In a solenoid, if number of turns per unit length is doubled, then self inductance will become:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 times of its initial value
The self-inductance of a long solenoid of length \( l \), cross-section \( A \) and \( n \) turns per unit length is
\( \displaystyle L = \mu_0 n^{2} A l \)
\( n \) enters twice: the field inside, \( B = \mu_0 n I \), is proportional to \( n \), and the number of turns linked with this field, \( nl \), is also proportional to \( n \). So \( L \propto n^{2} \).
\( \displaystyle \frac{L'}{L} = \left(\frac{2n}{n}\right)^{2} = 4 \)
The self-inductance becomes 4 times its initial value.