Find the inductance of a unit length of two parallel wires, each of radius $a$, whose centres are at distance $d$ apart and carry equal currents in opposite directions. Neglect the flux within the wire.

• A

  $\frac{\mu_0}{2 \pi} \ln \left(\frac{d-a}{a}\right)$
• Correct answer
  B

  $\frac{\mu_0}{\pi} \ln \left(\frac{d-a}{a}\right)$
• C

  $\frac{3 \mu_0}{\pi} \ln \left(\frac{d-a}{a}\right)$
• D

  $\frac{\mu_0}{3 \pi} \ln \left(\frac{d-a}{a}\right)$

Solution

Since, the wires are infinite, so the system of these two wires can be considered as closed rectangle of infinite length and breadth equal to $d$. Flux through strip $\phi=\int_a^{d-a} \frac{\mu_0 I}{2 \pi r}(l d r)=\frac{\mu_0 I}{2 \pi} \ln \left(\frac{d-a}{a}\right)$
The other wire produces the same result, so the total flux through the rectangle $A B C D$ is $\phi_{\text {total }}=\frac{\mu_0 I I}{\pi} \ln \left(\frac{d-a}{a}\right)$ The total inductance of length 1 $L=\frac{I_{\text {total }}}{I}=\frac{\mu_0 I}{\pi} \ln \left(\frac{d-a}{a}\right)$ Inductance per unit length $=\frac{L}{l}=\frac{\mu_0}{\pi} \ln \left(\frac{d-a}{a}\right)$

Hence, the correct answer is option: 'B'.