Question

a. What is the magnitude of the torque on the current loop in FIGURE EX29.40?

b. What is the loop’s equilibrium orientation?

Short answer

a. The magnitude of the torque on the current loop is $1.25\times {10}^{-11}N·m$.

b. The equilibrium orientation is about the horizontal axis.

Step-by-step solution

Part (a) step 1 : Given Information

We need to find the loop’s equilibrium orientation.

Part (a) step 2 : Simplifiy

We have a current in a wire that exerts a magnetic field on the loop. A current-carrying wire produces a magnetic field and the Biot-Savart law enables us to calculate the magnitude and direction of this magnetic field at any point where the magnetic field due to the segment $∆\overrightarrow{S}$of the current-carrying wire is given by equation ($29.7$)

$B_{wire}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I_{wire}}{r}\left(1\right)$

where $\tau$is the distance between the wire and the loop which equals$2cm$. Now, we use the values for $l_{wire}andr$ to get the magnetic field of the wire as:

$B_{wire}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I_{wire}}{r}$

$$\begin{gathered} B_{wire}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I_{wire}}{r} \\ =\frac{(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A})\left(2\mathrm{A}\right)}{2\mathrm{\pi }(0.02\mathrm{m})} \\ =2\times {10}^{-5}T \end{gathered}$$

to find the torque, a current loop inside a magnetic field experience a torque that is exerted by the magnetic force and the loop starts to rotate. For a circle loop with a diameter 2mm, its torque is given by equation (29.28) in the form

$\tau =IA{B}_{wire}\sin \theta \left(2\right)$

Where $A$is the area of the loop and $I$is the current in the loop, not the wire. The loop has a diameter 2 mm, so we get its area by

$A=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2=\mathrm{\pi }{\left(\frac{2\times {10}^{-3}\mathrm{m}}{2}\right)}^2=3.14\times {10}^{-6}{\mathrm{m}}^2$

Now, we plug the values for $I,A,B_{wire}and\theta$into equation (2) we get $\tau$

$$\begin{gathered} \tau =IA{B}_{wire}\sin \theta \\ =(0.2A)(3.14\times {10}^{-6}{m}^2)(2\times {10}^{-5}T)\sin {90}^{°} \\ =1.25\times {10}^{-11}N·m \end{gathered}$$

Part (b) step 1: Given Information

We need to find the loop’s equilibrium orientation.

Part (b) step 2: Explanation

The magnetic that is exerted on the loop is applied with in the horizontal axis with an angle $90°$Therefore the loop is oriented on the x-axis.

Hence, the equilibrium orientation is about the horizontal axis