5.2 Magnetisation, Magnetic Intensity and Magnetic Properties of Materials
Put a material in a magnetic field and its atoms add a field of their own: the magnetisation $\mathbf M$. With the magnetic intensity $\mathbf H$ this gives $\mathbf B=\mu_0(\mathbf H+\mathbf M)$, and the sign and size of the susceptibility $\chi=M/H$ sort every material into diamagnetic, paramagnetic or ferromagnetic.
A scrapyard crane lifts a tangle of old cars with an electromagnet and drops the steel onto a pile, leaving aluminium, copper and plastic behind. A hospital MRI scanner can distinguish oxygen-rich from oxygen-poor blood because the two respond slightly differently to a strong field. In a laboratory in the Netherlands a live frog has been floated in mid-air inside a powerful magnet — not because the frog contains iron, but because water is very weakly repelled by a magnetic field. And the core of every transformer and electric motor is made of a special iron alloy chosen for how easily it can be magnetised and demagnetised.
All of these depend on how matter responds to a magnetic field. In Section 5.1 we treated a bar magnet as a set of lined-up atomic current loops. Here we make that idea quantitative. We define the magnetisation $\mathbf M$ — the magnetic moment per unit volume of a material — and the magnetic intensity $\mathbf H$, which is set by the currents we control, and show that $\mathbf B=\mu_0(\mathbf H+\mathbf M)$. The ratio $\chi=M/H$, the magnetic susceptibility, then sorts materials into three families: diamagnetic (weakly repelled), paramagnetic (weakly attracted) and ferromagnetic (strongly attracted, and able to stay magnetised). Finally we see how temperature affects each family and how a ferromagnet "remembers" its history.
Magnetisation and Magnetic Intensity
Every atom contains moving electrons, and in many atoms the electrons' orbital motion and spin leave the atom with a small magnetic moment — each atom is a tiny current loop. In an unmagnetised material these atomic moments either cancel inside each atom or point in random directions, so a piece of the material has no overall moment. A magnetic field can change that, lining the moments up (or, as we shall see, inducing new ones). The result is described by the moment per unit volume.
The solenoid with a core. A long solenoid with $n$ turns per unit length carrying current $I$ produces inside it, in vacuum, the field $B_0=\mu_0nI$ (Chapter 4). Now fill it with a material. The lined-up atomic loops inside the material cancel each other's currents in the interior, but at the surface of the rod they do not: the material behaves as if an extra current, $M$ amperes for every metre of its length, were circulating round its surface — exactly like an extra winding. (For the bar magnet of Section 5.1, this surface current was the "equivalent solenoid".) So the material adds a field $\mu_0M$ to the field of the real winding: $$B=\mu_0nI+\mu_0M.$$ The first term comes from the current we control; the second from the material's response. It is useful to name the first part separately.
In any material, $$\mathbf B=\mu_0\left(\mathbf H+\mathbf M\right).$$ $\mu_0\mathbf H$ is the contribution of the external (free) currents and $\mu_0\mathbf M$ that of the material. In vacuum $\mathbf M=\mathbf 0$ and $\mathbf B=\mu_0\mathbf H$.
Magnetic Susceptibility and Permeability
For most materials — everything except the ferromagnets — the magnetisation is simply proportional to the magnetic intensity that produces it: double the current in the solenoid and $M$ doubles. The constant of proportionality is a property of the material.
Substituting $\mathbf M=\chi\mathbf H$ in $\mathbf B=\mu_0(\mathbf H+\mathbf M)$: $$\mathbf B=\mu_0(1+\chi)\mathbf H=\mu_0\mu_r\mathbf H=\mu\mathbf H.$$ Filling a long solenoid completely with the material multiplies the field inside by $\mu_r$: $B=\mu_r B_0=\mu_0\mu_rnI$.
The magnetic permeability plays the role for magnetism that the relative permittivity (dielectric constant) played in Chapter 2 — with one important difference. Every dielectric reduces the electric field inside it ($K\gt1$ always), but a magnetic material may increase or decrease $\mathbf B$, depending on the sign of $\chi$. Typical values at room temperature:
| Material | $\chi$ | family |
|---|---|---|
| bismuth | $-1.66\times10^{-5}$ | diamagnetic |
| copper | $-9.8\times10^{-6}$ | diamagnetic |
| water | $-9.0\times10^{-6}$ | diamagnetic |
| oxygen gas (STP) | $+2.1\times10^{-6}$ | paramagnetic |
| aluminium | $+2.3\times10^{-5}$ | paramagnetic |
| platinum | $+2.9\times10^{-4}$ | paramagnetic |
| soft iron | $\sim10^{3}$ to $10^{5}$ | ferromagnetic |
For the first six, $|\chi|$ is a few parts in a hundred thousand or less: $\mu_r$ differs from $1$ only in the fifth or sixth decimal place, and such materials barely change the field at all. For iron, $\chi$ is enormous — and, as we shall see, not even a constant.
Diamagnetism and Paramagnetism
A material is
diamagnetic if $\chi$ is small and negative ($-1\le\chi\lt0$, typically $|\chi|\sim10^{-5}$; $\mu_r$ slightly less than $1$): it is magnetised against the applied field and is weakly repelled by a magnet;
paramagnetic if $\chi$ is small and positive ($0\lt\chi\ll1$; $\mu_r$ slightly greater than $1$): it is magnetised weakly along the field and is weakly attracted;
ferromagnetic if $\chi$ is large and positive ($\chi\gg1$, $\mu_r$ in the hundreds to hundreds of thousands): it is magnetised strongly along the field and strongly attracted. Iron, cobalt, nickel, gadolinium and many of their alloys and oxides are ferromagnetic.
How they behave. Hang a small rod of each kind between the poles of a strong magnet. In a non-uniform field a diamagnetic sample moves from the stronger towards the weaker part of the field, while paramagnetic and ferromagnetic samples move towards the stronger part. Inside the sample, field lines are pushed out slightly by a diamagnet, drawn in slightly by a paramagnet and crowded in strongly by a ferromagnet. If a paramagnetic liquid (a solution of a manganese or copper salt) is put in a narrow U-tube with one arm between the poles, the liquid in that arm rises; a diamagnetic liquid (water, alcohol) is pushed down. A freely hung rod of paramagnetic or ferromagnetic material turns to lie along the field; a diamagnetic rod settles at right angles to it.
Why diamagnetism? In a diamagnetic atom the electrons' magnetic moments cancel exactly, so the atom has no moment of its own. When a field is switched on, the electron motion inside every atom is altered so as to oppose the change (the same rule that governs induced currents — Lenz's law, Chapter 6), producing a small moment against the field. Every material has this response; it shows only when nothing stronger hides it. Bismuth, copper, lead, silicon, water, nitrogen and sodium chloride are diamagnetic, and diamagnetism hardly depends on temperature. The extreme case is a superconductor cooled below its critical temperature: it expels the field from its interior completely ($B=0$ inside, so $M=-H$, $\chi=-1$, $\mu_r=0$). This expulsion, the Meissner effect, lets a small magnet float above a superconductor, and is used in some magnetically levitated trains.
Why paramagnetism? In a paramagnetic material each atom does carry a permanent magnetic moment, but thermal agitation keeps the moments pointing in random directions. An applied field exerts a torque $\mathbf m\times\mathbf B$ on each (Section 5.1) and lines them up a little — only a little, because collisions keep scrambling them. Aluminium, sodium, calcium, platinum, oxygen and copper chloride are paramagnetic. Cooling the material weakens the agitation, so the same field produces more alignment, and a stronger field also helps. Pierre Curie found the rule experimentally.
For a paramagnetic material in a field $B_0=\mu_0H$ that is not too strong (so that the alignment is far from complete), the magnetisation is proportional to $B_0$ and inversely proportional to the absolute temperature: $$M=C\,\frac{B_0}{T},\qquad\text{equivalently}\qquad \chi=\frac{\mu_0C}{T}\ \propto\ \frac1T,$$ where $C$ is the Curie constant of the material. As $B_0/T$ becomes very large the magnetisation approaches a saturation value $M_s=nm$ ($n$ atoms per unit volume, each of moment $m$, all aligned) and Curie's law no longer holds.
Ferromagnetism and the Curie Temperature
In iron, cobalt, nickel and gadolinium the atomic moments do something no paramagnet can: neighbouring atoms interact so strongly (a quantum effect) that their moments line up spontaneously, even with no applied field. The alignment extends over regions called magnetic domains, typically a micrometre to a millimetre across and each containing some $10^{11}$ or more atoms all pointing the same way. A piece of unmagnetised iron is not unmagnetised inside: it is a patchwork of fully magnetised domains pointing in different directions, whose moments cancel overall.
An applied field changes the patchwork. Domains pointing roughly along $\mathbf H$ grow at the expense of their neighbours as the walls between them move, and at higher fields whole domains swing round to point along $\mathbf H$. A small $H$ therefore produces a huge $M$ — this is why $\chi$ for iron is of order $10^3$ to $10^5$ — until every domain points the same way and $M$ reaches its saturation value. Because the response is not proportional to $H$ (and even depends on how the sample was treated before), $\chi$ for a ferromagnet is not a constant of the material.
The Curie temperature. Heat a ferromagnet and the thermal agitation fights the alignment inside each domain. At a definite temperature the spontaneous alignment collapses altogether: a magnet heated above it loses its magnetism, and the material becomes an ordinary paramagnet.
Gadolinium's Curie temperature is close to room temperature: a piece of gadolinium is attracted by a magnet on a cold day and much less so on a warm one. Lowering $T$ towards $T_C$ from above makes $\chi$ grow without limit — a sign of the alignment that sets in at $T_C$.
1. Confusing $\mathbf B$, $\mathbf H$ and $\mathbf M$. $\mathbf B$ is the field (tesla); $\mathbf H$ and $\mathbf M$ are both in A/m. Inside a solenoid $H=nI$ whatever the core; the core changes $M$ and hence $B$.
2. Forgetting the sign of $\chi$. For a diamagnet $\chi\lt0$ and $\mu_r\lt1$; $\mathbf M$ points opposite to $\mathbf H$.
3. Applying Curie's law to the wrong material. $\chi\propto1/T$ is for paramagnets. Diamagnetism is almost independent of $T$; a ferromagnet obeys $\chi=C'/(T-T_C)$ only above $T_C$.
4. Thinking a ferromagnet above $T_C$ becomes diamagnetic. It becomes paramagnetic.
5. Treating $\chi$ of iron as a constant. For a ferromagnet $B$ depends on $H$ non-linearly and even on the history of the sample.
Hysteresis is not in the current CBSE/NCERT, NEET, JEE Main or JEE Advanced syllabus. It is included here only as background: it explains how permanent magnets and transformer cores work, and it appears in older papers and reference books. You will not be examined on it.
The hysteresis loop. Wind a coil round an unmagnetised iron ring and slowly increase the current, so that $H$ grows from zero. $B$ rises along the "initial magnetisation curve" and levels off at saturation. Now reduce $H$ back to zero: $B$ does not retrace its path — it stays high, because the domains do not all swing back, and the iron is now a magnet. To remove this left-over field you must drive $H$ negative. Carry on to large negative $H$, then back to large positive $H$, and $B$ traces a closed loop. This lagging of $B$ behind $H$ is called hysteresis ("lagging behind"). The retentivity (remanence) $B_r$ is the $B$ left when $H$ returns to zero after saturation; the coercivity $H_c$ is the reverse $H$ needed to bring $B$ back to zero. A material with small coercivity is magnetically soft (soft iron); one with large coercivity is magnetically hard (steels and permanent-magnet alloys).
Energy loss. Going once round the loop costs energy: the work done per unit volume of the core in one cycle equals the area of the B–H loop (units $\text{T}\cdot\text{A/m}=\text{J/m}^3$), and it appears as heat. For a nearly rectangular loop the area is about $4B_rH_c$.
Choosing materials. Electromagnets, transformer and motor cores are magnetised and demagnetised many times a second, so they need a soft material: low coercivity, a narrow loop (little heating), and high permeability — soft iron and silicon steel. Permanent magnets need high retentivity (a strong field left behind) and high coercivity (so that stray fields, knocks and moderate heating do not demagnetise them): hard steels, alnico, ferrites and neodymium–iron–boron alloys. In the model loops of Interactive 2 below, soft iron ($B_r\approx1.14$ T, $H_c=80$ A/m) loses about $4.8\times10^2$ J/m³ per cycle, the hard alloy ($B_r\approx1.10$ T, $H_c=5.0\times10^4$ A/m) about $2.6\times10^5$ J/m³ — some $500$ times more.
Exercises
- 1 A rod of volume $2.0\times10^{-5}\ \text{m}^3$ has a net magnetic moment of $0.80\ \text{A m}^2$. Find its magnetisation.
- 2 A long solenoid has $1500$ turns per metre and carries $2.0$ A. Find $H$ inside it and $B$ when it is empty.
- 3 A core inserted in the solenoid of part 2 acquires a magnetisation of $1.2\times10^6$ A/m. Find $B$ inside the core.
- 4 Show that $\mu_0H$ has the unit of tesla, and that $H$ and $M$ have the same unit.
- 5 Inside a magnetic core $B=0.80$ T and $H=600$ A/m. Find $M$.
- 6 A uniformly magnetised rod $10$ cm long with cross-section $1.0\ \text{cm}^2$ has $M=8.0\times10^5$ A/m. Find its magnetic moment.
- 7 Nickel has $9.1\times10^{28}$ atoms per cubic metre, each with a moment of about $5.6\times10^{-24}\ \text{A m}^2$. Find its saturation magnetisation.
- 8 Explain why $H$ inside a long solenoid does not depend on the material of its core, while $B$ does.
- 1 Aluminium has $\chi=2.3\times10^{-5}$. Find its relative permeability and permeability.
- 2 Find the relative permeability of bismuth ($\chi=-1.66\times10^{-5}$). Is it greater or less than $1$?
- 3 Find the magnetisation of aluminium in $H=5.0\times10^4$ A/m, and $B$ inside it.
- 4 A core with $\mu_r=800$ fills a solenoid in which $H=1000$ A/m. Find $B$ and $M$.
- 5 A sample in $H=2.0\times10^3$ A/m acquires $M=8.0\times10^{-3}$ A/m along $\mathbf H$. Find $\chi$ and $\mu_r$ and classify the material.
- 6 A sample in $H=1.0\times10^4$ A/m acquires $M=0.17$ A/m directed opposite to $\mathbf H$. Find $\chi$ and classify the material.
- 7 A superconductor expels the field completely. Find its $\chi$, $\mu_r$ and $M$ in terms of $H$.
- 8 By what percentage does a copper core ($\chi=-9.8\times10^{-6}$) change the field inside a long solenoid?
- 1 Classify materials with (a) $\chi=-2.2\times10^{-5}$, (b) $\chi=3.0\times10^{-4}$, (c) $\chi=5000$, (d) $\chi=-1$.
- 2 Samples of bismuth, aluminium and iron are brought near the pointed pole of a strong magnet. Describe the motion of each.
- 3 A paramagnet has $\chi=3.0\times10^{-4}$ at $300$ K. Find $\chi$ at $150$ K and at $600$ K.
- 4 In a fixed field a paramagnetic sample has $M=0.80$ A/m at $300$ K. Find $M$ at $200$ K in the same field.
- 5 Why does diamagnetic susceptibility hardly depend on temperature, while paramagnetic susceptibility does?
- 6 A paramagnetic liquid fills a U-tube, one arm of which lies between the poles of a strong electromagnet. What happens to the liquid level in that arm when the magnet is switched on? What if the liquid is water?
- 7 Above its Curie temperature of $1043$ K, iron obeys $\chi=C'/(T-T_C)$ with $C'\approx2.0$ K. Find $\chi$ at $1143$ K and $1243$ K. Compare with ordinary paramagnets.
- 8 Gadolinium has $T_C\approx293$ K. Describe its magnetic behaviour at $283$ K and at $303$ K.
- 1 Cobalt has $9.0\times10^{28}$ atoms per cubic metre, each of moment $1.6\times10^{-23}\ \text{A m}^2$. Find its saturation magnetisation and the largest contribution $\mu_0M_s$ that cobalt can make to $B$.
- 2 A magnetic domain in iron is a cube of side $1.0\ \mu\text{m}$. Using the data of Example 5, find the number of atoms in it and its magnetic moment when fully aligned.
- 3 Nickel has a Curie temperature of $631$ K. Above it, its susceptibility is $0.012$ at $681$ K. Using $\chi=C'/(T-T_C)$, find $C'$ and the susceptibility at $731$ K.
- 4 An iron rod in a long solenoid where $H=200$ A/m acquires a magnetisation of $4.0\times10^5$ A/m. Find its susceptibility, its relative permeability (in this field) and the field $B$ inside it.
- 5 Explain, using domains, why the magnetisation of iron saturates.
- 6 Modelling. A recycling plant uses a strong magnet above a conveyor belt to pull steel cans off a mixed stream that also contains aluminium cans and plastic bottles. Explain why only the steel is lifted, even though aluminium is attracted by the magnet too, and whether heating the stream to $1100$ K (hypothetically) would still allow the steel to be separated.
E1. Board · 3 marks Distinguish between diamagnetic, paramagnetic and ferromagnetic materials on the basis of (i) susceptibility, (ii) relative permeability, (iii) behaviour in a non-uniform magnetic field and (iv) the effect of temperature. Give one example of each.
E2. Board · Case study An electromagnet is made by winding $2000$ turns per metre on a long rod of a ferromagnetic alloy of relative permeability $500$ (assumed constant), and passing a current of $0.50$ A. (i) Find the magnetic intensity and the field in the rod. (2 marks) (ii) Find the magnetisation of the rod. (1 mark) (iii) The rod is replaced by an aluminium rod ($\chi=2.3\times10^{-5}$) with the same winding and current. By roughly what factor does the field fall? (1 mark)
E3. JEE Main A paramagnetic sample has magnetisation $0.60$ A/m at $27^\circ$C in a certain magnetic field. It is cooled to $-73^\circ$C and the field is tripled. Its magnetisation becomes (A) $1.2$ A/m (B) $1.8$ A/m (C) $2.7$ A/m (D) $4.05$ A/m.
E4. NEET Which of the following has a negative magnetic susceptibility? (A) aluminium (B) copper (C) platinum (D) nickel
E5. JEE Advanced (One or more correct.) (A) The susceptibility of a diamagnetic material is nearly independent of temperature. (B) In weak fields the susceptibility of a paramagnetic material is inversely proportional to its absolute temperature. (C) A ferromagnet heated above its Curie temperature becomes diamagnetic. (D) A superconductor that expels a magnetic field completely has relative permeability zero.
E6. NEET Above its Curie temperature $T_C$, a ferromagnetic material has susceptibility $0.010$ at $T_C+100$ K. Its susceptibility at $T_C+50$ K is (A) $0.0050$ (B) $0.010$ (C) $0.020$ (D) $0.040$.
Research question. How strong must a magnetic field and its gradient be to levitate a diamagnetic object — a drop of water, a piece of bismuth, a flake of graphite, or a frog?
Why it matters. Diamagnetic levitation lets researchers float living cells, protein crystals and water drops without touching them, to imitate weightlessness on the ground for biology and materials research — work done in high-field laboratories in Europe, the USA, China and Japan. Levitated graphite is used in ultra-sensitive force and vibration sensors. The same physics separates minerals and plastics by their susceptibility in recycling plants.
What to do.
- A small object of volume $V$ and susceptibility $\chi$ in a field $B$ that changes with height $z$ feels a vertical magnetic force $F=\dfrac{\chi V}{\mu_0}B\dfrac{dB}{dz}$ (take this result as given). For a diamagnet ($\chi\lt0$) above the centre of a magnet, where $B$ falls with height, this force points upwards. Show that levitation requires $B\left|\dfrac{dB}{dz}\right|=\dfrac{\mu_0\rho g}{|\chi|}$, independent of the object's size.
- Evaluate the required $B\,|dB/dz|$ (in $\text{T}^2/\text{m}$) for water ($\rho=1000\ \text{kg/m}^3$, $\chi=-9.0\times10^{-6}$), bismuth ($\rho=9780\ \text{kg/m}^3$, $\chi=-1.66\times10^{-5}$) and pyrolytic graphite ($\rho=2200\ \text{kg/m}^3$, $\chi\approx-4.5\times10^{-4}$).
- A large research magnet produces $16$ T at its centre, falling to near zero over about $10$ cm. Estimate the largest $B\,|dB/dz|$ it provides. Which of the three can it levitate? Can a living frog (mostly water) be floated?
- The surface field of a neodymium magnet is about $0.5$ T and falls by half in a few millimetres. Estimate its $B\,|dB/dz|$ and decide which material can levitate over a set of such magnets on a desk.
Tools: Calculator or spreadsheet. Optional: a sheet of pyrolytic graphite and a few neodymium cube magnets arranged in a checkerboard of alternating poles.
Hand in: Your derivation, a table of the required $B\,|dB/dz|$ for each material, your estimates for the two magnets, and a conclusion.
Going further. Why can a paramagnetic object not float stably above a magnet in the same way? (Hint: in which direction is the force on it, and what happens if it is displaced slightly?)
🧪 Self-check quiz
Q1. The SI unit of magnetisation $\mathbf M$ is
Q2. For a diamagnetic material the relative permeability is
Q3. A paramagnet obeying Curie's law has susceptibility $\chi_0$ at $300$ K. At $600$ K its susceptibility is
Q4. When a ferromagnetic material is heated above its Curie temperature it becomes
Q5. Inside a long solenoid with $n$ turns per metre carrying current $I$ and filled with a material of magnetisation $M$, the field is