For a coil of N turns and area A rotating at angular speed ω in a field B, the instantaneous value of the emf is ε = NBAω sin ωt (6.19), where NBAω is the maximum value of the emf, which occurs when sin ωt = ±1.
-- NCERT Class 12 Physics, Ch. 6, p. 171AC Generator
8 MCQs9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 8 Oct 2026
Try this first
What energy conversion does an ac generator perform?
- A.Electrical energy into mechanical energy.
- B.Chemical energy into electrical energy.
- C.Mechanical energy into electrical energy.
- D.Heat energy directly into mechanical energy.
Tap to see the answer
Answer: C. C is correct: NCERT states that an ac generator converts mechanical energy into electrical energy (NCERT Class 12 Physics, Chapter 6, page 171).
A is wrong: A is wrong because it describes a motor, which runs the conversion the other way.
B is wrong: B is wrong because chemical-to-electrical conversion is what a cell does; a generator turns a coil in a magnetic field.
D is wrong: D is wrong because a generator does not turn heat directly into mechanical work; the mechanical input comes from falling water or steam, and the output is electrical.
AC Generator, explained for NEET
The trap is the wrong moment for the peak. The emf of a rotating coil is largest when the coil's plane is parallel to the field, where the flux through it is zero, and it is zero when the flux is largest. An answer that pairs maximum flux with maximum emf has the two swapped.
An ac generator converts mechanical energy into electrical energy. A coil, the armature, is mounted on a rotor shaft and turned in a uniform magnetic field, with its axis of rotation perpendicular to the field. Its ends reach the external circuit through slip rings and brushes (NCERT Class 12 Physics, Chapter 6, page 171). The principle is that turning the coil changes its effective area, A cos θ, where θ is the angle between A and B (NCERT Class 12 Physics, Chapter 6, page 170).
For a coil turning at constant angular speed ω, θ = ωt and the flux is BA cos ωt. Faraday's law then gives the instantaneous emf of an N-turn coil, ε = NBAω sin ωt, and the maximum value ε₀ = NBAω occurs when sin ωt = ±1 (NCERT Class 12 Physics, Chapter 6, page 171).
NCERT states the extremum condition directly: the emf has its extreme value when θ = 90° or 270°, because the change of flux is greatest at those points. The polarity of the emf reverses periodically, so the current alternates (NCERT Class 12 Physics, Chapter 6, page 171). With ω = 2πn, the emf is ε = ε₀ sin 2πnt, where n is the frequency of revolution of the coil (NCERT Class 12 Physics, Chapter 6, page 171).
Hydro-electric, thermal and nuclear generators differ only in what supplies the mechanical energy that turns the armature (NCERT Class 12 Physics, Chapter 6, page 171). NEET has asked for the maximum current in a coil rotating in the earth's field, which is ε₀ = NBAω followed by Ohm's law.
Watch-out: ω is in rad s⁻¹; a rotation rate in revolutions per second must be multiplied by 2π first.
How do you solve a AC Generator question? A worked example
- 1
Given
A coil of N = 250 turns and area A = 0.040 m² rotates at n = 20 revolutions per second in a uniform field B = 0.30 T. The emf is zero at t = 0, so ε = ε₀ sin 2πnt.
- 2
Required
The maximum emf ε₀, and the emf at t = 1/120 s.
- 3
Concept
For a rotating coil the emf is ε = NBAω sin ωt, with ω = 2πn, so ε₀ = NBAω and ε = ε₀ sin 2πnt (NCERT Class 12 Physics, Chapter 6, page 171).
- 4
Formula
ε₀ = NBA × 2πn; ε = ε₀ sin(2πnt).
- 5
Substitution
ε₀ = 250 × 0.30 × 0.040 × 2π × 20; angle = 2π × 20 × (1/120).
- 6
Calculation
NBA = 250 × 0.30 × 0.040 = 3.0, and 2πn = 2 × 3.142 × 20 = 125.7 rad s⁻¹, so ε₀ = 3.0 × 125.7 = 377 V, which is 3.8 × 10² V to two significant figures. The angle is 2π × 20 / 120 = π/3, and sin(π/3) = 0.866, so ε = 377 × 0.866 = 326 V, which is 3.3 × 10² V. Here π = 3.142.
- 7
Final answer
The maximum emf is 3.8 × 10² V, and the emf at t = 1/120 s is 3.3 × 10² V.
- 8
Common trap
Putting n = 20 in place of ω in ε₀ = NBAω, which gives 60 V and misses the factor 2π. A second slip is taking the emf as maximum when the coil lies perpendicular to the field, where the flux is largest and the emf is zero.
- 9
Similar NEET-style question
A coil of 100 turns and area 0.050 m² rotates at 50 revolutions per second in a field of 0.40 T. What is the maximum emf? (Answer: ε₀ = NBA × 2πn = 100 × 0.40 × 0.050 × 2π × 50 = 2.0 × 314 = 628 V, which is 6.3 × 10² V.)
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Can you answer these AC Generator MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
What energy conversion does an ac generator perform?
Show answer and why every option is right or wrong
Answer: C. C is correct: NCERT states that an ac generator converts mechanical energy into electrical energy (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because it describes a motor, which runs the conversion the other way.
Why B is wrong: B is wrong because chemical-to-electrical conversion is what a cell does; a generator turns a coil in a magnetic field.
Why D is wrong: D is wrong because a generator does not turn heat directly into mechanical work; the mechanical input comes from falling water or steam, and the output is electrical.
MCQ 2Easy RecallPractice
In the simple ac generator described by NCERT, how are the ends of the rotating coil connected to the external circuit?
Show answer and why every option is right or wrong
Answer: A. A is correct: the ends of the coil are connected to the external circuit by means of slip rings and brushes (NCERT Class 12 Physics, Chapter 6, page 171).
Why B is wrong: B is wrong because a transformer changes the size of an alternating voltage and is not how the rotating coil is contacted.
Why C is wrong: C is wrong because a capacitor blocks steady current and plays no part in connecting the rotor to the circuit.
Why D is wrong: D is wrong because wires fixed round a rotating shaft would twist; sliding contacts are needed, and NCERT names slip rings and brushes.
MCQ 3Easy RecallPractice
An N-turn coil of area A rotates at constant angular speed ω in a uniform field B, with θ = ωt. Which expression gives the instantaneous emf?
Show answer and why every option is right or wrong
Answer: D. D is correct: the flux is BA cos ωt, and Faraday's law gives ε = NBAω sin ωt, with maximum value NBAω (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because it leaves out ω, which arises from differentiating cos ωt with respect to time.
Why B is wrong: B is wrong because differentiating cos ωt gives −ω sin ωt, so the emf follows a sine, not a cosine, when θ = 0 at t = 0.
Why C is wrong: C is wrong because it omits the area A; the flux depends on BA cos θ.
MCQ 4Direct ApplicationPractice
A coil of 200 turns and area 0.050 m² rotates at an angular speed of 100 rad s⁻¹ in a uniform field of 0.20 T. What is the maximum emf?
Show answer and why every option is right or wrong
Answer: B. B is correct: ε₀ = NBAω = 200 × 0.20 × 0.050 × 100 = 200 V (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because 200 × 0.20 × 0.050 = 2.0 omits the angular speed ω.
Why C is wrong: C is wrong because 2.0 × 2π × 100 treats the 100 as a frequency n and multiplies by 2π again, but the stem already gives ω in rad s⁻¹.
Why D is wrong: D is wrong because it doubles the result; no factor of 2 appears in ε₀ = NBAω.
MCQ 5Direct ApplicationPractice
The emf of a generator is ε = ε₀ sin 2πnt, with ε₀ = 120 V and n = 25 Hz. What is the emf at t = 1/200 s?
Show answer and why every option is right or wrong
Answer: D. D is correct: the angle is 2π × 25 × (1/200) = π/4, and sin(π/4) = 0.707, so ε = 120 × 0.707 = 85 V (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because 120 V is the peak value, reached only when the angle is π/2, not π/4.
Why B is wrong: B is wrong because 60 V is 120 × sin 30°; the angle here is π/4 (45°), not π/6.
Why C is wrong: C is wrong because the emf is zero only when the angle is a multiple of π; at π/4 the sine is not zero.
MCQ 6Direct ApplicationPractice
A coil rotates in a uniform magnetic field, and θ is the angle between the field B and the area vector A of the coil. At which orientation is the induced emf maximum?
Show answer and why every option is right or wrong
Answer: A. A is correct: NCERT states that the emf has its extremum value at θ = 90° or 270°, because the change of flux is greatest at those points (NCERT Class 12 Physics, Chapter 6, page 171).
Why B is wrong: B is wrong because at θ = 0° the flux BA cos θ is at its maximum, but it is momentarily not changing, so the emf is zero.
Why C is wrong: C is wrong because at 45° the emf is NBAω sin 45°, which is about 0.71 of the maximum, not the maximum.
Why D is wrong: D is wrong because ε = NBAω sin ωt varies with the angle, changing sign twice per revolution.
MCQ 7CalculationPractice
A coil of 100 turns and area 0.020 m² rotates at 50 rad s⁻¹ in a uniform field of 0.50 T. The total resistance of the circuit is 10 Ω. What is the maximum current in the coil?
Show answer and why every option is right or wrong
Answer: C. C is correct: ε₀ = NBAω = 100 × 0.50 × 0.020 × 50 = 50 V, and the maximum current is 50 V / 10 Ω = 5.0 A (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because 0.50 A would need a resistance of 100 Ω, or an emf one tenth as large.
Why B is wrong: B is wrong because 2.5 A halves the correct result; no factor of one half enters ε₀ = NBAω.
Why D is wrong: D is wrong because 50 is the maximum emf in volts; the current requires dividing by the 10 Ω resistance.
MCQ 8CalculationPractice
The coil of a generator is turned at twice the earlier angular speed, with N, A and B unchanged. What happens to the maximum emf and to the frequency of the emf?
Show answer and why every option is right or wrong
Answer: B. B is correct: ε₀ = NBAω is proportional to ω, so it doubles, and with ω = 2πn the frequency n = ω/2π doubles as well (NCERT Class 12 Physics, Chapter 6, page 171).
Why A is wrong: A is wrong because ε₀ = NBAω contains ω, so a faster rotation raises the peak emf.
Why C is wrong: C is wrong because the frequency of the emf is the frequency of revolution of the coil, n = ω/2π, and it rises with ω.
Why D is wrong: D is wrong because ε₀ is linear in ω, not quadratic; doubling ω doubles ε₀.
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What to remember before solving AC Generator questions
7 NCERT lines
As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos q, where q is the angle between A and B.
-- NCERT Class 12 Physics, Ch. 6, p. 170An ac generator converts mechanical energy into electrical energy. The basic elements of an ac generator are shown in Fig. 6.13. It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes.
-- NCERT Class 12 Physics, Ch. 6, p. 171When the coil is rotated with a constant angular speed w, the angle q between the magnetic field vector B and the area vector A of the coil at any instant t is q = wt (assuming q = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. (6.1), the flux at any time t is FB = BA cos q = BA cos wt From Faraday’s law, the induced emf for the rotating coil of N turns is then, d d – – (cos ) dt d B N NBA t t Φ ε ω = = Thus, the instantaneous value of the emf is ε ω ω = NBA sin t (6.19) where NBAw is the maximum value of the emf, which occurs when sin wt = ±1. If we denote NBAw as e0, then e = e0 sin wt (6.20)
-- NCERT Class 12 Physics, Ch. 6, p. 171Since the value of the sine fuction varies between +1 and –1, the sign, or polarity of the emf changes with time. Note from Fig. 6.14 that the emf has its extremum value when q = 90° or q = 270°, as the change of flux is greatest at these points. The direction of the current changes periodically and therefore the current is called alternating current (ac).
-- NCERT Class 12 Physics, Ch. 6, p. 171Since w = 2pn, Eq (6.20) can be written as e = e0sin 2p n t (6.21) where n is the frequency of revolution of the generator’s coil.
-- NCERT Class 12 Physics, Ch. 6, p. 171In commercial generators, the mechanical energy required for rotation of the armature is provided by water falling from a height, for example, from dams. These are called hydro-electric generators. Alternatively, water is heated to produce steam using coal or other sources. The steam at high pressure produces the rotation of the armature. These are called thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear power generators.
-- NCERT Class 12 Physics, Ch. 6, p. 171AC Generator: NEET previous year questions (PYQs) with answers
1 question from NEET 2022, answers verified against NTA official keys
NEET 2022
12 A
20.25 A
31.5 A
41 A
NTA Answer: Option 4(final)
Why the other options are wrong
- Option 1: 2 A would need twice the emf. The peak emf is NBAω, which 'is the maximum value of the emf': 1000 × 2 × 10⁻⁵ T × π(10 m)² × 2 rad s⁻¹ = 4π ≈ 12.56 V, and 12.56 V / 12.56 Ω = 1 A.
- Option 2: 0.25 A corresponds to an emf of only π V. NBAω = 1000 × 2 × 10⁻⁵ × 100π × 2 = 4π V, which 'is the maximum value of the emf', so the maximum current is 4π/12.56 = 1 A.
- Option 3: 1.5 A needs an emf of about 18.8 V. With ε = NBAω sin ωt the peak, reached 'when sin wt = ±1', is NBAω = 4π ≈ 12.56 V, so the maximum current is 12.56/12.56 = 1 A.
All 23 past-paper questions from Electromagnetic Induction and Alternating Currents →
More in Electromagnetic Induction and Alternating Currents: 4 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Physics Chapter 6, p.171 | Class 12 Physics Chapter 6, p.170
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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