PhysicsISC Class 12 · Magnetism and Matter

Magnetisation, Magnetic Intensity, Susceptibility and Relative Permeability

Updated 15 October 2026 · Fact-checked

Magnetisation M is the magnetic moment per unit volume of a material. Magnetic intensity H is defined by $H = \frac{B}{\mu_0} - M$, so $B = \mu_0(H + M)$. Susceptibility is $\chi_m = \frac{M}{H}$, and then $B = \mu_0\mu_rH$ with relative permeability $\mu_r = 1 + \chi_m$. For a solenoid, H = nI whatever the core.

H = nI depends only on the current; the core adds magnetisation M, so B grows by the factor μr = 1 + χ.

Illustration drawn for QuestionBuddy.

Understand Magnetisation, Susceptibility and Permeability

Magnetic field B and flux. The magnetic field $B$ is defined through the force on a moving charge, $F = q (v \times B)$ , not through a "unit pole" (the ISC syllabus insists on this). Unit: tesla (T); $1 T = 1 0^{4}$ gauss. Field lines of B always form closed loops (there are no magnetic monopoles), unlike electric field lines which start and end on charges.
Magnetic flux through an area: $ϕ = B \cdot A = B A cos θ$ , θ being the angle between $B$ and the normal to the area; unit weber (Wb), $1 T = 1 Wb m^{- 2}$ . For a non-uniform field, $ϕ = \int B \cdot d A$ . B is sometimes called flux density, but B is a vector and $\frac{ϕ}{A}$ a scalar, so $B = \frac{ϕ}{A}$ is not a proper definition.

Magnetisation M. When a material is placed in a field, its atomic moments (permanent or induced) produce a net magnetic moment. The magnetisation is the net magnetic moment per unit volume:
$M = \frac{m _{n e t}}{V}$ , unit A m⁻¹.

Magnetic intensity H. Inside a material, B has two contributions: one from the free (conduction) currents in coils and one from the material's magnetisation. Define
$H = \frac{B}{μ _{0}} - M \Rightarrow B = μ_{0} (H + M) .$
H is set by the free currents. Inside a long solenoid with n turns per metre carrying I, $H = n I$ (A m⁻¹) whether or not there is a core; the core changes M and so B.

Susceptibility. For most (linear) materials M is proportional to H:
$M = χ_{m} H,$
with $χ_{m}$ the dimensionless magnetic susceptibility: negative and tiny for diamagnets, positive and small for paramagnets, large for ferromagnets.

Permeability. Substituting, $B = μ_{0} (1 + χ_{m}) H = μ_{0} μ_{r} H = μ H$ , where
$μ_{r} = 1 + χ_{m} = \frac{μ}{μ _{0}}$
is the relative permeability and $μ = μ_{0} μ_{r}$ the permeability of the material. $μ_{r}$ tells how many times the material multiplies the field produced by a given H: about 1 for dia and para (slightly less or more), hundreds to thousands for ferromagnets.

Inside a solenoid with a core.

$H = n I$ .
$M = χ_{m} H$ .
$B = μ_{0} (H + M) = μ_{0} μ_{r} n I$ .

For iron, almost all of B comes from the magnetisation ( $\frac{μ _{0} M}{B} = \frac{χ _{m}}{1 + χ _{m}}$ , close to 1).

Comparison with electrostatics. $μ_{r}$ plays a role like the dielectric constant K: K reduces the field in a dielectric, but $μ_{r} > 1$ increases B in a para- or ferromagnet. Note too that the 2025 ISC 1-mark item classified substances by whether $μ_{r}$ is slightly below or slightly above 1.

A revolving electron as a dipole. An electron in a circular orbit of radius r with speed v is a current loop with $I = \frac{e v}{2 π r}$ , so its orbital moment is $m = I A = \frac{e v r}{2}$ . Such atomic moments, added up per unit volume, are what M measures.

Magnetisation, Susceptibility and Permeability: formulas and key facts

Magnetic flux: $ϕ = B \cdot A = B A cos θ$; Weber; 1 T = 1 Wb m⁻² = 10⁴ G.
Magnetisation: $M = \frac{m _{n e t}}{V}$; A m⁻¹.
Magnetic intensity: $H = \frac{B}{μ _{0}} - M$; Solenoid: H = nI.
Field in a material: $B = μ_{0} (H + M) = μ_{0} μ_{r} H$; Linear materials.
Susceptibility: $χ_{m} = \frac{M}{H}$; Dimensionless.
Relative permeability: $μ_{r} = 1 + χ_{m} = \frac{μ}{μ _{0}}$; $μ_{0} = 4 π \times 1 0^{- 7}$ T m A⁻¹.
Orbital moment of an electron: $m = \frac{e v r}{2}$; Current $\frac{e v}{2 π r}$ round area $π r^{2}$ .

How to solve Magnetisation, Susceptibility and Permeability questions

Most problems are a chain: free current → H → M → B.

1Find H from the free current: H = nI for a solenoid or toroid (n = turns per metre).
2Find χ (or μr) from the data, using $μ_{r} = 1 + χ_{m}$ .
3Find $M = χ_{m} H$ .
4Find $B = μ_{0} (H + M) = μ_{0} μ_{r} H$ .
5If B and H are given, work backwards: $μ_{r} = \frac{B}{μ _{0} H}$ , then χ and M.
6For flux, use the angle with the normal; convert cm² to m² and T to gauss if asked.

Magnetisation, Susceptibility and Permeability 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 cores are used in coils

Suppose a relay coil has 2000 turns per metre and carries 50 mA, wound on a soft iron core of relative permeability 1500.

Find H and B in the core, and compare B with the value for an air core.

Try it first, then show the solutionSolution

$H = n I = 2000 \times 0.05 = 100$ A m⁻¹.
$B = μ_{0} μ_{r} H = 4 π \times 1 0^{- 7} \times 1500 \times 100 \approx 0.19$ T.
Air core: $μ_{0} H \approx 1.26 \times 1 0^{- 4}$ T, 1500 times smaller.

Answer: H = 100 A m⁻¹; B ≈ 0.19 T, 1500 times the air-core field.

Memory hook: Same H, μr times the B.

Units of B and H

Imagine a hospital MRI scanner whose magnet produces a field of 1.5 T in the air of its bore.

Express the field in gauss and find the magnetic intensity H there.

Try it first, then show the solutionSolution

$1.5 T = 1.5 \times 1 0^{4} = 15000$ G.
In air M ≈ 0, so $H = \frac{B}{μ _{0}} = \frac{1.5}{4 π \times 1 0 ^{- 7}} \approx 1.19 \times 1 0^{6}$ A m⁻¹.

Answer: 15 000 G; about $1.2 \times 1 0^{6}$ A m⁻¹.

Memory hook: Tesla × 10⁴ = gauss.

Tricks to remember Magnetisation, Susceptibility and Permeability

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

B is H plus M, times μ₀
$B = μ_{0} (H + M)$ : "Bee equals mu-nought of Haitch plus Em".
Use it when: Linking the three fields.
One plus chi
$μ_{r} = 1 + χ$ : the "1" is vacuum, χ is the material's extra.
Use it when: Converting χ and μr.
H from the coil, M from the core
Free current sets H; the material decides M.
Use it when: Solenoid-with-core problems.
Ten thousand gauss
1 T = 10⁴ G.
Use it when: Unit conversions.

Magnetisation, Susceptibility and Permeability: solved examples

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

Example 1: Unit of magnetic intensity · 1 mark · MCQBoard level

The SI unit of magnetic intensity H is:
(a) T
(b) Wb
(c) A m⁻¹
(d) A m²

Try it first, then show the solutionSolution

H = nI: turns per metre × amperes.

Answer: (c).

Example 2: Field in a cored solenoid · 2 marks · Short answerBoard level

A solenoid with 800 turns per metre carries 1.5 A. Its core has relative permeability 250. Find H and B inside.

Try it first, then show the solutionSolution

$H = n I = 1200$ A m⁻¹. (1)
$B = μ_{0} μ_{r} H = 4 π \times 1 0^{- 7} \times 250 \times 1200 \approx 0.377$ T. (1)

Answer: H = 1200 A m⁻¹; B ≈ 0.38 T.

Example 3: Magnetisation and its share of B · 3 marks · Short answerHarder board level

For the solenoid above, find the susceptibility, the magnetisation of the core and the fraction of B that is due to the magnetisation.

Try it first, then show the solutionSolution

$χ_{m} = μ_{r} - 1 = 249$ . (1)
$M = χ_{m} H = 249 \times 1200 \approx 2.99 \times 1 0^{5}$ A m⁻¹. (1)
Fraction $\frac{μ _{0} M}{B} = \frac{M}{H + M} = \frac{249}{250} = 99.6%$ . (1)

Answer: χ = 249; M ≈ $3.0 \times 1 0^{5}$ A m⁻¹; 99.6% of B comes from the core.

Example 4: Magnetic flux · 2 marks · Short answerHarder board level

A uniform field of 0.25 T passes through a coil of area 30 cm² whose plane makes 30° with the field. Find the flux through it, and express the field in gauss.

Try it first, then show the solutionSolution

Angle with the normal = 90° − 30° = 60°: $ϕ = 0.25 \times 3 \times 1 0^{- 3} \times cos 60° = 3.75 \times 1 0^{- 4}$ Wb. (1)
$0.25 T = 2500$ G. (1)

Answer: $3.75 \times 1 0^{- 4}$ Wb; 2500 G.

Example 5: Working back from B and H · 3 marks · Short answerHOTS / competency

In a material placed in a field of intensity H = 1000 A m⁻¹, the magnetic field is found to be 0.628 T. Find μr, χ and M, and classify the material.

Try it first, then show the solutionSolution

$μ_{r} = \frac{B}{μ _{0} H} = \frac{0.628}{4 π \times 1 0 ^{- 7} \times 1000} \approx 500$ . (1)
$χ_{m} = 499$ ; $M = χ_{m} H \approx 4.99 \times 1 0^{5}$ A m⁻¹. (1)
μr ≫ 1: ferromagnetic. (1)

Answer: μr ≈ 500, χ ≈ 499, M ≈ $5.0 \times 1 0^{5}$ A m⁻¹; ferromagnetic.

Example 6: Testing a toroid core · 4 marks · Case-basedHOTS / competency

Suppose an engineer winds 500 turns on a ring-shaped core of mean circumference 0.5 m, passes 0.4 A and measures B = 0.6 T in the core.
(i) Find H. (1 mark)
(ii) Find the relative permeability. (1 mark)
(iii) Find the susceptibility and magnetisation. (1 mark)
(iv) What would B be if the core were removed (air inside)? (1 mark)

Try it first, then show the solutionSolution

(i) n = $\frac{500}{0.5} = 1000$ m⁻¹; H = nI = 400 A m⁻¹.
(ii) $μ_{r} = \frac{0.6}{4 π \times 1 0 ^{- 7} \times 400} \approx 1194$ .
(iii) $χ_{m} \approx 1193$ ; $M \approx 1193 \times 400 \approx 4.77 \times 1 0^{5}$ A m⁻¹.
(iv) $B = μ_{0} H \approx 5.0 \times 1 0^{- 4}$ T.

Answer: (i) 400 A m⁻¹; (ii) ≈ 1190; (iii) χ ≈ 1190, M ≈ $4.8 \times 1 0^{5}$ A m⁻¹; (iv) ≈ $5.0 \times 1 0^{- 4}$ T.

Where students lose marks in Magnetisation, Susceptibility and Permeability

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

Mistake: Using the angle with the plane
Flux needs the angle between B and the normal to the area.
Wrong working
$ϕ = B A cos 30°$ when the plane makes 30° with B.
Right working
$ϕ = B A cos 60°$ .
Do this: Convert "angle with the plane" to "angle with the normal" first.
Usually costs: 1 mark.
Mistake: Mixing up μr and χ
Substituting μr where χ is needed changes M slightly for iron but completely for para- and diamagnets.
Do this: Write $χ_{m} = μ_{r} - 1$ explicitly.
Usually costs: 1 mark.
Mistake: Classifying by μr wrongly
The ISC 2025 examiners noted confusion over μr slightly less or more than 1.
Do this: μr < 1 dia; slightly > 1 para; ≫ 1 ferro.
Usually costs: 1 mark.
Mistake: Defining B by a unit pole
The syllabus asks that B be defined from the force on a moving charge.
Do this: Use $F = q (v \times B)$ .
Usually costs: 1 mark in definition questions.

Exam tips

Write units with every answer: T, A m⁻¹, Wb.
Use n in turns per metre, not total turns, for H = nI.
Keep χ and μr to sensible significant figures; for iron they are large, so the difference between them hardly matters.
Mention that B lines form closed loops when contrasting B with E.

Practice questions on Magnetisation, Susceptibility and Permeability

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
Relative permeability and susceptibility are related by:
(a) $μ_{r} = χ_{m}$
(b) $μ_{r} = 1 + χ_{m}$
(c) $μ_{r} = 1 - χ_{m}$
(d) $μ_{r} = \frac{1}{χ _{m}}$
Hint
$B = μ_{0} (H + M)$ .
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
Express a field of 0.03 T in gauss.
Hint
1 T = 10⁴ G.
Answer
300 G.
Q3 · 2 marks · Short answerHarder board level
A material has μr = 1.00005. Find χ, classify it, and find M when H = $4 \times 1 0^{4}$ A m⁻¹.
Hint
χ = μr − 1.
Answer
χ = $5 \times 1 0^{- 5}$ (paramagnetic); M = 2 A m⁻¹.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In vacuum, $B = μ_{0} H$ .
Reason (R): H is determined by the free currents only.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
What makes M zero?
Answer
(b): A follows because M = 0 in vacuum; R is true but is not the reason.
Q5 · 2 marks · Short answerHarder board level
A solenoid of 2000 turns and length 50 cm carries 0.25 A with an air core. Find H and B inside.
Hint
n = 4000 m⁻¹.
Answer
H = 1000 A m⁻¹; B = $μ_{0} H \approx 1.26 \times 1 0^{- 3}$ T.
Q6 · 3 marks · Short answerHOTS / competency
A core is magnetised to M = $8 \times 1 0^{5}$ A m⁻¹ by H = 800 A m⁻¹. Find χ, μr and B.
Hint
$B = μ_{0} (H + M)$ .
Answer
χ = 1000; μr = 1001; $B = 4 π \times 1 0^{- 7} \times 800800 \approx 1.01$ T.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a flat coil of 200 turns and area 40 cm² is placed in a uniform field of 0.5 T.
(i) Find the flux linkage when the field is along the coil's normal. (1 mark)
(ii) Find it when the normal makes 60° with the field. (1 mark)
(iii) Find it when the plane of the coil is parallel to the field. (1 mark)
(iv) Express the field in gauss and state why B, not $\frac{ϕ}{A}$ , defines the field. (1 mark)
Hint
Flux linkage = Nφ.
Answer
(i) $200 \times 0.5 \times 4 \times 1 0^{- 3} = 0.4$ Wb-turns. (ii) 0.2 Wb-turns. (iii) 0. (iv) 5000 G; B is a vector while flux per area is a scalar, so B is defined from $F = q v \times B$ .

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 = $\frac{m _{n e t}}{V}$ (A m⁻¹).
2.Relation between B, H and M?Show answer
$B = μ_{0} (H + M)$ .
3.H inside a solenoid of n turns per metre?Show answer
H = nI.
4.χ = 999: μr?Show answer
1000.
5.Unit of magnetic flux?Show answer
Weber (Wb).

Magnetisation, Susceptibility and Permeability: frequently asked questions

What is the difference between B and H?

H depends only on the free currents (for a solenoid, H = nI). B is the actual field, which includes the contribution of the material's magnetisation: B = \mu_0(H + M).

Why is relative permeability not like the dielectric constant?

A dielectric reduces the electric field inside it by K, but a paramagnetic or ferromagnetic core increases B by μr. Only diamagnets reduce B slightly.

Why can magnetic field lines not end anywhere?

There are no isolated magnetic poles (monopoles), so the lines of B always form closed loops, passing through a magnet from its S pole to its N pole inside.

What does the ISC 2027 syllabus need from this topic?

B defined from the force on a moving charge, magnetic flux and its units, the relation H = \frac{B}{\mu_0} - M, susceptibility and relative permeability, and their values for dia-, para- and ferromagnetic materials.

How was this tested recently?

In 2025 a 1-mark item gave relative permeabilities slightly below and above 1 and asked candidates to identify the materials.