Electromagnetic waves propagate much differently in conductors than they do in
dielectrics or in vacuum. If the resistivity of the conductor is sufficiently
low (that is, if it is a sufficiently good conductor), the oscillating
electric field of the wave gives rise to an oscillating conduction current
that is much larger than the displacement current. In this case, the wave
equation for an electric field \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en
propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x,
t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$
where \(\mu\) is the permeability of the conductor and \(\rho\) is its
resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max}
e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\).
Verify this by substituting E_y(x, t) into the above wave equation. (b) The
exponential term shows that the electric field decreases in amplitude as it
propagates. Explain why this happens. (\(Hint\): The field does work to move
charges within the conductor. The current of these moving charges causes
\(i^2R\) heating within the conductor, raising its temperature. Where does the
energy to do this come from?) (c) Show that the electric-field amplitude
decreases by a factor of 1/\(e\) in a distance \(1/k_C = \sqrt{2\rho/\omega\mu}\),
and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in
copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); permeability
\(\mu = \mu_0\)). Since this distance is so short, electromagnetic waves of this
frequency can hardly propagate at all into copper. Instead, they are reflected
at the surface of the metal. This is why radio waves cannot penetrate through
copper or other metals, and why radio reception is poor inside a metal
structure.
