Back EMF in Solenoid

[QwertyXP]

Consider an ideal solenoid (no resistance, no leakage reactance etc.) connected across an AC supply. The back EMF induced in it will be exactly equal and in opposite direction to the source voltage (which means that when a certain terminal of the AC supply is positive, the side of solenoid connected with it would also be positive, and vice versa).

My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out? It's like having having a circuit with only two batteries and terminals of similar polarities shorted with each other. The equation below doesn't appear to be balanced:
V(source)=Back EMF (which is equal to source) + CurrentxReactance
when back EMF is equal to source, the CurrentxReactance part should be zero!?

I've read quite a few explanations on the internet but have yet to fully understand what's happening here.

 

[duke37]

I do not know what you mean by leakage reactance. The solenoid will have reactance.

The current in an pure inductance will be out of phase with the voltage. So, when the voltage is at maximum, the current will be zero and will gradually build up. When the voltage has got to zero, the current will be a maximum.

When the voltage has reversed, the current will decrease so when the voltageis maximum negative, the current will have dropped to zero.

Power runs in and out of the solenoid which is stored in the magnetic field.

Plot a sine wave for the voltage and another for the current shifted at 90 degrees.

 

[BobK]

The back EMF induced in it will be exactly equal and in opposite direction to the source voltage

What makes do you think that?

Bob