= Solution
\b[Induction and its sign.] The Maxwell-Faraday equation is $\nabla\times\mathbf E=-\partial_t\mathbf B$. For a fixed loop, <Stokes theorem> gives
$$
\boxed{\mathcal E=\oint\mathbf E\cdot d\boldsymbol\ell
=-\frac{d}{dt}\int\mathbf B\cdot d\mathbf S
=-\frac{d\Phi}{dt}.}
$$
This is <Faraday's law of induction>. The orientation of the surface normal and positive circulation are linked by the <right-hand rule>. The minus sign is <Lenz's law>: the induced current's linked flux opposes the imposed flux change, rather than reinforcing it.
\b[The inductive current.] Let the loop normal make angle $\theta$ with the applied field, with positive current defined by that normal. The external <magnetic flux> is $\Phi_{\rm ext}=AB_0\sin\omega t\cos\theta$. With negligible <electrical resistance>, the circuit equation is $L\dot I=-\dot\Phi_{\rm ext}$, so total linked flux $LI+\Phi_{\rm ext}$ is constant. For zero initial trapped flux,
$$
\boxed{I=-\frac{AB_0}{L}\cos\theta\sin\omega t.}
$$
The printed result needs this initial-flux assumption; in general an additional constant current is allowed. The units are amperes because a weber divided by a henry is an ampere.
\b[Current and forces at one instant.] Take $B>0$ along $+x$ and $0<\theta<\pi/2$, so $I<0$ relative to the chosen normal. The loop <magnetic dipole moment> $\boldsymbol\mu=IA\mathbf n$ therefore has a component opposite the applied field. Each wire element experiences <Lorentz force> $d\mathbf F=I\,d\boldsymbol\ell\times\mathbf B$. Opposite sides have equal and opposite forces, giving zero net force in the uniform field but a nonzero couple.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-1-b6-induced-loop.png]
{title=Original top view of the loop's vertical sides at positive applied field; dots and crosses show vertical current, and the force pair increases the angle between normal and field}
The horizontal top and bottom wire segments also experience equal and opposite vertical forces. They cancel without adding torque about the vertical rotation axis; the displayed force pair supplies that torque.
\b[Average torque.] Since $\boldsymbol\tau=\boldsymbol\mu\times\mathbf B$, its generalized component for increasing $\theta$ is $\tau_\theta=-IAB\sin\theta$. Substitution gives
$$
\tau_\theta=\frac{A^2B_0^2}{L}\sin\theta\cos\theta\sin^2\omega t,
$$
and, using $\langle\sin^2\omega t\rangle=1/2$,
$$
\boxed{\langle\tau_\theta\rangle=
\frac{A^2B_0^2}{2L}\sin\theta\cos\theta
=\frac{A^2B_0^2}{4L}\sin2\theta.}
$$
The <orientation torque on a flux-conserving loop> drives its normal away from the field for the stated initial angle. Its magnitude has units joules, or newton metres.
\b[Free motion with large inertia.] Write $J$ for the mechanical <moment of inertia>. If its orientation changes little during each field cycle, the slow equation is $J\ddot\theta=\langle\tau_\theta\rangle$, with effective potential
$$
U_{\rm av}=\frac{A^2B_0^2}{4L}\cos^2\theta.
$$
The stable orientation is $\theta=\pi/2$, where the loop plane contains the applied field. Released from rest between $0$ and $\pi/2$, the loop accelerates toward that orientation, overshoots and librates around it. In the leading averaged, undamped model its turning angles are $\theta_i$ and $\pi-\theta_i$. It does not settle without a dissipative mechanism. For small librations the squared <angular frequency> is $A^2B_0^2/(2LJ)$.
\b[Negligible inductance and finite resistance.] Now the current is proportional to the induced electromotive force, not to the flux. Including slow rotation,
$$
I=-\frac1R\frac{d}{dt}(AB\cos\theta)
=-\frac AR\dot B\cos\theta+\frac ARB\sin\theta\,\dot\theta.
$$
The <rotational damping of a resistive loop in an oscillating field> follows from
$$
\tau_\theta=\frac{A^2}{R}B\dot B\sin\theta\cos\theta
-\frac{A^2}{R}B^2\sin^2\theta\,\dot\theta.
$$
At fixed orientation, the first term alternates at $2\omega$ and has zero mean. The loop can therefore rock slightly, but there is no leading mean aligning torque like that in the inductive case. Slow rotation gives
$$
\boxed{\langle\tau_\theta\rangle
=-\frac{A^2B_0^2}{2R}\sin^2\theta\,\dot\theta,}
$$
which opposes motion and produces <Joule heating>. Initial-phase transients can initiate motion, and the instantaneous torque is not zero. This description retains the large-inertia, slow-motion approximation; it does not assert an exact stationary solution at every starting angle.
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