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NCERT Class XII Chapter
Magnetism and Matter
Questions With Solutions

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Question : 7
Total: 25
A bar magnet of magnetic moment 1.5 J T–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 field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
Solution:  
Soln. (a) Work required to turn the dipole, W = MB [cos θi – cos θf]
(i) θi = 0° and θf = 90°
W = 1.5 × 0.22 [cos 0° – cos 90°] = 0.33 J
(ii) θi = 0° and θ = 180°, W = 1.5 × 0.22 [cos 0° – cos 180°] = 0.66 J
(b) Torque when θ = 90°, τ1 = MB sin 90° = 0.33 N m
Torque when θ = 180°, τ2 = MB sin 180° = 0
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