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A magnetic field B is confined to a region $r\leq a$ and points out of the paper (the z-axis), $r=0$ being the centre of the circular region. A charged ring (charge = Q) of radius $b,b>a$ and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time $\Delta t$ . Find the angular velocity $\omega$ of the ring after the field vanishes.

Answers (1)

The magnetic field decreases which induce an emf hence electric field around the ring. And therefore, the ring experiences a torque which produces a change in angular momentum.

As the magnetic field is brought to zero, the magnetic flux linked reduces to zero which is linked to the ring. This induces an emf in-ring and in turn an electric field E around the ring.

The induced emf =

$\text {Electric field E }\times \left ( 2\pi b\right )\left ( Because v=E\times d \right ) ....(i)$

By Faraday's law of EMI

$\left | \varepsilon \right |=\frac{d\phi }{dt}=A\frac{dB}{dt}$

$\left | \varepsilon \right |=\frac{B\pi a^{2}}{\Delta t}S ...(ii)$

From Eqs. (i) and (ii), we have

$2\pi b E=\varepsilon =\frac{B\pi a^{2}}{\Delta t}$

As we know the electric force experienced by the changed ring, $F_{e}=QE$

This force try to rotate the coil, and the torque is given by

$\text {Torque=b}\times\text {Force}$

$\tau =QEb=Q\left [ \frac{B\pi a^{2}}{2 \pi b \Delta t} \right ]b$

$\Rightarrow \tau =Q\frac{Ba^{2}}{2\Delta t}$

If $\Delta L$ is the change in angular momentum,

$\Delta L=Torque \times \Delta t = Q\frac{Ba^{2}}{2}$

Since initial angular momentum =0

and $Torque \times \Delta t=\text {change in angular momentum}$

Final angular momentum =

$mb^{2}\omega =\frac{QBa^{2}}{2}$

Where, $mb^{2}=I (\text {moment of inertia of ring })$

$\omega =\frac{QBa^{2}}{2mb^{2}}$

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