A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it.
A) 0.41 N ⋅ m
B) 0.14 N ⋅ m
C) 0.38 N ⋅ m
D) 0.27 N ⋅ m
E) 0.77 N ⋅ m

Question

Quizplus · Accepted Answer

The magnetic torque on a current loop in a magnetic field is given by:

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

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

For a circular loop of radius $r$ carrying a current $I$, the magnetic dipole moment is:

$\vec{\mu} = I \vec{A}$

where $\vec{A}$ is the area vector of the loop, which is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop.

In this problem, the circumference of the loop is given as 80 cm, so the radius is:

$r = \frac{C}{2 \pi} = \frac{80	ext{ cm}}{2\pi} \approx 12.7	ext{ cm}$

The current in the loop is given as 4.0 A, so the magnetic dipole moment is:

$\vec{\mu} = I \vec{A} = I \pi r^2 \hat{n}$

where $\hat{n}$ is a unit vector perpendicular to the plane of the loop, pointing in the direction of the current.

The magnitude of the magnetic field is given as 2.0 T, and the angle between the field and the plane of the loop is 20°, so the component of the magnetic field perpendicular to the loop is:

$B_\perp = B \sin 	heta = 2.0	ext{ T} \sin 20^\circ \approx 0.68	ext{ T}$

The torque on the loop is:

$	au = \vec{\mu} 	imes \vec{B} = I \pi r^2 \hat{n} 	imes B_\perp \hat{m}$

where $\hat{m}$ is a unit vector perpendicular to both $\hat{n}$ and the direction of the magnetic field, determined by the right-hand rule.

The magnitude of the torque is:

$|	au| = |\vec{\mu}| |\vec{B}_\perp| \sin 	heta = I \pi r^2 B_\perp \sin 	heta$

Substituting the given values, we get:

$|	au| = (4.0	ext{ A}) (\pi)(0.127	ext{ m})^2 (0.68	ext{ T}) \sin 20^\circ \approx 0.38	ext{ N} \cdot 	ext{m}$

Therefore, the answer is (C) 0.38 N ⋅ m.

A Current of 4