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 \( 4 \times 10^{-3}\,\mathrm{Wb} \). The self-inductance of the solenoid is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1 H
The self-inductance is the total flux linkage per unit current. The flux through each turn is given, so the total flux linkage is \( N \) times it.
\( \displaystyle L = \frac{N\,\Phi}{I} \)
\( N = 500 \)
\( \Phi = 4\times10^{-3}\,\mathrm{Wb} \)
\( I = 2\,\mathrm{A} \)
\( \displaystyle L = \frac{500 \times 4\times10^{-3}}{2} = \frac{2}{2} = 1\,\mathrm{H} \)