I'm studying the behaviour of superconductors in different cases, and I can't understand this.

We have a superconducting cylinder with a coaxial hole inside (vacuum). When the cylinder is cooled below it's critical temperature, and later a magnetic field is applied (case a), the superconductor will exclude the field even from the hole (B=0 inside the hole). However, if we first apply the magnetic field and then we cool down the cylinder (case b), the magnetic field will be excluded from the superconductor but not from the empty hole (as seen in the image below). Thus, depending on the procedure, we have two possible final states.

My question is, why is there this difference? Why doesn't the superconductor allow a magnetic field inside the hole in case a? It would still be excluding it from all the superconducting volume.

(Sorry, I don't know the source of the image, I took it from my professor's notes)

Field exclusion, expulsion and retention for different geometry superconductors following different procedures.


Since the magnetic field lines have to close themselves, when it transverses the superconductor it has to do it in a continuous fashion. This means, since the superconductor expels the field when in the SC state, the field gets trapped because it has no way of transverse the cylinder ring without opening the magnetic field lines (some geometric imagination is useful here ;).

Hope it's clear enough.

  • $\begingroup$ Yes, I see why it gets trapped, but my question is why doesn't the exact same thing happen when the magnetic field is applied after the phase transition from normal to superconductor qnd then turned off? $\endgroup$ – Ajayu Jan 12 '14 at 18:01
  • $\begingroup$ Because as I said, it has to go to the hollow part through the SC part first, and it has to be done continuously without opening the magnetic field lines which cannot happen since the SC part expels the field not letting it go through. You would see it happening in a very thin cylindrical shell (thinner than the penetration depth). $\endgroup$ – Ignacio Vergara Kausel Jan 12 '14 at 18:05
  • $\begingroup$ You are absolutely right! I just got what you mean. Thank you very much! ^^ $\endgroup$ – Ajayu Jan 12 '14 at 18:09

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