Show, using the vacuum Maxwell equations, that for any volume $V$ with surface $S$,

$$\frac{d}{d t} \int_{V}\left(\frac{\epsilon_{0}}{2} \mathbf{E} \cdot \mathbf{E}+\frac{1}{2 \mu_{0}} \mathbf{B} \cdot \mathbf{B}\right) d V=\int_{S}\left(-\frac{1}{\mu_{0}} \mathbf{E} \times \mathbf{B}\right) \cdot \mathbf{d} \mathbf{S}$$

What is the interpretation of this equation?

A uniform straight wire, with a circular cross section of radius $r$, has conductivity $\sigma$ and carries a current $I$. Calculate $\frac{1}{\mu_{0}} \mathbf{E} \times \mathbf{B}$ at the surface of the wire, and hence find the flux of $\frac{1}{\mu_{0}} \mathbf{E} \times \mathbf{B}$ into unit length of the wire. Relate your result to the resistance of the wire, and the rate of energy dissipation.