MHT-CET Physics · Electromagnetic Induction

Self-Inductance, the Energy an Inductor Stores, and Inductors Together

A coil's own current links flux through it, Nφ = LI; the constant L is its self-inductance, it opposes any change in that current with an e.m.f. L dI/dt, it is set by the coil's geometry (for a solenoid μ₀N²A/l), and it stores energy ½LI² in its magnetic field.

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Why this matters

43 PYQs, 4 HARD — the largest page in the chapter. Fourteen are the definition — L from flux and current, the slope of a φ–I graph (asked six times), the unit, e = L dI/dt, and a voltage across an inductor in a circuit; seventeen are how a solenoid's L depends on turns, length, area and core, including the length of wire needed to make one; twelve are energy ½LI² and inductors in series and parallel. Three cards.

Concept 1 of 3: What Self-Inductance Is

A current through a coil links flux Nφ with it, and the flux is proportional to the current: Nφ = LI. So on a graph of φ against I each coil is a straight line whose SLOPE is its inductance — steepest line, largest L. Change the current and the linked flux changes, inducing e = −L dI/dt, which fights the change; a plot of e against dI/dt is therefore a straight line through the origin with slope −L. Its unit is V·s/A, the henry. Because the stored energy is ½LI², L is numerically twice the work done in setting up the flux for a unit current.

Definition

$N\phi = LI$ ⇒ $L = \dfrac{N\phi}{I}$ ; on a $\phi$ – $I$ graph L is the slope.
$e = -L\dfrac{dI}{dt}$ : current reversed from 5 A to −5 A in 0.5 s with e = 2 V ⇒ L = 0.1 H.
Unit: $\dfrac{\text{V} \cdot \text{s}}{\text{A}}$ = henry.
L is twice the work done establishing the flux for unit current (from $W = \tfrac{1}{2}LI^2$ ).
In a circuit, going along the current: a resistor drops IR, an inductor drops $L\dfrac{dI}{dt}$ (negative when the current is falling), a cell drops or gains its e.m.f.

Self-inductance

N\phi = LI, \qquad e = -L\frac{dI}{dt}

Worked example

A 200-turn coil carries 2 A and each turn links 5 × 10⁻⁴ Wb. Its self-inductance, and the e.m.f. if the current falls to zero in 0.01 s?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q14Easy

Example 1 · Electromagnetic Induction · Self-Inductance, Energy Stored, and LR Circuit

When a current of 1 A is passed through a coil of 100 turns, the flux associated with it is

2.5 \times 10^{-5}

Wb/turn. The self inductance of the coil in millihenry is

Forgetting the N in Nφ = LI

The flux 'per turn' must be multiplied by the number of turns. 1500 × 2.8 × 10⁻²/35 is 1.2 H; leaving out N gives 0.8 mH.

Drill 13 more on what self-inductance is

Concept 2 of 3: Self-Inductance of a Solenoid

Inside a long solenoid B = μ₀NI/l, so each turn links μ₀NIA/l and all N turns link μ₀N²IA/l: L = μ₀μᵣN²A/l. It depends on the coil, never on the current. Written with turns per unit length n = N/l it is μ₀n²Al, the form to use when n is held fixed. So L goes as N², as A (r²), as 1/l for fixed N, and up by the core's μᵣ. Scale every linear dimension by k at fixed n and L grows k³.

Definition

$L = \dfrac{\mu_0\mu_r N^2 A}{l} = \mu_0\mu_r n^2 A l$ ; per unit length $\mu_0 n^2 \dfrac{\pi d^2}{4} = \mu_0\pi\left(\dfrac{nd}{2}\right)^2$ .
Turns: N × 2 ⇒ L × 4; N × 3 ⇒ L × 9; 600 → 500 turns ⇒ L × 25/36.
Size at fixed n: all lengths × 2 ⇒ L × 8; × 3 ⇒ × 27. Equal N, lengths and radii both 1 : 3 ⇒ L 1 : 3.
Core: μᵣ = 1000 with turns cut to a tenth ⇒ L × 10.
L rises if l decreases or A increases; the current does not change it.
Wire needed for a solenoid of length l and inductance L: $\sqrt{\dfrac{4\pi lL}{\mu_0}}$ (1 m, 1 mH ⇒ 100 m).

Solenoid

L = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_0\mu_r n^2 A l

Worked example

A coil of 0.2 H has its turns halved and an iron core of μᵣ = 400 inserted. New L?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q8Moderate

Example 2 · Electromagnetic Induction · Self-Inductance, Energy Stored, and LR Circuit

An air-cored coil has self-inductance of 0.1 H. A soft iron core of relative permeability 1000 is introduced and the number of turns is reduced to

\frac{1}{10}

th. The value of self-inductance is now

Holding N fixed when the question holds n fixed

'Turns per unit length stays the same' means n is fixed, so L = μ₀n²Al grows with BOTH A and l. With N fixed, L = μ₀N²A/l falls as l grows.

Drill 16 more on self-inductance of a solenoid

Concept 3 of 3: Energy in an Inductor, and Inductors in Series and Parallel

Building up current I in an inductor stores energy ½LI² in its field; halve the current and the energy falls to a quarter. Inductors combine like resistors: in series they add, in parallel their reciprocals add. So a coil cut in two and the halves put in parallel gives L/4. When a circuit carrying current through an inductor is broken, that stored energy must go somewhere — a capacitor across the switch soaks it up if ½CV² ≥ ½LI².

Definition

$U = \tfrac{1}{2}LI^2$ ; also $U = \tfrac{1}{2}N\phi I$ when given flux and turns.
Series $L = L_1 + L_2 + \dots$ ; parallel $\dfrac{1}{L} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \dots$ .
A coil cut into halves in parallel: $\dfrac{L}{4}$ . Three equal in series: $U = \tfrac{3}{2}LI^2$ .
Same e.m.f. and resistance ⇒ same current, so energies go as L (10 H : 10 mH = 1000 : 1).
Spark suppression: $C \ge \dfrac{LI^2}{V^2}$ (1 H, 1 A, 500 V ⇒ 4 μF).

Stored energy

U = \tfrac{1}{2}LI^2

Worked example

Two 40 mH inductors in parallel carry 3 A in total. Energy stored?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q22Easy

Example 3 · Electromagnetic Induction · Self-Inductance, Energy Stored, and LR Circuit

If current of 4 A produces magnetic flux of

3 \times 10^{-3}

Wb through a coil of 400 turns, the energy stored in the coil will be

Halving the energy when the current halves

Energy goes as I². Half the current stores a quarter of the energy, not half.

Drill 11 more on energy in an inductor, and inductors in series and parallel

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)

What Self-Inductance Is
Self-inductance
$N\phi = LI, \qquad e = -L\frac{dI}{dt}$
Self-Inductance of a Solenoid
Solenoid
$L = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_0\mu_r n^2 A l$
Energy in an Inductor, and Inductors in Series and Parallel
Stored energy
$U = \tfrac{1}{2}LI^2$

Watch out for (3)

Forgetting the N in Nφ = LI
→ What Self-Inductance Is
Holding N fixed when the question holds n fixed
→ Self-Inductance of a Solenoid
Halving the energy when the current halves
→ Energy in an Inductor, and Inductors in Series and Parallel

Test yourself on Electromagnetic Induction

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

AC Circuits — inductive reactance →