Electromagnetic induction > Magnetic flux conservation. Superconductors in magnetic field > Problem 11.5.23

Statement

$11.5.23.$ A copper ring of radius $r$ and mass $m$ hangs on a thread, making small torsional vibrations with a period of $T$. The inductance of the ring is $L$. How will the oscillation period of the ring change if it is placed in a horizontal homogeneous magnetic field of induction $B$ parallel to the plane of the ring in the equilibrium position? The moment of inertia of the ring relative to the axis passing through the diameter is equal to $J$. Ignore the resistance of the ring.

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

$$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$