NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm 2 and length = 20 cm . If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is μ 0 = 4 π × 10 - 7 T m A -1
Options
- A2.4 π × 10 - 4 H
- B2.4 π × 10 - 5 H
- C4.8 π × 10 - 4 H
- D4.8 π × 10 - 5 H
Correct answer
A. 2.4 π × 10 - 4 H
Step-by-step solution
Mutual inductance M = flux in ( 2 ) current in ( 1 ) = ϕ 2 i 1 = B 1 . A 2 N 2 i 1 = μ 0 N 1 i 1 l × A 2 N 2 i 1 = μ 0 N 1 . N 2 . A l A 1 = A 2 = A A = π r 1 2 = 10 cm 2 , l = 20 cm , N 1 = 300 , N 2 = 400 . M = μ 0 N 1 N 2 A l = 4 π × 10 - 7 × 300 × 400 × 10 × 10 - 4 0.20 = 2.4 π × 10 - 4 H .