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:
$$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$