The total magnetic induction at point O due to curved portion and straight portion in the following figure, will be

For arc, ${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi r}}\times \mathrm{\theta }$
${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}}{\mathrm{r}}[2\mathrm{\pi }-2\mathrm{\varphi }]=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}[\mathrm{\pi }-\mathrm{\varphi }]$
For straight conductor
$\begin{array}{l}{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi a}}[\sin \mathrm{\varphi }+\sin \mathrm{\varphi }];{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi }}\frac{\sin \mathrm{\varphi }}{\mathrm{a}}\\ \cos \mathrm{\varphi }=\frac{\mathrm{a}}{\mathrm{r}}\mathrm{\varphi },\mathrm{a}=\mathrm{rcos}\mathrm{\varphi };{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi }}\frac{\sin \mathrm{\varphi }}{\mathrm{rcos}\mathrm{\varphi }},\\ {\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{itan}\mathrm{\varphi }}{2\mathrm{\pi r}};\mathrm{B}={\mathrm{B}}_1+{\mathrm{B}}_2\\ \therefore \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}[\mathrm{\pi }-\mathrm{\varphi }+\tan \mathrm{\varphi }]\end{array}$