A flat circular loop carrying a current of $2.0 \mathrm {~A}$ is in a magnetic field of $3.5 \mathrm {~T}$. The loop has an area of $0.12 \mathrm {~m} ^ { 2 }$ and its plane is oriented at a $37 ^ { \circ }$ angle to the field. What is the magnitude of the magnetic torque on the loop?
A) $0.67 \mathrm {~N} \cdot \mathrm { m }$
B) $0.10 \mathrm {~N} \cdot \mathrm { m }$
C) $46 \mathrm {~N} \cdot \mathrm { m }$
D) $0.51 \mathrm {~N} \cdot \mathrm { m }$

Question

Quizplus · Accepted Answer

The magnetic torque on a current loop is given by $\tau = NIAB\sin\theta$, where $N$ is the number of turns in the loop, $I$ is the current in the loop, $A$ is the area of the loop, $B$ is the magnetic field strength, and $\theta$ is the angle between the normal to the loop and the magnetic field. In this problem, $N = 1$, $I = 2.0 \mathrm {~A}$, $A = 0.12 \mathrm {~m} ^ { 2 }$, $B = 3.5 \mathrm {~T}$, and $\theta = 37 ^ { \circ }$. Plugging these values into the equation, we get $\tau = (1)(2.0 \mathrm {~A})(0.12 \mathrm {~m} ^ { 2 })(3.5 \mathrm {~T})\sin(37 ^ { \circ }) = 0.67 \mathrm {~N} \cdot \mathrm { m }$.

A Flat Circular Loop Carrying a Current Of 2.0 A2.0 \mathrm {~A}2.0 A

A Flat Circular Loop Carrying a Current Of $2.0 \mathrm {~A}$