 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/15d/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

k \cdot E_{0} = 0

ω = c ∣ k ∣

, and

B = \frac{k \times E _{0}}{ω} sin (k \cdot x - ω t) .

Therefore

S = \frac{E _{0} \times ( k \times E _{0} )}{μ _{0} ω} sin^{2} (k \cdot x - ω t) = ϵ_{0} c ∣ E_{0} ∣^{2} sin^{2} (k \cdot x - ω t) k .

It is parallel to

k

, and its period average is

⟨ S ⟩ = \frac{1}{2} ϵ_{0} c ∣ E_{0} ∣^{2} k .

⟨ u_{E} ⟩ = ⟨ u_{B} ⟩ = \frac{1}{4} ϵ_{0} ∣ E_{0} ∣^{2} .

Hence $⟨ u ⟩ = ϵ_{0} ∣ E_{0} ∣^{2} /2$ and $⟨ S ⟩ = c ⟨ u ⟩ k$ , as summarized by energy density and flux of a plane electromagnetic wave.

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