Solution

The induction in each of the halves is modulo:

$$B_0 = \mu_0 i$$

Length of each camera:

$$l_0 = \frac{l-d}{2}$$

Let's say we shifted the plate by a small distance $x$. From the law of conservation of flow:

$$B_1 = \frac{B_0}{1+\frac{x}{l_0}} \approx B_0(1 - \frac{x}{l_0})$$

$$B_2 = \frac{B_0}{1- \frac{x}{l_0}} \approx B_0(1 + \frac{x}{l_0})$$

Now let's write down the energy density on each side, this will also be the magnetic pressure on the bridge.:

$$P_1 = \frac{B_1^2}{2 \mu_0} = \frac{B_0^2(1 - \frac{x}{l_0})^2}{2 \mu_0}$$

$$P_2 = \frac{B_2^2}{2 \mu_0} = \frac{B_0^2(1 + \frac{x}{l_0})^2}{2 \mu_0}$$

Hence the force acting on the partition:

$$F = ma = (P_2 - P_1)ah$$

$$F = ma = \frac{4\mu_0 i^2ahx}{l-d}$$

Solving the UGCC:

$$v = \frac{i}{\pi} \sqrt{\frac{\mu_0 ah}{m(l-d)}}$$

Savchenko has a cyclic frequency, that is, without dividing by $2\pi$