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Q.

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60o. 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=MB(cos θ1-cos θ2)=MB(cos 0o-cos 60o)=MB1-12=MB2and τ=MB sinθ=MB sin60o=MB32∴τ=MB23⇒τ=3W
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