Questions
A small bar magnet placed with its axis at with an external field of 0.06 T experiences a torque of . The minimum work required to rotate it from its stable to unstable equilibrium position is :
detailed solution
Correct option is B
Torque, τ=MBSin30∘=0.018⇒MB=0.036Work done is change in potential energy.W=MBcos0∘−cosπ=2MB=2(0.036)=0.072JTalk to our academic expert!
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If 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 an angle is
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