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

Define positive $\alpha$ by the right-hand rule about $+x$, starting from the loop in the $x$-$y$ plane at $\alpha =0$. Its magnetic dipole moment becomesThe corresponding $x$ component of magnetic dipole torque is ${\tau }_x=-m(B_z\sin \alpha +B_y\cos \alpha )$. Integrating ${\tau }_x=-dE/d\alpha$ gives the mechanical magnetic dipole potential energywhere the arbitrary additive constant is chosen as zero. Set $B_{\perp }=\sqrt{B_y^2+B_z^2}$. For $B_{\perp }>0$, a positive-amplitude form isEquivalently, ${\alpha }_0=atan2(B_y,-B_z)$ modulo $2\pi$. With this positive choice, minimum potential energy occurs at $\alpha ={\alpha }_0+\pi$, where the rotating magnetic dipole moment aligns with the projection of $\mathbf{B}$ into the $y$-$z$ plane. If $B_y=B_z=0$, the allowed torque vanishes, $E_0=0$ and ${\alpha }_0$ is arbitrary. Holding electric current constant includes exchange with its supply; differentiating only the isolated loop's stored magnetic energy would not give the appropriate mechanical potential energy.