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 teh 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.