Electromagnetic induction and EMF in wire

[adamg]

If you move a straight wire down through a magnet field, an emf is induced in the wire. If you then stop moving the wire, does the emf remain or does it disappear i.e do the electrons flow back? What about if you moved the wire out of the field?

 

[Doc Al]

Originally posted by adamg
If you then stop moving the wire, does the emf remain or does it disappear i.e do the electrons flow back?

The emf disappears. Yes, the electrons redistribute themselves along the wire.

What about if you moved the wire out of the field?

Same thing. The emf is due to the magnetic force on the charge carriers (electrons) in the wire. That force is [itex]F = q \vec{v} \times \vec{B}[/itex]; stop the motion or remove the field and the force (and emf) goes away.

 

[adamg]

(Thanks for the help) When the electrons flow back along the wire, doesn't that produce a force? the force will be in the same direction that originally created the emf i.e down. This force down will create a force on the electrons pushing them back i.e as in the creation of the emf. Is that correct?
The only possibility i could think up was that the final force mentioned above is not large enough to prevent the electrons flowing back to their original position??

 

[Doc Al]

Originally posted by adamg
When the electrons flow back along the wire, doesn't that produce a force? the force will be in the same direction that originally created the emf i.e down. This force down will create a force on the electrons pushing them back i.e as in the creation of the emf. Is that correct?

Is this what you mean. If the wire is stopped, will the charges moving back experience a force since they are moving in a magnetic field?

If that's what you mean, the answer is yes. But what direction is that force? It is not along the wire (like the emf was), it is sideways to it.

Let me know if I am missing your point.

 

[adamg]

When the wire is being moved down, the electrons accumulate at one end. When you stop the wire, the electrons move back. Using left hand, this creates a force down. So force on electrons is down, so wire will move down. This constitutes a current going up and the force created will try to push the electrons back (as when the wire was being moved down originally) But obviously the electrons continue to move back to their original position, so this final force must be smaller than the force created due to the emf between the ends of the wire. (?)

 

[Doc Al]

I think I understand your question now. Very interesting. It's difficult to answer, since I don't think it is physically realistic. (Perhaps I'm just confused!) As you slow the wire down, the moving charges will create a force pushing down---this force opposes your attempt to slow the wire. But the charge redistributes very quickly. If you could somehow stop the wire before the charge had a chance to move (not possible, I think) then I suppose the moving charge would create a movement of the wire and thus another induced emf (opposing the motion of the charge). But, yes, this would be a smaller emf (the emf depends on the speed of the wire). (I think energy conservation would be violated if it weren't.)

 

[adamg]

Oh, i think i understand. The charge starts to re-distribute when you start slowing it down, since the force on the electrons reduces, so by the time you actually stop it is already back in its starting position. Is that correct? Then, yes, my scenario is impossible i think.

 

[Doc Al]

Yes, that's the idea.