 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/5d/solution

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

In a vacuum with no charge or current, 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} .

\nabla \cdot B = 0

gives

k \cdot B_{0} = 0.

E_{0} = - \frac{c ^{2}}{ω} k \times B_{0},

and substituting this into Faraday's law gives the dispersion relation

ω^{2} = c^{2} ∣ k ∣^{2} .

k

, and the corresponding real electric field is

E (x, t) = Re [- \frac{c ^{2}}{ω} (k \times B_{0}) e^{i (k \cdot x - ω t)}] .

For incidence in the positive

x

-direction, choose the incident fields

B_{i} = B_{0} cos (k x - ω t) e_{z}, E_{i} = c B_{0} cos (k x - ω t) e_{y} .

The perfect conductor requires the tangential electric field to vanish at

x = 0

. The reflected wave therefore has

B_{r} = B_{0} cos (k x + ω t) e_{z}, E_{r} = - c B_{0} cos (k x + ω t) e_{y} .

x \leq 0

B = B_{i} + B_{r} = 2 B_{0} cos (k x) cos (ω t) e_{z},

so its tangential value at the surface is

B_{tangential} (0, t) = 2 B_{0} cos (ω t) e_{z} .

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