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

Question

Quizplus · Accepted Answer

The torque on a coil in a magnetic field is given by $\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)$, where $n$ is the number of turns, $I$ is the current, $A$ is the area of the coil, $B$ is the magnetic field, and $\theta$ is the angle between the normal to the coil and the magnetic field. Here, $n = 200$, $I = 7.0 \, \text{A}$, $B = 0.60 \, \text{T}$, $\theta = 30^\circ$, and the area $A = \pi \times (0.03 \, \text{m})^2$. Calculating gives $\tau = 200 \times 7.0 \times \pi \times (0.03)^2 \times 0.60 \times \sin(30^\circ) = 2.1 \, \text{N} \cdot \text{m}$.

A Flat Circular Coil of Wire Having 200 Turns and Diameter