| ⇦ | 
| ⇨ |
A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10⁻³ Wb. The self-inductance of the solenoid is
Options
(a) 1.0 henry
(b) 4.0 henry
(c) 2.5 henry
(d) 2.0 henry
Correct Answer:
1.0 henry
Explanation:
Total no. of turns in the solenoid,
N = 500 Currrent, I = 2A.
magnetic flux linked with each turn = 4 x 10⁻³ Wb
As, ? = LI or N ? = LI ⇒ L = N? / I
= 500 x 4 x 10⁻³ / 2 henry = 1 H
Related Questions:
- An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp
- In a uniform circular motion, work done in one complete rotation is
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- When the speed of electron beam used in Young’s double slit experiment is increased,
- An LCR circuit contains resistance of 100 ohm and a supply of 200 volt at 300 radian
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