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

The Poynting vector is the local electromagnetic energy flux density:Here $\mathbf{E}$ is the electric field and $\mathbf{H}$ the magnetic field intensity. For an oriented area element $d\mathbf{S}$, $\mathbf{N}\cdot d\mathbf{S}$ is the electromagnetic power crossing that element in its positive normal direction. Thus ${\int }_S\mathbf{N}\cdot d\mathbf{S}$ is outward power through a closed surface, in watts. The Poynting vector has units $\mathrm{W}{\mathrm{m}}^{-2}$, and its direction describes local electromagnetic energy transport.

The unit relative permittivity and vacuum-valued magnetic permeability imply $\mathbf{D}={ϵ}_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. The electromagnetic energy per unit volume isThe two contributions are the stored electric and magnetic energies. Although called $U$ in this paper, this quantity is an energy density; the total electromagnetic energy is ${\int }_{V}UdV$.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

Take the electric current along $+z$, and use cylindrical coordinates. A uniform current in a straight homogeneous wire hasThe local Ohm's law $\mathbf{J}=\sigma \mathbf{E}$ gives the longitudinal electric fieldFor a circular Amperian loop of radius $r\le a$, the enclosed electric current is $I_{\mathrm{enc}}=I{r}^2/a^2$. The magnetostatic Ampère law gives $2\pi r{H}_{\varphi }=I_{\mathrm{enc}}$, henceSince $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt is zero at the centre and points radially inward elsewhere. Its units are $\mathrm{W}{\mathrm{m}}^{-2}$. These are the interior fields in the long-wire approximation, neglecting end and contact effects.

The Ohmic heating per unit volume is $\mathbf{E}\cdot \mathbf{J}=J^2/\sigma$, uniform in this wire. Multiplying by its volume gives the dissipated powerwhere the electrical resistance is $R_{\mathrm{wire}}=L/(\sigma \pi a^2)$. The result is in watts; over a time interval $\Delta t$ the dissipated energy is $P_{\mathrm{loss}}\Delta t$.

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 $\nabla \cdot \mathbf{N}=r^{-1}{\partial }_r(r{N}_r)=-I^2/(\sigma {\pi }^2{a}^4)=-\mathbf{E}\cdot \mathbf{J}$.