If $\mathbf{E}(\mathbf{x}, t), \mathbf{B}(\mathbf{x}, t)$ are solutions of Maxwell's equations in a region without any charges or currents show that $\mathbf{E}^{\prime}(\mathbf{x}, t)=c \mathbf{B}(\mathbf{x}, t), \mathbf{B}^{\prime}(\mathbf{x}, t)=-\mathbf{E}(\mathbf{x}, t) / c$ are also solutions.

At the boundary of a perfect conductor with normal $\mathbf{n}$ briefly explain why

$$\mathbf{n} \cdot \mathbf{B}=0, \quad \mathbf{n} \times \mathbf{E}=0$$

Electromagnetic waves inside a perfectly conducting tube with axis along the $z$-axis are given by the real parts of complex solutions of Maxwell's equations of the form

$$\mathbf{E}(\mathbf{x}, t)=\mathbf{e}(x, y) e^{i(k z-\omega t)}, \quad \mathbf{B}(\mathbf{x}, t)=\mathbf{b}(x, y) e^{i(k z-\omega t)} .$$

Suppose $b_{z}=0$. Show that we can find a solution in this case in terms of a function $\psi(x, y)$ where

$$\left(e_{x}, e_{y}\right)=\left(\frac{\partial}{\partial x} \psi, \frac{\partial}{\partial y} \psi\right), \quad e_{z}=i\left(k-\frac{\omega^{2}}{k c^{2}}\right) \psi,$$

so long as $\psi$ satisfies

$$\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\gamma^{2}\right) \psi=0$$

for suitable $\gamma$. Show that the boundary conditions are satisfied if $\psi=0$ on the surface of the tube.

Obtain a similar solution with $e_{z}=0$ but show that the boundary conditions are now satisfied if the normal derivative $\partial \psi / \partial n=0$ on the surface of the tube.