 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/36b/a/solution

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

\nabla \cdot D = 0, \nabla \cdot B = 0,

\nabla \times E = - \frac{\partial B}{\partial t}, \nabla \times H = \frac{\partial D}{\partial t},

with

D = ϵ E

and

B = μ H

E = E_{0} e^{i (k \cdot x - ω t)}, B = B_{0} e^{i (k \cdot x - ω t)} .

k \cdot E_{0} = k \cdot B_{0} = 0,

B_{0} = \frac{1}{ω} k \times E_{0} .

k \times \frac{B _{0}}{μ} = - ω ϵ E_{0} .

Substituting the expression for

B_{0}

and using

k \times (k \times E_{0}) = - ∣ k ∣^{2} E_{0}

yields

\frac{∣ k ∣ ^{2}}{μ ω} E_{0} = ω ϵ E_{0} .

ω^{2} = \frac{∣ k ∣ ^{2}}{ϵ μ}, v = \frac{ω}{∣ k ∣} = \frac{1}{ϵ μ},

and equivalently

B_{0} = ϵ μ k \times E_{0} .

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