Figures Shows combination of two semi infinite long wires kept on Z and Y-axis. It carries constant current I as shown in figure. $\underset{Q}{\overset{P}{\int }}\overrightarrow{B}.\overrightarrow{dl}$  on line QP is  $x{\mu }_{0}i.$ Then fill the value of  $\frac{1}{x}$ _______. 

Co-ordinate of point P and Q are (a, 0, 0) and (0, a, 0) respectively 

$\underset{Q}{\overset{P}{\int }}\overrightarrow{B}.d\overrightarrow{\mathcal{l}}$  due to current in y axis is zero. As field due to this is r to line QP.
Now for semi-infinite wire along z axis:
If infinite wire
 
$\underset{Q}{\overset{P}{\int }}\overrightarrow{B}.d\overrightarrow{\mathcal{l}}=\frac{{\mu }_{0}i}{4}$  but field at points of QP due to semi-infinite wire will be half due to infinite
$\Rightarrow \underset{Q}{\overset{P}{\int }}\overrightarrow{B}.d\overrightarrow{\mathcal{l}}=\frac{{\mu }_{0}i}{8}=x{\mu }_{0}i$ 
 $\Rightarrow x=\frac{1}{8}$ Hence  $\frac{1}{x}=8$ Ans.