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If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is
Options
(a) 48.4 mH
(b) 200 mH
(c) 187.5 mH
(d) 320 mH
Correct Answer:
320 mH
Explanation:
Self inductance of a coil, L = μᵣμ₀n² Al
Where μᵣ is the relative permittivity of the coil, n is the number of terms per unit length, A is the area of cross-section, l is the length of the solenoid
L ∝ n²
Self inductance of 500 turns coil = 125 mH Therefore, L for the coil of 800 turns = [125 / (500)²] × 800² L = 320 mH
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A total charge of 5µC is distributed uniformly on the surface of the thin walled
- A galvanometer of 50 Ω resistance has 25 divisions. A current of 4×10⁻⁴ A gives
- An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp
- The potential energy of a system increases if work is done
- Source S₁ is producing, 10¹⁵ photons per second of wavelength 5000Å
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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