Basic Properties of
Electromagnetic Waves
Electromagnetic waves consist
of oscillating electric and
magnetic fields.
(site 1)
1. E( x , y , z, t )  B( x , y , z, t )
2. Both E and B are  to the direction of wave motion. At any
point in space at any time the direction of wave motion is parallel to E  B
3. The wave speed in vacuum is c = 2.9979  108 m / s
Regardless of the wavelength or state of motion of the source or observer.
4. B  E / c at each point in space at each time
The energy density in a light wave (energy
per unit volume) is given by
u( x, y, z, t )   0 E( x, y, z, t )
2
3
J
/
m
where  0 = 8.854  10-12
 V / m 2
This expression includes both the electric and magnetic field energy
Example:
An electromagnetic plane wave is moving in
the +y direction. It’s electric field is
polarized along the x-axis and has an
amplitude of 103 V/m. The wavelength is
550nm.
A. Find an expression for the electric field
We know that
2
2
7
k


1142
.

10
rad / m
9
 550  10 m
and that
2c
  2 f 
 k c  1.142  107 rad / m3  108 m / s

 3.426  1015 rad / s
so we can write
E (y, t) = (103 Vm ) cos( k y   t ) x
B. Give an expression for the magnetic field of this wave
103 mV
-6
B = E/c =

3.33

10
T
8 m
3  10 s
When E points along the + x - axis and the wave is moving
in the + y direction, we must have B pointing in the - z direction.
Therefore
B = 3.33  10-6 T cos( k y   t ) (  z )
  3.33  10-6 T cos( k y   t ) z
C. How much energy is contained in a cube of length 10nm
centered at the location x = 124 0 nm, y = 1180nm, z = 3412 nm
at time t = 1.5  10 -15 s ?
The energy density is
u( x , y , z, t )   0 E( x , y , z, t )
2
2
3


J
/
m
V


3
  8.854  10 -12
10



2
m
 V / m  

cos2 (1142
.
 10 7
rad
m
 1180  10 9 m - 3.426  1015
rad
s
 1.5  10 -15 s)
J
= 1.907  10
m3
so the total energy is u  volume =
-3
J
3
8
-27
-8
1.907  10

10
m

1.907

10
J
=
1.190

10
eV


3
m
-3
Consider an electromagnetic wave whose electric field is
given by E = E 0 cos( kx   t ) y
The average energy density at any point in space is the average value of
u( x , y , z, t )   0 E( x , y , z, t )
2
  0 E 20 cos2 ( kx   t )
1
The average value of cos ( ), averaged over one or more periods, is
2
1
1
2
2
cos ( ) 
so that cos ( kx   t ) 
2
2
2
u   0 E 20 cos2 ( kx   t )  21  0 E 20
Since the coordinates don' t enter into this expression,
it holds for every electromagnetic plane wave.
For example, for the wave discussed in the previous problem,
the average energy density is
2
3
1
J
/
m
 3 V
-12
-6 J
u   8.854  10
  4.427  10
2   10

 V / m
2
m
m3
The average energy in this box is
average energy = (  0 E ) A c t
1
2
2
0
The rate at which energy passes through the surface is
average energy
P =
 ( 21  0 E 20 ) A c
t
The average power per unit area is the intensity
P 1
I   2  0 c E 20
A