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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-

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

Correct answer: 7.2 × 10⁻² J

Solution

From torque, mB*sin30 = 0.018 so mB = 0.036. Work from stable (0 deg) to unstable (180 deg) is mB(cos0 - cos180) = 2mB = 0.072 J = 7.2 x 10^-2 J.

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