The angle formed by lines from each end of diameter to any point of the semicircle is $90°$.

Therefore, $\mathrm{CB} \perp \mathrm{DB}$ and $\mathrm{CA}\perp \mathrm{DA}$.

Magnetic field due to wire $\mathrm{AC}$,

$B_{\mathrm{AC}}=\frac{{\mathrm{\mu }}_{0}I}{4\mathrm{\pi }\left(2Rsin30°\right)}\left(cos30°-cos90°\right)=\frac{\sqrt{3}{\mathrm{\mu }}_{0}I}{8\mathrm{\pi }R}$

Magnetic field due to wire $\mathrm{BC}$,

$B_{\mathrm{BC}}=\frac{{\mathrm{\mu }}_{0}I}{4\mathrm{\pi }\left(2\mathit{R}sin60°\right)}\left(cos60°-cos90°\right)=\frac{{\mathrm{\mu }}_0\mathit{\text{I}}}{8\sqrt{3}\mathrm{\pi }\mathit{\text{R}}}=\frac{\sqrt{3}{\mathrm{\mu }}_0\mathit{\text{I}}}{24\mathrm{\pi }\mathit{\text{R}}}$

${\overrightarrow{\mathit{\text{B}}}}_{\text{net}}={\mathit{\text{B}}}_{\text{A}}-{\mathit{\text{B}}}_{\text{B}}$

       $=\frac{\sqrt{3}{\mathrm{\mu }}_{0}I}{8\mathrm{\pi }R}\left(1-\frac{1}{3}\right)$

       $=\frac{2\sqrt{3}{\mu }_{0}I}{24\pi R}=\frac{{\mu }_{\mathit{0}}I}{4\sqrt{3}\pi R}$