Question

a circular loop of wire with a radius of 20 cm lies in the xy - plane and carries a current of 2.0 a, counterclockwise when viewed from a point on the positive z - axis. its magnetic dipole moment is
o 2.5 a·m², in the negative z direction.
o 0.25 a·m², in the positive z direction.
o 2.5 a·m², in the positive z direction.
o 0.25 a·m², in the xy - plane.
o 0.25 a·m², in the negative z direction.
save for later
submit answer
28.8 the magnetic dipole moment

Explanation:

Step1: Recall magnetic - dipole moment formula

The magnetic - dipole moment of a current - carrying loop is given by $\vec{\mu}=IA\hat{n}$, where $I$ is the current, $A$ is the area of the loop, and $\hat{n}$ is the unit vector normal to the loop.

Step2: Calculate the area of the circular loop

The area of a circle is $A = \pi r^{2}$. Given $r = 20\ cm=0.2\ m$, then $A=\pi(0.2)^{2}= 0.04\pi\ m^{2}$.

Step3: Calculate the magnetic - dipole moment

Given $I = 2.0\ A$, then $\mu=IA$. Substituting the values, we get $\mu=2\times0.04\pi\ A\cdot m^{2}\approx 0.25\ A\cdot m^{2}$.

Step4: Determine the direction

Using the right - hand rule, for a counter - clockwise current in the $xy$ - plane (viewed from the positive $z$ - axis), the direction of the magnetic - dipole moment is along the positive $z$ - axis.

Answer:

$0.25\ A\cdot m^{2}$, in the positive $z$ direction.