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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
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- In a P-N junction
- Tangent galvanometer is used to measure
- Eight drops of a liquid of density ρ and each of radius a are falling through air
- In Young’s double slit experiment, the intensity of light coming from the first slit
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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