Answer

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**Hint:**An emf induced in a circuit due to the changes in the circuit's magnetic field is known as self-inductance. Here we have a coaxial cable consisting of two long cylinders of radius $a$ and $b$. We have to use ampere’s circuital law to find the self-inductance of the coaxial cable.

**Formula used:**

$B = \dfrac{{{\mu _0}I}}{{2\pi x}}$

where $B$ stands for the field,

${\mu _0}$ stands for the permeability of free space,

$I$ stands for the current flowing through the circuit,

$x$ stands for the radius of the cable.

$d\phi = BdA$

where $d\phi $ stands for the induced emf,

$B$ stands for the field, and

$dA$ stands for the small area considered.

**Complete step by step answer:**

The radius of the coil inside is given as $a$, and the coil surrounding has a radius $b$.

The field in the circuit is given by,

$B = \dfrac{{{\mu _0}I}}{{2\pi x}}$

The emf induced in a small area is given by,

$d\phi = BdA$

Substituting the value of $B$ in the above equation, we get

$d\phi = \dfrac{{{\mu _0}I}}{{2\pi x}}ldx$ (Considering a small rectangular area, $dA = ldx$)

To find the total induction, we have to integrate the above equation between the radii of the two cables, $a$ and $b$.

$d\phi = \int\limits_a^b {\dfrac{{{\mu _0}I}}{{2\pi x}}ldx} $

Taking the constants out,

$d\phi = \dfrac{{{\mu _0}Il}}{{2\pi }}\int\limits_a^b {\dfrac{{dx}}{x}} $

Integrating, we get

$\phi = \dfrac{{{\mu _0}Il}}{{2\pi }}\left[ {\ln x} \right]_a^b$

Applying the limits, we get

$\phi = \dfrac{{{\mu _0}Il}}{{2\pi }}\left[ {\ln b - \ln a} \right]$

This can be written as,

$\phi = \dfrac{{{\mu _0}Il}}{{2\pi }}\left[ {\ln b - \ln a} \right]$

When a current passes through a coil, flux is associated with the coil. The flux $\phi $is proportional to the current $I$ through the coil.

That is,

$\phi \propto I$

This can be written as,

$\phi = LI$

where $L$ is called the coefficient of self-induction or self-inductance.

Substituting this value for $\phi $,

$LI = \dfrac{{{\mu _0}Il}}{{2\pi }}\ln \dfrac{b}{a}$

Common terms on both sides are cancelled, we get

$L = \dfrac{{{\mu _0}l}}{{2\pi }}\ln \dfrac{b}{a}$

The self-inductance of a coaxial coil is thus, $L = \dfrac{{{\mu _0}l}}{{2\pi }}\ln \dfrac{b}{a}$

**Note:**The phenomenon by which a coil opposes the growth or decay of current through it by producing an emf. When $1A$ current passes through the coil, the coefficient of self-induction of a coil is numerically equal to the flux linked with the coil.

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