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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3 W

32W

answer is A.

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Detailed Solution

W=MBcos⁡θ1−cos⁡θ2=MBcos⁡0∘−cos⁡60∘=MB1−12=MB2and τ=MB sinθ=MBsin60o=MB32∴ τ=MB23⇒τ=3 W

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