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+### Statement
+
+$11.5.27.$ [Insert the problem statement]
+
+### Solution
+
+Before equilibrium is established, the flow changes according to the following law:
+$$\frac{d\Phi}{d t} =\frac{d(B - B_e)}{dt} \pi r^2$$
+Where $B_e$ is the interference generated by the motion of electrons.
+The vortex electric field is determined by the circulation theorem:
+$$E 2\pi r = \frac{d(B - B_e)}{dt} \pi r^2$$
+$$E = \frac{d(B - B_e)}{dt} \frac{r}{2}$$
+Let's write 2nd law:
+$$m\frac{d \upsilon}{dt} = \frac{d(B - B_e)}{dt} \frac{re}{2}$$
+From here:
+$$\upsilon = (B - B_e) \frac{re}{2m}$$
+Let's find the current density:
+$$j = \frac{n_e \upsilon dt Lh e}{Lh dt}$$
+$$j = n_e \upsilon e$$
+The equivalent linear current density is:
+$$i = n_e \upsilon e h$$
+Therefore:
+$$B_e = \mu_0 n_e \upsilon e h$$
+Substitute the velocity and find $B_e$:
+$$B_e = \frac{B}{1 + \frac{2m}{\mu_0 n_e e^2 h r}}$$
+From here, the desired induction inside:
+$$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]