Use the diagram for the next problem.
-A circular current loop lies in the xy plane and has radius R = 10 cm.The loop has 20 turns and carries a current I = 4 A.A magnetic field of strength 0.3 T is applied at an angle of $\theta$ = 30 $\circ$ to the loop.The potential energy of the magnetic dipole is
A)0.163 J
B)0.544 J
C)0.314 J
D)0.653 J
E)0.377 J

Question

Use the diagram for the next problem. 
-A circular current loop lies in the xy plane and has radius R = 10 cm.The loop has 20 turns and carries a current I = 4 A.A magnetic field of strength 0.3 T is applied at an angle of $\theta$ = 30 $\circ$ to the loop.The potential energy of the magnetic dipole is
A)0.163 J 
B)0.544 J 
C)0.314 J 
D)0.653 J 
E)0.377 J

Quizplus · Accepted Answer

The potential energy $U$ of a magnetic dipole in a magnetic field is given by $U = -\mu B \cos(\theta)$, where $\mu$ is the magnetic moment, $B$ is the magnetic field strength, and $\theta$ is the angle between the magnetic moment and the magnetic field. The magnetic moment $\mu$ is given by $NIA$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the loop. Given $N = 20$, $I = 4$ A, $R = 0.1$ m, and $B = 0.3$ T, with $\theta = 30^\circ$, we find $A = \pi R^2 = \pi (0.1)^2$, and thus $\mu = NIA = 20 \times 4 \times \pi (0.1)^2$. Substituting these values into the formula for $U$, and using $\cos(30^\circ) = \sqrt{3}/2$, we find $U = -\mu B \cos(30^\circ)$, which calculates to approximately $0.163$ J.

Use the Diagram for the Next Problem θ\thetaθ = 30 ∘\circ∘ to the Loop

Use the Diagram for the Next Problem $\theta$ = 30^$\circ$ to the Loop