| ⇦ |
| ⇨ |
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: - A very small circular loop of radius a is initially (at t=0) coplanar and concentric
- The current in the coil of inductance 5H decreases at the rate of 2 A/s
- Four indentical thin rods each of mass M and length l, form a square frame
- The ratio of the radii or gyration of a circular disc to that of a circular ring
- Bragg’s law for X-rays is
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A very small circular loop of radius a is initially (at t=0) coplanar and concentric
- The current in the coil of inductance 5H decreases at the rate of 2 A/s
- Four indentical thin rods each of mass M and length l, form a square frame
- The ratio of the radii or gyration of a circular disc to that of a circular ring
- Bragg’s law for X-rays is
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