1. EEE107
Electromagnetic Induction

2. Electromagnetic induction through flux cuttingS
bar magnet
coil
Galvanometer

3. Requirements for Induced EMF
For an emf to be induced in a conductor there must be a changing magnetic field; that is, relative motion between the conductor and the magnetic flux. In other words, the conductor must cut the magnetic flux or the magnetic flux must cut the conductor.
In addition, it can be demonstrated that a larger emf is produced if:
the magnet or coil is moved faster
the number of windings of the coil is increased
a stronger magnet is used
In other words, the induced emf depends on the strength of the magnetic field, the rate at which the flux cuts (or is cut by) the conductor and the number of turns in the coil.
The direction of the induced emf (direction of current flow) depends on the polarity of the magnet and its direction of movement relative to the coil.

4. Electromagnetic induction through flux linkage
Coil 1
Coil 2
ferromagnetic material
It may be observed that a momentary deflection of the galvanometer occurs when the current in coil 1 is switched on.
A momentary deflection in the opposite direction is observed when the current is switched off.
No deflection is observed when the current in coil 1 is steady.
From these observations, we may conclude that an emf is induced in coil 2 due to a changing current in coil 1.

5. Faraday’s Law of Electromagnetic Induction
Faraday’s law states that the emf, e, induced in a circuit is proportional to the rate of change of flux linkage, where flux linkage is the product of the magnetic flux through the circuit and the number of turns N affected by the flux.
Mathematically, we may say that:
e  d/dt(N) or
e  N.d/dt, since N is normally constant.
If the flux, , is measured in webers and t is measured in seconds, then Faraday’s law may be summarised as:

6. Lenz’s Law of Electromagnetic Induction
Faraday’s law gives the magnitude of the induced emf only. Its direction is given by Lenz’s law, which states that the direction of the induced emf is such that it opposes the cause which produces it.
For example, an increasing flux through a closed circuit will induce an emf (and hence a current in the circuit) that generates a magnetic field which tries to stop the increase in flux; that is, it opposes the applied field (cause) that produces the emf.
Conversely, a decreasing flux will induce an emf and hence a current that tries to increase the applied field, and stop it decreasing.
Lenz’s law can be combined with Faraday’s law into a single equation that gives both the magnitude and direction of induced emf:

7. Example Calculate the average emf induced in a coil of 500 turns when the flux linking it changes from 0.5mWb to 1mWb in 1 millisecond.
Solution
If the flux change is non-uniform and changes from an initial valuei to a final value (f in a time of t, to average rate of change of flux is
(f - i )/t
Thus the average induced emf is given by the expression:
e = -N(f - i)/t
e = -N(f - i)/t volts
where: N = 500 turns
f = 10-3 Wb
i = 0.5 × 10-3 Wb
t = s
e = -500(1 × × 10-3)/10-3
= -250V

8. Simple Transformer Primary windings (Np turns)~
Primary windings (Np turns)
Secondary windings (Ns turns)
Transformer core
Coil 1
Coil 2 ac
voltage source

9. Ideal Transformer primary input power = secondary output power
If we assume that the transformer is 100% efficient, i.e. there is no power loss, then the input (primary) electrical power must equal the output (secondary) power.
A transformer will never be 100% efficient as there will be power loss due to, e.g. heat generation due to the resistance of the windings and hysteresis effects.
For an ideal transformer we can say that
power in primary = power in secondary or
Ep  Ip = Es × Is
where Ip and Is are the primary and secondary currents respectively.
Therefore

10. Let Ep = peak value of voltage applied to primary
An alternating source of emf with peak value 200V is applied to the primary winding of a transformer. If the number of turns in the primary and secondary windings are 500 and 25 respectively, calculate the peak value of the voltage induced in the secondary winding. What is the value of the secondary current if the primary current is 5mA?
Let Ep = peak value of voltage applied to primary
Es = peak value of voltage applied to secondary
Np = number of turns on primary
Ns = number of turns on secondary
Es/Ep = Ns/Np
Es = Ns/Np × Ep
Therefore: Es = (25/500) × 200 = 10V
Is = Np/Ns × Ip = (500/25) × 5mA = 0.1A.