A long wire carrying a current $i$ is bent to form a plane angle $\theta$. The magnetic field at a point $O$ on the bisector of this angle situated at a distance $d$ from vertex $P$ is

Magnetic field due to both wires is in $\odot$ direction, so additive. Due to one wire,

$B=\frac{{\mu }_0}{4\pi }\frac{i}{r}\left(\sin \alpha +\sin \beta \right)$

$=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i}{d\sin {\displaystyle \frac{\theta }{2}}}\left[\cos \frac{\theta }{2}+1\right]$

$=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i}{d\sin {\displaystyle \frac{\theta }{2}}}\left[2{\cos }^2\frac{\theta }{4}-1+1\right]$

$=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{i}{d}\left[\frac{2{\cos }^2{\displaystyle \frac{\theta }{4}}}{2\sin {\displaystyle \frac{\theta }{4}}\cos {\displaystyle \frac{\theta }{4}}}\right]$

 $=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i\cot {\displaystyle \frac{\theta }{4}}}{d}$

$\therefore$Magnetic field due to both the wires will be double of it.