PhysicsCBSE Class 12 · Magnetism and Matter

Magnetisation and Magnetic Intensity Class 12: M, H, χ and μr

Updated 15 October 2026 · Fact-checked

Magnetisation M is the net magnetic moment per unit volume of a material. Magnetic intensity H is the part of the field set up by external currents (H = nI in a solenoid). They combine as B = μ₀(H + M). Susceptibility χ = M/H and relative permeability μr = 1 + χ describe how strongly a material responds.

The current sets H = nI; the material answers with M = χH; together they give B = μ₀(H + M), and μr = 1 + χ.

Illustration drawn for QuestionBuddy.

Understand Magnetisation and Magnetic Intensity

Put an iron rod inside a current-carrying solenoid and the field inside becomes hundreds of times stronger; put a piece of copper there and the field hardly changes. To describe this, we separate the field due to the current from the field due to the material.

Magnetisation M. Every atom can carry a tiny magnetic moment from the motion of its electrons. In a sample, these moments add as vectors. The magnetisation is the net magnetic moment per unit volume: $M = \frac{m _{n e t}}{V}$ . Its unit is A m² / m³ = A/m. M is zero if the atomic moments point at random and large if they are lined up.

Magnetic intensity H. In a long solenoid with n turns per metre and current I, the field without any core is $B_{0} = μ_{0} n I$ . Fill the solenoid with a material and the field becomes $B = B_{0} + B_{m}$ , where the material's contribution is $B_{m} = μ_{0} M$ . We define the magnetic intensity $H = \frac{B}{μ _{0}} - M$ , so that $B = μ_{0} (H + M)$ . H has the same unit as M (A/m). For a solenoid, $H = n I$ : it depends only on the current and the winding, not on the core material. So H represents the 'cause' (the external currents) and M the material's response.

Susceptibility. For most materials (not ferromagnets at high fields), M is proportional to H: $M = χ H$ . The dimensionless number $χ$ is the magnetic susceptibility, a measure of how readily the material is magnetised.

$χ$ small and positive: paramagnetic (M along H).
$χ$ small and negative: diamagnetic (M opposite to H).
$χ$ very large and positive (hundreds to thousands): ferromagnetic.

Permeability. Substituting $M = χH$ : $B = μ_{0} (1 + χ) H = μ_{0} μ_{r} H = μ H$ . The relative permeability $μ_{r} = 1 + χ$ is dimensionless; the permeability $μ = μ_{0} μ_{r}$ has the same unit as $μ_{0}$ (T m/A). $μ_{r}$ plays the role of the dielectric constant K in electrostatics. Only one of $χ$ , $μ_{r}$ and $μ$ is independent: given one, you can find the others. Diamagnets have $μ_{r}$ slightly less than 1, paramagnets slightly more than 1, ferromagnets much more than 1. A perfect diamagnet (a superconductor) has $χ = - 1$ , $μ_{r} = 0$ and B = 0 inside.

Magnetising current. The extra field $μ_{0} M$ from a core is the same as if an extra current $I_{m}$ flowed in the windings of the empty solenoid: $μ_{0} n I_{m} = μ_{0} M$ , so $I_{m} = \frac{M}{n}$ . For an iron core this can be hundreds of amperes, which is why iron cores make powerful electromagnets from small currents.

Effect of temperature (qualitative). Heat jostles the atomic moments and makes alignment harder, so the susceptibility of a paramagnet falls as temperature rises. A ferromagnet heated strongly enough loses its ordered domains and behaves as a paramagnet. Diamagnetism hardly depends on temperature.

Magnetisation and Magnetic Intensity: formulas and key facts

Magnetisation: $M = \frac{m _{n e t}}{V}$; Unit A/m.
Magnetic intensity in a solenoid: $H = n I$; Independent of the core material. Unit A/m.
Total field: $B = μ_{0} (H + M)$; μ₀H from currents, μ₀M from the material.
Susceptibility: $χ = \frac{M}{H}$; Dimensionless; + for para, − for dia, very large + for ferro.
Relative permeability and permeability: $μ_{r} = 1 + χ, μ = μ_{0} μ_{r}, B = μ H$; μ_r is the magnetic analogue of the dielectric constant K.
Magnetising current: $I_{m} = \frac{M}{n}$; Extra current that would give the same B without the core.

How to solve Magnetisation and Magnetic Intensity questions

Use this method for numericals on solenoid cores, susceptibility and permeability.

1Find H from the winding: H = nI (n in turns per metre).
2If μ_r (or χ) is given, get the other: μ_r = 1 + χ.
3Field: $B = μ_{0} μ_{r} H$ .
4Magnetisation: $M = χH = (μ_{r} - 1) H$ , or $M = \frac{B}{μ _{0}} - H$ .
5If B and H are given, find $μ_{r} = \frac{B}{μ _{0} H}$ first.
6For a sample of known moment, $M = \frac{m}{V}$ (V in m³).
7Classify the material from the sign and size of χ, and give units (A/m for M and H, T for B).

Magnetisation and Magnetic Intensity 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.

Why electromagnets have iron cores

Suppose a solenoid with 2000 turns per metre carries 0.5 A. You compare the field inside with no core and with a soft iron core of relative permeability 1000 (assume it stays in the range where μr is constant).

(a) Find H. (b) Find B without and with the core. (c) Find the magnetisation of the core. (d) What current would the empty solenoid need to give the same B?

Try it first, then show the solutionSolution

(a) $H = n I = 2000 \times 0.5 = 1000$ A/m (same with or without the core).
(b) Without core: $B_{0} = μ_{0} H \approx 1.26 \times 1 0^{- 3}$ T. With core: $B = μ_{r} μ_{0} H \approx 1.26$ T.
(c) $M = (μ_{r} - 1) H = 999 \times 1000 \approx 9.99 \times 1 0^{5}$ A/m.
(d) Total current needed $= \frac{B}{μ _{0} n} = μ_{r} I = 1000 \times 0.5 = 500$ A.

Answer: (a) 1000 A/m; (b) about 1.26 mT and 1.26 T; (c) about $1 0^{6}$ A/m; (d) 500 A, instead of 0.5 A with the core.

Memory hook: The current sets H; the iron multiplies B.

A weakly diamagnetic sample

Imagine a sample with susceptibility $- 9 \times 1 0^{- 6}$ placed in a strong laboratory field where H = $8 \times 1 0^{6}$ A/m.

(a) Find its magnetisation and its direction. (b) By what percentage is B inside the sample less than $μ_{0} H$ ? (c) Is the sample attracted into or pushed out of a strong-field region?

Try it first, then show the solutionSolution

$M = χH = - 9 \times 1 0^{- 6} \times 8 \times 1 0^{6} = - 72$ A/m, i.e. opposite to H.
$B = μ_{0} H (1 + χ)$ , so B is lower by $∣ χ ∣ \times 100% = 9 \times 1 0^{- 4} %$ .
M opposes H, so it is pushed from stronger towards weaker field regions (diamagnetic behaviour), very weakly.

Answer: (a) 72 A/m opposite to H; (b) by about 0.0009%; (c) pushed out, weakly.

Memory hook: Negative χ: the material argues back, gently.

Tricks to remember Magnetisation and Magnetic Intensity

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

H from the Hand (current), M from the Material
H = nI is what you control with the current; M is what the material adds.
Use it when: Separating H and M.
B is μ₀ times the sum
B = μ₀(H + M): 'Be μ₀ to Him and Me'.
Use it when: Recalling the relation.
One plus chi
μ_r = 1 + χ, like K for a dielectric.
Use it when: Converting between χ and μ_r.
Sign of χ = type
Minus small: dia. Plus small: para. Plus huge: ferro.
Use it when: Classifying materials.

Magnetisation and Magnetic Intensity: solved examples

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

Example 1: H, B and M for a cored solenoid · 3 marks · Short answerBoard level

A solenoid with 600 turns per metre carries 1.5 A. Its core has relative permeability 500. Find (a) H, (b) B and (c) M.

Try it first, then show the solutionSolution

(a) $H = n I = 600 \times 1.5 = 900$ A/m. (1)
(b) $B = μ_{r} μ_{0} H = 500 \times 4 π \times 1 0^{- 7} \times 900 \approx 0.565$ T. (1)
(c) $M = (μ_{r} - 1) H = 499 \times 900 \approx 4.49 \times 1 0^{5}$ A/m. (1)

Answer: (a) 900 A/m; (b) about 0.565 T; (c) about $4.49 \times 1 0^{5}$ A/m.

Example 2: Relating χ, μr and μ · 2 marks · Very short answerBoard level

Starting from $B = μ_{0} (H + M)$ and $M = χ H$ , show that $μ_{r} = 1 + χ$ . What are the units of M, H, χ and μ?

Try it first, then show the solutionSolution

$B = μ_{0} (H + χH) = μ_{0} (1 + χ) H$ . Writing $B = μ_{0} μ_{r} H$ gives $μ_{r} = 1 + χ$ . (1)
M and H: A/m; χ and μ_r: no unit; μ = μ₀μ_r: T m/A (same as μ₀). (1)

Answer: μ_r = 1 + χ; M and H in A/m, χ dimensionless, μ in T m/A.

Example 3: From susceptibility to permeability · 3 marks · Short answerHarder board level

A material has susceptibility $2.3 \times 1 0^{- 5}$ . (a) Classify it. (b) Find its relative permeability and permeability. (c) In a field H = $1 0^{5}$ A/m, find its magnetisation.

Try it first, then show the solutionSolution

(a) χ small and positive: paramagnetic. (1)
(b) $μ_{r} = 1 + 2.3 \times 1 0^{- 5} = 1.000023$ ; $μ = μ_{0} μ_{r} \approx 1.2567 \times 1 0^{- 6}$ T m/A. (1)
(c) $M = χH = 2.3 \times 1 0^{- 5} \times 1 0^{5} = 2.3$ A/m, along H. (1)

Answer: (a) paramagnetic; (b) 1.000023 and about $1.2567 \times 1 0^{- 6}$ T m/A; (c) 2.3 A/m along H.

Example 4: Magnetisation of a sample · 1 mark · Very short answerHarder board level

A small iron sample of volume 2 cm³ acquires a net magnetic moment of 1.6 A m² when placed in a field. Find its magnetisation.

Try it first, then show the solutionSolution

$M = \frac{m}{V} = \frac{1.6}{2 \times 1 0 ^{- 6}} = 8 \times 1 0^{5}$ A/m. (1)

Answer: $8 \times 1 0^{5}$ A/m.

Example 5: Magnetising current · 5 marks · Long answerHOTS / competency

For the cored solenoid of the first example (n = 600 per metre, I = 1.5 A, μ_r = 500):
(a) Explain what is meant by the magnetising current. (1)
(b) Find it. (2)
(c) Check that the empty solenoid carrying I + I_m gives the same B as the cored one. (2)

Try it first, then show the solutionSolution

(a) It is the extra current that, flowing in the windings of the empty solenoid, would produce the same field as the core's magnetisation. (1)
(b) $μ_{0} n I_{m} = μ_{0} M$ gives $I_{m} = \frac{M}{n}$ . (1)
$I_{m} = \frac{4.49 \times 1 0 ^{5}}{600} \approx 749$ A. (1)
(c) $B = μ_{0} n (I + I_{m}) = 4 π \times 1 0^{- 7} \times 600 \times (1.5 + 748.5) = 4 π \times 1 0^{- 7} \times 600 \times 750$ . (1)
$= 0.565$ T, the same as $μ_{r} μ_{0} n I$ , since $I + I_{m} = μ_{r} I = 750$ A. (1)

Answer: (a) the extra winding current equivalent to the core; (b) about 749 A; (c) B ≈ 0.565 T both ways.

Example 6: Two materials in the same field · 4 marks · Case-basedHarder board level

Read the passage and answer.

Two rods, A with χ = $1.5 \times 1 0^{- 4}$ and B with χ = $- 2 \times 1 0^{- 5}$ , are placed in turn in a region where H = $1 0^{4}$ A/m.
(a) Find the magnetisation of each, with its direction relative to H. (1)
(b) Classify A and B. (1)
(c) Compare the field B inside A with $μ_{0} H$ . (1)
(d) Which rod is pulled towards the stronger part of a non-uniform field? (1)

Try it first, then show the solutionSolution

(a) $M_{A} = 1.5$ A/m along H; $M_{B} = - 0.2$ A/m, opposite to H. (1)
(b) A is paramagnetic, B is diamagnetic. (1)
(c) $B = μ_{0} (H + M) = μ_{0} H (1 + 1.5 \times 1 0^{- 4})$ , larger than $μ_{0} H \approx 0.012566$ T by only 0.015%. (1)
(d) A, the paramagnet; B is pushed weakly towards weaker field. (1)

Answer: (a) 1.5 A/m along H and 0.2 A/m against H; (b) A paramagnetic, B diamagnetic; (c) only 0.015% larger; (d) A.

Where students lose marks in Magnetisation and Magnetic Intensity

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

Mistake: Changing H when a core is inserted.
Thinking the core changes everything.
Wrong working
With core, H = μ_r nI
Right working
H = nI with or without core; B = μ_r μ₀ nI
Do this: H depends only on the winding and current.
Usually costs: 1 to 2 marks.
Mistake: Writing M = μ_r H.
Mixing up μ_r with χ.
Do this: M = χH = (μ_r − 1)H.
Usually costs: The final mark (or an error of H).
Mistake: Giving units to χ or μ_r.
They look like 'permeability'.
Do this: χ and μ_r are pure numbers; only μ has units (T m/A).
Usually costs: ½ mark.
Mistake: Using total turns N for n.
n is turns per unit length.
Do this: n = N/L, in turns per metre.
Usually costs: The answer mark.
Mistake: Writing Curie's law in a temperature answer.
Older notes include it.
Do this: For 2026-27 describe the effect of temperature qualitatively; Curie's law is not in the syllabus.
Usually costs: Time; examiners mark the qualitative idea.

Exam tips

Define M, H, χ and μ_r in one line each, with units; definitions are a quick 2 marks.
Show B = μ₀(H + M) and derive μ_r = 1 + χ in two lines when asked how they are related.
In numericals, compute H first, then B, then M; check that B/μ₀ − H equals M.
A 2026 sample paper (Q27(II)) set a 3-mark question on these quantities.
Link the sign of χ to the type of material; it connects this topic to dia-, para- and ferromagnetism.

Practice questions on Magnetisation and Magnetic Intensity

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
The SI unit of magnetisation is
(a) A m² (b) A/m (c) T (d) T m/A
Hint
Moment per unit volume.
Answer
(b) A/m.
Q2 · 2 marks · Very short answerBoard level
Define magnetic susceptibility. What is its sign for diamagnetic and for paramagnetic substances?
Hint
χ = M/H.
Answer
Susceptibility is the ratio of magnetisation to magnetic intensity, χ = M/H, a measure of how easily a material is magnetised. It is small and negative for diamagnetic and small and positive for paramagnetic substances.
Q3 · 1 mark · Very short answerHarder board level
A material has susceptibility 599. Find its relative permeability and its permeability.
Hint
μ_r = 1 + χ.
Answer
$μ_{r} = 600$ ; $μ = 600 \times 4 π \times 1 0^{- 7} \approx 7.54 \times 1 0^{- 4}$ T m/A.
Q4 · 1 mark · Very short answerHarder board level
A rod 10 cm long with cross-section 1 cm² has a magnetic moment of 4 A m². Find its magnetisation.
Hint
V = 10⁻⁵ m³.
Answer
$M = \frac{4}{0.1 \times 1 0 ^{- 4}} = 4 \times 1 0^{5}$ A/m.
Q5 · 3 marks · Short answerHarder board level
With a core in a solenoid, B = 0.8 T when H = 1000 A/m. Find μ_r, χ and M.
Hint
μ_r = B/(μ₀H).
Answer
$μ_{r} = \frac{0.8}{4 π \times 1 0 ^{- 7} \times 1000} \approx 637$ ; $χ = μ_{r} - 1 \approx 636$ ; $M = χH \approx 6.36 \times 1 0^{5}$ A/m.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): The magnetic intensity H inside a long solenoid does not depend on the material of its core.
Reason (R): H = nI depends only on the number of turns per unit length and the current.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which quantity carries the material's effect?
Answer
(a). The material changes M and hence B, not H.
Q7 · 2 marks · Very short answerHOTS / competency
A superconductor expels magnetic field completely from its interior. Find its susceptibility and relative permeability, and say what type of magnetic material it is.
Hint
B = 0 inside means H + M = 0.
Answer
B = μ₀(H + M) = 0 gives M = −H, so χ = −1 and μ_r = 1 + χ = 0. It is a perfect diamagnet.

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.Define magnetisation.Show answer
Net magnetic moment per unit volume, M = m_net/V.
2.Magnetic intensity inside a solenoid?Show answer
H = nI.
3.Relation between B, H and M?Show answer
B = μ₀(H + M).
4.Relation between μ_r and χ?Show answer
μ_r = 1 + χ.
5.Unit of H?Show answer
A/m.
6.χ and μ_r of a perfect diamagnet?Show answer
χ = −1, μ_r = 0.

Magnetisation and Magnetic Intensity: frequently asked questions

What is the difference between B and H?

H is the magnetic intensity due to external currents alone (H = nI in a solenoid), measured in A/m. B is the total magnetic field, including the material's contribution: B = μ₀(H + M), measured in tesla.

Why is μ_r called the analogue of the dielectric constant?

Just as K tells by what factor a dielectric changes things in electrostatics (ε = Kε₀, C = KC₀), μ_r tells by what factor a material changes the magnetic relation, μ = μ₀μ_r and B = μ₀μ_rH. Both are dimensionless ratios.

Can susceptibility be negative?

Yes. Diamagnetic materials have small negative susceptibility: their magnetisation points opposite to H, so they are weakly repelled by magnets.

Is Curie's law part of the 2026-27 syllabus?

No. The syllabus asks for the effect of temperature on magnetic properties qualitatively: paramagnetic susceptibility falls with rising temperature, and a ferromagnet heated enough behaves as a paramagnet.

What is magnetising current?

The extra current that would have to flow in the windings of an empty solenoid to produce the same extra field μ₀M that the core produces. It equals M/n.