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+### Statement
+
+$11.5.23.$ [Insert the problem statement]
+
+### Solution
+
+The oscillation period of a conventional torsional pendulum with a small amplitude is:
+
+$$T = 2\pi \sqrt{\frac{J}{k}}$$
+
+Where $k$ is the torsional stiffness of the thread, we express it:
+
+$$k = \frac{4 \pi^2 J}{T^2}$$
+
+The moment acting on the frame from the side of the magnetic field is equal to:
+
+$$M = \mu \times B \approx \mu B$$
+
+where $\mu = \pi r^2i$ is the magnetic moment of the frame.
+
+The magnetic flux through the frame is:
+
+$$\Phi = L I = B \pi r^2 \varphi$$
+
+$$I = \frac{B \pi r^2}{L} \varphi$$
+
+Let's make up the equation of oscillations in 2ZN for rotational motion:
+
+$$J \beta = k = \frac{4 \pi^2 J}{T^2} \varphi + \frac{B^2 \pi^2 r^4}{L} \varphi$$
+
+The period is similar to the usual fluctuations, after the transformations:
+
+$$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]