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Question: Answered & Verified by Expert
A small bar magnet is placed with its axis at 30o with an external magnetic 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:
PhysicsMagnetic Properties of MatterJEE MainJEE Main 2020 (04 Sep Shift 1)
Options:
  • A 6.4×102 J
  • B 9.2×103 J
  • C 7.2×102 J
  • D 11.7×103 J
Solution:
2334 Upvotes Verified Answer
The correct answer is: 7.2×102 J

τ=MBsinθ=0.018

M=0.018B sin θ=0.0180.06×0.5=0.64 Am2

W=ΔU=UfUi

=MB cos 180MB cos0

=2MB

=2×0.6×0.06

=0.072 J

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