Verified Solution • 2025–26 Session moderate

QJEEM20Ph683c: JEE Main 2020 Physics - Class 12 JEEM20Ph683c Solution: A small bar magnet placed with its axis at 30 o with an…

A small bar magnet placed with its axis
at 30o 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 :

Multiple Choice Options

  • A 9.2 ×\times 10-3 J
  • B 6.4 ×\times 10-2 J
  • C 7.2 ×\times 10-2 J
  • D 11.7 ×\times 10-3 J

Step-by-Step Verified Solution

Torque on a bar magnet :

τ=MB sin⁡θ\tau = MB\,\sin \theta

Here, θ\theta = 30º, I = 0.018 N-m, B = 0.06 T

0.018=M×0.06×0.50.018 = M \times 0.06 \times 0.5

⇒M=0.6 Am2\Rightarrow M = 0.6\,A{m^2}

W=Uf−UiW = {U_f} - {U_i}

=MB(cos⁡θi−cos⁡θf)= MB(\cos {\theta _i} - \cos {\theta _f})

=0.6×0.06(1−(−1))= 0.6 \times 0.06(1 - ( - 1))

=7.2×10−2 J= 7.2 \times {10^{ - 2}}\,J

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