Find the self-inductance of a solenoid with n turns per unit length

Question

Derive the expression for the self-inductance of a solenoid of length $l$ , cross-sectional area $A$ , with $n$ turns per unit length.

(JEE Main 2022, similar pattern)

Solution — Step by Step

For an ideal solenoid carrying current $I$ :

B = \mu_0 n I

This field is uniform inside the solenoid and zero outside (for an ideal, infinitely long solenoid).

Total number of turns: $N = nl$

Flux through one turn: $\phi = BA = \mu_0 n I A$

Total flux linkage: $N\phi = nl \cdot \mu_0 n I A = \mu_0 n^2 l A I$

L = \frac{N\phi}{I} = \frac{\text{total flux linkage}}{\text{current}}

L = \frac{\mu_0 n^2 l A I}{I}

\boxed{L = \mu_0 n^2 l A}

If written in terms of total turns $N$ (where $n = N/l$ ):

L = \frac{\mu_0 N^2 A}{l}

Why This Works

Self-inductance measures how much flux a coil generates through itself per unit current. A solenoid is efficient at this because each turn’s field threads through all the other turns, creating a large total flux linkage.

The $n^2$ dependence is key: one factor of $n$ comes from the field ( $B = \mu_0 n I$ ) and the other from the number of turns the flux links through ( $N = nl$ ). Doubling the turn density quadruples the inductance.

The formula $L = \mu_0 n^2 l A$ shows that inductance increases with length $l$ and area $A$ . A fatter, longer solenoid with more closely packed turns has higher inductance.

Alternative Method — Using energy stored

The energy stored in a solenoid is $U = \frac{B^2}{2\mu_0} \times \text{Volume} = \frac{(\mu_0 nI)^2}{2\mu_0} \cdot lA = \frac{\mu_0 n^2 I^2 lA}{2}$ .

Since $U = \frac{1}{2}LI^2$ : $\frac{1}{2}LI^2 = \frac{\mu_0 n^2 I^2 lA}{2}$ , giving $L = \mu_0 n^2 lA$ .

The two forms $L = \mu_0 n^2 lA$ and $L = \mu_0 N^2 A / l$ are both correct — just different variables. JEE questions typically give $n$ (turns per unit length), so use the first form. CBSE questions often give total turns $N$ , so use the second. Read the question carefully to avoid confusion.

Common Mistake

Students sometimes write $L = \mu_0 n l A$ (missing one factor of $n$ ). Remember: the flux linkage involves $N\phi$ , not just $\phi$ . You multiply the flux per turn ( $\mu_0 nIA$ ) by the number of turns ( $nl$ ), giving $n^2$ in the final expression. One $n$ for the field strength, one $n$ for the number of turns — both matter.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using energy stored

Common Mistake

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