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Q.

A bar magnet of magnetic moment 1.5J/T is along the direction of the uniform magnetic field of 0.22T. The work done in turning the magnet opposite to the field direction and the torque required to keep in that position are:

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a

0.66J and 0.66 N-m

b

0.33J and 0.33N-m

c

0.33J and 0 N-m

d

0.66J and 0 N-m

answer is D.

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Detailed Solution

$W = MB (1 - \cos θ) W = 1 .5 \times 0 .22 (1 - \cos 180) = 0 .66 J τ = MBsin 180 ° = 0$

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A bar magnet of magnetic moment 1.5J/T is along the direction of the uniform magnetic field of 0.22T. The work done in turning the magnet opposite to the field direction and the torque required to keep in that position are: