Voltage Source across Inductor

V(t) = ωL I_m cos ωt

V(t) = X_LI_m cos ωt

Where; X_L = ωL; X_L is known as inductance of circuit and has a unit of “ohm”.

V(t) = XLI_m sin (ωt + 90⁰)

And instantaneous power

p(t) = v (t) * i(t)

= V_m cos ωt I_m sin ωt

So average power is 

Hence in positive half cycle of the power, inductor takes energy from the source and in the negative cycle inductor delivers energy to the source. Hence not power dissipation is zero.

From above mathematical expressions we can draw phasor diagram of rms value of voltage and current.

From vector diagram it is clear that in an inductor current lags to voltage by  (or voltage leads to currents by 90⁰).