From Meissner effect we know that the magnetic field $\vec{B}$ is zero inside the superconductor. Since $\vec{B}=0$ inside the superconductor (ignoring the tiny penetration effect for the moment), from $\vec{B}=\mu_0(\vec{H}+\vec{M})$, we get that the magnetic susceptibility $\chi$ is given by $\chi=M/H=-1$. Hence, a superconductor behaves as a perfect diamagnet.

However, from Faraday's law $\vec{\nabla}\times\vec{E}=-\frac{\partial}{\partial t}\vec{B}$, it can only be shown that the magnetic field $\vec{B}$ inside a superconductor is constant. And in fact, it is possible to lock magnetic field lines inside a superconductor if the magnetic field were applied before the material was cooled below $T_c$. This seems to suggest $\vec{B}\neq0$ inside the superconductor!

How is the conclusion from the second paragraph consistent with the first (i.e., existence of Meissner effect)?


First off, there's no contradiction between Faraday's law and the Meissner effect. Faraday's law says $\mathbf{B}$ is constant in time in a superconductor, and the Meissner effect says that constant is zero.

There are two scenarios where you can get a magnetic field "inside" a superconductor. The first is that if you have a superconductor with a hole, such as a ring, you can put a flux through the hole. This doesn't violate the Meissner effect because there's no superconducting material in the hole, and Faraday's law ensures this flux is constant in time.

The second scenario is that if you apply a strong enough external magnetic field to a Type-II superconductor, it can be thermodynamically favorable for non-superconducting regions to form inside the superconductor, called flux vortices. They act just like the hole in a ring, allowing flux to pass through.

  • $\begingroup$ If $\vec{B}$ is constant why is it expelled from the interior? This clearly shows that $\vec{B}$ is changing $\endgroup$ – mithusengupta123 Oct 8 '18 at 11:07
  • $\begingroup$ @mithusengupta123 It gets expelled as the superconductivity turns on, so I'm not quite sure. The dynamics during that are quite complicated. $\endgroup$ – knzhou Oct 8 '18 at 11:17

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