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Q.

A magnetic needle lying parallel to a magnetic field requires ‘W’ unite of work to turn it through 60°. The torque required to maintain the needle in the portion is

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a

3w

b

3w

c

2w

d

2w

answer is B.

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Detailed Solution

Case 1: W=MBcos00−cos600=MB2Case 2:τ=MB sin600τ=MB ×32⇒τ=3w
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