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+### Statement
+
+$11.5.24.$ [Insert the problem statement]
+
+### 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 $2\pi$
+
+#### Answer
+
+[Insert a concise answer or boxed result]