PhysicsISC Class 12 · Electromagnetic Induction

Self-Induction and Mutual Induction: Inductance of Solenoids

Updated 15 October 2026 · Fact-checked

Self-induction is the emf induced in a coil when its own current changes: $\phi = LI$ and $\varepsilon = -L\frac{dI}{dt}$. Mutual induction is the emf induced in one coil by a changing current in another: $\phi_2 = MI_1$, $\varepsilon_2 = -M\frac{dI_1}{dt}$. For a long solenoid $L = \frac{\mu_0N^2A}{l}$; for coaxial solenoids $M = \frac{\mu_0N_1N_2A}{l}$.

Understand Self-Inductance and Mutual Inductance

Self-induction. A current in a coil produces a magnetic field and hence a flux through the coil itself. The total flux linked with the coil (N times the flux through one turn) is proportional to the current:
$N ϕ = L I .$
L is the coefficient of self-induction or self-inductance. When the current changes, the linked flux changes and, by Faraday's law, an emf appears in the coil itself:
$ε = - \frac{d ( N ϕ )}{d t} = - L \frac{d I}{d t} .$
By Lenz's law this emf opposes the change of current (not the current itself): it is called the back emf. An inductor makes a current rise and fall gradually, much as mass makes velocity change gradually; it is "electrical inertia".

Definitions and unit. $L = \frac{N ϕ}{I}$ (flux linked per unit current), or $L = \frac{∣ ε ∣}{d I / d t}$ (emf per unit rate of change of current). Unit: henry, 1 H = 1 Wb A⁻¹ = 1 V s A⁻¹. A coil has a self-inductance of 1 H if a current changing at 1 A s⁻¹ induces an emf of 1 V in it.

Self-inductance of a long solenoid (derivation). Length l, N turns, cross-section A, carrying current I. Inside, $B = μ_{0} n I = \frac{μ _{0} N I}{l}$ (uniform). Flux through one turn: $ϕ = B A = \frac{μ _{0} N I A}{l}$ . Flux linked with all N turns:
$N ϕ = \frac{μ _{0} N ^{2} A I}{l} = L I \Rightarrow L = \frac{μ _{0} N ^{2} A}{l} = μ_{0} n^{2} A l .$
With a core of relative permeability $μ_{r}$ , $L = \frac{μ _{0} μ _{r} N ^{2} A}{l}$ . L depends on the geometry and the core, not on the current. Doubling N (same length) makes L four times; doubling the length with the same N halves L.

Mutual induction. Two coils near each other: a current $I_{1}$ in coil 1 sends flux through coil 2. The flux linked with coil 2 is proportional to $I_{1}$ :
$N_{2} ϕ_{2} = M I_{1}, ε_{2} = - M \frac{d I _{1}}{d t} .$
M is the mutual inductance (unit henry), defined as $M = \frac{N _{2} ϕ _{2}}{I _{1}}$ or $M = \frac{∣ ε _{2} ∣}{d I _{1} / d t}$ . It is the same whichever coil carries the current ( $M_{12} = M_{21}$ ).

Mutual inductance of two long coaxial solenoids (derivation). An inner solenoid (2) of $N_{2}$ turns and cross-section A is wound over (or inside) the middle of a long outer solenoid (1) of $N_{1}$ turns and length l. Current $I_{1}$ in the outer gives $B_{1} = \frac{μ _{0} N _{1} I _{1}}{l}$ inside. Flux linked with the inner coil:
$N_{2} ϕ_{2} = N_{2} B_{1} A = \frac{μ _{0} N _{1} N _{2} A}{l} I_{1} \Rightarrow M = \frac{μ _{0} N _{1} N _{2} A}{l} = μ_{0} n_{1} N_{2} A .$
M depends on the numbers of turns, the common area, the separation and relative orientation of the coils, and the core (× $μ_{r}$ ).

Where these matter. Self-induction gives sparks when switches open in inductive circuits, the delay in a fluorescent lamp's choke and the opposition of an inductor to a.c. Mutual induction is the working principle of the transformer, induction coils and wireless charging.

Self-Inductance and Mutual Inductance: formulas and key facts

Self-inductance: $N ϕ = L I, ε = - L \frac{d I}{d t}$; Back emf opposes the change.
Unit: $1 H = 1 Wb A^{- 1} = 1 V s A^{- 1}$; Henry.
Long solenoid: $L = \frac{μ _{0} N ^{2} A}{l} = μ_{0} n^{2} A l$; With a core × $μ_{r}$ .
Mutual inductance: $N_{2} ϕ_{2} = M I_{1}, ε_{2} = - M \frac{d I _{1}}{d t}$; $M_{12} = M_{21}$ .
Coaxial solenoids: $M = \frac{μ _{0} N _{1} N _{2} A}{l} = μ_{0} n_{1} N_{2} A$; A = area of the inner coil.
Average emf: $∣ ε ∣ = L \frac{Δ I}{Δ t}$; Uniform change.

How to solve Self-Inductance and Mutual Inductance questions

Decide first whether the changing current is in the same coil (L) or another coil (M).

1Same coil: use $ε = L \frac{Δ I}{Δ t}$ ; another coil: $ε_{2} = M \frac{Δ I _{1}}{Δ t}$ .
2Convert milliseconds to seconds and mH or µH to H.
3For a solenoid, find N, A (m²) and l (m), then $L = \frac{μ _{0} N ^{2} A}{l}$ (×μr with a core).
4For coaxial solenoids, use the turns per metre of the long outer coil and the area of the inner coil.
5Give the direction of the emf by Lenz's law: it opposes the change in current.
6For ratio questions, write how L or M depends on N, A, l and μr.

Self-Inductance 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.

Back emf at a switch

Suppose a motor winding of self-inductance 2 H carries 1.5 A, and a switch in its circuit opens so that the current falls to zero in about 5 ms.

Estimate the emf induced across the winding. Why may a spark jump across the switch contacts?

Try it first, then show the solutionSolution

$∣ ε ∣ = L \frac{Δ I}{Δ t} = 2 \times \frac{1.5}{0.005} = 600$ V.
This large back emf appears across the opening contacts, enough to ionise the air in the small gap.

Answer: About 600 V, which can make the air between the contacts spark.

Memory hook: Fast change of current, big back emf.

Effect of a core on inductance

Imagine a choke coil that has an inductance of 5 mH with an air core, into which an iron core of relative permeability 400 is pushed.

What is its new inductance?

Try it first, then show the solutionSolution

$L \propto μ_{r}$ for the same winding.
$L = 400 \times 5 = 2000$ mH = 2 H.

Answer: 2 H.

Memory hook: Core multiplies L by μr.

Tricks to remember Self-Inductance and Mutual Inductance

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

L is "electrical inertia"
Like mass resists change of velocity, L resists change of current.
Use it when: Explaining back emf.
N squared for L
L ∝ N² (both the field and the linking grow with N); M ∝ N₁N₂.
Use it when: Scaling questions.
Self vs mutual
Self: "my current, my emf". Mutual: "your current, my emf".
Use it when: Naming the phenomenon.
Henry = volt-second per amp
From $ε = L \frac{d I}{d t}$ .
Use it when: Units.

Self-Inductance and Mutual Inductance: solved examples

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

Example 1: What L does not depend on · 1 mark · MCQBoard level

The self-inductance of a long solenoid does NOT depend on:
(a) the number of turns
(b) the area of cross-section
(c) the current in it
(d) the permeability of its core

Try it first, then show the solutionSolution

$L = \frac{μ _{0} μ _{r} N ^{2} A}{l}$ contains no current.

Answer: (c).

Example 2: Finding L from an emf · 2 marks · Short answerBoard level

The current in a coil changes uniformly from 3 A to 1 A in 0.05 s, inducing an emf of 12 V. Find its self-inductance.

Try it first, then show the solutionSolution

$\frac{Δ I}{Δ t} = \frac{2}{0.05} = 40$ A s⁻¹. (1)
$L = \frac{12}{40} = 0.3$ H. (1)

Answer: 0.3 H.

Example 3: Inductance of a solenoid · 3 marks · Short answerHarder board level

A solenoid 40 cm long has 800 turns of radius 2 cm. Find its self-inductance with an air core and with an iron core of relative permeability 600.

Try it first, then show the solutionSolution

$A = π (0.02)^{2} \approx 1.257 \times 1 0^{- 3}$ m². (1)
$L = \frac{4 π \times 1 0 ^{- 7} \times 80 0 ^{2} \times 1.257 \times 1 0 ^{- 3}}{0.4} \approx 2.53 \times 1 0^{- 3}$ H. (1)
Iron core: $600 \times 2.53$ mH ≈ 1.52 H. (1)

Answer: ≈ 2.53 mH (air); ≈ 1.52 H (iron).

Example 4: A coil inside a long solenoid · 3 marks · Short answerHarder board level

A small coil of 300 turns and area 5 cm² is placed coaxially inside a long solenoid of 2000 turns per metre. Find the mutual inductance and the emf in the small coil when the solenoid current changes at 100 A s⁻¹.

Try it first, then show the solutionSolution

$M = μ_{0} n_{1} N_{2} A = 4 π \times 1 0^{- 7} \times 2000 \times 300 \times 5 \times 1 0^{- 4}$ . (1)
$M \approx 3.77 \times 1 0^{- 4}$ H. (1)
$ε = M \frac{d I}{d t} \approx 3.77 \times 1 0^{- 2}$ V. (1)

Answer: $M \approx 0.38$ mH; emf ≈ 38 mV.

Example 5: Emf from a current graph · 3 marks · Short answerHOTS / competency

The current in a 0.2 H coil rises uniformly from 0 to 5 A in 0.1 s, stays at 5 A until t = 0.3 s, then falls uniformly to zero by t = 0.35 s. Find the induced emf in each interval and its sense relative to the current.

Try it first, then show the solutionSolution

0–0.1 s: $ε = 0.2 \times \frac{5}{0.1} = 10$ V, opposing the rising current. (1)
0.1–0.3 s: the current is steady, so ε = 0. (1)
0.3–0.35 s: $ε = 0.2 \times \frac{5}{0.05} = 20$ V, now in the same direction as the current (trying to keep it going). (1)

Answer: 10 V (opposing), 0, 20 V (aiding the falling current).

Example 6: A solenoid with a pick-up coil · 4 marks · Case-basedHOTS / competency

Suppose a long air-cored solenoid has 1000 turns, length 0.5 m and cross-section $2 \times 1 0^{- 3}$ m², and a 50-turn pick-up coil is wound tightly over its middle.
(i) Find the self-inductance of the solenoid. (1 mark)
(ii) Find the back emf when its current of 4 A is switched off in 2 ms. (1 mark)
(iii) Find the mutual inductance of the pair. (1 mark)
(iv) Find the emf in the pick-up coil during the switch-off. (1 mark)

Try it first, then show the solutionSolution

(i) $L = \frac{4 π \times 1 0 ^{- 7} \times 1 0 ^{6} \times 2 \times 1 0 ^{- 3}}{0.5} \approx 5.03 \times 1 0^{- 3}$ H.
(ii) $ε = L \frac{Δ I}{Δ t} = 5.03 \times 1 0^{- 3} \times 2000 \approx 10$ V.
(iii) $M = \frac{μ _{0} N _{1} N _{2} A}{l} = \frac{4 π \times 1 0 ^{- 7} \times 1000 \times 50 \times 2 \times 1 0 ^{- 3}}{0.5} \approx 2.51 \times 1 0^{- 4}$ H.
(iv) $ε_{2} = M \frac{Δ I}{Δ t} \approx 2.51 \times 1 0^{- 4} \times 2000 \approx 0.50$ V.

Answer: (i) ≈ 5.0 mH; (ii) ≈ 10 V; (iii) ≈ 0.25 mH; (iv) ≈ 0.50 V.

Where students lose marks in Self-Inductance and Mutual Inductance

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

Mistake: Naming the wrong phenomenon
In ISC 2024 candidates wrote "induction", "self-inductance" or "mutual induction" instead of self-induction.
Do this: Self-induction is the phenomenon; self-inductance is the coefficient L.
Usually costs: 1 mark.
Mistake: Ignoring milliseconds
Also in 2024: times in ms were used as seconds.
Wrong working
$\frac{Δ I}{Δ t} = \frac{4}{10}$ for 10 ms.
Right working
$\frac{4}{0.01} = 400$ A s⁻¹.
Do this: Convert to SI before dividing.
Usually costs: 1 mark.
Mistake: Wrong formula or unit for L
The 2025 examiners reported wrong formulae for self-inductance and wrong units.
Do this: Use $ε = L \frac{d I}{d t}$ ; the unit is henry (H).
Usually costs: 1 mark.
Mistake: Using the outer coil's area for M
Only the flux through the inner coil links both.
Do this: Use the area of the smaller (inner) coil in $M = μ_{0} n_{1} N_{2} A$ .
Usually costs: 1 mark.

Exam tips

In derivations, write B inside the solenoid, then flux per turn, then total flux linkage, then L.
State that L and M depend only on geometry and core, not on current.
Mention Lenz's law when explaining the sign in $ε = - L \frac{d I}{d t}$ .
Keep µ₀ = 4π × 10⁻⁷ and multiply by π² carefully when A = πr².

Practice questions on Self-Inductance 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 · Very short answerBoard level
A changing current in one coil induces an emf in a neighbouring coil. Name this phenomenon, and name the phenomenon when the emf is induced in the same coil whose current changes. (A variation on ISC 2024.)
Hint
Same coil or another coil?
Answer
Mutual induction; self-induction.
Q2 · 2 marks · Short answerBoard level
When the current in a coil rises steadily from 2 A to 8 A in 0.05 s, an emf of 24 V is induced in it. Find its self-inductance. (A variation on ISC 2025.)
Hint
$∣ ε ∣ = L \frac{d I}{d t}$ .
Answer
$\frac{d I}{d t} = \frac{6}{0.05} = 120$ A s⁻¹, so $L = \frac{24}{120} = 0.2$ H.
Q3 · 2 marks · Short answerHarder board level
An emf of 50 mV is induced in a coil when the current in a neighbouring coil changes by 2 A in 0.1 s. Find the mutual inductance and the flux linked with the first coil when the neighbour carries 3 A.
Hint
$N ϕ = M I$ .
Answer
M = 2.5 mH; flux linkage = $7.5 \times 1 0^{- 3}$ Wb.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An inductor opposes a change in current but not a steady current.
Reason (R): The induced emf in an inductor is proportional to the rate of change of current.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
dI/dt = 0 for steady current.
Answer
(a).
Q5 · 1 mark · Very short answerHarder board level
A long solenoid has inductance L. What is its inductance if (a) a core of μr = 50 is inserted, (b) its length is doubled keeping the same N and area?
Hint
$L = \frac{μ _{0} μ _{r} N ^{2} A}{l}$ .
Answer
(a) 50L; (b) $\frac{L}{2}$ .
Q6 · 3 marks · Short answerHOTS / competency
A long solenoid of 500 turns and length 0.5 m surrounds a coaxial 200-turn coil of cross-section $1 \times 1 0^{- 3}$ m². Find M. If instead a current is passed through the 200-turn coil, is the mutual inductance different?
Hint
n₁ = 1000 m⁻¹.
Answer
$M = 4 π \times 1 0^{- 7} \times 1000 \times 200 \times 1 0^{- 3} \approx 2.51 \times 1 0^{- 4}$ H. No: $M_{12} = M_{21}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a 50 mH coil carries a current that rises uniformly from 0 to 2 A in 20 ms, stays steady, and is then reversed uniformly to −2 A in 10 ms.
(i) Find the emf while the current rises. (1 mark)
(ii) What is the emf while it is steady? (1 mark)
(iii) Find the emf during the reversal. (1 mark)
(iv) In which interval is the back emf largest, and why? (1 mark)
Hint
ΔI for the reversal is 4 A.
Answer
(i) $0.05 \times \frac{2}{0.02} = 5$ V. (ii) 0. (iii) $0.05 \times \frac{4}{0.01} = 20$ V. (iv) During the reversal: the rate of change of current is largest (400 A s⁻¹).

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.Define self-inductance.Show answer
Flux linked per unit current, $L = \frac{N ϕ}{I}$ ; or emf per unit rate of change of current.
2.Self-inductance of a long solenoid?Show answer
$\frac{μ _{0} N ^{2} A}{l}$ .
3.L = 0.5 H, dI/dt = 20 A s⁻¹: emf?Show answer
10 V.
4.Mutual inductance of coaxial solenoids?Show answer
$\frac{μ _{0} N _{1} N _{2} A}{l}$ .
5.SI unit of inductance?Show answer
Henry (V s A⁻¹).

Self-Inductance and Mutual Inductance: frequently asked questions

Why is the back emf called "back"?

It acts against the change that produces it (Lenz's law). When the current is increasing, the induced emf opposes the source; when it is decreasing, it tries to keep the current going.

Does an inductor affect a steady d.c. current?

Only through its resistance. Once the current is steady, dI/dt = 0 and there is no back emf.

Why does L depend on N²?

More turns make a stronger field (∝ N), and that field links more turns (another factor N).

Is energy stored in an inductor in the ISC 2027 syllabus?

The syllabus lists self-induction, L, ε = −L dI/dt, the henry, the inductance of a solenoid, mutual induction and M of coaxial solenoids; this page sticks to those.

How was this tested recently?

In 2024 a 5-mark question asked for the name of self-induction and a value of L; in 2025 a part of a 5-mark question asked for L from an emf and a falling current.