(a) State Faraday's law of electromagnetic induction. (b) Derive an expression for the self-inductance of an a

Question

Q. (a) State Faraday's law of electromagnetic induction.

(b) Derive an expression for the self-inductance of an air-filled long solenoid of length

l

and cross-sectional area

A

having

N

turns.

50\ \text{cm}

, with one end pivoted, is rotated with angular speed of

60\ \text{rpm}

in a uniform magnetic field of

4.0\ \text{mT}

directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

(OR)

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.

(b) The ratio of the number of turns in the primary to the secondary of an ideal transformer is

1 : 5

. If

5\ \text{kW}

power at

200\ \text{V}

is supplied to the primary, find

(i) current in the primary, and

(ii) output voltage.

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 — Transformer Principle

Transformer Principle: From Intuition to Precision

Imagine you have a water pipe with a narrow section and a wide section. Water flows through the narrow part fast but with low pressure; through the wide part it flows slow but with high pressure. The total amount of water (flow × pressure) stays the same. A transformer does something similar — but for electricity.

A transformer takes AC power at one voltage and current, and delivers nearly the same power at a different voltage and current. If voltage goes up, current must come down, and vice versa. The total power (voltage × current) is almost unchanged — minus a tiny loss.

The Core Idea: Mutual Induction

Two coils of wire are placed near each other, usually wound around a common iron core. When AC flows through the first coil (the primary), it creates a changing magnetic field. That changing field passes through the second coil (the secondary) and induces a voltage across it. This is mutual induction — a changing current in one coil induces a voltage in a neighbouring coil.

The iron core is crucial: it guides the magnetic field from one coil to the other with very little leakage, making the transfer efficient.

Note

A transformer works only with AC. A steady DC current produces a constant magnetic field, which induces nothing in the secondary coil. Change is essential.

The Precise Statement

For an ideal transformer (no energy losses), the relationship between primary and secondary voltages and currents is:

\frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \text{and} \quad \frac{I_s}{I_p} = \frac{N_p}{N_s}

where:

$V_p$ , $V_s$ = primary and secondary voltages
$I_p$ , $I_s$ = primary and secondary currents
$N_p$ , $N_s$ = number of turns in primary and secondary coils

V_p I_p = V_s I_s

What This Means

If the secondary has more turns than the primary ( $N_s > N_p$ ), the secondary voltage is higher — this is a step-up transformer. Current in the secondary is correspondingly lower.

If the secondary has fewer turns ( $N_s < N_p$ ), the secondary voltage is lower — a step-down transformer. Current in the secondary is higher.

Watch out

A step-up transformer raises voltage but lowers current. It does not create energy. The product $V \times I$ stays constant (ignoring losses). Many beginners think a step-up transformer "amplifies" power — it does not.

Why the Turns Ratio Works

The voltage induced in each turn of a coil is the same (because the same changing magnetic flux links every turn). So the total induced voltage is proportional to the number of turns:

V_p \propto N_p, \quad V_s \propto N_s

Dividing gives the ratio. For current, conservation of power forces the inverse relationship.

A Real Transformer: Small Losses …

Start your 14-day free trial to unlock the full solution →

(a) Faraday: $\varepsilon=-\frac{d\Phi_B}{dt}$ ; solenoid self-inductance $L=\frac{\mu_0N^2A}{l}$ ; the rotating rod develops $\varepsilon=\frac12B\omega r^2=\pi\times10^{-3}\approx3.14\times10^{-3}$ V.

(b) A step-up transformer works on mutual induction with $\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$ ; for the given data $I_p=25$ A and $V_s=1000$ V.

Labelled diagram of a step-up transformer: a primary coil of Np turns and a secondary coil of Ns turns (Ns > Np) wound on a common soft-iron core, the primary connected to an AC source and the secondary connected to the load, so that Vs > Vp.

Part (a)

(a) Faraday's law of electromagnetic induction

The magnitude of the induced emf in a circuit equals the rate of change of magnetic flux linkage through it:

\varepsilon=-\frac{d\Phi_B}{dt}\qquad(\text{or }\varepsilon=-N\frac{d\Phi_B}{dt}\text{ for }N\text{ turns}).

The negative sign (Lenz's law) shows the induced emf opposes the change that produces it.

(b) Self-inductance of a long air-cored solenoid

For a solenoid of length $l$ , area $A$ , $N$ turns carrying current $I$ , the (uniform) interior field is

B=\mu_0\frac{N}{l}I.

Flux through one turn: $\Phi=BA=\mu_0\frac{N}{l}IA$ . Total flux linkage:

N\Phi=\frac{\mu_0N^2A}{l}I.

Since $N\Phi=LI$ ,

L=\frac{\mu_0N^2A}{l},

which depends only on the geometry and turn number.

(c) Induced emf in the rotating rod

A rod pivoted at one end sweeps out area as it turns; an element at distance $x$ moves with speed $\omega x$ , contributing $d\varepsilon=B(\omega x)\,dx$ . Integrating over the length $r$ :

\varepsilon=\int_0^r B\omega x\,dx=\frac12B\omega r^2.

With $r=0.50$ m, $\omega=60\ \text{rpm}=2\pi$ rad/s, $B=4.0\times10^{-3}$ T: …

Start your 14-day free trial to unlock the full solution →

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 63 on this concept.

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. …
Start your 14-day free trial to see the full solution →
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 $90^\circ$ . …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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:

$\Phi=BA\cos\theta$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. An ideal transformer has 500 turns in the primary and 5000 turns in the secondary. If the primary be connected to a 6 V battery, then the secondary voltage is$
(a) 0
(b) 0.6 V
(c) 60 V
(d) 6 V
›Reveal solutionSolution

A transformer needs a changing current/flux to work. A DC battery gives a constant current, so once steady state is reached the secondary voltage is 0.

A transformer works on the principle of mutual induction: the emf induced in the secondary is

$e_s = -M\dfrac{di_p}{dt}$

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Why cannot a transformer be used to step up direct current (D.C.)?$
›Reveal solutionSolution

No changing flux, no induced EMF — a transformer needs AC to work at all.

A transformer operates on the principle of mutual induction: a time-varying current in the primary coil produces a time-varying magnetic flux in the core, which links the secondary coil and induces an EMF in it, given by $\varepsilon_2 = -N_2\dfrac{d\Phi}{dt}$ . With a constant DC current in the primary, the flux in the core, once established, remains steady (constant) — its rate of change $d\Phi/dt$ is zero in the steady state. Since the induced EMF depends entirely on this rate of change, no EMF (and hence no stepped-up voltage) is induced in the secondary for steady DC, so a t …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. Transformer cores are usually laminated. This is to reduce energy loss due to$
(a) flux leakage
(b) winding resistance
(c) eddy currents
(d) hysteresis
›Reveal solutionSolution

(c) eddy currents. The changing flux in the core induces circulating (eddy) currents in the solid metal, which dissipate energy as heat ( $\propto$ resistance path). Laminating the core with thin i …

Start your 14-day free trial to see the full solution →

Q.(a) State Faraday's law of electromagnetic induction.

Part (a)

(a) Faraday's law of electromagnetic induction

(b) Self-inductance of a long air-cored solenoid

(c) Induced emf in the rotating rod

🎓Unlock everything free for 14 days

Unlock everything free for 14 days