Concept understanding — Self-Inductance of a Solenoid
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:
E=−LdtdI
where E is the induced back emf, and dtdI 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:
L=μ0n2Al
L=μ0n2Al
Let's unpack each symbol:
μ0 — permeability of free space (4π×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=μ0n2Al come from?
Start with the magnetic field inside a long solenoid:
B=μ0nI
The magnetic flux through one turn is BA=μ0nIA. For all N=nl turns, the total flux linkage is:
Φtotal=N⋅BA=(nl)(μ0nIA)=μ0n2AlI
By definition, self-inductance is the constant of proportionality between flux linkage and current:
Φtotal=LI
Comparing, you get:
L=μ0n2Al
Note
This formula assumes an ideal solenoid — infinitely long, with a uniform field inside and zero field outside. Real solenoids are close approximations if l≫A.
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. …
For a long solenoid of N turns, length l and cross-sectional area A carrying current I, the field inside is B=μ0nI (where n=N/l), so the flux linked per turn is Φ1=BA, and the total flux linkage is NΦ1=μ0n2AlI. Since self-inductance is defined …
By computing the magnetic flux linked with all the turns of a current-carrying solenoid and comparing it to the defining relation NΦ=LI, the self-inductance of a long solenoid is found to be L=μ0n2Al.
Setup
Consider a long solenoid of length l, cross-sectional area A, having a total of N turns of wire closely wound, so that the number of turns per unit length is n=N/l. Let a current I flow through the solenoid.
Magnetic field inside the solenoid
For a long (ideal) solenoid, the magnetic field is uniform inside and (approximately) zero outside, with magnitude
B=μ0nI
where μ0 is the permeability of free space.
Flux linkage
The magnetic flux passing through each single turn of the solenoid (of area A) is
Φ1=BA=μ0nIA
Since the solenoid has N=nl turns, and (to a good approximation for a long solenoid) each turn links essentially the same flux Φ1, the total flux linkage (the sum of flux through every turn) is
NΦ1=(nl)(μ0nIA)=μ0n2AlI
Self-inductance
The self-inductance L of a coil is defined through the relation between the total flux linkage and the current producing it:
NΦ1=LI
Comparing this with the expression obtained above,
LI=μ0n2AlI
L=μ0n2Al
Since n=N/l, this can equivalently be written as …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
Showing the 12 most recent of 35 on this concept.
CBSE 2026Set ANNUAL1 markMCQ
Q.When the current changes from +2 A to −2 A in 0.05 second in a coil, an e.m.f. of 8 V is induced in it. The coefficient of self-induction of the coil is
(a) 0.1 henry
(b) 0.2 henry
(c) 0.4 henry
(d) 0.8 henry
›Reveal solutionSolution
Using e=LΔI/Δt with ΔI=4 A and Δt=0.05 s gives L=0.1 H.
The current changes from +2 A to −2 A, so the magnitude of the change is
Self-inductance L makes a circuit resist a change in the current flowing through it (via a back-emf = -L dI/dt), analogous to inertia resisting a change in velocity.
Whenever the current through a coil tends to change, the induced back-emf (from self-induction) opposes that change (Lenz's law). This behaviour - opposing change rather than opposing the current itself - is directly analogous to mechanical inertia, which oppos …
Self-inductance is defined via the induced emf a coil produces in itself when its own current changes - it behaves like the 'electrical inertia' of the circuit, but it is quantified through that induced emf.
When the current I through a coil changes, the magnetic flux linked with the coil changes too, and by Faraday's law this changing self-flux induces an emf in the SAME coil that opposes the change in current (Lenz's law). This self-induced emf defines the self-inductance L of the coil:
Inductance L relates induced emf to the rate of change of current, so its unit works out to volt-second per ampere - named the henry.
From emf = -L*(dI/dt), we get L = emf / (dI/dt), so the unit of L is volt / (ampere/second) = volt.second/ampere. This combination is given the special SI name henry (H), after Joseph Henry. Weber is the u …
Q.Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average e.m.f. of 100 V induced, give an estimate of the self-inductance of the circuit.
(a) L = 4 H
(b) L = 20 H
(c) L = 40 H
(d) L = 2 H
›Reveal solutionSolution
Using ∣ε∣=LdtdI, the self-inductance works out to 2 H.
Given: Current changes from Ii=5.0 A to If=0.0 A in Δt=0.1 s, with average induced emf ∣ε∣=100 V.
Q.When a ________ rod is inserted into a coil, its self-inductance increases.
›Reveal solutionSolution
ferromagnetic (soft iron) Self-inductance L=μrμ0n2Al. Inserting a ferromagnetic (soft iron) core has a large relative permeability μr≫1, which greatly increases the mag …
Q.The self-inductance of a solenoid depends on
(A) The current flowing through its medium
(B) The number of turns per unit length
(C) The length of the solenoid
(D) Both (B) and (C)
›Reveal solutionSolution
Self-inductance depends on the solenoid's geometry (turns per unit length and length/area), not on the current.
For a solenoid the self-inductance is
L=μ0n2Al
where n is the number of turns per unit length, A the cross-sectional area and l the length. So it depends on the number of turns per unit length and the length (and area) — both geometric factors.
Q.The self-inductance of a coil is 5 henry. A current of 1 ampere changes to 2 amperes within 5 seconds through the coil. The value of the induced e.m.f. is
Self-inductance L is defined by the induced EMF equation EMF = −L(dI/dt); its SI unit is the henry (H).
When the current through a coil changes, the changing magnetic flux linked with the coil itself induces an EMF in it (self-induction). This is written as:
EMF = −L (dI/dt)
where L is the coefficient of self-inductance of the coil. Rearranging, L = −EMF/(dI/dt), so the unit of L is volt/(ampere/second) = volt·second/ampere. This combination is given the special SI name henry (H), where 1 H = 1 V·s/A.