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Electromagnetic Induction question

2008 · Q132
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Electromagnetic Induction question

2008 · Q132

NEETPhysicsElectromagnetic InductionMCQ+4 / −1
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 ×\times× 10−-−3 Wb. The self-inductance of the solenoid is
  1. A
    1.0 henry
  2. B
    4.0 henry
  3. C
    2.5 henry
  4. D
    2.0 henry
View written solutionFree

Correct answer: A

Total number of turns in the solenoid,

N = 500 Current, I = 2A.

Magnetic flux linked with each turn = 4 × 10–3 Wb

Self inductance of coil :

L = NϕI=500×4×1032{{N\phi } \over I} = {{500 \times 4 \times {{10}^3}} \over 2}INϕ​=2500×4×103​ = 1 H

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