Show that voltage in an inductor leads the current by `pi//2` rad for a pure inductor

Explanation

Figure below show the circuit in which voltage source
`V = V_(0) sin omega t` ...(i)
is applied to pure inductor (zero resistance) coil of inductance L

As a current through the inductor varies and opposing induced emf is generated in it and is given by `-L(d i)/(d t)`
From kirchhoffs loop rule
`V_(0) sin omega t - L(d i)/(d t) = 0`
or,
`d i= (V_(0))/(L) sin omega d t`
integrating on both the side we get
`i = -(V_(0))/(omega L) cos omega t + C`
Where, C is the constant of integration. this integration constant has dimensions of current and is independent of time. Since force has an emf which oscillates symmetrically about zero, the current first and also oscillates symmetrically about zero so there is no time independent component of current that exits. Thus constant = 0
So we have
`i = (-V_(0))/(omega L) cos omega t`
`= (V_(0))/(omega L) sin(omega t - (pi)/(2))`
`i = i_(0) sin (omega t - (pi)/(2))` ...(ii)
where,
`i_(0) = (V_(0))/(omega L)` ....(iii)
is the peak value of current
From the instantaneous values of current and voltage (equation ii and i) we see that in pure inductive circuit the current lags behind emf by a phase angle of `pi//2`