What Self-Inductance Means
In mutual induction, an emf appears in one coil because the current in a neighbouring coil changes. Self-induction is the same effect turned inward: an emf is induced in a single isolated coil when the flux through it changes because the current through that very coil is varying. NCERT §6.7.2 states the phenomenon directly — emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil.
The current in the coil produces a magnetic field; that field threads the coil's own turns and produces a flux linkage. Because the field is proportional to the current, so is the flux linkage. Change the current and you change the flux, and a self-induced emf appears that, by Lenz's law, opposes the change.
The current $I$ generates a field $B$ whose flux links every turn. Varying $I$ varies this self-flux, inducing a back emf in the same coil.
The Defining Equation: NΦ = LI
For a coil of $N$ turns, the flux linkage $N\Phi_B$ is proportional to the current $I$. Writing the proportionality constant as $L$ gives the defining relation of self-inductance:
$$N\Phi_B = L I \qquad \Longrightarrow \qquad L = \frac{N\Phi_B}{I}$$
Here $L$ is the self-inductance (also called the coefficient of self-induction). It is a scalar with dimensions $[\mathrm{M\,L^2\,T^{-2}\,A^{-2}}]$. NCERT stresses a point NEET loves to test: inductance depends only on the geometry of the coil and on the permeability of the medium — never on the current itself.
| Quantity | Symbol / Relation | Note |
|---|---|---|
| Defining relation | NΦ = LI | Flux linkage proportional to current |
| Self-induced emf | ε = −L dI/dt | Opposes change in current (Lenz) |
| SI unit | henry (H) | 1 H = 1 V·s·A⁻¹ = 1 ohm-second |
| Dimensions | [M L² T⁻² A⁻²] | Scalar quantity |
| Depends on | geometry, permeability | Not on current or emf |
Back Emf and Electrical Inertia
Differentiating the defining relation and applying Faraday's law gives the self-induced emf. For fixed geometry $L$ is constant, so:
$$\varepsilon = -\frac{d(N\Phi_B)}{dt} = -L\,\frac{dI}{dt}$$
The negative sign is Lenz's law in action: the self-induced emf always opposes any change — increase or decrease — of current in the coil. For this reason it is called the back emf. NIOS §19.2.1 adds a useful consequence: since an infinite emf is impossible, the current through an inductor cannot change instantaneously.
L depends on geometry, not on current
Because the formula reads $L = N\Phi/I$, students sometimes conclude that a larger current means a larger inductance, or that switching off the current makes $L$ zero. Both are wrong. The flux $\Phi$ scales with $I$, so the ratio is fixed by the coil's turns, area, length and core material. Doubling the current doubles the flux and the energy, but leaves $L$ unchanged.
Remember: $L$ is set by geometry and permeability. The same coil has the same $L$ whether it carries 1 A, 5 A, or no current at all.
NCERT frames self-inductance as the electromagnetic analogue of mass. Just as mass resists changes in velocity, self-inductance resists changes in current; work must be done against the back emf to establish a current, and that work is stored as magnetic energy. This "electrical inertia" picture underlies both the energy formula and the gradual current build-up in circuits.
Inductance of a Solenoid
The self-inductance can be computed exactly for simple geometries. For a long solenoid of cross-sectional area $A$, length $l$, with $n$ turns per unit length (total turns $N = nl$), the interior field is $B = \mu_0 n I$, neglecting edge effects. The flux linkage is:
$$N\Phi_B = (nl)(\mu_0 n I)(A) = \mu_0 n^2 A l\, I$$
Dividing by $I$ gives the standard solenoid result. If the core is filled with a material of relative permeability $\mu_r$:
$$L = \mu_0 n^2 A l \qquad\text{and with a core}\qquad L = \mu_r\,\mu_0 n^2 A l$$
| Increase… | Effect on L | Reason |
|---|---|---|
| Turns per unit length n | L ∝ n² | Both field and linkage scale with n |
| Cross-sectional area A | L ∝ A | Larger flux per turn |
| Length l (n fixed) | L ∝ l | More total turns |
| Insert soft-iron core | L ×= μᵣ | High relative permeability |
The $n^2$ dependence is worth internalising: doubling the winding density quadruples the inductance, while inserting a ferromagnetic core can raise $L$ by a factor of hundreds through $\mu_r$.
The same solenoid geometry governs how two coils couple. See Mutual Inductance for how flux from one coil links another.
Energy Stored in an Inductor
Establishing a current means doing work against the back emf. If resistive losses are ignored and only the inductive effect is considered, the rate of work done is $\dfrac{dW}{dt} = \varepsilon\, I = L I \dfrac{dI}{dt}$. Integrating from zero to the final current $I$:
$$W = \int_0^I L I\, dI = \frac{1}{2} L I^2 \qquad\Longrightarrow\qquad U = \frac{1}{2} L I^2$$
This stored magnetic potential energy is the exact analogue of the kinetic energy $\tfrac{1}{2}mv^2$, with $L$ standing in for mass $m$ and $I$ for velocity $v$ — reinforcing the "electrical inertia" reading.
Stored energy grows as the square of the current. Doubling $I$ quadruples $U$, because $U = \tfrac{1}{2}LI^2$.
Energy is ½LI², not LI²
The factor of one-half comes from the integration of $L I\,dI$, exactly as for $\tfrac{1}{2}CV^2$ in a capacitor or $\tfrac{1}{2}mv^2$ in mechanics. Dropping the half is the single most common slip on energy-stored problems. Also watch the units: with $L$ in henry and $I$ in ampere, $U$ comes out in joule directly.
Remember: $U = \tfrac{1}{2}LI^2$. Energy ∝ $I^2$ (not $I$), and ∝ $L$ (not $L^2$).
Growth of Current in an LR Circuit
Connect a solenoid (inductance $L$, resistance $R$) to a battery through a switch and you have an LR circuit. From the instant the switch closes, the rising current induces a back emf $\varepsilon = -L\,dI/dt$ that opposes the increase, so the current cannot jump straight to the steady value $\varepsilon_0/R$. Instead it climbs gradually toward that value.
NIOS §19.2.2 defines the inductive time constant $\tau = L/R$: the time for the current to reach about two-thirds of its steady value. A larger $L$ produces a larger back emf and a longer build-up; this is why a bulb in series with a large inductor brightens slowly, and why switching off a circuit with a big inductor can produce a spark from the sudden back emf.
The current rises gradually toward $\varepsilon_0/R$, reaching about two-thirds of it in one time constant $\tau = L/R$.
Worked Examples
A long solenoid has 1000 turns. When a current of 4 A flows, the flux linked with each turn is $4\times10^{-3}$ Wb. Find its self-inductance.
Using $L = \dfrac{N\Phi}{I} = \dfrac{1000 \times 4\times10^{-3}}{4} = \dfrac{4}{4} = 1\ \text{H}$. The self-inductance is $1$ henry. (This is NEET 2016 Q.148.)
An inductor of inductance $4\ \mu\text{H}$ carries a current of $2\ \text{A}$. Find the stored magnetic energy.
$U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}\times 4\times10^{-6}\times(2)^2 = 8\times10^{-6}\ \text{J} = 8\ \mu\text{J}$. (This matches the structure of NEET 2023 Q.23.)
Self-Inductance in one screen
- Definition: $N\Phi = LI$; $L$ is the flux linkage per unit current.
- Back emf: $\varepsilon = -L\,dI/dt$; always opposes the change in current (Lenz).
- Depends on: geometry and permeability only — never on the current.
- Solenoid: $L = \mu_0 n^2 A l$; with a core, $L = \mu_r\mu_0 n^2 A l$ (so $L \propto n^2$).
- Energy: $U = \tfrac{1}{2}LI^2$ — the magnetic analogue of $\tfrac{1}{2}mv^2$.
- Unit: henry (H) = V·s·A⁻¹ = ohm-second; current cannot change instantaneously.
- LR circuit: current builds up gradually toward $\varepsilon_0/R$ with time constant $\tau = L/R$.