Q.

A copper ring carrying a current is placed in a uniform magnetic field in such a way that the ring experiences maximum possible torque of 2 N-m. Now the ring is rotated through an angle of 60o about a diameter which is perpendicular to the field direction. In this position torque experienced by the ring will be

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a

1 N-m

b

0.5 N-m

c

1.732 N-m

d

1.414 N-m

answer is A.

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Detailed Solution

For the position shown in figure, torque experienced by the ring is maximum and ςmax⁡=MBsin⁡90o=MB when the ring is rotated through an angle of 60o about the diameter PQ, the angle between M→ and B→ is 30o∴ς=MBsin⁡30o=ςmax⁡×12=2×12N−m=1N−m
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