# Sliding conductor on a metal rail, with perpendicular B-field

1. Sep 24, 2011

### serverxeon

1. The problem statement, all variables and given/known data

In the following diagram, the brown metallic conductor rod is given a slight push to the left.
The black lines are all conducting wires, with the entire setup placed in a perpendicular B-field. Ignore friction.

What will happen?

1) The rod slowly come to a stop?
2) The rod continues to move at constant speed?
3) The rod comes to a stop and reverses direction?

[PLAIN]http://img52.imageshack.us/img52/9687/magqj.png [Broken]

3. The attempt at a solution

1) I think this should be the answer, as kinetic energy has to be converted to electrical current (heating of the wires). The motion has got to die down.
However, I cannot identify the retardation force that acts on the rod which is necessary for the motion to stop.

2) I don't think it's gonna move forever as argued above.

3) I seriously doubt this man. Intuition, somehow.

Last edited by a moderator: May 5, 2017
2. Sep 25, 2011

### PeterO

As the conductor moves to the left, the area of the loop reduces. That means the amount of flux threading the loop reduces. That reduction in flux means a current is induced in the loop. Once that current is flowing through the moving wire, a force is induced on that wire [F = B.I.l ] that is the force that stops the wire.

Last edited by a moderator: May 5, 2017
3. Sep 25, 2011

### serverxeon

Hmm, but,

the direction of the induced current in this case is anti-clockwise.
Therefore, the direction of the current in the rod, is upwards.
Wont the force on the rod be to the left?! that's not retardation?

4. Sep 25, 2011

### PeterO

Induced Forces ALWAYS oppose change so one of those direction reckonings is backwards. See if you can work out which one.

5. Sep 25, 2011

### serverxeon

oh! sorry a slipped on my part.
so it's clockwise induced current. B-force is therefore on the right.
I see. thanks.