Question

FIGURE P29.73 is an edge view of a 2.0 kg square loop, 2.5 m on each side, with its lower edge resting on a frictionless, horizontal surface. A 25 A current is flowing around the loop in the direction shown. What is the strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown?

Short answer

The strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium is$0.16T$.

Step-by-step solution

Given information

The mass of the square loop is $m=2.0kg$

The length of each side of the square loop is $L=2.5m$

Current flowing around the loop is $I=25A$

The angle made by a loop with horizontal is$\theta =25°$

Torque

At equilibrium $\underset{}{\sum \tau =0}$.

The magnitude of the torque tends to rotate the loop clockwise due to the normal force of the table. Compute it around a horizontal axis through the center of the loop parallel to the bottom edge touching the surface.

$$\begin{gathered} {\tau }_{table}=rF\sin \theta \\ =\left(\frac{L}{2}\right)mg\sin \theta \end{gathered}$$

The magnitude of the magnetic torque around the same axis is

$$\begin{gathered} {\tau }_m=\mu B\sin \theta \\ =I{L}^{2}B\sin \theta \end{gathered}$$

Set the magnitudes of the two torques equal to each other.

$$\begin{gathered} {\tau }_{table}={\tau }_m \\ \left(\frac{L}{2}\right)mg\sin \theta =I{L}^{2}B\sin \theta \\ B=\frac{mg}{2IL} \end{gathered}$$

Magnetic field

Substitute the given values into the equation obtained in step 2

$$\begin{gathered} B=\frac{2.0\times 9.8}{2\times 25\times 2.5} \\ =0.16T \end{gathered}$$

Therefore, the strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium is$0.16T$.