JEE Main2015PhysicsElectromagnetic InductionActual
When the current in a coil changes from 5 A to 2 A in 0 .1 s , an average voltage of 50   V is produced. The self-inductance of the coil is
Options
- A1 .67 H
- B6 H
- C3 H
- D0 .67 H
Correct answer
A. 1 .67 H
Step-by-step solution
Magnetic flux ϕ = L I ⇒ ∆ ϕ ∆ t = L ∆ I ∆ t ∴ ε ind average = ∆ ϕ ∆ t = L ∆ I ∆ t ⇒ 50 = L × 5 - 2 0.1 ⇒ 5 3 = L ⇒ L = 1.67 H