A flat circular loop of wire is in a uniform $\vec { B }$ field of 0.30 T.The diameter of the loop is 1.0 m,and a 2.0-A current flows in it.What is the magnitude of the magnetic torque on the loop when the plane of the loop is parallel to the $\vec { B }$ field?

Question

Quizplus · Accepted Answer

The magnetic torque on a current loop can be found using the equation:

$$\vec{	au} = \vec{m} 	imes \vec{B}$$

where $\vec{m}$ is the magnetic dipole moment of the current loop and $\vec{B}$ is the magnetic field.

For a flat circular loop of wire, the magnetic dipole moment is given by:

$$\vec{m} = I \vec{A} \hat{n}$$

where $I$ is the current, $\vec{A}$ is the area vector of the loop, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop.

When the plane of the loop is parallel to the magnetic field, the area vector is perpendicular to the magnetic field, so the angle between $\vec{m}$ and $\vec{B}$ is 90 degrees. This means that the magnitude of the magnetic torque is simply:

$$	au = mB$$

Substituting in the given values, we get:

$$	au = (2.0	ext{ A})(\pi(0.5	ext{ m})^2)(0.30	ext{ T}) = 0.47	ext{ N $\cdot$ m}$$

Therefore, the best choice is C.

A Flat Circular Loop of Wire Is in a Uniform B⃗\vec { B }B

A Flat Circular Loop of Wire Is in a Uniform $\vec { B }$