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=MBsin90o=MB when the ring is rotated through an angle of 60o about the diameter PQ, the angle between M→ and B→ is 30o∴ς=MBsin30o=ςmax×12=2×12N−m=1N−m