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What's the direction of induced emf here?

  1. Jun 1, 2014 #1

    har

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    This thing has confused me for long .I have not come across any satisfactory answers on my own or while browsing through the internet.
    Everybody has seen the example of a rod moving while on parallel rails, with magnetic field perpendicular to the plane of loop formed by rod and rails.
    When the rod is an insulator, no current flows,direction of induced emf is easily understood.
    Now, consider the case in which rod is an conductor with some resistance . The induced emf across the rod according to faraday's law and voltage across it according to v = IR are in opposing directions.
    As,shown in the thumb nail ,if i attatch voltmeter across ab(Va-Vb) ends will it be positive ?
    if yes, then explain how drop caused by IR is not negative acrosss ab?
     

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    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2
    You're confusing the Emf that comes from Faraday's law of induction with a voltage drop that comes from Ohms law.
     
  4. Jun 1, 2014 #3

    Philip Wood

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    Gold Member

    It removes one annoying complication to suppose that the charge carriers in the conductor are positive rather than negative. Having mastered the argument in terms of positive carriers, you can then adapt it to negative carriers.

    The emf is in the direction b to a through the rod (that is counterclockwise around the circuit as drawn). This is because the carriers are forced through the rod in the direction b to a, and during their passage work is done on them, indirectly by means of the work done by the external agency pushing the rod to the right, mediated by the magnetic field.

    a is at a positive potential relative to b. It is easy to understand this by imagining cd to be open-circuit. Then positive carriers will pile up at the top end of the rod, leaving the bottom end negatively charged. The p.d. arises from from the charge imbalance. The consequent potential gradient across the rod means there is a downward electric field and a downward force on the charge carriers. [Within a split second of starting to move the rod, this force will balance the upward magnetic force and the carriers will stop moving up the rod.]

    If cd is not open-circuit, then a will still be at a positive potential relative to b. This is because if we move the rod at a steady speed, carriers will quickly reach a terminal velocity, when there is no net force on them. For a carrier inside the rod, this will be when…

    Magnetic force – Electric field force – resistive force due to collisions = 0

    [Strictly these 'forces' are magnitudes of forces; the directions are taken care of by the signs. Sorry of this is considered poor form; I don't know how au fait the original poster is with vector notation.]

    Multiplying each term by the rod length (to get the work done on a charge carrier moving through the whole rod), and dividing by the charge on a carrier, we get

    EMF – PD – Ir = 0

    Here r is the rod resistance. If it's negligible, then |p.d. between a and b| = |emf|.

    Outside the rod there is no emf and p.d - IR = 0

    in which R is the resistance of the outside circuit, bcda.
     
    Last edited: Jun 2, 2014
  5. Jun 6, 2014 #4

    Philip Wood

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    You haven't said if you're any clearer. Are you?
     
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