= Solution
In vacuum <magnetostatics>, the two relevant <Maxwell equations> are
$$
\nabla\cdot\mathbf B=0,\qquad \nabla\times\mathbf B=\mu_0\mathbf J.
$$
Orient the unit normal $\mathbf n$ from the minus side to the plus side. Integrating <Gauss's law for magnetism> over a pillbox straddling the <current sheet> and shrinking its thickness gives $\mathbf n\cdot(\mathbf B_+-\mathbf B_-)=0$.
To obtain the tangential condition, choose an arbitrary tangent unit vector $\mathbf t$ and an infinitesimal rectangle with its long sides of length $\ell$ parallel to $\mathbf t$ on the two sides of the sheet. Orient its surface normal as $\mathbf n\times\mathbf t$, so that its plus-side long edge is traversed along $\mathbf t$. The line <integral> in <Ampère's law> tends to $\ell\mathbf t\cdot(\mathbf B_+-\mathbf B_-)$. The enclosed <electric current> tends to $\ell\mathbf s\cdot(\mathbf n\times\mathbf t)$. Hence
$$
\mathbf t\cdot(\mathbf B_+-\mathbf B_-)=\mu_0\mathbf t\cdot(\mathbf s\times\mathbf n).
$$
Because this holds for every tangent direction and the normal jump is zero,
$$
\boxed{\mathbf n\times(\mathbf B_+-\mathbf B_-)=\mu_0\mathbf s.}
$$
Here $\mathbf s\cdot\mathbf n=0$, as required for a <surface current density>.
The <force on a current sheet from the mean magnetic field> is
$$
\boxed{\mathbf f_S=\mathbf s\times\frac{\mathbf B_++\mathbf B_-}{2}.}
$$
The mean removes the sheet's equal and opposite local self-fields, leaving the field that acts on its current through the <Lorentz force>. Using a one-sided limiting <magnetic field> would incorrectly include the sheet's own force.
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