Two long parallel wires have radii equal to a and are at distance d measured between their centres as shown. Their far ends are connected together so that the wires form two sides of the same loops. For a section far from the ends, if the flux within the wires is neglected, then the self-inductance per meter is 

Let us calculate the coefficient of self-induction for this arrangement, we need to find the flux between the wires. Let us consider a strip of width dy and length x as shown in figure.   
   
Magnetic field at the strip due to current in two wires, 
    $B=\frac{{\mu }_{0}I}{2\pi }(\frac{1}{y}+\frac{1}{d-y})\otimes$ 
Area of the strip, dA = x dy 
Hence the flux through the strip, $d\varphi =\frac{{\mu }_{0}l}{2\pi }(\frac{l}{y}+\frac{l}{d-y})xdy$         
Flux linked with the area between two wires in a length x will be  
 $\begin{array}{l}\varphi \simeq \frac{{\mu }_{0}Ix}{2\pi }[\underset{a}{\overset{d-a}{\int }}\frac{dy}{y}+\underset{a}{\overset{d-a}{\int }}\frac{dy}{d-y}]\\ \simeq \frac{{\mu }_{0}Ix}{2\pi }[\ln \left(\frac{d-a}{a}\right)-\ln \left(\frac{d-d+a}{d-a}\right)]\\ \simeq \frac{{\mu }_{0}Ix}{\pi }\ln \left(\frac{d-a}{a}\right)\end{array}$