 Induced current in a tilted resistive loop

Choose the positive loop normal and its positive circulation by the right-hand rule. A spatially uniform external magnetic field of fixed direction and signed magnitude $B_0\cos \omega t$, at angle $\theta$ to that normal, gives magnetic flux $\Phi =\pi r^2{B}_0\cos \theta \cos \omega t$. Neglecting the loop's self-inductance, Faraday's law of induction and Ohm's law give the displayed electric current. At a quarter-cycle the external field is zero but its normal component is decreasing most rapidly, so the current produces a field in the direction of the initial normal flux, as required by Lenz's law. With appreciable self-inductance the circuit instead obeys $L\dot{I}+RI=-{\dot{\Phi }}_{\mathrm{ext}}$.