Physics Class 12 Part I…
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Ac Generator

🎓 Class 12 Physics CBSE Theory Ch 6 – Electromagnetic Induction ⏱ ~14 min
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Ac Generator

6.10 AC Generator

Faraday's law has a beautiful technological consequence: rotate a coil in a magnetic field and you obtain a continuous, alternating EMF - the basis of every AC generator in the world's power stations.

6.10.1 Construction

Essential parts of a simple AC generator:

  • Field magnets — produce a uniform magnetic field B.
  • Armature — a rectangular coil of N turns wound on a soft-iron core, free to rotate about a horizontal axis perpendicular to B.
  • Slip rings — two metal rings rotating with the armature; each connected to one end of the coil.
  • Carbon brushes — stationary contacts pressing against the slip rings; they tap off the induced EMF without entanglement.
  • Prime mover — turbine, internal-combustion engine, etc., which mechanically rotates the armature.
N S slip ring slip ring load R Armature coil rotates with ω
Fig 6.7 Schematic of an AC generator. Coil rotation in field B yields a sinusoidal EMF tapped through slip rings.

6.10.2 Working and EMF expression

If the coil has N turns, area A, and rotates at angular speed ω in uniform B, the flux through one turn is

\[\Phi(t) = BA\cos(\omega t)\]

By Faraday's law:

\[\varepsilon(t) = -N\dfrac{d\Phi}{dt} = NBA\omega\sin(\omega t) = \varepsilon_0 \sin(\omega t)\]

The peak EMF is

\[\varepsilon_0 = NBA\omega = NBA(2\pi\nu)\]

where ν is the rotation frequency in hertz. India's grid operates at ν = 50 Hz (so ω ≈ 314 rad/s). The output is sinusoidal, alternating sign every half period - hence "alternating current".

t ε +ε₀ −ε₀ T = 1/ν
Fig 6.8 The output of an AC generator: ε(t) = ε₀ sin(ωt). Period T = 2π/ω = 1/ν.

Worked Example 6.10 - Generator design

Example 6.10 L3 Apply

A coil of 100 turns and area 0.10 m² rotates at 50 revolutions per second in a uniform 0.40 T field. Find the peak EMF.

ω = 2π × 50 = 100π rad/s ≈ 314 rad/s.

ε₀ = NBAω = 100 × 0.40 × 0.10 × 314 = 1256 V ≈ 1.26 kV.

Worked Example 6.11 - Frequency from peak EMF

Example 6.11 L4 Analyse

An armature with NBA = 0.50 V·s produces a peak voltage of 100 V. Find ω and frequency.

ω = ε₀/(NBA) = 100/0.50 = 200 rad/s.

ν = ω/(2π) = 200/(2π) ≈ 31.8 Hz.

6.11 Applications of Electromagnetic Induction

ApplicationUnderlying principle
Electric generators (AC and DC)Sinusoidal EMF from rotating coils.
Transformers (next chapter)Mutual induction between primary and secondary windings.
Induction motorsRotating magnetic fields induce currents in rotor → torque.
Induction cooktopsEddy currents heat the iron base of the pan; glass top stays cool.
Metal detectorsEddy currents in metal change the inductance of a search coil.
Magnetic braking (trains)Eddy currents in conducting rails dissipate kinetic energy.
Wireless / inductive chargingMutual inductance between charger and device coils.
Microphones (dynamic)Diaphragm motion in a magnet induces audio-frequency EMF.
Electric guitar pickupsVibrating steel string changes flux in a magnet-coil pickup.
Card readers / RFIDReader coil's changing flux induces current in the card's coil to power its chip.

6.11.1 The induction motor in plain words

Three sets of stator coils carry three-phase AC; their fields combine to a magnetic field that rotates in space at the supply frequency. This rotating field induces currents in the rotor (a "squirrel cage" of conducting bars). By Lenz's law these currents experience torques that drag the rotor along. No brushes, no slip rings, almost zero maintenance — over 90% of industrial motors world-wide are induction motors.

Interactive: AC Waveform Generator L3 Apply

Vary N, B, A, ω - the displayed sinusoidal trace updates with peak ε₀ = NBAω.

ε₀ = 1256 V  |  ω = 314 rad/s
Activity 6.4 - Build a hand-cranked generatorL5 Evaluate
  1. Wind 200 turns of insulated copper wire on a small rectangular frame (e.g. 5 × 8 cm).
  2. Mount it between two strong magnets so it can rotate freely.
  3. Connect the leads (via slip rings or directly while the coil swings) to a small LED.
  4. Spin the coil by hand.
Predict: How does brightness of the LED depend on the cranking speed? At what minimum speed does the LED first light up?

Faster crank ⇒ greater ω ⇒ ε₀ = NBAω increases ⇒ brighter LED. The LED only lights once ε₀ exceeds its forward-voltage threshold (~2 V for red, ~3 V for blue). Below that no current flows. This is the principle behind dynamo bicycle lights and emergency hand-crank radios.

Competency-Based Questions L1-L6

A small AC generator has a 200-turn coil of area 0.040 m² rotating at 50 Hz in a 0.30 T field, supplying a 100-Ω load.
1. The peak EMF is approximately: L3 Apply
  • (a) 60 V
  • (b) 240 V
  • (c) 753 V
  • (d) 30 V
(c) ε₀ = NBAω = 200 × 0.30 × 0.040 × 2π × 50 ≈ 753 V.
2. The peak current through the load is: L3 Apply
  • (a) 0.6 A
  • (b) 2.4 A
  • (c) 7.5 A
  • (d) 0.3 A
(c) I₀ = ε₀/R ≈ 753/100 ≈ 7.5 A.
3. State two reasons why AC is preferred over DC for grid power transmission. L2 Understand
(i) AC voltage can be stepped up/down efficiently with transformers (mutual induction), keeping I²R losses small in long-distance lines. (ii) AC generators (alternators) are simpler than DC generators - no commutator, just slip rings - hence cheaper and more reliable.
4. Why is the frequency of the EMF the same as the rotational frequency of the coil? L4 Analyse
The flux through the coil is Φ = BA cos(ωt). This completes one full cycle each time the coil rotates by 2π. Hence ε = −dΦ/dt also has period 2π/ω; the EMF frequency exactly equals the rotational frequency.
5. Sketch a project plan to convert a stationary bicycle into a 5-W phone charger using EMI principles. List required components and key safety/efficiency considerations. L6 Create
Components: small permanent-magnet alternator (or dynamo) coupled to bike wheel, bridge rectifier, smoothing capacitor, voltage-regulator IC (5 V output), USB connector, fuse. Sizing: pedaling at ~120 rpm with gear-up to ~1500 rpm at the alternator typical produces ~6-12 V AC; rectify and regulate to clean 5 V/1 A. Safety: fuse before regulator; thermal protection; cable strain relief. Efficiency: large-area magnets, low-resistance windings, soft-iron stator, lubricated bearings.

Assertion-Reason Pairs L4 Analyse

Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.

Assertion: The peak EMF of an AC generator increases with rotation speed.
Reason: ε₀ = NBAω, directly proportional to ω.
(A). Reason directly explains assertion.
Assertion: The induction cooktop heats the cooking vessel but not the glass top.
Reason: The vessel base is iron; eddy currents are induced in it. Glass is non-conducting.
(A). True; reason explains assertion.
Assertion: An AC generator delivers a steady DC current.
Reason: The flux through the coil is constant.
(D). Both statements are false: the EMF is sinusoidal (alternating), and the flux varies as cos(ωt).

Frequently Asked Questions - Ac Generator

What is the main concept covered in Ac Generator?
In NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction), "Ac Generator" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Ac Generator useful in real-life applications?
Real-life applications of "Ac Generator" from NCERT Class 12 Physics Chapter 6 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Ac Generator?
Key formulas in "Ac Generator" (NCERT Class 12 Physics Chapter 6 Electromagnetic Induction) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 6?
NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction) is structured so each part builds on the previous one. "Ac Generator" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Ac Generator?
CBSE board questions from "Ac Generator" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Ac Generator" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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