Electromagnetic Induction

A voltage (e.m.f.) is induced in a conductor whenever:

• The conductor MOVES through a magnetic field (cutting field lines), OR
• The magnetic field through it CHANGES (e.g. switching a nearby current on/off, or the magnet moving).

No motion or no change → no induced voltage. A magnet held STATIONARY inside a coil induces nothing. The coil and magnet must have RELATIVE motion, or the field must vary in time.

Factors affecting the SIZE of the induced voltage (spec 6.15).

• Speed of relative motion — faster → larger voltage.
• Magnetic field strength — stronger magnet → larger voltage.
• Number of turns on the coil — more turns → larger voltage.
• Cross-sectional AREA of the coil — larger → larger voltage.
• Iron core inserted in the coil — concentrates the field → larger voltage.

The underlying principle (not on 4PH1, but useful for understanding): induced voltage is proportional to the RATE OF CHANGE of magnetic flux through the coil.

Direction of induced voltage. Lenz's law (not named on spec, but its effect appears): the induced voltage drives a current whose magnetic field OPPOSES the change that produced it. This is why pushing a magnet into a coil and pulling it out give opposite-sign induced voltages.

An AC generator (also called an alternator) converts rotational kinetic energy into electrical energy by electromagnetic induction.

Method 1 — coil in a magnetic field. A rectangular coil is rotated by an external source (water, steam, wind turbine) between the poles of two permanent magnets. As the coil rotates, the magnetic flux through it changes continuously, inducing an alternating voltage in the coil. Slip rings + carbon brushes carry the AC out to the external circuit.

Method 2 — magnet inside a coil. A magnet is rotated inside a stationary coil. Used in modern car alternators, where rotating coils + slip rings would be mechanically harder. The induced voltage in the stationary coil is again AC.

Factors affecting the size of the induced voltage (same as basic induction):

• Faster rotation.
• Stronger magnets.
• More turns on the coil.
• Larger coil area.
• Soft iron core through the coil.

The induced voltage varies SINUSOIDALLY with time — it is zero when the coil sits parallel to the field (flux at maximum but not changing) and maximum when the coil sits perpendicular (flux changing fastest as it passes through zero). This is why mains electricity is a 50 Hz sine wave: the generators in power stations rotate at 50 Hz.

A transformer changes the size of an ALTERNATING voltage. It consists of two coils wound on a shared SOFT IRON CORE:

• Primary coil ($N_p$ turns) — connected to the input AC supply ($V_p$).
• Secondary coil ($N_s$ turns) — connected to the output load ($V_s$).

How it works.

1. Alternating current in the primary creates an ALTERNATING magnetic flux in the iron core.
2. The flux is channelled through the core to the secondary coil (the iron concentrates the field).
3. The alternating flux through the secondary INDUCES an alternating voltage in it.
4. The induced voltage has the SAME frequency as the primary, but a DIFFERENT magnitude depending on the turns ratio.

Turns ratio (spec 6.19P). For an IDEAL transformer:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

• More turns on secondary than primary ($N_s > N_p$) → step-UP transformer ($V_s > V_p$).
• Fewer turns on secondary ($N_s < N_p$) → step-DOWN transformer ($V_s < V_p$).

Critical point. Transformers work ONLY on AC. A DC input would produce a steady current → steady flux → no induced voltage in the secondary. This is one of the main reasons mains electricity is distributed as AC.

An IDEAL (100% efficient) transformer transfers all input power to its output:

$V_p I_p = V_s I_s \quad \Leftrightarrow \quad P_p = P_s$

This means stepping UP the voltage steps DOWN the current by the same factor, and vice versa.

Why this matters for the national grid. Power lost in transmission cables is:

$P_{\text{loss}} = I^2 R$

where $R$ is the cable's resistance. Note the SQUARED dependence on $I$. To transmit the same power $P = VI$ with much smaller $I^2 R$ losses, you want to make $V$ very LARGE and $I$ correspondingly small.

Power station to consumer chain.

1. Generator outputs (say) 25 kV.
2. Step-up transformer at the station boosts this to 400 kV (the UK grid).
3. Power travels along high-voltage cables across the country with LOW current → low $I^2 R$ heating loss.
4. Step-down transformers at substations near consumers reduce the voltage in stages: 400 kV → 33 kV → 11 kV → 230 V at the home socket.

Without transformers, mains electricity would have to travel at the generation voltage (a few kV) → much higher current → most of the energy would be wasted as heat in the cables.

Edexcel-favourite question style. "Calculate the percentage of power lost in a 50 km cable of resistance 2 Ω that carries 800 MW at (a) 25 kV, (b) 400 kV." Use $P = VI$ to find $I$, then $I^2 R$ to find $P_{\text{loss}}$, then percentage. Result: (a) ~50% lost; (b) ~0.2% lost.