Question

A rectangular loop of wire with a cross-sectional area of 0.812 m2 carries a current of 3.302 A. The loop is free to rotate about an axis that is perpendicular to a uniform magnetic field strength of 4.768 T. The plane of the loop is initially at an angle of 65.562o to the direction of the magnetic field. What is the magnitude of the torque on the loop ?

Answer 1

The magnetic moment μ of a loop that carries a current A whose cross-sectional area is A is given by:

$\mu=IA$

On the other hand, the torque on a current loop with magnetic moment μ placed in a magnetic field B is:

$\tau=\mu B\sin\theta$

Where θ is the angle between the vectors μ and B.

The direction of μ is determined using the right hand rule for the current circulating in the loop.

Draw a diagram to visualize the situation:

Since the given angle is measured from the plane of the loop to the direction of the magnetic field, and the angle θ relevant for the calculation is the complementary of the given angle, then we should use the cosine of the given angle.

Then, replace μ=IA, I=3.302A, A=0.812m^2, B=4.768T and sinθ=cos(65.562º) to find the magnitude of the torque on the loop:

$\begin{gathered} \tau=\mu B\sin\theta \\ =IAB\cos(90º-\theta) \\ =(3.302A)(0.812m^2)(4.768T)\cdot\cos(65.562º) \\ =5.28887871...Nm \end{gathered}$

The area has a precision of 3 significant figures. Therefore, the torque on the loop is approximately:

$\tau=5.29Nm$