| ⇦ | 
| ⇨ |
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:
- The phase difference between two points seperated by 0.8 m in a wave of frequency
- The magnetic field in a certain region of space is given by B =8.35×10⁻² Î T.
- The ‘rad’ is the correct unit used to report the measurement of
- Two metal spheres of radii 0.01m and 0.02m are given a charge of 15mC and 45mC
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
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