MHT-CET Physics · Magnetic Materials

Magnetisation, Susceptibility and Permeability

Inside a material the field is B = μ₀(H + M), where H is set by the currents and M, the magnetisation, is the material's own dipole moment per unit volume; M = χH defines the susceptibility, so B = μ₀(1 + χ)H and the relative permeability is μᵣ = 1 + χ.

All Magnetic Materials notes MHT-CET Physics strategy

Why this matters

13 PYQs, none HARD. Nine connect B, H, M, χ and μ — the relative permeability from a susceptibility, the absolute permeability, the percentage rise in a toroid's field when it is filled. Four compute a magnetisation from a moment and a volume or from a solenoid's current. Two cards.

Concept 1 of 2: B, H, M, χ and μ

H is what the free currents supply: nI inside a solenoid. The material responds with a magnetisation M = χH, and the total field is B = μ₀(H + M) = μ₀(1 + χ)H. So μᵣ = 1 + χ and μ = μ₀(1 + χ). A susceptibility of 5499 means μᵣ = 5500. Filling a toroid with a material raises B by the fraction χ. Paramagnets have a small positive χ, diamagnets a small negative one, ferromagnets a very large one. μ = B/H can also be read off a flux measurement: B = Φ/A.

Definition

$B = \mu_0(H + M)$ , $M = \chi H$ , $\dfrac{B}{H} = \mu_0(1 + \chi)$ .
$\mu_r = 1 + \chi$ ; $\mu = \mu_0(1 + \chi)$ (χ = 599 ⇒ $2.4\pi\times10^{-4}$ ).
Filled toroid: B rises by $\chi \times 100\%$ .
From flux: $\mu = \dfrac{\Phi/A}{H}$ ( $2.4\times10^{-5}$ Wb on 0.4 cm² at 500 A/m ⇒ $1.2\times10^{-3}$ ).

Field in a material

B = \mu_0(H + M) = \mu_0(1 + \chi)H, \qquad \mu_r = 1 + \chi

Worked example

A material has susceptibility 2 × 10⁻³ and is placed in H = 1000 A/m. Magnetisation and field B?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q43Easy

Example 1 · Magnetic Materials · Magnetisation, Susceptibility, Permeability, and B-H-M Relations

The relation between total magnetic field (B), magnetic intensity (H), permeability of free space

\mu_0

and susceptibility

(\chi)

Taking μᵣ = χ

μᵣ = 1 + χ. For iron with χ = 5499, μᵣ = 5500; the option 5499 is the trap.

Quoting (1 + χ) as the percentage rise

A toroid filled with a material of susceptibility χ has B multiplied by (1 + χ), so the RISE is χ × 100%.

Drill 8 more on b, h, m, χ and μ

Concept 2 of 2: Computing a Magnetisation

Magnetisation is dipole moment per unit volume: M = m/V. For a bar of known mass, get the volume from the density first. In a solenoid core, H = nI and M = (μᵣ − 1)nI, very nearly μᵣnI when μᵣ is large.

Definition

$M = \dfrac{m}{V}$ (6 A m² in 4 cm × 2 cm² ⇒ $7.5\times10^5$ A/m).
From mass: $V = \dfrac{\text{mass}}{\rho}$ (2.4 A m², 66 g, 7700 kg/m³ ⇒ $2.8\times10^5$ A/m).
Solenoid core: $M = (\mu_r - 1)nI$ (400/m, 0.5 A, μᵣ = 400 ⇒ $8\times10^4$ A/m).

Magnetisation

M = \frac{m}{V} = (\mu_r - 1)\,nI

Worked example

A solenoid of 1000 turns/m carries 2 A around an iron core with μᵣ = 1001. Magnetisation?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q23Moderate

Example 2 · Magnetic Materials · Magnetisation, Susceptibility, Permeability, and B-H-M Relations

A solenoid having 400 turns per metre has a core of a material with relative permeability 400. When a current of 0.5 A is passed through it, the magnetization of the core material in A m

^{-1}

is nearly

Leaving cm² and cm in the volume

4 cm × 2 cm² is 8 × 10⁻⁶ m³. Mixed units put the answer off by powers of ten, and the options are spaced that way.

Using μᵣ where μᵣ − 1 belongs

M = (μᵣ − 1)nI. With μᵣ = 5000 the difference is negligible, but with a weakly magnetic core, using μᵣ counts the vacuum part of the field as magnetisation.

Drill 3 more on computing a magnetisation

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

B, H, M, χ and μ
Field in a material
$B = \mu_0(H + M) = \mu_0(1 + \chi)H, \qquad \mu_r = 1 + \chi$
Computing a Magnetisation
Magnetisation
$M = \frac{m}{V} = (\mu_r - 1)\,nI$

Watch out for (4)

Taking μᵣ = χ
→ B, H, M, χ and μ
Quoting (1 + χ) as the percentage rise
→ B, H, M, χ and μ
Leaving cm² and cm in the volume
→ Computing a Magnetisation
Using μᵣ where μᵣ − 1 belongs
→ Computing a Magnetisation

Test yourself on Magnetic Materials

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Magnetic Materials chapter test Drill all 13 questions