If you by "AC source", you mean an AC Voltage source, then the inductor is not applying a back EMF; it is being driven to the voltage of the the Voltage source. Lets say that your AC voltage source is an ideal voltage source in that has no internal resistance and is capable of providing whatever current is drawn by the load. And lets say that its voltage at any instant in time is: V{sin(Ωt)} then the impedance of the Inductor is: Z = LΩ But the inductor has a phase angle associated with it of π/2 (+90 degrees). This is because the voltage across an inductor must follow this equation: V = L(di/dt) Here is how you find the current: V{sin(Ωt)} = L (di/dt) V/L{sin(Ωt)} = di/dt di = V/L{sin(Ωt)}dt i = V/L⌠sin(Ωt)dt (Note: I am using the symbol ⌠ to indicate integration) i = V/L{-cos(Ωt)/Ω} observing that -cos(Ωt) = sin(Ωt-π/2) we conclude that the current will flow with a magnitude of V/L and phase shifted by -π/2 as follows: i = (V/{LΩ}){sin(Ωt-π/2)} This can, also, be observed using ohms law with phase angle notation: V@0 = (I@-π/2)Z(@π/2) This is read as; "The voltage at a phase angle of 0 equals the current at a phase angle of minus pi over two multiplied by the impedance at an angle of plus pi over two." The most important thing that you should learn from this is that the inductor does not have some mysterious voltage (that you called back EMF) that opposes the AC Voltage source. Instead, it is being driven to the same voltage as the voltage source and that the current that flows must obey the impedance form of Ohm's Law.

The inductor does not apply an equal countervoltage. Energy is stored in the creation of a magnetic field, and the creation of the magnetic field is dependent on the current through the wire. It can be confusing to think of this as a system "starting" initially with 0 voltage. The simple explanation for current with an inductor assumes that current has been alternating for quite some time. Let E = -L di/dt, so that Enet = IR + L di/dt, and Enet = V Sin(wt), and assuming no resistance, then Vsin(wt) = L di/dt, and this gives you i as a function of time (more properly, Enet = Vsin(wt + phi), but that doesn't really change the way things are done). You can imagine this scenario doesn't make sense realistically, because in this case i is not 0 at t = 0, when Enet ought to equal 0. However, this could be solved to a certain degree of approximation: E = -Ldi/dt, then Edt = -Ldi, or di = (E/L)dt. Now let's assume E increases linearly with time (initially) at some rate m, then E = mt, and initial i = 1/2*m/Lt^2. Letting E be a typical sine taylor series, i becomes M/L* (1/2 t^2 - 1/4! t^4 + ...), which might seem somewhat wierd (it's -cos(t) +1), but which would approach the normal solution with a damping force (i.e. the damping force would be "biased" towards one direction until it balances). This "damping force" would have the form -RI