91Ó°ÊÓ

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

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

$\mathit{B}\mathbf{=}\frac{\mathit{μ}\mathit{I}\mathrm{ϕ}}{\mathbf{4}\mathit{π}\mathit{R}}$

For a straight conductor consider the formulas:

$\mathit{d}\stackrel{\mathbf{→}}{\mathit{B}}\mathbf{=}\frac{{\mathit{μ}}_{\mathbf{0}}\mathit{I}\mathit{d}\stackrel{\mathbf{→}}{\mathit{s}}\mathbf{×}\hat{\mathit{r}}}{\mathbf{4}\mathit{π}{\mathit{r}}^{\mathbf{2}}}$

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

Formulae:

$$\begin{gathered} \mathit{B}\mathbf{=}\frac{\mathbf{μ\pm õÏ•}}{\mathbf{4}\mathbf{Ï€¸é}} \\ \mathit{B}\mathbf{=}\frac{\mathbf{μ\pm õ»å²õ}\mathbf{}\mathbf{²õ¾\pm ²Ôθ}\mathbf{}}{\mathbf{4}{\mathbf{Ï€°ù}}^{\mathbf{2}}} \end{gathered}$$

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

$B_1=\frac{\mathrm{μ}Idssin\mathrm{θ}}{4\mathrm{π}{r}^2}$

$\mathrm{θ}=0°$

So,

$B_1=0$

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

$B_2=\frac{\mathrm{μ}Idssin\mathrm{θ}}{4\mathrm{π}{r}^2}$

Here $\mathrm{θ}=0°$

So

$B_2=0$

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

$$\begin{gathered} B_3=\frac{\mathrm{μ}I\mathrm{ϕ}}{4\mathrm{π}R} \\ {B}_3=\frac{\left(4\mathrm{π}×{10}^{-7}{\text{N/A}}^{\text{2}}\right)×34.8×{10}^{-3}\text{A}×\mathrm{π}}{4\mathrm{π}×9.26×{10}^{-2}\text{m}} \\ {B}_3=1.18×{10}^{-7}\text{T} \end{gathered}$$

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

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

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

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