 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/12b/b/solution

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

Both divergences vanish, so

ρ = 0

. Faraday's and Ampere's curl equations respectively require

k E_{0} = ω B_{0}, k B_{0} = ω E_{0} .

For positive constants these imply

ω = k, E_{0} = B_{0}, ρ = 0 .

P = E_{0}^{2} x cos^{2} (k x - ω t) .

Only the two

x

-faces of the box contribute to its outward flux. With

a = π / (2 k)

\int_{S} P \cdot d S = L^{2} E_{0}^{2} [cos^{2} (ka - ω t) - cos^{2} (ω t)] = - L^{2} E_{0}^{2} cos (2 ω t) .

The electromagnetic energy inside the box is

U (t) = L^{2} E_{0}^{2} \int_{0}^{a} cos^{2} (k x - ω t) d x,

and direct differentiation, using

ω = k

, gives

\frac{d U}{d t} = L^{2} E_{0}^{2} cos (2 ω t) .

Thus

\int_{S} P \cdot d S = - \frac{d U}{d t},

confirming the integral identity when

J = 0

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