Question

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

• A. $\frac{{\mu }_0}{2\pi }$ ln $\left(\frac{d-a}{a}\right)$
• 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 $\left(\frac{d-a}{a}\right)$

Answer 1

Answer: (B)

Since the wires are infinite, so the system of these two wires can be considered as a closed rectangle of infinite length and breadth equal to $d$ Flux through the strip of area $ldr$, due to current flowing in one wire is given by
$\varphi ={\int }_a^{d-a}\frac{{\mu }_{0}I}{2\pi r}\left(ldr\right)$
$=\frac{{\mu }_{0}Il}{2\pi }ln\left(\frac{d-a}{a}\right)$
The other wire produces the same result, so the total flux through the dotted rectangle is
${\varphi }_{total}=\frac{{\mu }_{0}Il}{\pi }ln\left(\frac{d-a}{a}\right)$
The total inductance of length $l$,
$L=\frac{{\varphi }_{total}}{I}=\frac{{\mu }_{0}l}{\pi }ln\left(\frac{d-a}{a}\right)$
Inductance per unit length $=\frac{L}{l}=\frac{{\mu }_0}{\pi }$ ln $\left(\frac{d-a}{a}\right)$