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A magnetic field B is confined to a region `r le` a and points out of the paper (the z-axis), r = 0 being the centre of the cicular region. A charged ring (charge = Q) of radius b,`bgt a` and mass m lie in the x-y plane with its centre at 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. |
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Answer» If E is the electric field generated aroung the charged ring of radius b, then as `e = (d phi)/(dt)` `oint E.dl = (B. pi a^(2))/(Delta t)` `E. 2 pi b = (B pi a^(2))/(Delta t)` or `Eb = (B a^(2))/(2(Delta t))` Torque acting on the ring `tau = b xx force = b xx Q xx E` Using (i), `tau = (Q B a^(2))/(2 (Delta t))` If `Delta L` is change in angluar momentum of the charged ring, then as `tau = (Delta L)/(Delta t) = (L_(2) - L_(1))/(Delta t) :. L_(2) - L_(1) = tau (Delta t) = (Q B a^(2) (Delta t))/(2 (Delta t)) = (Q B a^(2))/(2)` As initial angular momentum, `L_(1) = 0 :. L_(2) = (Q B a^(2))/(2) = I omega = (mb^(2)) omega` `omega = (Q B a^(2))/(2 mb^(2))` This is the angular velocity of the ring after the field vanishes. |
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