A bar magnet having a magnetic moment of 2×10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6×10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:

Question

Accepted Answer

0.6 J. The work done in rotating a magnetic dipole in a magnetic field is given by W = MB(1 - cosθ). Substituting M = 2×10⁴ J T⁻¹, B = 6×10⁻⁴ T, and θ = 60°, we get W = (2×10⁴)(6×10⁻⁴)(1 - cos60°) = 0.6 J.

A bar magnet having a magnetic moment of 2×10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6×10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:

Solution

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