Magnetisation and magnetic intensity

Slide a rod of iron into a solenoid and the field inside can jump to hundreds of times its original strength. Slide in a rod of bismuth instead and the field barely flinches — if anything, it weakens by a whisker. Same coil, same current, same applied field, wildly different outcomes, because the material sitting inside is not a passive bystander. Every atom in that rod carries a tiny magnetic dipole moment, born of orbiting and spinning electrons, and when an external field arrives those countless atomic dipoles react — some snap into line with the field, some line up against it, most barely notice. Add up all their moments and divide by the volume, and you get a single vector that captures exactly how the material has responded: the magnetisation $\vec M$ . This article builds the compact vocabulary that turns that vague word "responds" into three precise, measurable quantities — magnetisation $\vec M$ , magnetic intensity $\vec H$ , and the total field $\vec B$ — and the equally precise numbers, susceptibility $\chi$ and relative permeability $\mu_r$ , that summarise a material's personality in one glance. You will see why $\vec B$ , $\vec H$ , and $\vec M$ are not three names for the same thing but three genuinely different roles in one short equation, why iron and bismuth sit at opposite ends of the same scale, and how a handful of algebra lets you compute any one of these quantities the moment you know two others — exactly the skill CBSE numericals test, and exactly the skill you will lean on again the moment you meet dia-, para-, and ferromagnetic materials by name in the next topic. By the end you should be able to look at a bald statement like "a material has relative permeability 400" and immediately picture what that means physically — a field a few hundred times stronger than an empty coil alone would produce, and a material any electrical engineer would reach for to build a transformer core.

Key takeaways

Magnetisation $\vec M$ is the net magnetic dipole moment per unit volume that a material develops when placed in a field: $\vec M = \vec m_{net}/V$ , unit A/m.
Magnetic intensity $\vec H$ is the external, material-independent magnetising field set purely by the currents you supply from outside; unit A/m, same unit as $M$ .
The total field inside the material combines both contributions: $\vec B = \mu_0(\vec H + \vec M)$ , with $B$ measured in tesla.
Susceptibility $\chi = M/H$ is a dimensionless number that tells you how strongly, and in which direction, a material magnetises — small negative for diamagnetic, small positive for paramagnetic, large positive for ferromagnetic materials.
Relative permeability $\mu_r = 1 + \chi$ packages the whole response into one multiplier: $\vec B = \mu_0\mu_r\vec H = \mu\vec H$ , where $\mu = \mu_0\mu_r$ is the absolute permeability.
A quick sanity check on any answer: if $\chi$ is a small fraction of a per cent, $B$ should sit close to the vacuum value $\mu_0 H$ ; if $\mu_r$ is in the hundreds, $B$ should be dramatically larger than $\mu_0 H$ .
Key idea: placing a material in an external magnetic intensity H makes its atomic dipoles develop a magnetisation M = χH, and the field actually measured inside the material is B = μ₀(H + M) = μ₀μᵣH, with χ and μᵣ = 1 + χ summarising how strongly the material responds.

The core insight

The magnetic intensity $\vec H$ is the external cause set by free currents, the magnetisation $\vec M = \chi\vec H$ is the material's proportional response, and the total field measured inside the material is their combined effect, $\vec B = \mu_0(\vec H+\vec M) = \mu_0\mu_r\vec H$ .

Where it fits in the syllabus

This topic sits inside Unit V, Magnetism and Matter, of the CBSE Class 12 Physics syllabus, right after the bar magnet, its equivalent-solenoid picture, and the (qualitative) axial and equatorial dipole-field formulae. Board wording keeps the surrounding dipole-field and torque material strictly qualitative, but the definitions of $\vec M$ , $\vec H$ , $\chi$ , and $\mu_r$ , and the relation $\vec B = \mu_0(\vec H+\vec M)$ , are the quantitative core that examiners do test with short numericals. This article stays tightly inside that scope: it does not touch Earth's magnetism (removed from the 2025-26 syllabus), does not attempt a quantitative hysteresis-loop energy calculation, and does not treat Curie's law $\chi \propto 1/T$ as a formula to be applied — temperature's effect on magnetic behaviour is examined only at the qualitative "heat weakens the alignment" level, which the next topic (Magnetic properties of materials) picks up in full.

The physical picture

Picture a long current-carrying solenoid with nothing inside it but empty space. Ampere's-law reasoning (already familiar from the previous topic) tells you the field inside depends only on the current and the number of turns per unit length — call the field this bare current alone would produce the magnetic intensity $\vec H$ . It is entirely a statement about the free current you control from outside; it does not know or care whether a material is present. Now slide a rod of some material into the solenoid, keeping the current fixed. The atoms of the rod are themselves tiny magnetic dipoles, and the applied $\vec H$ nudges them to line up (fully or partially) along the field direction. Add up every atomic dipole moment inside a volume $V$ and divide by that volume, and you get the magnetisation:

\vec M = \frac{\vec m_{net}}{V}

where: $\vec m_{net}$ is the vector sum of all atomic magnetic dipole moments inside the sample and $V$ is the sample's volume. Because each atomic moment has SI unit A·m² and volume has unit m³, magnetisation carries unit A/m — exactly the same unit as $H$ , which is a useful memory hook: $M$ and $H$ live on the same footing, one external, one internal. The field a magnetometer or a Hall probe would actually measure inside the rod is not $H$ alone and not $M$ alone — it is their combined contribution, scaled into tesla by $\mu_0$ :

\vec B = \mu_0(\vec H + \vec M)

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space, the same constant that appears throughout electromagnetism. Physically, this equation says the total field is built from two layers: the layer you would have gotten anyway from the bare current ( $\mu_0\vec H$ ), plus an extra layer contributed by the material's own aligned atomic dipoles ( $\mu_0\vec M$ ). Pull the rod out and $M\to0$ , leaving $B=\mu_0H$ — the vacuum-solenoid answer you already know. How strongly a given material magnetises in response to a given $H$ is captured by a single dimensionless number, the magnetic susceptibility:

\vec M = \chi\, \vec H

where: $\chi$ (chi) is negative and tiny (of order $-10^{-5}$ ) for diamagnetic materials, whose induced dipoles oppose the field; positive and tiny (of order $10^{-3}$ to $10^{-5}$ ) for paramagnetic materials, whose dipoles align weakly with the field; and positive and large (hundreds to thousands) for ferromagnetic materials, whose dipoles cooperate strongly and align almost completely even in a modest field. Substituting $M=\chi H$ into $B=\mu_0(H+M)$ collapses the whole picture into one multiplier acting directly on $H$ :

\vec B = \mu_0(1+\chi)\vec H = \mu_0\mu_r\vec H = \mu\vec H

where: $\mu_r = 1+\chi$ is the relative permeability (how many times bigger $B$ is compared to the vacuum value $\mu_0H$ , for the same applied $H$ ) and $\mu=\mu_0\mu_r$ is the material's absolute permeability, again in T·m/A. For a diamagnetic material $\mu_r$ sits a shade below 1; for a paramagnetic material it sits a shade above 1; for a ferromagnetic material like iron it can run into the hundreds or thousands, which is exactly why iron cores are used to concentrate magnetic field in transformers and electromagnets.

Figure

A bar chart comparing the applied intensity H, the material response M, and the resulting field B for three materials, showing a diamagnetic bar with a tiny negative dip, a paramagnetic bar with a tiny positive rise, and a ferromagnetic bar where the material response towers far above the applied intensity

Derivation ARelative permeability from susceptibility

The goal is to show, using nothing but algebraic substitution, that the two ways a material's magnetic response is quoted — susceptibility $\chi$ and relative permeability $\mu_r$ — are not independent numbers but two names for the same physics, tied together by the single relation $\mu_r = 1+\chi$ , and to see exactly how that relation falls straight out of the two defining equations already established: $\vec B=\mu_0(\vec H+\vec M)$ and $\vec M=\chi\vec H$ .

Step 1. Start from the defining relation between the total field and its two contributors, $\vec B = \mu_0(\vec H+\vec M)$ . This equation is exact for any linear magnetic material and simply states that the measured field is $\mu_0$ times the sum of the applied intensity and the material's magnetisation.

Step 2. Replace $\vec M$ using its own definition, $\vec M=\chi\vec H$ , so that everything on the right is written in terms of $\vec H$ alone. Substituting gives $\vec B = \mu_0(\vec H+\chi\vec H)$ .

Step 3. Factor $\vec H$ out of the bracket. Since both terms inside the parentheses carry a common factor of $\vec H$ , this becomes $\vec B = \mu_0(1+\chi)\vec H$ . Nothing has been assumed here beyond ordinary algebra — no new physics has entered.

Step 4. Name the bracket. Define $\mu_r \equiv 1+\chi$ ; then the equation reads $\vec B = \mu_0\mu_r\vec H$ , which is precisely the form used to define relative permeability in the first place — the number by which $\vec H$ must be multiplied (after scaling by $\mu_0$ ) to get the true field inside the material. Comparing the two boxed forms side by side, $\vec B=\mu_0(1+\chi)\vec H$ and $\vec B=\mu_0\mu_r\vec H$ , forces $\mu_r=1+\chi$ as the only consistent identification.

Step 5. Read off the absolute permeability as a bookkeeping convenience: writing $\mu=\mu_0\mu_r$ lets the whole relation collapse to the shortest possible form, $\vec B=\mu\vec H$ , which looks exactly like the vacuum relation $B=\mu_0H$ but with $\mu_0$ replaced by the material's own $\mu$ . This is why $\mu_r=1$ (equivalently $\chi=0$ ) recovers empty space exactly, why $\mu_r$ slightly below 1 signals diamagnetism, slightly above 1 signals paramagnetism, and $\mu_r$ in the hundreds or thousands signals ferromagnetism.

Worked examples

Board level - susceptibility and field from a measured magnetisation

A sample of magnetic material has magnetic intensity $H = 2000\ \text{A/m}$ applied to it and develops a magnetisation of $M = 6\ \text{A/m}$ . Find (a) the magnetic susceptibility $\chi$ of the material, and (b) the resulting field $B$ inside the material. Take $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ .

Step 1. Find $\chi$ from its defining ratio. $\chi = \dfrac{M}{H} = \dfrac{6}{2000} = 3\times10^{-3}$ . Both $M$ and $H$ are in A/m, so the ratio is correctly dimensionless.

Step 2. Find $B$ using $B=\mu_0(H+M)$ . $B = 4\pi\times10^{-7}\times(2000+6) = 4\pi\times10^{-7}\times2006$ .

Step 3. Evaluate the arithmetic. $4\pi\times10^{-7} \approx 12.57\times10^{-7}$ , so $B \approx 12.57\times10^{-7}\times2006 \approx 2.52\times10^{-3}\ \text{T}$ .

Answer: $\chi = 3\times10^{-3}$ (small and positive, so this is a paramagnetic material) and $B \approx 2.52\times10^{-3}\ \text{T}$ inside the sample.

Competitive level - relative permeability, amplification, and the material's own contribution

A rod of material has relative permeability $\mu_r = 400$ . It is placed in an external magnetic intensity $H = 1500\ \text{A/m}$ . Find (a) the total field $B$ inside the rod, (b) the field that would exist at the same point with the rod removed, (c) the magnetisation $M$ the rod develops, and (d) comment on what kind of material this is likely to be.

Step 1. Compute the absolute permeability. $\mu = \mu_0\mu_r = 4\pi\times10^{-7}\times400 \approx 5.03\times10^{-4}\ \text{T}\,\cdot\,\text{m/A}$ .

Step 2. Find $B$ using $B=\mu H$ . $B = 5.03\times10^{-4}\times1500 \approx 0.754\ \text{T}$ .

Step 3. Find the vacuum-only field for comparison, $B_{vac} = \mu_0 H = 4\pi\times10^{-7}\times1500 \approx 1.885\times10^{-3}\ \text{T}$ . The ratio $B/B_{vac} = 0.754/1.885\times10^{-3} \approx 400$ , exactly $\mu_r$ — confirming that relative permeability is literally the amplification factor over the empty-space answer.

Step 4. Find the magnetisation directly, using $\chi = \mu_r-1 = 399$ and $M=\chi H$ . $M = 399\times1500 = 5.985\times10^{5}\ \text{A/m}$ . As a check, $B=\mu_0(H+M) = 4\pi\times10^{-7}\times(1500+598{,}500) = 4\pi\times10^{-7}\times6\times10^{5} \approx 0.754\ \text{T}$ — matching Step 2, confirming the whole calculation is self-consistent.

Step 5. Interpret. A relative permeability of $400$ , with the material's own magnetisation nearly $400$ times larger than the applied intensity itself, is far beyond anything a dia- or paramagnetic substance shows (those stay within a whisker of $\mu_r=1$ ). Answer: $B \approx 0.754\ \text{T}$ , versus a vacuum value of only $1.885\times10^{-3}\ \text{T}$ at the same $H$ ; $M \approx 5.99\times10^{5}\ \text{A/m}$ ; the rod behaves as a ferromagnetic material, such as iron. Timing note: in a three-mark JEE-style numerical, do not recompute $\mu_0H$ from scratch twice — calculate it once, then get $B$ either from $\mu H$ or from $\mu_0(H+M)$ and use the other route purely as a thirty-second consistency check before you commit to the final answer.

Common traps

Confusing $\vec B$ (the total, measured field inside the material) with $\vec H$ (the applied, material-independent magnetising intensity) — they are related by $B=\mu_0(H+M)$ , never simply equal to one another.
Mixing up units: $M$ and $H$ both carry SI unit A/m, while $B$ carries unit tesla (T); plugging A/m values directly in for $B$ , or vice versa, silently wrecks the arithmetic.
Assuming $\chi$ is always positive — diamagnetic materials have a small negative $\chi$ , meaning their $\vec M$ points opposite to $\vec H$ , and $B$ is then a shade below the vacuum value.
Forgetting the one-line relation $\mu_r=1+\chi$ and trying to look up or re-derive $\mu_r$ from scratch, instead of just adding 1 to a given $\chi$ (or subtracting 1 the other way).
Treating $\mu_r$ close to 1 (dia-/paramagnetic) and $\mu_r$ in the hundreds (ferromagnetic) as differing only in degree — they reflect a genuinely different physical mechanism, and mixing the two into one "all materials behave a little bit magnetically" sentence loses the marks for classification questions.

Common trapH is not a lesser cousin of B — it is a completely different, material-independent quantity

The single most common conceptual slip in this topic is treating $\vec H$ as just " $\vec B$ before you multiply by $\mu_0$ ," as though $H$ were some diluted or incomplete version of the real field. It is not. $\vec H$ is fixed entirely by the free currents supplied from outside — put any material you like inside the same solenoid at the same current, and $\vec H$ does not change at all. What changes is $\vec M$ , the material's own response, and therefore $\vec B$ , the field that actually results. Swapping a diamagnetic rod for a ferromagnetic one in the same coil leaves $H$ untouched but can send $B$ up by a factor of a thousand. Whenever a question gives you "the applied field," check carefully whether it means $H$ (set by current alone) or $B$ (the total, material-dependent result) — using the wrong one in $B=\mu_0(H+M)$ or $B=\mu H$ produces answers that are wrong by exactly the factor $\mu_r$ .

You should be able to…

Define magnetisation $\vec M$ and magnetic intensity $\vec H$ , quoting the correct SI unit (A/m) for both.
State and correctly apply $\vec B = \mu_0(\vec H+\vec M)$ , keeping straight which symbol is in tesla and which are in A/m.
Define susceptibility $\chi = M/H$ and relative permeability $\mu_r = 1+\chi$ , and connect both to the absolute permeability $\mu=\mu_0\mu_r$ .
Derive, from $B=\mu_0(H+M)$ and $M=\chi H$ , the compact relation $B=\mu_0\mu_r H$ , and reproduce that short chain of algebra on demand.
Compute any one of $B$ , $H$ , $M$ , $\chi$ , or $\mu_r$ given any two of the others, and sanity-check the answer by comparing it with the vacuum value $\mu_0H$ .
Use the sign and rough size of $\chi$ (or $\mu_r$ ) to say, at a glance, whether a material is behaving as diamagnetic, paramagnetic, or ferromagnetic.

Ask the tutor: Want a mixed drill that hands you random pairs of B, H, M, chi, and mu-r and asks you to fill in the rest, the way a board numerical would?

Why do M and H have the same unit but B has a different one?

Magnetisation is a dipole-moment density: total dipole moment (unit A·m squared) divided by volume (unit m cubed), which works out to A/m. Magnetic intensity is defined through an Ampere's-law-type relation to free current per unit length, which also comes out in A/m — that shared unit is not a coincidence, it reflects that H and M are added directly inside the same bracket in B = mu-zero(H + M). B only becomes a field in tesla after that sum is multiplied by mu-zero, which is what actually converts an "amps per metre" quantity into the units you measure with a Hall probe or a compass. So remember: H and M are measured on the same ruler; B lives on a different one, reached only after the mu-zero conversion.

Is chi ever exactly zero, and what would that mean?

Chi equal to zero means the material develops no magnetisation at all in response to H, so B reduces exactly to the vacuum value mu-zero H. No real material is exactly like this — every substance has some atomic-scale magnetic response, even if it is extremely small — but vacuum itself, by definition, behaves as chi equal to zero, mu-r equal to 1. In the syllabus, materials are sorted by how far chi sits from zero and in which direction: diamagnetic materials have a tiny negative chi, paramagnetic materials a tiny positive chi, and ferromagnetic materials a large positive chi, so chi is really a one-number summary of "how far from vacuum does this material behave."

Do I need to know why some materials are diamagnetic and others paramagnetic for this specific topic?

Not for this topic specifically — the mechanism (whether individual atoms have a net dipole moment to begin with, and how electron orbits respond to an applied field) belongs to the next topic, Magnetic properties of materials, where dia-, para-, and ferromagnetic behaviour is explained and illustrated with examples. Here the job is narrower and more computational: given any two of B, H, M, chi, and mu-r, be able to find the rest using B = mu-zero(H + M), M = chi H, and mu-r = 1 + chi, and use the sign and size of chi as a quick pointer to which family a material probably belongs to.

Is Curie's law, chi proportional to 1 over T, something I need to apply in a numerical?

No — the 2025-26 CBSE syllabus keeps temperature's effect on magnetic properties at a qualitative level only: warmer paramagnetic materials magnetise less readily because thermal jostling disrupts the alignment of atomic dipoles, and ferromagnetic materials lose their strong magnetism altogether above a characteristic temperature. You should be able to state that heating generally weakens magnetic ordering, but you will not be asked to plug numbers into chi is proportional to 1 over T as a formal formula — treat any question phrased that way as testing the qualitative idea, not the proportionality as an equation to solve.