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Why is there an induced EMF in this case?

  1. May 24, 2015 #1
    Hey people

    I was wondering why is an emf induced in a loop which moves perpendicular to a uniform magnetic field. I mean, say the loop is a square of area r^2, and moves in the xy plane, where there is a uniform magnetic field B in the z direction. Why will an emf be induced? The flux remains constant right? So if d(phi)/dt is 0, why is an EMF still induced??

  2. jcsd
  3. May 24, 2015 #2


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    Staff: Mentor

    Does the loop remain completely inside the region where the B field exists? Or, to put it another way, does the B field always "fill" the loop? If this is true, then you are correct, there should be no induced EMF. Can you describe the example or situation that claims there is an EMF?
  4. May 24, 2015 #3
    Thanks a lot!

    Yes, the loop does remain completely within the field. It was just "coming" to my mind that an emf will be induced.... however, I think it was because of a flawed intuition!...... Anyway, will it be right to conclude that there will be no net EMF in the loop, but there will be equal and opposite EMFs (given by Blv) in the opposite sides of the loop?

    And on a related note: If I move a straight piece of conductor in a uniform magnetic field, again an EMF is induced (proved using B*l*v I think). However, where is the flux change in that? I mean, there is no flux to begin with, since there is no loop- so how can we prove the EMF using Faradays Law?
  5. May 24, 2015 #4
    If a straight conducting rod is moving in a uniform magnetic field, in a direction perpendicular to the field, the induced emf is not due to a change in flux. It is because charges collect at the edges because of the Lorentz force. This is called motional emf.
  6. May 25, 2015 #5
    But it is due to a change in flux. The conducting rod forms only one part of the circuit that forms a complete loop.
  7. May 25, 2015 #6
    No, it is due to charges moving in a magnetic field. Change in flux is a rule of thumb. Notice that if the charges all move in one direction (i.e. down, not clockwise) then the emfs of each half of the loop will cancel. You'll just have some charges collecting at the top/bottom of the loop.
  8. May 25, 2015 #7
    I have no idea what any of this means. Apparently Faradays Law [itex]\oint E \cdot s +d\Phi_{B}/dt[/itex] is a "rule of thumb".
  9. May 25, 2015 #8


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    Staff: Mentor

    Faraday's Law in integral form requires a closed path. It can't be applied to an open path.
  10. May 25, 2015 #9
    Yes, that's why I used the [itex]\oint[/itex]. The open path wasn't made clear to me in the post. Thanks for clarifying.
  11. May 25, 2015 #10
    What this means is that the physics is simply about moving charges in magnetic fields.

    Yes, apparently Faraday's Law is a rule of thumb, since it only describes conveniently closed paths. Do you think the physics breaks down if I only look at each half of the path independantly?
  12. May 26, 2015 #11
    Again, say what you mean. I'm not a mind reader. Do you have anything to add beyond what jtbell said?
  13. May 26, 2015 #12
    I suppose I am not being eloquent enough for you. My apologies.

    I am trying to make you get it, namely that Faraday's Law is a rule of thumb for predicting the net result of the Lorentz forces on the charged particles in a closed loop of conducting material. This is not much beyond what jtbell and Chandra said, but it seemed like you were still having difficulty in accepting it.

    To re-illustrate the example I used: Say you have a uniform magnetic field pointing out of the page everywhere. If you drag a loop of conducting wire that is normally oriented to it towards the right, there will be a downward force on the positive charges, and an upward force on the negative ones, as per the Lorentz force law.

    The charges will migrate up and down respectively, establishing an electric field that will balance the effect of the magnetic field and movement, i.e. charges will migrate until the Lorentz force F = q [ E + v x B ] is equal to zero.

    So you will have a motional emf as Chandra said, a voltage difference between the top and bottom of the loop. You will not have an emf around the circuit though, since the endpoints of any closed path will have the same voltage. This is the meaning of Faraday's Law, as jtbell said: it only applies to the whole closed loop -- when flux is constant as in this case there is no net emf around the loop.

    I hope that helps!
  14. May 26, 2015 #13
    If it is a complete loop, and the loop is entirely within the uniform magnetic field, moving the loop in a direction perpendicular to the magnetic field does not change the flux, and there is no EMF. EMF is there only if the flux changes, as stated in Faraday's law.
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