Question

Find the inductance between the area enclosed by the two long parallel wires for a unit length, 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. $\frac{{\mu }_0}{2\pi }\ln \left(\frac{d-a}{a}\right)$
• 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

The correct option is B $\frac{{\mu }_0}{\pi }\ln \left(\frac{d-a}{a}\right)$


Since, the wires are of infinite length, 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 $(l\cdot dr),$ 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

${\varphi }_{\text{total}}=2\varphi =\frac{{\mu }_{0}Il}{\pi }\ln \left(\frac{d-a}{a}\right)$

$\therefore$ Flux for unit length of wires is,

${\varphi }_{\text{total}}=\frac{{\mu }_{0}Il}{\pi }\ln \left(\frac{d-a}{a}\right)$


The total inductance,

$L=\frac{{\varphi }_{\text{total}}}{I}=\frac{{\mu }_{0}l}{\pi }\ln \left(\frac{d-a}{a}\right)$

The total inductance of unit length,

$L=\frac{{\mu }_0}{\pi }\ln \left(\frac{d-a}{a}\right)$

Hence, option $\left(\text{B}\right)$ is correct.