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A small bar magnet placed with its axis at $30^{\circ}$ with an external field of 0.06T experiences a torque of 0.018Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:

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$6.4 \times 10^{- 2} J$

$9.2 \times 10^{- 3} J$

$7.2 \times 10^{- 2} J$

$11.7 \times 10^{- 3} J$

answer is C.

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Detailed Solution

$I = MBsin θ$

$\Rightarrow 0.018 = M \times 0.06 \times \sin 30^{\circ}$

$\Rightarrow M = 0.6 A - m^{2}$

$work done W = MB ({cosθ}_{1 -} {cosθ}_{2}); θ_{1} = 0^{0} and θ_{2} = 180^{0}$

$∴ W = 2 MB$

$W = 2 \times 0.6 \times 0.06$

$W = 7.2 \times 10^{- 2} J$

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