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Show that the time period (T) of oscillations of a freely suspended magnetic dipole of magneticmoment (m) in a uniform magnetic field (B) is given by T=2 pi sqrt((I)/(mB)) , where I is moment of inertia of the magnetic dipole. |
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Answer» Solution :Let a small magnetic needle of magnetic moment `vecm` be FREELY suspended in a uniform magnetic field `vecB` so that in equilibrium positive magnet comes to rest along the direction of `vecB`. If the magnetic needle is rotated by a small angle `theta` from its equilibrium position magnet comes to rest along the direction of `vecB`. if the magnetic needle is rotated by a small angle `theta` from its equilibrium and then released , a restoring TORQUE acts on the magnet, where Restoring torque`vectau = vecm xx vecB` or `tau = - m B sin theta` If I be the moment of inertia of magnetic needle about the axis of suspension, then `tau = I alpha = I (d^2 theta)/(dt^2)` Hence, in equilibrium state, we have `I = (d^2 theta)/(dt^2) = - m B sin theta` If `theta ` is small then `sin theta to theta` and we get ,br>`I (d^2 theta)/(dt) = - MB theta ` or `(d^2 theta)/(dt^2) = - (mB)/(I) theta` As here angular acceleration is directly proportional to angular displacement and direction towards the equilibrium position, motion of the magnetic needle is simple harmonic motion, and Angular frequency of SHM `omega = SQRT((mB)/(I))` `therefore ` Time PERIOD of oscillation `T= (2pi)/(omega) = 2pi sqrt((I)/(mB))`. 
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