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Loop Leaving a B-Field

  1. Feb 17, 2005 #1
    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.
  2. jcsd
  3. Feb 17, 2005 #2

    James R

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    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.
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