In figure shows an $\mathrm{AC}$ source connected to an inductor.

Inductor has negligible resistance. Thus, the circuit is a purely inductive $\mathrm{AC}$ circuit. Let the voltage across the source be $\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega t$. Using the Kirchhoff's loop rule,

$\mathrm{V}-\mathrm{L} \frac{d \mathrm{I}}{d t}=0$ where $-\mathrm{L} \frac{d \mathrm{I}}{d t}$ is the self induced $emf$.

$\therefore \mathrm{V}=\mathrm{L} \frac{d \mathrm{I}}{d t}$

$\therefore \frac{d \mathrm{I}}{d t}=\frac{\mathrm{V}}{\mathrm{L}}$

But $\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega t$

$\therefore \frac{d \mathrm{I}}{d t}=\frac{\mathrm{V}_{\mathrm{m}} \sin \omega t}{\mathrm{~L}}$

$\therefore d \mathrm{I}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{L}} \sin \omega d t$

… $(1)$

where $\mathrm{L}$ is the self inductance. Equation $(1)$ indicates that current $\mathrm{I}(t)$ as a function of time, must be such that its slope $\frac{d \mathrm{I}}{d t} \sin e$ is a sinusoidally varying quantity with the same phase as the source voltage and an amplitude given by $\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{L}}$.

To obtain the current integrate equation $(1)$ with respect to time,

$\therefore \int d \mathrm{I}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{L}} \int \sin \omega t d t$

$\therefore \mathrm{I}=-\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{L}} \times \frac{\cos \omega t}{\omega}+\text { constant }$

Here, integration constant has the dimension of current and is time independent.

At $t=0$, time $I=0$

$\therefore 0=0+$ constant

$\therefore$ Constant $=0$

$\therefore$ From equation $(2)$,

$\therefore I=-\frac{V_{m}}{L \omega} \cos \omega t$