= Solution
To first order in the applied fields, the carrier acceleration gives $m\dot{\mathbf v}=q\mathbf E$. The <current density> $\mathbf j=nq\mathbf v$ consequently obeys the first <London equation>
$$
\dot{\mathbf j}=\frac{nq^2}{m}\mathbf E=\epsilon_0\omega_p^2\mathbf E.
$$
Taking its curl and using <Faraday's law> yields
$$
\frac{\partial}{\partial t}\left(\nabla\times\mathbf j+\epsilon_0\omega_p^2\mathbf B\right)=0.
$$
For an initially field-free, current-free preparation, the integration constant is zero. Equivalently, impose the second <London equation> appropriate to the <Meissner state>, $\nabla\times\mathbf j=-\epsilon_0\omega_p^2\mathbf B$. In the final static state, <Ampère's law> and $\nabla\cdot\mathbf B=0$ then give
$$
-\nabla^2\mathbf B=\mu_0\nabla\times\mathbf j
=-\frac{\omega_p^2}{c^2}\mathbf B,\qquad
\boxed{\lambda_L=\frac c{\omega_p}=\sqrt{\frac{m}{\mu_0nq^2}}}.
$$
For a planar half-space $z>0$ with a tangential surface field, the bounded solution is
$$
\boxed{\mathbf B(z)=\mathbf B(0)e^{-z/\lambda_L}}.
$$
The <plasma-frequency form of London penetration depth> therefore shows that an applied field is exponentially small several penetration depths inside the material.
The <magnetic-history constant in an undamped plasma> explains the required qualification. Undamped carrier motion alone fixes a time-independent value of $\nabla\times\mathbf j+\epsilon_0\omega_p^2\mathbf B$; it does not force that value to vanish. A <perfect conductor> can retain a pre-existing field. The additional <London equation> describes <London screening> independently of that arbitrary ideal-conductor history. The original London paper explicitly distinguishes its description from acceleration theory.<The electromagnetic equations of the supraconductor by Fritz London and Heinz London>[pp. 71–74: acceleration equation, its integration freedom and the additional superconducting relation]{ref}
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