 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/5b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

\nabla \cdot E = 0

\nabla^{2} E - \frac{1}{c ^{2}} \partial_{t}^{2} E = 0, c = (μ_{0} ε_{0})^{- 1/2} .

The plane wave solves this when

ω = c ∣ k ∣

and

k \cdot E_{0} = 0

B = Re (\frac{k \times E _{0}}{ω} e^{i (k \cdot x - ω t)}) .

For the stated wave, propagation is along

+ x

(0, 1, 1) / 2

. Thus

B = \frac{E _{0}}{c} (0, - 1, 1) \frac{c o s ( k x - ω t )}{2},

S = \frac{E _{0}^{2}}{μ _{0} c} cos^{2} (k x - ω t) e_{x}, ⟨ S ⟩ = \frac{E _{0}^{2}}{2 μ _{0} c} e_{x} .

The Poynting vector is electromagnetic energy flux, and its average is the wave intensity.

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