Q.Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.

Self-Inductance of a Solenoid: From Intuition to Formula

Imagine you push a heavy door. It doesn't resist your push once it's moving — but it does resist you trying to change its speed suddenly. That resistance to change is inertia. A solenoid carrying current behaves the same way: it "wants" to keep its current steady, and fights any attempt to change it.

This property is called self-inductance. The solenoid generates a back emf that opposes the change in its own current — not the current itself, but the change in current. That's the core idea.

Why does a solenoid oppose current changes?

A solenoid is a long coil of wire. When current flows through it, it produces a magnetic field inside. If you try to increase the current, the magnetic field strengthens. But a changing magnetic field induces an emf in the coil itself (Faraday's law). By Lenz's law, this induced emf opposes the change that caused it — so it pushes back against the rising current.

If you try to decrease the current, the field weakens, and the induced emf tries to keep the current flowing. The solenoid acts like an electrical "flywheel."

The precise statement

Self-inductance $L$ is defined by the relation:

where $\mathcal{E}$ is the induced back emf, and $\frac{dI}{dt}$ is the rate of change of current. The negative sign tells you the emf opposes the change.

For a solenoid, $L$ depends only on its geometry and the core material — not on the current. The formula is:

Let's unpack each symbol:

• $\mu_0$ — permeability of free space ($4\pi \times 10^{-7}$ H/m). It's a universal constant that tells you how strongly a vacuum responds to magnetic fields.
• $n$ — number of turns per unit length (turns/m). More turns per metre means a stronger field per ampere, so more inductance.
• $A$ — cross-sectional area of the solenoid (m²). A wider coil encloses more magnetic flux.
• $l$ — length of the solenoid (m). Longer solenoid means more total turns, hence more inductance.

Where does $L = \mu_0 n^2 A l$ come from?

Start with the magnetic field inside a long solenoid:

The magnetic flux through one turn is $BA = \mu_0 n I A$. For all $N = nl$ turns, the total flux linkage is:

By definition, self-inductance is the constant of proportionality between flux linkage and current:

Comparing, you get:

What does a larger $L$ mean?

A solenoid with high $L$ strongly resists changes in current. If you try to switch the current on quickly, the back emf is large, so the current rises slowly. If you short-circuit the solenoid, the current doesn't drop instantly — it decays gradually.

This is why inductors are used in filters, chokes, and timing circuits. They smooth out current variations. …