 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-4/9b/a/solution

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

\dot{r}

\frac{d}{d t} (\frac{1}{2} m ∣ \dot{r} ∣^{2}) = \dot{r} \cdot F = - \dot{r} \cdot \nabla ϕ + \dot{r} \cdot (\dot{r} \times B) = - \frac{d}{d t} ϕ (r) .

Consequently

E = \frac{1}{2} m ∣ \dot{r} ∣^{2} + ϕ (r) is constant .

\dot{r} \times B

does no work, because it is perpendicular to the instantaneous velocity. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of

B

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