b
In a pure inductive circuit current always

lags behind the emf by \(\frac{\pi}{2}\)

If \(v(t)=v_{0} \sin \omega t\)

then \(\mathrm{I}=\mathrm{I}_{0} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)\)

Now, given \(v(t)=100 \sin (500\, t)\)

and \(\mathrm{I}_{0}=\frac{\mathrm{E}_{0}}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \,\mathrm{H}]\)

\(\mathrm{I}_{0}=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)\)

\(\mathrm{I}_{0}=-10 \cos (500 \mathrm{t})\)