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Question

# A bar magnet of magnetic moment 1.5JT−1 lies aligned with the direction of a uniform magnetic field of 0.22 T.(A) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:(i) normal to the field direction , (ii) opposite to the filed direction?(B) What is the torque on the magnet in case (i) and (ii)

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Solution

## It is given that magnetic moment is M=1.55JT−1 B=0.22T (A) Amount of work required by an external torque to turn the magnet so as to align its magnetic moment (1) Normal to the field direction is W1=−MB[cos90−cos0] =MB =1.55×0.22 =0.341N−m (2) Amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to field direction. W2=−MB[cos180−cos0] W2=2MB W2=0.682N−m (B) Torque acting on case (1) τ=MBsin90 =0.341J Torque acting in case (2) τ=MBsin180 =0J

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