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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
Related Questions: - A particle moves in a circle of radius 5 cm with constant speed and time period
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle moves in a circle of radius 5 cm with constant speed and time period
- Wb/m² is equal to
- A solenoid has length 0.4 cm, radius 1 cm and 400 turns of wire. If a current
- In R-L-C series circuit, the potential differences across each element is 20 V.
- The current in the coil of inductance 5H decreases at the rate of 2 A/s
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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