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?

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$i=34.8\text{mA}=34.8\times {10}^{-3}\text{A}$

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:

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

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, $1.18\times {10}^{-7}\text{T}$.

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