 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-4/5d/b/solution

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

B_{3} = 0

, and the wave equation requires

ω^{2} = c^{2} k^{2} .

For propagation in the positive

z

direction, Faraday's law gives

E = \frac{ω}{k} (B_{2}, - B_{1}, 0) cos (k z - ω t) .

Thus $E$ , $B$ , and the propagation direction are mutually perpendicular.

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