Question

A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field b parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is

Answer 1

Answer:

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Answer 2

Given:

• Mass of ring $= m$
• Radius $= R$
• Charge $= q$

To Find:

• The angular velocity acquired by the ring is , If the magnetic field is suddenly switched off.

Solution:

We know that,

                     $\int Edl = \frac{d\phi}{dt}$

​               $\int Edl =\pi r^2\frac{dB}{dt}$

Let $\lambda = \frac{q}{ 2\pi R }$  be the charge per unit length of the ring

The torque is given by  $N=\int \lambda Edl$

Angular momentum $=I\omega= \lambda Ndt$

On solving the above set of sequential equations

                    $R \lambda \pi^2 B=I \omega$

                 $\frac{R \pi R^2 Bq}{2\pi R} = mR^ 2 \omega$

                     $\frac{ Bq}{2} = m \omega$

                    $\frac{ Bq}{2m} = \omega$

Thus, Angular velocity $(\omega ) = \frac{ Bq}{2m}$

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