Questions
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through .What is the torque needed to maintain the needle in that position?
detailed solution
Correct option is C
In the case of a dipole in a magnetic field. W=MBcosθ1-cosθ2and Γ=MBsinθhere θ1=00 θ2=60° W=MB(1-cosθ)⇒W=MB2sin2θ/2 ∵(1-cosθ)=2sin2θ/2Dividing , ΓW=MBsinθMB2sin2θ/2=2sinθ2cosθ22sin2θ/2 ⇒ΓW=cotθ2⇒Γ=Wcot30°⇒Γ=3WTalk to our academic expert!
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If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B, the work done in rotating the magnet through an angle is
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