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)?


2 Answers 2


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 $\endgroup$ Commented Oct 8, 2018 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
    Commented Oct 8, 2018 at 11:17

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 $B\neq 0$ inside the superconductor!

Wrong! The magnetic field is not locked inside a superconductor. You are confusing between a superconductor and a perfect conductor. Superconductors (SC) have two mutually independent properties, namely, zero resistivity and perfect diamagnetism. One does not follow from the other.

What happens in a superconductor? Go to the figure of page 19 of this pdf.

Take a superconducting material above $T_c$. First, you cool it below $T_c$ such that it enters the superconducting phase. Now if you apply a magnetic field, you'll see that the magnetic field lines are repelled away by the sample.

Now, do it in reverse order i.e., you take the same superconducting material above $T_c$. First, apply a magnetic field which can pass right through the sample. Now, cool it below $T_c$. You'll again discover that the field lines are completely expelled from the interior so you get back the previous configuration with no fields inside.

Therefore, a SC does not remember that in what order it was subjected to a magnetic field. The two operations i.e. first apply H and then cool below $T_c$ and first cool below $T_c$ and then apply H, leads to the same final configuration of the SC with zero field inside. SC has no memory of the initial condition and these two experimental operations "commute". However, for a perfect conductor, fields can get locked inside.

Conclusion The magnetic field is always zero inside a SC at $T=0$.

From Ohm's law $\textbf{E}=\rho\textbf{J}$, the zero resistivity property (for finite current density ${\bf J}$) only dictates that $\textbf{E}=0$ inside. Now, from Faraday's law, ${\bf \nabla}\times\textbf{E}= - \frac{\partial\textbf{B}}{\partial t}$, it only implies that $\textbf{B}$ inside is a time-independent constant which could be zero or nonzero. Now, $\textbf{B}=0$ inside a SC follows from the other independent property of the SC that it is a perfect diamagnet. Therefore, two mutually independent properties i.e., (i) zero resistivity and (ii) perfect diamagnetism, together define a superconducting state!


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