## Solution to 2001 Problem 61

 We are given the magnitude of the electric field vector as a function of time, but not its direction. In fact, we cannot know its direction, but we do know that it is in a plane parallel to the conductor's surface. This follows from the general formula for an electromagnetic plane wave propagating in the $\hat{\mathbf{k}}$ direction:\begin{align*}\mathbf{E}\left(\mathbf{r},t \right) &= E_0 \cos \left(\mathbf{k} \cdot \mathbf{r} - \omega t + \delta \ri...where $\hat{\mathbf{n}}$ and $\hat{\mathbf{k}}$ are orthogonal. In our case, we know that $\mathbf{k} = k\hat{\mathbf{x}}$ and that $\delta = 0$. Therefore, the above equations reduce to\begin{align*}\mathbf{E}\left(\mathbf{r},t \right) &= E_0 \cos \left(k x - \omega t\right) \hat{\mathbf{n}}\\\mathbf{B}\...For the incoming wave $k > 0$ and for the reflected wave $k < 0$. The magnitude of $k$, which is determined by the speed of light in vacuum and the angular frequency $\omega$ must be the same for both the transmitted and the reflected waves, because $\omega$ must be the same. 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. $E_t = 0$ implies that the electric fields of the incident and reflected waves will be reversed. This implies $\hat{\mathbf{n}}_{ref} = -\hat{\mathbf{n}}_{inc}$ and that $E_{0,inc} = E_{0,ref}$. The transverse component of the electric field in the reflected and the incident waves is the only component. So, because the transverse electric field vector is $0$ at $x= 0$, the total electric field vector must also be zero at $x = 0$. The magnetic field vector of the reflected wave can be found using $\hat{\mathbf{n}}_{ref} = -\hat{\mathbf{n}}_{inc}$, $E_{0,inc} = E_{0,ref}$, and $k_{ref} = - k_{inc}$:\begin{align*}\mathbf{B}\left(\mathbf{r},t \right) =\frac{E_{0,inc}}{c}\cos \left(-k_{inc} x - \omega t \right) \left( -k_{in... So, the magnetic field vector of the reflected wave is not reversed and the total magnetic field vector at $x = 0$ is simply two times the incident magnetic field vector evaluated at $x = 0$. Therefore,\begin{align*}\mathbf{B}_{total}\left(x = 0,t \right) = 2 \mathbf{B}_{inc}\left(x = 0,t \right) = 2 \frac{E_{0,inc}}{c}\cos \...The magitude of the total magnetic field vector at $x = 0$ is therefore,\begin{align*}\left|\mathbf{B}_{total}\left(x = 0,t \right) \right| = 2 \frac{E_{0,inc}}{c}\left|\cos \left(\omega t \right)\...The notation of the problem uses $E_0$ in place of $E_{0,inc}$. So, to summarize the magnitudes of the total electric field and total magnetic field immediately to the left of the conductor are $0$ and $2 \frac{E_{0}}{c}\cos \left|\left(\omega t \right)\right|$, respectively. This is very similar to the answer in (C), except the absolute value signs are missing there. Apparently, the answer in (C) is the component of the magnetic field vector in the $k_{inc}/\left|k_{inc}\right| \hat{\mathbf{x}} \times \hat{\mathbf{n}}$ direction. Answer (C) is clearly the best one though, so it should be chosen.