In Fig. 29-40, two semicircular arcs have radii ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{\text{cm}}$and ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{\text{cm}}$carry current $\mathit{i}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{281}\mathbf{}\mathbf{}\mathbf{\text{A}}$ and have the same center of curvature C. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at C?

1. Radius is$R_2=7.80\text{cm}$
2. Radius is$R_1=3.15\text{cm}$
3. Current is$i=0.281\text{A}$
4. Figure 29-40 is two semi-circular arcs.

Use the concept of magnetic field at the center of circular arc. Using the equation, find the magnetic field at the given point. Using right hand rule determine the direction of the magnetic field.

Formulae:
$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i\varphi }}{\mathbf{4}\mathbf{\pi R}}$

The magnitude of the magnetic field at point C:

The magnetic fields due to both arcs are opposite due to opposite direction of currents.

The total magnetic field will be the vector addition of both. So,

$\frac{{\mu }_{0}i\varphi }{4\pi R_1}-\frac{{\mu }_{0}i\varphi }{4\pi R_2}$

Substitute the values and solve further as:

$B=\frac{{\mu }_{0}i\varphi }{4\pi }\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

The angle is$\varphi =\pi$

$$\begin{gathered} B=\frac{4\pi \times {10}^{-7}i\pi }{4\pi }\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ B=i\pi \times {10}^{-7}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \end{gathered}$$

.

Substitute the values and solve:

role="math" localid="1663096149133" $$\begin{gathered} B=\left(0.281\text{A}\right)\left(3.14\right)\times {10}^{-7}\left(\frac{1}{0.0315\text{m}}-\frac{1}{0.0780\text{m}}\right) \\ B=1.67\times {10}^{-6}\text{T} \end{gathered}$$

Hence the magnetic field is, $1.67\times {10}^{-6}\text{T}$.

Direction of the magnetic field at point C:

By using the right hand rule, we can get the direction of the magnetic field which is into the page.