 Solution

Choose a positive unit normal $\hat{\mathbf{n}}$ to the loop and choose positive circulation by the right-hand rule. Write the applied magnetic field as $\mathbf{B}(t)=B_0\cos (\omega t)\hat{\mathbf{b}}$, where $\hat{\mathbf{b}}\cdot \hat{\mathbf{n}}=\cos \theta$. Because the field is uniform, the external magnetic flux isFaraday's law of induction gives the induced electromotive force, and Ohm's law gives the current through the loop's electrical resistance:This is the induced current in a tilted resistive loop with the self-generated magnetic-flux contribution neglected as requested. The factor has units $\mathrm{Wb}{\mathrm{s}}^{-1}{\mathrm{\Omega }}^{-1}=\mathrm{A}$. A field parallel to the loop plane has $\cos \theta =0$ and induces no current in this uniform-field model.

For $\cos \theta >0$, the initial normal flux is positive. At $\omega t=\pi /2$, the external magnetic field is instantaneously zero but its normal component is decreasing, so $d{\Phi }_B/dt<0$ and $I>0$. The current is counterclockwise when viewed from the positive-normal side. Its own magnetic field points along $+\hat{\mathbf{n}}$ through the loop, preserving the initial normal flux in accordance with Lenz's law.

For $\cos \theta <0$, both circulation and the induced normal-field direction reverse. The current follows the rate of change of magnetic flux; the quarter-cycle current is nonzero even though the external field itself vanishes then.