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

Use the source orientation, anticlockwise viewed from $+z$. The position and oriented line element areThe Lorentz force on the electric current element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$. Evaluating the cross product givesEach component integrates to zero around $0\le \theta <2\pi$, so ${\mathbf{F}}_{\mathrm{net}}=0$. Equivalently, uniform magnetic flux density permits $I\oint d\mathbf{l}\times \mathbf{B}=I(\oint d\mathbf{l})\times \mathbf{B}=0$. Uniformity is essential; a nonuniform magnetic field can exert a net force on a magnetic dipole.

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.

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.