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Faraday disc paradox; is there an equivalent situation with E fields?

  1. Nov 3, 2014 #1

    sophiecentaur

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    The Faraday paradox is a hard one to get ones head around and I was wondering if there is a similar scenario, involving an E field where there is / or is not an induced emf?
     
  2. jcsd
  3. Nov 4, 2014 #2
    Do you mean situation where variable E-field induces emf or B-field induces emf?
     
  4. Nov 4, 2014 #3

    sophiecentaur

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    The Faraday 'paradox' is because moving the disc or magnet produces an emf, or not - depending. The Field seems to behave as if it has its own reference frame (I think). So I was looking for an example in which the E field appears to have its own frame, too, I guess, so that there could be the equivalent of emf/ no emf. I am being vague because I really have no idea of a scenario - there is often a sort of E-H symmetry and I thought there may be something here.
     
  5. Nov 4, 2014 #4
    The key to understanding the Faraday disc lies in understanding Arago's magnetism of rotation, where a copper disc is rotated and is seen to affect a nearby magnet. This shows that merely spinning a copper disc generates a magnetic field.

    We now know that an electric current in copper is caused by motion of mobile charge carriers (electrons) and this induces an observable magnetic field proportional to the rate at which these electrons move past the observer.

    So causing electrons to move past the observer by moving the copper itself is equivalent to causing electrons to move through the copper due to an applied electric field gradient.

    In a Faraday disc there is the further complication of the Lorentz force.

    If the electric current in the copper disc is radial i.e. from the centre of the disc out to the edge, in the presence of a magnetic field which has a component parallel to the axis of rotation of the disc then the electrons will experience a force which will accelerate them tangentially in the plane of the disc and at right angles to both the current and the external magnetic field.

    It doesn't make any difference if the magnet is fixed to the disc or if the disc rotates freely and the magnet is fixed, all that matters is that the magnetic flux is perpendicular to the motion of the disc and the current.
     
  6. Nov 4, 2014 #5
    Famous "paradox" involving reference frames and magnetic/electric fields is this one
    Albeit there's no induced emf (steady state fields )
     
  7. Nov 4, 2014 #6
    finally someone made a thread like this , ive been thinking about it too many times even here on PF a while ago , sophie probably wouldnt't remeber as it was under a different name. Blackadder is great , thanks sophie for the recomendation :)

    about the faraday disc itself I think sophie you already know how it works , due to the difference in tangential speed from the inner portion of the disc to the outer one the b field makes a net electron flow from center to outside or vice versa depending on the applied b field polarity , the thing i couldnt get my heada round too is why rotating just the magnet doesnt produce any current while spinning the magnet together wioth the copper disc synchronously, produces the same current as when the magnet is stationary while the disc spins with respect to the magnet, really strange. the only logical thing that comes to mind is that the b field of a permanent magnet is stationary and for an even geometry magnet the same at all places so whether one spins the magnet or keeps it stationary doesnt make a difference in the field lines nor the pole orientation nor the field strength so basically the copper disc sees a stationary field no matter whether the magnet spins or not. could you please explain more about your imagined E field situation with respect to the faraday disc , it would be very interesting to know your thoughts more , thank you :)
     
  8. Nov 4, 2014 #7

    sophiecentaur

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    That isn't what I'm looking for. I am looking for an Electrical equivalent of the apparent Faraday disc paradox. I don't care if a disc is involved or not or whether there needs to be rotation. I have a feeling it is a piece of nonsense, actually but I was hoping for some informed input.
    The way the Faraday Disc works is no in doubt (is it?).
     
  9. Nov 4, 2014 #8
    are you thinking in terms of whne the magnet is stationary and the copper disc is stationary no current induced even though the static b field is present and then as soon as the copper disc starts to rotate a current is induced , so you wanted this situation where you have a E field but in state A its just there having no effect even though the field lines are there and in state B it induces a current or sort of ?
     
  10. Nov 8, 2014 #9

    rude man

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    This isn't my understanding. The Arago magnetism of rotation requires the presence of a magnetic field in addition to a spinning non-ferrous disc like copper. The basic phenomenon is eddy currents induced by the B field.

    To my knowledge, a copper disc rotating by itself does not generate a magnetic field. The Cu atoms are charge-neutral. The electrons are bound to the nuclei unless an E field separates them to cause currents to flow, or a B field to apply the Lorenz force to them.
    [/quote]

    http://physics.kenyon.edu/EarlyApparatus/Electricity/Aragos_Disk/Aragos_Disk.html
     
  11. Nov 8, 2014 #10
    its not a paradox he just couldnt prove it
     
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