What happens from the point when we apply potential difference across an inductor? I am in a serious doubt about it.
Consider a battery of emf E and we connect it to an inductor. Initially the switch is open, now we close the switch. My question is: What mechanism happens just after closing the switch?
When we close the switch, the electric field produced in the conductor by the battery causes the electron to flow in the inductor. As the electrons flow inside the inductor, the flux changes and an emf is induced, my question is that how is this induced emf Ldi/dt is equal to the external emf e, not simply by saying Kirchoff voltage law but by the mechanism happening that it should be E only?
 A: What exactly do you mean when you say "but by the mechanism happening that it should be E only"? I am not sure I completely understand the question, but I will try to answer what I figured from it.
When you apply an EMF across the inductor, a current flows across it (which is dependent on the reactance of the inductor). Now this current flow creates a magnetic field. In an ideal current, with no resistive element, this magnetic field stores the energy supplied by the battery, and hence the induced EMF must equal the external EMF (for where else can the energy go: remember, no resistive element). Also, the magnetic field depends on the current, which in turn depends on the external EMF and the reactance. If you were to write the equations  for the inductor current and magnetic field keeping these relations in mind, you can yourself derive that induced EMF equals the external EMF.
I hope this answers your query. If not, feel free to post your query.
A: (1) if you accept that the voltage across an inductor is proportional to the time rate of change of current through the inductor (the physics behind this result are well known)
and
(2) if you accept that a voltage source of voltage E produces a voltage across a connected inductor equal to E (what else would it do?)
then what is the source of your doubt?  Do you reject (1) and/or (2)?
A: As we close switch term di/dt will have large value so at that instant back EMF Ldi/dt will also large due to this back EMF rate of current will decrease (current will not get 0) and due to this back EMF will also be decreased. And this will happen so on. In last di/dt tends to 0 and also EMF (L R circuit)
