A circular, 20-turn coil has a radius of 5.0 cm. What is the magnitude of the magnetic moment ) of the coil when it carries a current of 2.5 A?

Question

A circular, 20-turn coil has a radius of 5.0 cm. What is the magnitude of the magnetic moment  ) of the coil when it carries a current of 2.5 A?

Quizplus · Accepted Answer

The magnetic moment of a current loop is given by the formula:

$\boldsymbol{\mu} = \boldsymbol{I} 	imes \boldsymbol{A}$

where $\boldsymbol{I}$ is the current vector and $\boldsymbol{A}$ is the area vector of the loop.

For a circular loop, the area vector points perpendicular to the plane of the loop and has a magnitude equal to the area of the loop:

$|\boldsymbol{A}| = \pi r^2$

where $r$ is the radius of the loop.

The current vector for a loop is defined to circulate around the loop in the direction determined by the right-hand rule (curl your fingers in the direction of the current, and your thumb points in the direction of the magnetic moment vector).

For a circular loop with a current circulating uniformly around its perimeter, the magnetic moment vector points along the axis of the loop and has a magnitude given by:

$|\boldsymbol{\mu}| = I \pi r^2$

Substituting the given values, we get:

$|\boldsymbol{\mu}| = (2.5\ \mathrm{A})(\pi)(0.05\ \mathrm{m})^2 = 0.39\ \mathrm{A \cdot m}$

Therefore, the answer is (D) 0.39 A · m.

A Circular, 20-Turn Coil Has a Radius of 5