(a) Derive an expression for the induced emf developed when a coil of N turns, and area of cross-section A, is

Question

Figure — Figure — CBSE 2019 55/2/1 Q25 — FigureFigure — CBSE 2019 55/2/1 Q25

N

(b) A wheel with 100 metallic spokes each

0\cdot5

m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth's magnetic field. If the resultant magnetic field at that place is

4\times10^{-4}

T and the angle of dip at the place is

30^\circ

, find the emf induced between the axle and the rim of the wheel.

(OR)

(a) Derive the expression for the magnetic energy stored in an inductor when a current

I

develops in it. Hence, obtain the expression for the magnetic energy density.

(b) A square loop of sides 5 cm carrying a current of

0\cdot2

A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown. Calculate

(i) the resultant magnetic force, and

(ii) the torque, if any, acting on the loop.

CBSECBSE Class XII Board 2019Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Electromagnetic Induction

Electromagnetic Induction

Electromagnetic induction is the phenomenon in which a changing magnetic flux through a circuit produces an electromotive force (emf) — and hence a current, if the circuit is closed. It is the single idea behind generators, transformers, inductors, and the entire AC power grid.

The Central Discovery

Michael Faraday found (1831) that a current is induced in a coil not when a magnet sits still near it, but only while the magnet moves — that is, only while the magnetic flux linked with the coil is changing. A steady magnet, however strong, induces nothing.

Magnetic Flux

The key quantity is magnetic flux $\Phi_B$ through a surface of area $A$ in a field $\vec{B}$ :

\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta

where $\theta$ is the angle between $\vec{B}$ and the area's normal. Its SI unit is the weber (Wb), where $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$ .

Flux can change in three distinct ways, and any of them induces an emf:

the field strength $B$ changes,
the area $A$ of the loop changes,
the orientation $\theta$ changes (a coil rotating in a field — the basis of the generator).

Faraday's Law

The induced emf equals the negative rate of change of flux. For a coil of $N$ turns:

\mathcal{E} = -N\,\frac{d\Phi_B}{dt}

The faster the flux changes, the larger the emf. This is why a magnet dropped quickly through a coil gives a bigger deflection than one moved slowly.

Lenz's Law — the Minus Sign

The negative sign expresses Lenz's law: the induced current flows in the direction that opposes the change producing it. Push a magnet's north pole toward a coil, and the coil's near face becomes a north pole to repel it; pull it away, and the face becomes a south pole to attract it. This is simply energy conservation — you must do work against this opposition, and that work becomes the electrical energy of the induced current.

Motional emf

A special, very useful case: a conducting rod of length $l$ moving with speed $v$ perpendicular to a field $B$ sweeps out area and develops an emf

\mathcal{E} = Blv

Here the emf arises because the free charges in the rod experience a magnetic force $q\vec{v}\times\vec{B}$ , which drives them along the rod. …

Why this formula?

Electromagnetic Induction

Electromagnetic induction is the effect discovered by Faraday: a changing magnetic flux through a circuit drives an induced EMF (and hence a current). The key word is changing — a steady field, however strong, induces nothing.

Magnetic flux

Flux measures how many field lines thread a surface bounded by the loop:

\Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta

It can change three ways: by changing $B$ , by changing the area $A$ , or by rotating the loop (changing $\theta$ ).

Faraday's law

The induced EMF equals the rate of change of flux:

\mathcal{E} = -\frac{d\Phi_B}{dt}

For a coil of $N$ turns, $\mathcal{E} = -N\,\dfrac{d\Phi_B}{dt}$ . The EMF depends on how fast the flux changes, not on the flux itself — a slow change gives a small EMF, a rapid change a large one.

Lenz's law — the minus sign …

Part (b)Concept understanding — Magnetic Energy Density

Magnetic Energy Density

A magnetic field stores energy. Building up a current in an inductor requires work against the induced back-emf, and that work is stored in the field around it. The magnetic energy density is the energy stored per unit volume of the field.

Energy stored in an inductor

When the current in an inductor of self-inductance $L$ grows, the induced back-emf is $\varepsilon = -L\,dI/dt$ . The work done by the source to push charge $dq = I\,dt$ against this emf is

dW = L\,I\,dI

Integrating from $0$ to the final current $I$ ,

U = \int_0^I L\,I\,dI = \tfrac{1}{2}LI^2

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

From inductor to field: the energy density

Take a long solenoid with $n$ turns per unit length, cross-sectional area $A$ and length $l$ . Its self-inductance is $L = \mu_0 n^2 A l$ , and the field inside is $B = \mu_0 n I$ , so $I = B/(\mu_0 n)$ . The stored energy becomes

U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(\mu_0 n^2 A l)\left(\frac{B}{\mu_0 n}\right)^2 = \frac{B^2}{2\mu_0}\,(Al)

Since $Al$ is the volume occupied by the field, the energy per unit volume is

u_B = \frac{U}{Al} = \frac{B^2}{2\mu_0}

u_B = \frac{B^2}{2\mu_0}

Although derived for a solenoid, this result is general: wherever a magnetic field $B$ exists, it carries an energy density $B^2/2\mu_0$ , measured in $\text{J/m}^3$ .

Comparison with the electric field

The electric field stores energy at density $u_E = \tfrac{1}{2}\varepsilon_0 E^2$ . The magnetic analogue $u_B = B^2/2\mu_0$ has the same structure, and together they give the energy carried by electromagnetic waves. …

Why this formula?

Magnetic Energy Density

Building up a magnetic field costs work: as current rises, the induced back-emf (Lenz's law) opposes it, and you must push against that opposition. That work is not lost — it is stored in the magnetic field itself. Magnetic energy density $u_B$ is how much of this energy sits in each cubic metre of the field.

$u_B = \frac{B^2}{2\mu_0} \quad\text{(vacuum)},\qquad u_B = \frac{B^2}{2\mu}\ \text{(medium)}$

Deriving it from a solenoid

The energy stored in an inductor carrying current $I$ is a standard result:

U = \tfrac{1}{2} L I^2

For a long solenoid ( $n$ turns per unit length, area $A$ , length $l$ ):

Field inside: $B = \mu_0 n I$
Inductance: $L = \mu_0 n^2 A l$

Substituting and eliminating $I = B/(\mu_0 n)$ :

U = \tfrac{1}{2}(\mu_0 n^2 A l)\left(\frac{B}{\mu_0 n}\right)^2 = \frac{B^2}{2\mu_0}\,(A l)

Since $Al$ is exactly the volume $V$ where the field lives, the energy per unit volume is:

u_B = \frac{U}{V} = \frac{B^2}{2\mu_0}

Equivalent forms, using $B=\mu H$ , are $u_B = \tfrac{1}{2}\mu H^2 = \tfrac{1}{2}BH$ .

Why it makes sense …

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Part (a): A coil rotating in a field gives $\varepsilon=NBA\omega\sin\omega t$ ; the spoked wheel gives $\varepsilon=\tfrac12 B_H\omega l^2=\sqrt3\,\pi\times10^{-4}\approx5.4\times10^{-4}$ V. Part (b): Magnetic energy density $u=B^2/2\mu_0$ ; the square loop feels a net force $\approx6.7\times10^{-9}$ N toward the wire and zero torque.

Part (a)

Induced emf in a rotating coil

A coil of $N$ turns, each of area $A$ , rotates at constant angular speed $\omega$ in a uniform field $B$ . Taking the angle between $\vec B$ and the coil normal as $\theta=\omega t$ , the flux linkage is

\Phi=NBA\cos\omega t.

By Faraday's law,

\varepsilon=-\frac{d\Phi}{dt}=-NBA\frac{d}{dt}(\cos\omega t)=NBA\omega\sin\omega t.

\varepsilon=NBA\omega\sin\omega t,\qquad \varepsilon_0=NBA\omega.

The emf is sinusoidal — the principle of the AC generator.

Wheel with metallic spokes

Only the component of the Earth's field perpendicular to the plane of rotation induces emf; the wheel rotates in a plane normal to the horizontal component, so

B_H=B\cos\delta=4\times10^{-4}\times\cos30^\circ=4\times10^{-4}\times\frac{\sqrt3}{2}=2\sqrt3\times10^{-4}\ \text{T}.

Angular speed:

\omega=2\pi\times\frac{120}{60}=4\pi\ \text{rad/s}.

Each spoke is a rod of length $l=0.5$ m rotating about one end, giving motional emf $\tfrac12 B_H\omega l^2$ . The 100 spokes share the same two terminals (axle and rim), i.e. they are in parallel, so the net emf equals that of a single spoke:

$\varepsilon=\tfrac12 B_H\omega l^2=\tfrac12(2\sqrt3\times10^{-4})(4\pi)(0.25)=\sqrt3\,\pi\times10^{-4}\ \text{V}\approx5.44\times10^{-4}\ \text{V}.$ …

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

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CBSE 2026Set 55/2/11 markMCQ
$Q. A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top, (A) the induced current will flow in the anticlockwise direction. (B) the induced current will flow in the clockwise direction. (C) no induced current will flow in the solenoid. (D) the magnet will fall with a constant velocity.$
›Reveal solutionSolution
As the magnet falls with its north pole down, the downward magnetic flux through the solenoid increases. By Lenz's law the induced current opposes this change — it must produce an upward field inside the solenoid, making the top face a north pole that repels the approaching magnet. That requires an anticlockwise current as seen from above. The correct option is (A).

Why this approach works

Electromagnetic induction is about change: a current is induced in the solenoid only because the flux through it is changing as the magnet falls. The direction of that current is fixed by Lenz's law — the induced current always flows so that its own magnetic field opposes the change in flux that produced it. This is not an arbitrary rule; it is energy conservation. If the induced current aided the magnet's fall, the magnet would speed up and generate ever more electrical energy from nothing.

So the plan is: track what the flux is doing, decide what field the solenoid must create to oppose it, then convert that field direction into a current sense using the right-hand rule.

Step-by-step reasoning

Set up the situation.

The solenoid stands vertically on the table. The magnet is dropped along its axis from above, north pole downward. The observer looks down from the top.

What is the flux doing?

Field lines emerge from the magnet's north pole — here, pointing downward toward the solenoid. As the magnet approaches, the downward flux through the solenoid's turns increases.

What must the induced current do?

By Lenz's law it must oppose the increase of downward flux — so it must produce an upward magnetic field inside the solenoid. Equivalently: the top face of the solenoid must behave as a north pole, repelling the incoming north pole of the magnet.

Convert the field direction into a current sense.

Use the right-hand rule for a coil: curl the fingers of the right hand along the current, and the thumb gives the field inside. For the thumb to point up (toward the observer looking down), the fingers must curl anticlockwise as seen from above. So the induced current is anticlockwise for the top observer.

Tip

Quick pole check: the solenoid must repel the approaching north pole, so its top face is a north pole. Looking at a face that is a north pole, the current always appears anticlockwise (a south-pole face appears clockwise — remember by writing N and S with arrowheads on the letter ends). Same conclusion. …
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CBSE 2026Set 55/3/11 markMCQ
$50$
›Reveal solutionSolution
The induced emf is found from Faraday’s law: the change in magnetic flux divided by the time taken. The flux changes from maximum to zero as the loop rotates by $90^\circ$ , giving an average emf of $2.5$ V — but the options are in the range $0.25$ – $1.0$ V, so we must check the calculation carefully. The correct value is $2.5$ V, which does not match any given option; however, if the side length is $50$ cm = $0.5$ m, area $= 0.25$ m², flux change $= 3.0 \times 0.25 = 0.75$ Wb, time $= 0.3$ s, emf $= 0.75/0.3 = 2.5$ V. None of the options are correct as stated.

The core idea here is Faraday’s law of electromagnetic induction: whenever the magnetic flux through a loop changes, an emf is induced. The flux depends on three things — the field strength $B$ , the area $A$ of the loop, and the angle $\theta$ between the field and the normal to the loop. When you rotate the loop, you change $\theta$ , and that changes the flux. The induced emf is the rate of change of flux.

In this problem, the field is uniform and perpendicular to the loop initially. That means the initial angle between the field and the normal is $0^\circ$ , so the flux is maximum. After a $90^\circ$ rotation, the plane of the loop is parallel to the field, so the flux becomes zero. The change in flux is simply the initial flux minus zero.

Let’s work it out step by step.

Find the area of the loop.

Side length $= 50$ cm $= 0.5$ m.

Area $A = (0.5)^2 = 0.25$ m².

Initial magnetic flux.

Flux $\Phi = B A \cos\theta$ .

Initially $\theta = 0^\circ$ , so $\cos 0 = 1$ .

$\Phi_i = 3.0 \times 0.25 \times 1 = 0.75$ Wb.

Final magnetic flux.

After $90^\circ$ rotation, $\theta = 90^\circ$ , $\cos 90 = 0$ .

$\Phi_f = 3.0 \times 0.25 \times 0 = 0$ Wb.

Change in flux.

$\Delta\Phi = \Phi_f - \Phi_i = 0 - 0.75 = -0.75$ Wb.

The magnitude of the change is $0.75$ Wb.

Average induced emf.

By Faraday’s law, $|\mathcal{E}| = \left|\frac{\Delta\Phi}{\Delta t}\right|$ .

$\Delta t = 0.3$ s.

$|\mathcal{E}| = \frac{0.75}{0.3} = 2.5$ V.

Watch out

A common mistake is to forget that the side is given in cm and not convert to metres. If you use $50$ cm as $50$ m, you get an absurdly large area and emf. Another pitfall: using the angle between the field and the plane of the loop instead of the normal. Here, the field is perpendicular to the plane initially, so the normal is parallel to the field — that’s $\theta = 0^\circ$ , not $90^\circ$ . …
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CBSE 2026Set V11 markMCQ
$Q. The working principle of an A.C. generator is :$
(a) mutual induction
(b) eddy currents
(c) self induction
(d) electromagnetic induction
›Reveal solutionSolution

(d) electromagnetic induction …

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CBSE 2026Set A1 markMCQ
$Q. An example of natural electromagnetic induction is (A) radio (B) television (C) battery charging (D) lightning strike$
›Reveal solutionSolution

Lightning involves huge, rapidly changing currents/fields that induce emf in nearby conductors — natural electromagnetic induction.

Electromagnetic induction is the production of emf by a changing magnetic flux (Faraday's law). A lightning strike carries an enormous, rapidly varying current, producing a fast-changing magnetic field that induces emf/current in nearby loops and conductors — a natu …

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CBSE 2026Set A1 markMCQ
$Q. If magnetic field is same but the area of the loop is increased, then the flux (A) increases (B) decreases (C) becomes zero (D) remains unchanged$
›Reveal solutionSolution

Magnetic flux Φ = BA cosθ; with B constant, a larger area gives greater flux.

The magnetic flux through a loop is:

$\Phi=BA\cos\theta$

…

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CBSE 2026Set ANNUAL1 mark
$Q. What is electromagnetic induction?$
›Reveal solutionSolution

Any change of magnetic flux through a circuit produces an EMF in that circuit - this is electromagnetic induction.

Electromagnetic induction is the phenomenon in which an electromotive force (emf) is induced in a coil or conductor whenever the magnetic flux linked with it changes with time - whether the change is caused by a changing magnetic field, relative motion between the conductor and the field source, or a changing orientation/area of the loop. If the circuit is closed, this induced emf drives an induced current. It is quantitatively described by Faraday's law, EMF = -d(phi)/dt …

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CBSE 2026Set ANNUAL1 mark
$Q. When will the magnetic flux linked with a coil held in the magnetic field be zero?$
›Reveal solutionSolution

Flux is zero whenever the field lines lie entirely in the plane of the coil.

Magnetic flux linked with a coil is $\Phi = BA\cos\theta$ , where $\theta$ is the angle between the coil's area vector (normal) and the magnetic field $\vec{B}$ . This is zero when $\cos\theta=0$ , i.e. $\theta=90°$ — meaning the normal to the coil is perpendicular to $\vec{B}$ , which is the same as saying the field lines lie entirely within (parallel to) the plane of the coil, pa …

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Magnetic flux. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Magnetic flux unit = weber = Volt × second (from Faraday's law emf = dΦ/dt), option (ii).

Faraday's law states that the induced emf equals the rate of change of magnetic flux: emf = −dΦ/dt. Rearranging, Φ = emf × time (dimensionally). Since emf is in volts and time in seconds, magneti …

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CBSE 2025Set X11 markMCQ
$Q. Consider the following statements : Statement - 1: A.C. Generator works on the principle of electromagnetic induction Statement - 2: In an A.C. Generator, as the armature is rotated in a uniform magnetic field, the magnetic flux linked with the coil changes which induces an emf in the coil. Among the above two statements :$
(a) Both Statements are true
(b) Both Statements are false
(c) Statement-1 is true and Statement-2 is false
(d) Statement-1 is false and Statement-2 is true
›Reveal solutionSolution

(a) Both Statements are true. An A.C. generator works on electromagnetic induction (Statement 1). As the armature coil rotates in a uniform magnetic field, the flux $\phi = NBA\cos\omega t$ linked …

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CBSE 2025Set D1 markMCQ
$Q. Which of the following devices is based on the principle of electromagnetic induction? (A) Voltmeter (B) Electric motor (C) Electric generator (D) Ammeter$
›Reveal solutionSolution

The electric generator is based on electromagnetic induction.

An electric generator rotates a coil in a magnetic field. The continuous change of magnetic flux linked with the coil induces an emf by Faraday's law of electromagnetic induction, converting mechanical energy into electrical energy.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The laws of electromagnetic induction have been used in the construction of :$
(a) voltmeter
(b) ammeter
(c) electric motor
(d) generator
›Reveal solutionSolution

An electric generator converts mechanical energy into electrical energy by rotating a coil in a magnetic field, directly applying Faraday's law of electromagnetic induction.

Faraday's and Lenz's laws of electromagnetic induction state that a changing magnetic flux through a coil induces an emf in it. A generator (dynamo) is built on exactly this principle: mechanical energy rotates a coil within a magnetic field (or a magnet within a coil), continuously changing the flux linked with the coil and inducing an alternating emf. A voltmeter and ammeter are measuring instruments based on the magnetic effect of current, and an el …

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CBSE 2024Set 55/2/11 markMCQ
$\vec{B} = (1{\cdot}0\,\hat{i} + 0{\cdot}5\,\hat{j})$
›Reveal solutionSolution
Magnetic flux is the dot product of the field and the area vector; here $\Phi = \vec{B} \cdot \vec{A} = \vec{B} \cdot (A\,\hat{n})$ , which gives $31.4\,\mu\text{Wb}$ .

Why the dot product?

Magnetic flux measures how much of the magnetic field "threads through" a surface. Not all field lines contribute equally: only the component of $\vec{B}$ perpendicular to the surface matters. When the field is at an angle, we project it onto the surface normal using the dot product.

The flux through a flat surface is
$\Phi = \vec{B} \cdot \vec{A}$
where $\vec{A} = A\,\hat{n}$ is the area vector—magnitude $A$ (the area) pointing along the outward normal $\hat{n}$ .

Step-by-step calculation

Find the area of the coil. The coil is circular with radius $r = 10\,\text{cm} = 0.1\,\text{m}$ .

$A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\,\text{m}^2$

Write the area vector. The outward normal is $\hat{n} = 0.6\,\hat{i} + 0.8\,\hat{j}$ (already a unit vector since $0.6^2 + 0.8^2 = 1$ ), so

$\vec{A} = A\,\hat{n} = 0.01\pi\,(0.6\,\hat{i} + 0.8\,\hat{j})\,\text{m}^2$

Express the magnetic field in SI units.

Given $\vec{B} = (1.0\,\hat{i} + 0.5\,\hat{j})\,\text{mT} = (1.0\,\hat{i} + 0.5\,\hat{j}) \times 10^{-3}\,\text{T}$ .

Compute the dot product $\vec{B} \cdot \vec{A}$ .

$\Phi = \vec{B} \cdot \vec{A} = (1.0\,\hat{i} + 0.5\,\hat{j}) \times 10^{-3} \cdot 0.01\pi\,(0.6\,\hat{i} + 0.8\,\hat{j})$
The dot product of the unit vectors: …
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Q.(a) Derive an expression for the induced emf developed when a coil of NNN turns, and area of cross-section AAA, is rotated at a constant angular speed ω\omegaω in a uniform magnetic field BBB.

Part (a)

Induced emf in a rotating coil

Wheel with metallic spokes

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Q.(a) Derive an expression for the induced emf developed when a coil of $N$ turns, and area of cross-section $A$ , is rotated at a constant angular speed $\omega$ in a uniform magnetic field $B$ .

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