JEE Main Physics · Chapter 16

Electromagnetic Induction and Alternating Current

Faraday’s and Lenz’s laws, inductance, LCR circuits, resonance, power factor, generators and transformers.

NCERT Class 12JEE Main: 1–2 Qs per shift 9 min read

1. Chapter at a Glance

Electromagnetic induction (EMI) is the production of an emf when magnetic flux through a circuit changes. It is the working principle of generators, transformers and induction cooktops. Alternating current (AC) follows naturally: rotating coils produce sinusoidal voltages, and circuits with resistors, inductors and capacitors respond in special ways. JEE Main usually asks two questions from this unit.

What JEE asks most
  • Faraday's and Lenz's laws; motional emf of rods.
  • Self and mutual inductance; energy in an inductor; LR time constant.
  • Series LCR circuits: impedance, phase, power factor, resonance.
  • Transformers and AC generators.

2. Faraday's and Lenz's Laws

  • Magnetic flux Φ = B·A = BA cosθ (weber).
  • Faraday's law: induced emf ε = −N dΦ/dt.
  • Lenz's law: the induced current opposes the change that produces it (the minus sign). It follows from conservation of energy.
  • Motional emf: a rod of length l moving at v perpendicular to B: ε = Blv. A rod rotating about one end: ε = ½Bωl².
  • Charge that flows due to a flux change: Q = NΔΦ/R (independent of how fast it changes).
  • Eddy currents: loops of current induced in bulk metal; used in magnetic braking and induction furnaces, reduced by laminating cores.

3. Self and Mutual Inductance

  • Self-inductance: NΦ = LI, ε = −L dI/dt. For a solenoid L = μ₀n²Al.
  • Mutual inductance: N₂Φ₂ = MI₁, ε₂ = −M dI₁/dt. M ≤ √(L₁L₂).
  • Energy stored in an inductor: U = ½LI²; energy density in a field B²/2μ₀.
  • LR circuit: current grows as I = I₀(1 − e−t/τ) with time constant τ = L/R.

4. Alternating Current

  • V = V₀ sin ωt; rms value Vrms = V₀/√2 (household 220 V rms means V₀ ≈ 311 V). The average over a full cycle is zero.
  • Resistor: V and I in phase. Inductor: I lags V by 90°, reactance XL = ωL. Capacitor: I leads V by 90°, reactance XC = 1/ωC.
  • Series LCR: Z = √(R² + (XL − XC)²), tanφ = (XL − XC)/R.
  • Average power P = VrmsIrms cosφ; power factor cosφ = R/Z. A pure L or C consumes no average power ("wattless current").
  • Resonance: XL = XC ⇒ ω₀ = 1/√(LC); Z = R (minimum), current maximum, power factor 1. Quality factor Q = ω₀L/R = (1/R)√(L/C).
  • LC oscillations: energy swings between capacitor and inductor at ω₀ = 1/√(LC).
I VR VL VC V (resultant)φVL − VC
Series LCR phasors: V = √(VR² + (VL − VC)²) and tanφ = (VL − VC)/VR.

5. AC Generator and Transformer

  • AC generator: a coil of N turns and area A rotating at ω in field B gives ε = NBAω sin ωt; peak emf NBAω.
  • Transformer (works only on AC): Vs/Vp = Ns/Np. For an ideal transformer VpIp = VsIs; efficiency η = Pout/Pin.
  • Step-up transformers raise voltage and lower current, reducing I²R losses in long-distance transmission.
  • Losses: copper (I²R), eddy currents (reduced by laminated core), hysteresis (use soft iron), flux leakage.

6. Formula Master Table

QuantityFormula
Faradayε = −N dΦ/dt
Motional emfε = Blv; rotating rod ½Bωl²
Inductorε = −L dI/dt; U = ½LI²
LR time constantτ = L/R
rmsVrms = V₀/√2
ReactancesXL = ωL; XC = 1/ωC
ImpedanceZ = √(R² + (XL − XC)²)
PowerP = VrmsIrms cosφ
Resonancef₀ = 1/(2π√LC)
TransformerVs/Vp = Ns/Np

7. Mind Map

EMI & AC
Induction
  • Faraday, Lenz
  • Motional emf
  • Eddy currents
Inductance
  • L, M
  • ½LI²
  • LR growth
AC circuits
  • rms values
  • XL, XC, Z
  • Resonance, power factor
Devices
  • Generator
  • Transformer

8. Tips and Tricks

Shortcuts
  • "CIVIL": in a Capacitor I leads V; V leads I in an inductor (L).
  • At resonance, VL and VC can each be much larger than the supply voltage, but they cancel.
  • For a rod on rails with resistance R: I = Blv/R, retarding force B²l²v/R, power B²l²v²/R.
  • Transformer: voltage ratio = turns ratio; current ratio is its inverse (ideal case).
Mistakes students make
  • Adding VR, VL and VC arithmetically — they add as phasors.
  • Thinking a transformer works on DC.
  • Using peak values in the power formula without dividing by 2.

9. Important One-liners

  • Lenz's law is a consequence of conservation of energy.
  • Average power in a purely inductive or capacitive circuit is zero.
  • At resonance, the impedance of a series LCR circuit equals R.
  • Transformer cores are laminated to reduce eddy-current losses.
  • An inductor opposes changes in current; a capacitor opposes changes in voltage.
  • Household 220 V AC is the rms value; the peak is about 311 V.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working.

Average level

N1Average

A 100-turn coil of area 0.01 m² is perpendicular to a 0.5 T field. The field falls to zero in 0.1 s. Find the average induced emf.

Show answer & solution

Concept: ε = NΔΦ/Δt.

  1. ΔΦ per turn = 0.5 × 0.01 = 5 × 10⁻³ Wb.
  2. ε = 100 × 5 × 10⁻³/0.1 = 5 V.

Answer: 5 V

N2Average

A 1 m rod moves at 5 m s⁻¹ perpendicular to a 0.4 T field. Find the emf across its ends.

Show answer & solution

Concept: ε = Blv.

  1. ε = 0.4 × 1 × 5.
  2. = 2 V.

Answer: 2 V

N3Average

A 0.5 H inductor carries 4 A. Find the energy stored. If it is in series with a 10 Ω resistor, what is the time constant?

Show answer & solution

Concept: U = ½LI²; τ = L/R.

  1. U = ½ × 0.5 × 16 = 4 J.
  2. τ = 0.5/10 = 0.05 s.

Answer: 4 J; 0.05 s

N4Average

An AC voltage is V = 311 sin(100πt) V. Find its rms value and frequency.

Show answer & solution

Concept: Vrms = V₀/√2; ω = 2πf.

  1. Vrms = 311/1.414 ≈ 220 V.
  2. ω = 100π ⇒ f = 50 Hz.

Answer: 220 V; 50 Hz

N5Average

At 50 Hz, find the reactance of (a) a 0.1 H inductor, (b) a 100 μF capacitor.

Show answer & solution

Concept: XL = 2πfL; XC = 1/(2πfC).

  1. XL = 2π × 50 × 0.1 ≈ 31.4 Ω.
  2. XC = 1/(2π × 50 × 10⁻⁴) ≈ 31.8 Ω.

Answer: 31.4 Ω; 31.8 Ω

Advanced level

N6Advanced

A series LCR circuit has R = 30 Ω, XL = 80 Ω and XC = 40 Ω, connected to 100 V rms. Find the impedance, current, phase angle, power factor and average power.

Show answer & solution

Concept: Z = √(R² + (XL − XC)²).

  1. Z = √(30² + 40²) = 50 Ω.
  2. Irms = 100/50 = 2 A.
  3. tanφ = 40/30 ⇒ φ ≈ 53° (current lags, inductive).
  4. cosφ = R/Z = 0.6.
  5. P = I²R = 4 × 30 = 120 W (= 100 × 2 × 0.6).

Answer: 50 Ω; 2 A; 53°; 0.6; 120 W

N7Advanced

A series circuit has L = 0.1 H, C = 10 μF and R = 10 Ω. Find the resonant frequency and the quality factor.

Show answer & solution

Concept: f₀ = 1/(2π√LC); Q = (1/R)√(L/C).

  1. LC = 0.1 × 10⁻⁵ = 10⁻⁶ ⇒ √LC = 10⁻³.
  2. f₀ = 1/(2π × 10⁻³) ≈ 159 Hz.
  3. Q = (1/10)√(0.1/10⁻⁵) = 0.1 × 100 = 10.

Answer: about 159 Hz; 10

N8Advanced

A transformer has 100 primary and 2000 secondary turns, with 220 V on the primary. The secondary delivers 0.5 A. If the efficiency is 90%, find the secondary voltage and the primary current.

Show answer & solution

Concept: Vs = VpNs/Np; Pin = Pout/η.

  1. Vs = 220 × 20 = 4400 V.
  2. Pout = 4400 × 0.5 = 2200 W.
  3. Pin = 2200/0.9 ≈ 2444 W.
  4. Ip = 2444/220 ≈ 11.1 A.

Answer: 4400 V; about 11.1 A

N9Advanced

A generator coil of 100 turns and area 0.02 m² rotates at 50 revolutions per second in a 0.5 T field. Find the peak and rms emf.

Show answer & solution

Concept: ε₀ = NBAω with ω = 2πf.

  1. ω = 2π × 50 = 100π rad s⁻¹.
  2. ε₀ = 100 × 0.5 × 0.02 × 100π = 100π ≈ 314 V.
  3. εrms = 314/√2 ≈ 222 V.

Answer: about 314 V; about 222 V

N10Advanced

A 0.5 m rod slides at a steady 4 m s⁻¹ on frictionless rails in a 0.5 T field; the rails are joined by a 2 Ω resistor. Find the emf, current, the force needed to keep it moving, and verify energy conservation.

Show answer & solution

Concept: ε = Blv; the induced current feels a retarding force BIl (Lenz).

  1. ε = 0.5 × 0.5 × 4 = 1 V.
  2. I = 1/2 = 0.5 A.
  3. Required force = BIl = 0.5 × 0.5 × 0.5 = 0.125 N.
  4. Mechanical power = Fv = 0.125 × 4 = 0.5 W; heat = I²R = 0.25 × 2 = 0.5 W ✓.

Answer: 1 V; 0.5 A; 0.125 N; 0.5 W in = 0.5 W heat

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Lenz's law is a consequence of the conservation of:

  1. Charge
  2. Momentum
  3. Energy
  4. Mass
Show solution

If the induced current aided the change, energy would be created from nothing; opposing it ensures work is needed.

Answer: (C)

Q2Easy

The average power consumed by a pure inductor in an AC circuit is:

  1. VrmsIrms
  2. Zero
  3. ½V₀I₀
  4. Infinite
Show solution

φ = 90°, cosφ = 0, so P = 0.

Answer: (B)

Q3Medium

At resonance in a series LCR circuit:

  1. Z is maximum
  2. Z = R and current is maximum
  3. Power factor is zero
  4. Current lags by 90°
Show solution

XL = XC, so Z = R (minimum) and current is maximum; power factor = 1.

Answer: (B)

Q4Hard

A copper disc swinging between the poles of a magnet stops quickly. The main reason is:

  1. Air friction
  2. Eddy currents
  3. Gravitational pull
  4. Electrostatic attraction
Show solution

Changing flux through the disc induces eddy currents; by Lenz's law they oppose the motion (magnetic damping).

Answer: (B)

Q5Numerical

The current in a 2 H coil changes from 5 A to 1 A in 0.2 s. Find the magnitude of the self-induced emf in volts.

Show solution

ε = L|ΔI|/Δt = 2 × 4/0.2 = 40 V.

Answer: 40

Q6Numerical

The rms voltage of an AC supply is 100√2 V. Find the peak voltage in volts.

Show solution

V₀ = √2 × Vrms = 200 V.

Answer: 200

12. Quick Revision Summary

Remember before the exam
  • ε = −NdΦ/dt; Lenz opposes change; motional Blv.
  • U = ½LI²; τ = L/R.
  • Vrms = V₀/√2; XL = ωL; XC = 1/ωC.
  • Z = √(R² + (XL − XC)²); P = VrmsIrmscosφ.
  • Resonance ω₀ = 1/√LC; transformer V ∝ N.
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