# How is Meissner effect consistent with the frozen field lines inside the superconductor?

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.
• If $\vec{B}$ is constant why is it expelled from the interior? This clearly shows that $\vec{B}$ is changing – mithusengupta123 Oct 8 '18 at 11:07