Magnetism and Matter: Class 12 Physics · Chapter 5

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

Published 28 Sep 2026 · Updated 28 Sep 2026

110 min
2 simulations
17 practice questions
33 MCQs

What you will learn in Magnetism and Matter

Describe the field of a bar magnet with field lines and use their four properties to judge whether a field-line picture is possible
Explain with Ampère's circulating currents why a bar magnet behaves like a solenoid, and derive the far axial field of a finite solenoid
Calculate the torque and potential energy of a magnetic dipole in a uniform field and identify stable and unstable equilibrium
Find the axial and equatorial fields of a short magnet from the electric-dipole analogy
State Gauss's law for magnetism and connect it with the absence of magnetic monopoles
Define magnetisation, magnetic intensity, susceptibility and permeability, and calculate H, M and B inside a material
Classify materials as diamagnetic, paramagnetic or ferromagnetic and explain each at the atomic level, including domains and the Curie temperature

Original companion text aligned to NCERT Physics Class 12, Chapter 5, pp. 136–153. Not affiliated with or endorsed by NCERT.

5.1 Introduction

Open a map app on your phone and turn slowly on the spot: the little arrow on the screen turns with you. A sensor chip a few millimetres across is measuring the Earth's magnetic field. Magnetism reaches far beyond such gadgets. Electrons and atomic nuclei behave as tiny magnets, the Earth and the Sun carry magnetic fields, and even the thin gas between the stars of a galaxy is threaded by weak fields spread over thousands of light years.

People met magnetism long before they could explain it. Greek writers described a black iron ore, now called magnetite or lodestone, that tugs at iron; the district of Magnesia, where it was dug up, gave us the word magnet. Chapter 4 traced every magnetic field you met there to charges in motion. Here we turn to magnets themselves, and to the way each kind of matter answers when a magnetic field is applied (चुंबकत्व एवं द्रव्य).

Earlier classes gave you a handful of rules about magnets. Here they are, grouped under four headings; the rest of the chapter explains where each comes from.

Direction. Hang a bar magnet from a thread and it swings until it lies roughly north–south. The end that ends up facing geographic north is named the magnet's north pole (N) and the other end its south pole (S). A magnet does this because the Earth itself is magnetised: near the ground its field runs roughly from the geographic south towards the geographic north.
Forces. Two N poles repel, two S poles repel, and an N pole and an S pole attract: like poles repel, unlike poles attract.
Pairs. No one has ever produced a lone N or S pole, a magnetic monopole. However you break a magnet, every fragment has both poles.
Materials. Permanent magnets are made from iron and a few metals like it, cobalt and nickel, or from their alloys and compounds; most other substances respond to a magnet only feebly.

3/4Each piece is again a complete magnet with its own N and S.

Fig. 5.1Cutting a magnet never separates its poles: every piece, down to a single atom, is a complete magnet with an N and an S.

A horizontal bar magnet coloured blue at the S end and orange at the N end, with short strokes of iron filings at both ends. Step 1: the whole magnet with its two poles. Step 2: it is cut in the middle; the two halves move apart and each shows its own S half and N half, with filings clustering at all four ends. Step 3: each half is cut again, giving four small magnets, each labelled S then N. Step 4: a magnifier below the last piece shows a single atom as an electron orbit with an arrow for its magnetic moment pointing along the magnet, with S on its left and N on its right.

Test the first two rules in the simulation below. Magnet A hangs from a thread, so it can swing freely; magnet B can be dragged, turned and flipped. The grey arrow marks geographic north.

SimulationA hung magnet and a hand-held magnet

A stage with the Earth's field switched on and a North arrow in the corner. Magnet A hangs on a thread at the left and swings until it settles; magnet B, a weaker bar magnet, stands upright on the right with its N end at the bottom and can be dragged, turned or flipped. Arrows show whether the two magnets attract or repel, and field lines can be switched on.

Fig. 5.2Magnet A turns freely on its thread; drag, turn or flip magnet B to see how A responds.

Try this

Press Play, leave B where it is and watch A settle; it takes a few seconds. Which way does A's N end point, compared with the North arrow?
Drag B towards A until B's N end is close beside A's N end. Which way does A swing, and which of A's ends ends up facing B?
Press Flip B and bring B back to the same spot. What is different now?

Timeline

From lodestone to superconductors

c. 600 BCE
Lodestone described
Greek writers record that lodestone attracts iron.
1088
Needle compass
The Chinese scholar Shen Kuo describes a magnetised needle that points south–north.
1600
The Earth is a magnet
William Gilbert's De Magnete argues that the whole Earth is a giant magnet.
1820
Currents make fields
Oersted sees a current turn a compass; Ampère proposes that all magnetism comes from circulating currents.
1845
Diamagnetism
Faraday finds that bismuth and many other substances are weakly repelled by a strong magnet.
1895
Curie's measurements
Pierre Curie shows how paramagnetism weakens with temperature and how iron loses its magnetism when hot.
1907
Domains
Pierre Weiss explains ferromagnetism with regions of lined-up atomic magnets.
1933
Meissner effect
Meissner and Ochsenfeld find that a superconductor pushes magnetic field out of itself.

The first rule deserves a closer look (Fig. 5.3). If the Earth's field outside points from the south towards the north, then the magnetic pole near the geographic north attracts the N end of a compass. So it must be, magnetically, a south pole.

3/3The field looks like that of a bar magnet tilted from the axis, with its S end towards the north. Compass needles line up along the field lines.

Fig. 5.3Outside the Earth the field runs from the southern hemisphere to the northern one, as if a tilted bar magnet with its S end under the Arctic sat inside.

A circle represents the Earth with a dashed vertical rotation axis labelled geographic N at the top and geographic S at the bottom, and a dashed equator. Step 1 shows only the Earth and its axis. Step 2 adds field lines that leave the southern half of the Earth, loop out through space and re-enter the northern half, with arrows pointing towards the north on the outer parts of the loops. Step 3 adds a dotted magnetic axis tilted about ten degrees from the rotation axis, an imaginary bar magnet along it with its S end towards geographic north and its N end towards geographic south, and three compass needles on the surface lining up with the field: level at the magnetic equator, with the N end tipping down into the ground in the northern hemisphere and up out of it in the southern hemisphere.

Warm-up

1 of 3

A freely hung bar magnet settles with its N end towards geographic north. What kind of magnetic pole lies under the Arctic?

No pole at all, only the rotation axisA north magnetic pole, since it is in the northA south magnetic pole, since it attracts the N end

Show answer

Answer (C) A south magnetic pole, since it attracts the N end

Unlike poles attract. The magnet's N end is pulled towards the Arctic, so the pole there behaves as a south pole. Calling it the 'north magnetic pole' names its place on the map, not its magnetic kind.

5.2 The bar magnet

Tap a card dusted with iron filings while a short bar magnet lies underneath, and a pattern springs up. Each filing is magnetised by the field and swivels into line with it, and neighbouring filings join end to end into chains. The chains are thickest at the two ends of the magnet and arch from one end round to the other. Put this beside the drawing of an electric dipole's field from Chapter 1 and the likeness is hard to miss. The ends of the magnet act like the two opposite charges, and the whole magnet is a magnetic dipole (चुंबकीय द्विध्रुव).

Try this

Plot field lines with a compass

You’ll need

a bar magnet
a small plotting compass
a sheet of A4 paper and a pencil
sticky tape

Predict first

Where do you expect your plotted lines to be closest together?

near the middle of the magnetnear the two ends of the magnetequally spaced everywhere

Tape the paper to a table well away from iron objects. Put the magnet in the middle, trace its outline and mark N and S.
Place the compass just beside one corner of the N end. Mark a dot at the tip of the needle's N end.
Move the compass so that the tail of its needle sits on that dot, and mark a new dot at the tip. Repeat until you reach the magnet or the edge of the paper.
Join the dots with a smooth curve and draw an arrow showing the way the needle pointed.
Start from other points round the magnet until you have eight to ten lines.

Can’t do it now? Try the simulation · Fig. 5.5

5.2.1 The magnetic field lines

A magnetic field line (चुंबकीय क्षेत्र रेखा) is a curve drawn so that, at every point, it runs along the direction of $𝐁$ there. A set of such lines turns an invisible vector field into a picture you can read at a glance (Fig. 5.4). Four rules govern them.

They are closed loops. Outside a magnet the lines run from N to S; inside the magnet they carry on from S back to N. No line starts or stops anywhere. (Contrast the electric dipole of Chapter 1, whose lines have ends: on the charges.)
The tangent gives the direction. At any point, $𝐁$ lies along the tangent to the line through that point, pointing the way the arrow runs. A small compass placed there lines up with this tangent.
Crowding gives the strength. Where lines are packed tightly the field is strong; where they spread apart it is weak. That is why the lines bunch near the poles.
They never cross. At a crossing, $𝐁$ would have to point two ways at once, which is impossible because the field has one definite value at each point.

1/4Every line is a closed loop: out of N, round the outside, into S and back through the magnet.

Fig. 5.4Field lines of a bar magnet: closed loops whose tangent gives the direction of

𝐁

, whose crowding gives its strength, and which never cross.NCERT Fig. 5.2 · p. 138

A horizontal bar magnet, S on the left and N on the right, with blue field lines computed from a model of the magnet. The lines leave the N end, loop round above and below, enter the S end and continue inside the magnet from S to N. Step 1 highlights one complete loop in orange and notes that every line is a closed loop. Step 2 dims the lines, highlights one of them, and marks a point P on it with a small compass whose needle and a black tangent arrow labelled B point along the line. Step 3 draws two equal square windows: the one beside the N end is crossed by four lines and labelled strong, the one far away at the top left is crossed by one line and labelled weak. Step 4 shows an inset with two lines crossing, circled and labelled impossible.

The compass method of the activity above is the simplest way to draw these lines; iron filings show the whole pattern at once but cannot show which way the lines point. The simulation lets you do both.

SimulationMapping the field of a bar magnet

A bar magnet in the middle of the stage with its field lines drawn and a small compass you can drag anywhere. Toggles switch on iron filings and a grid of compass needles. Readouts give the direction of the compass needle and whether the field at the compass is weak, moderate or strong.

Fig. 5.5Drag the compass along the lines, and switch between field lines, iron filings and a grid of compasses.

Try this

Switch on Iron filings and switch off Field lines. Do the filings trace the same curves the lines did?
Drag the compass slowly along one field line from the N end, round the outside, to the S end. How does the needle turn as it goes?
Put the compass right against the N end, then drag it slowly straight out to the edge of the stage. How does the Field at compass reading change, and where does it change fastest?

Quick check

1 of 3

Inside a bar magnet, which way do its field lines run?

From the S end to the N endFrom the N end to the S endThey do not exist inside the magnet

Show answer

Answer (A) From the S end to the N end

Outside, the lines go from N to S; to close the loops they must return through the magnet from S to N. They do not stop at the poles.

5.2.2 Bar magnet as an equivalent solenoid

In Chapter 4 you found that a flat current loop has a magnetic moment $𝐦 = I 𝐀$ , and that a solenoid of $N$ turns has $m = N I A$ along its axis. Far away, the field of such a solenoid looks just like that of a bar magnet (Fig. 5.6). Ampère went a step further and suggested that every magnet owes its magnetism to tiny circulating currents inside it. Today we know what those currents are: the orbital motion and the spin of electrons in atoms.

3/3An electric dipole gives the same far pattern, but between the charges its lines run from $+$ to $-$ and they end on the charges.

Fig. 5.6Far away, a bar magnet, a solenoid and an electric dipole make the same pattern of lines. Inside,

𝐁

runs from S to N along

𝐦

, but

𝐄

runs from

+

-

, against

𝐩

.NCERT Fig. 5.2 · p. 138

Three versions of one picture share the same outer field lines. Step 1: a short bar magnet, S on the left and N on the right, with blue field lines that loop from N round to S and continue inside the magnet from S to N, shown by orange arrowheads pointing right. Step 2: the magnet is replaced by a solenoid of orange windings; the field lines are identical, including the rightward arrows inside. Step 3: the source becomes an electric dipole, a negative charge on the left and a positive charge on the right, with green electric field lines; far away they have the same shape as before, but between the charges the lines run from plus to minus, so the orange arrow inside points left.

How can countless tiny loops add up to a solenoid? Fig. 5.7 shows the end face of a magnet divided into small cells, each with an atomic current circulating the same way. Wherever two cells touch, their currents run in opposite directions and cancel. Only the currents along the outer edge survive, and together they form one big loop running round the surface. A magnet is a stack of such slices, so it behaves like a solenoid wound on its own surface. Ampère's picture also explains the cutting experiment of Fig. 5.1: a coil cut in two is simply two shorter coils.

3/4Only the current round the edge survives: one big loop, like one turn of a coil.

Fig. 5.7Neighbouring atomic currents cancel inside a magnet, leaving a current round its surface. The magnet is, in effect, a solenoid.

Step 1: a square end face of a magnet divided into a four by four grid, each cell holding a small blue current loop circulating anticlockwise. Step 2: the shared edge between two neighbouring cells is circled in red and labelled cancel, because the two currents there run in opposite directions. Step 3: the inner loops fade and a thick orange loop runs anticlockwise round the whole perimeter, the surviving surface current. Step 4: the end view slides left and a side view appears on the right: a bar coloured S on the left and N on the right, wrapped in orange windings with downward current arrows on the front, and a black arrow inside pointing from S to N; the end view is labelled as the N face.

DerivationFar axial field of a finite solenoid

Take a solenoid of length $2 l$ and radius $a$ , with $n$ turns per unit length carrying a current $I$ . Put its centre at the origin and find the field at a point P on its axis a distance $r$ from the centre, far away: $r ≫ l$ and $r ≫ a$ .

A thin slice of width $d x$ at position $x$ holds $n d x$ turns, so it is a ring carrying current $n I d x$ . A ring of radius $a$ produces, at a distance $z$ along its axis, the field $μ_{0} I a^{2} / 2 (a^{2} + z^{2})^{3 / 2}$ (Chapter 4). Here $z = r - x$ , and because $r - x ≫ a$ we can drop $a^{2}$ beside $z^{2}$ :

d B = \frac{μ_{0} n I a^{2} d x}{2 (r - x)^{3}}

Every slice gives a field along the axis in the same direction, so we add them from $x = - l$ to $x = + l$ :

B = \frac{μ_{0} n I a^{2}}{2} \int_{- l}^{l} \frac{d x}{(r - x)^{3}} = \frac{μ_{0} n I a^{2}}{4} [\frac{1}{(r - l)^{2}} - \frac{1}{(r + l)^{2}}] = \frac{μ_{0} n I a^{2} r l}{(r^{2} - l^{2})^{2}}

Far away $r ≫ l$ , so $(r^{2} - l^{2})^{2} \approx r^{4}$ and $B \approx μ_{0} n I a^{2} l / r^{3}$ . The solenoid has $n (2 l)$ turns, each of area $π a^{2}$ , so its magnetic moment is $m = n (2 l) I π a^{2}$ , that is, $n I a^{2} l = m / 2 π$ . Therefore

Far axial field of a solenoid or a bar magnet

B = \frac{μ_{0}}{4 π} \frac{2 m}{r^{3}}

(5.1)

Two things follow. First, a magnetometer moved along the axis of a real bar magnet finds this same $1 / r^{3}$ fall-off once it is a few magnet-lengths away, with the field pointing away from the N end, so the solenoid model passes an experimental test. Second, it gives us a working definition: a bar magnet is assigned the moment $m$ of whichever solenoid would reproduce its far field. That $m$ can then be measured, for example by timing a needle's swings in a known field (Section 5.2.3).

Example

A solenoid posing as a bar magnet

A solenoid $6.0 c m$ long has 400 turns of radius $1.2 c m$ and carries $1.5 A$ . (a) Find its magnetic moment. (b) Use (5.1) to find the field on its axis $30 c m$ from its centre. (c) An exact calculation for a finite solenoid gives $2.05 \times 10^{- 6} T$ at that point. How good is the approximation, and why?

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

Quick check

1 of 3

A coil of 250 turns, each of area

4.0 c m^{2}

, carries

0.60 A

. What is its magnetic moment?

A m²

Show answer

Answer 0.06 A m²

m = N I A = 250 \times 0.60 \times 4.0 \times 10^{- 4} = 0.060 A m^{2}

. Remember to convert

c m^{2}

m^{2}

5.2.3 The dipole in a uniform magnetic field

Pivot a small magnet, such as a compass needle, in a uniform field $𝐁$ (Fig. 5.8). Its N end is pushed along the field and its S end is pushed the opposite way by an equal amount, so the net force is zero. The two pushes act along one line only when the needle is parallel or antiparallel to the field; otherwise they form a couple, and the needle feels a torque that turns it towards $𝐁$ .

2/4 $θ = 60^{\circ}$ : the forces form a couple, and the torque $m B \sin 60^{\circ}$ turns the needle back towards the field.

Fig. 5.8In a uniform field the forces on the two ends of a needle cancel, but they form a couple unless the needle lies along the field. The energy curve

U = - m B \cos θ

has its valley at

0^{\circ}

and its peak at

180^{\circ}

.NCERT Fig. 5.3(b) · p. 138

Left: horizontal dashed field lines pointing right and a needle pivoted at its centre, orange at the N end and white at the S end. Green arrows show equal and opposite forces on the two ends, the one on the N end along the field and the one on the S end against it. Right: a graph of potential energy U against the angle theta, with the axis marked from minus 180 to plus 180 degrees: a cosine-shaped valley with its lowest point at zero, and a dot marking the current angle. When the needle is tilted, a short dashed line from the pivot marks the field direction and an arc marks theta. Step 1: the needle lies along the field; the forces act along one line, the dot sits at the bottom of the valley, labelled stable. Step 2: the needle is turned to 60 degrees; the forces form a couple and an orange curved arrow shows the torque turning it back towards the field; the dot is halfway up. Step 3: the needle is at 90 degrees; the torque is largest and the dot is at U equals zero. Step 4: the needle points against the field; the forces again act along one line, the torque is zero, and the dot sits at the top of the curve, labelled unstable.

DerivationTorque and potential energy of a magnetic dipole

For a current loop you showed in Chapter 4 that the torque in a uniform field is $𝐦 \times 𝐁$ . A bar magnet has the same moment as its equivalent solenoid, so the same result holds for it:

Torque on a magnetic dipole

𝝉 = 𝐦 \times 𝐁, τ = m B \sin θ

(5.2)

where

$θ$: the angle between $𝐦$ and $𝐁$

The torque is largest, $m B$ , when the needle is at right angles to the field, and zero when it lies along the field or against it. It always acts to reduce $θ$ : it is a restoring torque.

To find the potential energy, turn the needle slowly from an angle $θ_{0}$ to $θ$ with an outside torque that just balances the magnetic one. The work done is stored as potential energy:

U (θ) - U (θ_{0}) = \int_{θ_{0}}^{θ} m B \sin θ^{'} d θ^{'} = - m B \cos θ + m B \cos θ_{0}

Only differences in potential energy can be measured, so we may choose where $U = 0$ . Choosing $U = 0$ when the needle is at right angles to the field, $θ_{0} = 90^{\circ}$ , removes the constant:

Potential energy of a magnetic dipole

U = - m B \cos θ = - 𝐦 \cdot 𝐁

(5.3)

$U$ is lowest, $- m B$ , at $θ = 0$ . A needle along the field is in stable equilibrium: any small turn raises its energy, and the torque brings it back. $U$ is highest, $+ m B$ , at $θ = 180^{\circ}$ . A needle pointing against the field is in unstable equilibrium: the torque there is also zero, but the slightest nudge grows.

DerivationSmall oscillations: timing a needle to measure B or m (extension)

Turn the needle a little away from the field and let go. If $I$ is its moment of inertia about the pivot (not a current here), Newton's second law for rotation gives

I \frac{d^{2} θ}{d t^{2}} = - m B \sin θ \approx - m B θ (θ small, in radians)

This is the equation of simple harmonic motion with angular frequency $ω = \sqrt{m B / I}$ , so the needle swings with period

Period of small oscillations

T = 2 π \sqrt{\frac{I}{m B}}

Time the swings in a known field and you can find $m$ . Time the same needle in an unknown field and you can find $B$ , since $B \propto 1 / T^{2}$ . This is the arrangement that the torque figure shows: a needle free to swing in a steady field.

Fig. 5.9Set the angle, the field and the magnetic moment; read the torque and energy; release the needle and time its swings.

Example

Timing a needle

A magnetised needle with $m = 5.0 \times 10^{- 2} A m^{2}$ and moment of inertia $8.0 \times 10^{- 7} k g m^{2}$ can swing in a horizontal plane. Where the horizontal part of the Earth's field is $3.5 \times 10^{- 5} T$ : (a) find its period of small oscillations. (b) Near an electromagnet the same needle swings with a period of $1.2 s$ . Estimate the total horizontal field there, assuming it is uniform over the needle.

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

Example 5.1NCERT Example 5.1 · p. 139

Four puzzles about poles

(a) The flat ring magnet from a loudspeaker is magnetised through its thickness, so one flat face is N and the other is S. It is snapped across a diameter into two C-shaped halves. What poles does each half have?

(b) A compass needle floats on a cork in a bowl of water. It turns to face north, but the cork does not drift northwards. When a small fridge magnet is held at the edge of the bowl, the cork drifts towards the magnet as well as turning. Explain the difference.

(c) An iron ring is magnetised all the way round its circumference, like the core of a toroid carrying a current. Where are its poles?

(d) The N pole of a bar magnet is brought up to each end of a steel rod in turn, and both ends are attracted. (i) Does this show that the rod is magnetised? (ii) A second rod repels the N pole at one of its ends. What does that tell you, and where along this second rod would a small hanging paper clip be pulled most strongly?

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

(b) The Earth's field is practically uniform over the needle, so the pushes on its two ends are equal and opposite: a torque but no net force. Near the fridge magnet the field changes strongly along the needle. The needle turns so that its nearer end is the pole opposite to the magnet's face, and that nearer end sits in a much stronger field than the far end. The attraction on the near end outweighs the repulsion on the far end, so there is a net force towards the magnet.
(d) (i) No; in fact it suggests the rod is not magnetised. The magnet induces an opposite pole in whichever end is nearer, so unmagnetised steel is attracted by either pole. A magnetised rod would have one end that repels the N pole, unless the bar magnet is strong enough to re-magnetise it. Attraction proves nothing; only repulsion shows that both objects are magnets. (ii) Repulsion needs like poles, so the second rod is a magnet with an N pole at that end. A clip is pulled hardest near its two ends, the poles, and hardly at all near its middle.

Quick check

1 of 4

At what angle between

𝐦

and

𝐁

is the torque on a magnet largest?

0^{\circ}

45^{\circ}

90^{\circ}

180^{\circ}

Show answer

Answer (C) $90^{\circ}$

τ = m B \sin θ

is largest when

\sin θ = 1

. At

0^{\circ}

and

180^{\circ}

the torque is zero, even though

180^{\circ}

is the highest-energy position.

5.2.4 The electrostatic analog

Set the results so far beside those for an electric dipole in Chapter 1. The torque $𝐩 \times 𝐄$ becomes $𝐦 \times 𝐁$ , the energy $- 𝐩 \cdot 𝐄$ becomes $- 𝐦 \cdot 𝐁$ , and the far axial field $2 p / 4 π ε_{0} r^{3}$ becomes $μ_{0} 2 m / 4 π r^{3}$ . Every magnetic formula follows from its electric partner by one simple swap of symbols, so you can reuse all the electric-dipole results you already know.

DerivationFields of a short magnet from the electric-dipole results

Make the replacements

𝐄 \to 𝐁, 𝐩 \to 𝐦, \frac{1}{4 π ε_{0}} \to \frac{μ_{0}}{4 π}

For an electric dipole, at a distance $r$ much larger than its size, the field on the perpendicular bisector (the equatorial line) is $- 𝐩 / 4 π ε_{0} r^{3}$ , and on the axis it is $2 𝐩 / 4 π ε_{0} r^{3}$ . The swap gives the fields of a magnet of size $l ≪ r$ :

Equatorial field of a short magnet

𝐁_{E} = - \frac{μ_{0}}{4 π} \frac{𝐦}{r^{3}}

(5.4)

Axial field of a short magnet

𝐁_{A} = \frac{μ_{0}}{4 π} \frac{2 𝐦}{r^{3}}

(5.5)

Take the size of (5.5) and you recover (5.1): the solenoid calculation and the dipole analogy agree, and the analogy also supplies the direction. At equal distances the axial field is twice the equatorial field; it points along $𝐦$ , while the equatorial field points opposite to $𝐦$ (Fig. 5.10).

4/4The field lines confirm the directions: along $𝐦$ beyond the N end, against $𝐦$ beside the magnet.

Fig. 5.10At the same distance

r

, the axial field of a short magnet is twice the equatorial field. It points along

𝐦

; the equatorial field points against

𝐦

.NCERT p. 140

A short bar magnet, S on the left and N on the right, with faint dipole field lines. A point on the axis lies a distance r to the right of the magnet and another point on the equatorial line lies the same distance r above it, both joined to the magnet by dashed lines. Step 1 shows only the magnet and the two points. Step 2 adds a long orange arrow at the axial point pointing right, along the magnet's moment, labelled B sub A. Step 3 adds a shorter orange arrow at the equatorial point pointing left, opposite to the moment, labelled B sub E. Step 4 shows both arrows and a box stating the ratio 2 to 1.

Table 5.1Electric and magnetic dipoles side by sideNCERT Table 5.1 · p. 141

	Electric dipole	Magnetic dipole (short magnet)
Constant in the field law	$1 / 4 π ε_{0}$	$μ_{0} / 4 π$
Dipole moment and its unit	$𝐩$ , in $C m$	$𝐦$ , in $A m^{2}$
Field on the axis ( $r ≫ l$ )	$2 𝐩 / 4 π ε_{0} r^{3}$	$μ_{0} 2 𝐦 / 4 π r^{3}$
Field on the equatorial line	$- 𝐩 / 4 π ε_{0} r^{3}$	$- μ_{0} 𝐦 / 4 π r^{3}$
Torque in a uniform field	$𝐩 \times 𝐄$	$𝐦 \times 𝐁$
Energy in a uniform field	$- 𝐩 \cdot 𝐄$	$- 𝐦 \cdot 𝐁$
Net force in a uniform field	zero	zero
Field lines inside the source	from $+ q$ to $- q$ (against $𝐩$ )	from S to N (along $𝐦$ )
Flux out of a surface round one end	$q / ε_{0}$	zero

Example 5.2NCERT Example 5.2 · p. 141

A compass needle beside a magnet

A small bar magnet A with moment $m_{A} = 0.20 A m^{2}$ is fixed on a bench with its N end pointing east. A compass needle C with moment $m_{C} = 0.050 A m^{2}$ can only turn on its pivot, and the compass is placed at one of four spots, each $r = 5.0 c m$ from the centre of A: $C_{1}$ on A's axis to the east, needle pointing east; $C_{2}$ on the axis to the west, needle pointing west; $C_{3}$ on the perpendicular bisector to the north, needle pointing west; $C_{4}$ on the bisector to the south, needle pointing north. Treat A and C as point dipoles and ignore the Earth's field, which is several times weaker here.

Fig. 5.11Magnet A (centre) and the four spots for the compass C, all

r = 5.0 c m

from A. The orange end of each needle is its N end.

A dashed horizontal line marks the axis of magnet A and a dashed vertical line its perpendicular bisector; small arrows in the corner show north (up) and east (right). A short bar magnet A sits at the centre with its S half on the west and its N half on the east. Four round compasses stand the same distance r from its centre. On the axis, compass C1 to the east has its needle pointing east and C2 to the west has its needle pointing west. On the bisector, C3 to the north has its needle pointing west and C4 to the south has its needle pointing north.

(a) Find the field of A, magnitude and direction, at each spot. (b) Find the potential energy of the needle at each spot and rank the four from lowest to highest. (c) At which spots is the needle in equilibrium, and is each equilibrium stable or unstable? (d) At the remaining spot, find the torque on the needle and the way it turns.

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

Example

Keeping a compass honest

A bar magnet with $m = 1.2 A m^{2}$ lies on the same bench as a compass. The horizontal part of the Earth's field is $4.0 \times 10^{- 5} T$ . How far from the magnet must the compass be for the magnet's field to be at most 5 % of the Earth's if the compass is (a) on the magnet's axis, (b) on its equatorial line? Treat the magnet as short.

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

Quick check

1 of 3

At the same distance from a short magnet, how does the axial field compare with the equatorial field?

Equal in size and in directionHalf as large and pointing the same wayTwice as large and pointing the opposite way

Show answer

Answer (C) Twice as large and pointing the opposite way

B_{A} = μ_{0} 2 m / 4 π r^{3}

and

B_{E} = μ_{0} m / 4 π r^{3}

, so

B_{A} = 2 B_{E}

𝐁_{A}

points along

𝐦

and

𝐁_{E}

against it.

5.3 Magnetism and Gauss's law

Wrap an imaginary closed bag round the N end of a magnet and count the field lines that cross it (Fig. 5.12). For an electric dipole, the same bag round the positive charge has more lines leaving than entering, and Gauss's law of Chapter 1 turns that surplus into the enclosed charge, $q / ε_{0}$ . Is there a magnetic surplus to match?

2/4Lines leave through the outside of the surface and enter where it cuts through the magnet: as many enter as leave.

Fig. 5.12Every magnetic field line that leaves a closed surface re-enters it somewhere, so the magnetic flux out of any closed surface is zero. Electric lines that start on a charge inside simply leave.NCERT Fig. 5.5 · p. 142

Left: a bar magnet with its field lines, including the lines running from S to N inside it. Step 1: a dashed purple closed surface surrounds the N end, cutting through the magnet; a small patch on it is magnified on the right, showing the patch, its outward normal n and a field vector B at an angle theta to the normal. Step 2: every place where a line crosses the surface is marked, green dots where lines leave and orange squares where they enter; the lines leave through the outer part and enter through the part inside the magnet. A tally reads leaving 6, entering 6, net flux zero. Step 3: a larger surface encloses the whole magnet; again 6 lines leave on the right and 6 enter on the left, net flux zero. Step 4: on the right, a positive charge sits inside a dashed circle with eight radial electric field lines, all leaving: leaving 8, entering 0.

Draw any closed surface, even one that slices through the N end of a magnet. Every field line that leaves it is part of a closed loop, so it must come back in somewhere. Round the N end, lines leave through the outer part of the surface and enter through the part that cuts the magnet, where the lines run from S to N. The number of lines leaving always equals the number entering.

To state this exactly, split the closed surface $S$ into small patches and give each patch an area vector $Δ 𝐒$ pointing outwards. The magnetic flux through one patch is $Δ ϕ_{B} = 𝐁 \cdot Δ 𝐒$ , positive where the field leaves and negative where it enters. Adding over all the patches gives zero:

Gauss's law for magnetism

ϕ_{B} = \sum_{all patches} Δ ϕ_{B} = \sum_{all patches} 𝐁 \cdot Δ 𝐒 = 0

(5.6)

LawGauss's law for magnetism

Whatever closed surface you choose, as much magnetic flux enters it as leaves it: $\oint_{S} 𝐁 \cdot d 𝐒 = 0$ .

The electric version, $\sum 𝐄 \cdot Δ 𝐒 = q / ε_{0}$ , has a charge on the right because charges are where electric lines start and stop. The magnetic version has zero on the right because nature provides nothing for $𝐁$ lines to start from or end on. The smallest source of a magnetic field is therefore not a ‘magnetic charge’ but a current loop, whose N face and S face always come together.

Example 5.3NCERT Example 5.3 · p. 143

Spot the impossible field lines

The six sketches below show lines of some field. For each one, decide whether it could show magnetic field lines. If it cannot, say which rule it breaks, and whether it could instead show electrostatic field lines.

1/2Decide for each sketch before you look at the verdicts.

Fig. 5.13Which of these could be magnetic field lines? The second step shows the verdicts.

Six panels. A: straight lines radiating outwards in all directions from a small ball. B: concentric circles round a wire carrying current out of the page, with arrows running anticlockwise. C: a bar magnet with lines outside running from N to S and lines inside the magnet also drawn from N to S. D: two lines that cross each other. E: a solenoid with nearly straight parallel lines inside that fan out at the ends and curve back outside. F: a single closed oval loop in empty space with nothing passing through it. In the second step, B and E get green ticks and A, C, D and F get orange crosses.

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

C. The outside is correct, but inside the magnet the lines are drawn from N to S. Then lines leave the N end both outside and inside, and the flux out of a surface round the N end is not zero. Inside a magnet $𝐁$ runs from S to N. Not possible for $𝐁$ . It is, however, a correct electrostatic picture: a bar whose right end carries positive charge and left end negative charge has $𝐄$ running from $+$ to $-$ both outside and inside it, just like the electric dipole of Fig. 5.6.

Example 5.4NCERT Example 5.4 · p. 144

Four questions about field lines and moments

(a) A beam of electrons is fired at right angles to the lines of a uniform magnetic field. Does it travel along the lines? What does this say about calling them lines of force?

(b) Imagine that a magnetic monopole of strength $q_{m}$ is discovered and trapped at the centre of a sphere of radius $R$ . How would Gauss's law for magnetism read for that sphere, and what field would you expect on its surface?

(c) A horseshoe magnet floats inside an orbiting spacecraft, far from any other magnet. Each arm sits in the field produced by the other arm. Will the magnet start to drift or to spin?

(d) A neutron has no net charge, yet it has a magnetic moment. How can that be, when magnetic fields come from moving charges?

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

Quick check

1 of 3

A closed surface is made of a curved dome and a flat base. The magnetic flux leaving through the dome is

+ 3.6 \times 10^{- 5} W b

. What is the outward flux through the base, in webers?

Show answer

Answer -0.000036 Wb

The total outward flux is zero, so the base must carry

- 3.6 \times 10^{- 5} W b

: the same flux enters through it.

5.4 Magnetisation and magnetic intensity

A steel sewing needle and a steel girder can be magnetised equally thoroughly, yet the girder's total magnetic moment is millions of times larger, simply because it is bigger. To describe the material rather than one particular lump, we divide by the volume. The moment itself comes from the atoms: their orbiting and spinning electrons act as tiny current loops, and the atomic moments inside a sample add as vectors.

Definition

Magnetisation

The net magnetic moment per unit volume of a material, $𝐌 = 𝐦_{net} / V$ . It is a vector measured in $A m^{2} / m^{3} = A / m$ , with dimensions $[L^{- 1} A]$ .

ExampleA rod of volume $10 c m^{3}$ whose atomic moments add up to $2.0 A m^{2}$ has $M = 2.0 / (1.0 \times 10^{- 5}) = 2.0 \times 10^{5} A / m$ .

Magnetisation

𝐌 = \frac{𝐦_{net}}{V}

(5.7)

To see how magnetisation changes the field, fill a long solenoid with the material (Fig. 5.14).

3/4The moments line up. Inside they cancel in neighbouring pairs, leaving a magnetising current $I_{m}$ round the core's surface.

Fig. 5.14Filling a solenoid with a magnetised core adds a field

μ_{0} 𝐌

to the field

μ_{0} n I

of the winding, as if the core carried extra current round its surface.NCERT p. 145

A long coil drawn as slanted orange windings with downward current arrows on the front turns. Step 1: the coil is empty and three blue field lines run through it to the right. Step 2: a pink core fills the coil; small red arrows for atomic magnetic moments point in random directions, so the field is unchanged. Step 3: the atomic arrows all turn to point along the field, and dashed purple strokes with downward arrows appear on the surface of the core, the magnetising current that acts like extra windings. Step 4: the field inside now has many more lines, and the atomic arrows fade into the background.

DerivationB, H and M inside a filled solenoid

An empty long solenoid with $n$ turns per unit length carrying a current $I$ has, inside it,

B_{0} = μ_{0} n I

(5.8)

Now fill it with a material. The field inside becomes the field of the winding plus a contribution $𝐁_{m}$ from the magnetised material:

𝐁 = 𝐁_{0} + 𝐁_{m}

(5.9)

Why should $𝐁_{m}$ be proportional to $𝐌$ ? When the atomic loops line up, the currents of neighbouring loops cancel inside the material, just as in Fig. 5.7, and only a current running round the surface survives. A slice of the core of thickness $d x$ and cross-section $A$ has magnetic moment $M A d x$ , which is exactly what a surface current $M d x$ running round it would give. So $M$ is a current per unit length, playing the same part as $n I$ in the winding, and by the same reasoning that gave (5.8):

𝐁_{m} = μ_{0} 𝐌

(5.10)

Now introduce a quantity that can be worked out from the free currents alone, without knowing anything about the material. This is the magnetic intensity $𝐇$ , defined by

Magnetic intensity

𝐇 = \frac{𝐁}{μ_{0}} - 𝐌

(5.11)

so that the total field is

𝐁 = μ_{0} (𝐇 + 𝐌)

(5.12)

In the filled solenoid $B = μ_{0} n I + μ_{0} M$ , so (5.11) gives $H = n I$ . The magnetic intensity depends only on the free current, not on the core. It has the same unit as $M$ , the ampere per metre.

How much magnetisation does a given $𝐇$ produce? For many materials $𝐌$ is simply proportional to $𝐇$ :

Magnetic susceptibility

𝐌 = χ 𝐇

(5.13)

The constant of proportionality, a pure number $χ$ , is called the magnetic susceptibility (चुंबकीय प्रवृत्ति). Its sign and size sort materials into families, as Section 5.5 shows; a negative $χ$ means that $𝐌$ points against $𝐇$ . Putting (5.13) into (5.12) gives

𝐁 = μ_{0} (1 + χ) 𝐇

(5.14)

Permeability

𝐁 = μ_{0} μ_{r} 𝐇 = μ 𝐇

(5.15)

where

$μ_{r} = 1 + χ$: the relative permeability (a pure number)
$μ = μ_{0} μ_{r}$: the permeability of the material, in the same unit as $μ_{0}$

The relative permeability $μ_{r}$ is the magnetic cousin of the dielectric constant $K$ in electrostatics. Because $μ_{r} = 1 + χ$ and $μ = μ_{0} μ_{r}$ , a data table need list only one of $χ$ , $μ_{r}$ or $μ$ for each material.

Definition

Magnetic intensity

The vector $𝐇 = 𝐁 / μ_{0} - 𝐌$ , measured in $A / m$ . It is set by the free currents alone: inside a long solenoid $H = n I$ , whatever the core.

Definition

Magnetic susceptibility

The ratio $χ = M / H$ for a material magnetised in proportion to $H$ : negative for diamagnets, slightly positive for paramagnets, very large for ferromagnets.

InteractiveSolenoid core lab

Fig. 5.15Choose a core, the winding and the current; watch

H

M

and

B

, and the graph of

M

against

H

Example 5.5NCERT Example 5.5 · p. 146

A ferrite ring

A ring of ferrite with mean radius $8.0 c m$ is wound evenly with 400 turns (a Rowland ring). With $0.50 A$ in the winding, a search coil shows that the field in the ferrite is $0.60 T$ . Find (a) $H$ in the ferrite, (b) $μ_{r}$ and $χ$ , (c) $M$ , and (d) the magnetising current $I_{m}$ : the extra current the same winding would need, with no ferrite, to give the same $B$ .

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

Example

How small is diamagnetism?

A bismuth rod ( $χ = - 1.66 \times 10^{- 4}$ ) fills a long solenoid with 2000 turns per metre carrying $5.0 A$ . Find $H$ and $M$ in the rod and the fractional change in $B$ that the bismuth causes.

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

Quick check

1 of 4

A long solenoid has 1500 turns per metre and carries

0.40 A

. What is

H

inside it, in A/m?

A/m

Show answer

Answer 600 A/m

H = n I = 1500 \times 0.40 = 600 A / m

, whatever material fills the solenoid.

5.5 Magnetic properties of materials

Three quite different behaviours hide behind the sign and size of $χ$ , and they give the three families of Table 5.2: diamagnetic, paramagnetic and ferromagnetic. The values in Table 5.3 show how feeble the first two are, often around $10^{- 5}$ in size. Yet the sign of that tiny number decides whether a sample is pushed out of a strong field or drawn into it.

Table 5.2Three kinds of magnetic materialNCERT Table 5.2 · p. 147

	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility $χ$	$- 1 \leq χ < 0$ , usually near $- 10^{- 5}$	$0 < χ ≪ 1$ , usually $10^{- 5}$ to $10^{- 3}$	$χ ≫ 1$ , hundreds to many thousands
Relative permeability $μ_{r}$	$0 \leq μ_{r} < 1$	just above 1	$μ_{r} ≫ 1$
Permeability $μ$	$μ < μ_{0}$	$μ$ slightly above $μ_{0}$	$μ ≫ μ_{0}$
Field inside a sample	slightly weaker	slightly stronger	much stronger
In a non-uniform field	moves towards the weaker field	moves weakly towards the stronger field	pulled strongly towards the stronger field
Effect of temperature	almost none	$χ$ falls as the temperature rises	becomes paramagnetic above a critical temperature
Examples	bismuth, copper, gold, silver, water, nitrogen, most plastics	aluminium, platinum, magnesium, tungsten, oxygen gas, copper sulfate	iron, cobalt, nickel, gadolinium, alnico

Table 5.3Typical susceptibilities near room temperature (SI units, approximate)

Material	Kind	$χ$
bismuth	diamagnetic	$- 1.7 \times 10^{- 4}$
water	diamagnetic	$- 9.0 \times 10^{- 6}$
copper	diamagnetic	$- 9.7 \times 10^{- 6}$
air	paramagnetic	$+ 3.7 \times 10^{- 7}$
aluminium	paramagnetic	$+ 2.2 \times 10^{- 5}$
platinum	paramagnetic	$+ 2.7 \times 10^{- 4}$
soft iron	ferromagnetic	a few thousand (depends on $H$ )

Fig. 5.16 shows the same idea with field lines. A diamagnetic sample nudges the lines aside, a paramagnetic one draws a few extra lines through itself, and a ferromagnet funnels them in strongly. In a non-uniform field these differences turn into forces: paramagnets and ferromagnets are pulled towards the region of stronger field, while diamagnets are pushed towards the weaker field.

3/4A ferromagnet gathers the lines into itself: the field inside is many times larger.

Fig. 5.16How a sample changes a uniform field: pushed aside by a diamagnet, drawn in slightly by a paramagnet, funnelled in by a ferromagnet and completely expelled by a superconductor. The first two effects are hugely exaggerated.NCERT Fig. 5.7 · p. 147

A round sample sits in a uniform field whose lines run from left to right; the lines are calculated for a long cylinder. Step 1, diamagnetic: the lines bend slightly away from the sample and fewer of them pass through it; the effect is exaggerated. Step 2, paramagnetic: the lines bend slightly towards the sample and a little more pass through it; also exaggerated. Step 3, ferromagnetic: the lines are strongly funnelled into the sample, so almost all the lines in the picture pass through it. Step 4, superconductor: no line enters the sample; all of them curve round it.

5.5.1 Diamagnetism

Hang a small piece of bismuth near a strong magnet and it drifts slightly away, whichever pole you use. Materials that behave like this are diamagnetic (प्रतिचुंबकीय पदार्थ). The field inside them is reduced, but only a little: for water or copper, by about $0.001 %$ .

Where does the repulsion come from? In a diamagnetic atom the electrons are paired so that their magnetic moments cancel; the atom has no permanent moment. Switching on a field changes the magnetic flux through each electron's orbit, and by Lenz's law (Chapter 6) the induced effect opposes that change. The induced electric field circulates one way round both orbits, so it drags back one electron and hurries the other along (Fig. 5.17). The two moments no longer cancel: the atom gains a small moment opposite to the field, and so it is repelled.

2/3By Lenz's law the electron whose moment points along $𝐁$ slows down, and the other speeds up.

Fig. 5.17In a diamagnetic atom two opposite orbital moments cancel. An applied field slows one electron and speeds up the other, leaving a small moment opposite to

𝐁

Two elliptical electron orbits seen slightly from above, each with a nucleus at its centre and an electron on the front of the orbit. In the left orbit the electron moves to the left along the front and its magnetic moment arrow points up; in the right orbit the electron moves to the right and its moment arrow points down. Step 1: no field; the two moment arrows are equal, so the net moment is zero. Step 2: an upward field is switched on, shown by dashed arrows at the sides; the left electron slows down, its velocity arrow and upward moment shrink, while the right electron speeds up and its downward moment grows. Step 3: a purple arrow between the orbits shows the resulting net moment pointing down, opposite to the field.

Since the effect comes from orbiting electrons, which every atom has, no substance escapes it. It is simply drowned out wherever para- or ferromagnetism is present, and shows up on its own only in materials whose atoms have no permanent moment: bismuth, copper, silver, gold, lead, water, nitrogen gas, common salt and most organic compounds, for example.

Cool a disc of lead below about $7.2 K$ and lower a small magnet towards it: the magnet stops and floats in mid-air. The lead has become a superconductor. Its electrical resistance has vanished, and it now keeps a weak magnetic field out of its interior altogether, so $B = 0$ inside, $χ = - 1$ and $μ_{r} = 0$ . This perfect diamagnetism is the Meissner effect. Why zero resistance and zero internal field should arrive together puzzled physicists for decades. The answer came from quantum mechanics in 1957: in the theory of Bardeen, Cooper and Schrieffer (BCS theory, Nobel Prize 1972), electrons pair up below the critical temperature and move as one coordinated whole. Superconducting coils power MRI scanners and particle accelerators, and they lift magnetically levitated trains.

Try this

Push a grape with a magnet

You’ll need

two grapes of about the same size
a plastic drinking straw
a bottle cap or a pencil tip as a pivot
a strong neodymium magnet
a cardboard box to keep off draughts

Predict first

What will happen when you bring the magnet close to one grape?

The grape is slowly pushed awayThe grape is pulled towards the magnetNothing at all happens

Push a grape onto each end of the straw and balance the straw on the pivot so that it turns freely. Put the box round it to block draughts.
Wait until the straw is still. Bring one pole of the magnet slowly to within a few millimetres of one grape, without touching it.
Hold the magnet there for 20 to 30 seconds and watch. Then repeat with the other pole.

Quick check

1 of 3

A small piece of bismuth hangs near one pole of a strong magnet. What happens?

It is pushed away from the pole, whichever pole it isIt is pulled towards the poleIt is pushed away from N but pulled towards S

Show answer

Answer (A) It is pushed away from the pole, whichever pole it is

Bismuth is diamagnetic: its induced moment always opposes the applied field, so it moves towards the weaker field whichever pole is used. Attraction to one pole and repulsion from the other is how a permanent magnet behaves.

5.5.2 Paramagnetism

Aluminium, platinum, oxygen and many compounds of iron, copper or manganese, such as copper sulfate, are weakly attracted by a strong magnet. In these paramagnetic substances (अनुचुंबकीय पदार्थ) every atom, ion or molecule is itself a tiny magnet, usually because some of its electrons are unpaired.

With no field present, the atomic moments point every which way, jostled by thermal motion, and they cancel on average (Fig. 5.18). An applied field tries to line them up while heat keeps scrambling them, so at room temperature only a small extra fraction points along the field. The net effect is a small magnetisation along $𝐇$ . Aluminium ( $χ = 2.2 \times 10^{- 5}$ ), for instance, raises the field inside it by just $0.002 %$ . Weak as it is, this is enough to draw the sample towards the stronger part of a non-uniform field.

2/3The field wins only slightly against thermal agitation: a weak magnetisation along $𝐁_{0}$ .

Fig. 5.18Thermal motion scrambles the permanent moments of a paramagnet. A field lines them up a little; a stronger field or a lower temperature lines them up more, until the magnetisation saturates.NCERT Fig. 5.7 · p. 148

A green rectangle stands for the sample and holds a grid of 24 small red arrows, one for each atomic moment, with a vertical meter labelled M on the right. Step 1: no field; the arrows point in random directions and the meter reads zero. Step 2: a field arrow pointing right appears below the sample; the arrows are still scattered but lean slightly towards the right, and the meter shows a small magnetisation. Step 3: strong field and low temperature; nearly all arrows point to the right and the meter is almost full.

For fields that are not too strong, the susceptibility is inversely proportional to the absolute temperature, $χ = C / T$ , where $C$ is a constant for the material; this is Curie's law. It cannot hold for ever. In fields of several tesla at a few kelvin almost every atomic moment already points along the field, so $M$ has nothing left to gain: it levels off at its saturation value, the moment of one atom times the number of atoms per unit volume.

Quick check

1 of 3

Why does a paramagnetic sample show no magnetisation when there is no external field?

Its atoms carry no magnetic moments at allIts atomic moments are locked in opposite pairsHeat keeps its atomic moments pointing at random

Show answer

Answer (C) Heat keeps its atomic moments pointing at random

The atoms do have permanent moments, but with no field to favour one direction, thermal agitation leaves them random and their sum is zero. Atoms with no moments, or with moments locked in cancelling pairs, would make the material diamagnetic.

5.5.3 Ferromagnetism

What makes iron so different from aluminium, when the atoms of both carry permanent moments? In iron, cobalt, nickel and gadolinium each atomic moment feels its neighbours through a purely quantum-mechanical coupling called exchange, which has no classical explanation. The coupling favours neighbours pointing the same way, so the moments line up by themselves, with no applied field, across regions called domains. That is why these metals are pulled strongly by a magnet and can become permanent magnets themselves.

A domain is typically a few micrometres to a fraction of a millimetre across; even a cube one micrometre on a side holds about $10^{11}$ iron atoms. Inside each domain the magnetisation is complete, but in an unmagnetised piece of iron neighbouring domains point in different directions and the sample has no net moment (Fig. 5.19). You can actually see domains. Spread a drop of liquid carrying very fine magnetic powder on a polished iron surface, and the powder gathers along the walls between domains; under a microscope the walls can be watched sliding about as a magnet approaches.

2/6Domains lying along the field grow as their walls move.

Fig. 5.19Domains in a ferromagnet: they grow and turn in an applied field, relax in soft iron, stay aligned in a hard magnet, and vanish above the Curie temperature.NCERT Fig. 5.8 · p. 149

A rectangular piece of ferromagnet divided into irregular domains, each coloured by the direction of its arrow, with a magnetisation meter labelled M on the right. Step 1, no field: twelve domains point up, down, left and right in equal measure and the meter reads zero. Step 2, weak field to the right: the right-pointing domains have grown at the expense of their neighbours and the meter reads about half. Step 3, strong field: the whole sample is one domain with every arrow to the right and the meter is full. Step 4, field removed from soft iron: the domains return to a mixed pattern and the meter falls to nearly zero. Step 5, field removed from a hard magnet: the sample stays mostly aligned, with a few small reversed domains, and the meter stays high. Step 6, heated past the Curie point: the domains vanish and small atomic arrows point in random directions; the meter reads zero.

Apply a weak field and the walls between domains start to slide: domains that already point roughly along the field swallow their neighbours. A stronger field swings the remaining domains round, until the sample is close to a single domain magnetised along the field and $M$ approaches saturation. Because a small $H$ produces such a large $M$ , $μ_{r}$ reaches thousands (above $10^{5}$ for some special alloys). The same huge magnetisation is why a nail or a paper clip is snatched towards the pole of a magnet, where the field is strongest.

Switch the field off and the two kinds of ferromagnet part company. In a soft ferromagnet such as annealed iron the domain walls slide easily, so the domains fall back into a disordered pattern and almost no magnetisation remains. That suits electromagnets and transformer cores, which must follow a changing current. In a hard ferromagnet the walls are pinned by tiny grains and impurities, so the aligned pattern is locked in. Alnico (iron alloyed with aluminium, nickel and cobalt), neodymium–iron–boron and the mineral lodestone are hard; they are the stuff of permanent magnets in speakers, motors, fridge doors and compass needles.

Compare

Hard and soft ferromagnets

Hard

keeps most of its magnetisation when the field is removed
hard to magnetise and hard to demagnetise
alnico, lodestone, neodymium–iron–boron
permanent magnets: compass needles, speakers, motors

Soft

loses almost all its magnetisation when the field is removed
easy to magnetise and to demagnetise
soft iron, silicon steel, ferrites
electromagnets, relays, transformer cores

Heat destroys the order. As a ferromagnet warms, thermal agitation disturbs the alignment within its domains and the magnetisation falls, slowly at first and then steeply. Above a critical temperature, the Curie temperature (क्यूरी ताप), the domains disappear and the material becomes paramagnetic. For iron this happens at about $1043 K$ (770 °C) and for nickel at about $627 K$ (354 °C). For gadolinium it is only about $293 K$ (20 °C), so gadolinium is ferromagnetic only on a cool day.

Quick check

1 of 4

Put these stages of magnetising an unmagnetised piece of iron in order.

Domains point in many directions and the net magnetisation is zero
The sample is nearly one domain and the magnetisation saturates
A stronger field turns the remaining domains towards the field
A weak field makes domains lying along it grow as their walls move

Show answer

Answer

Domains point in many directions and the net magnetisation is zero
A weak field makes domains lying along it grow as their walls move
A stronger field turns the remaining domains towards the field
The sample is nearly one domain and the magnetisation saturates

Wall motion comes first because it needs less energy; turning whole domains needs a stronger field; saturation comes last.

Table 5.4Magnetic quantities at a glanceNCERT p. 151

Quantity	Symbol	Scalar or vector	SI unit	Dimensions	Useful relation
permeability of free space	$μ_{0}$	scalar	$T m / A$	$[M L T^{- 2} A^{- 2}]$	$μ_{0} / 4 π = 10^{- 7} T m / A$
magnetic field	$𝐁$	vector	$T$ (tesla)	$[M T^{- 2} A^{- 1}]$	$1 T = 10^{4} G$
magnetic moment	$𝐦$	vector	$A m^{2}$	$[L^{2} A]$	$m = N I A$ for a coil
magnetic flux	$ϕ_{B}$	scalar	$W b$ (weber)	$[M L^{2} T^{- 2} A^{- 1}]$	$1 W b = 1 T m^{2}$
magnetisation	$𝐌$	vector	$A / m$	$[L^{- 1} A]$	moment per unit volume
magnetic intensity	$𝐇$	vector	$A / m$	$[L^{- 1} A]$	$𝐁 = μ_{0} (𝐇 + 𝐌)$
magnetic susceptibility	$χ$	scalar	none	none	$𝐌 = χ 𝐇$
relative permeability	$μ_{r}$	scalar	none	none	$μ_{r} = 1 + χ$
permeability	$μ$	scalar	$T m / A$	$[M L T^{- 2} A^{- 2}]$	$𝐁 = μ 𝐇$ , $μ = μ_{0} μ_{r}$

Recap

Magnetism and Matter: key points and summary

Magnetic poles come in pairs: cutting a magnet gives smaller complete magnets, and no magnetic monopole has ever been found (Section 5.1).
Magnetic field lines are closed loops (running from S to N inside a magnet); their tangent gives the direction of $𝐁$ , their crowding its strength, and they never cross (Section 5.2.1).
A bar magnet behaves like a solenoid: atomic current loops add up to a current round its surface. Far out on the axis both give $B = μ_{0} 2 m / 4 π r^{3}$ (Section 5.2.2).
In a uniform field a dipole feels no net force but a torque $𝝉 = 𝐦 \times 𝐁$ . Its energy $U = - 𝐦 \cdot 𝐁$ is lowest along the field (stable) and highest against it (unstable) (Section 5.2.3).
A needle of moment of inertia $I$ swings with period $T = 2 π \sqrt{I / m B}$ , which can be used to measure $m$ or $B$ (Period of small oscillations).
Swapping $𝐄 \to 𝐁$ , $𝐩 \to 𝐦$ and $1 / 4 π ε_{0} \to μ_{0} / 4 π$ turns electric-dipole results into magnetic ones: $𝐁_{A} = μ_{0} 2 𝐦 / 4 π r^{3}$ and $𝐁_{E} = - μ_{0} 𝐦 / 4 π r^{3}$ (Section 5.2.4).
Gauss's law for magnetism: over any closed surface the flux of $𝐁$ adds up to zero, because there are no magnetic monopoles to act as sources (Section 5.3).
Magnetisation $𝐌$ is magnetic moment per unit volume. The magnetic intensity $𝐇 = 𝐁 / μ_{0} - 𝐌$ is fixed by free currents ( $H = n I$ in a solenoid), so $𝐁 = μ_{0} (𝐇 + 𝐌)$ (Section 5.4).
For a linear material $𝐌 = χ 𝐇$ and $𝐁 = μ_{0} μ_{r} 𝐇 = μ 𝐇$ , with $μ_{r} = 1 + χ$ ; any one of $χ$ , $μ_{r}$ , $μ$ fixes the others ((5.15)).
Diamagnets ( $χ$ small and negative) are pushed towards weaker fields because an applied field induces atomic moments opposing it; a superconductor is a perfect diamagnet with $χ = - 1$ (Section 5.5.1).
Paramagnets ( $χ$ small and positive) have permanent atomic moments that a field partly aligns against thermal agitation; $χ$ rises as the temperature falls (Section 5.5.2).
Ferromagnets ( $χ ≫ 1$ ) contain domains that grow and turn in a field. Hard ferromagnets stay magnetised and soft ones do not; above the Curie temperature all become paramagnetic (Section 5.5.3).

Flash cards · 8

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Warm-up3 questions
5.2.1 The magnetic field lines3 questions
5.2.2 Bar magnet as an equivalent solenoid3 questions
5.2.3 The dipole in a uniform magnetic field4 questions
5.2.4 The electrostatic analog3 questions
5.3 Magnetism and Gauss's law3 questions
5.4 Magnetisation and magnetic intensity4 questions
5.5.1 Diamagnetism3 questions
5.5.2 Paramagnetism3 questions
5.5.3 Ferromagnetism4 questions

Glossary · 15

Key terms (English–Hindi)

Curie temperatureक्यूरी तापkyuri taap: The temperature above which a ferromagnet loses its domains and becomes paramagnetic: about $1043 K$ for iron.First seen in the chapter →

diamagneticप्रतिचुंबकीयpratichumbakiya: Having a small negative $χ$ : weakly magnetised against the field and pushed towards weaker fields.First seen in the chapter →
domainडोमेनdomen: A region of a ferromagnet in which all the atomic moments point the same way.First seen in the chapter →

ferromagneticलौहचुंबकीयlauhchumbakiya: Having a very large $χ$ because atomic moments line up in domains; strongly attracted and able to become a permanent magnet.First seen in the chapter →

magnetic dipoleचुंबकीय द्विध्रुवchumbakiya dvidhruv: A pair of equal and opposite poles, or a small current loop, whose far field falls as $1 / r^{3}$ ; a bar magnet is one.First seen in the chapter →
magnetic field lineचुंबकीय क्षेत्र रेखाchumbakiya kshetra rekha: A curve whose tangent at every point gives the direction of $𝐁$ ; the lines form closed loops and never cross.First seen in the chapter →
magnetic intensityचुंबकीय तीव्रताchumbakiya teevrata: $𝐇 = 𝐁 / μ_{0} - 𝐌$ , set by free currents alone; $H = n I$ in a long solenoid.First seen in the chapter →
magnetic momentचुंबकीय आघूर्णchumbakiya aaghurn: The vector $𝐦$ that measures the strength and direction of a magnetic dipole; $m = N I A$ for a coil, in $A m^{2}$ .First seen in the chapter →
magnetic monopoleचुंबकीय एकल ध्रुवchumbakiya ekal dhruv: A hypothetical isolated N or S pole. None has ever been found, which is why $\oint 𝐁 \cdot d 𝐒 = 0$ .First seen in the chapter →
magnetic poleचुंबकीय ध्रुवchumbakiya dhruv: Either end of a magnet, where its field lines leave (N pole) or enter (S pole) and where its pull is strongest.First seen in the chapter →
magnetic susceptibilityचुंबकीय प्रवृत्तिchumbakiya pravritti: The dimensionless ratio $χ = M / H$ : how strongly a material is magnetised by a given $H$ .First seen in the chapter →
magnetisationचुंबकनchumbakan: The net magnetic moment per unit volume of a material, $𝐌$ , in $A / m$ .
Meissner effectमाइसनर प्रभावmaisnar prabhav: The complete expulsion of a weak magnetic field from a superconductor, which behaves as a perfect diamagnet with $χ = - 1$ .First seen in the chapter →

paramagneticअनुचुंबकीयanuchumbakiya: Having a small positive $χ$ due to permanent atomic moments: weakly pulled towards stronger fields.First seen in the chapter →

relative permeabilityआपेक्षिक चुंबकशीलताaapekshik chumbakasheelata: $μ_{r} = 1 + χ = μ / μ_{0}$ , the factor by which a linear material multiplies $μ_{0} H$ .First seen in the chapter →

Magnetism and Matter: Class 12 Physics · Chapter 5

What you will learn in Magnetism and Matter

5.1 Introduction

5.2 The bar magnet

5.2.1 The magnetic field lines

5.2.2 Bar magnet as an equivalent solenoid

5.2.3 The dipole in a uniform magnetic field

5.2.4 The electrostatic analog

5.3 Magnetism and Gauss's law

5.4 Magnetisation and magnetic intensity

5.5 Magnetic properties of materials

5.5.1 Diamagnetism

5.5.2 Paramagnetism

5.5.3 Ferromagnetism

Magnetism and Matter: key points and summary

Practice questions with answers (NCERT “Exercises”)

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Key terms (English–Hindi)

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