| ⇦ | 
| ⇨ |
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: - When the angle of incidence is 60° on the surface of a glass slab, it is found
- The drift velocity does not depend upon
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
- A body travels 200 cm in the first two seconds and 220 cm in the next 4 secs
- A particle moves along a circle of radius 20/π m with constant tangential acceleration
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When the angle of incidence is 60° on the surface of a glass slab, it is found
- The drift velocity does not depend upon
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
- A body travels 200 cm in the first two seconds and 220 cm in the next 4 secs
- A particle moves along a circle of radius 20/π m with constant tangential acceleration
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