 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/8d/b/solution

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

In Cartesian components,

\frac{\partial L}{\partial r ˙ _{i}} = m \overset{r}{˙}_{i} + q A_{i}

and

\frac{d}{d t} \frac{\partial L}{\partial r ˙ _{i}} = m \overset{r}{¨}_{i} + q \partial_{t} A_{i} + q \overset{r}{˙}_{j} \partial_{j} A_{i}, \frac{\partial L}{\partial r _{i}} = - q \partial_{i} ϕ + q \overset{r}{˙}_{j} \partial_{i} A_{j} .

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

Using

E = - \nabla ϕ - \partial_{t} A, B = \nabla \times A,

and

[\overset{r}{˙} \times B]_{i} = \overset{r}{˙}_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})

, this becomes the Lorentz force

m \overset{r}{¨} = q (E + \overset{r}{˙} \times B) .

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