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

Explanation

Since magnetic flux is a function of time, induced emf in the passive element 'L' is `(-L(di)/(dt))`.
Applying the voltage law we write.
`v_(m)sinomegat-L(di)/(dt)=0`
i.e., `di=(v_(m))/(L)sinomegatdt`.
On integrating both sides we get,
`intdi=int(v_(m))/(L)sinomegatdt`
i.e., `i=(v_(m))/(Lomega)(-cosomegat)+` constant
For symmetric oscillation, integration constant = 0
`therefore i=i_(m)sin(omegat-pi//2)`
and `i_(m)=v_(m)//Lomega`.
Inductive reactance (ac resistance offered by an inductor) = `omegaL=2pifL`
Hence `X_(L)propfandX_(L)propL`.
We also note that the voltage across the inductor leads the current by `pi//2` radian.