Question

A rectangular loop of wire with a cross-sectional area of 0.118 m2 carries a current of 0.679 A. The loop is free to rotate about an axis that is perpendicular to a uniform magnetic field strength of 0.179 T. The plane of the loop is initially at an angle of 22.09o to the direction of the magnetic field. What is the magnitude of the torque on the loop ?

Answer 1

Given:

The area of the rectangular loop is,

$A=0.118\text{ m}^2$

The current in the loop is,

$I=0.679\text{ A}$

The magnetic field is of strength

$B=0.179\text{ T}$

The plane of the loop is at an angle

$\theta=22.09\degree$

To find:

The magnitude of the torque on the loop

Explanation:

The torque on a current-carrying loop is,

$\begin{gathered} \tau=IABsin\theta \\ =0.679\times0.118\times0.179\times sin22.09\degree \\ =5.39\times10^{-3}\text{ N.m} \end{gathered}$

Hence, the torque is

$5.39\times10^{-3}\text{ N.m}$