Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 17E / Solution

In SI units, the two curl Maxwell equations areUse the vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ to obtainIntegration over the fixed volume and the divergence theorem give Poynting's theorem:The first term is the rate of change of stored electromagnetic energy. The surface integral is the outward energy flux carried by the Poynting vector. The right side is minus the power delivered by the field to matter; for an Ohmic conductor it is minus the Ohmic heating rate. Thus energy lost from the field either flows out or does work on charges.

For the specified transverse plane wave, both divergence Maxwell equations hold because $E_y$ depends only on $x,t$. Faraday's equation requiresHere we take the magnetic wave associated with the electric wave, without an additional static background field. Ampère's vacuum equation then gives $k^2={\mu }_0{ϵ}_0{\omega }^2$, equivalentlyThus all four vacuum Maxwell equations are satisfied. The Poynting vector is $\mathbf{P}=(k{E}_0^2/({\mu }_0\omega )){\cos }^2(kx-\omega t){\mathbf{e}}_x$, and averaging over a period givesFor propagation in the positive $x$ direction, $\omega =ck$ and this is $E_0^2{\mathbf{e}}_x/(2{\mu }_{0}c)$. It represents the mean energy transmitted per unit area per unit time, the wave's time-averaged energy flux per unit area, directed along its propagation.