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
Answer & explanation
Correct answer: option 4
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