 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/36a/e/solution

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

Suppose

ε_{ij} e_{j} = ε_{e} e_{i}

. Taking the scalar product of the equation in part (d) with

k

gives

k \cdot e = 0

(- k^{2} + ω^{2} μ ε_{e}) e = 0,

k^{2} = ω^{2} μ ε_{e}

. From part (d),

H = \frac{B}{μ} = \frac{k \times e}{μ ω} sin (k \cdot x - ω t) .

Consequently

N = E \times H = \frac{s i n ^{2} ( k \cdot x - ω t )}{μ ω} e \times (k \times e) = \frac{∣ e ∣ ^{2} s i n ^{2} ( k \cdot x - ω t )}{μ ω} k .

Thus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction $k$ .

 New to topics? Read the docs here!