Magnetisation, Intensity & Susceptibility

Every electrical machine on Earth is a $\mu_r$ story. The transformer humming in your street substation channels its flux through grain-oriented electrical steel with effective $\mu_r$ in the tens of thousands — the design rule "flux follows iron" is Hopkinson's magnetic circuit at work, and it is why a transformer is mostly iron by weight. Electric-vehicle motor designers spend their careers at the edge of the ledger: pushing iron toward saturation ($M \to M_s$) squeezes maximum torque from every kilogram, and the design software plots $B = \mu_0(H+M)$ curves for candidate steels the way you plotted graphs in practicals. At the opposite end of the table, the feeble responders get their own industry. MRI machines must keep their 1.5–3 T field uniform to parts per million across a patient — but a patient is mostly water ($\chi = -9\times10^{-6}$), and even that whisper of diamagnetism smears images; "shimming" coils are engineering aimed at the fifth decimal place of $\mu_r$. And magnetically shielded rooms wrap brain-signal labs in mu-metal — a nickel-iron alloy bred for colossal $\mu_r$ — so field lines prefer the walls to the interior, leaving a magnetic silence a hundred thousand times deeper than the street outside.

Picture the empty solenoid first, exactly as Solenoids & Toroids built it: $n$ turns per metre, current $I$, a uniform field $\mu_0 n I$ filling the bore. Every part of that expression is yours — you chose the winding, you set the current. Nothing about it can change unless you act. Now slide in the iron rod. Bar Magnets & the Dipole Picture told you what iron is made of: atoms that are each a tiny current loop with a magnetic moment. In the unmagnetised rod those moments point in scrambled directions and cancel — a crowd of compasses in a shaken box. Your coil's field is the organising instruction: moment by moment, the atomic magnets swing toward alignment. And an aligned atomic magnet is a source — each one adds its own small field, in the same direction as yours. The rod has become a magnet nested inside your coil, and the bore now holds two fields stacked: the one you built, and the one the iron volunteered. The bookkeeping problem is immediate: the measured field $B$ inside the rod is one number, but it has two authors. Physics answers with two ledger columns. The magnetic intensity $H$ tracks the coil's column — it is defined so that, for a solenoid, $H = nI$ whatever sits in the bore. The magnetisation $M$ tracks the material's column — the total atomic moment the rod has lined up, per unit volume. The measured field is the sum of the columns, each converted to tesla by the same $\mu_0$.

Look hard at the ledger bar in the figure: in an iron-cored coil, your winding contributes 0.2 % of the final field. The other 99.8 % is the iron's own atomic magnetism, unlocked by your 0.2 %. That is why every electromagnet, every transformer, every relay coil in Household Circuits' wiring has an iron heart — the core is not an accessory, it is the amplifier. One caution before the formulas, because it guards the whole module: the material does not amplify your current, and $\mu_r$ is not a "gain" applied to electrons. Your current and the field it makes are untouched. The material adds its own, separate contribution. The distinction feels pedantic until the exam asks "what happens to $H$ when the core is inserted?" — answer: nothing. $H$ is the coil's column of the ledger, and only the coil writes there.

Start with the material's column. Each atom carries a magnetic moment (Current Loops as Magnetic Dipoles gave the currency: the ampere–metre²). When a field partially aligns them, a volume $V$ of material acquires a net moment $m_{\text{net}}$. The useful measure is the density:

where: $M$ is the magnetisation in A·m²/m³ = A/m — net moment per unit volume. It is the alignment census: $M = 0$ for a shaken crowd of atomic compasses, and $M$ maxes out (saturates) when every last moment points the same way. For iron that ceiling is enormous — about $1.7 \times 10^6$ A/m — which is the deep reason iron cores work. Now the coil's column. For a solenoid of $n$ turns per metre carrying $I$:

where: $H$ is the magnetic intensity (also "magnetising field" or "magnetic field strength"), also in A/m — and that shared unit with $M$ is no accident: both columns of the ledger are moment-per-volume ledger entries, one bookkept from your free current, one from the material's bound atomic currents. In an empty bore, $B_0 = \mu_0 n I = \mu_0 H$: the vacuum field is just $H$ converted to tesla. Note what equation (2) does not contain: any property of the core. Insert iron, remove it, swap in bismuth — $H$ stays $nI$. The measured field inside the material stacks the two columns, each wearing the same conversion factor:

where: $B$ is the real, compass-deflecting, force-producing field in tesla; $\mu_0 = 4\pi \times 10^{-7}$ T·m/A is the familiar constant from Fields of Straight Wires & Circular Loops. Equation (3) is the master ledger and it is always true, for any material in any state — it is really just the definition of $M$ doing its job. Everything after this point is about the shortcut available when the material is well-behaved. The shortcut: for most materials at ordinary fields, the response is proportional to the push. Double $H$, double $M$:

where: $\chi$ (chi) is the magnetic susceptibility — a pure, dimensionless number, since $M$ and $H$ share units. Its sign and size are the material's magnetic personality: $\chi \approx -9 \times 10^{-6}$ for water (feebly opposing), $+2.2 \times 10^{-5}$ for aluminium (feebly assisting), and effectively in the thousands for soft iron (massively assisting). Why those three families exist is exactly the next module's business — Dia-, Para- & Ferromagnetism reads personalities; this module runs the arithmetic. Feed (4) into (3) and the ledger collapses into one multiplication:

where: $\mu_r$ is the relative permeability — the single most quotable number of the module, because $B = \mu_r B_0$: the core multiplies the empty-bore field by $\mu_r$. The "+1" in $\mu_r = 1+\chi$ is your coil's own contribution keeping its seat at the table; $\chi$ is the material's added chorus. Some books also define the absolute permeability $\mu = \mu_0\mu_r$ (unit T·m/A), so that $B = \mu H$ mirrors the vacuum law with $\mu_0$ upgraded.

The dictionary between χ and μᵣ

The two response numbers carry identical information; exams switch between them freely, so read this table until the translation is automatic.

Two honesty notes the table is hiding. First, a ferromagnet's "$\chi$" is not a fixed constant — it depends on how hard you push and on the sample's history (the full story is Domains, Hysteresis & Choosing Magnetic Materials, two modules on). Board questions sidestep this by giving you an effective $\mu_r$; use it as a snapshot. Second, $\mu_r < 1$ is real physics, not a misprint: diamagnets genuinely thin the field threading them. The snapshot picture is worth one graph. Plot the measured $B$ against the dial $H$ for an iron core, and the "multiply by $\mu_r$" story shows its expiry date: a steep, nearly straight climb while the iron still has moments to align, then a knee, then a gentle slope where the iron is spent and only your coil's vacuum-rate contribution keeps growing. Every claim in this module lives on the steep stretch.

Board level — the full ledger for an iron-cored solenoid (CBSE/ISC pattern, 5 marks)

A solenoid with 2000 turns per metre carries a current of 1.5 A. Its core is a material of relative permeability $\mu_r = 600$. Find (a) the magnetic intensity $H$, (b) the field $B_0$ the solenoid would have when empty, (c) the field $B$ with the core in place, (d) the susceptibility $\chi$ and the magnetisation $M$ of the core, and (e) verify the ledger $B = \mu_0(H+M)$.

Step 1. Part (a). The intensity is the coil's entry, core-blind: $H = nI = 2000 \times 1.5 = \mathbf{3000\ \text{A/m}}$.

Step 2. Part (b). Empty bore: $B_0 = \mu_0 H = 4\pi\times10^{-7} \times 3000 = \mathbf{3.77 \times 10^{-3}\ \text{T}}$ — about 3.8 mT, some 75 Earth-fields.

Step 3. Part (c). The core multiplies: $B = \mu_r B_0 = 600 \times 3.77\times10^{-3} = \mathbf{2.26\ \text{T}}$ — thoroughly respectable MRI-class field from a bench coil, courtesy of the material.

Step 4. Part (d). Susceptibility from the dictionary: $\chi = \mu_r - 1 = \mathbf{599}$. Magnetisation from linear response: $M = \chi H = 599 \times 3000 = \mathbf{1.80 \times 10^{6}\ \text{A/m}}$.

Step 5. Part (e). Ledger check: $\mu_0(H + M) = 4\pi\times10^{-7}\,(3000 + 1.797\times10^{6}) = 4\pi\times10^{-7} \times 1.800\times10^{6} = \mathbf{2.26\ \text{T}}$ ✓. Notice the shape of the sum: $M$ outweighs $H$ six hundred to one — the core is writing 99.8 % of the field.

Competitive level — from coil specs to the rod's total magnetic moment (ISC/olympiad-flavoured, multi-concept)

A solenoid wound with 2000 turns per metre carries 2.0 A. A soft-iron rod of volume $2.0 \times 10^{-5}\ \text{m}^3$ fills its bore; under these conditions the iron behaves linearly with $\mu_r = 500$. Find (a) $B$ inside the rod, (b) the magnetisation $M$, and (c) the total magnetic dipole moment acquired by the rod. (d) A student claims the rod now out-magnetises a strong laboratory bar magnet of moment 5 A·m². Is she right?

Step 1. Coil's column: $H = nI = 2000 \times 2.0 = 4000$ A/m. Empty-bore field would be $\mu_0 H = 5.03\times10^{-3}$ T.

Step 2. Part (a). $B = \mu_0 \mu_r H = 4\pi\times10^{-7} \times 500 \times 4000 = \mathbf{2.51\ \text{T}}$.

Step 3. Part (b). $\chi = \mu_r - 1 = 499$, so $M = \chi H = 499 \times 4000 = \mathbf{2.0 \times 10^{6}\ \text{A/m}}$ (three-figure value $1.996\times10^6$).

Step 4. Part (c). Magnetisation is moment per volume — so invert the definition, equation (1): $m_{\text{net}} = M \times V = 1.996\times10^{6} \times 2.0\times10^{-5} = \mathbf{39.9\ \text{A}\,\cdot\,\text{m}^2 \approx 40\ \text{A}\,\cdot\,\text{m}^2}$.

Step 5. Part (d). Yes, and it isn't close: $40$ A·m² is eight times the bar magnet's 5 A·m². A fist-sized helping of organised iron is a formidable magnet — which is why relays, electromagnets and magnetic cranes are iron-core devices, not bare coils. Timing note: 3–4 minutes. The only place candidates stall is part (c) — the reflex is to hunt for a new formula, but it is just the definition $M = m/V$ read backwards. If a question gives volume, expect to convert between $M$ and total moment.

$H$ is core-blind — and examiners know you forget. "The core is inserted; what happens to $H$, $B$, and $M$?" Answer: $H$ unchanged ($= nI$), $M$ jumps from ~0 to $\chi H$, $B$ jumps by the factor $\mu_r$. Any answer that changes $H$ contradicts $H$'s definition.

$\mu_r = 1 + \chi$, not $\chi$. For iron-family numbers the slip is invisible (599 vs 600), which is exactly why examiners test it on weak magnets, where $\chi = -9\times10^{-6}$ and $\mu_r = 0.999991$ are very different claims. State both explicitly.

Units: A/m for both $M$ and $H$; tesla only after $\mu_0$ touches them. Writing "$M = 2\times10^6$ T" is a two-way unit crime. $\chi$ and $\mu_r$ carry no units at all — if your $\chi$ has units, you divided the wrong pair.

The JEE-Main/NEET honesty line. The formal $M$–$H$–$\chi$–$\mu_r$ chain was removed from the JEE-Main syllabus in the 2024 reduction and does not appear in the current NEET syllabus either — this module's arithmetic is CBSE board-exam and ISC territory. But the personalities ($\chi$ signs, sizes and temperature behaviour of dia/para/ferro materials) remain firmly on both: that craft lives in Dia-, Para- & Ferromagnetism, next.

Constants discipline. $\mu_0 = 4\pi\times10^{-7}$ T·m/A — keep it symbolic until the last line; the $4\pi$ loves to cancel (as in $4\pi\times10^{-7} \times \frac{1}{4\pi} \times 10^{7} = 1$ patterns). Boards accept $\mu_0 \approx 1.257\times10^{-6}$ if you must go decimal.

Ferromagnetic $\mu_r$ is a snapshot, not a constant. If a question hands you "$\mu_r$ of iron = 500," use it — but never derive future behaviour from it (double the current does not reliably double $B$ in real iron; saturation intervenes). The one-line honesty: linear formulas describe iron only over the stretch where the given $\mu_r$ was measured.

Texts. NCERT Physics Class 12, Ch. 5 (Magnetism and Matter — magnetisation of materials) · ISC Class 12 Physics, magnetism unit (the full B–H–M–χ–μᵣ chain) · HC Verma, Concepts of Physics Vol. 2, Ch. 37 · Resnick–Halliday–Walker, Ch. 32 (magnetism of matter). For the university-level treatment of the same ledger — bound currents, the H-field's fine print, demagnetising effects — see Magnetization, Magnetic Susceptibility and Magnetic Permeability (UG deep dive).

Provenance. All results standard. A verifier should recompute: $H = 2000\times1.5 = 3000$ A/m; $B_0 = \mu_0 H = 3.77$ mT; $B = 600 B_0 = 2.26$ T; $M = 599\times3000 = 1.80\times10^6$ A/m and $\mu_0(H{+}M) = 2.26$ T ✓; competitive example $H = 4000$ A/m, $B = 2.51$ T, $M = 1.996\times10^6$ A/m, $m = MV = 39.9$ A·m²; iron ceiling $\mu_0 M_s \approx 4\pi\times10^{-7}\times1.7\times10^6 = 2.1$ T.

Prerequisites. Solenoids & Toroids · Bar Magnets & the Dipole Picture. Leads to. Dia-, Para- & Ferromagnetism (why χ has three personalities) · Domains, Hysteresis & Choosing Magnetic Materials (what happens when linearity dies).