Electromagnetic Induction: Class 12 Physics · Chapter 6

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

105 min
3 simulations
16 practice questions
29 MCQs

What you will learn in Electromagnetic Induction

Describe how the experiments of Faraday and Henry show that only a changing magnetic flux induces an emf
Calculate the magnetic flux through a surface and the emf produced by its rate of change using Faraday's law
Apply Lenz's law to find the direction of induced currents and explain it through conservation of energy
Derive and use the motional emf of a sliding rod and of a rod rotating about one end
Define mutual and self-inductance, derive them for long solenoids and find the energy stored in an inductor
Explain how an ac generator produces a sinusoidal emf and calculate its peak value and frequency

Original companion text aligned to NCERT Physics Class 12, Chapter 6, pp. 154–176. Not affiliated with or endorsed by NCERT.

6.1 Introduction

Tap a metro card on a gate reader, set a phone down on a wireless charger, or watch a bicycle dynamo lamp brighten as you pedal harder. Each of these works because a changing magnetic field can push charges around a circuit. By 1820 Oersted had shown one half of the link between electricity and magnetism: a current deflects a nearby compass needle. The obvious next question was the reverse one (Fig. 6.1). If currents make magnetic fields, can magnetic fields make currents?

3/3Pushing the magnet into the coil lights the lamp: the converse effect found in 1831.

Fig. 6.1Electricity and magnetism are linked both ways: a current makes a magnetic field, and a changing magnetic field makes a current.

Two panels side by side. The left panel, labelled 1820, shows a copper wire running north to south over a compass. The right panel, labelled 1831, shows a coil joined to a small lamp, with a bar magnet lying beside the coil. Step 1: no current flows in the wire, so the compass needle points north; the magnet is at rest and the lamp is dark. Step 2: a current flows north along the wire, shown by an arrow, and the compass needle swings round towards the west. Step 3: the needle stays turned, and now the magnet is pushed into the coil, shown by an arrow, and the lamp glows.

The answer came in 1831–32, independently from Michael Faraday in London and Joseph Henry in Albany, New York, and it had a twist. A steady field does nothing; a field that changes through a closed coil drives a current around it. This process is called electromagnetic induction (वैद्युतचुंबकीय प्रेरण). It is at work in almost every power station, transformer, induction cooktop and electric-guitar pickup in use today.

In this chapter we make the idea precise. We define magnetic flux, state Faraday's law for the size of the induced emf and Lenz's law for its direction, and see how both follow the conservation of energy. We then study conductors moving through fields, the inductance of coils and the energy they store, and finally the ac generator that supplies the mains in your home.

Timeline

From a twitching compass to the power grid

1820
Oersted: a current turns a compass
A compass needle swung whenever a nearby wire carried a current, showing that currents create magnetic fields.
1831
Faraday's iron ring
With two coils wound on one iron ring, a brief current appeared in one coil only at the moments the current in the other was switched on or off.
1831–32
Henry's coils
Working with powerful electromagnets in Albany, Joseph Henry observed induction independently and discovered self-induction.
1834
Lenz's rule
Emil Lenz stated the rule that fixes the direction of every induced current.
1880s
Practical ac machines
Nikola Tesla and others developed alternating-current generators and motors that made long-distance power supply possible.
1902
Hydroelectricity at Shivanasamudra
One of the first large hydroelectric stations in Asia began sending power from the Kaveri river in Karnataka to the Kolar gold fields.

Warm-up

1 of 2

What did Oersted observe in 1820?

A moving magnet drives a current around a coilA magnet held still beside a coil keeps a current flowing in itA current-carrying wire deflects a nearby compass needle

Show answer

Answer (C) A current-carrying wire deflects a nearby compass needle

Oersted found that currents create magnetic fields, which turn a compass needle. The converse effect, a moving magnet driving a current, was Faraday's and Henry's discovery ten years later.

6.2 The experiments of Faraday and Henry

Three simple experiments contain the whole story. In each, a coil is joined to a sensitive centre-zero galvanometer, whose needle swings left or right depending on the direction of the current. Watch what makes the needle move — and, just as important, what does not.Every coil in this chapter is wound from insulated wire, so the current follows the turns instead of short-circuiting across them.

Experiment 6.1NCERT Experiment 6.1 · p. 155

A magnet and a coil

You’ll need

a coil of 300–500 turns of insulated copper wire
a strong bar magnet (a neodymium magnet works best)
a centre-zero galvanometer, or a digital multimeter on its lowest current range
two connecting leads

Predict first

You push the N pole of the magnet into the coil and then hold it still inside. What does the needle do while the magnet is held still?

It stays deflected as long as the magnet is insideIt swings to the other side and stays thereIt returns to zero and stays there

Connect the coil to the galvanometer and lay it on the bench with its axis horizontal.
Push the N pole of the magnet quickly into the coil and watch the needle.
Hold the magnet still inside the coil for a few seconds.
Pull the magnet out, first slowly and then quickly.
Repeat with the S pole leading. Finally, keep the magnet fixed and move the coil onto it.

Table 6.1What did the needle do?

What you do	Needle: left, right or none?	Big or small swing?
N pole pushed in
Magnet held still inside
N pole pulled out slowly
N pole pulled out quickly
S pole pushed in
Coil moved onto the still magnet

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 6.3

2/5The N pole moves towards the coil: the needle swings to one side.

Fig. 6.2The needle moves only while the magnet moves relative to the coil.NCERT Fig. 6.1 · p. 155

A horizontal coil on the right is joined to a centre-zero galvanometer below it, and a bar magnet lies on the same axis to the left with its N pole facing the coil. Step 1: the magnet is far away and at rest, and the needle points to zero. Step 2: the magnet moves towards the coil, shown by an arrow, and the needle swings to the right. Step 3: the magnet rests with its N end inside the coil and the needle is back at zero. Step 4: the magnet moves away from the coil and the needle swings to the left. Step 5: the magnet is pushed in fast, with speed lines behind it, and the needle swings much further to the right.

SimulationMagnet and coil

A bar magnet that you can drag towards, into and away from a coil joined to a centre-zero galvanometer. Field lines and a flux readout show how much field threads the coil, and the needle swings only while that flux is changing.

Fig. 6.3Drag the magnet and watch the flux readout and the galvanometer together.

Try this

Hold the magnet still inside the coil. What do the flux readout and the needle show?
Pull the magnet out slowly, then quickly. Which gives the bigger swing, and does the total change in flux differ?
Raise the number of turns and repeat the same push. How does the swing change?

Experiment 6.2NCERT Experiment 6.2 · p. 155

A coil in place of the magnet

You’ll need

the coil and galvanometer from Experiment 6.1
a second, smaller coil that fits beside or inside the first
one or two 1.5 V cells

Predict first

The small coil carries a steady current from the cell. You hold it still beside the big coil. What does the galvanometer show?

No deflection at allA steady deflectionA deflection that slowly grows

Connect the small coil to the cell so that a steady current flows in it.
Move the small coil towards the big coil, then away from it, and watch the needle.
Hold the small coil fixed and move the big coil instead.

Experiment 6.3NCERT Experiment 6.3 · p. 156

Nothing moves, yet the needle kicks

You’ll need

two coils placed end to end on a common axis
a cell and a tapping key (push switch) for the first coil
the galvanometer for the second coil
a soft-iron rod long enough to pass through both coils

Predict first

Nothing moves in this set-up. When you press the key, what will the needle in the second coil do?

Nothing, because nothing movesKick briefly and return to zeroDeflect and stay deflected while the key is held

Connect coil P to the cell through the key, and coil S to the galvanometer. The coils must not touch.
Press the key and watch the needle. Keep the key pressed for a few seconds.
Release the key and watch again.
Slide the soft-iron rod through both coils and repeat.

Table 6.2Needle readings

Moment	What the needle does
Key just pressed
Key held down
Key just released
Key pressed with the iron rod in place

Your entries are saved automatically.

2/5The current in P grows from zero, and S gets a brief kick.

Fig. 6.4Coil S responds only while the current in coil P is changing.NCERT Fig. 6.3 · p. 156

Coil P on the left is joined to a cell and a key; coil S on the right is joined to a galvanometer. Below them, a blue graph shows the current in P against time and an orange graph shows the galvanometer reading. Step 1: the key is open, both graphs are flat at zero and the needle is at zero. Step 2: the key is pressed, the current in P rises steeply and the reading of G shows a short upward spike; the needle swings right. Step 3: the key is held, the current in P is flat and the reading of G is zero. Step 4: the key is released, the current in P drops to zero and G shows a short downward spike; the needle swings left. Step 5: an iron rod passes through both coils and the spikes in the reading of G are much taller.

What do the three experiments share? In the first two, a magnet or a current-carrying coil moves, so the field at the coil changes as the source comes nearer or goes farther. In the third, the source stays put but its strength changes when the key is pressed or released. The needle never responds to a field that is simply there, however strong; it responds only to a field that is changing through the coil. Section 6.3 turns ‘the field through the coil’ into a quantity we can measure.

6.3 Magnetic flux

Hold a hoop out in steady rain. If the rain falls straight down and the hoop is level, it catches the most drops each second. Tilt the hoop and it catches fewer; hold it upright and almost none pass through. The catch depends on how hard it is raining, on the size of the hoop and on its tilt. Magnetic flux (चुंबकीय फ्लक्स) measures how much magnetic field passes through a surface in exactly this three-part way.

Definition

Magnetic flux

For a flat surface of area $A$ in a uniform field $𝐁$ , the magnetic flux is $Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ$ , where $θ$ is the angle between $𝐁$ and the normal to the surface.

ExampleA loop of area $0.02 m^{2}$ facing a $0.5 T$ field squarely has $Φ_{B} = 0.01 W b$ .

Here $𝐀$ is the area vector of the surface: its magnitude is the area and it points along the normal, just as in the definition of electric flux in Chapter 1. Fig. 6.5 shows three cases. With the loop facing the field squarely ( $θ = 0$ ) the flux has its largest value, $B A$ . Tilted so that $θ = 60^{\circ}$ , it catches half as much. Edge-on ( $θ = 90^{\circ}$ ) no field line passes through, and the flux is zero.

2/3 $θ = 60^{\circ}$ : half as many lines cross, $Φ_{B} = \frac{1}{2} B A$ .

Fig. 6.5The flux depends on the angle between

𝐁

and the loop's area vector.NCERT Fig. 6.4 · p. 157

Ten horizontal field lines run from left to right. A flat circular loop, drawn edge-on as a thin green ellipse, sits in the middle with a red arrow for its area vector. Step 1: the loop faces the field, its area vector points along the field, and eight highlighted lines cross it. Step 2: the loop is tilted so that the area vector makes 60 degrees with the field; an angle arc is marked and only four lines cross. Step 3: the loop lies along the field, its area vector points straight up at 90 degrees, and no field lines cross it.

Magnetic flux (uniform field, flat surface)

Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ

(6.1)

where

$θ$: the angle between $𝐁$ and the normal to the surface

If the field varies from place to place, or the surface is curved, divide the surface into small patches $d 𝐀_{i}$ , each small enough that the field $𝐁_{i}$ over it is effectively uniform. The total flux is the sum of the contributions of all the patches:

Φ_{B} = 𝐁_{1} \cdot d 𝐀_{1} + 𝐁_{2} \cdot d 𝐀_{2} + \dots = \sum_{all} 𝐁_{i} \cdot d 𝐀_{i}

(6.2)

Flux is a scalar, and it can be negative: when the field crosses a surface against the chosen normal, $\cos θ < 0$ . Its SI unit is the weber (Wb), with $1 W b = 1 T m^{2}$ .

Quick check

1 of 3

A flat loop of area

0.020 m^{2}

lies in a uniform field of

0.50 T

. The field makes an angle of

60^{\circ}

with the normal to the loop. What is the flux through the loop, in webers?

Show answer

Answer 0.005 Wb

Φ_{B} = 0.50 \times 0.020 \times \cos 60^{\circ} = 5.0 \times 10^{- 3} W b

. Using

\sin 60^{\circ}

by mistake gives

8.7 \times 10^{- 3} W b

6.4 Faraday's law of induction

Faraday's great step was to see that all three experiments of Section 6.2 obey one rule once they are described in terms of flux. In the first two, the source of the field moves, so the flux through the coil joined to the meter changes. The experiment with the key needs no motion at all. The current in coil P cannot jump from zero to its final value; for a fraction of a second after the key is pressed it builds up, and so does the flux that P sends through S. That brief build-up is the kick. Once the current settles, the flux through S is constant and the meter reads zero. Opening the key lets the flux collapse, a change in the opposite sense, so the kick is reversed.

LawFaraday's law of electromagnetic induction

Whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The magnitude of the emf equals the rate at which the flux changes.

Faraday's law

ε = - \frac{d Φ_{B}}{d t}

(6.3)

The minus sign tells us the direction of the emf, and hence of the current in a closed circuit; Section 6.5 explains it. In a tightly wound coil of $N$ turns, every turn links the same flux, and the emfs of the turns add up like cells in series:

Faraday's law for a coil of N turns

ε = - N \frac{d Φ_{B}}{d t}

(6.4)

The product $N Φ_{B}$ is called the flux linkage of the coil. Winding more turns is therefore the simplest way to get a bigger emf from the same change of field. If the whole circuit has resistance $R$ , the induced current is $I = ε / R$ .

Because $Φ_{B} = B A \cos θ$ , there are three independent ways to change it, and each one induces an emf. You can change the field strength $B$ . You can change the area $A$ enclosed by the circuit, by stretching, squeezing or sliding part of it. Or you can change the angle $θ$ by turning the circuit. Fig. 6.6 shows each in turn. Faraday's experiments use the first way, motional emf in Section 6.6 uses the second, and the ac generator in Section 6.8 uses the third.

2/4Doubling $B$ doubles the flux to $2 Φ_{0}$ . While $B$ is changing, an emf is induced.

Fig. 6.6Three ways to change the flux: change

B

, change

A

or change

θ

. An emf is induced only while the change is happening.

A panel of crosses shows a magnetic field pointing into the page, with a green rectangular loop in it and a vertical gauge on the right showing the flux through the loop. Step 1: the loop as it is; the gauge is half full, marked Phi nought. Step 2: extra crosses appear, the field is doubled and the gauge fills to two Phi nought. Step 3: the field is back to normal and the loop has shrunk to half its area; the gauge drops to half of Phi nought. Step 4: the loop is turned by 60 degrees about a vertical dashed axis, so it looks half as wide, and the gauge again shows half of Phi nought.

Example 6.1NCERT Example 6.1 · p. 158

Making the kick bigger

You repeat Experiment 6.2, moving a small current-carrying coil towards a large coil joined to a galvanometer, but the needle barely twitches. (a) Suggest three changes that would give a clearly larger deflection. (b) The galvanometer is taken away for repair. How could you still show that a current is induced, and even which way it flows?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.2NCERT Example 6.2 · p. 158

A loop in a fading field

A rectangular wire loop measuring $20 c m \times 15 c m$ has a resistance of $0.80 Ω$ . A uniform field of $0.25 T$ makes an angle of $60^{\circ}$ with the normal to the loop. The field is reduced steadily to zero in $0.50 s$ . Find the induced emf and current.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.3NCERT Example 6.3 · p. 159

Flipping a coil in the Earth's field

A flat circular coil of 600 turns and radius $12 c m$ has a resistance of $5.0 Ω$ . It stands upright with its face towards the north, so the horizontal component of the Earth's field, $B_{H} = 3.6 \times 10^{- 5} T$ , passes straight through it. The coil is turned through $180^{\circ}$ about a vertical diameter in $0.30 s$ . Estimate the average emf and current induced in it.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A 50-turn coil sits in a field that grows steadily, so that the flux through each turn rises from

2.0 m W b

8.0 m W b

0.10 s

. What is the size of the induced emf, in volts?

Show answer

Answer 3 V

∣ ε ∣ = N Δ Φ_{B} / Δ t = 50 \times (6.0 \times 10^{- 3}) / 0.10 = 3.0 V

. Leaving out the factor

N

gives the emf of one turn,

0.060 V

6.5 Lenz's law and conservation of energy

Faraday's law gives the size of the induced emf. Its sign — the direction in which an induced current flows — follows from a rule stated in 1834 by the Baltic German physicist Emil Lenz.

LawLenz's law

An induced current always circulates in the sense whose own magnetic field works against the change of flux that gave rise to it.

Notice that Lenz's law opposes the change, not the flux itself. If the flux through a loop is growing, the induced current makes a field against the growing field, trying to hold the flux down. If the flux is shrinking, the induced current makes a field along the existing one, trying to prop the flux up. The minus sign in (6.3) is the mathematical form of this rule.

A handy way to apply it uses the poles of a current loop. Look at one face of a loop: if the current circulates anticlockwise as you see it, that face is a north pole; if clockwise, a south pole. In Fig. 6.7, the N pole of a magnet approaches a ring. The flux through the ring grows, so the near face of the ring must become a north pole and push the magnet back: seen from the magnet, the current is anticlockwise. Pull the magnet away and the near face becomes a south pole, pulling the magnet back: the current is now clockwise.

1/3The flux grows, the near face becomes N, and the approaching magnet is pushed back.

Fig. 6.7The induced current always acts to oppose the motion that causes it.NCERT Fig. 6.6 · p. 160

A copper ring stands on the left and a bar magnet lies on the same axis to the right with its N pole facing the ring; below the ring, a face-on view shows the current direction as seen from the magnet. Step 1: the magnet moves in; the ring face near the magnet is labelled N and the far face S, a green arrow shows the magnet pushed back, and the face-on view shows an anticlockwise current. Step 2: the magnet moves out; the near face is now S, the green arrow shows the magnet pulled back, and the current is clockwise. Step 3: an impossible case where the magnet moves in but the near face is S; the force arrow points towards the ring, the magnet speeds up, and a cross marks this as energy from nothing.

Why opposition and not help? Run the thought experiment the other way round. If an approaching N pole met a south face, the ring would pull the magnet in. The magnet would arrive faster, the flux would change faster, the induced current would grow, and the pull would get stronger still: a runaway loop. The magnet's kinetic energy and the heat in the ring would both keep rising while nobody supplied any energy. No such free-energy machine has ever worked, because energy is conserved. The only direction of induced current consistent with that law is the one that resists the change.

So in a real ring, moving the magnet in or out always takes effort. The energy you spend reappears first as the energy of the induced current and then as heat in the ring's resistance. Seen this way, Lenz's law is simply energy conservation written for induced currents.

SimulationLenz's law

A magnet moves along the axis of a coil while arrows on the turns show the direction of the induced current and labels show the poles it creates on the coil. Moving the magnet in or out shows the induced current always opposing the change in flux.

Fig. 6.8The arrows show the induced current; compare them with the direction in which the magnet moves.

Try this

Move the N pole towards the coil. Looking from the magnet's side, is the current clockwise or anticlockwise, and which pole faces the magnet?
Now move the magnet away. Predict the new direction of the arrows before you look.
Move the magnet very slowly and then very quickly. Does the direction change, or only the size of the current?

Try this

A magnet that refuses to fall

You’ll need

a strong neodymium magnet
a copper or aluminium pipe about 50 cm long, a little wider than the magnet
a steel bolt or iron nail of similar size

Predict first

You drop the nail and then the magnet down the vertical pipe. Which takes longer to reach the bottom?

The nailBoth take the same timeThe magnet

Hold the pipe upright over a cushion or a folded cloth.
Drop the nail through the pipe and time its fall.
Drop the magnet through the pipe and time its fall.
If you can, look down the pipe while the magnet is falling.

Can’t do it now? Try the simulation · Fig. 6.8

Fig. 6.9 follows a rectangular loop as it slides at constant speed through a region of uniform field. The flux through the loop changes only while the loop is crossing an edge of the region, so only then is an emf induced. On the way in, the flux grows and the current is anticlockwise. Wholly inside, the flux is steady and nothing is induced. On the way out, the flux falls and the current turns clockwise. Because a straight edge crosses the boundary at a steady rate, the emf stays constant while it lasts, a result we derive in Section 6.6.

2/4The flux grows at a steady rate, so a constant emf drives an anticlockwise current.

Fig. 6.9An emf appears only while the loop crosses an edge of the field region.NCERT Fig. 6.8 · p. 162

A dashed square region of crosses shows a field into the page. A green rectangular loop slides from left to right through it, and two graphs below plot the flux through the loop and the induced emf against the position of the loop, with a dashed guide line joining the loop's front edge to both graphs. Step 1: the loop is to the left of the region, and markers on both graphs sit at zero. Step 2: the loop is half inside; the flux graph is rising, the emf graph shows a constant block below the axis labelled anticlockwise, and a curved arrow in the loop turns anticlockwise. Step 3: the loop is wholly inside; the flux is flat at its largest value and the emf is zero. Step 4: the loop is half out on the right; the flux is falling, the emf is a block above the axis labelled clockwise, and the arrow in the loop turns clockwise.

Example 6.4NCERT Example 6.4 · p. 161

Loops crossing the edge of a field

A uniform magnetic field points out of the page inside the dashed square. Four conducting loops move in the directions of the arrows. For each loop, decide whether a current is induced and, if so, whether it is clockwise or anticlockwise as you see it.

Fig. 6.10NCERT Fig. 6.7 · p. 161

A dashed square filled with dots marks a field pointing out of the page. Loop a, a rectangle, is half inside the left edge and moving right into the field. Loop b, a hexagon, sticks out through the top edge and is moving upward out of the field. Loop c, a circle, lies wholly inside the field and moves to the right. Loop d, a rectangle, is half inside the right edge and moving left into the field.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.5NCERT Example 6.5 · p. 161

Four quick judgements

(a) A closed wire loop lies at rest inside the very strong, steady field of a hospital MRI magnet. Is a current induced in it?
(b) A student claims that sliding a closed loop through the steady electric field between two charged plates will drive a current round it, ‘just as a magnet would’. Is she right while the loop is completely between the plates? While it is halfway out?
(c) Two identical square loops leave a region of uniform magnetic field at the same constant speed, with the field perpendicular to both. Loop 1 moves with one side parallel to the edge of the field; loop 2 is turned through $45^{\circ}$ , so it leaves corner first. For which loop is the induced emf constant while it leaves?
(d) A circular loop lies in the plane of this page with a small capacitor inserted in its top side: plate P is on the left and plate Q on the right. The N pole of a bar magnet, held in front of the page and pointing at it, is pushed towards the loop. Which plate becomes positive?

Try it yourself first. Then reveal the solution one step at a time.

6.6 Motional electromotive force

So far the field has been changing. Now keep the field uniform and steady, and move part of the circuit through it instead. In Fig. 6.11, a metal rod of length $l$ slides without friction along two parallel rails that are joined at the left end through a resistor $R$ . A uniform field $𝐁$ points into the page, perpendicular to the plane of the rails. Rod, rails and resistor form a closed loop whose area grows as the rod moves to the right with speed $v$ .

2/4As the rod moves right the enclosed area grows, and a current $I$ flows anticlockwise.

Fig. 6.11Sliding the rod changes the area of the circuit, and hence the flux through it.NCERT Fig. 6.10 · p. 162

Two horizontal rails are joined at the left end through a resistor R, and a copper rod of length l lies across them, in a field shown by crosses pointing into the page. The area enclosed by the circuit is shaded green. Step 1: the rod is at distance x from the left end and the flux is B l x. Step 2: the rod has slid further right, the shaded area is larger, and arrows show an anticlockwise current I, flowing up the rod. Step 3: plus signs gather at the top of the rod and minus signs at the bottom, with a green arrow showing the force q v B on positive charges pointing up the rod. Step 4: a purple arrow shows the magnetic force I l B on the rod pointing left, against the motion, and a black arrow labelled your push points right.

DerivationMotional emf from Faraday's law

Let $x$ be the distance of the rod from the closed end. The loop encloses area $l x$ , so the flux through it is

Φ_{B} = B l x

Only $x$ changes, at the rate $d x / d t = v$ . By (6.3), the size of the induced emf is

Motional emf

∣ ε ∣ = ∣ \frac{d Φ_{B}}{d t} ∣ = B l \frac{d x}{d t} = B l v

(6.5)

Lenz's law fixes the direction. The growing into-page flux is opposed by an anticlockwise current, which flows up the rod. The top of the rod acts like the positive terminal of a cell of emf $B l v$ .

An emf produced by moving a conductor through a steady field is called a motional emf (गतिक विद्युत वाहक बल). We can reach the same result without mentioning flux at all, by looking at the charges inside the moving rod.

DerivationThe same result from the Lorentz force

Every free charge $q$ in the rod travels to the right with the rod at speed $v$ . The magnetic force on it, $q 𝐯 \times 𝐁$ , has magnitude $q v B$ and points along the rod, towards the top end for a positive charge. The force is the same at every point along the rod.

Carrying the charge from the bottom of the rod to the top, this force does work $W = q v B l$ . Since emf is work done per unit charge,

ε = \frac{W}{q} = \frac{q v B l}{q} = B l v

This matches (6.5). Once a current flows, each charge also drifts along the rod, and that drift gives it a second magnetic force, directed across the rod. Added up over the whole rod, this is the force $I l B$ on the current that we meet next.

The force picture also settles the energy account. The induced current $I$ flows up the rod, which lies in the field, so the rod feels a magnetic force of size $I l B$ . That force points to the left, against the motion, just as Lenz's law requires. To keep the rod moving at constant speed you must push with an equal force, delivering power $I l B v$ . Since $I = B l v / R$ , this power equals $I^{2} R$ , the rate at which the resistor heats up.

Example

Power in a sliding-rod generator

A rod $0.50 m$ long slides at a steady $4.0 m / s$ along rails in a uniform field of $0.30 T$ perpendicular to the rails. The total resistance of the circuit is $1.2 Ω$ . Find (a) the induced emf, (b) the current, (c) the force needed to keep the rod moving steadily, and (d) the power delivered by that force. Compare (d) with the rate of heating.

Try it yourself first. Then reveal the solution one step at a time.

Now turn the situation around: hold the conductor still and change the field, as in Experiment 6.3. A charge at rest feels no magnetic force, because the full Lorentz force on a charge is

𝐅 = q (𝐄 + 𝐯 \times 𝐁) = q 𝐄 when 𝐯 = 0

(6.6)

So whatever drives the induced current in a stationary coil must be an electric field: a magnetic field that changes with time produces an electric field around it. This induced electric field is not like the electrostatic field of Chapter 1. Its field lines form closed loops instead of starting and ending on charges. It also does net work on a charge carried once around a loop, and that work per unit charge is the induced emf. Oersted had shown that moving charges push on magnets; Faraday showed the reverse, that a changing magnetic field pushes on charges that are standing still.

A conductor need not move in a straight line. In Fig. 6.12, a metal rod of length $R$ turns with angular speed $ω$ about one end, like the hand of a clock, in a uniform field parallel to the axis of rotation. Its free end slides around a circular metal ring. Different parts of the rod now move at different speeds: a point at distance $r$ from the axis moves at $v = ω r$ , so the outer parts contribute most of the emf.

3/3In time $t$ the rod sweeps a sector of area $\frac{1}{2} R^{2} θ$ with $θ = ω t$ , which gives $ε = \frac{1}{2} B ω R^{2}$ .

Fig. 6.12A rod turning about one end in a uniform axial field.NCERT Fig. 6.11 · p. 164

A metal ring encloses a grid of dots showing a field out of the page. A copper rod runs from the centre O to the ring and turns anticlockwise, shown by a curved arrow labelled omega. Step 1: the rod points to the right and its length R is labelled. Step 2: the rod has turned 40 degrees and three orange velocity arrows along it grow longer with distance from O, labelled v equals omega r. Step 3: the rod has turned 70 degrees and the sector it has swept is shaded, with its area one half R squared theta and the emf one half B omega R squared.

Example 6.6NCERT Example 6.6 · p. 163

A spinning rod

A metal rod $0.80 m$ long is pivoted at the centre of a metal ring of radius $0.80 m$ and turns at 20 revolutions per second in the plane of the ring. A uniform field of $0.60 T$ acts parallel to the axis of rotation. Find the emf between the centre and the ring (a) by adding up the motional emfs along the rod and (b) from the rate at which the rod sweeps out area.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.7NCERT Example 6.7 · p. 165

The spokes of a bicycle wheel

A cyclist rides due east, so the front wheel turns in a vertical east–west plane. The horizontal component of the Earth's field, $B_{H} = 3.5 \times 10^{- 5} T$ , points north, at right angles to that plane. The wheel has 32 metal spokes, each $0.30 m$ long, and turns at 180 revolutions per minute. What emf appears between the axle and the rim?

Try it yourself first. Then reveal the solution one step at a time.

In the real world

Beyond the syllabus: eddy currents

Induction does not need a wire loop. When a solid piece of metal moves through a magnetic field, or sits in a changing one, currents swirl around inside it like eddies in a stream. These eddy currents (भँवर धाराएँ) obey Lenz's law, so they oppose the motion or the change that causes them.

1/2Large eddy currents circulate in the solid plate and stop it within a few swings.

Fig. 6.13Eddy currents brake a solid plate strongly; slots break them up.

A metal plate hangs from a pivot as a pendulum and swings through a magnet pole face, shown as a dashed square of crosses on the plate. Dashed outlines show how far the plate swings. Step 1: the plate is solid, two large orange eddy-current loops circulate in it in opposite senses, and the dashed outlines show only a small swing, labelled stops quickly. Step 2: the plate has vertical slots, the eddy currents are broken into small narrow loops between the slots, and the dashed outlines show a wide swing, labelled keeps swinging.

The drag they produce is put to work in the electromagnetic brakes of some trains and in the damping that makes a galvanometer needle settle quickly. Induction cooktops heat a steel pan with eddy currents driven by a rapidly changing field, and metal detectors sense the eddy currents set up in hidden metal. Where eddy currents only waste energy, as in the iron cores of transformers and motors, the core is built from thin insulated sheets called laminations, which break up the current loops just as the slots do in the pendulum.

Quick check

1 of 3

A straight rod

0.40 m

long moves at

5.0 m / s

perpendicular both to its length and to a uniform field of

0.25 T

. What emf is induced between its ends, in volts?

Show answer

Answer 0.5 V

ε = B l v = 0.25 \times 0.40 \times 5.0 = 0.50 V

6.7 Inductance

Every coil that carries a current sends magnetic flux through its own turns and through any coil near it. If the coils keep their shape and position, doubling the current doubles the field everywhere, and so doubles each of these fluxes: $Φ_{B} \propto I$ . For a tightly wound coil of $N$ turns, all threaded by the same flux, the quantity that enters Faraday's law is the flux linkage $N Φ_{B}$ , so we can write

N Φ_{B} \propto I ⟹ N Φ_{B} = (constant) \times I

The constant of proportionality is the inductance (प्रेरकत्व). It is fixed by geometry and materials alone: the size, shape, number of turns and spacing of the coils, and whatever fills their core. Capacitance works the same way, since a parallel-plate capacitor's value is set by its plates, their gap and the dielectric between them. Inductance is a scalar. Its SI unit is the henry (H), where $1 H = 1 W b / A = 1 V s / A$ , and its dimensions are $[M L^{2} T^{- 2} A^{- 2}]$ . The flux may come from a neighbouring coil or from the coil's own current; we take these two cases in turn.

6.7.1 Mutual inductance

Take two long coaxial solenoids of the same length $l$ , one inside the other (Fig. 6.14). The inner solenoid $S_{1}$ has radius $r_{1}$ , $n_{1}$ turns per unit length and $N_{1} = n_{1} l$ turns in all; the outer solenoid $S_{2}$ has radius $r_{2}$ , $n_{2}$ turns per unit length and $N_{2}$ turns. A current $I_{2}$ in $S_{2}$ sends a flux $Φ_{1}$ through each turn of $S_{1}$ . We define the mutual inductance (अन्योन्य प्रेरकत्व) $M_{12}$ of $S_{1}$ with respect to $S_{2}$ by

N_{1} Φ_{1} = M_{12} I_{2}

(6.7)

2/3A current in $S_{2}$ fills all of $S_{2}$ with a uniform field $B_{2}$ , but only the area $π r_{1}^{2}$ counts for the flux through $S_{1}$ .

Fig. 6.14Whichever coil carries the current, the flux linking the other coil passes through the area

π r_{1}^{2}

.NCERT Fig. 6.12 · p. 166

A side view shows a long outer solenoid S2, drawn as two rows of blue wire cross-sections, and a narrower inner solenoid S1, drawn as two rows of orange wire cross-sections, both of length l. On the right, an end view shows two concentric circles. Step 1: the radii r1 and r2 and the length l are labelled and no current flows. Step 2: current flows in the outer coil, blue field arrows fill the whole inside of S2, dots fill both circles of the end view, and the inner disc of area pi r1 squared is shaded. Step 3: current flows in the inner coil only, orange field arrows appear only inside S1, and dots appear only inside the inner circle of the end view.

DerivationMutual inductance of two long coaxial solenoids

Ignore the edge effects at the ends, a good approximation when $l ≫ r_{2}$ . A current $I_{2}$ in $S_{2}$ makes a uniform field $B_{2} = μ_{0} n_{2} I_{2}$ inside it. Each turn of $S_{1}$ has area $π r_{1}^{2}$ , so the flux linkage with $S_{1}$ is

N_{1} Φ_{1} = (n_{1} l) (π r_{1}^{2}) (μ_{0} n_{2} I_{2}) = μ_{0} n_{1} n_{2} π r_{1}^{2} l I_{2}

(6.8)

Comparing with (6.7),

M_{12} = μ_{0} n_{1} n_{2} π r_{1}^{2} l

(6.9)

Now reverse the roles. Let a current $I_{1}$ flow in $S_{1}$ and define $M_{21}$ by

N_{2} Φ_{2} = M_{21} I_{1}

(6.10)

The inner solenoid's field $B_{1} = μ_{0} n_{1} I_{1}$ exists only inside $S_{1}$ . Every turn of $S_{2}$ surrounds $S_{1}$ , but field passes through it only over the area $π r_{1}^{2}$ , so

N_{2} Φ_{2} = (n_{2} l) (π r_{1}^{2}) (μ_{0} n_{1} I_{1}) ⟹ M_{21} = μ_{0} n_{1} n_{2} π r_{1}^{2} l

(6.11)

The two results are identical:

M_{12} = M_{21} = M

(6.12)

We proved $M_{12} = M_{21}$ only for two long coaxial solenoids, but it holds for any pair of circuits, whatever their shapes, and that makes it a powerful short cut. Example 6.8 uses it. The field of a large coil at a small coil near its centre is easy to write down, whereas the field that the small coil spreads through the large one varies from point to point and would be very hard to add up. Because the two mutual inductances are equal, working out the easy one gives the hard one for free.

If the space inside the solenoids is filled with a material of relative permeability $μ_{r}$ , such as soft iron, the field and hence the mutual inductance are multiplied by $μ_{r}$ : $M = μ_{r} μ_{0} n_{1} n_{2} π r_{1}^{2} l$ . For any pair of coils, $M$ also depends on their separation and orientation. Coils close together on a common axis link much more flux than coils far apart or set at right angles.

Now return to Experiment 6.3. With $N_{1} Φ_{1} = M I_{2}$ and a current $I_{2}$ that varies with time, Faraday's law gives the emf induced in coil 1:

ε_{1} = - \frac{d (N_{1} Φ_{1})}{d t} = - M \frac{d I_{2}}{d t}

A changing current in one coil induces an emf in its neighbour, in proportion to the mutual inductance and to how fast the current changes. Transformers, wireless phone chargers and the charging stands of electric toothbrushes all rely on this.

Example 6.8NCERT Example 6.8 · p. 167

A small coil at the centre of a large one

A small circular coil of radius $r_{1} = 1.0 c m$ and $N_{1} = 50$ turns lies at the centre of a large circular coil of radius $r_{2} = 20 c m$ and $N_{2} = 200$ turns. The coils are coaxial and lie in the same plane. (a) Obtain an expression for their mutual inductance, and evaluate it. (b) Find the emf induced in the small coil when the current in the large coil changes at $50 A / s$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check: mutual inductance

1 of 3

A changing current in coil A induces an emf of

2.0 m V

in a nearby coil B. If the same changing current is sent through coil B instead, the emf induced in coil A is

2.0 m V

, the same as beforelarger, if A has more turns than Bsmaller, if A is the inner coilzero

Show answer

Answer (A) $2.0 m V$ , the same as before

The mutual inductance is the same both ways,

M_{12} = M_{21}

, so the same rate of change of current gives the same emf. The numbers of turns are already built into

M

and cannot make the two directions differ.

6.7.2 Self-inductance

A coil does not need a neighbour to show induction. The current in a coil sends flux through its own turns, so when that current changes, the coil's own flux linkage changes and an emf appears in the coil itself. This is self-induction (स्वप्रेरण). The flux linkage is proportional to the coil's own current:

N Φ_{B} = L I

(6.13)

We call the constant $L$ the self-inductance, or coefficient of self-induction, of the coil. Faraday's law then gives

Self-induced emf

ε = - \frac{d (N Φ_{B})}{d t} = - L \frac{d I}{d t}

(6.14)

The minus sign is Lenz's law once more. Try to raise the current and the coil pushes back against the rise; try to cut the current and the coil pushes forward, trying to keep it flowing (Fig. 6.15). Because it always fights the change, the self-induced emf is called a back emf.

1/3While $I$ increases, the self-induced emf pushes against it.

Fig. 6.15The self-induced emf

- L d I / d t

fights every change in the current, up or down.

A coil drawn as a row of loops carries a current I shown by a blue arrow; below it an orange arrow shows the self-induced emf. Under the coil are two graphs against time: the current ramps up, stays level, then ramps down, and the emf is a negative block, then zero, then a positive block. Step 1: the rising part of the current is highlighted and the orange emf arrow points against the current, labelled emf opposes. Step 2: the level part is highlighted, there is no emf arrow, and the label reads no induced emf. Step 3: the falling part is highlighted and the orange emf arrow points along the current, labelled emf aids.

DerivationSelf-inductance of a long solenoid

A solenoid of length $l$ and cross-sectional area $A$ has $n$ turns per unit length, so $n l$ turns in all. Ignoring end effects, a current $I$ produces a uniform field $B = μ_{0} n I$ inside it. The flux linkage is

N Φ_{B} = (n l) (μ_{0} n I) (A) = μ_{0} n^{2} A l I

L = \frac{N Φ_{B}}{I} = μ_{0} n^{2} A l

(6.15)

If the solenoid is wound on a core of relative permeability $μ_{r}$ , the field for a given current grows by the factor $μ_{r}$ , and so does $L$ :

L = μ_{r} μ_{0} n^{2} A l

(6.16)

As with mutual inductance, $L$ is set by the size and shape of the coil and by what fills its core, not by the current it carries. Because $L \propto n^{2}$ , winding twice as many turns on the same former makes $L$ four times as large, and a soft-iron core, with $μ_{r}$ in the hundreds or thousands, raises it enormously.

Example

How big is a henry?

A solenoid $40 c m$ long has 1000 turns, each of cross-sectional area $5.0 c m^{2}$ , with air inside. (a) Find its self-inductance. (b) A current of $2.0 A$ in it is switched off and falls to zero in $5.0 m s$ . What average emf is induced? (c) Repeat (b) for the same solenoid with a soft-iron core of $μ_{r} = 800$ .

Try it yourself first. Then reveal the solution one step at a time.

The back emf gives a coil a kind of electrical inertia. A loaded trolley is hard to start and hard to stop; a coil with a large $L$ makes the current in it hard to start and hard to stop. To build up a current, the source must keep doing work against the back emf. That work is not lost. It is stored in the magnetic field of the coil, much as the work done in speeding up a trolley is stored as its kinetic energy.

DerivationEnergy stored in an inductor

Suppose that at some instant the current in a coil of inductance $L$ is $I$ and is rising at the rate $d I / d t$ . Ignoring resistance, the source must drive the current against the back emf of size $L d I / d t$ , so it does work at the rate

\frac{d W}{d t} = ∣ ε ∣ I = L I \frac{d I}{d t}

The total work needed to raise the current from zero to $I$ is

Energy stored in an inductor

W = \int_{0}^{I} L I^{'} d I^{'} = \frac{1}{2} L I^{2}

(6.17)

This energy is given back when the current falls to zero. The formula mirrors $\frac{1}{2} m v^{2}$ for kinetic energy, with $L$ playing the part of $m$ and $I$ the part of $v$ .

Compare

Mass and inductance

Mechanics

Mass $m$ resists changes in velocity
Velocity $v$
Force $F = m d v / d t$
Kinetic energy $\frac{1}{2} m v^{2}$

Circuits

Inductance $L$ resists changes in current
Current $I$
Back emf of size $L d I / d t$
Magnetic energy $\frac{1}{2} L I^{2}$

When two nearby coils both carry changing currents, the flux linkage of coil 1 has two parts, one from its own current and one from its neighbour's: $N_{1} Φ_{1} = L_{1} I_{1} + M_{12} I_{2}$ . The emf induced in coil 1 is then

ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}

Example 6.9NCERT Example 6.9 · p. 169

Energy in a magnetic field

(a) Express the energy stored in a long solenoid using only its field $B$ and its dimensions, and deduce the energy per cubic metre of field. (b) An MRI magnet produces a field of $1.5 T$ over a volume of about $0.20 m^{3}$ . Estimate the energy it stores. (c) What electric field would store energy at the same density, and could air withstand it?

Try it yourself first. Then reveal the solution one step at a time.

Energy per unit volume of a magnetic field

u_{B} = \frac{B^{2}}{2 μ_{0}}

(6.18)

The solenoid was only a convenient place to derive (6.18). Wherever there is a magnetic field, in the gap of a magnet, around a power cable or in a beam of light, it stores $B^{2} / 2 μ_{0}$ joules in every cubic metre. The electric counterpart, $\frac{1}{2} ε_{0} E^{2}$ , is just as general, and both grow as the square of the field strength.

6.8 AC generator

Nearly all the electrical energy you use begins in a generator, a machine that converts mechanical energy into electrical energy by electromagnetic induction. It uses the third of our three ways of changing flux: a coil turns in a magnetic field, so the angle between the field and the coil's area vector keeps changing. Practical alternating-current machines owe a great deal to the Serbian-American inventor Nikola Tesla, and a single modern generator can deliver hundreds of megawatts.

Fig. 6.16 builds up a simple ac generator (प्रत्यावर्ती धारा जनित्र). A coil of $N$ turns, called the armature, is mounted on a shaft between the poles of a magnet, with the axis of rotation perpendicular to the field. A turbine or engine turns the shaft. Each end of the coil is fixed to its own metal slip ring, which turns with the shaft. A pair of stationary carbon brushes presses on the rings and carries the current out to the load.

4/4Wires from the brushes carry the alternating current to a load, such as a lamp.

Fig. 6.16The main parts of a simple ac generator.NCERT Fig. 6.13 · p. 170

An N pole block at the top and an S pole block at the bottom with field lines running down between them. Step 1: only the magnet and its field are shown. Step 2: a rectangular armature coil appears on a horizontal shaft between the poles, with a curved arrow at the end of the shaft showing that it rotates. Step 3: the two ends of the coil run along the shaft to two slip rings to the right of the magnet, the nearer ring joined to one end and the farther ring to the other, with a small brush pressing on each ring from below. Step 4: wires run from the two brushes to a lamp on the right, which glows.

DerivationThe emf of a rotating coil

Let the coil have area $A$ and turn at a constant angular speed $ω$ , with its area vector along $𝐁$ at $t = 0$ . The angle between them at time $t$ is $θ = ω t$ , so the flux through each turn is

Φ_{B} = B A \cos θ = B A \cos ω t

By Faraday's law, the emf of the $N$ -turn coil is

ε = - N \frac{d Φ_{B}}{d t} = - N B A \frac{d}{d t} (\cos ω t)

ε = N B A ω \sin ω t

(6.19)

The largest value, reached whenever $\sin ω t = \pm 1$ , is the peak emf $ε_{0} = N B A ω$ , so that

ε = ε_{0} \sin ω t, ε_{0} = N B A ω

(6.20)

With $ω = 2 π ν$ , where $ν$ is the number of revolutions per second,

ε = ε_{0} \sin 2 π ν t

(6.21)

2/5The coil is edge-on: the flux is zero but changing fastest, so the emf reaches its peak $+ ε_{0}$ . Current flows into the page in the round side.

Fig. 6.17The emf is largest when the flux is zero, and zero when the flux is largest.NCERT Fig. 6.14 · p. 171

On the left, an end view shows a coil, drawn as a copper bar with a round wire cross-section at one end and a square one at the other, turning clockwise between an N pole above and an S pole below, with a green arrow for its area vector. On the right, a graph plots the emf as an orange sine curve and the flux as a dashed blue cosine curve against the angle from 0 to 360 degrees, with dots marking the current values. Step 1: at 0 degrees the coil is level, facing the field; the flux dot is at its maximum and the emf dot is at zero. Step 2: at 90 degrees the coil is upright with the round side at the top; the flux is zero, the emf dot is at its positive peak, and a cross in the round side and a dot in the square side show the current flowing into and out of the page. Step 3: at 180 degrees the coil is level again, the flux is at its negative maximum and the emf is zero. Step 4: at 270 degrees the coil is upright the other way, with the round side at the bottom now marked with a dot, so the current in the coil has reversed, and the emf is at its negative peak. Step 5: at 360 degrees the coil is back where it started and one full cycle of the emf is complete.

Fig. 6.17 brings out a point that often surprises learners: the emf is largest when the flux through the coil is zero ( $θ = 90^{\circ}$ and $270^{\circ}$ ), and zero when the flux is largest. Faraday's law cares about the slope of the flux curve, and a cosine is steepest where it crosses zero. Every half turn the emf changes sign, so the current it drives through the load keeps reversing. Such a current is an alternating current (ac); the next chapter develops the tools for handling it, including its average values.

SimulationAC generator

A rectangular coil turns between two magnet poles at a speed you can change. A live graph plots the induced emf against time together with the flux through the coil, so you can see where the emf peaks, where it passes through zero and where it changes sign.

Fig. 6.18Change the speed and watch both the height and the spacing of the emf peaks.

Try this

Watch the moment the coil faces the poles squarely, when the flux is largest. What is the emf at that instant?
Double the rpm. What happens to the peak emf, and to the number of cycles each second?
Halve the number of turns at the original speed. Which feature of the graph changes, and which stays the same?

Power-station generators follow the same principle, with engineering refinements. In large machines it is the electromagnets that rotate, while the heavy output coils stay still; this avoids carrying huge currents through sliding contacts. What differs from one station to another is how the rotor is turned. Kudankulam in Tamil Nadu raises steam with nuclear fuel, most Indian thermal stations raise it by burning coal, Bhakra Nangal lets falling water spin its turbines, and a wind turbine uses moving air. Generators in India run in step with the grid at $50 H z$ , while some countries, including the USA, use $60 H z$ .

Example 6.10NCERT Example 6.10 · p. 172

A micro-hydro generator

A micro-hydro plant beside a mountain stream drives a generator whose armature has 150 turns, each of area $0.040 m^{2}$ . The armature turns at 25 revolutions per second in a uniform field of $0.20 T$ . (a) Find the peak emf. (b) What is the frequency of the ac, and how fast would the coil have to turn to supply ac at the Indian mains frequency?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A generator coil of 200 turns, each of area

0.030 m^{2}

, turns at 50 revolutions per second in a field of

0.15 T

. What is its peak emf, in volts?

Show answer

Answer 282.74 V

ε_{0} = N B A ω = 200 \times 0.15 \times 0.030 \times (2 π \times 50) \approx 283 V

. Using 50 in place of

ω = 2 π \times 50

gives a value

2 π

times too small.

Table 6.3Quantities met in this chapterNCERT p. 174

Quantity	Symbol	SI unit	Dimensions	Key relation
Magnetic flux	$Φ_{B}$	weber (Wb)	$[M L^{2} T^{- 2} A^{- 1}]$	$Φ_{B} = 𝐁 \cdot 𝐀$
Induced emf	$ε$	volt (V)	$[M L^{2} T^{- 3} A^{- 1}]$	$ε = - N d Φ_{B} / d t$
Mutual inductance	$M$	henry (H)	$[M L^{2} T^{- 2} A^{- 2}]$	$ε_{1} = - M d I_{2} / d t$
Self-inductance	$L$	henry (H)	$[M L^{2} T^{- 2} A^{- 2}]$	$ε = - L d I / d t$
Magnetic energy density	$u_{B}$	$J / m^{3}$	$[M L^{- 1} T^{- 2}]$	$u_{B} = B^{2} / 2 μ_{0}$

Flash cards · 8

Quick recall

1 / 8

Recap

Electromagnetic Induction: key points and summary

Faraday and Henry found that a current is induced in a closed coil only while the magnetic field through it changes, whether by motion or by switching (Section 6.2).
The magnetic flux through a flat surface is $Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ$ , a scalar measured in webers (Section 6.3).
Faraday's law: the emf induced in an $N$ -turn coil is $ε = - N d Φ_{B} / d t$ , and a closed circuit of resistance $R$ carries a current $ε / R$ (Section 6.4).
Flux can be changed by changing $B$ , the area $A$ or the angle $θ$ , and each way induces an emf (Fig. 6.6).
Lenz's law: the induced current opposes the change in flux that produces it, as the conservation of energy demands (Section 6.5).
A rod of length $l$ moving with speed $v$ at right angles to a uniform field has a motional emf $B l v$ ; a rod of length $R$ turning about one end gives $\frac{1}{2} B ω R^{2}$ (Section 6.6).
A magnetic field that changes with time creates an electric field, which drives the induced current in a stationary conductor ((6.6)).
Inductance is flux linkage per unit current, $N Φ_{B} / I$ . It depends only on geometry and core material and is measured in henries (Section 6.7).
Mutual inductance: $ε_{1} = - M d I_{2} / d t$ with $M_{12} = M_{21}$ ; for coaxial solenoids $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ (Section 6.7.1).
Self-inductance: $ε = - L d I / d t$ opposes every change of current; for a long solenoid $L = μ_{r} μ_{0} n^{2} A l$ (Section 6.7.2).
A coil carrying a current $I$ stores energy $\frac{1}{2} L I^{2}$ ; any magnetic field stores $B^{2} / 2 μ_{0}$ per unit volume ((6.18)).
An ac generator turns a coil in a magnetic field to give $ε = N B A ω \sin ω t$ , converting mechanical energy into electrical energy (Section 6.8).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercises 6.1	Practice Q1
NCERT Exercises 6.2	Practice Q2
NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5	Practice Q5
NCERT Exercises 6.6	Practice Q6
NCERT Exercises 6.7	Practice Q7
NCERT Exercises 6.8	Practice Q8

Quick check: MCQs on Electromagnetic Induction

Warm-up2 questions
6.2 The experiments of Faraday and Henry4 questions
6.3 Magnetic flux3 questions
6.4 Faraday's law of induction3 questions
6.5 Lenz's law and conservation of energy4 questions
6.6 Motional electromotive force3 questions
Quick check: mutual inductance3 questions
Quick check: self-inductance4 questions
6.8 AC generator3 questions

Glossary · 14

Key terms (English–Hindi)

ac generatorप्रत्यावर्ती धारा जनित्रpratyavarti dhara janitra: A machine that turns a coil in a magnetic field, converting mechanical energy into an alternating emf.First seen in the chapter →

back emf: The self-induced emf in a coil, which opposes any change in the current through it.First seen in the chapter →

eddy currentsभँवर धाराएँbhanvar dharaen: Loops of induced current that circulate inside a solid piece of metal when the flux through it changes.First seen in the chapter →
electromagnetic inductionवैद्युतचुंबकीय प्रेरणvaidyut-chumbakiya preran: The production of an emf in a circuit whenever the magnetic flux through it changes.First seen in the chapter →

Faraday's lawफैराडे का नियमFaraday ka niyam: The emf induced in a circuit equals minus the rate of change of the flux linked with it, $ε = - N d Φ_{B} / d t$ .First seen in the chapter →
flux linkage: The total flux through all the turns of a coil; $N Φ_{B}$ when each of its $N$ turns links the same flux $Φ_{B}$ .First seen in the chapter →

henryहेनरीhenri: The SI unit of inductance: $1 H = 1 W b / A = 1 V s / A$ .First seen in the chapter →

inductanceप्रेरकत्वpreraktva: Flux linkage per unit current, fixed by the geometry of the coils and the material of their core.First seen in the chapter →

Lenz's lawलेंज़ का नियमLenz ka niyam: An induced current circulates so that its own field works against the change of flux that caused it.First seen in the chapter →

magnetic fluxचुंबकीय फ्लक्सchumbakiya flux: A measure of the magnetic field passing through a surface; $Φ_{B} = B A \cos θ$ for a flat surface in a uniform field.First seen in the chapter →
motional emfगतिक विद्युत वाहक बलgatik vidyut vahak bal: The emf of a conductor moving through a magnetic field; $B l v$ for a rod moving at right angles to $𝐁$ .First seen in the chapter →
mutual inductanceअन्योन्य प्रेरकत्वanyonya preraktva: Flux linkage with one coil per unit current in a neighbouring coil; it gives $ε_{1} = - M d I_{2} / d t$ .First seen in the chapter →

self-inductanceस्वप्रेरकत्वsvapreraktva: Flux linkage with a coil per unit current in the same coil; it gives $ε = - L d I / d t$ .First seen in the chapter →

weberवेबरvebar: The SI unit of magnetic flux: $1 W b = 1 T m^{2}$ .First seen in the chapter →

Electromagnetic Induction: Class 12 Physics · Chapter 6

What you will learn in Electromagnetic Induction

6.1 Introduction

6.2 The experiments of Faraday and Henry

6.3 Magnetic flux

6.4 Faraday's law of induction

6.5 Lenz's law and conservation of energy

6.6 Motional electromotive force

6.7 Inductance

6.7.1 Mutual inductance

6.7.2 Self-inductance

6.8 AC generator

Electromagnetic Induction: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Electromagnetic Induction

Key terms (English–Hindi)

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