$l$; $L \approx \mu _{0}l$ is the magnetic inductance; $s$ is the cross section of superconducting wires; $n _{s}$ is the density of the Cooper pairs; $\lambda _{L} = (m/\mu _{0}q^{2}n_{s})^{0.5} $ is the London penetration depth.

Two permitted states $n$ and $n+1$ have equal energy at $\Phi = (n+0.5) \Phi _{0} $ according to (2) and thus equal probability $P \propto \exp (-E _{k}/k _{B}T) $. The probability of other permitted states is negligible and $P(n+1) \approx 1 - P(n)$ at $\Phi \approx (n+0.5) \Phi _{0} $ when $\Phi _{0}^{2}/2L_{k} = I _{p,A}\Phi _{0} \gg k _{B}T$. The probability of the $n$ state may be described with the relation $$P(n)  \approx  \frac{1}{1+\exp \epsilon 2\frac{\delta \Phi}{\Phi _{0}}} \eqno {(3)}$$ at the magnetic flux $\Phi = (n+0.5) \Phi _{0} + \delta \Phi $, when $\delta \Phi \ll \Phi _{0}$. Here $ I _{p,A} = \Phi _{0}/2L_{k}$ is the persistent current (1) at $|n - \Phi /\Phi _{0}| = 1/2$ and $\epsilon = I _{p,A}\Phi _{0}/k _{B}T $. The probability (3) changes from $P(n) \approx 1$ to $P(n) \approx 0$ in a region of the magnetic flux from $\delta \Phi \approx  -\Phi _{0}/2\epsilon $ to $\delta \Phi \approx  \Phi _{0}/2\epsilon $. This 