= Finite-thickness Meissner screening
{c}
{title2=$B_x(z)=B_0\cosh(z/\lambda_L)/\cosh(d/(2\lambda_L))$}
To isolate finite size from <quantum vortex> penetration, keep the same weak-field <London equations> but replace the <half-space> by an infinite slab $-d/2\leq z\leq d/2$. Put equal tangential <magnetic fields> $B_0\widehat{\mathbf x}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed <current density> are
$$
B_x(z)=B_0\frac{\cosh(z/\lambda_L)}{\cosh(d/(2\lambda_L))},\qquad
J_y(z)=\frac{B_0}{\mu_0\lambda_L}\frac{\sinh(z/\lambda_L)}{\cosh(d/(2\lambda_L))}.
$$
The <hyperbolic cosine> profile has equal boundary values, while the <electric currents> on the two sides have opposite signs. In particular,
$$
\frac{B_x(0)}{B_0}=\frac{1}{\cosh(d/(2\lambda_L))},\qquad
\frac{\overline B_x}{B_0}=\frac{1}{dB_0}\int_{-d/2}^{d/2}B_x(z)\,dz
=\frac{2\lambda_L}{d}\tanh\!\left(\frac{d}{2\lambda_L}\right).
$$
For the stipulated thickness $d=2\lambda_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a <Meissner state>. For $d\gg\lambda_L$, the center <magnetic field> is exponentially small and the mean is about $2\lambda_LB_0/d$, reflecting the two surface layers. For $d\ll\lambda_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their <demagnetizing field> and edge geometry accounted for; the slab calculation only isolates the effect of thickness.
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