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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a
11.7×10−3 J
b
7.2×10−2 J
c
9.2×10−3 J
d
6.4×10−2 J
answer is B.
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Detailed Solution
Torque, τ=MBSin30∘=0.018⇒MB=0.036Work done is change in potential energy.W=MBcos0∘−cosπ=2MB=2(0.036)=0.072J