Tamil Nadu HSE Second Year Physics study materials
Tamil Nadu HSE Second Year Physics study resources

Tamil Nadu HSE Second Year Physics: Electromagnetic Induction and Alternating Current

Tamil Nadu HSE Second Year Physics · Work through your exam, one topic at a time.

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Electromagnetic Induction and Alternating Current

Electromagnetic Induction and Alternating Current covers induced emf, Lenz's law, self and mutual inductance, transformers, and series AC circuits with phasors. Expect one derivation, usually emf of a rotating coil or resonance, plus a phasor-based numerical.

Core rules

  • Flux Φ = BA cos θ; induced emf ε = −dΦ/dt. Faraday's law gives the magnitude, Lenz's law the direction: the induced current opposes the change producing it.
  • Motional emf in a rod moving at speed v is ε = Blv. A coil of N turns rotating at angular speed ω gives ε = NBAω sin ωt, with peak ε₀ = NBAω.
  • Self-inductance: ε = −L dI/dt; energy stored U = ½LI². Mutual inductance: ε₂ = −M dI₁/dt.
  • Ideal transformer: V_s/V_p = N_s/N_p = I_p/I_s; power and frequency are unchanged.
  • AC values: I_rms = I₀/√2, V_rms = V₀/√2. Reactances X_L = ωL and X_C = 1/ωC.
  • Series LCR: Z = √(R² + (X_L − X_C)²), tan φ = (X_L − X_C)/R, and P = V_rms I_rms cos φ.
  • Resonance when X_L = X_C, i.e. ω = 1/√(LC); Z is then minimum (= R) and current maximum.

Worked example

A series circuit has R = 30 Ω and X_L = 40 Ω across 200 V rms. Find Z, the current and the power. Z = √(30² + 40²) = 50 Ω; I = 200/50 = 4.0 A; cos φ = R/Z = 0.60; P = 200 × 4.0 × 0.60 = 480 W. Note that P is not VI = 800 W; the power factor must be included.

Traps

Using peak values where rms is required; claiming power dissipation in a pure inductor or capacitor, where average power is zero; saying a transformer changes frequency; reading Lenz's law as opposing the flux rather than the change in flux; and slipping on resonance units.