A long straight wire carrying electric current ${}^{\text{'}}i^{\text{'}}$ is bent at its mid point to form an angle of $45°$ as shown in the figure. Magnetic field at a point P at a distance d from the point Q of bending is ________

Given data:

The midpoint of a long, straight wire that is carrying current I has been bent to form a 45-degree angle.

Formula used: $B=\frac{{\mu }_{0}d}{4\pi d}\left[\sin {\varphi }_1+\sin {\varphi }_2\right]$

Detailed Solution:

Based on the given question, it can be inferred that the point in question is located on the axis of the straight part OA of the wire. The magnetic field at point P is zero due to part OA of the wire. However, for part OC, the magnetic field at point P can be determined using the ΔOPN formula.

$\therefore d=R\cos {45}^0$

As both ends O and C of the wire are situated on the same side of the normal PN,

So,

${\varphi }_1=-{45}^0$ and $\varphi ={90}^0$

$B=\frac{{\mu }_{0}d}{4\pi d}\left[\sin {\varphi }_1+\sin {\varphi }_2\right]$

$\begin{array}{l}\Rightarrow B=\frac{{\mu }_{0}i}{4\pi R\cos {45}^0}\left[\sin \left(-{45}^0\right)+\sin \left({90}^0\right)\right]\\ \Rightarrow B=\frac{{\mu }_{0}i\sqrt{2}}{4\pi R}\left[-\sin \left({45}^0\right)+1\right]\\ \Rightarrow B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi R}\left[1-\frac{1}{\sqrt{2}}\right]\\ \Rightarrow B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi R}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]\\ \Rightarrow B=\frac{{\mu }_{0}i}{4\pi R}[\sqrt{2}-1]\\ \therefore B=\frac{{\mu }_{0}i}{4\pi R}[\sqrt{2}-1]\end{array}$