Past exam of the physics 2 paper of the University of Oxford / 2022 / 10 / b

The elemental torque about the origin is $d\mathbf{\tau }=\mathbf{r}\times d\mathbf{F}$. Substitution of part (a) givesOnly its $x$ component drives the allowed rotation about the $x$ axis. Using ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ and ${\int }_0^{2\pi }\sin \theta \cos \theta d\theta =0$, the projected torque isThe magnetic dipole moment is $\mathbf{m}=I\pi a^2\hat{\mathbf{z}}$, so the full electromagnetic torque isIts $x$ component agrees exactly. The $y$ component is a real electromagnetic torque, but the bearing constraining motion to the $x$ axis supplies a cancelling reaction torque. For an allowed infinitesimal rotation $d\mathbf{\alpha }=\hat{\mathbf{x}}d\alpha$, only $\mathbf{\tau }\cdot d\mathbf{\alpha }={\tau }_{x}d\alpha$ does mechanical work.