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A bar magnet of magnetic moment 1.5 JT^–1 lies aligned with the direction of a uniform magnetic field of 0.22 T. Then 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:

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0.66 J, 0.38 J

0.45 J, 0.25 J

0.55 J, 0.88 J

0.33 J, 0.66 J

answer is A.

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Detailed Solution

$W = MB (1 - \cos θ) W_{1} = 1.5 \times 0.22 (1 - 0) = 0.33 J (normal) W_{2} = 1.5 \times 0.22 (1 + 1) = 0.66 J (opposite)$

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