= Meissner screening current
{c}
{title2=$J_y(z)=-B_0e^{-z/\lambda_L}/(\mu_0\lambda_L)$}
In <SI> units, the <magnetostatic Ampère law> gives the <current density> supporting <Planar London screening>:
$$
\mathbf J=\frac{1}{\mu_0}\nabla\times\mathbf B
=-\frac{B_0}{\mu_0\lambda_L}e^{-z/\lambda_L}\widehat{\mathbf y}.
$$
Here $\mu_0$ is the <vacuum permeability>. The sign follows from $(\nabla\times\mathbf B)_y=\partial_zB_x$ in the chosen coordinates. Thus $B_0>0$ requires <electric current> along $-\widehat{\mathbf y}$. Its own <magnetic field> opposes the imposed <magnetic field> inside the material. The <electric current> is distributed over a finite layer; its depth <integral> is the effective <surface current density>
$$
\mathbf K=\int_0^\infty\mathbf J(z)\,dz
=-\frac{B_0}{\mu_0}\widehat{\mathbf y}.
$$
The units are $\mathrm{A\,m^{-2}}$ for $\mathbf J$ and $\mathrm{A\,m^{-1}}$ for $\mathbf K$. Shrinking $\lambda_L$ at fixed $B_0$ increases the surface <current density>, while this integrated <surface current density> stays fixed. The <current sheet> approximation is the large-length-scale limit of this layer, not a claim that an actual sample carries an infinitely thin <electric current>. It reproduces the <magnetic-field jump across a surface current> between the exterior $B_0\widehat{\mathbf x}$ and a field-free interior, with outward <unit normal> $-\widehat{\mathbf z}$.
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