 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/36a/d/solution

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

k \times e = ω b .

H = B / μ

and

D_{i} = ε_{ij} E_{j}

, gives

k \times \frac{b}{μ} = - ω ε e .

Eliminating

b = (k \times e) / ω

[k \times (k \times e)]_{i} + ω^{2} μ ε_{ij} e_{j} = 0 .

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