 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-3/17b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

\nabla \times (\nabla \times E) = - \partial_{t} (\nabla \times B) = - μ_{0} ε_{0} \partial_{t}^{2} E .

Since the electric divergence vanishes, the double-curl identity makes the left side

- Δ E

Δ E - \frac{1}{c ^{2}} \partial_{t}^{2} E = 0, c = \frac{1}{μ _{0} ε _{0}} .

The same calculation interchanging the fields gives the wave equation for

B

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