A long wire carrying a current i is bent to create a plane angle α...

The bent wire can be considered as two semi-infinite straight wires joined at the vertex.

Let the point on the angle bisector be $P$, located at a distance $x$ from the vertex. The perpendicular distance $d$ from $P$ to each semi-infinite wire is given by:

$d = x \sin$\left(\dfrac{\alpha}{2}\right)

For each semi-infinite wire, the magnetic field at $P$ is determined by the angles subtended by its ends at $P$ with respect to the perpendicular. One end extends to infinity, subtending an angle $\theta_2 =$\dfrac{\pi}{2}. The other end is at the vertex, subtending an angle $\theta_1 =$\dfrac{\pi}{2} - \dfrac{\alpha}{2}.

The magnetic field due to one semi-infinite wire is:

$B_1 =$\dfrac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$<br><br>Substituting the values of$d$,$\theta_1$, and$\theta_2$:<br><br>$B_1 = $\dfrac{\mu_0 i}{4\pi x \sin\left(\dfrac{\alpha}{2}\right)} \left[\sin \left(\dfrac{\pi}{2} - \dfrac{\alpha}{2}\right) + \sin\left(\dfrac{\pi}{2}\right)\right]$

$B_1 = \dfrac{\mu_0 i}{4\pi x \sin \left(\dfrac{\alpha}{2}\right)}$ \left(\$cos\dfrac{\alpha}{2} +$1\right)$<br><br>By the right-hand rule, the magnetic fields produced by both semi-infinite segments at point$P$point in the same direction. The total magnetic field$B is the sum of the fields from both wires:

$B = 2B_1 =$\dfrac{\mu_0 i}{2\pi x \sin \left(\dfrac{\alpha}{2}\right)}$$\left(1 + \$cos\dfrac{\alpha}{2}\right)<br><br>Using the half-angle trigonometric identities$1 + \$cos\dfrac{\alpha}{2} =$2\$\cos^2\dfrac{\alpha}{4} and$\$sin\dfrac{\alpha}{2} =$2\$sin\dfrac{\alpha}{4}\cos\dfrac{\alpha}{4}:<br><br>$B = $\dfrac{\mu_0 i}{2\pi x} \left(\dfrac{2\cos^2\dfrac{\alpha}{4}}{2\sin\dfrac{\alpha}{4}\cos\dfrac{\alpha}{4}}\right)$

$B = \dfrac{\mu_0 i}{2\pi x} \cot\dfrac{\alpha}{4}$

Answer: $\dfrac{\mu_0 i}{2\pi x} \cot\dfrac{\alpha}{4}$