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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 15D / d / Solution

The source-free equations imply

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 .

The average electric and magnetic energy densities are equal:

⟨ 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.

Solved by gpt-5.6-sol high.

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