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Are the usual interface conditions for electromagnetic fields, i.e. $$\mathbf{n}_{12}\times(\mathbf{E}_1-\mathbf{E}_2) = 0,$$ $$\mathbf{n}_{12}\bullet(\mathbf{D}_1-\mathbf{D}_2) = \sigma_s,$$ $$\mathbf{n}_{12}\bullet(\mathbf{B}_1-\mathbf{B}_2) = 0,$$ $$\mathbf{n}_{12}\times(\mathbf{H}_1-\mathbf{H}_2) = \mathbf{J}_s, $$ valid at the interface if one of the media is a superconductor? I would assume that they are since they are derived from Maxwell's equations. Is this correct?

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Magnetic field is expelled from a superconductor by the Meissner effect, so the normal magnetic field goes to zero at the surface. For a type-I superconductor, a strong enough magnetic field will destroy the superconducting state. In a type-II superconductor a partial magnetic field can penetrate.

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  • $\begingroup$ The Meissner effect says that the magnetic field decays exponentially into the superconductor, so the normal component of the magnetic field should still be continuous at the interface, shouldn't it? $\endgroup$ – ECA18 Jun 14 at 21:09
  • $\begingroup$ @ECA18, the London penetration depth is small for most superconductors (around the order of a few hundred nm), so the surface current quickly cancels the internal field. A superconductor acts like a nearly perfect diamagnetic material. $\endgroup$ – amateurAstro Jun 14 at 21:25

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