InterviewSolution
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InterviewSolution
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A bar magnet of magnetic moment m and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field B. What would be the similar period T' for each piece ? |
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Answer» Solution :Time period of bar magnet in magnetic FIELD B `T= 2pi sqrt((I)/( MB) ) ""…(1)` where M = magnetic moment of magnet B = uniform magnetic field Here `I= (ml^(2) )/( 12)` where m = mass of magnet l = LENGTH of magnet I = moment of INERTIA of magnet When magnet is cut into two equal pieces, perpendicular to length then moment of inertia I. of each piece of magnet about an axis perpendicular to the length passing through its centre is, `therefore I. = (m)/(2) ((l)/(2))^(2) .(1)/(12)` `I. = (ml^(2) )/( 12 xx 8 ) = (1)/(8) ""...(2)` Original magnetic moment `M = (2l) q_(m)` DIPOLE moment of each piece `M. = ((2l)/( 2))q_m` `M. = (M)/(2)""...(3)` Period of each piece, `T. = 2pi sqrt((I.)/( M.B))` `= 2pi sqrt((I//8)/( (M//2)B) ) ""` [ From EQU. (2) and (3) ] `= 2pi sqrt((1)/(4) (1)/(MB) )` `= (1)/(2) 2pi sqrt((I)/(MB))` `T.= (T)/(2)""` [From equation (1)] |
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