 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/ii/solution

Differentiate

u^{E} = \frac{1}{2} \nabla \times E

\partial_{t} E = c^{2} \nabla \times B

, so

\partial_{t} u^{E} = \frac{c ^{2}}{2} \nabla \times (\nabla \times B) .

\nabla \times (\nabla \times B) = \nabla (\nabla \cdot B) - \nabla^{2} B

. Since

\nabla \cdot B = 0

\frac{\partial u ^{E}}{\partial t} = - \frac{c ^{2}}{2} \nabla^{2} B .

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