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 :

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11.7×10−3 J

7.2×10−2 J

9.2×10−3 J

6.4×10−2 J

answer is B.

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Detailed Solution

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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