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Current in a Superconducting Loop

  1. Sep 13, 2009 #1
    Suppose that, a sudden current pulse has been generated (by means of a Magnet) in a Super-Conducting Loop (or coil). It is commonly said that the current remain there forever. Is is true?? I mean will the current remain in the loop undiminished till t=infinity? But theoretically it isn't possible right?
     
  2. jcsd
  3. Sep 13, 2009 #2
    by theoretically not possible I do not mean about the finite small resistance (however small it might be) and instead the radiation effects of the accelerating electrons in the Coil (since it is not a straight conductor)
     
  4. Sep 13, 2009 #3
    It is not possible to induce a current loop in a superconductor (except in type II where loops can be forced in). In persistent mode magnets, a short section of conductor is made normal and a voltage is applied, current is pumped in, and then the conductor is made superconducting again. See Meissner Effect in
    http://en.wikipedia.org/wiki/Meissner_effect
     
  5. Sep 13, 2009 #4
    The superconductor itself is in the form of a ring and if we throw a magnet through the center of this ring current should inevitably be formed in the ring right??
    I am sorry for framing the question that way. But the point in the question was not that at all. If there is a dc current in a circular superconducting ring, will it die out due to radiation effects or will it persist forever??
     
  6. Sep 14, 2009 #5

    ZapperZ

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    It will persist "forever".

    In fact, this phenomenon of persistent current has also been discovered in normal metals under very specific conditions:

    http://physics.aps.org/articles/v1/7

    Zz.
     
  7. Sep 14, 2009 #6

    Vanadium 50

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    You don't have radiation from the ring. To get radiation you need a changing moment, and the ring's moments are all constant. For a perfect circle, you have a magnetic dipole moment that's non-zero and all the other moments are zero.
     
  8. Sep 14, 2009 #7
    hello ashokakand-
    I think the current is on the surface of the ring, not in the superconductor of the ring.
    Bob S
     
  9. Sep 14, 2009 #8
    The electrons moving in the Ring are not moving in a straight line (on an average). In other words they are accelerating. Any accelerating charge should radiate em field. This is a standard result in classical electrodynamics. So invariably these electrons also should radiate em energy. Shouldn't they?

    Please correct me if I am wrong.
     
  10. Sep 14, 2009 #9

    Dale

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    Yes, this is another example of how classical physics is wrong. In a similar manner classically electrons orbiting a nucleus should continuously radiate thereby losing energy and falling into the nucleus.

    Like atomic orbitals, a real understanding of superconduction requires quantum mechanics. In QM you accurately conclude that in each case the electrons are in a state where they do not accelerate (i.e. they are in a quantum "cloud" that is spread out over the entire atom or superconducting current ring) and therefore do not radiate. Weird, but that is how the math falls out.
     
  11. Sep 14, 2009 #10
    They do. This is called synchrotron radiation. A 10 GeV electron in a 1 Tesla field radiates x-rays. But the radiation energy varies as gamma-cubed. See
    http://geant4.web.cern.ch/geant4/UserDocumentation/UsersGuides/PhysicsReferenceManual/BackupVersions/V9.1/html/node49.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  12. Sep 14, 2009 #11

    f95toli

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    No, they don't. Read Dalespams reply.
    The reason is that we are not talking about normal electrons here; some of the properties of a condenstate in a superconducting ring are very different from what you would expect from a system of normal electrons. The fact that that normal electrons radiate when accelerated in a ring is irrelevant.

    Also, introducing a current in a superconducting ring is not very difficult; all you need is some form of inductive coupling. You only need a heatswitch if you want do e.g. drive a solenoid from a dc-source.
     
  13. Sep 14, 2009 #12
    I agree that you can get a current in a superconducting ring, if you use a heat switch. If you try to induce a current in a superconducting ring without a heatswitch (e.g., by dB/dt), the current is actually on the surface of the superconductor. If you can induce 10-GeV electrons in a superconducting ring (see my previous post) they will radiate.
     
  14. Sep 14, 2009 #13

    Vanadium 50

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    I've got to disagree with a lot of people here.

    What is the radiation from a purely classical charged ring spinning around its axis? I maintain that it is zero.

    Radiation requires a changing multipole moment. This system has exactly one non-zero moment, the magnetic dipole moment, and it's static.

    The argument that a single accelerating charge radiates neglects a very important fact: this is not a single accelerating charge. Suppose I had a rotating plate, and I place a charge at 0 degrees. This plate radiates. Now, I put another charge at 180, and now I no longer have any electric dipole radiation. I do still have quadrupole radiation, so I place charges at 90 and 270. Now the highest surviving moment is electric octopole, which can be zeroed out with charges every 45 degrees. In the limit of a continuous ring of charge, there is no radiation: radiation from any point on the ring is canceled by radiation from other points on the ring.

    This has nothing to do with superconductivity. It's true if you simply glued charges to a ring. It's all in Chapter 9 of Jackson.

    Now, you might ask, "yes, but we know charges aren't continuous: they're discrete - doesn't that wreck your argument?" Yes, but they are really, really close to discrete: you have millions of electrons, so it all cancels except for the million-pole terms, and those radiate very, very little power. Possibly the age of the universe per photon.
     
  15. Sep 14, 2009 #14

    f95toli

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    I am not sure what you are talking about here. 10 GeV electrons? Presumably we are talking about inducing current in a solid here; i.e. a loop made of e.g. niobium cooled to a sufficently low temperature (<9 K) for it to be superconducting; not an accelerator where free electrons (or in this case Cooper pairs) are flying around.

    Yes, if you try to to induce a current by using a fast pulse the current will flow on the surface; but that is a technical detail that can be solved by simply using a low enough frequency or a superconductor much thinner than the penetration depth (i.e. make the loop by etching a structure in a thin film).
    If you REALLY want to control the current you can just start incorporating Josephson junctions in the loop; if you use three junctions (two identical junctions and one somewhat smaller) and bias the loop at half a flux quanta you can even get a current flowing in both directions at once (this is how a phase qubit works).

    The point is that there are lots of different ways of inducing a current in a superconducting loop; but the details of how it is actually done has nothing whatsoever to do with the OPs question.

    Vanadium: Good point.
     
  16. Sep 14, 2009 #15

    ZapperZ

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    The original question has NOTHING to do with charges going around in circles. I don't know why this thread is going along that path.

    Supercurrent in a loop is nowhere near being similar to that scenario. As has been mentioned already, because of long-range coherence, the supercurrent is everywhere at once!

    Zz.
     
  17. Sep 14, 2009 #16

    Dale

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    Very nice argument. I am going to shamelessly steal it whenever this question crops up again.
     
  18. Sep 14, 2009 #17
    Yes, This is a valid point. But that brings another question. If the conductor is of some arbitrary shaped closed loop (neither a circle nor any closed figure with any particular symmetry, it can be non-planar also). Is it correct to say that the argument still holds? Please give reasons also??
     
  19. Sep 15, 2009 #18

    Vanadium 50

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    Calculate the multipole moments. Do they change with time? Then you have radiation. Simple as that.

    Again, note that this is a purely classical result. You don't need to know any properties of superconductors.
     
  20. Sep 15, 2009 #19

    Dale

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    To be specific, you don't need to know any properties of superconductors to understand why the loop does not radiate. Even a loop of resistive current will not radiate (other than black body, of course).

    You still need to know the QM properties of superconductors to explain superconductivity itself.
     
  21. Sep 15, 2009 #20

    Vanadium 50

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    Provided you keep the current going with a source. If you let it die out, you will have a changing magnetic dipole, which will radiate.

    Of course. And even then, it's not easy. I understand people get Nobel prizes for figuring it out.
     
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