Question

A square current loop 5.5 cm on each side carries a 470 mA current. The loop is in a 0.70 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30 degrees away from the field direction.
What is the magnitude of the torque on the current loop? 2 sig fig.

Answer 1

Concepts and reason

The concept used in this question is torque in magnetic field.

Firstly, find the magnetic moment of the loop. Then, find the angle between the magnetic moment and the magnetic field. Finally, find torque acting on the loop.

Fundamentals

The magnetic moment is a property of a magnet which interacts with the applied field to give a mechanical moment (due to which the magnet aligns itself in the applied field direction.).

The expression for the magnetic moment of a current carrying loop is as follows:

$\mu = IA$

Here, $\mu$ is the magnetic moment, I is the current and A is the area of the current carrying loop.

The torque acting on the loop to align it in the direction of the given magnetic field is given as follows:

$\tau = \mu \times B$

Here, B is the magnetic field.

The torque acting on the loop to align it in the direction of the given magnetic field is given as follows:

$\tau = \mu \times B$

Substitute IA for $\mu$ in the above expression and simplify it.

$\begin{array}{c}\\\tau = IA \times B\\\\ = IAB\sin \theta \\\end{array}$

Here, $\theta$ is the angle between the normal of the area and the magnetic field.

The area of the loop can be determined as follows:

$A = {a^2}$

Here, a is the side of the square loop.

Substitute ${a^2}$ for A in equation $\tau = IAB\sin \theta$.

$\tau = I{a^2}B\sin \theta$

Substitute 470 mA for I, 5.5 cm for a, 0.70 T for B, and ${30^{\rm{o}}}$ for $\theta$ in the above expression.

$\begin{array}{c}\\\tau = \left( {470{\rm{ mA}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ A}}}}{{1{\rm{ mA}}}}} \right){\left( {5.5{\rm{ cm}}} \right)^2}{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {0.70{\rm{ T}}} \right)\sin {30^{\rm{o}}}\\\\ = 5.0 \times {10^{ - 4}}{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans:

The magnitude of torque is $5.0 \times {10^{ - 4}}{\rm{ N}} \cdot {\rm{m}}$.