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 angle θ is
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a
MB(1−cosθ)
b
MBsinθ
c
MBcosθ
d
MB
answer is A.
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Detailed Solution
The amount of work done to rotate the magnet from θ1 and θ2 W=MB(cosθ1−cosθ2) Put θ1=00, θ2=θ ⇒W=MB(1−cosθ)