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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/7a/solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 7A / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-29

In vacuum the Maxwell equations are

\nabla \cdot E = 0, \nabla \cdot B = 0, \nabla \times E = - \frac{\partial B}{\partial t}, \nabla \times B = \frac{1}{c ^{2}} \frac{\partial E}{\partial t} .

Substitution of the complex plane electromagnetic wave amplitudes into these equations gives

k \cdot E_{0} = 0, B_{0} = \frac{k \times E _{0}}{ω}, ω^{2} = c^{2} ∣ k ∣^{2} .

Thus the electric field must be transverse to

k

; taking the positive-frequency branch gives

ω = c ∣ k ∣

.

For

E_{0} = E (1, i, 0)

,

k = k z

, and

φ = k z - ω t

, the real fields are

E (x, t) = E (cos φ, - sin φ, 0), B (x, t) = \frac{E}{c} (sin φ, cos φ, 0) .

The electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector is

S = \frac{1}{μ _{0}} E \times B = \frac{E ^{2}}{μ _{0} c} z .

The Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

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