### Electromagnetic Waves

```Electromagnetic Waves
In one dimension, the wave equations for E and B are
G
G
G
G
∂2 E
∂2 E
∂2 B
∂2 B
= µoε o 2 and
= µoε o 2 .
∂z 2
∂t
∂z 2
∂t
The solutions to these equations are given by
G
G
G
G
E ( z , t ) = Eo cos(kz − ω t ) and B ( z , t ) = Bo cos(kz − ω t )
where k = 2π/λ and ω = 2πf.
Let’s check one of these (say, for E) by direct substitution:
G
The left side becomes, after 2 derivatives with respect to z: Eo (− k 2 ) cos(kz − ω t )
G
while the right side becomes, after two time derivatives: µoε o Eo (−ω 2 ) cos(kz − ω t ) .
For these to be equal we require that
− k 2 = µoε o (−ω 2 ) or k = ω / c ,
but this means that 2π/λ = 2πf/c or λf = c, an equation which is always true for waves. This
equation simply means that the speed of the wave, c, equals one wavelength divided by the
period T (with T = 1/f) – and is always true for all types of periodic waves.
These solutions for E and B represent traveling waves moving along the z- axis with speed c.
What else can we say about the direction of the vector electric and magnetic fields?
Since both of them have zero divergence in our region of empty space, this means that
∂Ex ∂E y ∂Ez
+
+
= 0 , but Ex and Ey are independent of x and y and so we have that
∂x
∂y
∂z
∂Ez ( z , t )
= 0 , or that Ez = 0 – similarly, Bz=0 – so that these waves are transverse, having no
∂z
component along the direction of motion.
Also we can use Faraday’s law to find a relationship between Eo and Bo, the amplitudes
of the electric and magnetic fields:
ˆj
iˆ
kˆ
G
∇× E =
∂
∂
∂x
∂y
Ex ( z ) E y ( z )
∂E
∂E
∂
= − y iˆ + x ˆj
∂z
∂z
∂z
0
G
∂B
∂B
∂B
= − x iˆ − y ˆj , so
and −
∂t
∂t
∂t
equating x and y components we have that
∂E y
∂Bx
⇒ k ( Eo ) y sin(kz − ω t ) = −ω ( Bo )x sin(kz − ω t )
∂z
∂t
∂By
∂E
+ x =−
⇒ − k ( Eo ) x sin(kz − ω t ) = −ω ( Bo ) y sin(kz − ω t )
∂z
∂t
These give us the result that
− k ( Eo ) y = ω ( Bo )x
−
=−
+ k ( Eo )x = ω ( Bo ) y
which can be written more simply, using k/ω = 1/c, and the definition of cross product, as
G 1
G
Bo = kˆ × Eo
c
G G
G
G
We conclude that the vectors E , B, and kˆ are all mutually perpendicular with E and B lying in
the tranverse plane (plane perpendicular to the direction of the wave, along the z-axis). Also the
magnitude of Bo is given as (1/c)Eo. We conclude that
(
)
Maxwell’s equations have led to the prediction that electromagnetic waves
• Travel at speed c, related to the permittivity and permeability constants
• Are transverse
• Have electric and magnetic fields perpendicular to each other
• Have magnetic fields that are (1/c) times weaker than electric fields
The picture above summarizes our ideas on electromagnetic waves that we have obtained
from Maxwell’s equations. This is the special case in which the E field is vertical and the B field
is horizontal – known as a vertically polarized electromagnetic wave.
Below is a table summarizing the different types of electromagnetic waves, distinguished
on the basis of their frequency or wavelength.
```