JEE Main Magnetism, EMI and EM Waves: Electromagnetic Induction and Alternating Currents – Chapter Test

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Difficulty: 10 medium · 10 hard

Electromagnetic Induction and Alternating Currents MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. The energy stored in an inductor of inductance 0.5 H carrying a steady current of 2 A is:

   0.5 H प्रेरकत्व वाले एक प्रेरक में, जिसमें 2 A की स्थिर धारा बह रही है, संचित ऊर्जा है:

   • A. 4 J
   • B. 0.5 J
   • C. 1 J
   • D. 2 J
   Solution

   Energy stored = ½LI² = ½ × 0.5 × 2² = 1 J.

   संचित ऊर्जा = ½LI² = ½ × 0.5 × 2² = 1 J।

2. Q2. Arrange the following discoveries/developments in electromagnetism in correct chronological order (earliest first): P. Oersted's discovery of the magnetic effect of an electric current Q. Faraday's (and Henry's) discovery of electromagnetic induction R. Lenz's law S. Maxwell's formulation of the electromagnetic field equations

   वैद्युतचुंबकत्व में निम्नलिखित खोजों/विकासों को सही कालानुक्रमिक क्रम में व्यवस्थित कीजिए (सबसे पहले वाला पहले): P. ओर्स्टेड द्वारा विद्युत धारा के चुंबकीय प्रभाव की खोज Q. फैराडे (तथा हेनरी) द्वारा वैद्युतचुंबकीय प्रेरण की खोज R. लेंज़ का नियम S. मैक्सवेल द्वारा वैद्युतचुंबकीय क्षेत्र समीकरणों का सूत्रीकरण

   • A. P, Q, S, R
   • B. Q, P, R, S
   • C. P, Q, R, S
   • D. P, R, Q, S
   Solution

   Oersted (1820) came first, followed by Faraday and Henry's induction experiments (1831), then Lenz's law (1834), and finally Maxwell's field equations (1865).

   ओर्स्टेड (1820) सबसे पहले आया, उसके बाद फैराडे और हेनरी के प्रेरण प्रयोग (1831), फिर लेंज़ का नियम (1834), और अंत में मैक्सवेल के क्षेत्र समीकरण (1865)।

3. Q3. An ideal step-down transformer converts 2200 V to 220 V. If the primary winding has 1000 turns, the number of turns in the secondary winding is:

   एक आदर्श अपचायी (step-down) ट्रांसफॉर्मर 2200 V को 220 V में परिवर्तित करता है। यदि प्राथमिक कुंडलन में 1000 फेरे हैं, तो द्वितीयक कुंडलन में फेरों की संख्या है:

   • A. 100
   • B. 10000
   • C. 200
   • D. 10
   Solution

   For an ideal transformer Ns/Np = Vs/Vp = 220/2200 = 1/10, so Ns = 1000/10 = 100 turns.

   आदर्श ट्रांसफॉर्मर के लिए Ns/Np = Vs/Vp = 220/2200 = 1/10, अतः Ns = 1000/10 = 100 फेरे।

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