PhysicsISC Class 12 · Magnetism and Matter

Diamagnetic, Paramagnetic and Ferromagnetic Materials, Curie's Law and Curie Temperature

Updated 15 October 2026 · Fact-checked

Diamagnetic materials have small negative susceptibility and are weakly repelled by magnets; paramagnetic materials have small positive susceptibility and are weakly attracted; ferromagnetic materials have very large positive susceptibility and are strongly attracted. For paramagnets, Curie's law gives $\chi_m = \frac{C}{T}$. Above its Curie temperature a ferromagnet becomes paramagnetic.

Compare the arrows: diamagnets get tiny opposite moments, paramagnets align partly, ferromagnets align strongly with the field.

Illustration drawn for QuestionBuddy.

Understand Magnetic Materials and Curie's Law

Where magnetism in matter comes from. Electrons orbiting and spinning give atoms magnetic dipole moments. How these moments behave in an external field decides the class of a material. The key quantities are the magnetisation M (magnetic moment per unit volume), the susceptibility $χ_{m} = \frac{M}{H}$ and the relative permeability $μ_{r} = 1 + χ_{m}$ .

Diamagnetic substances (bismuth, copper, water, silver, gold, most organic compounds).

Atoms have no net permanent moment; the orbital and spin moments cancel.
An external field induces small moments opposite to the field (by changing the electrons' orbital motion), so M is opposite to H: $χ_{m}$ small and negative (about $- 1 0^{- 5}$ ), $μ_{r}$ slightly less than 1.
Field lines are slightly expelled (less dense inside). In a non-uniform field the material moves from stronger to weaker field: weak repulsion. A small rod sets perpendicular to the field.
$χ_{m}$ is independent of temperature. A superconductor is a perfect diamagnet ( $χ_{m} = - 1$ ).

Paramagnetic substances (aluminium, platinum, sodium, oxygen, copper sulphate).

Atoms have permanent moments, but they point randomly because of thermal agitation, so there is no net moment without a field.
An external field partly aligns them: M is along H, $χ_{m}$ small and positive (about $1 0^{- 5}$ to $1 0^{- 3}$ ), $μ_{r}$ slightly more than 1.
Field lines are slightly concentrated; weak attraction; a rod sets parallel to the field.
Alignment fights thermal disorder, so $χ_{m}$ depends on temperature.

Curie's law. For a paramagnet (in weak fields, away from saturation),
$χ_{m} = \frac{C}{T}, equivalently M = C \frac{B}{μ _{0} T} (M \propto \frac{B}{T}),$
where T is the absolute temperature and C the Curie constant. Halving T doubles $χ_{m}$ . Curie's law is a law about paramagnets; it is not the same thing as the Curie temperature, which the 2024 ISC examiners saw candidates confuse.

Ferromagnetic substances (iron, cobalt, nickel, gadolinium, alloys such as alnico).

Strong interactions line up neighbouring atomic moments in small regions called domains (about 1 mm or less), each magnetised to saturation; in an unmagnetised sample the domains point in random directions.
An external field makes favourably aligned domains grow and others rotate, giving a very large M along H: $χ_{m}$ very large (hundreds to thousands), $μ_{r} ≫ 1$ .
Field lines crowd into the material; strong attraction; rod sets parallel to the field. M depends on the past history of magnetisation (hysteresis).

Curie temperature. On heating, thermal agitation eventually breaks up the domain alignment. Above a definite temperature $T_{C}$ , the Curie temperature, a ferromagnet loses its ferromagnetism and becomes paramagnetic. So "the Curie temperature of soft iron is 770 °C" means that at and above 770 °C soft iron behaves as a paramagnetic substance. It does not mean the iron melts, and its crystal change is not what is being described.

Summary table.

Property	Dia	Para	Ferro
$χ_{m}$	small, negative	small, positive	very large, positive
$μ_{r}$	slightly < 1	slightly > 1	≫ 1
In a non-uniform field	weak repulsion	weak attraction	strong attraction
Rod in uniform field	perpendicular	parallel	parallel
Temperature	independent	$χ_{m} = \frac{C}{T}$	para above $T_{C}$

Magnetic Materials and Curie's Law: formulas and key facts

Susceptibility: $χ_{m} = \frac{M}{H}$; Dimensionless.
Relative permeability: $μ_{r} = 1 + χ_{m}$; < 1 dia, > 1 para, ≫ 1 ferro.
Curie's law: $χ_{m} = \frac{C}{T}$; Paramagnets; T in kelvin.
Magnetisation of a paramagnet: $M \propto \frac{B}{T}$; Weak fields, far from saturation.
Comparing temperatures: $χ_{1} T_{1} = χ_{2} T_{2}$; Same paramagnet.

How to solve Magnetic Materials and Curie's Law questions

Classification questions read the sign and size of χ or μr; Curie's law questions are a simple inverse proportion.

1From μr: below 1 is diamagnetic, slightly above 1 paramagnetic, much greater than 1 ferromagnetic. From χ: negative, small positive, very large.
2Describe behaviour: repelled or attracted, and the rod's orientation.
3For Curie's law, convert all temperatures to kelvin first.
4Use $χ_{1} T_{1} = χ_{2} T_{2}$ (or $M \propto \frac{B}{T}$ when both field and temperature change).
5For Curie temperature statements, say the ferromagnet becomes paramagnetic at and above that temperature.
6Explain behaviour in terms of atomic moments: none (dia), random but alignable (para), aligned domains (ferro).

Magnetic Materials and Curie's Law in real life

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.

Diamagnetism of water

Suppose a strong permanent magnet is brought close to one end of a light straw carrying a water-filled grape at that end, balanced on a pivot, and suppose water has $χ_{m} \approx - 9 \times 1 0^{- 6}$ .

What happens to the grape, and what are the relative permeability and class of water?

Try it first, then show the solutionSolution

$μ_{r} = 1 + χ_{m} \approx 0.999991$ : slightly less than 1.
Water is diamagnetic, so the grape is pushed towards the weaker field: it moves slowly away from the magnet.

Answer: The grape is gently repelled; water is diamagnetic with $μ_{r}$ just below 1.

Memory hook: Dia = "away"; para = "towards" (weakly); ferro = "towards" (strongly).

Curie temperature

Imagine a blacksmith heating an iron horseshoe magnet to bright red heat, above its Curie temperature (suppose 770 °C), and letting it cool on a bench away from any magnet.

Will it still be a magnet afterwards? Explain.

Try it first, then show the solutionSolution

Above the Curie temperature thermal agitation destroys the alignment within domains: the iron becomes paramagnetic.
On cooling with no external field, domains re-form pointing in random directions, so there is no net magnetisation.

Answer: No: it is demagnetised.

Memory hook: Above $T_{C}$ , ferro turns para.

Tricks to remember Magnetic Materials and Curie's Law

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

Minus, small plus, huge plus
χ for dia, para, ferro: negative, small positive, huge positive.
Use it when: Classifying from data.
Dia is "across"
A diamagnetic rod lies across (perpendicular to) the field; para and ferro lie along it.
Use it when: Orientation questions.
Colder para, stronger para
Curie's law: χ ∝ 1/T.
Use it when: Temperature questions.
Law vs temperature
Curie's law → paramagnets; Curie temperature → ferromagnets turning para.
Use it when: Avoiding the 2024 confusion.

Magnetic Materials and Curie's Law: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Negative susceptibility · 1 mark · MCQBoard level

Which of the following has a small negative magnetic susceptibility?
(a) Aluminium
(b) Copper
(c) Nickel
(d) Platinum

Try it first, then show the solutionSolution

Copper is diamagnetic; aluminium and platinum are paramagnetic; nickel is ferromagnetic.

Answer: (b).

Example 2: Curie's law · 2 marks · Short answerBoard level

A paramagnetic material has susceptibility $4.5 \times 1 0^{- 4}$ at 250 K. Find its Curie constant and its susceptibility at 375 K.

Try it first, then show the solutionSolution

$C = χ T = 4.5 \times 1 0^{- 4} \times 250 = 0.1125$ K. (1)
$χ = \frac{0.1125}{375} = 3.0 \times 1 0^{- 4}$ . (1)

Answer: C = 0.1125 K; $χ = 3.0 \times 1 0^{- 4}$ .

Example 3: Field and temperature both change · 2 marks · Short answerHarder board level

A paramagnetic sample has magnetisation M at 300 K in a field B. Find its magnetisation when the field is doubled and the temperature is raised to 450 K (assume Curie's law holds).

Try it first, then show the solutionSolution

$M \propto \frac{B}{T}$ . (1)
$M^{'} = M \times \frac{2 B}{B} \times \frac{300}{450} = \frac{4}{3} M$ . (1)

Answer: $\frac{4}{3} M$ .

Example 4: Explaining the three classes · 3 marks · Long answerHarder board level

Explain, in terms of atomic magnetic moments, why (a) a diamagnetic substance is repelled by a magnet, (b) a paramagnetic substance is weakly attracted and its susceptibility falls as temperature rises, (c) a ferromagnetic substance is strongly attracted.

Try it first, then show the solutionSolution

(a) Atoms have no permanent moment; the field induces small moments opposite to itself, so M is opposite to H and the substance moves to the weaker field. (1)
(b) Atoms have permanent moments, randomly oriented; the field partly aligns them along itself (weak attraction). Thermal agitation opposes the alignment more at higher T, so χ falls: $χ \propto \frac{1}{T}$ . (1)
(c) Moments are already aligned within domains; the field grows and rotates domains, giving a very large M along H. (1)

Answer: Opposing induced moments (dia); partial alignment of permanent moments against thermal disorder (para); domain growth and rotation (ferro).

Example 5: Why does temperature not affect diamagnetism? · 2 marks · Short answerHOTS / competency

A student claims that heating a diamagnetic sample should reduce its susceptibility, just as for a paramagnet. Is she right? Explain.

Try it first, then show the solutionSolution

Diamagnetism comes from moments induced in every atom by the field (a change in electron orbits), not from aligning existing moments. (1)
Thermal agitation randomises orientations, but there are no permanent moments to disorder; the induced effect is the same at any temperature, so χ stays (almost) constant. She is wrong. (1)

Answer: No: diamagnetic susceptibility is essentially independent of temperature.

Example 6: Identifying materials from measurements · 4 marks · Case-basedHOTS / competency

Suppose a student measures the susceptibilities of three samples. A: $- 2.2 \times 1 0^{- 5}$ at all temperatures tested. B: $3.6 \times 1 0^{- 4}$ at 300 K and $5.4 \times 1 0^{- 4}$ at 200 K. C: several thousand at room temperature, falling sharply to a small positive value above 1043 K.
(i) Classify A, B and C. (1 mark)
(ii) Show that B obeys Curie's law. (1 mark)
(iii) Predict B's susceptibility at 120 K. (1 mark)
(iv) What is special about 1043 K for C? (1 mark)

Try it first, then show the solutionSolution

(i) A diamagnetic, B paramagnetic, C ferromagnetic.
(ii) $χ T = 3.6 \times 1 0^{- 4} \times 300 = 0.108$ and $5.4 \times 1 0^{- 4} \times 200 = 0.108$ : constant, so $χ \propto \frac{1}{T}$ .
(iii) $χ = \frac{0.108}{120} = 9.0 \times 1 0^{- 4}$ .
(iv) It is C's Curie temperature: above it C loses its ferromagnetism and becomes paramagnetic.

Answer: (i) dia, para, ferro; (ii) χT = 0.108 K in both cases; (iii) $9.0 \times 1 0^{- 4}$ ; (iv) Curie temperature.

Where students lose marks in Magnetic Materials and Curie's Law

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

Mistake: Describing a crystal change for the Curie temperature
The ISC 2024 examiners saw answers about fcc changing to bcc, melting, or para becoming dia.
Do this: State the magnetic change: ferromagnetic → paramagnetic at and above $T_{C}$ .
Usually costs: 1 mark.
Mistake: Confusing Curie temperature with Curie's law
Also noted in 2024.
Do this: Curie's law: χ = C/T for paramagnets. Curie temperature: where ferromagnetism disappears.
Usually costs: 1 mark.
Mistake: Mixing up μr values
In 2025 candidates could not classify substances with μr slightly below and slightly above 1.
Wrong working
μr < 1 means paramagnetic.
Right working
μr < 1: diamagnetic; slightly > 1: paramagnetic.
Do this: Use $μ_{r} = 1 + χ_{m}$ and the sign of χ.
Usually costs: 1 mark.
Mistake: Celsius in Curie's law
Using 27 and −73 instead of 300 K and 200 K gives nonsense.
Do this: Always convert to kelvin.
Usually costs: 1 mark.

Exam tips

Give one example of each class when asked to distinguish them.
In explanations, mention atomic moments (absent, random, aligned in domains).
Quote typical magnitudes of χ to show understanding.
Draw field lines bending away from (dia) or into (para, ferro) a sample if a sketch is asked for.

Practice questions on Magnetic Materials and Curie's Law

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
According to Curie's law, the susceptibility of a paramagnetic substance is proportional to:
(a) T
(b) $T^{2}$
(c) $\frac{1}{T}$
(d) independent of T
Hint
Thermal agitation opposes alignment.
Answer
(c).
Q2 · 1 mark · Very short answerBoard level
The Curie temperature of nickel is about 358 °C. What happens to a magnetised nickel bar heated to 400 °C, and how does it behave above that temperature? (A variation on ISC 2024.)
Hint
Ferromagnetic below, paramagnetic above.
Answer
Above 358 °C thermal agitation destroys the domain alignment: the bar loses its ferromagnetism (and its magnetisation) and behaves as a paramagnetic substance.
Q3 · 2 marks · Short answerHarder board level
A paramagnetic sample has susceptibility $2.4 \times 1 0^{- 4}$ at 27 °C. Find its susceptibility at −73 °C.
Hint
Use kelvin: 300 K and 200 K.
Answer
$χ = 2.4 \times 1 0^{- 4} \times \frac{300}{200} = 3.6 \times 1 0^{- 4}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The susceptibility of a diamagnetic substance is independent of temperature.
Reason (R): Diamagnetism arises from moments induced opposite to the field, not from aligning permanent moments that thermal agitation could disturb.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Compare with paramagnets.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Sample X has relative permeability 0.99998 and sample Y has 1.0004. Find the susceptibility of each, name their classes and say how small rods of each set in a strong uniform field.
Hint
$μ_{r} = 1 + χ_{m}$ .
Answer
$χ_{X} = - 2 \times 1 0^{- 5}$ : diamagnetic (rod sets perpendicular to the field); $χ_{Y} = 4 \times 1 0^{- 4}$ : paramagnetic (rod sets parallel to the field).
Q6 · 2 marks · Short answerHOTS / competency
At 4 K in a field of 2 T, a paramagnetic salt reaches 30% of its saturation magnetisation and obeys Curie's law. Estimate the percentage at 16 K in a field of 6 T.
Hint
$M \propto \frac{B}{T}$ .
Answer
$30% \times \frac{6}{2} \times \frac{4}{16} = 22.5%$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose four small rods P, Q, R, S are tested between the poles of a strong electromagnet. P is pulled in strongly and stays slightly magnetised when the current is switched off. Q is pulled in weakly, and more weakly when warmed. R is pushed out weakly whatever its temperature. S is pulled in strongly at room temperature but only weakly after being heated to bright red heat.
(i) Classify P, Q and R. (1 mark)
(ii) Which law describes Q's behaviour? (1 mark)
(iii) Explain S's behaviour. (1 mark)
(iv) In which of the four would field lines inside be less dense than outside? (1 mark)
Hint
Look for temperature dependence.
Answer
(i) P ferromagnetic, Q paramagnetic, R diamagnetic. (ii) Curie's law, χ ∝ 1/T. (iii) S is ferromagnetic but above its Curie temperature it becomes paramagnetic. (iv) R (diamagnetic).

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.Sign of χ for a diamagnet?Show answer
Negative.
2.State Curie's law.Show answer
The susceptibility of a paramagnet is inversely proportional to absolute temperature: $χ_{m} = \frac{C}{T}$ .
3.What happens to a ferromagnet above its Curie temperature?Show answer
It becomes paramagnetic.
4.χ = 3 × 10⁻⁴ at 300 K: χ at 150 K?Show answer
$6 \times 1 0^{- 4}$ .
5.Give one example each of dia, para and ferro materials.Show answer
Bismuth (or copper); aluminium (or platinum); iron (or nickel).

Magnetic Materials and Curie's Law: frequently asked questions

Is every material diamagnetic?

Every material has a diamagnetic response, but in paramagnetic and ferromagnetic materials the effect of permanent atomic moments is far larger and hides it.

Why does a diamagnetic rod set perpendicular to the field?

It tends to move towards weaker field. Lying across the field puts its ends in the weaker region, so it turns to that position.

What are domains?

Small regions in a ferromagnet in which all the atomic moments are aligned. Magnetising a ferromagnet means growing and rotating these domains, not creating moments.

What does the ISC 2027 syllabus require here?

Properties of diamagnetic, paramagnetic and ferromagnetic substances with examples, their susceptibility and relative permeability, Curie's law for paramagnets, and the change of a ferromagnet to a paramagnet at the Curie temperature.

How was this tested recently?

In 2024 a 1-mark item asked for the meaning of the Curie temperature of soft iron; in 2025 a 1-mark item classified materials by their relative permeability.