5 ms·

Meissner effect: https://en.wikipedia.org/wiki/Meissner_effect https://en.wikipedia.org/wiki/Meissner_effect

lordnacho 3y ago
Yes but why does expelling the magnetic field make it hover stably instead of just falling out of the side of the field it's in?
- robinduckett 3y ago
  My laymen’s interpretation is that superconductors perfectly expel the magnetic field without distorting the field and so can sit in a position that is effectively locked and held in position. I could be completely wrong however.
  - bhattid 3y ago
    There's a noticeable distortion of the magnetic field near the superconductor. It's this distortion that keeps the superconductor stable - it's caused currents that appear on the surface which match and cancel* the surrounding field. *some of the magnetic field penetrates onto the surface of the superconductor, but internally, there's 0 magnetic field within the superconductor.
- bazodedo 3y ago
  https://en.wikipedia.org/wiki/Flux_pinning https://en.wikipedia.org/wiki/Flux_pinning
- space_fountain 3y ago
  I think that for true hovering demos the trick is to cool the super conductor to below its critical point while it’s already inside the magnetic field. Then any way you move it will immediately induce a reacting current in the superconductor that will move it back into place. Exactly the same place since it’s a super conductor and has 0 resistance. I think in these demos though it is the fact that one side is heavier and dragging that’s leading it to stay in place
- zrezzed 3y ago
  Afaik, your intuition is basically right! A superconductor exhibiting only the Meissner effect won't hover ("pin"). It's a defect (where small amounts of the field aren't expelled) that allows for the pinning to happen: https://en.wikipedia.org/wiki/Type-II_superconductor#Flux_pinning https://en.wikipedia.org/wiki/Type-II_superconductor#Flux_pi...
- bhattid 3y ago
  When a superconductor interacts with a magnetic field, currents generate at the surface of the superconductor that will produce their own magnetic field, which cancels the external one. The superconductor doesn't fall out to the sides because the field gets cancelled and there's no net force acting on the conductor. There can still be torque (i.e. rotations) for type I superconductors, and type II superconductors when they're fully superconducting. I'm not familiar with how the specific dynamics work though - I'm guessing it's related to gravity?

Yes but why does expelling the magnetic field make it hover stably instead of just falling out of the side of the field it's in?

robinduckett 3y ago
My laymen’s interpretation is that superconductors perfectly expel the magnetic field without distorting the field and so can sit in a position that is effectively locked and held in position. I could be completely wrong however.
- bhattid 3y ago
  There's a noticeable distortion of the magnetic field near the superconductor. It's this distortion that keeps the superconductor stable - it's caused currents that appear on the surface which match and cancel* the surrounding field. *some of the magnetic field penetrates onto the surface of the superconductor, but internally, there's 0 magnetic field within the superconductor.
bazodedo 3y ago
https://en.wikipedia.org/wiki/Flux_pinning https://en.wikipedia.org/wiki/Flux_pinning
space_fountain 3y ago
I think that for true hovering demos the trick is to cool the super conductor to below its critical point while it’s already inside the magnetic field. Then any way you move it will immediately induce a reacting current in the superconductor that will move it back into place. Exactly the same place since it’s a super conductor and has 0 resistance. I think in these demos though it is the fact that one side is heavier and dragging that’s leading it to stay in place
zrezzed 3y ago
Afaik, your intuition is basically right! A superconductor exhibiting only the Meissner effect won't hover ("pin"). It's a defect (where small amounts of the field aren't expelled) that allows for the pinning to happen: https://en.wikipedia.org/wiki/Type-II_superconductor#Flux_pinning https://en.wikipedia.org/wiki/Type-II_superconductor#Flux_pi...
bhattid 3y ago
When a superconductor interacts with a magnetic field, currents generate at the surface of the superconductor that will produce their own magnetic field, which cancels the external one. The superconductor doesn't fall out to the sides because the field gets cancelled and there's no net force acting on the conductor. There can still be torque (i.e. rotations) for type I superconductors, and type II superconductors when they're fully superconducting. I'm not familiar with how the specific dynamics work though - I'm guessing it's related to gravity?