Past exam of the theoretical physics I course of the University of Cambridge / 2012 / Paper 1 / 3 / a / Solution

The cylindrical entries of $\mathbf{A}$ are physical vector components, so $\mathbf{v}\cdot \mathbf{A}=\mu r^2\dot{\theta }$, not $\mu r\dot{\theta }$. The Lagrangian isIts three equations areFor $r>0$, the angular equation can be expanded as $r\ddot{\theta }+2\dot{r}\dot{\theta }+(2e\mu /m)\dot{r}=0$. Directly computing the fields gives $\mathbf{B}=2\mu \hat{\mathbf{z}}$ and $\mathbf{E}=-2\lambda z\hat{\mathbf{z}}$, agreeing with the equations above.