 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/14e/a/solution

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

\frac{\partial L}{\partial r ˙} = m \dot{r} + q A .

m \overset{r}{¨}_{i} + q (\partial_{t} A_{i} + \overset{r}{˙}_{j} \partial_{j} A_{i}) = q (- \partial_{i} ϕ + \overset{r}{˙}_{j} \partial_{i} A_{j}) .

Thus

m \overset{r}{¨}_{i} = q [- \partial_{i} ϕ - \partial_{t} A_{i} + \overset{r}{˙}_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})] .

The last term is the

i

th component of

v \times (\nabla \times A)

, so

m \ddot{r} = q (E + v \times B),

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