 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/36d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Differentiating

g = ϵ_{0} E \times B

\partial_{t} g_{j} = \partial_{i} [ϵ_{0} (E_{i} E_{j} - \frac{1}{2} E^{2} δ_{ij}) + \frac{1}{μ _{0}} (B_{i} B_{j} - \frac{1}{2} B^{2} δ_{ij})] - (ρ E + J \times B)_{j} .

Moving the divergence to the left identifies

a = - ϵ_{0}, b = \frac{ϵ _{0}}{2}, c = - \frac{1}{μ _{0}}, d = \frac{1}{2 μ _{0}} .

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