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JEE Main Magnetism, EMI and EM Waves – Topic Test

JEE Main Magnetism, EMI and EM Waves – Topic Test is an online topic test in the JEE Main pattern, covering Magnetism, EMI and EM Waves. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.

30
Questions
30
Minutes
30
Marks
−0.25
Negative marking

Topics in this test

TopicQuestions
Magnetism, EMI and EM Waves30

Difficulty: 15 medium · 15 hard

Magnetism, EMI and EM Waves MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

  1. Q1. Which of the following is NOT equivalent to the SI unit of self-inductance, the henry?

    निम्नलिखित में से कौन-सा स्व-प्रेरकत्व के SI मात्रक, हेनरी, के तुल्य नहीं है?

    • A. V·s/A
    • B. Wb/A
    • C. C/V
    • D. Ω·s
    Solution

    1 H = 1 Wb/A = 1 V·s/A = 1 Ω·s (since Ω = V/A). The coulomb per volt (C/V) is the farad, the unit of capacitance, not inductance.

    1 H = 1 Wb/A = 1 V·s/A = 1 Ω·s (क्योंकि Ω = V/A)। कूलॉम प्रति वोल्ट (C/V) फैराड है, जो धारिता का मात्रक है, प्रेरकत्व का नहीं।

  2. Q2. 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।

  3. Q3. In a plane electromagnetic wave propagating in vacuum, the peak value of the electric field is 48 V m⁻¹. The peak value of the magnetic field is (c = 3 × 10⁸ m s⁻¹)

    निर्वात में संचरित एक समतल विद्युतचुंबकीय तरंग में, विद्युत क्षेत्र का शिखर मान 48 V m⁻¹ है। चुंबकीय क्षेत्र का शिखर मान है (c = 3 × 10⁸ m s⁻¹)

    • A. 1.6 × 10⁻⁷ T
    • B. 1.6 × 10⁻⁸ T
    • C. 1.6 × 10⁻⁶ T
    • D. 48 × 10⁸ T
    Solution

    For an EM wave in vacuum E₀ = cB₀, so B₀ = 48 / (3 × 10⁸) = 1.6 × 10⁻⁷ T.

    निर्वात में एक विद्युतचुंबकीय तरंग के लिए E₀ = cB₀, अतः B₀ = 48 / (3 × 10⁸) = 1.6 × 10⁻⁷ T।

Frequently asked questions

How many questions are in the JEE Main Magnetism, EMI and EM Waves – Topic Test?

30 questions for 30 marks, to be answered in 30 minutes.

Is there negative marking in the JEE Main Magnetism, EMI and EM Waves – Topic Test?

Yes — 0.25 marks are deducted for each wrong answer, as in the real exam pattern.

Which topics does the JEE Main Magnetism, EMI and EM Waves – Topic Test cover?

Magnetism, EMI and EM Waves (30 questions).

Is the JEE Main Magnetism, EMI and EM Waves – Topic Test free?

It is part of the JEE Main 2027 Mock Test Series (104 tests, ₹299). You can first try Basic Concepts & Atomic Structure – Topic Test for free.

More tests in the JEE Main 2027 Mock Test Series

See the full JEE Main 2027 Mock Test Series → · JEE Main: pattern, notification & syllabus →

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