Question: In Fig. 29-77, a closed loop carries current $\mathbf{200}\mathbf{}\mathbf{mA}$. The loop consists of two radial straight wires and two concentric circular arcs of radii $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$and $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. The angle is role="math" localid="1662809179609" $\mathit{\theta }\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{4}}\mathit{r}\mathit{a}\mathit{d}$. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at the center of curvature P?

1. Loop carries a current of $200\mathrm{mA}$
2. Radius$r_1=2.00\mathrm{m}$
3. Radius $r_2=4.00\mathrm{m}$
4. The angle is $\theta =\frac{\mathrm{\pi }}{4}\mathrm{rad}$

We use the formula for magnetic field at the center of a circular arc, and it depends on current, radius of circular arc, and central angle.

Formula:

$\mathit{B}\mathbf{=}\frac{\mathbf{\mu }\mathbf{i}\mathbf{\Phi }}{\mathbf{4}\mathbf{\pi R}}$

The central angle is as follows-

$$\begin{gathered} \varnothing =2\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ =\frac{7\mathrm{\pi }}{4}\mathrm{rad}. \end{gathered}$$

Now magnetic field due to the outer wire is as follows

$$\begin{gathered} B_1=\frac{\mu i\varnothing }{4\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200\times {10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{\pi }}{4}}\right)}{4\mathrm{\pi }\times 4} \\ =2.74\times {10}^{-8}\mathrm{T} \end{gathered}$$

From the right hand rule, this field is directed out of page.

Now magnetic field due to inner wire is as follows

$$\begin{gathered} B_2=\frac{\mu i\varnothing }{4\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200\times {10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{\pi }}{4}}\right)}{4\mathrm{\pi }\times 2} \\ =5.49\times {10}^{-8}\mathrm{T} \end{gathered}$$

From the right hand rule, this field is directed into the page.

Net field is as follows

role="math" localid="1662811190294" $$\begin{gathered} B=B_2-B_1 \\ =5.49\times {10}^{-8}\mathrm{T}-2.74\times {10}^{-8}\mathrm{T} \\ =2.75\times {10}^{-8}\mathrm{T} \end{gathered}$$

Net field is directed into the page because field by the inner arc is greater than field by the outer arc.