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A flat insulating disc of radius 'a' carries an excess charge on its surface is of surface charge density sigma C//m^(2). Consider disc to rotate around the axis passing through its centre and perpendicular to its plane with angular speed omega rad/s. If magnetic field vecB is directed perpendicular to the rotation axis, then find the torque acting on the disc. |
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Answer» Solution :Suppose the DISC is PLACED in xy-plane and is rotated about the z-axis. Consider an annular ring of radius r and of thickness dr, the charge on this ring `DQ=sigma(2pirdr)` As the ring rotates with angular VELOCITY `omega`, so the CURRENT `i=(dq)/(dt)=(sigma(2pidr))/((2pi)/(omega))=sigmaomegardr` The torque on the current loop `vectau=ivecAxxvecB`.  Hence the torque on this annular ring `dvectau=i(dvecAxxvecB)` `=sigmaomegardr(pir^(2)Bsin90^(@))=pisigmaomegar^(3)Bdr` and `tau=pisigmaomegaBint_(0)^(a)r^(3dr)=(pisigmaomegaBa^(4))/4` |
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