InterviewSolution
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InterviewSolution
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A magnetic field B is confined to a region rle 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 gt a and mass m lies in the xy-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 Deltat. Find the angular velocity omega of the ring after the field vanishes. |
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Answer» Solution :In `Deltat` time flux linked with ring becomes zero from its maximum VALUE which induces emf in ring electric field is also induced in ring. Induced emf = Electric field x `(2pib) [ because V=Ed]` But induced emf `EPSILON=(Deltaphi)/(Deltat)=(pia^2B-0)/(Deltat)=(Bpia^2)/(Deltat)` `therefore (Bpia^2)/(Deltat)=F(2pib)` `therefore E=(Ba^2)/(2bDeltat)` Force produced on ring `F=QE=(QBa^2)/(2bDeltat)` Torque produced on ring , `T= rF sin theta =(b Q Ba^2)/(2bDeltat)=(QBa^2)/(2Deltat)` Now torque = RATE of change in angular momentum, `tau=(DELTAL)/(Deltat)` `=(mvb-0)/(Deltat)` `(QBa^2)/(2Deltat)=(mb^2omega)/(Deltat)` `therefore omega=(QBa^2)/(2mb^2)` |
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