Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 10B / c / Solution

The equation is $m\ddot{x}=-\nabla V+q\dot{x}\times B$; the magnetic force does no work. For $B=\beta \hat{r}/r^2$, the triple-product identity gives $\dot{L}=q\beta d\hat{r}/dt$, so $J=L-q\beta \hat{r}$ is conserved. Since $J\cdot \hat{r}=-q\beta$, the trajectory lies on a cone with $\cos \alpha =-q\beta /|J|$.