Question

In Fig 29-39 two circular arcs have radii $\mathbf{}\mathbf{a}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{cm}}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{b}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{\text{cm}}$, subtend angle $\mathbf{\theta }\mathbf{=}\mathbf{74}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{,}$carry current $\mathit{i}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{411}\mathbf{}\mathbf{\text{A}}$, and share the same center of curvature P.What are the (a)magnitude and (b) direction (into or out of the page) of the net magnetic field at P?

Short answer

1. The magnitude is $B=1.02\times {10}^{-7}\text{T}$.
2. The direction of the magnetic field is out of the page.

Step-by-step solution

Given

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 $\theta =74.0°$
4. Current is$i=0.411\text{A}$
5. Figure 29-39 of the circular arcs.

Determine the magnetic field as:

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.

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

(a) Calculate the net magnetic field at P

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:

$B=\frac{{\mu }_{0}i\varphi }{4\pi R}+\frac{{\mu }_{0}i\varphi }{4\pi R}$

Substitute the values and solve as:

$$\begin{gathered} B=\frac{{\mu }_{0}i\varphi }{4\pi b}-\frac{{\mu }_{0}i\varphi }{4\pi a} \\ B=\frac{{\mu }_{0}i\varphi }{4\pi }\left(\frac{1}{b}-\frac{1}{a}\right) \\ B=\frac{4\pi \times {10}^{-7}i\varphi }{4\pi }\left(\frac{1}{b}-\frac{1}{a}\right) \\ B=i\varphi \times {10}^{-7}\left(\frac{1}{b}-\frac{1}{a}\right) \end{gathered}$$

Substitute the values and solve as:

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

Since, $\pi \text{rad}={180}^0$

$B=1.02\times {10}^{-7}\text{T}$

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

(b) Calculate the direction (into or out of the page) of the net magnetic field atP

Direction of the magnetic field at point P:

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