(i) Define mutual inductance of a pair of coils. Write its SI unit. (ii) A long solenoid of radius $R$ and length $L$ has $n$ turns per unit length. A circular loop of radius $r\,(<R)$ is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop. OR Two long straight parallel conductors $A$ and $B$ carrying steady currents $I_a$ and $I_b$ in the same direction are separated by a distance $d$. Deduce the expressions for the force acting on length $L$ of conductor $B$ due to conductor $A$ and show it in a figure. Write the expression for the force acting on length $L$ of conductor $A$ due to conductor $B$ and show that it follows Newton's third law.

(i) Define mutual inductance of a pair of coils. Write its SI unit. (ii) A long solenoid of radius R and lengt

Question

Q. (i) Define mutual inductance of a pair of coils. Write its SI unit.

(ii) A long solenoid of radius

R

and length

L

has

n

turns per unit length. A circular loop of radius

r\,(<R)

is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.

(OR)

A

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Part (a)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) …

Part (b)Concept understanding — Force Between Parallel Wires

Force Between Parallel Current-Carrying Wires

Imagine two long, straight wires placed side by side, each carrying an electric current. You already know that a current-carrying wire creates a magnetic field around it. And you know that a wire placed in a magnetic field experiences a magnetic force. So here, each wire sits inside the magnetic field created by the other wire. That is the whole story — each wire feels a force because of the other wire's magnetic field.

The direction of that force — attraction or repulsion — depends on whether the currents flow in the same direction or opposite directions.

The Intuition

Take two wires with currents in the same direction. Use the right-hand thumb rule: for wire 1, the magnetic field lines circle around it. At the location of wire 2, that field points in a particular direction. Now apply the right-hand rule for force on a current-carrying wire (Fleming's left-hand rule works too): the current in wire 2, crossed with the field from wire 1, gives a force toward wire 1. The same reasoning from wire 2's perspective gives a force on wire 1 toward wire 2. So they attract.

If the currents are opposite, the field directions reverse, and the forces point away from each other — they repel.

Tip

A quick memory aid: Same direction → Attract; Opposite direction → Repel. This is the opposite of what you might guess from electric charges, where like charges repel. Don't mix them up.

The Precise Statement

For two long, straight, parallel wires separated by a distance $d$ , carrying steady currents $I_1$ and $I_2$ , the magnitude of the force per unit length on either wire is:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

where $\mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2$ is the permeability of free space.

The force is attractive if the currents are in the same direction, repulsive if they are opposite.

Where Does This Formula Come From?

Wire 1 produces a magnetic field at the location of wire 2. The magnitude of that field is:

B_1 = \frac{\mu_0 I_1}{2 \pi d}

This field is perpendicular to wire 2. The magnetic force on a length $L$ of wire 2 carrying current $I_2$ in a perpendicular field $B_1$ is:

F = I_2 L B_1

Substitute $B_1$ :

F = I_2 L \cdot \frac{\mu_0 I_1}{2 \pi d}

Divide both sides by $L$ to get force per unit length:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

That is the entire derivation — two simple steps: field from one wire, then force on the other.

Important

This formula assumes the wires are infinitely long (or at least very long compared to $d$ ) and thin. It gives the force per unit length, which is constant along the wires.

The Definition of the Ampere

This effect is so fundamental that it defines the SI unit of current. One ampere is defined as the constant current which, when flowing through two infinitely long, straight, parallel wires of negligible cross-section placed one metre apart in vacuum, produces a force of exactly $2 \times 10^{-7}$ newtons per metre of length between them. …

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Loop inside a long solenoid: $M=\mu_0 n\pi r^2$ (henry). Parallel wires: each feels $F=\dfrac{\mu_0 I_a I_b L}{2\pi d}$ , attractive for same-direction currents, equal and opposite (Newton's third law).

Part (a) — mutual inductance of a solenoid and a coaxial loop

(i) Mutual inductance $M$ is the flux linkage produced in one coil per unit current in another: $M=\dfrac{N_2\Phi_{21}}{I_1}$ , with induced emf $\varepsilon_2=-M\dfrac{dI_1}{dt}$ . Its SI unit is the henry (H) ( $=\text{Wb/A}=\text{V·s/A}$ ).

(ii) Inside a long solenoid of $n$ turns per unit length carrying $I_s$ , the field is uniform:

B=\mu_0 n I_s.

A coaxial single-turn loop of radius $r<R$ intercepts this field over its area $\pi r^2$ :

\Phi_{loop}=B\,\pi r^2=\mu_0 n I_s\,\pi r^2.

Therefore

$M=\frac{\Phi_{loop}}{I_s}=\mu_0 n\pi r^2=\frac{\mu_0 N\pi r^2}{L}\quad(n=N/L).$ …

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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 14 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Two long parallel wires each carrying a current of 1 A in the same direction, are placed 1 m apart. The force of attraction between them is$
(a) 2 x 10^7 N/m
(b) 2 x 10^-4 N/m
(c) 2 x 10^-7 N/m
(d) 4 x 10^-7 N/m
›Reveal solutionSolution

Two parallel current-carrying wires attract if their currents are in the same direction; the force per unit length is mu_0I1I2/(2pid).

Each current-carrying wire produces a magnetic field around it, and this field exerts a force on the other current-carrying wire (F = I*L x B). The standard result for the force per unit length between two long straight parallel wires carrying currents I1 and I2, separated by a distance d, is

F/L = mu_0 * I1 * I2 / (2 * pi * d)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): Two infinitely long straight conductors carrying current in the same direction attract each other. Reason (R): The net magnetic field at a point exactly halfway between two infinitely long straight conductors carrying current in the same direction is zero.$
(a) Both Assertion and Reason are true, and reason is the correct explanation
(b) Both Assertion and Reason are true, but the Reason is not the correct explanation
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
›Reveal solutionSolution

Both statements are individually correct, but the field being zero at the midpoint is not why the two wires attract each other.

Checking the Assertion: Two infinitely long straight parallel conductors carrying current in the SAME direction do attract each other. Each wire sits in the magnetic field created by the other wire, and using $\vec F = I\vec L \times \vec B$ (or the right-hand/Fleming's left-hand rule), the force on each wire due to the other's field points towards the other wire. So the Assertion is TRUE.

Checking the Reason: Take the two wires along the $y$ -axis at $x=-a$ and $x=+a$ , both carrying current $I$ in the $+y$ direction. At the midpoint (origin), using $\vec B = \dfrac{\mu_0 I}{2\pi r}\hat\phi$ with $\hat\phi = \hat I \times \hat r$ : the field due to the left wire points in $+\hat y'$ s perpendicular direction (say $+\hat z$ ), while the field due to the right wire (displacement now in $-\hat x$ from that wire) points in the opposite transverse direction ( $-\hat z$ ). Since both wires are equidistant and carry equal current, these two fields are equal in magnitude and opposite in direction — they cancel exactly. So the net field at the midpoint IS zero when the currents …

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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 D1 markMCQ
$Q. Dimensional formula of permeability is (A) [MLT^-2 A^-2] (B) [MLT^2 A^-2] (C) [MLT^2 A^2] (D) [MLT^-2 A]$
›Reveal solutionSolution

Using the force per unit length between two wires, μ₀ works out to dimensions [M L T⁻² A⁻²].

The force per unit length between two parallel current-carrying wires is

$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Solving for μ₀:

$\mu_0 = \frac{2\pi d\,(F/\ell)}{I_1 I_2}$

…

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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 2025Set ANNUAL1 markMCQ
$Q. Assertion: The turns of a spring come close to each other, when current is passed through it. Reason: It is because, the turns of a spring carry current in same direction and hence attract each other.$
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not a correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
›Reveal solutionSolution

Adjacent turns of a current-carrying spring act like parallel wires carrying current in the same direction, which attract each other by the magnetic force between parallel currents — so the coils are pulled together.

Two straight parallel conductors carrying currents in the SAME direction attract each other (force per unit length $F/l = \mu_0 I_1 I_2 / 2\pi d$ , attractive for like-directed currents, repulsive for opposite). A spring is essentially a coil of many closely-spaced turns; each turn carries current in the same sense as its neighbours. Treating adjacent turns as parallel current-carrying wires, they attract each other, so the spring's turns are pulled closer together ( …

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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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CBSE 2024Set 55/1/11 markMCQ
$Q. For question 15, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. Assertion (A) : Two long parallel wires, freely suspended and connected in series to a battery, move apart. Reason (R) : Two wires carrying current in opposite directions repel each other.$
›Reveal solutionSolution
Connected in series, the two freely suspended parallel wires carry equal currents in opposite directions — the current goes out along one wire and returns along the other. Antiparallel currents repel, so the wires move apart. Both statements are true and the Reason is exactly why the wires separate. The correct option is (A).

The physical setup: what does "in series" mean here?

When two long parallel wires hang freely side by side and are joined in series to a battery, there is a single current path: the current leaves the battery, travels along the first wire, crosses over at the far end, and comes back along the second wire to the battery. Because the second wire carries the return current, the two adjacent wires carry equal currents in opposite directions — antiparallel currents.

Force between the wires

Each wire sits in the magnetic field created by the other. For two long parallel wires a distance $d$ apart carrying currents $I_1$ and $I_2$ , the force per unit length on either wire is
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$
with the standard direction rule: parallel (same-direction) currents attract; antiparallel (opposite-direction) currents repel. You can check this with $\vec{F} = I\vec{L} \times \vec{B}$ : for opposite currents, the field of wire 1 at wire 2 gives a force on wire 2 pointing away from wire 1, and by Newton's third law wire 1 is pushed away from wire 2 with equal magnitude.

Evaluating the statements

Assertion (A): "Two long parallel wires, freely suspended and connected in series to a battery, move apart." As shown above, the series connection makes the currents antiparallel, the wires repel, and — being freely suspended — they move apart. True. …
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CBSE 2024Set A1 markMCQ
$Q. The nature of electron beams moving with uniform velocity in the same direction will be (A) converging (B) diverging (C) parallel (D) none of these$
›Reveal solutionSolution
Like charges repel electrostatically; this force exceeds the magnetic attraction at ordinary speeds, so the beams diverge.

Two parallel electron beams experience two effects:

As parallel currents in the same direction, the magnetic force is attractive.

As streams of like (negative) charges, the electrostatic force is repulsive. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two long parallel wires each carrying a current of 1 A in the same direction, are placed 1 m apart. The force of attraction between them is$
(a) 2 x 10^-7 N/m
(b) 2 x 10^-4 N/m
(c) 1 x 10^-7 N/m
(d) 4 x 10^-7 N/m
›Reveal solutionSolution

Two parallel current-carrying wires attract each other (same direction) with a force per unit length given by mu0 I1 I2 / (2pid).

The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ , separated by distance $d$ , is

$\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$

Substituting $\mu_0 = 4\pi\times10^{-7}\ \text{T m/A}$ , $I_1 = I_2 = 1\ \text{A}$ , $d = 1\ \text{m}$ :

$\dfrac{F}{l} = \dfrac{4\pi\times10^{-7}\times 1\times 1}{2\pi\times 1} = 2\times10^{-7}\ \text{N/m}$

…

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CBSE 2023Set ANNUAL1 mark
$Q. Fill in the blank: The force between two parallel current carrying conductors (flowing in the same direction) is __________.$
›Reveal solutionSolution

Two parallel current-carrying conductors carrying current in the same direction attract each other.

Each current-carrying conductor sets up a magnetic field around itself (by the Biot-Savart/Ampere law), and the other conductor, carrying current in that field, experiences a force $F = BIL$ (via $\vec F = I\vec L\times\vec B$ ). Working out the directions with the right-hand rule shows that when the currents flow in the same direction, the force on each conductor points toward the other - i.e. the conductors …

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Q.(i) Define mutual inductance of a pair of coils. Write its SI unit.

Part (a)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) …

Part (b)Concept understanding — Force Between Parallel Wires

Force Between Parallel Current-Carrying Wires

The Intuition

The Precise Statement

Where Does This Formula Come From?

The Definition of the Ampere

Mutual inductance and force between parallel currents

Part (a) — solenoid and coaxial loop

Part (a) — mutual inductance of a solenoid and a coaxial loop

The physical setup: what does "in series" mean here?

Force between the wires

Evaluating the statements

Q.(i) Define mutual inductance of a pair of coils. Write its SI unit.

Part (a) — mutual inductance of a solenoid and a coaxial loop

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