The external circuit establishes and drives the inductor current. That current produces a magnetic field in the inductor as described by Ampère’s law. That magnetic field produces the back EMF as described by Faraday’s law. When those two laws are combined, the equation relating voltage and current in an ideal inductor becomes:

[math]v=L\cfrac{di}{dt}.\tag*{}[/math]

An inductor consists of a coil of wire and the inductance relates the magnetic flux to the current that produces that flux:

Imagine the coil consisting of two ideal elements: a resistance [math]R[/math] and an inductance [math]L[/math]

The terminal voltage [math]v_T[/math] of the inductor is equal to the resistive voltage drop from Ohm’s law and the back EMF voltage across the inductance:

[math]v_T=v_R+v.\tag*{}[/math]

Now, apply a constant voltage [math]V_s[/math] to the inductor terminals by closing a switch at time [math]t=0[/math]:

The current which drives the inductor can be found by using Kirchhoff’s voltage law to sum the voltages around the closed loop after the switch is closed:

[math]-V_s+Ri+v=0\tag*{}[/math]

and using the inductor eqaution, this becomes a linear 1st order differential equation:

[math]L\cfrac{di}{dt}+Ri=V_s,\tag*{}[/math]

which has the solution

[math]i=\cfrac{V_s}{R}(1-e^{-t/\tau}),\tag*{}[/math]

where [math]\tau=L/R[/math] is the natural circuit time constant of the inductor. The back EMF voltage is then

[math]v=L\cfrac{di}{dt}=V_se^{-t/\tau}.\tag*{}[/math]

Plotting the voltage and current solutions provides a clear answer to the question:

After the switch closes, the inductor current begins to increase. This produces a magnetic flux that changes with time. As the flux changes, a back EMF voltage is produced. Initially, the back EMF is equal to the applied voltage but as time increases, the back EMF across the inductor tends towards zero as the current in the circuit tends towards [math]V_s/R[/math]. Hence, it is the external circuit which establishes and drives the inductor current.