Figures shows combination of two semi-infinite long wires kept on Z and Y-axis. It carries constant current $\mathrm{I}$ as shown in figure. Calculate $\underset{Q}{\overset{P}{\int }}\overrightarrow{B}·\overrightarrow{dl}$  on line $QP$. Coordinates of points $P$ and $Q$ are $(\mathrm{a},0,0)$ and $(0, \mathrm{a}, 0)$respectively. If $\underset{Q}{\overset{P}{\int }}\overrightarrow{B}.\overrightarrow{dl}$  is $\frac{{\mu }_0\mathrm{I}}{K}$, then the value of $\text{'}K\text{'}$is

 

Due to wire along Y axis, $\int \overrightarrow{B}·\overrightarrow{dl}=0$. It is because at each point of the path, magnetic field and dl vector are perpendicular to each other.

For the wire along Z axis: Consider a point, say M, on the line $PQ$, such that angle between line OM and the line joining O to the midpoint of $PQ$ is $\theta$ . 

Magnetic field at the point M is $B=\frac{{\mu }_0\mathrm{I}}{4\mathrm{\pi r}}$, where $r$ is the distance of the point M from the point O. The length of a small differential element, MN, along $PQ$ around M can be written as $\mathit{\text{dl}}=\frac{\text{rdθ}}{\text{cosθ}}$

So, at M, $\overrightarrow{\text{B}}·\overrightarrow{\text{dl}}=\mathit{\left(}\frac{{\mathit{\text{μ}}}_{\mathit{\text{0}}}\text{I}}{\mathit{\text{4πr}}}\mathit{\right)}\left(\frac{\text{rdθ}}{\text{cosθ}}\right)\cos \theta =\frac{{\mathit{\text{μ}}}_{\mathit{\text{0}}}\text{I}}{\mathit{\text{4π}}}d\theta$

So, ${\int }_{\frac{-\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\overrightarrow{\text{B}}·\overrightarrow{\text{dl}}=\frac{{\mathit{\text{μ}}}_{\mathit{\text{0}}}\text{I}}{\mathit{\text{4π}}}\frac{\pi }{\mathit{2}}\mathit{=}\frac{{\mathit{\text{μ}}}_{\mathit{\text{0}}}\text{I}}{\mathit{8}}$