Question

A loop of current-carrying wire has a magnetic dipole moment of 5ᴇ−4 A·m^2. The moment initially is aligned with a 0.5-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:

a. 0
b.2.5ᴇ−4 J
c.−2.5ᴇ−4 J
d.1.0ᴇ−3 J
e.−1.0ᴇ−3 J

Answer 1

The work done to rotate a current-carrying loop with a magnetic dipole moment of 5⋅⁴ A⋅m² from being aligned with to being perpendicular to a 0.5-T magnetic field is 2.5⋅⁴ J.

Step-by-step explanation:

To determine the work done to rotate a current-carrying loop from being aligned with a magnetic field to being perpendicular to it, we use the formula for the potential energy of a magnetic dipole in a magnetic field: U = -μ ⋅ B.

The work done (W) to rotate the magnetic dipole moment from parallel (0 degrees) to perpendicular (90 degrees) to the field is the change in potential energy, so W = U_final - U_initial. The initial orientation (parallel to the field) has the lowest potential energy, which is U_initial = -μB, and the final orientation (perpendicular to the field) has zero potential energy, U_final = 0, as the magnetic dipole moment is perpendicular to the field.

Substituting the given values (μ = 5−4 A⋅m² and B = 0.5 T), we calculate the work done: W = 0 - (-5−4 ⋅ 0.5) = 2.5−4 J. Thus, the work done to rotate and hold the loop in the perpendicular orientation is 2.5−4 J.