In an a.c. circuit with pure capacitance 'C' and a.c. source $\mathrm{E}=\mathrm{E}_0 \sin \omega \mathrm{t}$, the equation of instantaneous current is given by

The instantaneous current in a pure capacitor relates to the voltage by $I = C \frac{dE}{dt}$.

Given $E = E_0 \sin \omega t$, differentiating yields $\frac{dE}{dt} = E_0 \omega \cos \omega t$.

Substituting into the current equation gives $I = C E_0 \omega \cos \omega t$.

Using the identity $\cos \theta = \sin (\theta + \frac{\pi}{2})$, this becomes $I = E_0 \omega C \sin (\omega t + \frac{\pi}{2})$.

This corresponds to option B, with the correct peak current magnitude $E_0 \omega C$ and phase shift $\frac{\pi}{2}$.