Question

A square current loop $5.0cm$on each side carries a$500mA$ current. The loop is in a $1.2T$ uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is $30°$° away from the field direction. What is the magnitude of the torque on the current loop?

Short answer

The magnitude of the torque on the current loop is$7.5\times {10}^{-4}N·m$.

Step-by-step solution

Given Information

We need to find the magnitude of the torque on the current loop.

Simplify

The torque $\tau$represents the magnitude of the force multiplied by the distance between the axis and the line of the action. A current loop inside a magnetic field experiences a torque that is exerted by the magnetic force and the loop starts to rotate. For a square loop with $l$, its torque is given by equation ($29.28$) in the form :

$\tau =(I{l}^2)B\sin \theta$

Where $I^2$represents the area of the loop, $l$is the current in the loop and $\theta$is the angle between the magnetic field and the magnetic dipole moment of the loop.

Calculation

When the dipole moment is parallel to the magnetic field, the torque is zero and it is maximum when the dipole moment is perpendicular to the magnetic field. Now, we plug the values for $I,l,Band\theta$into equation ($1$) to get as $\tau$.

$$\begin{gathered} \tau =(I{l}^2)B\sin \theta \\ =(500\times {10}^{-3}A){(0.05m)}^2(1.2T)\sin {30}^{°} \\ =7.5\times {10}^{-4}N.m \end{gathered}$$