PhysicsISC Class 12 · Chapter 5

Magnetism and Matter ISC Class 12: Notes, Concepts and Solved Questions

Updated 14 October 2026 · Fact-checked

Magnetism and Matter is Chapter 5 of ISC Class 12 Physics. It treats a current loop and a bar magnet as magnetic dipoles, the torque on them in a uniform field, magnetic flux, the quantities B, H, M, susceptibility and relative permeability, diamagnetic, paramagnetic and ferromagnetic materials, Curie's law, electromagnets and the hysteresis loop.

What Magnetism and Matter is about

Chapter 4 showed that a current loop behaves like a tiny magnet with dipole moment $m = N I A$ . This chapter extends the idea to bar magnets and to matter itself.

A bar magnet behaves like a solenoid: its field lines leave the north end, curve round and re-enter at the south end, then continue inside the magnet from S to N, forming closed loops because there are no magnetic monopoles. The field on the axial (end-on) and equatorial (broadside-on) lines of a short magnet has the same form as for an electric dipole, but the 2027 syllabus treats these fields as qualitative only, with no derivations. A magnetic dipole in a uniform field feels a torque $τ = m \times B$ , and an orbiting electron has an orbital magnetic moment.

Magnetic flux $ϕ = B \cdot A$ is measured in weber; B is in tesla ( $1 T = 1 0^{4}$ gauss). Inside a material we use the magnetising field H, the magnetisation M (magnetic moment per unit volume) and $B = μ_{0} (H + M)$ , i.e. $H = \frac{B}{μ _{0}} - M$ . The susceptibility $χ_{m} = M / H$ and relative permeability $μ_{r} = 1 + χ_{m}$ classify materials: diamagnetic ( $χ < 0$ , tiny; weakly repelled), paramagnetic ( $χ > 0$ , small; weakly attracted; $χ \propto 1/ T$ by Curie's law) and ferromagnetic ( $χ$ very large; strongly attracted; becomes paramagnetic above the Curie temperature).

Finally, electromagnets and permanent magnets are chosen from the B-H (hysteresis) loop (qualitative only): soft iron (high permeability, low coercivity, thin loop) for electromagnets and transformer cores; steel or alloys (high retentivity and high coercivity) for permanent magnets. The Earth's magnetic field and magnetic elements were removed from the 2027 syllabus.

This chapter shares 16 theory marks with Chapters 4, 6 and 7; it mainly supplies one-mark items and short reasoning questions on magnetic materials.

Weightage: Magnetism and Matter is part of the unit Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carries 16 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 3 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Current loop and bar magnet as magnetic dipoles: A current loop has magnetic moment $m = N I A$ (A m²), and far away its field has the same form as that of a short bar magnet of moment m. A bar magnet is therefore equivalent to a solenoid: both have field lines that emerge from one end (north), loop round outside and pass back through the inside, forming closed loops. Iron filings or a plotting compass show the same pattern for both. The magnetic moment of a bar magnet points from its S pole to its N pole inside the magnet. An electron orbiting a nucleus is also a current loop, with orbital magnetic moment $μ_{l} = \frac{e v r}{2}$ (opposite to its angular momentum, because its charge is negative).; Worked example
A solenoid of 500 turns, cross-section $2 \times 1 0^{- 4} m^{2}$ , carries 0.5 A. Find its magnetic moment and the torque on it when its axis makes $3 0^{\circ}$ with a uniform field of 0.2 T.
Try it first, then show the solutionSolution
$m = N I A = 500 \times 0.5 \times 2 \times 1 0^{- 4} = 0.05 A m^{2}$
$τ = m B sin θ = 0.05 \times 0.2 \times 0.5 = 5 \times 1 0^{- 3} N m$
Answer: $m = 0.05 A m^{2}$ ; $τ = 5 \times 1 0^{- 3} N m$
Field of a bar magnet on its axis and equator (no derivation): For a short magnet of moment m, at distance r large compared with its length: end-on (axial) position $B = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}}$ , along $m$ ; broadside-on (equatorial) position $B = \frac{μ _{0}}{4 π} \frac{m}{r ^{3}}$ , opposite to $m$ . These mirror the electric dipole results, with $\frac{μ _{0}}{4 π}$ in place of $\frac{1}{4 π ε _{0}}$ and m in place of p. The 2027 syllabus says 'qualitative only, no derivations': know the formulas, the 2 : 1 ratio and the $1/ r^{3}$ fall-off, and draw the field lines. In a uniform field the magnet feels zero net force but a torque $τ = m \times B$ , which tends to align it with the field.; Worked example
A short bar magnet has moment $2 A m^{2}$ . Find the field it produces 10 cm from its centre (a) on its axis and (b) on its equatorial line.
Try it first, then show the solutionSolution
(a) $B = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}} = 1 0^{- 7} \times \frac{2 \times 2}{( 0.1 ) ^{3}} = 4 \times 1 0^{- 4} T$
(b) $B = 1 0^{- 7} \times \frac{2}{( 0.1 ) ^{3}} = 2 \times 1 0^{- 4} T$
Answer: (a) $4 \times 1 0^{- 4} T$ along $m$ (b) $2 \times 1 0^{- 4} T$ opposite to $m$
Magnetic field lines and magnetic flux: The magnetic field B is defined through $F = q (v \times B)$ , not through force on a 'unit pole'. Its lines form closed loops (outside a magnet from N to S, inside from S to N), never cross and are crowded where B is strong. Electric field lines, by contrast, start on positive charges and end on negative charges. Magnetic flux through an area: $ϕ = B \cdot A = B A cos θ$ (θ between B and the normal); for non-uniform B, $ϕ = \int B \cdot d A$ . Unit: weber (Wb) = T m². B is sometimes called flux density ( $ϕ / A$ when B ⊥ area), but this is not a defining equation, since B is a vector and φ/A a scalar. $1 T = 1 0^{4} g a u ss$ .; Worked example
A uniform field of 0.4 T passes through a loop of area $50 c m^{2}$ at $6 0^{\circ}$ to the loop's normal. Find the flux. Express B in gauss.
Try it first, then show the solutionSolution
$A = 50 \times 1 0^{- 4} = 5 \times 1 0^{- 3} m^{2}$
$ϕ = B A cos 6 0^{\circ} = 0.4 \times 5 \times 1 0^{- 3} \times 0.5 = 1 \times 1 0^{- 3} W b$
$0.4 T = 0.4 \times 1 0^{4} = 4000 g a u ss$
Answer: $1 \times 1 0^{- 3} W b$ ; 4000 G
B, H, M, susceptibility and relative permeability: When a material is placed in a magnetising field H (set up, for example, by a solenoid: $H = n I$ , unit A m⁻¹), its atomic dipoles give it a magnetisation M = net magnetic moment per unit volume (A m⁻¹). The total field is $B = μ_{0} (H + M)$ , so $H = \frac{B}{μ _{0}} - M$ . Susceptibility $χ_{m} = \frac{M}{H}$ (no unit) measures how easily the material is magnetised. Then $B = μ_{0} (1 + χ_{m}) H = μ_{0} μ_{r} H = μ H$ , with relative permeability $μ_{r} = 1 + χ_{m}$ . Diamagnetic: $χ_{m} < 0$ (small), $μ_{r} < 1$ . Paramagnetic: $χ_{m} > 0$ (small), $μ_{r} > 1$ . Ferromagnetic: $χ_{m} ≫ 1$ , $μ_{r} ≫ 1$ .; Worked example
A solenoid with 1000 turns per metre carries 2 A. Its core has relative permeability 400. Find H, M and B in the core.
Try it first, then show the solutionSolution
$H = n I = 1000 \times 2 = 2000 A m^{- 1}$
$χ_{m} = μ_{r} - 1 = 399$ ; $M = χ_{m} H = 399 \times 2000 = 7.98 \times 1 0^{5} A m^{- 1}$
$B = μ_{0} μ_{r} H = 4 π \times 1 0^{- 7} \times 400 \times 2000 = 1.01 T$
Answer: $H = 2000 A m^{- 1}$ , $M = 7.98 \times 1 0^{5} A m^{- 1}$ , $B \approx 1.0 T$
Diamagnetic, paramagnetic and ferromagnetic substances: The cause of magnetisation is the magnetic moment of the atoms, ions or molecules. Diamagnetic (bismuth, copper, water, gold, superconductors): atoms have zero net moment; an applied field induces a small moment opposite to it. B lines become slightly less dense inside; the material is weakly repelled and a small bar sets perpendicular to the field. $χ$ is small, negative and independent of temperature. Paramagnetic (aluminium, platinum, sodium, oxygen): atoms have a small permanent moment, randomly oriented; the field aligns them slightly along it. Lines become slightly denser; weak attraction; a bar sets parallel to the field; $χ$ small and positive, $χ \propto 1/ T$ (Curie's law). Ferromagnetic (iron, cobalt, nickel, gadolinium): atoms have large moments aligned in domains; the field grows aligned domains, giving very large magnetisation along the field. Lines become much denser; strong attraction; $χ$ very large and depends on T in a complicated way. Above the Curie temperature a ferromagnet becomes paramagnetic.; Worked example
Classify materials with relative permeabilities (a) 0.99998, (b) 1.0004 and (c) 2000. How would a small rod of each set itself in a strong uniform field?
Try it first, then show the solutionSolution
(a) $μ_{r} < 1$ , $χ = - 2 \times 1 0^{- 5}$ : diamagnetic; sets perpendicular to the field.
(b) $μ_{r}$ slightly > 1, $χ = 4 \times 1 0^{- 4}$ : paramagnetic; sets parallel to the field.
(c) $μ_{r} ≫ 1$ : ferromagnetic; sets parallel to the field and is strongly attracted.
Answer: (a) diamagnetic, perpendicular (b) paramagnetic, parallel (c) ferromagnetic, parallel
Curie's law and the Curie temperature: For a paramagnetic substance, thermal agitation opposes the alignment of atomic dipoles, so the susceptibility falls as temperature rises: $χ_{m} = \frac{C}{T}$ (Curie's law, T in kelvin, C the Curie constant). Diamagnetic susceptibility does not depend on temperature. For a ferromagnet, heating disturbs the domain alignment; at the Curie temperature $T_{C}$ the domain structure breaks down and the substance becomes paramagnetic. So 'the Curie temperature of iron is 770 °C' means iron is ferromagnetic below 770 °C and paramagnetic above it. (Do not confuse Curie's law, which is about paramagnets, with the Curie temperature, which is about ferromagnets.); Worked example
A paramagnetic salt has susceptibility $2.8 \times 1 0^{- 4}$ at 300 K. Find its susceptibility at 200 K.
Try it first, then show the solutionSolution
Curie's law: $χ \propto \frac{1}{T}$ , so $χ_{2} = χ_{1} \frac{T _{1}}{T _{2}}$
$χ_{2} = 2.8 \times 1 0^{- 4} \times \frac{300}{200} = 4.2 \times 1 0^{- 4}$
Answer: $4.2 \times 1 0^{- 4}$
Electromagnets, permanent magnets and the hysteresis loop: An electromagnet is a soft-iron core inside a current-carrying coil: it is strongly magnetised while the current flows and loses most of its magnetism when it stops. Its strength increases with the number of turns, the current, and the permeability of the core (and a smaller air gap between poles). A B-H loop (hysteresis loop, qualitative only) shows that when H is reduced to zero after magnetising a ferromagnet, some B remains: the retentivity (remanence). The reverse H needed to bring B to zero is the coercivity. The loop's area represents energy lost as heat per cycle. Choice of material: permanent magnets need high retentivity and high coercivity (steel, alnico); electromagnets and transformer cores need high permeability, low coercivity and a thin loop (low hysteresis loss), so soft iron is used.; Worked example
Material P has retentivity 1.2 T and coercivity 50 A m⁻¹, with a narrow loop. Material Q has retentivity 0.9 T and coercivity $5 \times 1 0^{4} A m^{- 1}$ , with a wide loop. Which would you choose for (a) a transformer core and (b) a permanent magnet? Give reasons.
Try it first, then show the solutionSolution
(a) Transformer core is taken round the loop 50 times a second: it needs a narrow loop (little energy loss) and low coercivity, so P.
(b) A permanent magnet must keep its magnetism and resist demagnetisation: it needs high coercivity (and good retentivity), so Q.
Answer: (a) P (soft, narrow loop) (b) Q (hard, high coercivity)

Physics in real life: Magnetism and Matter

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electromagnets with a ferromagnetic core

Suppose a scrapyard crane lifts old cars with an electromagnet: a coil wound on a soft-iron core.

Suppose the coil has 1000 turns per metre and carries 2 A, and the iron core has relative permeability 500. Find B in the core and compare it with the field without the core.

Try it first, then show the solutionSolution

$H = n I = 2000 A m^{- 1}$ ; without core $B_{0} = μ_{0} H = 2.51 \times 1 0^{- 3} T$
With core $B = μ_{0} μ_{r} H = 500 \times 2.51 \times 1 0^{- 3} = 1.26 T$

Answer: About 1.26 T, 500 times stronger than the air-cored coil; switching off the current releases the load because soft iron keeps little magnetism.

Memory hook: Soft iron: strong while switched on, weak once switched off.

Curie's law

Imagine physicists cooling a paramagnetic salt to very low temperatures and using its susceptibility as a thermometer.

Suppose the salt's susceptibility is $3 \times 1 0^{- 4}$ at 300 K. What will it be at 4 K?

Try it first, then show the solutionSolution

$χ \propto 1/ T$ : $χ_{2} = 3 \times 1 0^{- 4} \times \frac{300}{4}$
$χ_{2} = 0.0225$

Answer: 0.0225, 75 times larger; the cold salt magnetises far more easily because thermal agitation is weaker.

Memory hook: Cool a paramagnet, and its dipoles fall into line.

Perfect diamagnetism

Suppose you watch a small magnet floating above a superconducting disc cooled in liquid nitrogen.

A superconductor behaves as a perfect diamagnet with $χ_{m} = - 1$ . If the magnet sets up H = 2000 A m⁻¹ at the disc, find M and B inside the superconductor.

Try it first, then show the solutionSolution

$M = χH = - 1 \times 2000 = - 2000 A m^{- 1}$ (opposite to H)
$B = μ_{0} (H + M) = μ_{0} (2000 - 2000) = 0$

Answer: M = −2000 A m⁻¹; B = 0 inside. The field is pushed out, so the magnet is repelled and floats.

Memory hook: Chi of minus one: no field gets in.

Topics in this chapter

Magnetism and Matter formula sheet

Magnetic moment of a current loop: $m = N I A$; Unit A m².
Orbital magnetic moment: $μ_{l} = \frac{e v r}{2}$; Opposite to the electron's angular momentum.
Short bar magnet: axial field (no derivation): $B_{a x ia l} = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}}$; Along $m$ .
Short bar magnet: equatorial field (no derivation): $B_{e q} = \frac{μ _{0}}{4 π} \frac{m}{r ^{3}}$; Opposite to $m$ ; $B_{a x ia l} = 2 B_{e q}$ at equal r.
Torque on a magnetic dipole: $τ = m \times B, τ = m B sin θ$; Net force zero in a uniform field.
Magnetic flux: $ϕ = B \cdot A = B A cos θ, ϕ = \int B \cdot d A$; Unit weber; $1 T = 1 W b m^{- 2} = 1 0^{4} G$ .
Magnetising field of a solenoid: $H = n I$; Unit A m⁻¹.
B, H and M: $B = μ_{0} (H + M), H = \frac{B}{μ _{0}} - M$; M = magnetic moment per unit volume.
Susceptibility and permeability: $χ_{m} = \frac{M}{H}, μ_{r} = 1 + χ_{m}, B = μ_{0} μ_{r} H$; Dia: $χ < 0$ ; para: small $χ > 0$ ; ferro: $χ ≫ 1$ .
Curie's law: $χ_{m} = \frac{C}{T}$; Paramagnets only; T in kelvin.

Magnetism and Matter: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Field lines of a bar magnet and of a solenoid

'Draw the magnetic field lines of a bar magnet' or 'show that a bar magnet is equivalent to a solenoid'.

Draw

Bar magnet: a rectangle marked N and S; lines leave N, curve round outside and enter S; continue them inside the magnet from S to N so each line is closed.
Solenoid: a coil with current arrows; draw the same pattern, with straight parallel lines inside and the outside lines curving from one end to the other.
Mark the direction of $m$ (S to N inside the magnet; by the right-hand rule for the solenoid).

Label

N, S
closed field lines with arrows
$m$
current direction in solenoid

Watch out: Show the lines inside the magnet; leaving them out suggests that the lines begin and end on the poles, which is wrong.

Behaviour of dia, para and ferro materials in a uniform field

Comparing the three classes; examiners expect the pattern of field lines through each sample.

Draw

Three boxes, each with parallel field lines entering from the left.
Diamagnetic sample: lines bend round and thin out inside it.
Paramagnetic sample: lines slightly crowd into it. Ferromagnetic sample: lines strongly crowd into it.

Label

external field $B_{0}$
sample: dia / para / ferro
less dense / more dense / much more dense

Watch out: Add one line under each box: repelled weakly / attracted weakly / attracted strongly.

B-H (hysteresis) loop

Qualitative questions on retentivity, coercivity and the choice of material.

Draw

Axes: H horizontal, B vertical. Start at the origin and draw the magnetisation curve rising to saturation.
Reduce H to zero: the curve meets the B-axis above the origin (retentivity). Reverse H: B reaches zero at the coercivity on the negative H-axis.
Complete the closed loop symmetrically through negative saturation; draw a narrow loop for soft iron and a wide loop for steel.

Label

B, H axes
saturation
retentivity (remanence)
coercivity
soft iron (narrow)
steel (wide)

Watch out: The loop area is the energy lost per cycle per unit volume; mention it when justifying a transformer core.

How the board asks Magnetism and Matter

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Classify the material (Section A)	From μr or χ values, from behaviour in a field (repelled, attracted, rod perpendicular or parallel), or an odd-one-out list.	μr < 1: dia; slightly > 1: para; ≫ 1: ferro. Dia sets perpendicular, para and ferro parallel.
Meaning of a statement	'Curie temperature of iron is 770 °C', 'susceptibility of a material is −2 × 10⁻⁵', 'relative permeability is 1000'.	Translate into behaviour: what the material does and in which temperature range.
B, H, M numericals	Find M, B or μr for a core in a solenoid; use Curie's law to find χ at another temperature.	H = nI first; then $M = χH$ and $B = μ_{0} (H + M)$ .
Torque on a magnetic dipole	Torque on a magnet or solenoid at an angle to B; angle at which torque is a given fraction of maximum.	$τ = m B sin θ$ ; find m = NIA for a solenoid.
Hysteresis and choice of materials	Why soft iron for electromagnets and transformer cores, steel for permanent magnets; reading retentivity and coercivity from a loop.	Link each use to retentivity, coercivity and loop area.
Comparison tables (3 or 5 marks)	Three differences between dia, para and ferromagnetic substances, with examples.	Use the same headings for all three: atomic moment, χ, μr, behaviour in field, temperature dependence, examples.

Tricks to remember Magnetism and Matter

Short memory aids. Say each one aloud once, then use it on a question the same day.

Dia-Para-Ferro: Down-Positive-Far
Diamagnetic χ goes Down (negative); Paramagnetic χ is Positive but small; Ferromagnetic χ is Far bigger.
Use it when: Classifying from χ or μr.
Dia stands across
A diamagnetic rod 'disagrees' and stands across (perpendicular to) the field; para and ferro lie along it.
Use it when: Behaviour of suspended rods.
Curie's law vs Curie temperature
Law = para (χ ∝ 1/T). Temperature = ferro (the point where ferro turns para).
Use it when: Avoiding the confusion examiners report.
Hard magnets hold on
Permanent magnet: High retentivity, High coercivity (HH). Core: low coercivity, thin loop.
Use it when: Material choice questions.
B = μ0(H + M)
Total field = vacuum part plus the material's own part: 'B is H plus M, times μ-nought'.
Use it when: Any B-H-M numerical.

Magnetism and Matter: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Which material is repelled? · 1 mark · MCQBoard level

A small rod of which of these materials, suspended in a strong uniform magnetic field, sets itself perpendicular to the field?
(a) Aluminium
(b) Bismuth
(c) Nickel
(d) Platinum

Try it first, then show the solutionSolution

Only a diamagnetic rod sets perpendicular. Aluminium and platinum are paramagnetic, nickel is ferromagnetic, bismuth is diamagnetic.

Answer: (b) Bismuth

Example 2: Curie temperature · 1 mark · Assertion-reasonBoard level

Assertion: Above its Curie temperature, iron behaves as a paramagnetic substance.
Reason: Thermal agitation above the Curie temperature destroys the alignment of magnetic moments within the domains.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Both statements are true; loss of domain alignment is exactly why iron stops being ferromagnetic.

Answer: (a)

Example 3: Magnetisation in a solenoid core · 2 marks · Short answerHarder board level

A long solenoid with 1000 turns per metre carries 2 A. A core of relative permeability 400 fills it. Find the magnetisation of the core and the field B inside. [2]

Try it first, then show the solutionSolution

$H = n I = 2000 A m^{- 1}$ ; $χ = μ_{r} - 1 = 399$ ; $M = χH = 7.98 \times 1 0^{5} A m^{- 1}$
$B = μ_{0} (H + M) = 4 π \times 1 0^{- 7} \times 8.0 \times 1 0^{5} = 1.01 T$

Answer: $M = 7.98 \times 1 0^{5} A m^{- 1}$ ; $B \approx 1.0 T$

Example 4: Solenoid as a bar magnet · 3 marks · Short answerHarder board level

(a) Explain why a current-carrying solenoid behaves like a bar magnet. (b) A solenoid of 500 turns and cross-section $2 \times 1 0^{- 4} m^{2}$ carries 0.5 A. It is placed with its axis at θ to a 0.2 T field. Find the maximum torque, and the angle at which the torque is half the maximum. [3]

Try it first, then show the solutionSolution

(a) Each turn is a current loop with moment NIA along the axis; the loops' fields add inside and the external pattern matches a bar magnet's, with one end acting as N and the other as S. Like a magnet, it aligns with an external field.
(b) $m = 500 \times 0.5 \times 2 \times 1 0^{- 4} = 0.05 A m^{2}$ ; $τ_{ma x} = m B = 0.01 N m$ (axis perpendicular to B).
$τ = \frac{1}{2} τ_{ma x} \Rightarrow sin θ = 0.5 \Rightarrow θ = 3 0^{\circ}$ (or $15 0^{\circ}$ ).

Answer: $τ_{ma x} = 0.01 N m$ ; θ = 30° or 150°

Example 5: Hysteresis and material choice · 3 marks · Short answerHarder board level

Sketch a B-H loop for a ferromagnetic material and mark retentivity and coercivity on it. Use the loop to explain why soft iron is used in transformer cores and steel in permanent magnets. [3]

Try it first, then show the solutionSolution

Loop: rises to saturation; at H = 0, B is still positive (retentivity); B becomes zero at a negative H (coercivity); symmetric closed loop.
Transformer core: magnetised and demagnetised every cycle; soft iron has low coercivity and a narrow loop, so little energy is lost as heat per cycle.
Permanent magnet: must keep its magnetism and resist stray fields; steel has high coercivity and good retentivity (a wide loop).

Answer: Soft iron: narrow loop, low coercivity, small loss. Steel: high coercivity and retentivity.

Example 6: Comparing the three classes of materials · 5 marks · Long answerHarder board level

(a) Give three differences between diamagnetic, paramagnetic and ferromagnetic substances, with one example of each. [3]
(b) The susceptibility of a paramagnetic material is $2.8 \times 1 0^{- 4}$ at 300 K. Find it at 200 K, and its relative permeability at 200 K. [2]

Try it first, then show the solutionSolution

Atomic moment: zero (dia); small, random (para); large, aligned in domains (ferro).
Susceptibility: small negative (dia); small positive (para); very large positive (ferro). Behaviour: weakly repelled; weakly attracted; strongly attracted.
Temperature: dia χ independent of T; para χ ∝ 1/T; ferro becomes para above the Curie temperature. Examples: bismuth; aluminium; iron.
(b) $χ_{200} = 2.8 \times 1 0^{- 4} \times \frac{300}{200} = 4.2 \times 1 0^{- 4}$
$μ_{r} = 1 + χ = 1.00042$

Answer: (b) $χ = 4.2 \times 1 0^{- 4}$ , $μ_{r} = 1.00042$

Example 7: A perfect diamagnet · 3 marks · Short answerHOTS / competency

A superconductor expels magnetic field completely from its interior. (a) What are its susceptibility and relative permeability? (b) Sketch the field lines of a uniform field near a superconducting sphere. (c) Why does a magnet float above it? [3]

Try it first, then show the solutionSolution

(a) B = 0 inside: $μ_{0} (H + M) = 0 \Rightarrow M = - H$ , so $χ = - 1$ and $μ_{r} = 1 + χ = 0$ .
(b) Field lines bend round the sphere; none pass through it, and they crowd at its sides.
(c) The superconductor's induced magnetisation opposes the magnet's field, so it repels the magnet; the repulsion balances the magnet's weight.

Answer: χ = −1, μr = 0; lines go round, not through; repulsion supports the magnet

Example 8: Curie's law with a temperature change · 2 marks · Short answerHOTS / competency

The magnetisation of a paramagnetic sample in a fixed field is M at 27 °C. At what temperature will it be 1.5 M in the same field? [2]

Try it first, then show the solutionSolution

At fixed H, $M = χH \propto \frac{1}{T}$ ; T must be in kelvin: 27 °C = 300 K.
$\frac{M _{2}}{M _{1}} = \frac{T _{1}}{T _{2}} \Rightarrow 1.5 = \frac{300}{T _{2}} \Rightarrow T_{2} = 200 K = - 73^{\circ} C$

Answer: 200 K (−73 °C)

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Identifying three unknown rods

Suppose a student has three rods X, Y and Z of equal size. She hangs each between the poles of a strong electromagnet and records: X sets itself perpendicular to the field and is pushed slightly out; Y sets along the field and is pulled in slightly; Z sets along the field and is pulled in strongly. Their relative permeabilities, in some order, are 1.00002, 0.99983 and 1500.

(i) · 1 mark
Match each rod to its relative permeability.
Answer
Answer: X: 0.99983; Y: 1.00002; Z: 1500
Repelled and perpendicular means diamagnetic (μr < 1); weak attraction means paramagnetic; strong attraction means ferromagnetic.
(ii) · 1 mark
Find the magnetisation of Y when H = 5000 A m⁻¹.
Answer
Answer: 0.1 A m⁻¹
$χ = 1.00002 - 1 = 2 \times 1 0^{- 5}$ ; $M = χH = 2 \times 1 0^{- 5} \times 5000 = 0.1 A m^{- 1}$
(iii) · 1 mark
Rod Y is cooled from 300 K to 150 K. What happens to its susceptibility?
Answer
Answer: It doubles, to $4 \times 1 0^{- 5}$ .
Curie's law: $χ \propto 1/ T$ ; halving T doubles χ.
(iv) · 2 marks
Rod Z is heated well above its Curie temperature. How does its behaviour in the field change, and which rod would it then resemble?
Answer
Answer: It loses its strong attraction and becomes weakly attracted: it resembles Y.
Above the Curie temperature thermal agitation destroys domain alignment, so the ferromagnet becomes paramagnetic, with small positive χ.

A solenoid that acts as a compass

Suppose a solenoid of 500 turns and cross-section $2 \times 1 0^{- 4} m^{2}$ , carrying 0.5 A, is hung so it can turn freely in a horizontal uniform field of 0.2 T.

(i) · 1 mark
Find its magnetic moment.
Answer
Answer: $0.05 A m^{2}$
$m = N I A = 500 \times 0.5 \times 2 \times 1 0^{- 4} = 0.05 A m^{2}$
(ii) · 1 mark
What is the torque when its axis is perpendicular to the field?
Answer
Answer: 0.01 N m
$τ = m B = 0.05 \times 0.2 = 0.01 N m$
(iii) · 1 mark
In which position does it come to rest, and is the net force on it zero?
Answer
Answer: With its axis (and $m$ ) along B; net force zero.
Torque vanishes when $m ∥ B$ ; in a uniform field the forces on its ends cancel.
(iv) · 2 marks
Using the formula for a short magnet, estimate the field it produces at a point 20 cm along its axis (ignore the external field).
Answer
Answer: $1.25 \times 1 0^{- 6} T$
$B = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}} = 1 0^{- 7} \times \frac{2 \times 0.05}{( 0.2 ) ^{3}} = 1.25 \times 1 0^{- 6} T$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The relative permeability of a substance is 0.99983. (a) Name its class. (b) How will a small rod of it set itself in a strong uniform magnetic field? (2 marks)

Step	Marks
(a) μr < 1, so χ is negative: diamagnetic	1
(b) Perpendicular to the field (weakly repelled from stronger field regions)	1

Define magnetic susceptibility and relative permeability, and derive the relation between them. (3 marks)

Step	Marks
$χ_{m} = M / H$ : ratio of magnetisation to magnetising field	1
$μ_{r} = μ / μ_{0}$ , where $B = μ H$	1
$B = μ_{0} (H + M) = μ_{0} (1 + χ_{m}) H \Rightarrow μ_{r} = 1 + χ_{m}$	1

Compare diamagnetic, paramagnetic and ferromagnetic substances (three points with examples), and state Curie's law. (5 marks)

Step	Marks
Atomic magnetic moment: zero / small permanent / large, in domains	1
Susceptibility and μr: small negative, μr < 1 / small positive, μr > 1 / very large	1
Behaviour in a field: weakly repelled, perpendicular / weakly attracted, parallel / strongly attracted, parallel	1
Examples: bismuth or copper / aluminium or platinum / iron, nickel or cobalt	1
Curie's law: for a paramagnet $χ_{m} = C / T$	1

Where students lose marks in Magnetism and Matter

The slips examiners see most often, and what to write instead.

Mistake: Misreading the meaning of Curie temperature
CISCE examiners' comments, 2024: explaining 'Curie temperature for soft iron is 770 °C', candidates described a change of crystal lattice, said the iron melts, said it becomes diamagnetic, or confused it with Curie's law.
Wrong working
At 770 °C soft iron melts and loses its magnetism.
Right working
Below 770 °C soft iron is ferromagnetic; above it, it becomes paramagnetic.
Do this: Curie temperature: ferro → para. Curie's law: χ ∝ 1/T for para.
Usually costs: 1 mark.
Mistake: Classifying by μr the wrong way round
CISCE examiners' comments, 2025: given one material with μr slightly below 1 and another slightly above 1, candidates called the second ferromagnetic instead of paramagnetic.
Wrong working
μr slightly more than 1: ferromagnetic.
Right working
Slightly > 1: paramagnetic. Ferromagnetic needs μr ≫ 1 (hundreds or thousands).
Do this: Learn the three ranges together: < 1, slightly > 1, ≫ 1.
Usually costs: 1 mark.
Mistake: Drawing bar-magnet field lines that stop at the poles
Diagrams without lines inside the magnet suggest magnetic monopoles; the syllabus stresses that B lines form closed loops.
Wrong working
Lines drawn only outside, starting at N and ending at S.
Right working
Lines continue inside from S to N, forming closed loops.
Do this: Always complete the loop inside the magnet.
Usually costs: 1 mark.
Mistake: Defining B through force on a unit pole
The ISC syllabus says B is defined by $F = q (v \times B)$ and not by force on a unit pole.
Wrong working
B is the force on a unit north pole.
Right working
B is defined so that a charge q moving with velocity v feels $F = q (v \times B)$ ; unit tesla.
Do this: Use the moving-charge definition.
Usually costs: 1 mark.
Mistake: Choosing the wrong material for a transformer core
Candidates often pick the material with the higher retentivity, forgetting that a core is cycled continuously.
Wrong working
Steel, because it keeps its magnetism.
Right working
Soft iron: high permeability, low coercivity, narrow loop, so low hysteresis loss.
Do this: Ask: does the use need the magnetism to stay (permanent magnet) or to follow the current (core)?
Usually costs: 1-2 marks.

Practice questions on Magnetism and Matter

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of magnetic flux is:
(a) tesla
(b) weber
(c) gauss
(d) henry
Hint
Flux = B × area.
Answer
(b) weber
Q2 · 1 mark · MCQBoard level
The susceptibility of a paramagnetic substance is:
(a) small and negative
(b) small and positive
(c) large and positive
(d) zero
Hint
Weak attraction.
Answer
(b) small and positive
Q3 · 1 mark · MCQBoard level
A material for a permanent magnet should have:
(a) low retentivity and low coercivity
(b) high retentivity and low coercivity
(c) high retentivity and high coercivity
(d) low retentivity and high coercivity
Hint
It must keep its magnetism.
Answer
(c)
Q4 · 1 mark · Very short answerBoard level
The Curie temperature of nickel is 358 °C. What does this statement mean?
Hint
Ferro to para.
Answer
Nickel is ferromagnetic below 358 °C and becomes paramagnetic above it.
Q5 · 2 marks · Short answerBoard level
Classify substances with susceptibilities (a) $- 1.6 \times 1 0^{- 4}$ , (b) $2.1 \times 1 0^{- 5}$ , (c) 1200. Which one is attracted most strongly by a magnet?
Hint
Sign and size of χ.
Answer
(a) diamagnetic (b) paramagnetic (c) ferromagnetic; (c) is attracted most strongly.
Q6 · 2 marks · Short answerHarder board level
A field of 0.4 T crosses a loop of area $50 c m^{2}$ at $6 0^{\circ}$ to its normal. Find the flux through the loop. What angle gives zero flux?
Hint
θ is from the normal.
Answer
$ϕ = 0.4 \times 5 \times 1 0^{- 3} \times cos 6 0^{\circ} = 1 \times 1 0^{- 3} W b$ ; zero flux when the field is parallel to the plane of the loop (θ = 90°).
Q7 · 2 marks · Short answerHarder board level
A short bar magnet has moment $2 A m^{2}$ . Using the standard formulas, find the field at 10 cm on its axis and on its equatorial line. How does each change if the distance is doubled?
Hint
$B \propto 1/ r^{3}$ .
Answer
$4 \times 1 0^{- 4} T$ (axis) and $2 \times 1 0^{- 4} T$ (equator); doubling r divides each by 8 (to $5 \times 1 0^{- 5}$ and $2.5 \times 1 0^{- 5} T$ ).
Q8 · 2 marks · Short answerHOTS / competency
A U-tube contains a liquid; one limb is placed between the poles of a strong electromagnet. Explain what happens to the liquid level in that limb if the liquid is (a) paramagnetic, (b) diamagnetic.
Hint
Which way does each class move in a non-uniform field?
Answer
(a) Paramagnetic liquid is attracted towards the stronger field, so the level rises in the limb between the poles. (b) Diamagnetic liquid is pushed towards the weaker field, so the level falls there.
Q9 · 3 marks · Short answerHarder board level
List three factors on which the strength of an electromagnet depends. A coil with 1000 turns per metre carrying 2 A has a core of μr = 500. Find B inside.
Hint
$B = μ_{0} μ_{r} n I$ .
Answer
Number of turns per unit length, current, and permeability of the core (also the air gap). $B = 4 π \times 1 0^{- 7} \times 500 \times 2000 = 1.26 T$ .
Q10 · 3 marks · Short answerHarder board level
A rod of susceptibility 599 is placed in a field H = 1500 A m⁻¹. Find M, μr and B in the rod.
Hint
$B = μ_{0} (H + M)$ .
Answer
$M = 599 \times 1500 = 8.99 \times 1 0^{5} A m^{- 1}$ ; $μ_{r} = 600$ ; $B = 4 π \times 1 0^{- 7} \times 9.0 \times 1 0^{5} = 1.13 T$ .
Q11 · 3 marks · Short answerHarder board level
A solenoid of moment $0.05 A m^{2}$ is in a uniform field of 0.2 T. Find the torque on it when its axis makes (a) 0°, (b) 30°, (c) 90° with the field. In which position is it in stable equilibrium?
Hint
$τ = m B sin θ$ .
Answer
(a) 0 (b) $5 \times 1 0^{- 3} N m$ (c) 0.01 N m. Stable at 0° (moment along the field).
Q12 · 5 marks · Long answerHarder board level
(a) Explain diamagnetism, paramagnetism and ferromagnetism in terms of the magnetic moments of atoms. [3]
(b) State Curie's law. A paramagnetic sample has χ = $3 \times 1 0^{- 4}$ at 300 K. Find χ at 4 K. [2]
Hint
Zero, small random, large aligned moments.
Answer
(a) Dia: atoms have zero net moment; a field induces a small moment opposite to it. Para: atoms have small permanent moments, randomly oriented; a field partly aligns them along it, against thermal agitation. Ferro: large atomic moments are aligned within domains; a field grows and rotates domains along it, giving very large M. (b) $χ = C / T$ ; $χ_{4} = 3 \times 1 0^{- 4} \times \frac{300}{4} = 0.0225$ .
Q13 · 2 marks · Short answerHarder board level
Show that a current loop behaves as a magnetic dipole, and find the moment of a solenoid of 500 turns, area $2 \times 1 0^{- 4} m^{2}$ , carrying 0.5 A.
Hint
Compare the far axial field of a loop with that of a bar magnet.
Answer
Far along its axis a loop's field is $\frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$ with m = NIA, the same form as a short bar magnet's, and it feels torque $m \times B$ , so it is a magnetic dipole. Here $m = 500 \times 0.5 \times 2 \times 1 0^{- 4} = 0.05 A m^{2}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define magnetisation.Show answer
Net magnetic moment per unit volume of a material, M (A m⁻¹).
2.Relation between B, H and M?Show answer
$B = μ_{0} (H + M)$ .
3.Define magnetic susceptibility.Show answer
$χ_{m} = M / H$ .
4.State Curie's law.Show answer
The susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature: $χ = C / T$ .
5.What happens to a ferromagnet above its Curie temperature?Show answer
It becomes paramagnetic.
6.Define retentivity and coercivity.Show answer
Retentivity: B remaining when H is reduced to zero. Coercivity: reverse H needed to reduce B to zero.
7.Why do magnetic field lines form closed loops?Show answer
Because there are no magnetic monopoles; lines continue inside a magnet from S to N.
8.1 tesla in gauss?Show answer
$1 0^{4}$ gauss.

Best order to study it

Revise the current loop as a magnetic dipole (m = NIA) and the orbital moment of an electron.
Learn the bar magnet as an equivalent solenoid, its field lines and the axial and equatorial formulas (no derivations).
Learn torque on a magnetic dipole and magnetic flux, with units.
Learn H, M, B, χ and μr and practise the numericals.
Make a comparison table of dia, para and ferro; learn Curie's law and the Curie temperature.
Finish with electromagnets, the hysteresis loop and the choice of materials.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you show why a solenoid behaves like a bar magnet?
Can you draw closed field lines for a bar magnet?
Can you write the axial and equatorial field formulas for a short magnet and their ratio?
Can you use $τ = m B sin θ$ and $ϕ = B A cos θ$ ?
Can you use $B = μ_{0} (H + M)$ , $χ = M / H$ and $μ_{r} = 1 + χ$ ?
Can you classify materials from μr or χ and describe their behaviour in a field?
Can you state Curie's law and explain the Curie temperature without confusing them?
Can you sketch a hysteresis loop and choose materials for a core and a permanent magnet?

Magnetism and Matter: frequently asked questions

Do I need to derive the field of a bar magnet for ISC 2027?

No. The revised syllabus says the field on the axial and equatorial lines of a bar magnet is qualitative only, with no derivations. Learn the formulas and the 2 : 1 ratio.

Is the Earth's magnetic field in the 2027 syllabus?

No. Earth's magnetic field and the magnetic elements (declination, dip, horizontal component) were removed in the February 2026 revision.

Do I need to draw and explain the hysteresis loop?

Yes, qualitatively. You should sketch a B-H loop, mark retentivity and coercivity, and use them to choose materials for permanent magnets, electromagnets and transformer cores. No calculation of loop area is needed.

What is the difference between Curie's law and Curie temperature?

Curie's law (\chi = C/T) describes paramagnetic substances. The Curie temperature is the temperature above which a ferromagnetic substance becomes paramagnetic.

How is H different from B?

H is the magnetising field set up by free currents (H = nI in a solenoid, unit A m⁻¹). B is the total field including the material's contribution: B = \mu_0(H + M), unit tesla.

How much is this chapter worth?

It shares 16 of the 70 theory marks with Moving Charges and Magnetism, Electromagnetic Induction and Alternating Current; expect one-mark and short reasoning items on materials.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024