Does the Meissner effect preclude currents from the bulk of superconductors?

Question

The Meissner effect means that superconductors will spontaneously set up currents that expel magnetic fields from them. The Ampere-Maxwell law, $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$ implies that wherever there's a net current density in a finite sized region there will also be a magnetic field. Does this mean that all superconductor current are confined to the surface of the superconductor (exponentially decaying with the London penetration depth)?

Answer 1

Yes --- surface currents only. There is an interesting case when the superconductor is rotating at angular velocity $\Omega$ and the electrons perceive the Coriolis force as being like a magnetic field $``B''= -(2m_e/e) \Omega$ . So currents are set up, again at the surface, that produce real magetic field $B= +(2m_e/e) \Omega$ in the bulk of the superconductor that precisely cancels the Coriolis force field. This is the origin of the London magnetic moment: https://en.wikipedia.org/wiki/London_moment

Answer 2

Yes, the current is on the surface of the superconductor and it decays exponentially. The London equation reads $$\vec\nabla\times \vec J+\frac{ne^2}{m}\vec B=0,$$ where $n$ is the density of electrons. Take the curl of this equation and use the identity $$\vec\nabla\times(\vec\nabla\times\vec J)=\vec \nabla(\vec\nabla\cdot\vec J)-\nabla^2\vec J.$$ Since $\vec\nabla\cdot\vec J=\vec\nabla\cdot\frac{1}{\mu_0}\vec\nabla\times\vec B=0$, the equation for the current can be written as $$\nabla^2\vec J-\frac{\vec J}{\lambda_L^2}=0,\tag1$$ where $\lambda_L=\sqrt{\frac{m}{\mu_0ne^2}}$ is the London penetration depth. If the $z$ is the direction normal to the superconductor's surface then the solution for Eq. (1) is $$\vec J=\vec J(0)\exp{\left(-\frac{z}{\lambda_L}\right)}.$$