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 .

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64. τ = MB sin 30°, so MB = (0.016)/(0.5) = 0.032 J. Work from θ = 0 to 180°: W = 2MB = 0.064 J = 64 × 10⁻³ J. → 64

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 ______.

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