A flat circular coil of wire having 200 turns and diameter $6.0 \mathrm {~cm}$ carries a current of $7.0 \mathrm {~A}$. It is placed in a magnetic field of $0.60 \mathrm {~T}$ with the plane of the coil making an angle of $30 ^ { \circ }$ with the magnetic field. What is the magnitude of the magnetic torque on the coil?
A) $5.4 \mathrm {~N} \cdot \mathrm { m }$
B) $1.0 \mathrm {~N} \cdot \mathrm { m }$
C) $2.1 \mathrm {~N} \cdot \mathrm { m }$
D) $1.2 \mathrm {~N} \cdot \mathrm { m }$
E) $3.9 \mathrm {~N} \cdot \mathrm { m }$

Question

Quizplus · Accepted Answer

The torque ($\tau$) on a current-carrying coil in a magnetic field is given by $\tau = nIBA\sin(\theta)$, where $n$ is the number of turns, $I$ is the current, $B$ is the magnetic field strength, $A$ is the area of the coil, and $\theta$ is the angle between the magnetic field and the normal to the plane of the coil. Here, $n = 200$, $I = 7.0 \, \text{A}$, $B = 0.60 \, \text{T}$, $\theta = 30^\circ$, and the diameter of the coil is $6.0 \, \text{cm}$ so the radius $r = 3.0 \, \text{cm} = 0.03 \, \text{m}$. The area $A = \pi r^2 = \pi (0.03)^2 \, \text{m}^2$. Plugging these values into the formula gives $\tau = 200 \times 7.0 \times 0.60 \times \pi (0.03)^2 \times \sin(30^\circ)$, which calculates to approximately $2.1 \, \text{N} \cdot \text{m}$.

A Flat Circular Coil of Wire Having 200 Turns and Diameter