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Counter EMF

  1. Feb 2, 2012 #1
    In the example diagram I have provided how can the counter EMF produced by the conductor moving through the magnetic field be determined?

    I am assuming the field is uniform and that the velocity and current in the wire is constant. My logic tells me that there is no net flux cut by the conductor as the magnetic flux is constant although I know this not to be true.


    Also, in terms of a solenoid coil if the applied current is alternating with time therefore creating a varying magnetic flux is there any counter EMF produced in the solenoid? Again, my text book resources would suggest there is but I cannot understand why.

  2. jcsd
  3. Feb 2, 2012 #2
    To me your diagram indicates a wire with a current I flowing into the plane of the diagram (the blue + and the blue letter I) This means the wire must be connected to a battery with an emf.
    As a result of this current in the wire it is moving to the left with a velocity v. The force on the wire is given by BIL
    The wire moving with velocity v in the magnetic field will generate an induced emf given by
    BLv. This emf will oppose the change producing it.... the motion of the wire... and so will be in the opposite direction to the applied emf (causing the current I).
    This is Lenz's law and is an example of BACK emf.
    The logic is that the wire will reach a velocity when induced emf = applied emf
    for simplicity, I have assumed the resistance in the circuit supplying the current to the wire = zero.
  4. Feb 2, 2012 #3


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    Hi Dan

    wasnt sure how to answer the first query, I'm picking Lenz's Law will answer that question :)

    as for your second query, as long as there is a varying current flow there will be a back EMF produced
    NOTE -- BACK EMF-- not counter EMF and of course there will be eddy currents generated in the metal core of the solenoid.
    Once again Lenz's laws will probabaly cover that

  5. Feb 4, 2012 #4
    Ok i've done a bit of digging with regards to the first question.

    Turns out that the counter EMF is actually related somehow to the electromotive force on a moving charge as opposed to the mechanical force i.e. F=qVBsinθ

    So in other words because the wire carries current and is in a magnetic field it has a mechanical force acting in the same direction as v (left hand rule). But also because the charge in the wire has a velocity v it experiences an electromotive force equal to qVBsinθ so using the left hand rule again where charge q is assumed to be negative and so is in the direction of v the resultant electromotive force is in the opposite direction to the current.

    So really the confusion is, if there is a force on the conductor in the direction of v due to F=BIL which is a mechanical force yet the same principle (F=qvB) results in an electromotive force on the charge. When applying the Lorentz Law how do you distinguish between mechanical and electromotive? why doesn't the force due to the velocity of the wire result in a mechanical force to move the wire?
    Last edited: Feb 4, 2012
  6. Feb 4, 2012 #5
    the force on the wire to the left is the result of the CURRENT in the wire, (this current must be coming from a battery connected to the wire).
    This force causes the wire to accelerate.
    The Back (counter) emf comes about because the wire is moving and is given by emf = BLv (this expression comes from emf = rate of change of magnetic flux linkage)
    The wire will reach a velocity when the back emf = emf applied to the wire causing the current I. (assuming there is zero resistance in the wire circuit
  7. Feb 4, 2012 #6
    But in the example the current is constant and the magnetic field source is uniform so where is the flux linkage?
  8. Feb 4, 2012 #7
    The wire is moving to the left (first sentence in your post). The wire is cutting through the magnetic field (the vertical yellow lines)
    That is what gives rise to the changing magnetic flux linkage =BLv (standard text book formula in this aspect of physics)
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