A wire of cross-sectional area A forms three sides of a square and is free to rotate about axis OO1 . If the structure is deflected by an angle θ from the vertical when current i is passed through it in a magnetic field B acting vertically upwards and density of the wire is ρ , then the value of θ is given by
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a
2AρgiB=cotθ
b
2AρgiB=tanθ
c
AρgiB=sinθ
d
Aρg2iB=cosθ
answer is A.
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Detailed Solution
Let a is the side of square.Torque of current = MB=sin(90−θ)=MBcosθ=ia2BcosθMass of each side= m=ρAaTorque of gravity = 2mga2sinθ+mg a sinθ =2mg a sinθ=2ρAga2sinθNow, at equilibrium both torques should be samei.e., ia2Bcosθ=2ρAga2sinθ⇒ cotθ=2ρAgiB .