The magnetic moment of a bar magnet is 0.5 Am². It is suspended in a uniform magnetic field of 8 × 10⁻² T. The work done in rotating it from its most stable to most unstable position is: (a) 4 × 10⁻² J (b) 16 × 10⁻² J (c) 8 × 10⁻² J (d) Zero

Answer: (c). Most stable θ = 0°, most unstable θ = 180°: W = MB(1 - cos 180°) = 2MB. W = 2 × 0.5 × 8 × 10⁻² = 8 × 10⁻² J So the answer is (c).

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Field of a Magnet & Magnet in a Uniform Field

The magnetic moment of a bar magnet is $0.5\text{ Am}^2$ . It is suspended in a uniform magnetic field of $8 \times 10^{-2}\text{ T}$ . The work done in rotating it from its most stable to most unstable position is:

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Answer: (c)

Most stable $\theta = 0^\circ$ , most unstable $\theta = 180^\circ$ : $W = MB(1 - \cos 180^\circ) = 2MB$ .
$W = 2 \times 0.5 \times 8 \times 10^{-2} = 8 \times 10^{-2}\ \text{J}$
So the answer is (c).

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