Self and Mutual Inductance

Inductance is the electrical equivalent of inertia. Just as a heavy object resists changes in velocity, an inductor resists changes in current. This chapter sits inside Electromagnetic Induction in Class 12 — the physical machinery is the same Faraday-Lenz framework, but the questions test whether you can compute coefficients, predict transient behaviour, and handle coupled coils.

For JEE Main, expect 1–2 questions per year on inductors in LR/LC circuits. NEET asks 1 conceptual question. CBSE boards ask a 3-mark or 5-mark derivation almost every year.

What Inductance Means Physically

When current $I$ flows through a coil, it produces magnetic flux $\Phi$ linked with the coil itself. If $I$ changes, $\Phi$ changes, and Faraday’s law produces an induced EMF — opposing the change (Lenz). The coil resists current change. This self-induction effect is quantified by self-inductance $L$ :

\Phi = LI \quad \Rightarrow \quad \varepsilon_{\text{self}} = -L\dfrac{dI}{dt}

If two coils are placed near each other, current in coil 1 produces flux in coil 2. A changing $I_1$ induces an EMF in coil 2 — quantified by mutual inductance $M$ :

\Phi_{12} = M I_1 \quad \Rightarrow \quad \varepsilon_2 = -M\dfrac{dI_1}{dt}

SI unit of inductance: henry (H), where 1 H = 1 V·s/A = 1 Wb/A.

Key Terms & Definitions

Self-inductance ( $L$ ) — The flux linkage per unit current in the same coil. Depends only on geometry and the magnetic material inside.

Mutual inductance ( $M$ ) — The flux linkage in coil 2 per unit current in coil 1 (or vice versa — $M_{12} = M_{21}$ always).

Coefficient of coupling ( $k$ ) — How tightly two coils share flux. $M = k\sqrt{L_1 L_2}$ , with $0 \le k \le 1$ . $k = 1$ means every line of flux from coil 1 threads coil 2 (ideal transformer).

Back EMF — The EMF generated by self-induction; always opposes the cause that produced it.

Time constant ( $\tau$ ) of LR circuit — $\tau = L/R$ . Time for current to reach $\sim 63\%$ of its final value when switched on.

Core Formulas

Self-inductance of a long solenoid: $L = \mu_0 n^2 A \ell$ , where $n$ is turns per unit length, $A$ is cross-section, $\ell$ is length.

Self-inductance of a toroid (mean radius $R$ , $N$ turns): $L = \dfrac{\mu_0 N^2 A}{2\pi R}$ .

Mutual inductance of two coaxial solenoids: $M = \mu_0 n_1 n_2 A \ell$ (inner solenoid threading outer, $A$ = inner cross-section).

Energy stored in an inductor: $U = \tfrac{1}{2} L I^2$ .

Energy density in magnetic field: $u = \dfrac{B^2}{2\mu_0}$ .

LR circuit growth: $I(t) = I_0(1 - e^{-t/\tau})$ , $\tau = L/R$ .

LR circuit decay: $I(t) = I_0 e^{-t/\tau}$ .

Methods & Concepts

Deriving Solenoid Self-Inductance

Inside a long solenoid of $n$ turns per unit length carrying current $I$ :

B = \mu_0 n I

Flux through one turn: $\Phi_1 = BA = \mu_0 n I A$ . Total turns: $N = n\ell$ . Flux linkage:

N\Phi_1 = \mu_0 n^2 I A \ell

By definition $N\Phi_1 = LI$ , so

L = \mu_0 n^2 A \ell

This shows $L$ depends only on geometry and the medium (replace $\mu_0$ with $\mu_0\mu_r$ for a magnetic core). It does not depend on the current.

Inductors in Series and Parallel

If two inductors have negligible mutual coupling ( $M = 0$ ):

Series: $L_{\text{eq}} = L_1 + L_2$
Parallel: $\dfrac{1}{L_{\text{eq}}} = \dfrac{1}{L_1} + \dfrac{1}{L_2}$

If they are coupled, mutual inductance modifies these:

Series, fluxes adding: $L_{\text{eq}} = L_1 + L_2 + 2M$
Series, fluxes opposing: $L_{\text{eq}} = L_1 + L_2 - 2M$

The aiding/opposing depends on which way the windings face — this often appears in JEE Advanced.

Energy in an Inductor

Work done by the source to ramp current from 0 to $I$ against the back EMF:

W = \int_0^I L i \, di = \tfrac{1}{2} L I^2

This energy is stored in the magnetic field — and recovered when the current decays. Inductors are energy-storage elements, like capacitors.

Solved Examples

Easy (CBSE) — Solenoid inductance

Q. A solenoid of length 50 cm, cross-section 4 cm² has 500 turns. Find its self-inductance.

Solution. $n = 500/0.5 = 1000$ turns/m, $A = 4 \times 10^{-4}$ m², $\ell = 0.5$ m.

L = (4\pi \times 10^{-7})(1000)^2(4 \times 10^{-4})(0.5) \approx 2.51 \times 10^{-4} \text{ H} = 0.25 \text{ mH}

Medium (JEE Main) — Energy and time constant

Q. A 50 mH inductor is connected in series with a 5 Ω resistor and a 10 V battery. Find (a) steady-state current, (b) time constant, (c) energy stored at steady state.

Solution. $I_0 = V/R = 10/5 = 2$ A. $\tau = L/R = 0.05/5 = 0.01$ s = 10 ms. $U = \tfrac{1}{2}LI_0^2 = \tfrac{1}{2}(0.05)(4) = 0.1$ J.

Hard (JEE Advanced) — Coupled coils

Q. Two coils with $L_1 = 4$ H and $L_2 = 9$ H are coupled with $k = 0.5$ . Find $M$ and the equivalent inductance when connected in series with fluxes adding.

Solution. $M = k\sqrt{L_1 L_2} = 0.5 \times \sqrt{36} = 3$ H. $L_{\text{eq}} = L_1 + L_2 + 2M = 4 + 9 + 6 = 19$ H.

If the connection were flux-opposing: $L_{\text{eq}} = 4 + 9 - 6 = 7$ H. Same coils, very different equivalent values — depends entirely on winding direction.

Exam-Specific Tips

CBSE 5-mark: Derive $L = \mu_0 n^2 A\ell$ . Memorise the steps and the diagram. This appears at least every 3rd year.

JEE Main: Watch for the energy density $B^2/(2\mu_0)$ — verify by integrating over the volume of a solenoid: $\int u \, dV = (\mu_0 n^2 I^2 / 2) \cdot A\ell = \tfrac{1}{2}LI^2$ . Self-consistent.

NEET: Conceptual question on Lenz’s law — induced EMF opposes change. If current is decreasing, induced EMF tries to maintain it (same direction).

JEE Advanced: Coupled-coil problems with $k < 1$ and unequal inductances. Watch for the $k = 1$ ideal-transformer limiting case.

Common Mistakes to Avoid

1. Sign of back EMF. Use $\varepsilon = -L\, dI/dt$ when applying Kirchhoff’s voltage law. The minus is what makes Lenz’s law work.

2. Forgetting $n^2$ . $L \propto n^2$ , not $n$ . Doubling turns per unit length quadruples $L$ .

3. Using current in the formula for $L$ . $L$ depends only on geometry — it’s a constant for a given coil. Numerical $L$ values stay fixed across changing $I$ .

4. Wrong sign for series-coupled coils. Need to inspect winding directions or read the problem carefully.

5. Confusing $M$ with $L$ . Mutual inductance is between two distinct coils; self-inductance is one coil’s flux linkage with itself.

Practice Questions

Q1. Find $L$ of a 2000-turn toroid, mean radius 10 cm, cross-section 5 cm².

$L = \mu_0 N^2 A/(2\pi R) = (4\pi \times 10^{-7})(4 \times 10^6)(5 \times 10^{-4})/(0.628) \approx 4 \times 10^{-3}$ H = 4 mH.

Q2. Current 5 A in a 0.2 H inductor — find stored energy.

$U = \tfrac{1}{2}(0.2)(25) = 2.5$ J.

Q3. $L_1 = 2$ H, $L_2 = 8$ H, $M = 3$ H. Find $k$ .

$k = M/\sqrt{L_1 L_2} = 3/4 = 0.75$ .

Q4. What’s the EMF induced if current changes from 0 to 4 A in 0.02 s through a 0.5 H coil?

$|\varepsilon| = L \cdot dI/dt = 0.5 \times 200 = 100$ V.

Q5. In an LR circuit, after how many time constants does current reach 99% of $I_0$ ?

$0.99 = 1 - e^{-t/\tau} \Rightarrow t = \tau \ln 100 \approx 4.6 \tau$ .

Q6. Why does opening a switch in a circuit with a large inductor produce a spark?

Sudden interruption forces $dI/dt \to \infty$ , generating a huge back EMF (sometimes thousands of volts) that breaks down the air gap and causes a spark. This is why industrial relays have flyback diodes.

Q7. Two solenoids share a common axis; inner has 1000 turns/m, outer has 500 turns/m, common cross-section $5$ cm², common length 20 cm. Find $M$ .

$M = \mu_0 n_1 n_2 A \ell = (4\pi \times 10^{-7})(1000)(500)(5 \times 10^{-4})(0.2) \approx 6.28 \times 10^{-5}$ H.

Q8. A 5 H inductor stores 100 J. Find the current.

$I = \sqrt{2U/L} = \sqrt{40} \approx 6.32$ A.

FAQs

Is mutual inductance always equal in both directions?

Yes — $M_{12} = M_{21}$ , always. This is a deep theorem from electromagnetic reciprocity. So you only need one number $M$ for any pair of coils.

Why does an iron core increase inductance?

Iron has $\mu_r \gg 1$ , multiplying the magnetic field for the same current. Since $L \propto \mu$ , inductance scales accordingly. A typical iron core can boost $L$ by a factor of 1000 or more.

Does inductance depend on current?

In free space or with linear magnetic materials, no — $L$ is purely geometric. With ferromagnetic cores, $L$ can vary with current because the core saturates at high field.

What’s the difference between an inductor and a transformer?

A transformer is two (or more) inductors with strong mutual coupling, designed for AC voltage transformation. A single inductor stores energy and resists current changes. Both rely on the same underlying physics.

Why is the time constant $\tau = L/R$ ?

From the differential equation $L\, dI/dt + IR = V$ , the homogeneous solution is $e^{-Rt/L}$ . The exponential decay timescale is $L/R$ — that’s where the formula comes from.