Electromagnetic Induction — Flashcards
Madhya Pradesh Board · Class 12 · Physics
50 flashcards for Electromagnetic Induction (Madhya Pradesh Board Class 12 Physics) to test yourself on key terms and facts.
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State 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.
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