A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm$^2$ and 150 turns ($N_C$) is wound tightly around the centre of the solenoid. If a current of 3.0 A flows through the solenoid, calculate: (a) the magnetic flux density at the centre of the solenoid, (b) the flux linkage in the coil C, (c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (Take $\mu_0=4\pi\times10^{-7}$ T.m/A)

The solenoid has n=1500 turns/m and carries I=3.0 A; the coil C wound around its centre has $N_c=150$ turns and area $A_c=25\,\text{cm}^2=25\times10^{-4}\,\text{m}^2$. (a) Interior field of the solenoid: $$B=\mu_0nI=(4\pi\times10^{-7})(1500)(3.0)\approx5.655\times10^{-3}\,\text{T}\approx5.66\times10^{-3}\,\text{T}$$ (b) Flux linkage in coil C (the solenoid's field, uniform over C's smaller cross-section): $$N_c\Phi=N_cBA_c=(150)(5.655\times10^{-3})(25\times10^{-4})\approx2.121\times10^{-3}\,\text{Wb}\approx2.12\times10^{-3}\,\text{Wb}$$ …

A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm² and 150 turns (NC) is wound tigh

Numericals · Q15

^2

(a) the magnetic flux density at the centre of the solenoid,

(b) the flux linkage in the coil C,

(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (Take

\mu_0=4\pi\times10^{-7}

T.m/A)

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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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The solenoid has n=1500 turns/m and carries I=3.0 A; the coil C wound around its centre has $N_c=150$ turns and area $A_c=25\,\text{cm}^2=25\times10^{-4}\,\text{m}^2$ .

Interior field of the solenoid: $B=\mu_0nI=(4\pi\times10^{-7})(1500)(3.0)\approx5.655\times10^{-3}\,\text{T}\approx5.66\times10^{-3}\,\text{T}$
Flux linkage in coil C (the solenoid's field, uniform over C's smaller cross-section): $N_c\Phi=N_cBA_c=(150)(5.655\times10^{-3})(25\times10^{-4})\approx2.121\times10^{-3}\,\text{Wb}\approx2.12\times10^{-3}\,\text{Wb}$ …

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

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 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 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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CBSE 2024Set 55/1/11 markMCQ
$Q. For question 14, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. Assertion (A) : The mutual inductance between two coils is maximum when the coils are wound on each other. Reason (R) : The flux linkage between two coils is maximum when they are wound on each other.$
›Reveal solutionSolution
The mutual inductance between two coils depends on the flux linkage between them. Winding the coils on each other places them as close as possible, maximising the flux linkage and therefore the mutual inductance. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Mutual inductance $M$ between two coils is defined by the relation $M = \frac{N_2 \Phi_{21}}{I_1}$ , where $\Phi_{21}$ is the magnetic flux through coil 2 due to current $I_1$ in coil 1, and $N_2$ is the number of turns in coil 2. The key physical idea is simple: mutual inductance measures how effectively a changing current in one coil induces an emf in another. The closer the coils are, and the more their magnetic fields overlap, the larger the mutual inductance.

When two coils are wound directly on each other (like one layer of wire over another on the same core), almost every magnetic field line produced by one coil passes through the other coil. This gives the maximum possible flux linkage — the fraction of flux from one coil that threads the other is nearly 100%. If the coils were separated or placed at an angle, some flux would leak out, reducing the linkage and hence the mutual inductance.

Now let’s examine the statements step by step.

Assertion (A) says mutual inductance is maximum when coils are wound on each other. This is true. The mutual inductance depends on geometry, distance, and orientation. Winding one coil directly over the other gives the smallest possible separation and the best alignment, so the coupling coefficient $k$ (where $M = k\sqrt{L_1 L_2}$ ) approaches 1. That is the maximum possible value for a given pair of coils.

Reason (R) says flux linkage between two coils is maximum when they are wound on each other. This is also true. Flux linkage is the product of the number of turns and the magnetic flux passing through the coil. When coils are wound on each other, the magnetic field lines from one coil almost entirely pass through the other coil’s turns, giving the highest possible flux linkage. …
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Q.A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2^22 and 150 turns (NCN_CNC​) is wound tightly around the centre of the solenoid. If a current of 3.0 A flows through the solenoid, calculate:

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Q.A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm $^2$ and 150 turns ( $N_C$ ) is wound tightly around the centre of the solenoid. If a current of 3.0 A flows through the solenoid, calculate:

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