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Meissner effect and superconductor

  1. Nov 14, 2007 #1

    From what I understand the Meissner effect occurs when a current is induced in a superconductor, from a magnet, giving rise to a magnetic field from the superconductor that opposes the magnet.

    What stops the magnet from being completely repelled away from the superconductor when it is hovering over it ?

    Secondly why do superconductors exhibit this effect and not ordinary conductors?

    Help needed please

  2. jcsd
  3. Nov 14, 2007 #2


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    First of all, superconductors are NOT "perfect conductors"; their properties are quite different from e.g. very good metals and even a (hypothetical) metal with infinite conductivity would NOT be a superconductor. Hence, trying to understand superconductors by comparing them to normal metals is very likely to just be confusing.

    A levitating magnet WILL under ordinary circumstances be repelled from the superconductor. However, most demonstrations of the Meissner effect use the high temperature superconductor YBCO which is a type II superconductor. If a type II superconductor is already in a magnetic field when it goes through the superconducting transitions (i.e. when it is cooled) it will "trap" some of the magnetic flux forming "tubes" (pinned vortices with norma center) that goes through the whole superconductor. The vortices will always form in such a way that they minimize the energy of the system.
    These "tubes" are difficult to move meaning the magnet becomes "stuck" in exactly field configuration in which the superconductor was cooled, the magnet is essentially "resting" in an energy minimum. If you try to move the magnet sideways if feels like it is moving through something sticky, and when you let go it will usually move back into its original position (unless it can find another minum nearby).

    Now, if you cool the YBCO in zero field and THEN try to levitate the magnet there is no minimum and the magnet is repelled. The same is true for most ordinary superconductors (i.e. metals that become superconducting at low temperatures, e.g. aluminium) since they are type I superconductors which do NOT form flux tubes in this way.

    Cooling the superconductor+magnet in just the right way is actually quite tricky, it is neccesary to use a spacer between the two during cooldown or otherwise the magnet will just rest on top of the superconductor and not levitate.

    The reason why ordinary conductors do not exhibit this effect is simply that they are never diamagnetic which is what is needed to see this effect; they do not repell dc magnetic fields (copper, gold etc will not even perturb a dc magnetic field).
  4. Nov 14, 2007 #3
    That certainly contradicts my experience.

    I think of type-II as basically "only superconducting in a particular plane", hence while they repel a magnet approaching from above, they only resist/disipate sideways motion, hence they make for very stable levitation (but possibly inefficient maglev trains?)

    From what I've read about type-I superconductors, you just need a concave top surface in order that any magnet you try to levitate does not "slide" off (not sure if that's the right word when there isn't surface-contact to begin with).

    Now, those are of course behaviours of a perfect conductor, whilst the Meisner effect (flux expulsion) is something completely else. I'm pretty sure I also remember this working with a YBCO demonstration kit (the only problem being that sometimes the two would "stick" seemingly due to formation of ice, but a small nudge would suffice to provoke levitation), and that's supported by page 417 of "Introduction to High-Temperature Superconductivity" on google books.

    [In fact, for all the effort you went to in distinguishing superconductors from perfect conductors, strangely it seems that you have described phenomena of the latter rather than the former.]

    New question: it is http://www.physics.ubc.ca/~outreach/phys420/p420_96/bruce/meissner.html" that superconductors are perfect diamagnets (as opposed to perfect conductors). Are ordinary diamagnets conductive?
    Last edited by a moderator: Apr 23, 2017
  5. Nov 14, 2007 #4


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    A zero cooled type II superconductor while essentially behave like a type I superconductor at moderate fields; it is only when it is cooled in a magnetic field that the effects of vortices become important.

    Also, imagine that we have a material that we can "switch" between an insulating and perfect conducting state. Now, place this imaginary material in a magnetic field and "turn on" the perfect conductivity. What would happen (if it could be done) is that the magnetic field would be "frozen" inside the conductor, i.e. it would not be expelled. Moreover, the field would remain frozen there even if the source of the field was removed.
    This is of course not what happens with a superconductor (be it type I or II); when it goes through the superconducting transition it will try to expell all the field (which is why one can't consider a superconductor to be a "perfect conductor").
    However, in the case of a type II superconductor it is often more energetically favorable to create vortices with normal cores instead. This is what happens with YBCO (YBCO tablets also contain a lot of defects such as grain boundaries etc; meaning there are plenty of good pinning sites for the flux).

    I have demonstrated this with YBCO many times (when I was a PhD student I used to teach courses on solid state physics and low temperature physucs/superconductivity etc so I did it 2-3 times a week) and belive me, it is DEFINTLY possible to "fail" when it comes to making the magnet levitate.
    The different effects of zero field/in field cooling are also easy to demonstrate; if you pour the nitrogen on the YBCO tablet and then try to find a stable position for the magnet it simply won't work (I used to challenge my students to do this).

    Also, I am not sure what you mean by "planes"? While the cuprates are certainly layered convetional type II superconductors are not; hence these effects would happen in all directions for e.g. lead or niobium.

    edit: About diamagnesitm. Normal metals are extremely weak diamagnets. Some gases are diamagnetic but the effects are generally quite weak. Liquid oxygen is rather fun to play with since it is a diamagnetic liquid (you can actually see small droplets of oxygen forming when playing with a Meissner kit, there was even a rather fun paper in Nature about this a few years ago)
    Last edited: Nov 14, 2007
  6. Nov 14, 2007 #5
    How then do you explain http://www.physics.ubc.ca/~outreach/phys420/p420_96/bruce/ybco.html"? Conspiracy?

    Even if you're unwilling to accept my experience, if you want your own to be taken seriously then you should at least respond to my citation of a *textbook* (which you can conveniently access online) that directly contradicts statements of yours. :grumpy:
    Last edited by a moderator: Apr 23, 2017
  7. Nov 14, 2007 #6


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    No conspiracy. They simply cooled the superconductor with the magnet in place and then removed the magnet. Since the flux is "frozen" in place the field configuration stays the same. Hence, if you put the magnet back it ends up in the same place again.
    This is is what I meant when I wrote "and when you let go it will usually move back into its original position (unless it can find another minum nearby)" in my first post.

    Also, a magnet will certainly float above a type I superconductor. But unless you somehow "engineer" the shape of the field the position will be very unstable (whereas for a type II superconductor it is very stable). It might be possible if you get the shape right (it is impossible for two strong permanent ferromagnets, but they are of course not diamagnetic) but I have never seen it being done (but that could simply be because you need to go well below LHe temperatures to get below Tc for a type I superconductor).
  8. Nov 14, 2007 #7
    What I'm trying to understand is how perfect diamagnetism gives rise to the other properties that superconductors have, specifically the perfect conductivity?

    (As for liquid oxygen, http://www.newton.dep.anl.gov/askasci/phy99/phy99171.htm".)
    Last edited by a moderator: Apr 23, 2017
  9. Nov 14, 2007 #8


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    I (and many others) would argue that it is wrong to say that superconductors exhibit "perfect conductivity" (or are "perfect conductors"). It is much better to say that they exhibit zero electrical dc resistance and perfect diamagnetism.
    The reason is that if you write down the equations for a perfect conductor, deriving e.g. equations for the time depence of the B-field, you get the behaviour I described above; i.e. you do NOT get the Meissner effect (you should be able to find this is various books).

    This is a well known "problem" when modelling superconductors; you can't even get a good approximation of how they behave if you simply set the conductivity to a "a very high value" (or indeed infinity); the current- and field distributions are very different from what you would get in the "perfect conductor" that is often used in e.g. modelling of electrical circuits.
  10. Nov 14, 2007 #9
    You call that "no conspiracy" when the context of that animation (which explicitly refers to doing the exact same experiment http://www.physics.ubc.ca/~outreach/phys420/p420_96/bruce/copper.html" ordinary conductor) specifically implies they performed no such tricks beforehand? :uhh:

    What does it take? Why would books agree with me that YBCO will lift a magnet from its surface as it is cooled, if as you say that is false? Despite all these contradictions you give the impression you at some point knew a lot about superconductors, did you produce them yourself (since perhaps high impurities would explain your differing observations)?
    Last edited by a moderator: Apr 23, 2017
  11. Nov 14, 2007 #10


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    First of all, this experiment is impossible with a normal conductor (unless it is strongly diamagnetic and copper is not, there is even a theorem about this called Earnshaw's theorem). Also, if you read the page it says "cause the magnet to hesitate as it falls". It does not levitate.

    Secondly, I have done this experiment hundreds of times. Yes, if you did it with a type I superconductor the magnet would levitate (and fall off) but with YBCO the situation is different. Unless a spacer is used the field tends to "freeze" in a configuration where the magnet does not leviate freely, the best you can usually get is that the magnet "rests" on a corner.

    I have no idea why they have written the page the way they have. I suspect it is because they did not want to go into a lot of details about vortex-physics. You need to ask them, not me.

    And no, I did not produce the superconductors (as far as I remember we bought the kits from Poland).
    I spent 5 years (2000-2005) as a PhD student working on electrical transport properties (Josephson junctions and SQUIDs) of grain boundaries in YBCO but I only worked with thin films (on bicrystal substrates); over the past two years I have mainly been working with conventional superconductors (mainly niobium and aluminium) . I might have forgotten a few things about YBCO, but not THAT much.
    Also, I can assure you that any piece of YBCO of that size is most definitly polycrystalline meaning there are plenty of pinning sites for the flux so sample preparation does not really matter.
    I have also done exactly the same experiment as they show in the animation so I know how it is done (the fact that you can rotate the magnet as long as you do not alter the field configuration is quite neat, and can actually be used to make magnetic bearings).
  12. Nov 14, 2007 #11
    We can both take the point as conceded, since now you're only disagreeing in degree, and I've also now seen (on youtube with a larger and perhaps stronger magnet) how the flux (vortex) trapping effect can be strong enough to lift a superconductor. Perhaps the weight of the magnet is also relevant to whether the (Meissner) flux exclusion effect occurs (explaining why the smaller magnets often at least levitates up onto one edge and sometimes completely as in one of the animations I linked), much like the applied force (and geometry) seems to determine whether a strong magnet is completely repelled from a (zero field) superconductor (as from a perfect conductor) versus becoming locked into position (as due to vortex rearrangements)?

    Back to my question then, theoretically does perfect diamagnetism imply "zero electrical dc resistance"?
  13. Nov 14, 2007 #12
    Last edited: Nov 15, 2007
  14. Nov 27, 2007 #13

    What do you mean by "Imply"?

    Do you mean "leads to" or something more along the lines of "exists because of"?

    This is spinning my head a bit, especially since I have been taught that a superconductor = a perfect conductor + meisner effect. And that the meisner effect is what creates the diamagnetic moment in the SC.

    On your link provided they state

    "Perfect diamagnetism implies zero resistance that we have measured plus and added effect called "Flux Expulsion".

    Which throws in flux expulsion (meisner effect) at the end. Seems a bit circular to me.

    I know I've raised more questions, but I look forward to your comments.
  15. Nov 28, 2007 #14
    Are you sure Earnshaw's Theorem applies here? It simply states that an electrostatic trap is impossible. This theorem is a result of Possoin's Equation in an area of zero charge. Which reduces it to Laplace's Equation. Functions satisfying Laplace's Equation admit no maxima or minima and you have your proof. However, I thought that the arguement for the meisner effect in a perfect conducter (which I know nothing about) was that it also could maintain nondecaying currents. Thus allowing for a magnetic trap.
  16. Dec 1, 2007 #15
    If I may, I shall take a moment to respond to the question posed by the OP rather than getting embroiled in arguments about the nature of Type I and II superconductors and diamagnetism. In fact, the original question is significantly simpler to answer, as I understand it.

    The 'hovering' magnet does not continue to move up away from the superconductor because the field from the superconductor falls off as some function of space, just as any field will tend to where it is not held uniform between, for example, parallel plates/poles. The result is that the force on the magnet from the field of the superconductor reduces with height, eventually balancing the magnet's weight. At this point, the magnet will not continue to rise as there is no net force on it.

    There - it seems to be a question of Newton's laws and the nature of fields, not the nature of superconductors. The answer to the second question (why the effect doesn't occur in normal conductors) is however given by the nature of superconductors as you have discussed them.
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