Question

a loop of current - carrying wire has a magnetic dipole moment of 5.0×10⁻⁴ 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
○ -1.0×10⁻⁵ j.
○ 2.5×10⁻⁴ j.
○ -2.5×10⁻⁴ j.
○ 1.0×10⁻⁵ j.
○ 0 j.

Explanation:

Step1: Recall work - energy relation

The work done in rotating a magnetic dipole in a magnetic field is given by $W = \Delta U$. The potential energy of a magnetic dipole in a magnetic field is $U=-\vec{\mu}\cdot\vec{B}=-\mu B\cos\theta$.

Step2: Calculate initial and final potential energies

Initially, $\theta_1 = 0^{\circ}$, so $U_1=-\mu B\cos(0^{\circ})=-\mu B$. Finally, $\theta_2 = 90^{\circ}$, so $U_2=-\mu B\cos(90^{\circ}) = 0$.

Step3: Calculate the work done

$W=U_2 - U_1=0-(-\mu B)=\mu B$. Given $\mu = 5.0\times 10^{-4}\text{ A}\cdot\text{m}^2$ and $B = 0.50\text{ T}$, then $W=(5.0\times 10^{-4}\text{ A}\cdot\text{m}^2)\times(0.50\text{ T}) = 2.5\times 10^{-4}\text{ J}$.

Answer:

$2.5\times 10^{-4}\text{ J}$