Faraday's law-is voltage induction immediate/instantaneous?

Let's say we let a horizontal bar drop towards the ground inside a magnetic field coming out of the screen.

The charges inside the bar feel a force because they are moving inside a magnetic field. The positive ones gather at the left side of the bar and the negative ones on the right.

I know the voltage induced is $$V_{in}=BvL$$ This gives: $$\require{cancel} E\cancel{L}=Bu\cancel{L}=>\frac{F}{q}=Bu$$ $$F_=Buq$$ So the electric force is equal to the magnetic force on the charges. If that is the case how do the charges move?

Maybe I began this all wrong. I already assume that the voltage exists, that's why I get this result. I'm interested in how the voltage builds up. Because during that time there has to be a current.

Here it gets more confusing for me.That current will cause a new force opposing the bar's drop. But the bar is constantly accelerating causing the induced voltage to increase. That means more current and greater force on the bar.

When does this stop? Will there be a time when the bar stops accelerating?

And since I'll have an exam in some time , if I'm given a problem about this do I ignore the $$F=BIL$$ force due to the current? Will I consider that the only force on the bar is gravity?

• This sounds similar to the Hall Effect, did you look it up? – Ofek Gillon Jun 27 '17 at 13:01