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Falling superconductor in a magnetic field

  1. Oct 25, 2011 #1
    Hello.

    Recently I have been inspired by the classic "drop a NdFeB magnet down a copper tube" demonstration, and I have been thinking about a superconductor falling in an ambient magnetic field. I want to determine the terminal velocity, if one exists.

    Now, I'm interested only in the effect of the superconductivity. I realize that if I perform this experiment, that air drag and evaporation of refrigerant would have an effect, but pretend that the superconductor is insulated quite well and falling in a vacuum. It becomes clear quite quickly that using an "ideal" superconductor is not appropriate for modeling this, simply because a perfect superconductor (one without a critical current density) would always have a terminal velocity of zero. Since we don't observe superconductors floating around on the Earth's weak magnetic field, this is a non-physical solution.

    If I were to use, say, a superconductor of mass M with a critical magnetic field of Hc, in a weak magnetic field of strength B orthogonal to the direction of the fall, then when the superconducting sample begins to fall (Vo=0), then the only screening currents will be those required to exclude the B-field from the sample. As the velocity increases, the B-field will induce a current in the superconductor. In a perfect superconductor, even a small velocity and weak field would cause a large current which moves any energy associated with the velocity quickly into the magnetic field associated with the screening current. However, for a real-world type I or II superconductor, what happens? If the Hc is large enough relative to B and M is there a terminal velocity? If not, what happens?

    I have tried searching for an answer to this, and I found this interesting paper: arxiv.org/pdf/physics/0609141. I'm uncertain if this paper is implying that once the screening currents associated with moment through a particular field are established that no more energy can then be absorbed by the field (and thus there is no terminal velocity) or if that implication is just associated with the particular geometry (a superconducting tube around a magnet) that they are using.

    Any help, thoughts, ideas, etc. would be greatly appreciated.
     
  2. jcsd
  3. Oct 25, 2011 #2

    BruceW

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    hmmm. The paper was about a magnet falling through a superconducting pipe. If we imagine that we move along with the magnet, it would look like the superconducting pipe was falling upwards through an ambient magnetic field. This is now equivalent to the question you were asking about a superconductor falling through an ambient magnetic field, right? And in the paper, it found that the magnet simply fell in free-fall, accelerating at g. So from the the magnet's point of view, the superconductor would be falling at g.

    So the answer is that the superconductor would not be affected at all by the ambient magnetic field. Does this look right? I don't know for certain if it is ok to view the situation from another reference frame that is non-inertial with respect to the first one...
     
  4. Oct 26, 2011 #3
    I think that it is okay to switch the frames of reference so that the superconductor is moving and the magnet is not. Since we are simply considering forces it doesn't really matter which of the objects is accelerated. However, there is one other big difference between the a sample of superconductor falling in a magnetic field and a magnet falling into a superconducting tube, and that is the tube geometry. I'm not so sure that it is correct to consider the inside of the tube (away from the ends) as equivalent to an open uniform magnetic field. The magnetic field lines in each case look nothing alike.

    Thinking more about it, I have noticed that the arxiv paper purposely excludes any magnetodynamic effects (in other words, the magnet's velocity is not a factor in their calculations). Their justification for this is that they are calculating for an ideal superconductor, where the screening currents will always exists at any moment in time the same as if the magnet was not moving at all. Of course, this will result in no drag force, since by definition drag forces are in some way proportional to the velocity.

    Any magnetostatic calculation of a falling magnet will result in no drag force, since it is only magnetodynamic effects that cause the drag in the first place. I'm afraid that I don't know much about superconductor magnetodynamics, but I do know that using a magnetostatic approximation for a problem dependent on magnetostatic effects is likely to give the wrong answer!
     
  5. Oct 26, 2011 #4

    BruceW

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    Yeah, the geometry of the superconductor might be important. I am not very knowledgeable about superconductor physics. I should know how to calculate it for a perfect superconductor (if I find my lecture notes somewhere). But you're interested in a not-perfect superconductor, right? So I don't think I'd be much help apart from making a speculative guess.

    Its an interesting question, I hope someone replies who has worked it out before.
     
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