A common confusion in this topic is applying one temperature law to all magnetic materials. Diamagnets, paramagnets and ferromagnets respond to heat in three different ways, and NEET usually tests which law applies rather than the arithmetic.
Diamagnetism comes from induced orbital currents that oppose the applied field. It is present in all substances and is essentially independent of temperature: heating a diamagnetic sample leaves its small negative susceptibility unchanged.
Paramagnetism is a contest between field alignment and thermal agitation. Heating randomises the atomic dipoles, so magnetisation and susceptibility fall. NCERT Class 12 Physics (pre-2023 edition), Chapter 5 (Magnetism and Matter), page 193 gives Curie's law (the current book dropped it): at a fixed magnetising field the magnetisation is proportional to B₀/T, so χ ∝ 1/T. Two working consequences — χ₁T₁ = χ₂T₂ between any two states, and T is always in kelvin, never in degrees Celsius.
Ferromagnetism depends on ordered domains. Thermal agitation breaks that order, and above a material-specific Curie temperature T_c the domain structure collapses and the sample becomes paramagnetic (current book: page 149, which describes the change without naming T_c). Above T_c the susceptibility follows the Curie–Weiss form χ = C/(T − T_c), not the plain Curie form (NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 194). Below T_c it stays ferromagnetic, with susceptibility orders of magnitude above any paramagnet's.
In NEET these questions are short. You are given a susceptibility at one temperature and asked for another; or asked what a material becomes past T_c; or handed both T and T_c and expected to notice the denominator is a difference.
Watch-out: at T = 1.20 × 10³ K with T_c = 6.00 × 10² K, Curie–Weiss gives twice the plain-Curie value, not the same value. And a permanent magnet taken above T_c and cooled in zero field does not come back — its domains re-form in random directions.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
For a paramagnetic material obeying Curie's law, the magnetic susceptibility χ varies with the absolute temperature T as
Answer: C. Curie's law states that at a fixed magnetising field the magnetisation is proportional to B₀/T, so χ = C′/T — see NCERT Class 12 Physics (pre-2023 edition), Chapter 5 (Magnetism and Matter), page 193.
Why A is wrong: A is wrong because it inverts the law: heating increases thermal agitation and destroys dipole alignment, so χ falls as T rises.
Why B is wrong: B is wrong because temperature-independent susceptibility is the signature of diamagnetism, not paramagnetism.
Why D is wrong: D is wrong because only the first power of T appears in Curie's law; a plot of χ against 1/T is a straight line, not a plot of χ against 1/T².
A ferromagnetic specimen is held at a temperature well above its Curie temperature. Its magnetic behaviour there is that of
Answer: A. Above the Curie temperature the domain structure disintegrates and the material turns paramagnetic, as stated in NCERT Class 12 Physics, Chapter 5, page 149, which does not name the temperature; the name Curie temperature and its symbol Tc are in NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 194.
Why B is wrong: B is wrong because the atoms keep their permanent magnetic moments after the domains break up; losing domain order gives paramagnetism, not the induced-moment behaviour of a diamagnet.
Why C is wrong: C is wrong because the change at T_c is a loss of the ordered phase, not a gradual weakening — above T_c there is no spontaneous domain magnetisation left to reduce.
Why D is wrong: D is wrong because a paramagnet has a small positive susceptibility, not zero; it still magnetises weakly along the applied field.
Which class of magnetic material has a susceptibility that is essentially independent of temperature?
Answer: D. Diamagnetism arises from field-induced orbital currents rather than from aligning pre-existing moments, so thermal agitation has nothing to randomise and χ is essentially temperature-independent (the orbital-current origin of diamagnetism: NCERT Class 12 Physics, Chapter 5, page 148).
Why A is wrong: A is wrong because paramagnetic susceptibility follows Curie's law, χ ∝ 1/T, and therefore changes with temperature.
Why B is wrong: B is wrong because ferromagnetic behaviour weakens as the temperature approaches T_c and vanishes above it — strongly temperature-dependent.
Why C is wrong: C is wrong because above T_c the material is paramagnetic and its susceptibility follows χ = C/(T − T_c), again temperature-dependent.
A paramagnetic sample has susceptibility 6.0 × 10⁻⁴ at 3.00 × 10² K. Assuming Curie's law holds over this range, its susceptibility at 5.00 × 10² K is
Answer: B. Curie's law gives χ₁T₁ = χ₂T₂, so χ₂ = 6.0 × 10⁻⁴ × (3.00 × 10²)/(5.00 × 10²) = 3.6 × 10⁻⁴ (NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 193).
Why A is wrong: A is wrong because it multiplies by T₂/T₁ instead of T₁/T₂, treating χ as directly proportional to temperature — the sign of the dependence is reversed.
Why C is wrong: C is wrong because it assumes temperature independence, which belongs to diamagnets; a paramagnet's susceptibility must fall on heating.
Why D is wrong: D is wrong because it uses χ ∝ 1/T², giving 6.0 × 10⁻⁴ × (3/5)² = 2.2 × 10⁻⁴; Curie's law carries only the first power of T.
Curie's law for a paramagnetic sample is written χ = C′/T, with T the absolute temperature. The sample has χ = 2.0 × 10⁻⁵ at 2.50 × 10² K. The value of C′ is
Answer: D. Rearranging gives C′ = χT = (2.0 × 10⁻⁵)(2.50 × 10² K) = 5.0 × 10⁻³ K, the constant form of Curie's law in NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 193.
Why A is wrong: A is wrong because it divides χ by T instead of multiplying; C′ is the product χT, which is what stays fixed as the sample is heated or cooled.
Why B is wrong: B is wrong by a factor of ten — a decimal slip in 2.0 × 10⁻⁵ × 2.50 × 10²; check the exponent arithmetic (−5 + 2 = −3).
Why C is wrong: C is wrong because it reports the susceptibility itself rather than the Curie constant; χ is the quantity that changes with T, C′ is the one that does not.
A paramagnetic sample sits in a magnetising field of fixed magnitude. Its absolute temperature is then doubled, the field being unchanged. Its magnetisation becomes
Answer: A. At fixed B₀ the magnetisation is proportional to B₀/T, so doubling the absolute temperature halves M (NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 193).
Why B is wrong: B is wrong because heating randomises the atomic dipoles; alignment and therefore magnetisation must decrease, not increase.
Why C is wrong: C is wrong because temperature-independent behaviour belongs to diamagnets — the defining feature of paramagnetism is that M depends on T.
Why D is wrong: D is wrong because it applies a T² dependence and in the wrong direction; the dependence is the first power of T and inverse.
A ferromagnetic material has Curie temperature 6.00 × 10² K. At 1.20 × 10³ K its susceptibility follows the Curie–Weiss law χ = C/(T − T_c). The ratio of this susceptibility to the value the plain Curie law χ = C/T would predict at the same temperature is
Answer: C. Curie–Weiss gives C/(1.20 × 10³ − 6.00 × 10²) = C/(6.00 × 10²); plain Curie gives C/(1.20 × 10³). The ratio is (1.20 × 10³)/(6.00 × 10²) = 2.00, using the above-T_c form in NCERT Class 12 Physics (pre-2023 edition), Chapter 5, page 194.
Why A is wrong: A is wrong because it inverts the ratio — subtracting T_c makes the denominator smaller, so the Curie–Weiss susceptibility is larger than the plain-Curie value, not smaller.
Why B is wrong: B is wrong because it ignores T_c in the denominator; the two laws coincide only in the limit T ≫ T_c, and here T is only twice T_c.
Why D is wrong: D is wrong because it squares the ratio of temperatures; both expressions are first order in the denominator, so the ratio is (T)/(T − T_c) = 2.00.
A permanent bar magnet is heated well above the Curie temperature of its material and then cooled back to room temperature with no external field applied. On testing it at room temperature, one finds that it
Answer: B. Heating past T_c destroys the domain alignment that stored the permanent moment; on cooling in zero field the domains re-form in random orientations, so the net moment is essentially zero even though the material is ferromagnetic again (NCERT Class 12 Physics, Chapter 5, page 149).
Why A is wrong: A is wrong because returning below T_c restores the ability to hold domains, not the particular alignment that was destroyed — that alignment needs an applied field to be re-established.
Why C is wrong: C is wrong because the material's class does not change permanently; below T_c it is ferromagnetic again, merely unmagnetised.
Why D is wrong: D is wrong because nothing in heating selects a reversed direction; random re-formation of domains cancels the moment rather than flipping it.
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Given.
A paramagnetic salt has susceptibility χ₁ = 3.6 × 10⁻³ at T₁ = 2.00 × 10² K. It is warmed to T₂ = 3.00 × 10² K, the magnetising field being unchanged.
Required.
Its susceptibility at the higher temperature, and the factor by which its magnetisation changes.
Concept.
Paramagnetic alignment is opposed by thermal agitation, so susceptibility falls as absolute temperature rises — Curie's law, NCERT Class 12 Physics (pre-2023 edition), Chapter 5 (Magnetism and Matter), page 193. At a fixed magnetising field the magnetisation is proportional to the susceptibility, so both scale the same way.
Formula.
χ = C′/T, so between two states χ₁T₁ = χ₂T₂, giving χ₂ = χ₁ × (T₁/T₂).
Substitution.
χ₂ = (3.6 × 10⁻³) × (2.00 × 10² K)/(3.00 × 10² K).
Calculation.
(2.00 × 10²)/(3.00 × 10²) = 0.667, so χ₂ = 3.6 × 10⁻³ × 0.667 = 2.4 × 10⁻³. No exact constants enter this calculation: 2.00 × 10² K and 3.00 × 10² K are measured values written to three significant figures, so the two-significant-figure input 3.6 × 10⁻³ is what fixes the answer at two significant figures.
Final answer.
χ₂ = 2.4 × 10⁻³. The magnetisation falls to 2/3 — about 0.67 — of its earlier value.
Common trap.
Two ways to lose this mark. First, multiplying by T₂/T₁ instead of T₁/T₂, which reports a rise on heating and contradicts the physics. Second, feeding Celsius values into the ratio; Curie's law is written in absolute temperature, and a stem quoting −73 °C and 27 °C is testing exactly that conversion.
Similar NEET-style question.
A paramagnetic salt has susceptibility 4.8 × 10⁻³ at −73 °C. Assuming Curie's law, find its susceptibility at 127 °C. *(Convert first: 2.00 × 10² K and 4.00 × 10² K; χ₂ = 4.8 × 10⁻³ × 200/400 = 2.4 × 10⁻³.)*
Atoms have permanent dipoles aligning weakly with B. Aluminium, sodium, calcium, oxygen (at STP), copper chloride. χ > 0, small. χ and μr depend on the sample temperature as well as the material; as the field is increased or the temperature lowered, the magnetisation increases towards saturation.
-- NCERT Class 12 Physics, Ch. 5, p. 148Strong domain alignment. Iron, cobalt, nickel, gadolinium, etc.; relative magnetic permeability > 1000, chi very large. The ferromagnetic property depends on temperature: at high enough temperature a ferromagnet becomes a paramagnet, the domain structure disintegrating; the loss of magnetisation is gradual.
-- NCERT Class 12 Physics, Ch. 5, p. 149In the NEET syllabus; removed from current NCERT.
For a paramagnetic material the magnetisation is inversely proportional to the absolute temperature T: M = C B0/T (Eq. 5.20a), or equivalently chi = C mu0/T (Eq. 5.20b). This is Curie's law, after Pierre Curie; C is Curie's constant. So for a paramagnetic material both chi and mu_r depend on the material and, in a simple fashion, on the sample temperature.
-- NCERT Class 12 Physics (pre-2023 edition), Chapter 5, p. 193In the NEET syllabus; removed from current NCERT.
The ferromagnetic property depends on temperature: at high enough temperature a ferromagnet becomes a paramagnet, the domain structure disintegrating gradually - a phase transition. The temperature of transition from ferromagnetic to paramagnetic is the Curie temperature Tc; above it, in the paramagnetic phase, chi = C/(T - Tc) (Eq. 5.21). Table 5.4 lists Tc: cobalt 1394 K, iron 1043 K, Fe2O3 893 K, nickel 631 K, gadolinium 317 K.
-- NCERT Class 12 Physics (pre-2023 edition), Chapter 5, p. 194More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
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