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A loop of current-carrying wire has a magnetic dipole moment of 5.0 x 10–4 A·m2. The moment initially is aligned with a 0.50-T magnetic field. To rotate the loop so its dipole moment is perpendicular to the field and hold it in that orientation, you must do work of

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jolis1796

Answer:

W = 2.5*10⁻⁴ J  (by me)

Explanation:

μy = 5.0 x 10⁻⁴ A·m²

By = 0.5 T

∅ = 0º

If

μx = 5.0*10⁻⁴ A·m²

By = 0.5 T

∅ = 90º

We get ΔU as follows

ΔU = Uf - Ui

Ui = - μy*By*Cos ∅ = - 5.0*10⁻⁴ A·m²*0.5 T*Cos 0º = - 2.5*10⁻⁴ J

Uf = - μx*By*Cos ∅ = - 5.0*10⁻⁴ A·m²*0.5 T*Cos 90º = 0 J

Finally we use the equation

W = - ΔU = - (0 - (- 2.5*10⁻⁴)) J = - 2.5*10⁻⁴ J  (by the field)

W = 2.5*10⁻⁴ J  (by me)

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