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A bar magnet with magnetic moment 1.0 * 10⁴ J/T is free to rotate in a horizontal plane where a uniform horizontal magnetic field B = 4 * 10⁻⁵ T exists. What is the work done in slowly rotating the magnet from alignment parallel to the field to a position 60 deg from the field?

1. 0.2 J
2. 0.1 J
3. 0.4 J
4. 0.3 J

Correct answer: 0.2 J

Solution

The potential energy of a magnetic dipole in a field is U = -M*B*cos(theta). Initial state: parallel (theta = 0 deg), U_i = -MB. Final state: theta = 60 deg, U_f = -MB*cos(60) = -MB/2. Work done = U_f - U_i = -MB/2 - (-MB) = MB/2. W = (1.0*10⁴) * (4*10⁻⁵) / 2 = 0.4/2 = 0.2 J.

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