= Induced current in a tilted resistive loop
{title2=$I(t)=\pi r^2B_0\omega\cos\theta\sin(\omega t)/R$}
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> $\Phi=\pi r^2B_0\cos\theta\cos\omega t$. 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 $L\dot I+RI=-\dot\Phi_{\rm ext}$.
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