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.
   
 

[vk6zgo]

The Back EMF doesn't happen instantaneously.
Current has to flow through the coil,setting up a magnetic field before a Back EMF can be present.

This is a  basic characteristic of an Inductor.

[IanB]

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).

I don't think so.

My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out?

But they are not 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!?

But there should be no back EMF in that equation.

The correct equation for average AC values in an AC circuit is:

voltage = current x reactance

If you know two of these you can calculate the third.

The exact equation for an ideal inductor using instantaneous voltage and current is this:

v = L di/dt

v is the instantaneous voltage, L is the inductance and di/dt is the instantaneous rate of change of current.

Since this is a differential equation you have to solve it by integration. If you find the solution where v(t) is an AC sine wave, you will find the current i(t) is also a sine wave, but it lags the voltage by 90 degrees.

[Simon]

I suspect this may be a spammer laying down innocent threads ready to dump links and crap in their sig