Field from Half Wire

$\textcolor{#20A900}{Fun problem}$

Due to symmetry this problem can be converted into this $A(\sqrt{2}, -\sqrt{2})$
I will not show all the basic steps untill anybody asks me. @Karan Chatrath has also explained basic steps for this types of problem earlier.

Finally I reach
$\Large B_{x}=\int_{-1}^{+1} \int_{-\sqrt{1-x^{2}}}^{0} \frac{\sqrt{2}+y}{(2π((x-\sqrt{2})^{2}+(y+\sqrt{2})^{2})) }dydx \approx 0.0896954$ $\Large B_{y}=\int_{-1}^{+1} \int_{-\sqrt{1-x^{2}}}^{0} \frac{\sqrt{2}-x}{(2π((x-\sqrt{2})^{2}+(y+\sqrt{2})^{2})) }dydx \approx 0.114775$
$\Large B_{net}=\sqrt{B_{x}^{2} +B_{y}^{2}} \approx \boxed{\textcolor{#3D99F6}{0.145}}$

General guideline for solution. The tedious computations can be outsourced to Wolfram-Alpha. Consider a general point within the orange region $(x,y)$ . Position vector joining $(2,0)$ and that general point directed towards $(2,0)$ is:

$\vec{R} = (2-x) \ \hat{i} - y \ \hat{j}$ $\hat{R} = \frac{\vec{R}}{\lvert \vec{R} \rvert}$

The magnetic field due to an infinite current-carrying wire where current flows along the Z direction is:

$\vec{B} = \frac{\mu_o I}{2 \pi \lvert \vec{R} \rvert} \left(\hat{k} \times \hat{R}\right)$

Applying this formula to an area element in the given region yields:

$d\vec{B} = \frac{ dy \ dx}{2 \pi \sqrt{(2-x)^2+y^2}} \left(\frac{(2-x) \ \hat{i} - y \ \hat{j}}{\sqrt{(2-x)^2+y^2}}\right)$

$dB_x = \frac{1}{2 \pi } \ \frac{y \ dy \ dx}{\left((2-x)^2+y^2\right)} =f_1(x,y) \ dy \ dx$ $dB_y = \frac{1}{2 \pi } \ \frac{(2-x) \ dy \ dx}{\left((2-x)^2+y^2\right)}=f_2(x,y) \ dy \ dx$

$B_x = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1-x^2}}^{x} f_1(x,y) \ dy \ dx + \int_{1/\sqrt{2}}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f_1(x,y) \ dy \ dx$ $B_y = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1-x^2}}^{x} f_2(x,y) \ dy \ dx + \int_{1/\sqrt{2}}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f_2(x,y) \ dy \ dx$