# Self-inductance of a coaxial cable: A coaxial cable consists of a long cylinder of radius $a$ which is surrounded by a hollow coaxial cylinder of the radius $b$. Find the self-induction per unit length of such a cable.

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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.

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.