JEE Mains Physics · Electromagnetic Induction

Self and Mutual Inductance, Energy and LR Circuits

A coil opposes a change in its own current with an emf −L dI/dt and induces −M dI/dt in a neighbour; it stores energy ½LI², and in an LR circuit the current grows or decays with time constant L/R.

All Electromagnetic Induction notes JEE Mains Physics strategy

Why this matters

Twenty-six PYQs, fourteen of them multiple choice, and four from 2026. Seven are self-inductance: L from a back emf, from a solenoid's shape, or from what L depends on; nine are mutual inductance, seven of them a small loop at the centre of a large one; ten are about the energy an inductor stores and how the current grows in an LR circuit.

Concept 1 of 3: Self-inductance and the back emf

A coil's own current makes a flux through it, LI. If the current changes, that flux changes, and the coil induces an emf that fights the change: it pushes back when the current rises and keeps it going when it falls. L plays the part of mass for current, an inertia that resists any change. It depends only on the coil's shape and the material inside it.

Definition

$N\Phi = LI$ and $\varepsilon = -L\,\dfrac{dI}{dt}$ . For a size, $|\varepsilon| = L\,|\Delta I|/\Delta t$ .
$\Delta I$ across zero: from $-2\ \text{A}$ to $+3\ \text{A}$ is a change of 5 A, not 1 A.
Switching off a steady current $I_0 = E/R$ in time $\Delta t$ : average emf $L I_0/\Delta t$ , often far above the battery's emf.
Long solenoid of length l, N turns, $n = N/l$ , area A: $L = \mu_0 n^{2}Al = \mu_0 N^{2}A/l$ . Filled with a material: $\mu_0 \to \mu_r\mu_0$ .
L depends on geometry and on the permeability of the core, not on the current. Doubling N at fixed length makes L four times larger.
Work is done against the back emf while a current is built up; that work is the stored energy.

Self-inductance

\varepsilon = -L\frac{dI}{dt} \qquad L = \mu_0 n^{2}Al

Worked example

A solenoid

50\ \text{cm}

long has 2000 turns and a cross-section of

4\ \text{cm}^{2}

. Find its inductance, and the emf when its current rises steadily from

1\ \text{A}

3\ \text{A}

0.1\ \text{s}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q11Moderate

Example 1 · Electromagnetic Induction · Self and Mutual Inductance, Energy and LR Circuits

12\text{ }V

battery connected to a coil of resistance 6

\Omega

through a switch, drives a constant current in the circuit. The switch is opened in

1\text{ }ms

. The emf induced across the coil is

20\text{ }V

. The inductance of the coil is:

Change of current across zero

A current that goes from −2 A to +2 A changes by 4 A. Taking the difference of the sizes, zero, or forgetting the sign gives the wrong rate.

Using the battery emf as the induced emf

When a switch opens, the back emf is L × (steady current)/(switching time). It is usually many times the battery's emf; the battery only fixes the current, E/R.

Thinking L depends on the current

L = NΦ/I is a ratio fixed by the coil's turns, size and core. A larger current gives a larger flux but the same L.

Drill 5 more on self-inductance and the back emf

Concept 2 of 3: Mutual inductance of two coils

A current in one coil sends flux through a second, and the flux is proportional to the current: Φ₂ = MI₁. Change I₁ and the second coil gets an emf −M dI₁/dt. M is the same whichever coil carries the current, so always work out the flux through the coil where the field is easiest to know. For a small loop at the centre of a big one, the big loop's field is nearly uniform over the small loop.

Definition

$N_2\Phi_2 = MI_1$ ; $\varepsilon_2 = -M\,\dfrac{dI_1}{dt}$ . With currents in both: $\varepsilon_1 = -L_1\dfrac{dI_1}{dt} - M\dfrac{dI_2}{dt}$ .
Small loop of area a at the centre of a large coplanar loop: $M = \dfrac{B_{\text{centre}}}{I} \times a$ .
Field at the centre of a circle of radius b: $\dfrac{\mu_0 I}{2b}$ . At the centre of a square of side s: four sides, each $\dfrac{\mu_0 I}{4\pi(s/2)}\cdot 2\sin 45^{\circ}$ , total $\dfrac{2\sqrt2\,\mu_0 I}{\pi s}$ .
Coil of $N_2$ turns wound on a long solenoid (n turns per metre, area A): $M = \mu_0 n N_2 A$ .
Coils in series: $L = L_1 + L_2 + 2M$ when their fluxes aid, $L_1 + L_2 - 2M$ when they oppose. Always $M \le \sqrt{L_1L_2}$ .

Mutual inductance

\varepsilon_2 = -M\frac{dI_1}{dt} \qquad M = \frac{B_{\text{centre}}}{I}\,a_{\text{small}}

Worked example

A short coil of 50 turns is wound over the middle of a long solenoid with 2000 turns per metre and a cross-section of

6\ \text{cm}^{2}

. Find their mutual inductance, and the emf in the coil when the solenoid's current changes at

5\ \text{A/s}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q18Moderate

Example 2 · Electromagnetic Induction · Self and Mutual Inductance, Energy and LR Circuits

A circular current loop of radius

R

is placed inside square loop of side length

L\ (L \gg R)

such that they are co-planar, and their centers coincide. The permeability of free space is

\mu_{0}

. The mutual inductance between circular loop and square loop is

\_\_\_\_

Working out the flux through the wrong loop

M is the same both ways, but only the big loop's field is known simply over the other loop. Pass the current through the big loop and find the flux through the small one.

Field at a square's centre

Each side is a finite wire at distance s/2, giving μ₀I(2 sin 45°)/(4π · s/2). Four sides make 2√2μ₀I/(πs). Using the formula for an infinite wire overcounts.

Sign of 2M in series

Coils wound the same way, carrying current the same way round, add 2M. Coils wound in opposite senses subtract 2M. The winding, often shown only in the figure, decides.

Drill 9 more on mutual inductance of two coils

Concept 3 of 3: Energy stored in an inductor and current growth in an LR circuit

Building up a current against the back emf takes work, and that work is stored in the magnetic field: ½LI². It depends only on the final current, not on how it got there. In an LR circuit the inductor stops the current jumping: it grows towards E/R and gets about 63% of the way in one time constant, L/R. Kirchhoff's loop law still holds, with L dI/dt as one more voltage drop.

Definition

$U = \tfrac12 LI^{2}$ ; energy per unit volume $u = \dfrac{B^{2}}{2\mu}$ , with $\mu = \mu_r\mu_0$ in a filled solenoid.
Growth: $I = I_0(1 - e^{-t/\tau})$ , $I_0 = E/R$ , $\tau = L/R$ . Decay after the battery is removed: $I = I_0e^{-t/\tau}$ .
Energy goes as $I^{2}$ : a fraction f of the final energy needs $I = \sqrt f\,I_0$ .
Loop law: $E - L\dfrac{dI}{dt} - IR = 0$ . If the current is falling, dI/dt is negative and the inductor adds to the battery.
Rate of storing energy $\dfrac{dU}{dt} = LI\dfrac{dI}{dt} = I(E - IR)$ .
Equal inductors in parallel share a current equally; find each one's current before adding energies.

Inductor energy and LR growth

U = \frac12 LI^{2} \qquad I = \frac{E}{R}\left(1 - e^{-tR/L}\right) \qquad \tau = \frac{L}{R}

Worked example

A coil with

L = 0.5\ \text{H}

and

R = 5\ \Omega

is connected to a

20\ \text{V}

battery. Find the time constant, the final current and the final stored energy. How long does the current take to reach 75% of its final value?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q7Moderate

Example 3 · Electromagnetic Induction · Self and Mutual Inductance, Energy and LR Circuits

An inductor coil stores

64\text{ }J

of magnetic field energy and dissipates energy at the rate of

640\text{ }W

when a current of

8\text{ }A

is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds:

Time constant upside down

The time constant is L/R. A bigger inductance makes the current slower to grow; a bigger resistance makes it settle faster, to a smaller value.

Energy fraction taken as current fraction

Stored energy goes as I². Half the final energy needs I = I₀/√2, and half the final current stores only a quarter of the final energy.

Sign of dI/dt for a falling current

When the current is decreasing, L dI/dt is negative in E − L dI/dt − IR = 0, so the inductor's emf adds to the battery's and the current can exceed E/R for a moment.

Using μ₀ in a filled solenoid

With a core of relative permeability μr, the energy density is B²/(2μrμ₀). For the same B, a filled solenoid stores less energy per unit volume than an air-cored one.

Drill 9 more on energy stored in an inductor and current growth in an lr circuit

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Self-inductance and the back emf
Self-inductance
$\varepsilon = -L\frac{dI}{dt} \qquad L = \mu_0 n^{2}Al$
Mutual inductance of two coils
Mutual inductance
$\varepsilon_2 = -M\frac{dI_1}{dt} \qquad M = \frac{B_{\text{centre}}}{I}\,a_{\text{small}}$
Energy stored in an inductor and current growth in an LR circuit
Inductor energy and LR growth
$U = \frac12 LI^{2} \qquad I = \frac{E}{R}\left(1 - e^{-tR/L}\right) \qquad \tau = \frac{L}{R}$

Watch out for (10)

Change of current across zero
→ Self-inductance and the back emf
Using the battery emf as the induced emf
→ Self-inductance and the back emf
Thinking L depends on the current
→ Self-inductance and the back emf
Working out the flux through the wrong loop
→ Mutual inductance of two coils
Field at a square's centre
→ Mutual inductance of two coils
Sign of 2M in series
→ Mutual inductance of two coils
Time constant upside down
→ Energy stored in an inductor and current growth in an LR circuit
Energy fraction taken as current fraction
→ Energy stored in an inductor and current growth in an LR circuit
Sign of dI/dt for a falling current
→ Energy stored in an inductor and current growth in an LR circuit
Using μ₀ in a filled solenoid
→ Energy stored in an inductor and current growth in an LR circuit

Test yourself on Electromagnetic Induction

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