Derive B = μ₀nI inside a long solenoid using Ampere's law.

Ampere's Circuital Law — Field Inside a Solenoid

Question

A long solenoid has $n$ turns per unit length and carries current $I$ . Using Ampere’s Circuital Law, derive the magnetic field $B$ inside the solenoid.

This is a classic derivation that JEE Advanced and CBSE Class 12 both love — it tests whether you understand why we choose a particular Amperian loop, not just the final formula.

Solution — Step by Step

A solenoid is a tightly wound helix of wire. For a long solenoid, the field outside is essentially zero, and the field inside runs parallel to the axis (along $\hat{z}$ ). This is the physical picture we exploit.

We pick a rectangular loop $ABCD$ straddling the solenoid wall. Side $AB$ (length $L$ ) lies inside the solenoid along the axis. Side $CD$ lies outside. The two short sides $BC$ and $DA$ are perpendicular to the axis.

Why this shape? Because $\vec{B}$ is parallel to $AB$ and perpendicular to $BC$ , $DA$ . That kills off two sides immediately.

Ampere’s law states:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

We go around $ABCD$ :

Side AB (inside, field $= B$ , length $= L$ ): contribution $= BL$
Side CD (outside, field $\approx 0$ ): contribution $= 0$
Sides BC and DA ( $\vec{B} \perp d\vec{l}$ ): contribution $= 0$

So: $\oint \vec{B} \cdot d\vec{l} = BL$

The loop encloses the portion of the solenoid of length $L$ . Number of turns in this length $= nL$ . Each turn carries current $I$ , so:

I_{\text{enc}} = nLI

Plugging into Ampere’s law:

BL = \mu_0 \cdot nLI

\boxed{B = \mu_0 n I}

The $L$ cancels — beautiful. The field is uniform inside and independent of position along the axis.

Why This Works

The power of Ampere’s law is symmetry exploitation. We don’t compute the contribution of each tiny current loop separately (that would be Biot-Savart, and it’s brutal for a solenoid). Instead, we pick a loop where the integral is either trivially zero or trivially $BL$ .

The assumption that $B_{\text{outside}} = 0$ deserves scrutiny. For a real solenoid of finite length, the outside field isn’t exactly zero — but for a long solenoid (length $\gg$ radius), it’s negligible. JEE problems always specify “long solenoid” for exactly this reason.

Also notice that $B$ depends on $n$ (turns per unit length), not the total number of turns or the total length separately. A solenoid twice as long with twice as many turns has the same $B$ — because $n$ stays constant.

Alternative Method

You can also arrive at $B = \mu_0 n I$ by integrating the Biot-Savart result for a circular loop along the solenoid axis. Treat the solenoid as an infinite stack of rings, each carrying current $I$ , with $n\,dx$ turns in a length $dx$ .

The field on the axis of a single circular loop of radius $R$ at its centre is $\frac{\mu_0 I}{2R}$ . Summing over all rings with proper geometry and integrating from $-\infty$ to $+\infty$ gives the same $B = \mu_0 n I$ .

This method works, but takes two pages of integration. That’s why Ampere’s law is the preferred route in exams — same answer, one-tenth the effort.

JEE Advanced often asks you to find $B$ at the end of a semi-infinite solenoid. The answer there is $\frac{\mu_0 n I}{2}$ — exactly half. Think about it: only half the infinite solenoid contributes.

Common Mistake

Students write $I_{\text{enc}} = I$ instead of $I_{\text{enc}} = nLI$ . They forget that multiple turns pass through the Amperian loop. Each of the $nL$ turns is a separate wire carrying current $I$ , so the total enclosed current is their sum. Ampere’s law counts total current threading the loop — not current per wire.

This single error drops the $n$ from your answer, giving $B = \mu_0 I / L$ , which has wrong dimensions of $\mu_0 \times \text{current} / \text{length}$ … actually the same dimensions, which is why students don’t catch it immediately. But plugging in numbers will give a wildly wrong magnitude.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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