 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-4/12a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

F = q v \times B .

For nonrelativistic motion, the work-energy theorem gives

\frac{d K}{d t} = \frac{d}{d t} (\frac{1}{2} m ∣ v ∣^{2}) = m \dot{v} \cdot v = F \cdot v = q (v \times B) \cdot v = 0.

Consequently

K (t) = K (0)

. The reason is geometric: magnetic forces do no work because they are perpendicular to the instantaneous velocity. They can change its direction but not its magnitude. The same energy conclusion holds relativistically, using

d E / d t = F \cdot v

and subtracting the constant rest energy to obtain kinetic energy.

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