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