A long cylinder of radius $R$ is made from two different material. Its radius $r<r_0$ $(r_0<R)$ part is a material with superconducting transition temperature $T_1$, and its $r_0<r<R$ part is another material with superconducting transition temperature $T_2$ $(T_1<T_2)$.

Let us consider two different processes.

First, if we low down the temperature to $T$ $(T_1<T<T_2)$, and then add magnetic field $\vec{H}$ parallel to the axis of the cylinder.

Second, if we add magnetic field which $\vec{H}$ parallel to the axis of the cylinder, and then low down the temperature to $T$ $(T_1<T<T_2)$.

In the first case, we can easily know that only the surface at $r=R$ has superconducting current due to Meissner effect.

In the second case, two surface both have superconducting current. The current on the surface $r=r_0$ appears to keep the magnetic flux unchanged in the area $r<r_0$.

Are the currents on the surface at $r=R$ the same? I have two different answers and I don't know which one is right.

One answer is that the currents are the same in two case. Because of Meissner effect, $\vec{B}=0$ in superconductivity, the currents are the same in two case, because only current on the surface at $r=R$ can affect the magnetic field in the superconductivity.

Another answer is different by considering the change of magnetic flux. the variations of flux are different in two case, in the first is $B \pi R^2$ and the second case is $B \pi (R^2-r_0^2)$, so the currents are different.

  • $\begingroup$ That's an interesting question which I believe you're smart enough to resolve by yourself. It is just the problem of the London equation with a temperature dependent penetration length $\lambda_{1,2}\left(T_{1},T_{2}\right)$. You can find a lot of similar calculations in the book by London, Superfluids vol.1, macroscopic theory of superconductivity, 1961. Then it will be interesting to get your solution on these pages of course :-) $\endgroup$
    – FraSchelle
    Oct 16 '13 at 12:23
  • $\begingroup$ I forgot to say in my answer: IMHO conditions implicitly assume a type-I superconductor. Is it so, or I didn’t understand something? $\endgroup$ Aug 23 '14 at 8:18
  • 4
    $\begingroup$ This question's title could be much more descriptive. $\endgroup$
    – DanielSank
    Jan 1 '15 at 2:28

I hope the cylinder is, although long, not infinitely long? I mean the field must extend beyond the cylinder’s ends.

The problem is not well-formulated because the resulting state in the second case depends on how exactly the transition to superconductivity is effected.

If you cool the cylinder from inside, it will expunge the magnetic field outwards and ends with $B\pi r_0^2$ flux locked in the non-superconductive core, with the same (in average) field strength $B$ as before cooling. An extra $B\pi (R^2-r_0^2)$ flux will be pushed to the cylinder’s vicinity.

If you cool the cylinder from the surface, it will expunge the magnetic field towards the axis (into non-superconductive core) and the cylinder will contain the same $B\pi R^2$ flux as before cooling. In this scenario the magnetic field in the core will have the average strength $B\frac{R^2}{r_0^2}$, but there will be no field excess in the cylinder’s vicinity outside.

I am unsure about the current density at $r = R$; IMHO will depend more on the $z$ coordinate (along the axis) than on whichever of three cases will appear. It should be almost the same (sorry, forgot a necessary formula) in the middle of the cylinder, and will be stronger near its ends (where the field is diverted to avoid the cylinder) in all results except cooling-from-outside scenario of the of the second case where it will not depend on $z$.


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