Electromagnetic Induction
Madhya Pradesh Board · Class 12 · Physics
Flashcards for Electromagnetic Induction — Madhya Pradesh Board Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
Interactive on Super Tutor
Studying Electromagnetic Induction? Get the full interactive chapter.
Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for flashcards and more.
1,000+ Class 12 students started this chapter today
This is just one of 18+ visuals inside Super Tutor's Electromagnetic Induction chapter
Explore the full setState Faraday’s law of electromagnetic induction.
Answer
Faraday’s law states: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. Formula: ε = -dΦ_B/dt. For N turns: ε = -N dΦ_B/dt. The ne…
विद्युतचुंबकीय प्रेरण का Faraday का नियम लिखिए।
Answer
Faraday का नियम कहता है: किसी परिपथ में प्रेरित emf का परिमाण, उस परिपथ से होकर गुजरने वाले चुंबकीय फ्लक्स के परिवर्तन की समय दर के बराबर होता है। सूत्र: ε = -dΦ_B/dt. N चक्करों के लिए: ε = -N dΦ_B/dt…
Lenz का नियम लिखिए।
Answer
Lenz का नियम कहता है: प्रेरित emf की ध्रुवीयता ऐसी होती है कि वह ऐसी धारा उत्पन्न करने की प्रवृत्ति रखती है जो उस चुंबकीय फ्लक्स के परिवर्तन का विरोध करती है जिसने उसे उत्पन्न किया था। Faraday के नियम…
State Lenz’s law.
Answer
Lenz’s law states: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The negative sign in Faraday’s law represents this…
A magnetic flux through a loop changes from 0.8 Wb to 0.2 Wb in 0.3 s. Find the induced emf.
Answer
Given: Φ1 = 0.8 Wb, Φ2 = 0.2 Wb, Δt = 0.3 s. Formula: ε = |ΔΦ_B|/Δt. Step 1: ΔΦ_B = 0.2 - 0.8 = -0.6 Wb. Step 2: |ΔΦ_B| = 0.6 Wb. Step 3: ε = 0.6/0.3 = 2.0 V. Answer: 2.0 V.
किसी लूप से होकर चुंबकीय फ्लक्स 0.8 Wb से 0.2 Wb हो जाता है 0.3 s में। प्रेरित emf ज्ञात कीजिए।
Answer
दिया है: Φ1 = 0.8 Wb, Φ2 = 0.2 Wb, Δt = 0.3 s. सूत्र: ε = |ΔΦ_B|/Δt. चरण 1: ΔΦ_B = 0.2 - 0.8 = -0.6 Wb. चरण 2: |ΔΦ_B| = 0.6 Wb. चरण 3: ε = 0.6/0.3 = 2.0 V. उत्तर: 2.0 V.
एक कुंडली में 200 चक्कर हैं और प्रति चक्कर फ्लक्स 0.015 Wb s^-1 की दर से बदलता है। प्रेरित emf ज्ञात कीजिए।
Answer
दिया है: N = 200, dΦ_B/dt = 0.015 Wb s^-1. सूत्र: ε = -N dΦ_B/dt. चरण 1: ε = -200 × 0.015 V. चरण 2: ε = -3.0 V. प्रेरित emf का परिमाण = 3.0 V.
A coil has 200 turns and the flux per turn changes at 0.015 Wb s^-1. Find the induced emf.
Answer
Given: N = 200, dΦ_B/dt = 0.015 Wb s^-1. Formula: ε = -N dΦ_B/dt. Step 1: ε = -200 × 0.015 V. Step 2: ε = -3.0 V. Magnitude of induced emf = 3.0 V.
+42 more flashcards available
Practice AllFrequently Asked Questions
What are the important topics in Electromagnetic Induction for Madhya Pradesh Board Class 12 Physics?
How to score full marks in Electromagnetic Induction — Madhya Pradesh Board Class 12 Physics?
How many flashcards are available for Electromagnetic Induction?
Sources & Official References
Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.
More resources for Electromagnetic Induction
Practice Quiz
Test yourself with a quick quiz
Important Questions
Practice with board exam-style questions
Revision Notes
Key points for last-minute revision
Formula Sheet
All formulas in one place
Chapter Summary
Understand the chapter at a glance
Concept Maps
See how topics connect visually
Study Plan
Step-by-step plan to ace this chapter
Syllabus
What topics to cover
NCERT Solutions
Every textbook question solved step by step
For serious students
Get the full Electromagnetic Induction chapter — for free.
Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Madhya Pradesh Board Class 12 Physics.