Solution to 1992 Problem 36

The boundary conditions for time-varying electric and magnetic fields at the surface of a perfect conductor are:

\begin{align*}E_t &= 0 \\E_n &= \frac{\rho_s}{\epsilon_0} \\\left|B_t \right| &= \mu_0 \left| J_{s} \right| \\B_n...
where J_s is the surface current density and \rho_s is the surface charge density. %(See Walter Lewin's MIT 8.03 Lecture 16 video or Bekefi and Barratt Section 8.1).
From the first equation, we see that the electric field vector must be reversed, in order to cancel with the incident electric field vector. The Poynting vector
\begin{align*}\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}\end{align*}
always points in the direction of propagation of the electromagnetic wave. When the wave is reflected, the direction of \mathbf{S} must be reversed. Because the electric field vector is reversed, the magnetic field vector cannot be reversed, otherwise the Poynting vector would point in the incident direction. Therefore, answer (C) is correct.

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