Electromagnetic Induction and Alternating Current
What is examined
Electromagnetic Induction and Alternating Current tests Faraday's law, Lenz's law, motional emf, self- and mutual inductance, eddy currents, rms values, AC through R, L and C, series LCR resonance, power factor and the transformer.
Core rules
- Faraday: ε = −dΦ/dt; for N turns ε = −N dΦ/dt. Lenz's law gives the minus sign: induced effects oppose the change.
- Motional emf: ε = Bvl for a rod moving perpendicular to B; induced current I = Bvl/R.
- Self-inductance: Φ = LI, ε = −L dI/dt; solenoid L = μ₀n²Al. Mutual inductance: Φ₂ = MI₁.
- AC: I = I₀ sin ωt; I_rms = I₀/√2; average current over a full cycle is zero.
- Reactances: X_L = ωL, X_C = 1/(ωC). Impedance Z = √[R² + (X_L − X_C)²]; tan φ = (X_L − X_C)/R.
- Resonance: X_L = X_C, ω₀ = 1/√(LC), Z = R (minimum), current maximum. Q-factor = ω₀L/R.
- Power P = V_rms I_rms cos φ; pure L or C has zero average power.
- Transformer: V_s/V_p = N_s/N_p = I_p/I_s (ideal); frequency unchanged; step-up raises voltage and lowers current.
Worked example
A series circuit has L = 0.50 H and C = 20 μF. Find the resonant frequency. ω₀ = 1/√(LC) = 1/√(0.50 × 20×10⁻⁶) = 1/√(1.0×10⁻⁵) = 1/(3.16×10⁻³) ≈ 316 rad s⁻¹. f₀ = ω₀/2π ≈ 50 Hz. Reasoning: at resonance the reactances cancel, so the current is limited only by R.
Traps
Dropping the minus sign in Faraday's law; using peak values in power; thinking Z = R at resonance even when R = 0; and claiming a transformer changes frequency or power.