Uses of electromagnetic induction

In a nutshell

AC generators use Faraday's law to create alternating current. Transformers are devices that use electromagnetic induction to increase or decrease the voltage of an AC supply.

Equations

description	equation
Induced e.m.f.	$\varepsilon=-\dfrac{\Delta(BANcos\theta)}{\Delta t}$
Turn ratio equation	$\dfrac{n_s}{n_p}=\dfrac{V_s}{V_p}$

Variable definitions

quantity name	symbol	derived units	si base units
$magnetic\ flux$	$\phi$	$Wb$	$kg\ m^2\ s^{-2}\ A^{-1}$
$magnetic\ flux\ density$	$B$	$T$	$kg\ s^{-2}\ A^{-1}$
$area$	$A$	$m^2$	$m^2$
$number\ of \ coil\ turns$	$N$ / $n$
$e.m.f.$	$\varepsilon$	$V$	$kg\ m^2\ s^{-3}\ A^{-1}$
$time$	$t$	$s$	$s$
$voltage$	$V$	$V$	$kg\ m^2\ s^{-3}\ A^{-1}$

Alternating current generators

Alternating current (AC) is a type of electric current which periodically reverses direction.

Electromagnetic induction is used to create alternating currents in AC generators such as the one shown below:

Physics; Electromagnetism; KS5 Year 12; Uses of electromagnetic induction

1.	Slip rings
2.	Brushes
3.	Rotation of the coil
4.	Coil
5.	Circuit

It works on the principle of Faraday's law, which can be represented as:

$\varepsilon=-\dfrac{\Delta(BANcos\theta)}{\Delta t}$

As the coil moves in the magnetic field it changes the magnetic flux and generates a current. This then gets reversed as the coil rotates and so on. The values of the induced e.m.f. over time are shown in the following graph:

Physics; Electromagnetism; KS5 Year 12; Uses of electromagnetic induction

1.	Flux linkage
2.	Induced e.m.f
3.	Maximum rate of flux change
4.	No flux changing
5.	Max e.m.f
6.	No induced e.m.f
7.	e.m.f = - gradient of magnetic flux linkage graph

Step-up and step-down transformers

Transformers are devices that use electromagnetic induction to increase or decrease the voltage of an AC supply. A simple transformer is shown below:

Physics; Electromagnetism; KS5 Year 12; Uses of electromagnetic induction

When an alternating current is supplied to the primary coil it produces a varying magnetic flux in the iron core which then induces a current in the secondary coil. Depending on the number of turns in the coils, the output voltage can be increased or decreased according to the turn-ratio equation:

$\dfrac{n_s}{n_p}=\dfrac{V_s}{V_p}$

Where $n$ represents the number of turns in a coil and $V$ the voltage in it. The $p$ and $s$ indicate the primary (input) and secondary (output) coils.

A transformer can be set up in two ways:

Step-up transformer; Increases the output voltage. The number of turns in the secondary coil is higher than in the primary coil so the secondary voltage is higher than the primary voltage.
Step-down transformer; decreases the output voltage. The number of turns in the secondary coil is lower than in the primary coil so the secondary voltage is lower than the primary voltage.

Efficiency in transformers

A $100 \%$ efficient transformer would transfer all the power from the primary coil into the secondary coil, $P_p=P_s$ . Since $P=IV$ you can write:

$I_pV_p=I_sV_s$

$\dfrac{I_p}{I_s}=\dfrac{V_s}{V_p}$

This means that when increasing the output voltage, the current decreases and vice versa.

To increase the efficiency of a transformer, you can use low resistance wires or use a core made by layers of iron separated by an insulator, to minimise currents induced by the core itself.

Example

A transformer is used to step down a voltage of $230\ V$ to $10\ V$ . How many turns must the secondary coil have if the primary coil has $500$ turns?

Firstly, write down what you know:

$V_p=230 \\ V_s=10\ V \\n_p=500$ 

Next, write down the turn-ratio equation:

$\dfrac{n_s}{n_p}=\dfrac{V_s}{V_p}$ 

Rearrange it to find the number of turns in the secondary coil:

$n_s=\dfrac{V_s}{V_p}n_p$ 

Substitute the numbers and calculate the value:

$n_s=\dfrac{10}{230}\times500=22\ (2\ s.f.)$ 

The number of turns in the secondary coil must be $\underline{22}$ .