NEET UG Magnetism, EMI & AC: Moving Charges and Magnetism – Chapter Test

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Difficulty: 9 medium · 11 hard

Moving Charges and Magnetism MCQ questions with answers

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

1. Q1. A straight conductor of length 0.2 m carrying a current of 2 A is placed in a uniform magnetic field of 0.5 T such that the conductor makes an angle of 30° with the field. The magnitude of the force on the conductor is:

   0.2 m लंबाई का एक सीधा चालक, जिसमें 2 A की धारा बह रही है, 0.5 T के एक समान चुंबकीय क्षेत्र में इस प्रकार रखा गया है कि चालक क्षेत्र से 30° का कोण बनाता है। चालक पर लगने वाले बल का परिमाण है:

   • A. 0.4 N
   • B. 0.05 N
   • C. 0.2 N
   • D. 0.1 N
   Solution

   F = BIL sinθ = 0.5 × 2 × 0.2 × sin30° = 0.5 × 2 × 0.2 × 0.5 = 0.1 N.

   F = BIL sinθ = 0.5 × 2 × 0.2 × sin30° = 0.5 × 2 × 0.2 × 0.5 = 0.1 N।

2. Q2. An electron moves with a speed of 3×10⁷ m/s at right angles to a uniform magnetic field of 2 T. The magnitude of the magnetic force on the electron is:

   एक इलेक्ट्रॉन 2 T के एकसमान चुंबकीय क्षेत्र के समकोण पर 3×10⁷ m/s की चाल से गति करता है। इलेक्ट्रॉन पर चुंबकीय बल का परिमाण है:

   • A. 9.6×10⁻¹¹ N
   • B. 9.6×10⁻¹² N
   • C. 4.8×10⁻¹² N
   • D. 3.2×10⁻¹² N
   Solution

   F = qvB = (1.6×10⁻¹⁹)(3×10⁷)(2) = 9.6×10⁻¹² N, since the velocity is perpendicular to the field.

   F = qvB = (1.6×10⁻¹⁹)(3×10⁷)(2) = 9.6×10⁻¹² N, क्योंकि वेग क्षेत्र के लंबवत है।

3. Q3. The Biot–Savart law for the magnetic field dB produced by a current element I dl at a position r from it is best expressed as:

   किसी धारा अवयव I dl द्वारा उससे r स्थिति पर उत्पन्न चुंबकीय क्षेत्र dB के लिए बायो–सावर्ट नियम को सर्वोत्तम रूप से इस प्रकार व्यक्त किया जाता है:

   • A. dB = (μ₀/4π) I (dl × r)/r³
   • B. dB = (μ₀/4π) I (dl · r)/r³
   • C. dB = (μ₀/4π) I (dl × r)/r²
   • D. dB = (μ₀/4π) I (dl × r)/r
   Solution

   The Biot–Savart law states dB = (μ₀/4π) I (dl × r̂)/r², which equals (μ₀/4π) I (dl × r)/r³. The field varies inversely as the square of the distance.

   बायो–सावर्ट नियम के अनुसार dB = (μ₀/4π) I (dl × r̂)/r², जो (μ₀/4π) I (dl × r)/r³ के बराबर है। क्षेत्र दूरी के वर्ग के व्युत्क्रमानुपाती होता है।

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