PhysicsCBSE Class 12 · Electromagnetic Induction

Self and Mutual Inductance Class 12: Solenoids, Back EMF and Energy

Updated 15 October 2026 · Fact-checked

Mutual inductance M links two coils: a changing current in one induces ε₂ = −M dI₁/dt in the other. Self-inductance L links a coil with itself: ε = −L dI/dt opposes changes in its own current. For a long solenoid L = μ₀n²Al, for coaxial solenoids M = μ₀n₁n₂πr₁²l, and an inductor stores energy ½LI².

Understand Self and Mutual Inductance

Switch off a large electromagnet and a spark may jump across the switch. The coil fights the sudden fall in its own current. This is self-induction, and its cousin between two coils is mutual induction.

Mutual inductance. When current $I_{1}$ flows in coil 1, part of its magnetic flux passes through coil 2. The flux linkage of coil 2 is proportional to $I_{1}$ : $N_{2} Φ_{2} = M I_{1}$ . The constant M is the mutual inductance of the pair. If $I_{1}$ changes, Faraday's law gives an emf in coil 2: $ε_{2} = - M \frac{d I _{1}}{d t}$ . M depends on the number of turns, the sizes, the separation and relative orientation of the coils, and the material (core) between them; it does not depend on the current. Its SI unit is the henry: 1 H = 1 Wb/A = 1 V s/A. The value is the same whichever coil carries the current: $M_{12} = M_{21}$ .

Two long coaxial solenoids. Let the inner solenoid $S_{1}$ have radius $r_{1}$ and $n_{1}$ turns per metre, and the outer $S_{2}$ have $n_{2}$ turns per metre; both have length l. A current $I_{2}$ in the outer solenoid gives a uniform field $B_{2} = μ_{0} n_{2} I_{2}$ inside it. The flux through each turn of the inner one is $B_{2} π r_{1}^{2}$ , and it has $n_{1} l$ turns, so $N_{1} Φ_{1} = μ_{0} n_{1} n_{2} π r_{1}^{2} l I_{2}$ . Hence $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ . With a core of relative permeability $μ_{r}$ , $M = μ_{r} μ_{0} n_{1} n_{2} π r_{1}^{2} l$ .

Self-inductance. A coil's own current produces flux through its own turns: $N Φ = L I$ . If I changes, an emf is induced in the same coil: $ε = - L \frac{d I}{d t}$ . This back emf opposes the change of current (Lenz's law): it slows a rise and delays a fall. So an inductor is the electrical analogue of inertia (mass) in mechanics: it opposes changes in current, not current itself.

Long solenoid. With n turns per metre, length l and cross-section A, $B = μ_{0} n I$ , flux per turn $μ_{0} n I A$ , and nl turns: $N Φ = μ_{0} n^{2} A l I$ . So $L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$ , and $μ_{r}$ times this with a core. Doubling the turns (same length) makes L four times; an iron core multiplies it by $μ_{r}$ .

Energy stored in an inductor. To build up a current against the back emf, a source must do work: power $P = ∣ ε ∣ I = L I \frac{d I}{d t}$ . Integrating from 0 to I: $W = \int_{0}^{I} L I d I = \frac{1}{2} L I^{2}$ . This energy is stored in the magnetic field, like kinetic energy $\frac{1}{2} m v^{2}$ is stored in motion. For a solenoid, dividing by the volume Al gives the magnetic energy density $u_{B} = \frac{B ^{2}}{2 μ _{0}}$ (in vacuum), the magnetic match of $\frac{1}{2} ε_{0} E^{2}$ .

Why it matters. Mutual induction is the principle of the transformer and of induction coils. Self-induction explains the spark when an inductive circuit is broken, and inductors are used to smooth or control changing currents.

Self and Mutual Inductance: formulas and key facts

Mutual inductance: $N_{2} Φ_{2} = M I_{1}, ε_{2} = - M \frac{d I _{1}}{d t}$; M₁₂ = M₂₁; unit henry (H).
Coaxial solenoids: $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$; r₁ = radius of the inner solenoid; × μ_r with a core.
Self-inductance: $N Φ = L I, ε = - L \frac{d I}{d t}$; Back emf opposes the change in current.
Long solenoid: $L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$; × μ_r with a core.
Energy stored in an inductor: $U = \frac{1}{2} L I^{2}$; Stored in the magnetic field.
Magnetic energy density: $u_{B} = \frac{B ^{2}}{2 μ _{0}}$; In vacuum; use μ in place of μ₀ in a medium.

How to solve Self and Mutual Inductance questions

Use this method for inductance, induced emf and energy questions.

1Identify whether the emf is in the same coil (self, L) or in a neighbouring coil (mutual, M).
2For emf, find the rate of change of current: $\frac{Δ I}{Δ t}$ for a steady change, or differentiate I(t).
3Use $∣ ε ∣ = L \frac{d I}{d t}$ or $∣ ε ∣ = M \frac{d I _{1}}{d t}$ ; for flux linkage use $N Φ = L I$ or $M I_{1}$ .
4For a solenoid, write n = N/l (turns per metre) and A in m²; include μ_r if there is a core.
5For coaxial solenoids, use the cross-section of the inner one.
6For energy, $U = \frac{1}{2} L I^{2}$ ; for energy per volume, $\frac{B ^{2}}{2 μ _{0}}$ (or $\frac{B ^{2}}{2 μ}$ in a core).
7Check units: H for L and M, V for emf, J for energy.

Self and Mutual Inductance in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mutual induction in an ignition coil

Suppose a car's ignition coil has a mutual inductance of 2 H between its primary and secondary windings. A current of 4 A in the primary is switched off and falls to zero in about 1 ms.

(a) Estimate the average emf induced in the secondary. (b) Why does the current need to stop so quickly? (c) What is the source of the energy delivered to the spark?

Try it first, then show the solutionSolution

$ε = M \frac{Δ I}{Δ t} = 2 \times \frac{4}{1 0 ^{- 3}} = 8000$ V.
The emf is proportional to the rate of change of current; a fast cut-off gives a large rate and a high voltage, enough to make a spark across the plug gap.
The energy was stored in the magnetic field of the coil while the primary current flowed; it is released when the field collapses.

Answer: (a) about 8000 V; (b) a faster change gives a larger emf; (c) the magnetic energy stored in the coil.

Memory hook: Fast change in one coil, big kick in the other.

Self-induction and stored energy

Imagine a large lab electromagnet with self-inductance 5 H carrying 2 A. Someone opens the switch and the current falls to zero in about 10 ms.

(a) Find the energy stored in the magnet before switching off. (b) Estimate the back emf. (c) Why does a spark appear at the switch, and how can it be avoided?

Try it first, then show the solutionSolution

$U = \frac{1}{2} L I^{2} = \frac{1}{2} \times 5 \times 4 = 10$ J.
$ε = L \frac{Δ I}{Δ t} = 5 \times \frac{2}{0.01} = 1000$ V.
This large back emf appears across the opening switch gap, and the stored 10 J has to go somewhere, so the air breaks down and a spark jumps. Reducing the current slowly (for example through a resistor) lowers the rate of change and the back emf.

Answer: (a) 10 J; (b) about 1000 V; (c) the large back emf across the gap makes a spark; switch the current down gradually.

Memory hook: An inductor is a flywheel for current: stop it suddenly and it kicks.

Tricks to remember Self and Mutual Inductance

Short memory aids. Say each one aloud once, then use it on a question the same day.

L is electrical inertia
Mass resists changes in velocity; L resists changes in current. ½mv² ↔ ½LI².
Use it when: Understanding back emf and stored energy.
Mu-n-squared-A-l
L = μ₀ n² A l: 'Mu Never Squares Any Length' except n.
Use it when: Recalling the solenoid formula.
Mutual uses the inner area
For coaxial solenoids, flux is shared only through the smaller (inner) cross-section: πr₁².
Use it when: Avoiding the wrong radius.
B squared over two mu
Magnetic energy density B²/2μ₀ mirrors electric ½ε₀E².
Use it when: Energy density questions.

Self and Mutual Inductance: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Self-inductance from back emf · 2 marks · Very short answerBoard level

An emf of 5 V is induced in a coil when the current in it changes steadily at 100 A/s. Find its self-inductance and the flux linkage when the current is 2 A.

Try it first, then show the solutionSolution

$L = \frac{∣ ε ∣}{d I / d t} = \frac{5}{100} = 0.05$ H. (1)
$N Φ = L I = 0.05 \times 2 = 0.1$ Wb. (1)

Answer: 0.05 H; 0.1 Wb.

Example 2: Derivation: self-inductance of a long solenoid · 3 marks · Short answerBoard level

Derive an expression for the self-inductance of a long air-cored solenoid of length l, cross-sectional area A and n turns per unit length. How does it change with an iron core?

Try it first, then show the solutionSolution

With current I, the field inside is uniform: $B = μ_{0} n I$ . Flux through one turn: $μ_{0} n I A$ . (1)
Number of turns = nl, so total flux linkage $N Φ = (n l) (μ_{0} n I A) = μ_{0} n^{2} A l I$ . (1)
Since $N Φ = L I$ , $L = μ_{0} n^{2} A l$ . With a core of relative permeability $μ_{r}$ , $L = μ_{r} μ_{0} n^{2} A l$ . (1)

Answer: $L = μ_{0} n^{2} A l$ ; with a core, multiplied by μ_r.

Example 3: Derivation and use: mutual inductance of coaxial solenoids · 5 marks · Long answerHarder board level

(a) Two long coaxial solenoids of the same length l have n₁ and n₂ turns per unit length; the inner one has radius r₁. Derive their mutual inductance. (3)
(b) Evaluate it for n₁ = 1000 per metre, n₂ = 500 per metre, r₁ = 2 cm and l = 0.5 m. (2)

Try it first, then show the solutionSolution

(a) Pass current I₂ through the outer solenoid: inside it, $B_{2} = μ_{0} n_{2} I_{2}$ . (1)
This field passes through each turn of the inner solenoid, of area $π r_{1}^{2}$ ; it has $n_{1} l$ turns, so its flux linkage is $N_{1} Φ_{1} = (n_{1} l) (π r_{1}^{2}) (μ_{0} n_{2} I_{2})$ . (1)
$M = \frac{N _{1} Φ _{1}}{I _{2}} = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ (the same as M₁₂, as expected). (1)
(b) $M = 4 π \times 1 0^{- 7} \times 1000 \times 500 \times π \times (0.02)^{2} \times 0.5$ . (1)
$M \approx 3.95 \times 1 0^{- 4}$ H. (1)

Answer: (a) $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ ; (b) about $3.95 \times 1 0^{- 4}$ H.

Example 4: Energy in an inductor · 3 marks · Short answerHarder board level

(a) Find the energy stored in a 0.2 H inductor carrying 3 A. (b) What current would store 3.6 J? (c) Write the work-energy argument that gives U = ½LI².

Try it first, then show the solutionSolution

(a) $U = \frac{1}{2} \times 0.2 \times 9 = 0.9$ J. (1)
(b) $I = \frac{2 U}{L} = \frac{7.2}{0.2} = 6$ A. (1)
(c) Against the back emf the source supplies power $L I \frac{d I}{d t}$ ; work to raise the current from 0 to I is $\int_{0}^{I} L I d I = \frac{1}{2} L I^{2}$ . (1)

Answer: (a) 0.9 J; (b) 6 A; (c) integrating LI dI gives ½LI².

Example 5: Solenoid with an iron core · 5 marks · Long answerHOTS / competency

A solenoid of 500 turns, length 25 cm and cross-section 4 cm² has an iron core of relative permeability 400.
(a) Find its self-inductance. (2)
(b) The current rises steadily from 0 to 1 A in 0.01 s. Find the back emf. (1)
(c) Find the energy stored at 1 A, and check it with the energy density of the field in the core. (2)

Try it first, then show the solutionSolution

(a) Air-cored: $L_{0} = \frac{μ _{0} N ^{2} A}{l} = \frac{4 π \times 1 0 ^{- 7} \times 250000 \times 4 \times 1 0 ^{- 4}}{0.25} \approx 5.03 \times 1 0^{- 4}$ H. (1)
With the core: $L = μ_{r} L_{0} \approx 0.201$ H. (1)
(b) $ε = L \frac{Δ I}{Δ t} = 0.201 \times 100 \approx 20.1$ V. (1)
(c) $U = \frac{1}{2} L I^{2} \approx 0.1005$ J. (1)
Check: $B = μ_{r} μ_{0} n I = 400 \times 4 π \times 1 0^{- 7} \times 2000 \times 1 \approx 1.005$ T; $u = \frac{B ^{2}}{2 μ _{r} μ _{0}} \approx 1005$ J/m³; volume $= 4 \times 1 0^{- 4} \times 0.25 = 1 0^{- 4}$ m³; $U \approx 0.1005$ J. (1)

Answer: (a) about 0.201 H; (b) about 20.1 V; (c) about 0.10 J by both methods.

Example 6: Two coupled coils · 4 marks · Case-basedHarder board level

Read the passage and answer.

Two coils P and S are placed near each other with a mutual inductance of 0.05 H.
(a) The current in P falls steadily from 4 A to zero in 0.02 s. Find the emf in S. (1)
(b) Find the flux linked with S when P carries 2 A. (1)
(c) If S carries a changing current instead, will the same M apply to the emf in P? (1)
(d) Give two ways of increasing M without changing the number of turns. (1)

Try it first, then show the solutionSolution

(a) $ε_{S} = M \frac{Δ I _{P}}{Δ t} = 0.05 \times \frac{4}{0.02} = 10$ V. (1)
(b) $N_{S} Φ_{S} = M I_{P} = 0.05 \times 2 = 0.1$ Wb. (1)
(c) Yes: M₁₂ = M₂₁, so the same M gives the emf in P. (1)
(d) Bring the coils closer (or make them coaxial) and put a soft iron core through both. (1)

Answer: (a) 10 V; (b) 0.1 Wb; (c) yes, M is the same both ways; (d) bring them closer/coaxial, add an iron core.

Where students lose marks in Self and Mutual Inductance

The slips examiners see most often, and what to write instead.

Mistake: Using total turns N in place of n (turns per metre).
The formula L = μ₀n²Al uses n.
Wrong working
L = μ₀(500)² × 4 × 10⁻⁴ × 0.25
Right working
L = μ₀(2000)² × 4 × 10⁻⁴ × 0.25 (n = 500/0.25), or μ₀N²A/l
Do this: Write n = N/l before substituting.
Usually costs: The answer mark.
Mistake: Using the outer radius in M for coaxial solenoids.
Thinking the bigger coil sets the area.
Do this: Only the inner cross-section carries flux common to both coils.
Usually costs: 1 mark.
Mistake: Saying an inductor opposes current.
Mixing up opposition to current (resistance) with opposition to change.
Do this: An inductor opposes changes in current; a steady current flows through an ideal inductor freely.
Usually costs: 1 mark in reasoning items.
Mistake: Skipping the work step in the energy derivation.
Jumping to ½LI².
Do this: Show dW = LI dI and integrate from 0 to I.
Usually costs: 1 mark.
Mistake: Thinking M depends on the current.
Seeing MI₁ in the flux equation.
Do this: M depends only on geometry and the medium.
Usually costs: ½ to 1 mark.

Exam tips

Define L and M in two ways each (flux linkage per unit current; emf per unit rate of change of current).
For the coaxial-solenoid derivation, say which solenoid carries the current and why the inner area is used.
Self and mutual inductance questions appear in each of the 2023-2026 board paper sets we checked, for example 2023 (55/1/1 Q28), 2025 (55/1/1 Q24) and 2026 (55/1/1 Q33(a)).
Quote U = ½LI² with the work argument when asked to 'show' it.
Use the mass analogy (½mv² ↔ ½LI²) to explain why an inductor opposes sudden changes.

Practice questions on Self and Mutual Inductance

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of inductance, the henry, is equivalent to
(a) Wb A (b) V A/s (c) V s/A (d) T m²
Hint
ε = L dI/dt.
Answer
(c) V s/A (also Wb/A).
Q2 · 2 marks · Very short answerBoard level
Define self-inductance. On what factors does the self-inductance of a solenoid depend?
Hint
NΦ = LI.
Answer
Self-inductance is the flux linkage per unit current in the coil itself (NΦ = LI), or the back emf per unit rate of change of current. For a solenoid, L = μ_rμ₀n²Al: it depends on the turns per unit length, the cross-sectional area, the length and the core material.
Q3 · 2 marks · Very short answerHarder board level
An emf of 30 mV is induced in coil B when the current in coil A changes at 20 A/s. Find the mutual inductance and the flux linked with B when A carries 5 A.
Hint
M = ε/(dI/dt).
Answer
$M = \frac{0.03}{20} = 1.5 \times 1 0^{- 3}$ H; flux linkage $= M I = 7.5 \times 1 0^{- 3}$ Wb.
Q4 · 1 mark · Very short answerHarder board level
Find the energy density of a magnetic field of 0.5 T in air.
Hint
B²/2μ₀.
Answer
$u = \frac{0.25}{2 \times 4 π \times 1 0 ^{- 7}} \approx 9.95 \times 1 0^{4}$ J/m³.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An inductor opposes a change in the current through it, not the current itself.
Reason (R): The induced emf in an inductor is proportional to the rate of change of current.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
ε = −L dI/dt.
Answer
(a). A steady current gives dI/dt = 0 and no back emf.
Q6 · 2 marks · Very short answerHOTS / competency
The number of turns of a solenoid is doubled, keeping its length and cross-section the same. How do (a) its self-inductance and (b) the energy stored at the same current change?
Hint
L ∝ N² for fixed l and A.
Answer
(a) L = μ₀N²A/l becomes 4 times. (b) U = ½LI² also becomes 4 times.
Q7 · 2 marks · Very short answerHOTS / competency
For two long coaxial solenoids, the radius of the inner solenoid is doubled (it still fits inside the outer one), keeping all else the same. How does M change? Would doubling the outer radius change M?
Hint
Only the shared flux area counts.
Answer
M ∝ r₁², so it becomes 4 times. Changing the outer radius does not change M, since the flux linking both coils passes only through the inner cross-section.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Self-inductance of a long solenoid?Show answer
L = μ₀n²Al.
2.Mutual inductance of two coaxial solenoids?Show answer
M = μ₀n₁n₂πr₁²l.
3.Energy stored in an inductor?Show answer
½LI².
4.Emf induced by mutual induction?Show answer
ε₂ = −M dI₁/dt.
5.SI unit of inductance?Show answer
Henry (H) = Wb/A = V s/A.
6.Magnetic energy density?Show answer
B²/2μ₀.

Self and Mutual Inductance: frequently asked questions

What is the difference between self and mutual inductance?

Self-inductance relates the emf in a coil to the change of its own current. Mutual inductance relates the emf in one coil to the change of current in another nearby coil.

Why is M₁₂ equal to M₂₁?

It can be shown that the flux linkage of coil 1 due to a unit current in coil 2 equals that of coil 2 due to a unit current in coil 1. For coaxial solenoids both give μ₀n₁n₂πr₁²l.

Why is self-inductance called electrical inertia?

Just as mass opposes changes in velocity, self-inductance opposes changes in current. The energy ½LI² is the electrical counterpart of kinetic energy ½mv².

Why does an iron core increase inductance?

Iron (ferromagnetic) has a large relative permeability, so the field and flux for a given current are μ_r times larger, and L or M is μ_r times larger.

Where is the energy of an inductor stored?

In its magnetic field. For a solenoid, the energy per unit volume is B²/2μ₀, and multiplying by the volume gives ½LI².