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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)?

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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 $$\vec J=\vec J(0)\exp{\left(-\frac{z}{\lambda_L}\right)}.$$

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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

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