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# A bar magnet of magnetic moment 1.5 J T–1 lies aligned with thedirection of a uniform magnetic field of 0.22 T.(a) What is the amount of work required by an external torque toturn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?(b) What is the torque on the magnet in cases (i) and (ii)?

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Solution

## Given: The magnetic moment of the bar magnet is 1.5  JT −1 and the magnetic field strength is 0.22 T. a) Work done by an external torque on the bar magnet is given as, W=−mB( cos θ 2 −cos θ 1 )(1) Where, m is the magnetic moment of the bar magnet, B is the magnetic field, θ 1 is the initial angle between axis of the bar magnet and magnetic field and θ 2 is the final angle between axis of the bar magnet and magnetic field. (i) Since, the bar magnet is initially aligned in the same direction of the magnetic field and finally aligned normal to the magnetic field direction, so the initial angle will be 0° and the final angle will be 90°. By substituting the given values in equation (1), we get W 1 =−1.5×0.22( cos90°−cos0° ) =−1.5×0.22( 0−1 ) =0.33 J Thus, the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is 0.33 J. (ii) Since, the bar magnet is initially aligned in the same direction of the magnetic field and finally aligned opposite to the magnetic field direction, so the initial angle will be 0° and the final angle will be 180°. By substituting the given values in equation (1), we get W 2 =−1.5×0.22( cos180°−cos0° ) =−1.5×0.22( −1−1 ) =0.66 J Thus, the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction is 0.66 J. b) The torque acting on a bar magnet is given as, τ=mBsinθ(2) (i) Since, the magnetic moment is normal to the magnetic field, so the angle between these two will be 90°. By substituting the given values in equation (2), we get τ=1.5×0.22sin90° =0.33×1 =0.33 N-m Thus, the torque on the magnet when it is aligned normal to the magnetic field direction is 0.33 M-m. (ii) Since, the magnetic moment is opposite to the magnetic field direction, so the angle between these two will be 180°. By substituting the given values in equation (2), we get τ=1.5×0.22sin180° =0.33×0 =0 N-m Thus, the torque on the magnet when it is aligned opposite to the magnetic field direction is zero.

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