A very cold torsion pendulum

Considering the torsion spring, restoring torque is given by $\tau = -k \theta$ Here $\theta$ is the angle made by plane of the ring with the equilibrium position.

And thus the Time period of the oscillation of the ring is $T_{0} = 2 \pi \sqrt{\frac {J}{k}}$ where J is the moment of interia with respect to the vertical axis.

$\Rightarrow k=4 \pi^2 \frac{J}{T_{0}^2}$ …..(*)

The ring is made of Aluminium which, when cooled below its critical temperature and placed in a weak magnetic field, becomes perfectly diamagnetic (Meissner’s effect). Here, since the vertical hanging ring is oscillating, there is change in magnetic field. Therefore, the change in magnetic field induces current $I$ in the ring.

The magnetic flux through a superconducting ring is constant and its given by,

$LI +AB = constant$ where L is self-inductance A is the area enclosed by the ring and B is the magnetic field which is in this case horizontal to the plane of the ring when its in equilibrium (i.e $\theta=0$ ). One more thing to be observed is that, when the ring is at equilibrium position (i.e $\theta=0$ ) the magnetic flux through the ring and its area enclosed is zero.

Therefore, $LI+AB=0$ $\Rightarrow I=- \frac{AB}{L}$

Because of the diamagnetic property(repulsion due to induced magnetic field and weak magnetic field) there is torque applied(external torque) on the ring due to the magnetic field and the torque is given by $\tau_{external} = I \overrightarrow{A} \times \overrightarrow{B}$ Area enclosed by the ring is $A=\pi a^2$ where $a$ is the radius. Substituting for I and A

$\tau_{external} = - (\frac{\pi a^2B}{L})( \pi a^2)B sin \theta$ But since the ring under goes small torsional oscillations $sin \theta \approx \theta$

$\Rightarrow \tau_{external} = - \frac{ \pi ^2 a^4 B \theta}{L}$

The net torque on the ring is $\tau_{net} = \tau +\tau_{external}$

i.e $\tau_{net} = -k \theta - \frac{\pi ^2 a^4 B \theta}{L}$

This can be represented as

$J \alpha = -k \theta - \frac{\pi ^2 a^4 B \theta}{L}$

$-J \omega^2 \theta =-k \theta - \frac{\pi ^2 a^4 B \theta}{L}$

On simplyfing and substituting for k (*).

and $\omega = \frac{2 \pi}{T}$ where T is the new time period of oscillation.

We get $\frac{4 \pi^2}{T^2} = 4\pi^2 \frac{1}{T_{0}^2} + \frac{\pi ^2 a^4 B}{LJ}$ Simplying further, $\frac{1}{T^2} = \frac{1}{T_{0}^2} + \frac{ a^4 B}{4LJ}$ ....(**)

Here, $T = 2 \pi \sqrt{\frac {J}{k + \frac{\pi ^2 a^4 B}{L}}}$ which is less than $T_{0}$

Thus, $T$ can be represented as $T= T_{0} - \Delta T$ or $T= T_{0 }(1 - \frac{ \Delta T}{T{0}})$ where $\Delta$ T is the change in time period.

Considering the fact that

$\frac{\Delta T}{T_{0}} << 1$ ,

$T^2 = T_{0}^2(1- 2 \frac{\Delta T}{T_{0}})$ …..(**)

Substituting for $T^2$ in (**) and and simplifying we get,

$2\Delta T = \frac{T_{0}^3 a^4 B}{4LJ} - \frac{\Delta T}{T_{0}} \cdot \frac{ T_{0}^3 a^4 B}{4LJ}$ here the term $\frac{\Delta T}{T_{0}} \cdot \frac{ T_{0}^3 a^4 B}{4LJ}$ on the RHS can be ignored since its very comparativelysmall ,

$\Rightarrow \Delta T=\frac{ T_{0}^3 a^4 B}{8LJ}$

$\therefore \Delta T=(0.125) (\frac{ T_{0}^3 a^4 B}{LJ})$

Thus, the numerical coefficient C $= 0.125$

The magnetic flux through the ring must be constant when the ring becomes superconducting ( $R=0$ ). This follows from Faraday's law $\mathcal{E}=-\frac{d\Phi}{dt}= IR=0 \rightarrow \Phi=\textrm{const}.$ Since the magnetic field is parallel to the plane of ring when $\theta=0$ we have that $\Phi=0$ . On the other hand, for $\theta>0$ , we can write $LI+B A \sin(\theta)=\Phi=0.$ Therefore, a current equal to $I=-\frac{BA}{L} \sin(\theta)\approx -\frac{BA}{L}\theta$ must circulate through the ring in order to keep the flux constant. This current interacts with the external magnetic field resulting in a torque given by $\vec{\tau}= \vec{\mu} \times \vec{B} \rightarrow \tau=\mu B \cos(\theta)\approx \mu B$ where $\mu=I A$ is the magnetic moment of the ring. Thus, the new equation of motion for the ring is $\tau=-\kappa \theta - \frac{B^{2}A^{2}}{L}\theta=J \ddot{\theta} \rightarrow T=\frac{2\pi}{\sqrt{\frac{\kappa}{J}+\frac{B^{2}A^{2}}{JL}}}.$ Since $T_{0}=\frac{2\pi}{ \sqrt{\frac{\kappa}{J}}}$ and $A=\pi a^{2}$ we have that $T=\frac{T_{0}}{\sqrt{1+ \frac{a^{4}B^{2}T_{0}^2}{4 J L}}}\approx T_{0}- \frac{a^{4}T_{0}^3 B^{2}}{8 J L}\rightarrow C=\frac{1}{8}.$