A short bar magnet placed with its axis at 30° with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is α × 10⁻³ J. The value of α is ______. Numerical answer type. Work it out, then open the answer.

Answer: 64. B = 800 G = 0.08 T; τ = MBsin 30° ⇒ MB = (0.016)/(0.5) = 0.032 J Most stable to most unstable: W = 2MB = 0.064 J = 64 × 10⁻³ J So α = 64.

A short bar magnet placed with its axis at 30° with an external field… | JEE Main 2026 (24 Jan, Shift 1)

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Field of a Magnet & Magnet in a Uniform Field

A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of $0.016\text{ N.m}$ . The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3}\text{ J}$ . The value of $\alpha$ is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 64

$B = 800\ \text{G} = 0.08\ \text{T}$ ; $\tau = MB\sin 30^\circ \Rightarrow MB = \frac{0.016}{0.5} = 0.032\ \text{J}$
Most stable to most unstable: $W = 2MB = 0.064\ \text{J} = 64 \times 10^{-3}\ \text{J}$
So $\alpha = 64$ .

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