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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 super conductor exhibits perfect
- For a diatomic gas charge in internal energy for unit charge in temperature for
- A particle of mass M is suited at the centre of a spherical shell of same mass and radius
- A bomb of mass 30 kg at rest exploded into two pieces of masses 18 kg
- A carnot engine working between 300 K and 600 K has a work output of 800J per cycle.
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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