Faraday's and Lenz's Laws

​​In a nutshell

When a magnet moves near a coil it induces an e.m.f. due to a change in the magnetic flux. Faraday's law states that the magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linkage. Lenz's law states that the direction of the induced current is always such as to oppose the change that caused it.

Equations

Variable definitions

Electromagnetic induction

Since an electric current can produce magnetism, it follows that you can use magnetism to produce an electric current. This phenomenon is known as electromagnetic induction.

To induce an e.m.f. you need a coil and a magnet as shown below:

When both are stationary the voltmeter shows no readings. When the magnet is pushed towards the coil it induces an e.m.f. and when it is pushed away it induces a reverse e.m.f.,both of which are picked up by the voltmeter. These continuous motions create an alternate current in the coil.

Note: The poles will switch in the induced magnet as the permanent magnet moves back and forth due to the current switching directions (use the right-hand grip rule to help visualise this!). 

The reason for this induced e.m.f. is that when the magnet moves, it exerts a force on the electrons in the coil which move and create a current. The work done on the magnet is transferred into electrical energy thus, the faster the magnet moves the higher the induced e.m.f. .

Magnetic flux and magnetic flux linkage

The magnetic flux $$\phi$$ is a quantity defined as the total magnetic field passing through an area and is measured in weber $$Wb$$. It can be calculated with the formula:

​$$\phi=BAcos\theta$$​​

When the field is normal to the cross sectional-area ($$\theta=0\degree$$​), $$\phi=BA$$.

The magnetic flux linkage is the product between the number of turns in the coil $$N$$​ and the magnetic flux:

​$$magnetic\ flux\ linkage=N\phi$$​​

One can then write:

​$$N\phi=BANcos\theta$$​​

An e.m.f. is induced in a coil whenever there is a change in the magnetic flux.

Faraday's and Lenz's law

Faraday's law, states that the magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linkage. This can be written as:

​$$\varepsilon\propto\dfrac{\Delta(N\phi)}{\Delta t}$$​​

Where the $$\Delta$$ represent change in the variable in front of them.​

Lenz's law states that the direction of the induced current is always such as to oppose the change causing it. 

In other words, if you move the magnet towards the coil the induced current will move towards it and vice versa when moving the magnet away. This is because when the current is induced it will create a magnetic field of its own and in order to not violate the conservation of energy the induced field needs to oppose that of the magnet.

Lenz's law can be combined with Faraday's law in the form of a minus sign to give the formula for induced e.m.f.:

​$$\varepsilon=-\dfrac{\Delta(N\phi)}{\Delta t}$$​​

Note: the minus sign in the equation can be ignored most of the time as it is just a reminder for the conservation of energy.

Example

An e.m.f. of $$50\ mV$$ is induced in a coil with a cross-sectional area of $$5\times10^{-4}\ m^2$$ and $$500$$ turns. The magnet that caused the induction is placed such that its field lines are normal to the coil and moves in a time of $$0.5\ s$$. What is the magnetic flux density at the pole of the magnet?

Firstly, write down what you know:

$$\varepsilon=50\ mV=0.05\ V\newline A=5\times10^{-4}\ m^2\newline N=500\newline \Delta t=0.5\ s$$​​

Next, write down the equation for the magnetic flux linkage and the induced e.m.f.:

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

Substitute the magnetic flux linkage into the equation for the e.m.f.:

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

Since the field is normal to the coil ($$\theta=0\degree$$​) and you are looking for the magnetic flux density, $$cos(0)=1$$ and you can ignore the minus sign. Hence you can write:

$$\varepsilon=\dfrac{BAN}{\Delta t}$$​​

Now rearrange the formula for $$B$$:

$$B=\dfrac{\varepsilon\Delta t}{AN}$$​​

Substitute the numbers into the equation and calculate the magnetic flux density:

$$B=\dfrac{0.05\times0.5}{5\times10^{-4}\times500}=0.1\ T$$​​

The magnetic flux density at the pole of the magnet is $$\underline{0.1\ T}$$.