(a)(i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current I

Question

I

(ii) The currents passing through two inductors of self-inductances 10 mH and 20 mH increase with time at the same rate. Draw graphs showing the variation of: (I) the magnitude of the induced emf with the rate of change of current in each inductor; (II) the energy stored in each inductor with the current flowing through it.

(OR)

(b)(i) Define the term mutual inductance. Deduce the expression for the mutual inductance of two long coaxial solenoids of the same length having different radii and different numbers of turns.

(ii) The current through an inductor is uniformly increased from zero to 2 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at

t=10

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Part (a)Concept understanding — Self-Inductance of a Solenoid

Self-Inductance of a Solenoid: From Intuition to Formula

Imagine you push a heavy door. It doesn't resist your push once it's moving — but it does resist you trying to change its speed suddenly. That resistance to change is inertia. A solenoid carrying current behaves the same way: it "wants" to keep its current steady, and fights any attempt to change it.

This property is called self-inductance. The solenoid generates a back emf that opposes the change in its own current — not the current itself, but the change in current. That's the core idea.

Why does a solenoid oppose current changes?

A solenoid is a long coil of wire. When current flows through it, it produces a magnetic field inside. If you try to increase the current, the magnetic field strengthens. But a changing magnetic field induces an emf in the coil itself (Faraday's law). By Lenz's law, this induced emf opposes the change that caused it — so it pushes back against the rising current.

If you try to decrease the current, the field weakens, and the induced emf tries to keep the current flowing. The solenoid acts like an electrical "flywheel."

The precise statement

Self-inductance $L$ is defined by the relation:

\mathcal{E} = -L \frac{dI}{dt}

where $\mathcal{E}$ is the induced back emf, and $\frac{dI}{dt}$ is the rate of change of current. The negative sign tells you the emf opposes the change.

For a solenoid, $L$ depends only on its geometry and the core material — not on the current. The formula is:

L = \mu_0 n^2 A l

L = \mu_0 n^2 A l

Let's unpack each symbol:

$\mu_0$ — permeability of free space ( $4\pi \times 10^{-7}$ H/m). It's a universal constant that tells you how strongly a vacuum responds to magnetic fields.
$n$ — number of turns per unit length (turns/m). More turns per metre means a stronger field per ampere, so more inductance.
$A$ — cross-sectional area of the solenoid (m²). A wider coil encloses more magnetic flux.
$l$ — length of the solenoid (m). Longer solenoid means more total turns, hence more inductance.

Where does $L = \mu_0 n^2 A l$ come from?

Start with the magnetic field inside a long solenoid:

B = \mu_0 n I

The magnetic flux through one turn is $BA = \mu_0 n I A$ . For all $N = nl$ turns, the total flux linkage is:

\Phi_{\text{total}} = N \cdot BA = (nl)(\mu_0 n I A) = \mu_0 n^2 A l I

By definition, self-inductance is the constant of proportionality between flux linkage and current:

\Phi_{\text{total}} = L I

Comparing, you get:

L = \mu_0 n^2 A l

Note

This formula assumes an ideal solenoid — infinitely long, with a uniform field inside and zero field outside. Real solenoids are close approximations if $l \gg \sqrt{A}$ .

What does a larger $L$ mean?

A solenoid with high $L$ strongly resists changes in current. If you try to switch the current on quickly, the back emf is large, so the current rises slowly. If you short-circuit the solenoid, the current doesn't drop instantly — it decays gradually.

This is why inductors are used in filters, chokes, and timing circuits. They smooth out current variations. …

Part (b)Concept understanding — Mutual Inductance

Mutual Inductance: From Intuition to Definition

Imagine you have two separate coils of wire placed near each other. You connect one coil to a battery — current starts flowing through it. Now, something strange happens in the other coil, which isn't connected to anything: a voltage appears across its ends. That voltage can even light a small bulb for an instant.

This is mutual inductance in action. One circuit "feels" the changing current in another circuit, even though they are not physically connected.

The Core Intuition

The key idea is changing magnetic fields. When current flows through a coil, it creates a magnetic field around it. If that current changes (increases or decreases), the magnetic field also changes. This changing field reaches the second coil. And a changing magnetic field, by Faraday's law, induces an electromotive force (emf) in any nearby conductor.

So mutual inductance is simply: how effectively a change in current in one coil induces a voltage in another coil.

Note

Mutual inductance only works when the current is changing. A steady DC current produces a steady magnetic field, which induces nothing in the second coil. That's why the bulb lights only for an instant when you first connect the battery — the current is rising from zero.

The Precise Definition

Let's formalise this. Consider two coils: coil 1 and coil 2. Let $I_1$ be the current in coil 1. This current produces a magnetic flux $\Phi_{21}$ through coil 2 (the flux from coil 1 that passes through coil 2).

The mutual inductance $M$ (also written $M_{21}$ ) is defined as the constant of proportionality between the current $I_1$ and the flux it produces in coil 2:

\Phi_{21} = M \, I_1

Similarly, if current $I_2$ flows in coil 2, it produces a flux $\Phi_{12}$ through coil 1:

\Phi_{12} = M \, I_2

Important

The mutual inductance $M$ is the same for both directions. $M_{21} = M_{12} = M$ . This is a fundamental symmetry property.

Now, by Faraday's law, the induced emf in coil 2 due to a changing current in coil 1 is:

\mathcal{E}_2 = - \frac{d\Phi_{21}}{dt} = - M \frac{dI_1}{dt}

And the induced emf in coil 1 due to a changing current in coil 2 is:

\mathcal{E}_1 = - M \frac{dI_2}{dt}

The negative sign is Lenz's law — the induced emf opposes the change that produced it.

Units

The SI unit of mutual inductance is the henry (H), named after Joseph Henry. From the definition:

1 \, \text{H} = 1 \, \frac{\text{V} \cdot \text{s}}{\text{A}} = 1 \, \frac{\text{Wb}}{\text{A}}

One henry means that a current change of 1 ampere per second induces an emf of 1 volt in the other coil.

What Determines Mutual Inductance?

$M$ depends on:

Geometry: size, shape, number of turns of both coils
Relative position: how close they are and how they are oriented
Core material: if a magnetic material (like iron) is present, $M$ increases dramatically

For two coaxial solenoids of length $l$ , with $N_1$ and $N_2$ turns, and cross-sectional area $A$ , the mutual inductance is: …

Why this formula?

Mutual Inductance: Why the Formula Holds

Mutual inductance is a beautiful example of Faraday's Law in action — it describes how a changing current in one coil can induce an EMF in a nearby coil, without any direct electrical connection.

1. The Core Idea: Flux Linkage

Imagine two coils, Coil 1 and Coil 2, placed close together.

When a current $I_1$ flows in Coil 1, it creates a magnetic field $\vec{B}_1$ .
Some of the magnetic field lines from Coil 1 pass through Coil 2.
The total magnetic flux through Coil 2 due to $I_1$ is called the mutual flux:

\Phi_{21} = \text{flux through Coil 2 due to current in Coil 1}

Key insight: For a fixed geometry (coils not moving), the mutual flux is directly proportional to the current $I_1$ :

\Phi_{21} \propto I_1

Why? Because $\vec{B}_1$ itself is proportional to $I_1$ (Biot–Savart law), and the area of Coil 2 is fixed. So:

\Phi_{21} = M_{21} \, I_1

where $M_{21}$ is the mutual inductance (a constant depending on coil shapes, sizes, turns, and relative positions).

2. Why the EMF Formula Arises

Now, if $I_1$ changes with time, then $\Phi_{21}$ changes with time. By Faraday's Law, a changing flux induces an EMF in Coil 2:

\mathcal{E}_2 = -\frac{d\Phi_{21}}{dt}

Substitute $\Phi_{21} = M_{21} I_1$ :

\mathcal{E}_2 = -M_{21} \frac{dI_1}{dt}

That's the key formula. The negative sign (Lenz's law) tells us the induced EMF opposes the change in flux.

3. Symmetry: $M_{12} = M_{21}$

If we reverse the situation — current $I_2$ in Coil 2 induces flux $\Phi_{12}$ in Coil 1 — we get:

\Phi_{12} = M_{12} \, I_2

and

\mathcal{E}_1 = -M_{12} \frac{dI_2}{dt}

A deep result from energy conservation (or from the reciprocity theorem in electromagnetism) shows:

\boxed{M_{12} = M_{21} = M}

So we simply call it $M$ , the mutual inductance between the two coils.

4. The Complete Formula Set

Quantity	Expression	Why?
Mutual flux (Coil 2 due to Coil 1)	$\Phi_{21} = M I_1$	Proportionality from Biot–Savart
Induced EMF in Coil 2	$\mathcal{E}_2 = -M \frac{dI_1}{dt}$	Faraday's Law
Mutual flux (Coil 1 due to Coil 2)	$\Phi_{12} = M I_2$	Symmetry
Induced EMF in Coil 1	$\mathcal{E}_1 = -M \frac{dI_2}{dt}$	Faraday's Law

5. Physical Intuition (Exam-Ready) …

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Self-inductance stores $W=\tfrac12 LI^2$ ; emf– $(dI/dt)$ is a line of slope $L$ and energy– $I$ a parabola (both steeper/higher for 20 mH). Coaxial solenoids: $M=\dfrac{\mu_0 N_1N_2\pi r_1^2}{l}$ ; the inductor's flux at $t=10\ \text{s}$ is $0.05\ \text{Wb}$ .

Part (a) — self-inductance and energy

(i) Definition. The self-inductance $L$ of a coil is the flux linkage set up per unit current, $N\Phi=LI$ ; a changing current induces a back-emf $\varepsilon=-L\dfrac{dI}{dt}$ . SI unit: henry (H).

Energy to build up current $I$ . While the current grows, the source must push charge against the back-emf. The instantaneous power delivered is $P=\varepsilon_{applied}I=L\dfrac{dI}{dt}\,I$ , so

W=\int_0^{t} L\frac{dI}{dt}\,I\,dt=\int_0^{I} L\,I\,dI=\frac12 LI^2.

This energy is stored in the magnetic field of the coil.

W=\tfrac12 L I^2.

(ii) Graphs for $L_1=10\ \text{mH}$ and $L_2=20\ \text{mH}$ (same $dI/dt$ ).

(I) Induced emf vs rate of change of current: $|\varepsilon|=L\dfrac{dI}{dt}$ is linear through the origin with slope $L$ . The $20\ \text{mH}$ inductor gives the steeper line (twice the slope of the $10\ \text{mH}$ one). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 39 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. When the current changes from +2 A to -2 A in 0.05 second in a coil, an e.m.f. of 8 V is induced in it. The coefficient of self-induction of the coil is$
(a) 0.1 henry
(b) 0.2 henry
(c) 0.4 henry
(d) 0.8 henry
›Reveal solutionSolution

Using $e = L\,\Delta I/\Delta t$ with $\Delta I = 4$ A and $\Delta t = 0.05$ s gives $L = 0.1$ H.

The current changes from $+2$ A to $-2$ A, so the magnitude of the change is

$\Delta I = |(-2) - (2)| = 4\ \text{A}, \qquad \Delta t = 0.05\ \text{s}$

The magnitude of the self-induced emf is …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Self inductance is called$
(a) electric force
(b) electrical inertia
(c) electric pressure
(d) electric energy
›Reveal solutionSolution

Self-inductance L makes a circuit resist a change in the current flowing through it (via a back-emf = -L dI/dt), analogous to inertia resisting a change in velocity.

Whenever the current through a coil tends to change, the induced back-emf (from self-induction) opposes that change (Lenz's law). This behaviour - opposing change rather than opposing the current itself - is directly analogous to mechanical inertia, which oppos …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The self-inductance of a coil is measured by$
(a) Electrical inertia
(b) Electrical friction
(c) Induced emf
(d) Induced current
›Reveal solutionSolution

Self-inductance is defined via the induced emf a coil produces in itself when its own current changes - it behaves like the 'electrical inertia' of the circuit, but it is quantified through that induced emf.

When the current I through a coil changes, the magnetic flux linked with the coil changes too, and by Faraday's law this changing self-flux induces an emf in the SAME coil that opposes the change in current (Lenz's law). This self-induced emf defines the self-inductance L of the coil:

emf = -L * (dI/dt)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of inductance is$
(a) henry
(b) weber
(c) newton
(d) ohm
›Reveal solutionSolution

Inductance L relates induced emf to the rate of change of current, so its unit works out to volt-second per ampere - named the henry.

From emf = -L*(dI/dt), we get L = emf / (dI/dt), so the unit of L is volt / (ampere/second) = volt.second/ampere. This combination is given the special SI name henry (H), after Joseph Henry. Weber is the u …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In a solenoid number of turns per unit length are doubled, it's self-inductance:$
(a) Halved
(b) Doubled
(c) Remains constant
(d) Becomes four times
›Reveal solutionSolution

Self-inductance of a solenoid is proportional to the square of the number of turns per unit length, so doubling $n$ makes $L$ four times as large.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average e.m.f. of 100 V induced, give an estimate of the self-inductance of the circuit.$
(a) L = 4 H
(b) L = 20 H
(c) L = 40 H
(d) L = 2 H
›Reveal solutionSolution

Using $|\varepsilon| = L\left|\dfrac{dI}{dt}\right|$ , the self-inductance works out to 2 H.

Given: Current changes from $I_i = 5.0\text{ A}$ to $I_f = 0.0\text{ A}$ in $\Delta t = 0.1\text{ s}$ , with average induced emf $|\varepsilon| = 100\text{ V}$ .

Step 1 — rate of change of current:

$\left|\frac{\Delta I}{\Delta t}\right| = \frac{5.0 - 0.0}{0.1} = 50\ \text{A/s}$

…

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CBSE 2026Set SEM31 markMCQ
$Q. The dimensional formula of coefficient of mutual inductance is$
(a) [ ML²T⁻²I² ]
(b) [ ML²T⁻²I⁻² ]
(c) [ ML⁻²T²I² ]
(d) [ ML⁻²T⁻²I⁻² ]
›Reveal solutionSolution

Mutual inductance M satisfies EMF = M(dI/dt), so [M] = [EMF]·[time]/[current] = [ML²T⁻²I⁻²]. Option (b).

Step 1 — defining relation: The induced emf in the secondary is ε = M(dI/dt), so M = ε/(dI/dt).

Step 2 — dimensions of emf (a potential difference): [ε] = [ML²T⁻³I⁻¹].

Step 3 — dI/dt has dimensions [I T⁻¹].

Step 4 — divide: [M] = [ML²T⁻³I⁻¹]/[I T⁻¹] = [ML²T⁻²I⁻²].

…

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CBSE 2025Set X11 mark
$Q. When a ________ rod is inserted into a coil, its self-inductance increases.$
›Reveal solutionSolution

ferromagnetic (soft iron) Self-inductance $L = \mu_r \mu_0 n^2 A l$ . Inserting a ferromagnetic (soft iron) core has a large relative permeability $\mu_r \gg 1$ , which greatly increases the mag …

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CBSE 2025Set D1 markMCQ
$Q. The self-inductance of a solenoid depends on (A) The current flowing through its medium (B) The number of turns per unit length (C) The length of the solenoid (D) Both (B) and (C)$
›Reveal solutionSolution

Self-inductance depends on the solenoid's geometry (turns per unit length and length/area), not on the current.

For a solenoid the self-inductance is

$L = \mu_0 n^{2} A l$

where n is the number of turns per unit length, A the cross-sectional area and l the length. So it depends on the number of turns per unit length and the length (and area) — both geometric factors.

…

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CBSE 2025Set A1 mark
$Q. Write answer in one sentence: Write the SI unit of mutual inductance.$
›Reveal solutionSolution

The SI unit of mutual inductance is the henry (H).

Mutual inductance M between two coils is defined through $\varepsilon_2 = -M\dfrac{dI_1}{dt}$ , i.e., the emf induced in the secondary coil per unit rate of change of current in the primary coil. Its SI unit, the henry (H), is defined such that 1 H is the mutual inductance between two coils when a cu …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The self-inductance of a coil is 5 henry. A current of 1 ampere changes to 2 amperes within 5 seconds through the coil. The value of the induced e.m.f. is$
(i) 10 volts
(ii) 0.1 volt
(iii) 1 volt
(iv) 100 volts
›Reveal solutionSolution

emf = L dI/dt = 5 x (1 A / 5 s) = 1 V.

…

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CBSE 2024Set 55/5/11 markMCQ
$5\,\text{A/s}$
›Reveal solutionSolution
The mutual inductance $M$ is defined by the induced emf $\mathcal{E} = -M \frac{di}{dt}$ .

Using the given values: $\mathcal{E} = 2 \times 10^{-3}\,\text{V}$ , $\frac{di}{dt} = 5\,\text{A/s}$ , we get $M = 0.4 \times 10^{-3}\,\text{H} = 0.4\,\text{mH}$ .

The correct option is (A).

The idea is simple: mutual inductance tells you how effectively a changing current in one coil “induces” an emf in a neighbouring coil. The definition is direct — the induced emf in the second coil is proportional to the rate of change of current in the first coil, and the constant of proportionality is the mutual inductance $M$ .

The formula is:
$\mathcal{E} = -M \frac{di}{dt}$
The negative sign is Lenz’s law (direction of induced emf), but for magnitude we drop the sign.

Let’s work it out.

Write down what’s given

Rate of change of current in the first coil: $\frac{di}{dt} = 5\,\text{A/s}$

Induced emf in the second coil: $\mathcal{E} = 2\,\text{mV} = 2 \times 10^{-3}\,\text{V}$

We need $M$ .

Use the defining relation

From $\mathcal{E} = M \frac{di}{dt}$ (taking magnitude), we get:

$M = \frac{\mathcal{E}}{di/dt}$

Plug in the numbers

$M = \frac{2 \times 10^{-3}}{5} = 0.4 \times 10^{-3}\,\text{H}$
That’s $0.4$ millihenry. …
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Q.(a)(i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current $I$ in a coil of self-inductance $L$ .

Part (a)Concept understanding — Self-Inductance of a Solenoid

Self-Inductance of a Solenoid: From Intuition to Formula

Why does a solenoid oppose current changes?

The precise statement

Where does $L = \mu_0 n^2 A l$ come from?

What does a larger $L$ mean?

Part (b)Concept understanding — Mutual Inductance

Mutual Inductance: From Intuition to Definition

The Core Intuition

The Precise Definition

Units

What Determines Mutual Inductance?

Why this formula?

Mutual Inductance: Why the Formula Holds

1. The Core Idea: Flux Linkage

2. Why the EMF Formula Arises

3. Symmetry: $M_{12} = M_{21}$

4. The Complete Formula Set

5. Physical Intuition (Exam-Ready) …

Self-inductance and mutual inductance

Part (a) — self-inductance, energy, and graphs

Part (a) — self-inductance and energy

Q.(a)(i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current III in a coil of self-inductance LLL.

Part (a) — self-inductance and energy

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Q.(a)(i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current $I$ in a coil of self-inductance $L$ .

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