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# A bar magnet of magnetic moment M is divided into 'n' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength 2T and held making an angle 600 with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:MnJMn2JMn JMn2 J

A
MnJ
B
Mn2J
C
Mn J
D
Mn2 J
Solution
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#### Answer is ANet magnetic moment after cutting=MnWe know that, |U|=MBcosθIn equilibrium position, all the potential energy will be converted into kinetic energy i.e. KE=|U|Given Mnew=Mn,B=2T,θ=60∘Therefore, K.E=Mn×2×cos60∘=MnJ

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