 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/6/b/iv/solution

\nabla \cdot (B \times E) = E \cdot (\nabla \times B) - B \cdot (\nabla \times E) = \frac{1}{c ^{2}} E \cdot \partial_{t} E + B \cdot \partial_{t} B = \partial_{t} (\frac{∣ E ∣ ^{2}}{2 c ^{2}} + \frac{∣ B ∣ ^{2}}{2}) .

The divergence theorem therefore gives the required balance with

λ = \frac{1}{2} (\frac{∣ E ∣ ^{2}}{c ^{2}} + ∣ B ∣^{2}), \frac{\partial}{\partial t} \int_{V} λ d V = \oint_{A} (B \times E) \cdot d A .

An additive time-independent constant in

λ

is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently

\partial_{t} λ + \nabla \cdot (E \times B) = 0

. The order of the cross product accounts for the plus sign in the stated integral identity.

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