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Induced EMF in loop

  1. Mar 24, 2010 #1
    ok, if i have a rectangular loop parallel to the screen and in a uniform constant magnetic field pointing perpendicularly outwards of the screen, if i move the loop to the right of the screen, will there be an induced emf in the loop?

    my understanding is that there will be? because we have a force to the right , b-field outwards of the screen, and by fleming's left hand rule we get a current heading upwards of the screen?

    but why is the answer no emf? is it because of mutual cancellations within the loop or something? or is my concept totally wrong?

    thanks
     
  2. jcsd
  3. Mar 24, 2010 #2
    Was the answer referring to the emf or to the current?
     
  4. Mar 24, 2010 #3
    the answer was referring to the induced emf.

    also i realise, the current would be heading upwards on the left side of the loop, and also heading upwards on the right side of the loop. so wouldn't the current be warring each other?
    also, the top and bottom sides are parallel to the motion so i assume no induced current in them?
     
  5. Mar 24, 2010 #4

    Born2bwire

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    By pulling the loop, you are giving the charges in the loop a velocity. By the Lorentz law then, there is a force acting on the conduction electrons in the wire. So the top and bottom parts of the loop will also experience a force and hence an EMF.

    However, because the magnetic field through the loop is uniform, the net EMF is zero. This is due to the phenomenon that you already surmised. That is, the EMF induced on the right hand wire is cancelled out by the EMF induced on the left hand wire. Like wise for the top and bottom. So while a straight section of wire does experience an EMF, it is cancelled out by the section's mirror image on the other side of the current loop.

    Now if the magnetic field differs on opposite sides, that is the magnetic field is nonuniform, then we can experience a net EMF since the net force on opposite segments is no longer zero.
     
  6. Mar 24, 2010 #5
    so lets say the positive charges get pushed downwards by the induced force, so it would be like attaching 2 batteries in parallel?
    - _-
    | |
    | |
    | |
    +_+

    so postiive potential at the bottom, negative on top..

    so the emf othe right side is lets say 5V. then the emf of the left side is also 5 votes.

    so by kirchoff's loop rule, it will be 5+-5 = 0? so net emf = 0?

    btw, if we assume the top and bottom to be thin rods, then the charges would not be moved by the lorentz force right? so no induced emf in those portions?
     
  7. Mar 24, 2010 #6

    Born2bwire

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    As to your Kirchoff loop simplification, that is correct. There still exists an EMF, but the EMF from opposite sections of wire oppose each other when we sum it all up. This would be true for a closed loop of arbitrary shape but it is easier to show it with a rectangular loop.

    Yes, if we assume a rod of zero radius then the electrons would not move, but there would still be a force acting on the charges. What technically happens here is that the Lorentz force induces the negative electrons to try and separate from the positive ionic lattice of the wire. As they separate spatially, there arises a restorative Coulombic force between the electrons and the ions. So with a one-dimensional wire, the EMF is countered by the Coulombic attraction that keeps the charges from moving in a direction other than along the wire. So it exists for a fleeting moment as it does a momentary amount of work but gets cancelled out in steady-state. Whether or not you consider it to be an EMF is probably just a technical point, probably not considering that EMF usually refers to the net effect.
     
  8. Mar 24, 2010 #7
    The explanations in terms of the Lorentz force are correct, of course. It is also possible to think in terms of Faraday's Law to help understand this one.

    Consider the most general form of Faraday's Law in integral form as follows.

    [tex]
    EMF=\ointop_{\partial S} E \cdot dl=-{{d}\over{dt}}\Biggl(\int_S B \cdot ds\Biggr)

    [/tex]

    Here it is clear that if the magnetic field is constant, and if the loop is constant, there is no net flux change as the loop is moved. Hence, the EMF must be zero.
     
  9. Mar 25, 2010 #8
    ah i see. thanks!
     
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