Question

In Fig. 29-4, a wire forms a semicircle of radius $\mathit{R}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{26}\mathbf{}\mathbf{\text{cm}}$and two (radial) straight segments each of length $\mathit{L}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{1}\mathbf{\text{cm}}$. The wire carries current $\mathbf{}\mathit{i}\mathbf{=}\mathbf{34}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\text{mA}}$. What are the(a) magnitude and(b) direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C?

Short answer

1. The magnitude of the magnetic field is$B=1.18\times {10}^{-7}\text{T}$
2. The direction of the magnetic field is into the page.

Step-by-step solution

Given

1. Radius of circle is$R=9.26\text{cm}=9.26\times {10}^{-2}\text{m}$
2. Length of each straight segment isrole="math" localid="1663098315135" $L=13.1\text{cm}=13.1\times {10}^{-2}\text{m}$
3. Current is$i=34.8\text{mA}=34.8\times {10}^{-3}\text{A}$

Determine the formulas

The magnitude of the magnetic field Bat centre of circular arc, of radius R, and central angle$\mathit{\varphi }$, carrying a current Iis given as follows:

$\mathit{B}\mathbf{=}\frac{\mathit{\mu }\mathit{I}\varphi }{\mathbf{4}\mathit{\pi }\mathit{R}}$

For a straight conductor consider the formulas:

$\mathit{d}\overrightarrow{\mathit{B}}\mathbf{=}\frac{{\mathit{\mu }}_{\mathbf{0}}\mathit{I}\mathit{d}\overrightarrow{\mathit{s}}\mathbf{\times }\hat{\mathit{r}}}{\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}}$

Here,$\mathit{d}\overline{\mathit{B}}$is the magnetic field through the current carrying conductor,$\mathit{d}\overline{\mathbf{s}}$length of the element and$\hat{\mathbf{r}}$is unit vector that is a point from the element to the given point.

Formulae:

$$\begin{gathered} \mathit{B}\mathbf{=}\frac{\mathbf{\mu I\varphi }}{\mathbf{4}\mathbf{\pi R}} \\ \mathit{B}\mathbf{=}\frac{\mathbf{\mu Ids}\mathbf{}\mathbf{sin\theta }\mathbf{}}{\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}} \end{gathered}$$

(a) Calculate the magnitude of the net magnetic field at the semicircle’s center of curvature C

To calculate the magnetic field $B_1$due to current for left straight segment as follows:

$B_1=\frac{\mu Idssin\theta }{4\pi r^2}$

$\theta =0°$

So,

$B_1=0$

Calculate the magnetic field $B_2$due to current for a right straight segment as follows:

$B_2=\frac{\mu Idssin\theta }{4\pi r^2}$

Here $\theta =0°$

So

$B_2=0$

Calculate the magnetic field $B_3$due to current from circular section as follows

$$\begin{gathered} B_3=\frac{\mu I\varphi }{4\pi R} \\ {B}_3=\frac{\left(4\pi \times {10}^{-7}{\text{N/A}}^{\text{2}}\right)\times 34.8\times {10}^{-3}\text{A}\times \pi }{4\pi \times 9.26\times {10}^{-2}\text{m}} \\ {B}_3=1.18\times {10}^{-7}\text{T} \end{gathered}$$

So, the total magnetic field at the center of semicircle arc is

$$\begin{gathered} B=B_1+B_2+B_3 \\ =1.18\times {10}^{-7}\text{T} \end{gathered}$$

Hence the magnetic field is, $1.18\times {10}^{-7}\text{T}$.

(b) Calculate the direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C

Using the right hand rule, direction of magnetic field is into the page