Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 7D (Electromagnetism)

In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

The phase has wavevector $\mathbf{k}=k{\mathbf{e}}_z$. The electric field has fixed direction $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, so it has linear polarization at $45^{\circ }$ to the $x$ and $y$ axes, perpendicular to the wavevector. Substitution into the electromagnetic wave equation givesFor the usual positive-frequency convention, $\omega =c|k|$ and propagation is in the direction $\mathrm{sgn}(k){\mathbf{e}}_z$. If $k>0$, this reduces to $\omega =ck$. The following wave formulas assume the nonconstant case $\omega \ne 0$.

For the oscillatory magnetic field, Faraday's law gives $\mathbf{B}=(1/\omega )\mathbf{k}\times \mathbf{E}$. Indeed its time derivative equals $-\nabla \times \mathbf{E}$, and the dispersion relation makes it satisfy the remaining Maxwell equations. ThusFor $\omega =ck$, its amplitude is $E_0/c$. The magnetic field, electric field and direction of propagation are mutually perpendicular. A constant background magnetic field could be added; the expression above is the wave-associated field.

Since $\mathbf{E}\cdot {\mathbf{e}}_z=0$, the vector triple product yields $\mathbf{E}\times ({\mathbf{e}}_z\times \mathbf{E})=|\mathbf{E}{|}^2{\mathbf{e}}_z$. Consequently the Poynting vector isIts average over a full period uses $⟨{\cos }^2⟩=1/2$:This is the mean electromagnetic energy flux per unit area. For positive frequency it points in the propagation direction and has magnitude $E_0^2/(2{\mu }_{0}c)={ϵ}_{0}c{E}_0^2/2$.