A long solenoid has length \(l\), area of cross-section \(A\) and \(n\) turns per unit length. A current \(I\) in it sets up a field \(B=\mu_0nI\) inside. Using \(N\Phi_B=LI\), the self-inductance of the solenoid comes out as
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\mu_0n^{2}Al\)
Self-inductance is the flux linked with the coil per unit current. The solenoid has \( N=nl \) turns in all, and the flux through each turn is \( \Phi_B=\mu_0nIA \).
\( \begin{aligned} N\Phi_B &= (nl)(\mu_0nI)(A) \\ &= \mu_0n^{2}AlI \end{aligned} \)
\( \begin{aligned} L &= \frac{N\Phi_B}{I} \\ &= \mu_0n^{2}Al \end{aligned} \)
Option (1) \(\mu_0nAl\) — Incorrect. This takes the field as \( \mu_0I \) and drops \( n \). The field inside is \( \mu_0nI \), so \( n \) comes in twice.
Option (2) \(\mu_0n^{2}A/l\) — Incorrect. Here the length divides. The coil has \( nl \) turns in all, so \( l \) must multiply the flux of one turn.
Option (3) \(\mu_0nA\) — Incorrect. This is the flux through one turn for unit current. It leaves out the \( nl \) turns of the solenoid.
Option (4) \(\mu_0n^{2}Al\) — Correct. The flux of one turn, \( \mu_0nIA \), is multiplied by all \( nl \) turns and divided by \( I \).
NCERT (2026-27):
Self-inductance – “… The total flux linked with the solenoid is”
\( N\Phi_B=(nl)(\mu_0nI)(A)=\mu_0n^{2}AlI \)
“where nl is the total number of turns. Thus, the self-inductance is,”
\( \displaystyle L=\frac{N\Phi_B}{I}= \)\( \mu_0n^{2}Al \)\( \qquad (6.15) \)