A long solenoid has 500 turns. When a current of 2 A 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

1 H

The correct answer is Option (4) → 1 H

Given:

Number of turns: $N = 500$

Current: $I = 2$ A

Flux per turn: $\Phi = 4 \times 10^{-3}$ Wb

Self-inductance: $L = \frac{N \Phi}{I}$

$L = \frac{500 \times 4 \times 10^{-3}}{2}$

$L = \frac{2}{2} = 2$ H

Answer: $L = 1$ H