| ⇦ | 
| ⇨ |
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:
- All components of the electromagnetic spectrum in vacuum have the same
- A body travels 200 cm in the first two seconds and 220 cm in the next 4 secs
- Two resistors of 6Ω and 9Ω are connected in series to a 120 V source.
- Two solid discs of radii r and 2r roll from the top of an inclined plane
- Application of a forward bias to a P-N junction
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply