Skip to main content
Chapter 19 of 30
Practice Quiz

Electromagnetic Induction and Alternating Current

NIOS · Class 12 · Physics

Practice quiz for Electromagnetic Induction and Alternating Current — NIOS Class 12 Physics. MCQs and questions with answers to test your preparation.

45 questions35 flashcards5 concepts

Interactive on Super Tutor

Studying Electromagnetic Induction and Alternating Current? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for practice quiz and more.

1,000+ Class 12 students started this chapter today

A labeled diagram illustrating the concept of magnetic flux through a plane area placed in a uniform magnetic field, showing how the angle between the area vector and magnetic field lines affects the
Super Tutor

This is just one of 26+ visuals inside Super Tutor's Electromagnetic Induction and Alternating Current chapter

Explore the full set

Quick Quiz: Electromagnetic Induction and Alternating Current

0/4

Tap an answer to check it instantly. No sign-up needed for these 4.

1

A circular coil of 75 turns and radius 35 mm has its axis parallel to a uniform magnetic field. The field changes from 25 mT to 50 mT in 250 ms. What is the magnitude of the induced EMF in the coil?

2

A bar magnet is moved towards a stationary conducting ring. According to Lenz's Law, the induced current in the ring will:

3

A solenoid has length 1 m, cross-sectional area 3.14 × 10⁻² m² (radius 10 cm), and 1000 turns. What is its self-inductance? (μ₀ = 4π × 10⁻⁷ H/m)

4

In a series LCR circuit, at resonance frequency, which of the following is TRUE about the impedance Z?

45 Questions·
multiple choicemultiple correct

Sample Questions

1multiple choice
1 marks

An AC generator produces an EMF given by ε(t) = ε₀ sin(ωt). If the coil has N turns, area A, and rotates in a field B with angular velocity ω, what is the peak EMF ε₀?

Show answer

ε₀ = NBAω

Step 1: The magnetic flux through a rotating coil: φ(t) = NBA cos(ωt). Step 2: By Faraday's Law for N turns: ε = -N dφ/dt. Step 3: Differentiating: ε = -N × (-NBAω sin(ωt)) — wait, for a single-turn flux NBA cosωt, ε = NBAω sin(ωt). Step 4: Therefore, the peak (maximum) EMF is ε₀ = NBAω. Step 5: Option A (NBA) is just the maximum flux, not the EMF. Option C (NBA/ω) confuses multiplication and division. Option D adds an extra N, which is incorrect.

2multiple choice
1 marks

A transformer has 100 turns in its primary and 500 turns in its secondary. If the primary voltage is 120 V and primary current is 3 A, what is the secondary current? (Assume ideal transformer)

Show answer

0.6 A

Step 1: For an ideal transformer: Np/Ns = Ip/Is (turns ratio equals inverse current ratio). Step 2: Given Np = 100, Ns = 500, Ip = 3 A. Step 3: Is = Ip × (Np/Ns) = 3 × (100/500) = 3 × 0.2 = 0.6 A. Step 4: Verification using power: Primary power = 120 × 3 = 360 W. Secondary voltage = 120 × (500/100) = 600 V. Secondary power = 600 × 0.6 = 360 W ✓ (ideal transformer, no power loss). Step 5: Option A (15 A) multiplies instead of dividing. Option B (3 A) ignores the turns ratio. Option D (1.5 A) uses an incorrect ratio calculation.

3multiple choice
1 marks

In a purely capacitive AC circuit, if the supply voltage is V = V_m cos(ωt), what is the phase relationship between the current and voltage?

Show answer

Current leads voltage by 90°

Step 1: For a capacitor, charge q = CV = CV_m cos(ωt). Step 2: Current I = dq/dt = -CV_m ω sin(ωt). Step 3: Since -sin(ωt) = cos(ωt + 90°), the current can be written as I = CV_m ω cos(ωt + 90°). Step 4: This shows the current is ahead of the voltage by 90°, i.e., current LEADS voltage by 90°. Step 5: Option A describes an inductor (current lags). Option B describes a pure resistor. Option D (45°) is incorrect — the phase difference for a pure capacitor is always exactly 90°.

4multiple choice
1 marks

The rms value of the AC mains voltage in India is 220 V. What is the peak (maximum) voltage?

Show answer

311 V

Step 1: The relationship between peak and rms values: V_rms = V_m / √2. Step 2: Rearranging: V_m = V_rms × √2. Step 3: Substituting: V_m = 220 × √2 = 220 × 1.414 ≈ 311 V. Step 4: This is why AC mains at 220 V (rms) can be dangerous — the peak voltage reaches about 311 V. Step 5: Option A (220 V) confuses rms with peak. Option B (155.6 V) divides by √2 instead of multiplying. Option D (440 V) doubles the rms value, which is incorrect.

+41 more questions available

Practice All

Frequently Asked Questions

What are the important topics in Electromagnetic Induction and Alternating Current for NIOS Class 12 Physics?
Key topics in Electromagnetic Induction and Alternating Current include Chapter Overview: Electromagnetic Induction and AC, Electromagnetic Induction and AC — Complete Concept Map, Complete Mind Map of Electromagnetic Induction and AC Chapter. These are the concepts NIOS Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Electromagnetic Induction and Alternating Current — NIOS Class 12 Physics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

For serious students

Get the full Electromagnetic Induction and Alternating Current chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for NIOS Class 12 Physics.