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Potential difference between plates in a B-Field?

  1. Feb 14, 2005 #1
    How is potential difference found between two plates in a B-field? Any help is greatly appreciated.
     
  2. jcsd
  3. Feb 14, 2005 #2

    dextercioby

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    What potential are u talking about?Magnetic potential...?

    Daniel.
     
  4. Feb 14, 2005 #3
    Unfortunately, the magnetic field cannot be treated by the method of potential energy. This is because the magnetic field is not a conservative field (it depends on the speed you are going and the path you take, unlike a gravitational or electrostatic field).
     
  5. Feb 14, 2005 #4
    Well i'm given two plates and a B-field. a charge is placed inbetween the two plates and moves paralllel to the two plates until the end, where it then makes a downward circle turn. All the while, the charge is in a B-Field.

    the distance between the two plates is given
    the B-field is given
    the radius of the downward circle is given
    and the velocity of the charge is given.
     
  6. Feb 14, 2005 #5
    The question runs me through a series of questions actually, and I think they are all linked somehow.

    The first one asks what the sign of the charge of the particle is. And the answer I got was Negative, since the B-Field is going INTO the paper, and the velocity is to the right. Therefore, the charge SHOULD be moving upwards [right hand rule number one], but instead it moves downwards, therefore the charge is negative.

    The second question is indicate the direction of the electric field between the plates. My repsonse was the electric field moves from the bottom plate to the top plate, because the charge is being pushed downward by the B-Field [seen when the charge exits the two plates]. Since it should be moving downwards, and is not [the charge moves in a straight parallel line between the two plates]. Therefore, the plates electric field must be opposing the downward force with an upward force. Aka a force from the bottom plate to the top plate.

    And the third question asks whats the magnitude of the potential difference (V) between the plates, which is where I got stuck.

    On second thought. emf = BLv, where B = b-field, L = distance between the plates, and v = the velocity of the charge. Would this solve for the potential difference between the plates? If so, how come? any simple reason?
     
    Last edited: Feb 14, 2005
  7. Feb 14, 2005 #6

    dextercioby

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    Equate the electric force (in ansolute value) with the magnetic one and from there extract the value of E.Then V=Ed.

    Daniel.
     
  8. Feb 14, 2005 #7
    F = Eq = qvB. Therefore E = vB
    So.. V=Ed = vBd?
     
  9. Feb 14, 2005 #8

    dextercioby

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    Exactly.

    Daniel.
     
  10. Feb 14, 2005 #9
    Awsome, thanks so much.
     
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