Page 30 - Electrical Properties of Materials

Waves in the presence of an applied magnetic field: cyclotron resonance 13

Medium 1 Medium 2 Medium 3
Vacuum Conductor Vacuum
Incident wave Forward
travelling wave
Transmitted
wave Fig. 1.6
Incident electromagnetic wave
Reflected wave Backward
transmitted to medium 3. The
amplitude of the wave decays in
travelling wave
medium 2 but without any energy
ωτ < < 1 ω > ω p absorption taking place.

converted into heat. The energy balance in the most general case is

energy in the wave = energy in the transmitted wave
+ energy in the reflected wave
+ energy absorbed.

A good example of the phenomena enumerated above is the reflection of
radio waves from the ionosphere. The ionosphere is a layer which, as the name
suggests, contains ions. There are free electrons and positively charged atoms,
so our model should work. In a metal, atoms, and electrons are closely packed;
in the ionosphere, the density is much smaller, so that the critical frequency
ω p is also smaller. Its value is a few hundred MHz. Thus, radio waves below
this frequency are reflected by the ionosphere (this is why short radio waves
can be used for long-distance communication) and those above this frequency
are transmitted into space (and so can be used for space or satellite commu-
nication). The width of the ionosphere also comes into consideration, but at
the wavelengths used (it is the width in wavelengths that counts) it can well be
regarded as infinitely wide.

1.6 Waves in the presence of an applied magnetic field:
cyclotron resonance

In the presence of a constant magnetic field, the characteristics of electromagn-
etic waves will be modified, but the solution can be obtained by exactly the
same technique as before. The electromagnetic eqns (1.22) and (1.23) are still
valid for the a.c. quantities; the equation of motion should, however, contain
the constant magnetic field, which we shall take in the positive z-direction. The
applied magnetic field, B 0 , may be large, hence v × B 0 is not negligible; it is a
first-order quantity. Thus, the linearized equation of motion for this case is In order to satisfy this vector equa-
tion, we need both the v x and v y
components. That means that the
∂v v
m + = e(E E E + v × B 0 ). (1.56) current density, and through that
∂t τ
the electric and magnetic fields,
Writing down all the equations is a little lengthy, but the solution is not will also have both x and y com-
more difficult in principle. It may again be attempted in the exponential form, ponents.
and ∂/∂z and ∂/∂t may again be replaced by ik and –iω, respectively. All the