Past exam of the physics B course of the University of Cambridge / 2022 / Paper 1 / Section C / C8 / f

On the curved wire surface, the outward normal is $\hat{\mathbf{r}}$ and the Poynting vector has the constant radial value $N_r(a)=-I^2/(2\sigma {\pi }^2{a}^3)$. Its outward flux isThe Poynting vector is radial, so it has zero normal component on the two end faces in this approximation. This is therefore also the outward flux through the complete wire surface. Equivalently, positive electromagnetic power $P_{\mathrm{loss}}$ flows inward.

The fields are steady, so the stored electromagnetic energy does not change. Poynting's theorem consequently requires the incoming power to equal the Ohmic heating in part (e), exactly as found. The energy reaches the wire through the surrounding electromagnetic field. A local check is $\nabla \cdot \mathbf{N}=r^{-1}{\partial }_r(r{N}_r)=-I^2/(\sigma {\pi }^2{a}^4)=-\mathbf{E}\cdot \mathbf{J}$.