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Rotating disc in a rotating magnetic field

  1. Jun 26, 2017 #1
    1. The problem statement, all variables and given/known data :A metallic disc of radius R is placed perpendicular to a magnetic field B.Both the magnetic field and the disc rotate with an angular velocity w about the axis of the disc.Find the induced EMF between the center and the circumference


    2. Relevant equations
    Motional Emf=Bvl;
    Flux=B.ds;
    emf=-d(flux)/dt;

    3. The attempt at a solution
    I suspect that it is BwR^2/2 as the flux is changing but the solution states that as there is no relative motion between the two (disc and field) no EMF is induced.
     
  2. jcsd
  3. Jun 26, 2017 #2

    cnh1995

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    That is correct.
    Edit: I pictured the disc and the magnetic field rotating along a differet axis. As Tsny says, it is actually a Faraday disc. So the emf should not be zero.
     
    Last edited: Jun 26, 2017
  4. Jun 26, 2017 #3

    TSny

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    The Faraday Disc "Paradox" generates as much controversy as it does emf. :oldsmile:

    I think your answer for the emf is correct. It is not zero.

    The notion of a "rotating magnetic field" is suspect. I assume that what you mean is that the magnet that produces the B field is rotating with the disc, as in this picture

    faraday1.gif
    (Source of picture: https://magnetsperiod2.wikispaces.com/How+does+a+generator+make+electricity?)

    When the magnet is rotating, the B field at any point in the lab is time independent. I don't see any meaning in claiming that the B field is "rotating".

    The amount of emf generated in the picture above is the same whether or not the magnet is rotating. You just need to make sure the copper disk is rotating and the external parts of the circuit (brush, wires, etc.) are at rest in the lab. [Or, you could keep the copper disk at rest and rotate the external circuit around the shaft (not so easy to do) and you would still generate the emf.]
     
    Last edited: Jun 26, 2017
  5. Jun 26, 2017 #4
    Ok.got it.The force on Electrons ,q(v*B) is zero,as v,the relative velocity, is zero and therefore no EMF is generated.Therefore Emf=0.
     
  6. Jun 26, 2017 #5

    TSny

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    If the phrase "both the magnetic field and the disc rotate..." means that the magnet rotates with the disk, then an emf will be generated. That is, you could light a bulb in an external circuit.

    For a nice demonstration see


    In particular, at time 1:45 in the video, you can see that an emf is generated when both the magnet and disc rotate together.

    [As I mentioned in my previous post, I don't think it makes any sense to say that the magnetic field is rotating when the magnet is rotating.]
     
  7. Jun 26, 2017 #6
    ...No I dont think thats what it is.I guess a nice way to interpret this would be to stand in a uniform magnetic field B along z direction(perpendicular) and then turn round and round and rotate with angular velocity negative .then from your perspective the field is rotating with angular velocity w.
     
  8. Jun 27, 2017 #7

    TSny

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    Is the z-direction perpendicular to the ground?

    If so, then from my perspective, I don't see why I would say that the field is rotating relative to me. From my perspective, the B field at any point in my rotating frame of reference would be time independent in both direction and magnitude. I can't see what it means to say that B is rotating relative to me in this case.

    But, we don't need to pursue that further if you don't want to.

    What's important is to answer correctly whether or not there is an induced emf for the situation stated in the problem. To do this, we need to understand the scenario described in the problem. Can you describe an actual experimental setup where you would say that "the magnetic field rotates with the disc"? This might help in arriving at the answer.
     
  9. Jun 27, 2017 #8
    A moving or rotating magnetic field does makes sense.For example lets take a charge q moving with velocity v in a magnetic field B. The force acting on it is q(vXB).Now if we start moving in the direction of v the velocity that we observe changes.Therefore the force on the charge qvB also changes.The only way this makes sense is that the field is also moving in an opposite direction and that in the Lorrentz force's quation F=(vXB) v is the velocity of the charge relative to the field i.e. the velocity of the charge in a frame in which the field is stationary.
     
  10. Jun 27, 2017 #9

    TSny

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    Yes
    Yes, the magnetic force qvB changes. But, in your new frame of reference, there will now be an electric field E pointing in the direction of the magnetic force. So, there will be an additional electric force qE that acts on the charge. The net force in the new frame will be essentially the same as the net force in the original frame (if we assume that we are dealing with nonrelativistic speeds). This is covered in special relativity, which I don't know if you have studied yet.

    In the Lorentz force equation F=q(vXB), v is always the velocity of the particle relative to your frame of reference and B is the magnetic field as measured in your frame. You never need to consider "the velocity of the charge relative to the field", which I'm not sure has much meaning.
     
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