A long straight wire, carrying current I, is bent at its midpoint to form an angle of 45^o. The magnetic field at point P which is at distance R from the point of bending must be

For segment BC, perpendicular distance is

$\mathrm{r}=\mathrm{Rcos}\left({45}^{\circ }\right)$

B and C ends are on the same side of the perpendicular PN.

Therefore, $\alpha$ becomes negative and $\beta$ is positive.

Hence, $\alpha =-{45}^{\circ }\text{ and }\beta ={90}^{\circ }$

Using

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi r}}(\mathrm{Sin}\mathrm{\alpha }+\mathrm{Sin}\mathrm{\beta })$

$\Rightarrow \mathrm{B}=\frac{(\sqrt{2}-1){\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi R}}$

(out of the plane of paper)