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

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- Solution c

Solution

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:

\nabla \cdot (E \times H) = H \cdot (\nabla \times E) - E \cdot (\nabla \times H),

\nabla \times E = - μ_{0} \frac{\partial H}{\partial t}, \nabla \times H = J + ϵ_{0} \frac{\partial E}{\partial t} .

With the electromagnetic energy density of part (b), these give

\nabla \cdot N = - E \cdot J - μ_{0} H \cdot \partial_{t} H - ϵ_{0} E \cdot \partial_{t} E = - E \cdot J - \frac{\partial U}{\partial t} .

Integrate over a fixed volume

V

and apply the divergence theorem, choosing

d S

outward:

\oint_{S} N \cdot d S = - \int_{V} E \cdot J d V - \frac{d}{d t} \int_{V} U d V .

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.

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