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Questions  

A small bar magnet placed with its axis at  30° with an external field of 0.06 T experiences a torque of  0.018  Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

a
11.7×10−3 J
b
7.2×10−2 J
c
9.2×10−3 J
d
6.4×10−2 J

detailed solution

Correct option is B

Torque, τ=MBSin⁡30∘=0.018⇒MB=0.036Work done is change in potential energy.W=MBcos⁡0∘−cos⁡π=2MB=2(0.036)=0.072J

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