= Solution
Write the superconducting order parameter as $\psi=\sqrt{n_s}e^{i\vartheta}$, with carrier mass $m$. Minimal electromagnetic coupling gives mechanical momentum $\hbar\nabla\vartheta-q\mathbf A$, so the current density is
$$
\boxed{\mathbf j=\frac{n_sq}{m}
(\hbar\nabla\vartheta-q\mathbf A).}
$$
It also follows by substituting $\psi$ into $(q\hbar/m)\operatorname{Im}(\psi^*\nabla\psi)-(q^2/m)|\psi|^2\mathbf A$. In a gauge where the phase is constant, this reduces to the <London equation> $\mathbf j=-n_sq^2\mathbf A/m$.
Under $\mathbf A\mapsto\mathbf A+\nabla\chi$, the phase changes by $\vartheta\mapsto\vartheta+q\chi/\hbar$. The two gradient changes cancel, proving <gauge invariance> of the actual current. A formula omitting the phase term is valid only in its chosen gauge.
For a uniform superconductor without vortices, taking a curl gives $\nabla\times\mathbf j=-n_sq^2\mathbf B/m$. Combine this with the stationary <Ampère's law> $\nabla\times\mathbf B=\mu_0\mathbf j$ and $\nabla\cdot\mathbf B=0$ to obtain
$$
\nabla^2\mathbf B=\frac{\mathbf B}{\lambda_L^2},
\qquad \lambda_L=\sqrt{\frac{m}{\mu_0n_sq^2}}.
$$
For a <magnetic field> parallel to the planar interface, translation <symmetry> reduces the equation to $\mathbf B''=\mathbf B/\lambda_L^2$. Boundedness and decay as $z\to\infty$, together with the tangential boundary value, give the <Meissner effect>
$$
\boxed{\mathbf B(z)=\mathbf B_0e^{-z/\lambda_L},\qquad z>0.}
$$
Here $\lambda_L$ is the <London penetration depth>.
The parallel-field assumption is necessary for this one-dimensional Meissner solution. If the printed arbitrary constant $\mathbf B_0$ has a nonzero normal component, $\nabla\cdot\mathbf B=0$ forces that component to be constant across a translation-invariant half-space, whereas the <London equation> forces it to vanish. Thus no decaying planar Meissner state with that boundary value exists; one must change the geometry or allow a different magnetic state.
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