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

Question

Maryjane Graham

Physics

29 June, 20:16

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

+4

Answers (1)

Know the Answer?

Kasen Huynh · Answer 1

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)