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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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