oE2

1   B2
Magnetic field energy density:        uB     2
o

32-4
The energy in the slab volume is:

1               2 
1 B  2
U = u E  u B ( volume of slab)  u E  u B L2 x   2  o E 2  2
                    L x
               o 


The instantaneous power (rate of transfer of energy) is:

U
p       =
t

D.   When looking at the strength of the electromagnetic wave, the quantity that is usually measured is
its intensity. The intensity is defined as the average energy delivered per unit area per unit time =
average power delivered per unit area.

In this case, we will first find the instantaneous power delivered per unit area (called the Poynting

Vector, S ), and then find its average value. From above:

 EB Em Bm sin kx  t 
2
p   EB 2  1
S =            
  L  L2                            ,
L2
 o            o         o

which can be written in vector form as:
 
 E B
S     , where the Poynting vector points in the direction of energy travel.
o

The intensity of the electromagnetic radiation is the average value of the Poynting vector:

Intensity = Savg

32-5
1.   Example: The intensity of the light reaching the earth's surface from the sun is approximately
1.4 kW/m2. Find the amplitude of the electric and magnetic fields due to this light.

2.   Example: If all the electrical energy supplied to a 100 watt light bulb is changed to light, the
find the intensity of the electromagnetic wave and the amplitudes of the electric and magnetic
fields one meter from the light bulb.

III. Pressure exerted by EM radiation. When EM radiation hits a surface, pressure is exerted by that

A.   In order to do so, we need to find the momentum and the energy associated with the radiation:

Momentum = mv = (effective mass of the radiation) x (velocity of the radiation)

= mv = (meff)c

Energy            = U = (effective mass of the radiation) x (velocity of the radiation) 2

= U = (meff)c2

U
So the effective mass of the radiation, m eff          , and the momentum of the radiation becomes
c2
U
mv  m eff c      .
c

32-6
B.   Now find the pressure exerted by the radiation:

Average Force           Favg
Average pressure = Pavg =                                            .
Area               A

The average force can be found using the impulse-momentum equation:

U         U 
Favg t  mv final  mv initial                       ,
 c  final  c  initial

U         U 
          
 c  final  c  initial     U           U 
Favg                                                         .
t                 tc  final  tc  initial

So the average pressure becomes:

 U            U 
Pavg                             .
 Atc  final  Atc  initial

Since energy per unit time per unit area is the intensity or the Poynting vector, we get

S         S
Pavg                       .
 c  final  c  initial

Question: Is more pressure exerted when the radiation is absorbed or when it is reflected?

32-7

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