# Lecture 23 - Inductance

## Transformers

In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same.

$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$

## Mutual Inductance

In the example of a transformer we assumed that the flux induced in the secondary coil was equal to the flux in the primary coil.

In general for two coils the relationship between the flux in one coil due to the current in another is described by a parameter called the mutual inductance.

$\Phi_{21}$ is the magnetic flux in each loop of coil 2 created by the current in coil 1.

The total flux in the second coil is then $N_{2}\Phi_{21}$ and is related to the current in coil 1, $I_{1}$ by

$N_{2}\Phi_{21}=M_{21}I_{1}$

As, from Faraday's Law, the emf induced in coil 2 is $\mathcal{E}_{2}=-N_{2}\frac{d\Phi_{21}}{dt}$ so

$\mathcal{E}_{2}=-M_{21}\frac{dI_{1}}{dt}$

The mutual inductance of coil 2 with respect to coil 1, $M_{21}$ does not depend on $I_{1}$.

It does depend on factors such as the size, shape and number of turns in each coil, their position relative to each other and whether there is some ferromagnetic material in the vicinity.

In the reverse situation where a current flows in coil 2

$\mathcal{E}_{1}=-M_{12}\frac{dI_{2}}{dt}$

but in fact $M_{12}=M_{21}=M$

The mutual inductance is measured in Henrys ($\mathrm{H}$), $1\mathrm{H}=1\mathrm{\frac{Vs}{A}}=1\mathrm{\Omega s}$

## Mutual inductance of a solenoid and a coil

To calculate the mutual inductance of the above situation we consider the magnetic field of a solenoid with cross-sectional area $A$

$B=\mu_{0}\frac{N_{1}}{l}I_{1}$

The magnetic flux through the loose coil due to the current in the solenoid is thus

$\Phi_{21}=BA=\mu_{0}\frac{N_{1}}{l}I_{1}A$

and the mutual inductance is then

$M=\frac{N_{2}\Phi_{21}}{I_{1}}=\frac{\mu_{0}N_{1}N_{2}A}{l}$

## Self-inductance

Now consider a single coil to which a time varying current is applied.

A changing magnetic field is produced, which induces an emf in the coil, which opposes the change in flux.

The magnetic flux $\Phi_{B}$ passing through the coil is proportional to the current.

As we did for mutual inductance we can define a constant of proportionality between the current and the flux:

The self-inductance $L$

$N\Phi_{B}=LI$

The emf $\mathcal{E}=-N\frac{d\Phi_{B}}{dt}=-L\frac{dI}{dt}$

The self-inductance is also measured in henrys.

A component in a circuit that has significant inductance is shown by the symbol.

When we draw this symbol it implies an inductor with negligible resistance.

If the inductor has significant resistance we draw that as a resistor in series with the inductor.

A large inductor does however reduce the AC current flowing through a circuit because the back emf generated opposes the applied potential so that the total potential across the inductor is small, and the current is also small.

The degree to which an inductor opposes an AC current is called the reactance (more on that later!).

## Self-inductance of a solenoid

We can can calculate the self-inductance of a solenoid from it's field

$B=\mu_{0}\frac{NI}{l}$

The flux in the solenoid is

$\Phi_{B}=BA=\mu_{0}\frac{N_{1}IA}{l}$

so

$L=\frac{N\Phi_{B}}{I}=\frac{\mu_{0}N^{2}A}{l}$

## Coaxial cable inductance

To find the inductance we need to know the total flux that is generated by the current.

$\Phi_{B}=\int\vec{B}\cdot d\vec{A}$

From Ampere's law ($\oint\vec{B}\cdot d\vec{l}=\mu_{0}I$)

$B=\frac{\mu_{0}I}{2\pi r}$

The magnetic flux through a rectangle of width $dr$ and length $l$ at a distance $r$ from the center

$d\Phi_{B}=B(l\,dr)=\frac{\mu_{0}I}{2\pi r}l\,dr$

The total flux is

$\Phi_{B}=\int d\Phi_{B}=\frac{\mu_{0}Il}{2\pi}\int_{r_{1}}^{r_{2}}\frac{dr}{r}=\frac{\mu_{0}Il}{2\pi}\ln\frac{r_{2}}{r_{1}}$

so the inductance is

$L=\frac{\Phi_{B}}{I}=\frac{\mu_{0}l}{2\pi}\ln\frac{r_{2}}{r_{1}}$

and the inductance per unit length is

$\frac{L}{l}=\frac{\mu_{0}}{2\pi}\ln\frac{r_{2}}{r_{1}}$

## Energy stored in a magnetic field

When a time varying current $I$ is being carried in an inductor $L$ the power being supplied to the inductor is

$P=I\mathcal{E}=LI\frac{dI}{dt}$

The amount of work done in a time $dt$ is

$dW=P\,dt=LI\,dI$

so the work done in increasing the current from zero to $I$ is

$W=\int dW=\int_{0}^{I}LI\,dI=\frac{1}{2}LI^{2}$

The work done in going from zero to $I$ is equivalent to the energy stored in the magnetic field

$U=\frac{1}{2}LI^{2}$