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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 parallel plate capacitor as a uniform electric field E in the space between the plates
- In a nuclear reactor, the number of U²³⁵ nuclei undergoing fissions per second
- The de-broglie wavelength of an electron in 4th orbit is
- A galvanometer of 50 Ω resistance has 25 divisions. A current of 4×10⁻⁴ A gives
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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