91Ó°ÊÓ

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 $\mathrm{θ}=\frac{\mathrm{π}}{4}\mathrm{rad}$

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:

$\mathit{B}\mathbf{=}\frac{\mathbf{μ}\mathbf{i}\mathbf{Φ}}{\mathbf{4}\mathbf{Ï€¸é}}$

The central angle is as follows-

$$\begin{gathered} âˆ\dots =2\mathrm{Ï€}-\frac{\mathrm{Ï€}}{4} \\ =\frac{7\mathrm{Ï€}}{4}\mathrm{rad}. \end{gathered}$$

Now magnetic field due to the outer wire is as follows

$$\begin{gathered} B_1=\frac{\mathrm{μ}iâˆ\dots }{4\mathrm{Ï€¸é}} \\ =\frac{\left(4\mathrm{Ï€}×{10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200×{10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{Ï€}}{4}}\right)}{4\mathrm{Ï€}×4} \\ =2.74×{10}^{-8}\mathrm{T} \end{gathered}$$

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

Now magnetic field due to inner wire is as follows

$$\begin{gathered} B_2=\frac{\mathrm{μ}iâˆ\dots }{4\mathrm{Ï€¸é}} \\ =\frac{\left(4\mathrm{Ï€}×{10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200×{10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{Ï€}}{4}}\right)}{4\mathrm{Ï€}×2} \\ =5.49×{10}^{-8}\mathrm{T} \end{gathered}$$

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

Net field is as follows

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

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