# Faraday's Law and a metal loop

1. Jun 7, 2009

### robera1

1. The problem statement, all variables and given/known data
A metal loop is attached to an axle with a handle as shown. The north pole of a magnet is placed blow the loop and handle turned so that the looks rotates counterclockwise at a constant angular speed.

Suppose the loop (the one above) were replaced by a second loop that is identical to the first except for a small cut in it. The loop is rotated as before.
1. How does the maximum induced emf in the uncut loop compare to the maximum induced emf in the cut loop?
2. How does the maximum induced current in the uncut loop compare to the maximum induced current in the cut loop?

2. Relevant equations
This question is conceptual

3. The attempt at a solution
Well, for #2 I think that the maximum induced current in the cut loop will be smaller than in the uncut loop, because in the uncut loop the current continuously moves around the loop. This does not happen in the cut loop.
But as for #1, I don't really know how to relate current to emf.

Any suggestions on what to do?

2. Jun 7, 2009

### elect_eng

Faraday's Law says that the line integral of electric field (i.e. voltage potential) around a loop is equal to the negative rate of change of flux penetrating the area bounded by the loop (EMF due to flux change). The voltage potential will be the resistance of the loop times the current in the loop. However, the EMF due to flux change is the same in both cases. You correctly noted that the current is very small for the cut loop. The current in the uncut loop is very high since the conductor resistance is small. So basically, the EMF is the same in both cases and since EMF is equal to the potential, the current is low in once case (high resistance) and the current is high in the other case (low resistance). Basically, the cut wire acts like a perfect conductor with a very high resistance in the gap. So the potential in directly across the gap. The uncut wire has the potential drop is a distributed way across the whole wire.

Last edited: Jun 8, 2009