Electromagnetic induction is the production of an e.m.f. by a changing magnetic field. It reverses the ideas of Topics 17–18. Instead of using a current to make a magnetic field, a changing magnetic field is used to make a current. Almost all mains electricity, from the kettle at home to the MRT, is generated this way.
This topic covers Faraday's magnet-and-coil experiment, Lenz's law and why it follows from conservation of energy, and the factors that increase the induced e.m.f. It then applies these ideas to the simple a.c. generator, transformers and high-voltage power transmission, with the exam phrasing that earns the marks.
What these notes cover
- •Faraday's magnet-and-coil experiment
- •Lenz's law and the direction of the induced current
- •Lenz's law and conservation of energy
- •Factors that increase the induced e.m.f.
- •The simple a.c. generator: construction and slip rings
- •Generator output graph and doubling the rotation speed
- •Transformer structure and how a transformer works
- •Step-up and step-down transformers and the ideal transformer equations
- •High-voltage power transmission
- •Common mistakes and exam technique
- •Summary
Key definitions
- Electromagnetic induction
- A changing magnetic field through a circuit induces an e.m.f. in the circuit. If the circuit is closed, this induced e.m.f. drives an induced current around it.
- Lenz's law
- The induced e.m.f. (and hence the induced current) acts in a direction that opposes the change producing it.
- Centre-zero galvanometer
- A meter whose needle rests in the middle and can deflect either left or right, so it shows both the size and the direction of any current.
- Slip rings
- Two rings in an a.c. generator, one fixed to each end of the coil, which rotate with the coil. Through the carbon brushes they keep continuous contact with the external circuit and never reverse the connection.
- Transformer
- A device that changes the size of an alternating voltage. It consists of a primary coil, a secondary coil and a laminated soft-iron core on which both coils are wound.
Key formulas
Frequency and period
f = 1 ÷ T
f is the output frequency in Hz, equal to the number of revolutions of the generator coil per second. T is the period in s, the time for one revolution.
Turns ratio
Vₚ ÷ Vₛ = Nₚ ÷ Nₛ
Vₚ and Vₛ are the primary and secondary voltages in V. Nₚ and Nₛ are the numbers of turns (no unit). Step-up: Nₛ > Nₚ. Step-down: Nₛ < Nₚ.
Ideal transformer (100% efficient)
Vₚ × Iₚ = Vₛ × Iₛ
Iₚ and Iₛ are the primary and secondary currents in A. Power in equals power out, so when the voltage goes up the current goes down.
Transformer current ratio
Iₚ ÷ Iₛ = Nₛ ÷ Nₚ
This comes from combining the turns-ratio and power equations. The current ratio is the inverse of the turns ratio.
Cable power loss
Pₗₒₛₛ = I²R
I is the current in the cables, found from I = P ÷ V using the transmission voltage. R is the resistance of the cables themselves, not of the whole circuit.
Cable loss for a fixed power delivered
Pₗₒₛₛ = P²R ÷ V²
This comes from substituting I = P ÷ V into Pₗₒₛₛ = I²R. The loss is proportional to 1 ÷ V², so doubling the transmission voltage cuts the loss to a quarter.
Worked example
A step-down transformer converts the 240 V a.c. mains supply to 12 V a.c. for a desk lamp rated at 24 W. Assume the transformer is ideal. Find (a) the turns ratio Nₚ : Nₛ, (b) the current in the secondary coil when the lamp operates normally, and (c) the current drawn from the mains by the primary coil.
- 1(a) Using the turns-ratio equation, Nₚ ÷ Nₛ = Vₚ ÷ Vₛ = 240 V ÷ 12 V = 20, so Nₚ : Nₛ = 20 : 1. This is a step-down transformer, as expected, since the primary has more turns.
- 2(b) The lamp receives 24 W at 12 V, so using P = VₛIₛ, Iₛ = P ÷ Vₛ = 24 W ÷ 12 V = 2.0 A.
- 3(c) For an ideal transformer, input power equals output power, VₚIₚ = VₛIₛ, so Iₚ = (Vₛ × Iₛ) ÷ Vₚ = (12 V × 2.0 A) ÷ 240 V = 0.10 A.
- 4Check: the voltage was stepped down by a factor of 20 and the current was stepped up by the same factor of 20 (from 0.10 A to 2.0 A), so the power on both sides is 24 W.
Watch out: The current does not follow the turns ratio. The current ratio is the inverse, Iₚ ÷ Iₛ = Nₛ ÷ Nₚ. Write units at every substitution, because missing units lose marks far more often than wrong arithmetic.
Common mistakes
Saying a strong magnetic field induces an e.m.f.
Wrong: The magnet is very strong, so an e.m.f. is induced in the coil.
Right: An e.m.f. is induced only when the magnetic field linking the coil is changing, for example when there is relative motion between the magnet and the coil. A stationary magnet inside a coil, however strong, induces zero e.m.f. because the flux through the coil is constant.
Putting the peak e.m.f. at the wrong coil position
Wrong: The e.m.f. is maximum when the plane of the coil is perpendicular to the field, because the most field lines pass through it.
Right: The e.m.f. depends on the rate of cutting field lines. It is maximum when the coil plane is parallel to the field, because the sides cut field lines at the greatest rate. It is zero when the plane is perpendicular to the field, because the sides move along the field lines.
Using V² ÷ R with the transmission voltage
Wrong: Cables carry power at 25 kV, so the power lost in the cable is P = V² ÷ R = 25000² ÷ R.
Right: 25 kV is not the p.d. across the cable resistance. The p.d. across the cables is only the small drop IR. Find I = P ÷ V first, then use Pₗₒₛₛ = I²R, where R is the resistance of the cables.
Questions students ask
Why does a transformer not work with d.c.?
A steady d.c. supply gives a constant current in the primary coil, so the magnetic flux in the core is constant. With no change in flux linking the secondary coil, no e.m.f. is induced and the output is zero. A brief pulse of e.m.f. appears only at the instants of switching on and off.
Why is electricity transmitted at high voltage?
For the same power delivered (P = VI), transmitting at a higher voltage means a smaller current flows in the transmission cables. The power dissipated as heat in the cables is Pₗₒₛₛ = I²R, where R is the resistance of the cables. Since the current is much smaller and is squared in the formula, much less power is wasted as heat, so more power reaches the consumers.
Why is Lenz's law a consequence of conservation of energy?
The induced current opposes the change, so the person moving the magnet feels an opposing force and must do work against it. That work is converted into electrical energy. If the induced current helped the motion instead, the magnet would be pulled in faster and faster, creating energy from nothing, which is impossible.
What happens to an a.c. generator's output when the coil rotates twice as fast?
The peak e.m.f. doubles, because the magnetic field through the coil changes at twice the rate. The frequency also doubles and the period halves, because the coil makes twice as many revolutions per second. Examiners award separate marks for each change, so a sketch must show both.
In the full PDF
- •Diagrams: the induced poles on a coil as a N pole approaches, the doubled-speed generator graph and the power transmission chain
- •Further worked examples: Lenz's law pole prediction, an oscillating magnet, two dropped-magnet setups, doubling generator speed, and transmission losses at 1.0 kV vs 25 kV
- •The remaining four common mistakes: Lenz's law pole errors, slip rings vs split-ring commutator, stepping up a battery and the inverted current ratio
- •Transformer design explained: soft-iron core, laminations and eddy currents, and insulated coils
- •The full exam-technique checklist for Lenz's law, generator sketches and transformer calculations
- •One-page summary with all key formulae for night-before revision