In the case of an inductor

In the purely inductive circuit,

Input voltage, $V=V_m sin (\omega t)$ . . . . . . .(1)

The induced emf across the inductor

$e=-L\dfrac{dI}{dt}$

The induced emf in the circuit is equal and opposite to the applied voltage

$e=-V$

$V_msin\omega t=L\dfrac{dI}{dt}$

$dI=\dfrac{V_m}{L}sin\omega t$

$I=\int dI=\dfrac{V_m}{\omega L}(-cos\omega t)$

$I=\dfrac{V_m}{X_L}sin(\omega t-\pi /2)$

$I=I_msin(\omega t-\pi /2)$. . . . . .(2)

where, $I_m=\dfrac{V_m}{X_L}$

$X_L=\omega L$ (inductive reactants)

From equation (1) and (2)

Voltage leads the current by $90^0$.

The correct option is B.