A coil of 150 turns of wire is in the form of a circle, carries a current of 63 mA, and sits in the center of a region of uniform magnetic field of strength 1.5 T. The torque on the coil has a maximum value of 0.25 N·m. What is the radius of the coils?
A) 9.2 cm
B) 13 cm
C) 0.56 cm
D) 7.5 cm

Question

Quizplus · Accepted Answer

The maximum torque ($\tau_{max}$) on a current-carrying coil in a magnetic field is given by the formula $\tau_{max} = NIBA$, where $N$ is the number of turns, $I$ is the current, $B$ is the magnetic field strength, and $A$ is the area of the coil. The area of a circle is $A = \pi r^2$, where $r$ is the radius. Rearranging the formula for torque to solve for $r$, we get $r = \sqrt{\frac{\tau_{max}}{NIB\pi}}$. Substituting the given values ($\tau_{max} = 0.25$ N·m, $N = 150$, $I = 63$ mA or $0.063$ A, $B = 1.5$ T), we find $r = \sqrt{\frac{0.25}{150 \times 0.063 \times 1.5 \times \pi}}$, which calculates to approximately 7.5 cm.

A Coil of 150 Turns of Wire Is in the Form