A small bar magnet is placed with its axis at 30 o with an external magnetic field of 0 . 06   T…

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:
  1. 6.4×10−2 J
  2. 9.2×10−3 J
  3. 7.2×10−2 J
  4. 11.7×10−3 J

Solution

τ=MBsinθ=0.018

M=0.018B sin θ=0.0180.06×0.5=0.64 Am2

W=ΔU=Uf−Ui

=−MB cos 180∘−−MB cos0∘

=2MB

=2×0.6×0.06

=0.072 J

Asked in: JEE Main 2020 (04 Sep Shift 1)

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