Permanent magnetismMedium
Question
A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through 60o. The torque needed to maintain the needle in this position will be -
Options
A.√3 W
B.W
C.(√3 / 2) W
D.2W
Solution
W = MB(1-cos 60o) = 
τ = MB sin 60o = √3
= √3 W
τ = MB sin 60o = √3
= √3 WCreate a free account to view solution
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