The bar magnet as an equivalent solenoid
Cut a bar magnet into two and you get two smaller magnets, each with its own N and S. Keep cutting; you never get a single pole. Why? Because the magnetism comes from tiny current loops: electrons going round and spinning inside atoms.
In a magnet these tiny loops point the same way. Inside, their currents cancel between neighbours, but on the outer surface they add up to a current going round the bar, just like the turns of a solenoid. So a bar magnet and a solenoid make the same field pattern, and a solenoid with current has a north end and a south end.
Magnetic moment of a bar magnet: m, pointing from S to N inside the magnet (unit A m²).
Field of a bar magnet on its axis and equator
For a short magnet (distance r much bigger than its size) the field looks like that of a small current loop:
- On the axis (the line through N and S): B = (μ₀/4π) · 2m / r³, pointing along m.
- On the equator (the perpendicular line through the centre): B = (μ₀/4π) · m / r³, pointing opposite to m.
So at the same distance the axial field is twice the equatorial field. Both fall as 1/r³, much faster than a single charge's 1/r². These have the same form as the fields of an electric dipole, with p replaced by m and 1/4πε₀ by μ₀/4π.
Torque on a magnet in a uniform field
In a uniform field B, the N end is pushed along B and the S end against B. The two forces are equal and opposite, so there is no net force, but they form a couple:
τ = m × B, size τ = mB sinθ
θ is the angle between m and B. The torque tries to line m up with B, just as it does for a current loop (τ = NIAB sinθ).
Energy
Potential energy U = −mB cosθ. Lowest (−mB) when m is along B: the stable position. Highest (+mB) when m is opposite: unstable. The work needed to turn a magnet from θ₁ to θ₂ is mB(cosθ₁ − cosθ₂). A compass needle points north because of this torque.
Magnetic field lines
Field lines are a picture of B. Rules:
- They form closed loops: outside the magnet from N to S, inside from S to N. (Electric lines start and end on charges; magnetic lines have no start or end.)
- The tangent at any point gives the direction of B.
- Where lines are closer, the field is stronger.
- Two lines never cross, since B cannot have two directions at one point.
- There are no separate magnetic poles (no monopoles). So the total field flux through any closed surface is zero: Gauss's law for magnetism, ∮B·dA = 0.
Magnetisation, magnetic intensity and susceptibility
Magnetisation M = net magnetic moment per unit volume, M = mnet / V (unit A m⁻¹). It tells how magnetised a piece of material has become.
Put a material inside a long solenoid. The coil alone gives B₀ = μ₀nI. Define the magnetic intensity H = nI (unit A m⁻¹). The material adds its own part, so
B = μ₀ (H + M)
For most materials M grows in proportion to H: M = χH, where χ (chi) is the magnetic susceptibility (no unit). Then
B = μ₀(1 + χ)H = μ₀μᵣH = μH, with relative permeability μᵣ = 1 + χ.
Diamagnetic, paramagnetic and ferromagnetic materials
Diamagnetic
Atoms have no permanent magnetic moment. An outside field makes small opposing moments, so the material is weakly pushed away from strong-field regions. χ is small and negative (about −10⁻⁵), μᵣ slightly less than 1. Examples: bismuth, copper, lead, water, nitrogen. A superconductor pushes the field out completely (χ = −1, the Meissner effect).
Paramagnetic
Atoms have small permanent moments pointing randomly. A field lines some of them up, so the material is weakly pulled in. χ is small and positive (10⁻⁵ to 10⁻³), μᵣ slightly more than 1. Examples: aluminium, sodium, calcium, oxygen gas.
Ferromagnetic
Atomic moments line up together in small regions called domains (about 1 μm to 1 mm). In a field, domains that point along B grow, so the material is strongly pulled in. χ is very large (10³ or more). Examples: iron, cobalt, nickel, gadolinium. Soft ferromagnets (soft iron) lose magnetism when the field is removed; hard ones (alnico) keep it and make permanent magnets.
| Property | Dia | Para | Ferro |
|---|---|---|---|
| χ | small, − | small, + | very large, + |
| μᵣ | < 1 | > 1 (slightly) | ≫ 1 |
| In a non-uniform field | moves to weaker field | moves to stronger field | strongly to stronger field |
Effect of temperature
Heat makes atoms jiggle. This spoils the lining up of the atomic magnets.
- Diamagnetism hardly changes with temperature.
- Paramagnetism falls as temperature rises: Curie law χ = C / T (T in kelvin, C = Curie constant). Double the kelvin temperature, the susceptibility halves.
- Ferromagnetism disappears above the Curie temperature Tc: the domains break up and the material becomes paramagnetic. Above Tc, χ = C / (T − Tc). Iron: Tc ≈ 1043 K (770 °C); nickel ≈ 631 K; cobalt ≈ 1394 K.
Exam corner
Common board questions: compare dia, para and ferromagnetic materials (3 marks); properties of field lines (2 marks); torque on a magnet and its potential energy (2–3 marks); bar magnet as an equivalent solenoid (2 marks); numericals on τ = mB sinθ, M = m/V, χ = M/H and Curie law.
Key formulas and definitions
- Axial field (short magnet): B = (μ₀/4π) · 2m / r³
- Equatorial field: B = (μ₀/4π) · m / r³ (opposite to m)
- Torque: τ = m × B, τ = mB sinθ
- Potential energy: U = −mB cosθ; work to turn θ₁ → θ₂ = mB(cosθ₁ − cosθ₂)
- Magnetisation: M = m_net / V (A m⁻¹)
- B = μ₀(H + M); H = nI in a solenoid
- M = χH; μᵣ = 1 + χ; μ = μ₀μᵣ
- Curie law (paramagnets): χ = C / T; ferro above T_c: χ = C / (T − T_c)
Worked examples
1. A magnet of moment 2 A m² is at 30° to a 0.3 T field. Find the torque.
τ = mB sinθ = 2 × 0.3 × 0.5 = 0.3 N m.
2. A magnet at 30° to a 0.2 T field feels 0.06 N m. Find its magnetic moment.
m = τ / (B sinθ) = 0.06 / (0.2 × 0.5) = 0.6 A m².
3. A short magnet has m = 0.5 A m². Find B on its axis and on its equator at 10 cm.
Axis: B = 10⁻⁷ × 2 × 0.5 / (0.1)³ = 10⁻⁴ T. Equator: half of this, 5 × 10⁻⁵ T, opposite to m.
4. How much work turns a magnet (m = 1.5 A m²) from along a 0.2 T field to opposite it?
W = mB(cos 0° − cos 180°) = 1.5 × 0.2 × 2 = 0.6 J.
5. An iron rod of volume 5 × 10⁻⁶ m³ has a net magnetic moment of 4 A m². Find its magnetisation.
M = m / V = 4 / (5 × 10⁻⁶) = 8 × 10⁵ A m⁻¹.
6. In H = 2000 A m⁻¹ a material gets M = 3 A m⁻¹. Find χ and μᵣ, and name the type.
χ = M / H = 1.5 × 10⁻³; μᵣ = 1 + χ = 1.0015. Small and positive, so paramagnetic.
7. A paramagnetic salt has χ = 2.4 × 10⁻⁴ at 300 K. Find χ at 200 K.
Curie law χ ∝ 1/T: χ₂ = 2.4 × 10⁻⁴ × 300 / 200 = 3.6 × 10⁻⁴.
8. A solenoid (1000 turns/m, 2 A) has an iron core with μᵣ = 400. Find H, B and M.
H = nI = 2000 A m⁻¹. B = μ₀μᵣH = 4π × 10⁻⁷ × 400 × 2000 ≈ 1.0 T. M = (μᵣ − 1)H = 399 × 2000 ≈ 8 × 10⁵ A m⁻¹.
Common mistakes
- Saying field lines start at N and end at S. They continue inside from S to N and form closed loops.
- Using the equatorial formula on the axis. Axis has 2m, equator has m, at the same r.
- Mixing up signs: diamagnetic χ is negative, paramagnetic χ is positive, both small.
- Using °C in Curie law. Always change to kelvin: T = t + 273.