Integrating the $y$ component of <Ampère's law>,
$$
\partial_zB_x-\partial_xB_z=\frac{4\pi}{c}j_y,
$$
through the <current sheet> gives the <magnetic-field jump across a surface current>
$$
\boxed{\delta B_x(0^+)-\delta B_x(0^-)
=\frac{4\pi}{c}J_y
=-\frac{4\pi\Sigma u_0B_0}{c^2}}.
$$
Reflection in the wing plane reverses the tangential perturbation $\delta B_x$. If
$$
b=\frac{2\pi\Sigma u_0B_0}{c^2},
$$
the upper and lower boundary values on the wing are consequently
$$
\boxed{\delta B_x(x,0^+)=-b,
\qquad
\delta B_x(x,0^-)=b,
\qquad |x|<\frac L2}.
$$
Outside the wing, the corresponding boundary value is zero.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 3 / c / Solution

For a surface current density $\mathbf K$ on an interface with unit normal $\mathbf n$, <Ampère's law> gives
$$
\mathbf n\times(\mathbf B_+-\mathbf B_-)=\frac{4\pi}{c}\mathbf K
$$
in Gaussian units, or $\mu_0\mathbf K$ in SI units.

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