How Is Potential Difference Affected When a Loop Exits a Magnetic Field?

[thursdaytbs]

If there's a loop exiting a B-field, where half of it is in the b-field and half of it is out. How would a potential difference be found in the wire?

I said that V = BLv, (where L is the height of the loop, disregarding the width) and found V to be 3v. Although, there is one resistor on the loop as well, does this effect the potential difference?

Also, the B-field goes into the paper, and the B-field is moving to the right, out of the B-field. It asks for the direction of the current, and I said that the current moves clockwise because Lenz's law.

Can anyone confirm that the potentail difference is found through V=BLv when a loop is leaving a B-field, and that the L is the height of the loop, and the width of the loop can be disregarded since it's leaving the B-field constantly.
And, that in a B-field going into the paper, and a loop leaving it, the current is clockwise?

Any help's appreciated.

 

[James R]

It depends on the shape of the loop, and the direction of the magnetic field.

Whatever the case, the induced EMF around the loop is equal to the rate of change of magnetic flux (field strength times loop area, if the field lines are perpendicular to the loop) through the loop. That's Faraday's law.

Blv could turn out to be correct, depending on the configuration.

 

[wais]

Yes, your approach to finding the potential difference using V=BLv is correct. The L in this equation represents the length of the wire that is within the magnetic field, so the width of the loop can be disregarded as it is not within the field. The potential difference is also affected by the presence of a resistor in the circuit, as it will cause a voltage drop. However, in this scenario, the resistor is not included in the equation V=BLv.

As for the direction of the current, your reasoning based on Lenz's law is correct. The current will flow clockwise in this situation, as the changing magnetic field induces a current in the opposite direction to try to oppose the change.

Overall, your understanding of the concept is correct. Just make sure to include the resistor in your calculations if it is present in the circuit. Keep up the good work!