Rotating coils, alternating EMF, transformers and the generator effect
AQA A-Level Physics 7
📐Apply Faraday's law: ε = −dNΦ/dt
🔄Derive and use ε = BAN ω sin(ωt) for a rotating coil
📊Relate the flux linkage and EMF graphs (90° phase shift)
🔌Calculate transformer voltages and currents: Vs/Vp = Ns/Np
⚡Explain the ideal transformer and power conservation
🔬Investigate the transformer and AC generator
Faraday's Law — Quantitative
Faraday's Law: The magnitude of the induced EMF equals the rate of change of flux linkage.
ε = −dNΦ/dt = −N dΦ/dt
|ε| = ΔNΦ/Δt (for uniform rate of change)
ε = induced EMF (V)
N = number of turns
Φ = flux through each turn (Wb)
t = time (s)
The negative sign (Lenz's law) means the induced EMF drives a current that opposes the change in flux. In calculations, take the magnitude |ε| = ΔNΦ/Δt. The greater the rate of flux change, the greater the induced EMF.
Ways to increase induced EMF: faster rate of change of B or position, larger N, larger area A, stronger B.
The AC Generator (Rotating Coil)
When a rectangular coil of N turns, area A rotates at angular velocity ω in a uniform magnetic field B, the flux linkage varies sinusoidally:
NΦ = NBA cos(ωt) [maximum flux when coil is perpendicular to field]
The flux linkage is maximum when the coil plane is perpendicular to B (coil "face-on" to field). The EMF is maximum when the coil plane is parallel to B (coil "edge-on" to field — sides cutting field lines most rapidly). Flux and EMF are 90° (π/2) out of phase.
A common exam error: confusing when EMF is maximum and when flux is maximum. They are at opposite extremes: max EMF → zero flux change rate; max flux linkage → zero EMF (flux not changing at that instant).
Transformers
Transformer: A device that changes AC voltage using electromagnetic induction. An AC current in the primary coil creates a changing flux in the core, which induces an EMF in the secondary coil.
Turns ratio equation:
Vs/Vp = Ns/Np
For an ideal (100% efficient) transformer:
Power in = Power out: VpIp = VsIs
→ Vs/Vp = Ip/Is = Ns/Np
Vp = primary voltage (V), Vs = secondary voltage (V)
Np = primary turns, Ns = secondary turns
Ip = primary current (A), Is = secondary current (A)
Type
Ns vs Np
Vs vs Vp
Is vs Ip
Use
Step-up
Ns > Np
Vs > Vp
Is < Ip
Power transmission (high V, low I)
Step-down
Ns < Np
Vs < Vp
Is > Ip
Household supply (e.g. 230 V → 12 V)
Transformers only work with AC — a changing flux is needed to induce an EMF in the secondary. DC produces a constant flux (after initial switch-on) and no sustained induced EMF in the secondary.
Power Transmission and Efficiency
Electrical power is transmitted at high voltage to reduce energy losses in cables:
Power lost in cable: P = I²R
At higher voltage → lower current → less I²R loss
Efficiency of transformer:
η = (VsIs)/(VpIp) × 100%
Real transformers are not 100% efficient due to:
Eddy currents in the iron core (reduced by laminating the core with insulating layers)
Hysteresis losses — energy lost magnetising and demagnetising the core each cycle
Resistive heating in the coil wires (copper loss: I²R)
Flux leakage — not all flux from primary links with secondary
A generator has a coil of 300 turns, area 0.02 m², rotating at 50 rev s⁻¹ in a field of 0.40 T. Calculate: (a) the peak EMF; (b) the EMF at t = 5.0 ms after the coil is in the plane of maximum flux.
ε₀ = 753 V; at t = 5 ms, ε = 753 V (maximum EMF occurs at 1/4 period = 5 ms for 50 Hz)
A step-up transformer has 500 primary turns and 4000 secondary turns. The primary is connected to 230 V AC. (a) Find the secondary voltage. (b) If the secondary current is 0.5 A, find the primary current (ideal transformer).
1Vs/Vp = Ns/Np → Vs = 230 × (4000/500) = 230 × 8 = 1840 V
2Ideal: VpIp = VsIs → Ip = VsIs/Vp = 1840 × 0.5/230 = 920/230 = 4.0 A
Secondary voltage = 1840 V; Primary current = 4.0 A
A transformer has primary voltage 240 V and secondary voltage 12 V. The secondary current is 8.0 A. Calculate the efficiency if the primary current is 0.45 A.
A power station generates 60 MW at 25 kV. It feeds a transmission line of total resistance 2.0 Ω. Calculate the power loss in the line. Then calculate the loss if a step-up transformer raises the voltage to 400 kV before transmission.
2At 400 kV: I = 60×10⁶ / 400×10³ = 150 A. P_loss = 150² × 2.0 = 45000 W = 45 kW
Loss at 25 kV: 11.52 MW (19% of power). Loss at 400 kV: 45 kW (0.075%). High-voltage transmission reduces losses by a factor of (400/25)² = 256.
1. A transformer has 200 primary turns and 50 secondary turns, connected to a 240 V AC supply. What is the secondary voltage?
Vs = Vp × Ns/Np = 240 × 50/200 = 240 × 0.25 = 60 V. This is a step-down transformer (fewer secondary turns).
2. For a rotating coil generator, when is the induced EMF at its maximum?
EMF is maximum when the coil is "edge-on" (parallel to field) — the sides are cutting field lines at the fastest rate. At this position, flux linkage is zero but rate of change of flux is maximum.
3. An ideal transformer has Vp = 100 V, Np = 200, Ns = 500, and Is = 0.2 A. What is the primary current?
Ideal: VpIp = VsIs. Vs = 100 × 500/200 = 250 V. Ip = VsIs/Vp = 250 × 0.2/100 = 0.5 A. (Or: Ip/Is = Ns/Np → Ip = Is × Ns/Np = 0.2 × 500/200 = 0.5 A.)
4. Why do transformers only work with AC and not DC?
Faraday's law: ε = −dNΦ/dt. With DC, B is constant after switch-on → dΦ/dt = 0 → ε = 0. No sustained EMF is induced in the secondary. (There is a brief pulse when DC is switched on or off, but not a continuous voltage.)
5. A generator coil has peak EMF of 320 V and frequency 50 Hz. Write the equation for the instantaneous EMF and find the value at t = 3.0 ms.
1. A generator coil of 200 turns, area 80 cm², rotates at 25 rev s⁻¹ in a 0.45 T field. (a) Write the expression for flux linkage as a function of time, starting from the position of maximum flux. (b) Differentiate to find the EMF as a function of time. (c) Find the EMF when the flux linkage is half its maximum value.
(a) ω = 2π × 25 = 157.1 rad s⁻¹. A = 80 × 10⁻⁴ = 8.0 × 10⁻³ m². Maximum flux linkage: NΦ₀ = NBA = 200 × 0.45 × 8.0×10⁻³ = 0.720 Wb. NΦ = 0.720 cos(157.1t). (b) ε = −d(NΦ)/dt = 0.720 × 157.1 × sin(157.1t) = 113.1 sin(157.1t) ≈ 113 sin(157t) V. Peak EMF ε₀ = NBAω = 0.720 × 157.1 = 113 V. (c) NΦ = 0.360 Wb (half maximum). cos(157t) = 0.360/0.720 = 0.5 → 157t = π/3 rad. ε = 113 sin(π/3) = 113 × (√3/2) = 113 × 0.866 = 97.9 V. (Note: ε and NΦ are 90° out of phase, so max ε → zero NΦ and vice versa.)
2. A 400 kV transmission line carries power from a power station over a distance of 50 km. The line has resistance 0.05 Ω km⁻¹. The power transmitted is 80 MW. Calculate: (a) the line current; (b) the total line resistance; (c) the power loss; (d) the voltage drop across the line; (e) the voltage at the receiving end.
(a) I = P/V = 80×10⁶/400×10³ = 200 A. (b) R_line = 0.05 × 50 × 2 = 5.0 Ω (two conductors: go and return). (c) P_loss = I²R = 200² × 5.0 = 200000 W = 200 kW = 0.2 MW. (d) V_drop = IR = 200 × 5.0 = 1000 V = 1 kV. (e) V_receiving = 400 kV − 1 kV = 399 kV. The loss is 0.2/80 × 100% = 0.25% — very efficient. If transmitted at 25 kV: I = 3200 A, loss = 3200² × 5 = 51.2 MW (64% loss!) — impractical.
3. Explain how eddy current losses in a transformer core are reduced by laminating the core with thin insulated iron sheets. Include in your answer: why eddy currents form, how they cause energy loss, and why lamination reduces them.
The alternating current in the primary coil creates an alternating magnetic flux in the iron core. By Faraday's law, this changing flux induces EMFs within the bulk iron core itself. Because iron is a conductor, these EMFs drive currents in closed loops within the core — called eddy currents. These currents are large (iron has low resistance over large cross-sections) and dissipate energy as heat (P = I²R), reducing transformer efficiency. By laminating the core (thin sheets of iron insulated from each other), the cross-section available for eddy current loops is drastically reduced. Each lamination can only carry eddy currents within its own thin sheet. Since the effective resistance of each path is much higher (thinner cross-section, same or shorter path), the eddy current magnitude is greatly reduced. Since P ∝ I² and I is reduced by perhaps 100×, power loss is reduced by ≈10000×. The insulating varnish or oxide layer between laminations prevents currents flowing between sheets, ensuring the benefit is maintained.
Required Practical Investigating the Transformer
Objective: Verify the turns ratio equation Vs/Vp = Ns/Np and investigate transformer efficiency.
Equipment
Transformer with interchangeable coils (primary and secondary windings with known turns)
AC signal generator (e.g. 4 V, 50 Hz)
AC voltmeters for primary and secondary
Ammeters (to measure Ip and Is for efficiency)
Known load resistor for secondary
Method — Turns Ratio
Connect the primary coil to the AC supply. Set primary voltage Vp = 4 V.
Vary the secondary coil turns (or select different coil sets). Measure Vs for each Ns.
Plot Vs vs Ns — expect a straight line through the origin with gradient = Vp/Np.
Method — Efficiency
Connect primary to supply, secondary to a known load resistor.
Measure Vp, Ip, Vs, Is. Calculate Pin = VpIp and Pout = VsIs.
Calculate efficiency = Pout/Pin × 100%.
Vary the load and observe how efficiency changes.
Safety
Use low-voltage AC only (laboratory transformer with max 12 V). Do not connect mains AC to this experiment. Ensure all connections are secure before switching on. Do not touch live terminals.