A sinusoidal voltage $$V\left( t \right) = 100\sin \left( {500t} \right)$$     is applied across a pure inductance of $$L = 0.02\,H.$$   The current through the coil is:

Solution :
In a pure inductive circuit current always lags behind the emf by $$\frac{\pi }{2}.$$
If $$v\left( t \right) = {v_0}\sin \omega t\,{\text{then}}\,I = {I_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$$
Now, given $$v\left( t \right) = 100\sin \left( {500t} \right)$$
and $${I_0} = \frac{{{E_0}}}{{\omega L}} = \frac{{100}}{{500 \times 0.02}}\left[ {\because L = 0.02\,H} \right]$$
$${I_0} = 10\sin \left( {500t - \frac{\pi }{2}} \right) = - 10\cos \left( {500t} \right)$$