A bar magnet of magnetic moment 2 × 10⁴ J T⁻¹ can rotate freely in a horizontal plane. A horizontal magnetic field B = 6 × 10⁻⁴ T is applied. Calculate the work required to rotate the magnet from parallel to the field to 60° from the field.
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Question

A bar magnet of magnetic moment 2 × 10⁴ J T⁻¹ can rotate freely in a horizontal plane. A horizontal magnetic field B = 6 × 10⁻⁴ T is applied. Calculate the work required to rotate the magnet from parallel to the field to 60° from the field.

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G1 Physics · Accepted Answer

Step 1: The potential energy of a magnetic dipole in a uniform field is U = −MBcosθ. Step 2: For slow rotation by an external agent, W(ext) = ΔU = MB(cosθ₁ − cosθ₂). Step 3: Substitute θ₁ = 0°, θ₂ = 60°, M = 2 × 10⁴ J/T and B = 6 × 10⁻⁴ T. Step 4: W(ext) = (2 × 10⁴)(6 × 10⁻⁴)(1 − 1/2) = 6 J. Final answer: (C) 6 J.