Magnetisation, magnetic intensity and susceptibility
Magnetisation M is the magnetic moment per unit volume. With magnetic intensity H: B = μ₀(H + M), M = χH and μr = 1 + χ.
Magnetisation M
Each atom of a material can act as a tiny magnetic dipole. Usually these point in all directions, so their moments cancel. In a magnetic field, some of them line up and the sample gets a net magnetic moment. Magnetisation M is the net magnetic moment per unit volume: M = m_net/V. It is a vector, and its SI unit is A m⁻¹ (A m² divided by m³). M tells us how strongly the material itself is magnetised.
Magnetic intensity H
Put a material inside a long solenoid of n turns per metre carrying current I. Without the material, the field would be B₀ = μ₀nI. We call H = nI the magnetic intensity. It depends only on the current, not on the material. The material adds its own field, μ₀M. So the total field inside is B = μ₀(H + M), which gives H = B/μ₀ − M. The SI unit of H is also A m⁻¹. Remember: H is what you apply; M is how the material responds.
From H to B, step by step
- 1For most materials, M is proportional to H: M = χH. The constant χ (chi) is the magnetic susceptibility, a pure number.
- 2Put this into B = μ₀(H + M): B = μ₀(H + χH) = μ₀(1 + χ)H.
- 3★Define the relative permeability: μr = 1 + χ.
- 4Define the permeability of the material: μ = μ₀μr.
- 5★So, B = μ₀μrH = μH.
★ key step
Quantities and their units
| Quantity | Symbol | SI unit |
|---|---|---|
| Magnetic field | B | tesla (T) |
| Magnetic intensity | H | A m⁻¹ |
| Magnetisation | M | A m⁻¹ |
| Susceptibility | χ | None (a pure number) |
| Relative permeability | μr | None (a pure number) |
| Permeability | μ | T m A⁻¹, the same as μ₀ |
Worked example: an iron-core solenoid
A long solenoid has 1000 turns per metre and carries 1.0 A. Its core has relative permeability 800. Find (a) H, (b) B, (c) χ and (d) M. (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
- (a) H = nI = 1000 m⁻¹ × 1.0 A = 1.0 × 10³ A m⁻¹.
- (b) B = μ₀μrH = 4π × 10⁻⁷ × 800 × 1000 T.
- B = 0.32π T ≈ 1.0 T.
- (c) χ = μr − 1 = 800 − 1 = 799.
- (d) M = χH = 799 × 1000 A m⁻¹ ≈ 8.0 × 10⁵ A m⁻¹.
(a) 1.0 × 10³ A m⁻¹; (b) about 1.0 T; (c) 799; (d) about 8.0 × 10⁵ A m⁻¹
Worked example: what kind of material?
A small rod is placed where the magnetic intensity is 2000 A m⁻¹. Its magnetisation is found to be 0.40 A m⁻¹, in the same direction as H. Find χ and μr, and say what kind of material it is.
- Formula: χ = M/H.
- χ = 0.40/2000 = 2.0 × 10⁻⁴.
- μr = 1 + χ = 1.0002.
- χ is small and positive, so the material is paramagnetic.
χ = 2.0 × 10⁻⁴; μr = 1.0002; paramagnetic
The cores of transformers on electricity poles and of electromagnets are made of soft iron alloys with a very large μr. A small current then gives a strong field inside the core, so less copper and energy are needed.
1 marks: “Write the SI units of magnetic intensity and magnetisation.” Both are measured in A m⁻¹.
2 marks: “Define magnetic susceptibility. How is it related to relative permeability?” χ = M/H, the magnetisation produced per unit magnetic intensity; it has no unit. μr = 1 + χ.
3 marks: “A solenoid with 1000 turns per metre carries 2.0 A. Its core has μr = 500. Find H, B and M.” H = nI = 2000 A m⁻¹; B = μ₀μrH = 4π × 10⁻⁷ × 500 × 2000 T = 0.4π T ≈ 1.26 T; M = (μr − 1)H = 499 × 2000 ≈ 1.0 × 10⁶ A m⁻¹.
B = μ₀(H + M): H is what you apply, M is what the material adds. μr = 1 + χ.