Answer

Verified

385.5k+ views

**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.

Recently Updated Pages

Draw a labelled diagram of DC motor class 10 physics CBSE

A rod flies with constant velocity past a mark which class 10 physics CBSE

Why are spaceships provided with heat shields class 10 physics CBSE

What is reflection Write the laws of reflection class 10 physics CBSE

What is the magnetic energy density in terms of standard class 10 physics CBSE

Write any two differences between a binocular and a class 10 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail