A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be
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a
3 W
b
W
c
32W
d
2W
answer is A.
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Detailed Solution
W=MBcosθ1−cosθ2=MBcos0∘−cos60∘=MB1−12=MB2and τ=MB sinθ=MBsin60o=MB32∴ τ=MB23⇒τ=3 W
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A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be