91Ó°ÊÓ

1. Radius of outer arc is$a=13.5\text{cm}$
2. Radius of inner arc is$b=10.7\text{cm}$
3. Angle is $\mathrm{θ}=74.0°$
4. Current is$i=0.411\text{A}$
5. Figure 29-39 of the circular arcs.

Use the concept of magnetic field due to a circular arc. Using the equation of magnetic field at the center of the circular arc, find the magnetic field at the given point and from the right hand rule, find the direction of the magnetic field.

$\mathit{B}\mathbf{=}\frac{{\mathit{μ}}_{\mathbf{0}}\mathit{i}\mathrm{ϕ}}{\mathbf{4}\mathit{π}\mathit{R}}$

The magnitude of the magnetic field at point P:

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

Write the total magnetic field as:

$B=\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}R}+\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}R}$

Substitute the values and solve as:

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}b}-\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}a} \\ B=\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}}\left(\frac{1}{b}-\frac{1}{a}\right) \\ B=\frac{4\mathrm{π}×{10}^{-7}i\mathrm{ϕ}}{4\mathrm{π}}\left(\frac{1}{b}-\frac{1}{a}\right) \\ B=i\mathrm{ϕ}×{10}^{-7}\left(\frac{1}{b}-\frac{1}{a}\right) \end{gathered}$$

Substitute the values and solve as:

$B=0.411\text{A}×74.0^0×\left(\frac{\mathrm{π}\text{rad}}{{180}^0}\right)×{10}^{-7}\left(\frac{1}{0.107\text{cm}}-\frac{1}{0.135\text{cm}}\right)$

Since, $\mathrm{Ï€}\text{rad}={180}^0$

$B=1.02×{10}^{-7}\text{T}$

Hence the value of the magnetic field is, $1.02×{10}^{-7}\text{T}$.

Direction of the magnetic field at point P:

According to the right hand rule, the direction of magnetic field is out of page.