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The boundary condition $\textbf{n}\times \textbf{H}=\textbf{K}$ (A/m) is valid in the surface of a perfect conductor (where the conductivity $\sigma$ is infinity). Is it satisfied for the surface of a real conductor?

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2 Answers 2

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Yes, more or less. The boundary condition can be derived by taking a surface integral of Ampère's Law $\vec{\nabla} \times \vec{H} = \vec{J}_f + \partial \vec{D}/\partial t$ over the surface of a long, thin loop with two parallel sides, one running on each side of the surface. Such a loop is depicted on the left-hand side of the diagram below:

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Diagram stolen from this question.

Applying Stokes' theorem to the right-hand side of the integrated equation, and taking the limit as the "height" of the loop goes to zero, one can derive the boundary condition $\hat{n} \times \vec{H} = \vec{K}_f$. (The term involving the time derivatives of $\vec{D}$ vanishes in this limit.) Nothing in this derivation requires any assumptions about the conductivity of the media.

The one caveat is that you should only be counting "free" surface currents in this integral. If neither medium has significant magnetization, this distinction isn't really relevant, but if you are dealing with magnetized media then you'll need to be careful not to include the bound currents when applying your boundary condition. Alternately, you could derive a boundary condition that applies to the total surface current $\vec{K}$ by applying the same logic to the general version of Ampere's Law: $\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \partial \vec{E}/\partial t$.

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Some other approximate boundary conditions were derived for real conductors, e.g., the Leontovich boundary condition (https://www.researchgate.net/publication/225114265_On_the_Leontovich_boundary_condition_in_geoelectromagnetism)

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