 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2022/paper-1/section-c/c8/c/solution

\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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