I've been studying about inductors lately, so I'm going to tell you how inductors work based on what I understood, please correct me if I'm wrong.
As soon as the switch is closed (in a DC circuit with an inductor), there is an induced voltage (lets say E) which is produced to oppose the flow of current which is produced due to the voltage (V) of the battery. At this instant, the value of induced voltage is equal to the applied voltage i.e V=E (the direction of E is opposite of V). Now, since the net voltage in the circuit is 0 (because V=E and are in opposite directions) there will be no flow of current (so the current now, just after closing the switch is 0). We know that back EMF is produced only when there is a change in current/flux through the inductor. Now that there is no change in current/flux through inductor (as no net voltage) there is no need for the back EMF to be present, therefore it starts wearing off. Let’s just say dE amount of back EMF wore off / reduced in time dt. Now there is a very small potential difference dE between the applied voltage V and the reduced induced EMF (E-dE). This produces a current di. Now as soon as the current is produced, the inductor tries to oppose it by getting back up the induced EMF. BUT note that the new induced EMF won't be equal to the initial induced EMF i.e E, this time the induced EMF produced by inductor will be equal to E-dE Because since its the property of inductor to try and not change the current (i.e it neither tries to increase nor decrease current) in the circuit, and right now the current in the circuit as discussed before is di. So the new induced EMF produced such that it won't change this current di, to do so it has to maintain a constant potential difference equal to dE. So it brings back up the induced EMF to E-dE, so the potential difference between applied voltage and induced voltage is dE and current is di.
NOW that the current is constant (di), (and back EMF is produced ONLY when there is a change in current), the back EMF starts wearing off and this whole process repeats until back EMF value reaches 0 and the current in the circuit is purely due to applied voltage V and the inductor starts acting as a wire, where as in the beginning it was acting as an open switch as no current was passing (due to no net voltage).
Can any one please confirm what I understood about inductors is correct or not? If not, please help me understand in simple words and preferably not using complex mathematics and complex theories. (I'm just a teenager and just started learning about inductors.)