Electromagnetic Induction and Alternating Currents
Faraday's and Lenz's laws, motional EMF, self and mutual inductance, and then AC circuits with resistance, inductance and capacitance: reactance, impedance, resonance, power factor and the transformer. Code 151 typically pairs a derivation (EMF induced in a rotating coil, or average power in an AC circuit) with numericals on induced EMF and on series LCR impedance or resonance frequency.
Core rules
- Faraday's law: e = −dΦ/dt; for N turns, e = −N dΦ/dt. Lenz's sign is the energy-conservation statement.
- Motional EMF for a rod of length l moving with speed v perpendicular to B: e = Blv.
- Self-induced EMF e = −L dI/dt; energy stored in an inductor U = ½LI².
- AC: peak and rms values relate by I(rms) = I₀/√2; reactances X(L) = ωL, X(C) = 1/ωC; impedance Z = √(R² + (X(L) − X(C))²).
- Resonance when X(L) = X(C), giving ω = 1/√(LC); Z is minimum (= R) and current maximum.
- Average power P = V(rms)I(rms)cosφ, where cosφ = R/Z; a pure inductor or capacitor has cosφ = 0 (wattless current).
- Transformer: V(s)/V(p) = N(s)/N(p); step-up raises voltage and lowers current, and power is ideally conserved.
Worked example
A 0.5 m rod slides at 4 m s⁻¹ perpendicular to a 0.2 T field, in a circuit of resistance 2 Ω. Find the induced EMF, current and the force needed to keep it moving. e = Blv = 0.2 × 0.5 × 4 = 0.4 V; I = e/R = 0.4/2 = 0.2 A. The magnetic force opposing the motion is F = BIl = 0.2 × 0.2 × 0.5 = 0.02 N, so an equal external force of 0.02 N is required. Mechanical power supplied = Fv = 0.08 W, matching the electrical power I²R = 0.04 × 2 = 0.08 W - the energy balance required by Lenz's law.
Traps that cost marks
- Dropping the minus sign or ignoring the direction in Lenz's-law answers.
- Using peak values in power calculations instead of rms values.
- Forgetting that at resonance the voltages across L and C can each far exceed the supply voltage.
- Treating a transformer as able to increase power.