The bar magnet and its field
A compass needle and a fridge magnet both come from the same property: a magnetic dipole with a north and a south pole. Link to earlier classes : magnets and field lines from Class 7 and Class 10, and the electric dipole of Chapter 1.
A bar magnet behaves like a magnetic dipole with a magnetic moment m pointing from its south pole to its north pole. Cutting it in two never gives isolated poles: each piece is a smaller magnet. A current loop or solenoid has the same kind of field.
Rule 1: Magnetic field lines — They form closed loops: outside the magnet they run from north to south, inside from south to north. They never cross, and they are crowded where the field is strong.
- Easy : Iron filings sprinkled round a magnet gather along these lines.
- Medium : Since lines are closed loops, a bar magnet has no single source or sink of field, unlike an electric charge.
- Exam : The field lines of a solenoid and of a bar magnet look the same.
Rule 2: Field of a short bar magnet — On the axis, B = (μ₀ ÷ 4π) × 2m ÷ r³. On the equatorial line, B = (μ₀ ÷ 4π) × m ÷ r³. For a coil, m = N I A.
- Easy : m = 3.0 A m² and r = 0.20 m on the axis ⇒ B = 10⁻⁷ × 2 × 3 ÷ 0.008 = 7.5 × 10⁻⁵ T.
- Medium : At the same distance on the equatorial line, B = 3.75 × 10⁻⁵ T, which is half of the axial value.
- Exam : A coil of 100 turns, 0.5 A and area 2 × 10⁻³ m² has m = 0.1 A m².
In the figure : Field lines of a bar magnet: closed loops, leaving from the north pole and entering at the south pole.
Magnetic poles always occur in pairs; axial field = 2 × equatorial field at the same distance.
Saying that field lines start at the north pole and end at the south pole. Inside the magnet they continue, so the loops are closed.
Memory clue : "Outside N to S, inside S to N: always closed."
Spot the difference
| Electric dipole | Bar magnet |
|---|---|
| axial E = 2kp ÷ r³ | axial B = (μ₀ ÷ 4π) 2m ÷ r³ |
| lines start and end on charges | lines are closed loops |
Practice questions
- 1What happens when a bar magnet is cut into two pieces?
- 2Write the SI unit of the magnetic moment.
- 3In which direction do the field lines run inside a bar magnet?
- 4A bar magnet has m = 3.0 A m². Find its axial field at 0.20 m.
- 5Find its equatorial field at the same distance.
- 6A magnet with m = 2.0 A m² is considered. Find the axial field at 0.30 m.
- 7Find the magnetic moment of a coil with N = 100, I = 0.5 A and A = 2 × 10⁻³ m².
- 8What is the ratio of the axial to the equatorial field at the same distance?
Show answers (try them yourself first)
- 1Two smaller magnets, each with a north and a south pole.
- 2A m² (or J T⁻¹).
- 3From the south pole to the north pole.
- 4B = 7.5 × 10⁻⁵ T.
- 5B = 3.75 × 10⁻⁵ T.
- 6B = 10⁻⁷ × 2 × 2 ÷ 0.027 = 1.5 × 10⁻⁵ T.
- 7m = NIA = 0.10 A m².
- 82 : 1.
A magnet in a uniform field and Gauss's law for magnetism
A compass needle swings and settles along the north–south direction. Link to earlier classes : torque on a dipole from Chapters 1 and 4.
A magnetic dipole in a uniform field experiences a torque that turns it along the field. Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero, because there are no magnetic monopoles.
Rule 1: Torque and energy — τ = m B sin θ and U = − m B cos θ (θ = angle between m and B). The stable position is θ = 0°, where U is least.
- Easy : m = 0.5 A m², B = 0.4 T, θ = 30° ⇒ τ = 0.5 × 0.4 × 0.5 = 0.10 N m.
- Medium : At θ = 60°, U = −0.5 × 0.4 × 0.5 = −0.10 J.
- Exam : The work needed to turn the magnet from 0° to 180° is 2mB = 0.40 J.
Rule 2: Small oscillations and Gauss's law — A magnet displaced a little from the field direction oscillates with T = 2π √(I ÷ (m B)) (I = moment of inertia). For any closed surface, ∮ B · dS = 0.
- Easy : I = 2 × 10⁻⁵ kg m², m = 0.5 A m², B = 0.4 T ⇒ T = 2π √(2 × 10⁻⁵ ÷ 0.2) = 0.063 s.
- Medium : A magnet cut across its length into two equal parts has half the moment, m ÷ 2, in each part.
- Exam : As many field lines leave a closed surface as enter it; this is why B lines are closed loops, unlike E lines.
τ = mB sin θ; U = −mB cos θ; ∮ B·dS = 0 (no monopole).
Applying Gauss's law for electricity (q ÷ ε₀) to magnetism. The magnetic version has zero on the right side.
Memory clue : "Magnets never leave a net flux: what goes in comes out."
Spot the difference
| Law | Right side |
|---|---|
| Gauss (electric) | q ÷ ε₀ |
| Gauss (magnetic) | 0 |
Practice questions
- 1State Gauss's law for magnetism.
- 2What is the stable orientation of a magnet in a field?
- 3At what angle is the torque greatest?
- 4Find the torque on m = 0.5 A m² at 30° to a 0.4 T field.
- 5Find the potential energy of the same dipole at 60°.
- 6Find the work to turn it from 0° to 180°.
- 7Find the period of small oscillations if I = 2 × 10⁻⁵ kg m², m = 0.5 A m² and B = 0.4 T.
- 8A magnet of moment m is cut into two equal pieces across its length. What is the moment of each piece?
Show answers (try them yourself first)
- 1The net magnetic flux through any closed surface is zero.
- 2Along the field (θ = 0°).
- 390°.
- 4τ = 0.10 N m.
- 5U = −0.10 J.
- 6W = 2mB = 0.40 J.
- 7T = 2π√(I ÷ mB) ≈ 0.063 s.
- 8m ÷ 2 (the pole strength stays the same but the length halves).
Magnetisation and magnetic intensity
An iron nail placed in a coil becomes a much stronger magnet, which is the idea behind every electromagnet. Link to earlier classes : the field of a solenoid from Chapter 4.
When a material is placed in a magnetic field, its atomic dipoles line up and it acquires magnetisation M (magnetic moment per unit volume). The magnetic intensity H is the part of the field due to the external current alone. The ratio M ÷ H is the magnetic susceptibility χ.
Rule 1: The relations — B = μ₀ (H + M), M = χ H, μ_r = 1 + χ and B = μ₀ μ_r H = μ H (μ_r = relative permeability). For a solenoid, H = n I.
- Easy : n = 1000 m⁻¹ and I = 2 A ⇒ H = 2000 A m⁻¹.
- Medium : With an iron core of μ_r = 400, B = 4π × 10⁻⁷ × 400 × 2000 = 1.0 T.
- Exam : Here χ = 399, so M = χH = 7.98 × 10⁵ A m⁻¹, and B = μ₀(H + M) = 1.0 T, which agrees.
Rule 2: Units and meaning — H and M are both in A m⁻¹; χ and μ_r have no unit. A large positive χ means the material strengthens the field.
- Easy : In vacuum M = 0, so B = μ₀H.
- Medium : For a vacuum μ_r = 1 and χ = 0.
- Exam : The core multiplies B by about μ_r while H, set by the current, is unchanged.
B = μ₀(H + M); χ = M/H; μ_r = 1 + χ.
Taking H and B to be the same thing. H comes only from the external current.
Memory clue : "H is what you put in; M is what the material adds; B is the total (times mu-nought)."
Spot the difference
| Quantity | Meaning | Unit |
|---|---|---|
| H | field from external current | A m⁻¹ |
| M | moment per unit volume | A m⁻¹ |
| B | total field | T |
Practice questions
- 1Write the SI unit of magnetic intensity.
- 2Define magnetic susceptibility.
- 3What is B in vacuum in terms of H?
- 4Find H inside a solenoid with n = 1000 m⁻¹ and I = 2 A.
- 5Find B in the solenoid with an iron core of μ_r = 400.
- 6Find χ for that core.
- 7Find M for the core of question 5.
- 8Verify that μ₀(H + M) gives the same B as in question 5.
Show answers (try them yourself first)
- 1A m⁻¹.
- 2χ = M ÷ H, the ratio of magnetisation to magnetic intensity.
- 3B = μ₀H.
- 4H = nI = 2000 A m⁻¹.
- 5B = μ₀μ_rH ≈ 1.0 T.
- 6χ = μ_r − 1 = 399.
- 7M = χH = 399 × 2000 = 7.98 × 10⁵ A m⁻¹.
- 8μ₀(2000 + 798 000) = 4π × 10⁻⁷ × 8 × 10⁵ ≈ 1.0 T.
Dia-, para- and ferromagnetism
A steel needle becomes a lasting magnet, copper and wood are not attracted at all, and aluminium is attracted only very weakly. Link to earlier classes : magnetic and non-magnetic materials from Class 6.
Materials fall in three groups. Diamagnetic materials are weakly repelled (χ small and negative). Paramagnetic materials are weakly attracted (χ small and positive). Ferromagnetic materials are strongly attracted and can stay magnetised (χ large and positive).
Rule 1: Comparison of the three types — Diamagnetic (bismuth, copper, water): χ about −10⁻⁵, no permanent dipoles. Paramagnetic (aluminium, sodium, oxygen): χ about 10⁻⁵ to 10⁻³, and Curie's law χ = C ÷ T. Ferromagnetic (iron, nickel, cobalt): χ of 10³ or more, and they change to paramagnetic above the Curie temperature (about 1043 K for iron).
- Easy : A paramagnetic sample with χ = 2.0 × 10⁻⁵ at 300 K has χ = 4.0 × 10⁻⁵ at 150 K.
- Medium : A diamagnetic rod, hung in a field, turns to lie across the field and moves towards the weaker region.
- Exam : Above its Curie temperature, iron becomes paramagnetic.
Rule 2: Hysteresis, soft and hard magnets — The B–H graph of a ferromagnet is a loop, which shows that the magnetisation lags behind the field. Retentivity is the B left at H = 0 and coercivity is the reverse H needed to bring B to zero. The area of the loop is the energy lost per cycle.
- Easy : Soft iron has a narrow loop, low coercivity and high permeability, so it suits electromagnets, transformer cores and motors.
- Medium : Steel has a wide loop with high retentivity and coercivity, so it suits permanent magnets.
- Exam : Cores for AC machines are made of soft iron to waste little energy as heat in each cycle.
In the figure : A hysteresis loop showing retentivity (on the B axis) and coercivity (on the H axis).
Dia : χ < 0, repelled. Para : small χ > 0, weak attraction. Ferro : large χ > 0, strong, with hysteresis.
Saying that all metals are magnetic. Copper and aluminium are not ferromagnetic.
Memory clue : "Dia = Down-pushed (repelled); Para = Pulled weakly; Ferro = Fiercely pulled."
Spot the difference
| Type | χ | Behaviour in a field | Example |
|---|---|---|---|
| Diamagnetic | small, negative | weakly repelled | bismuth, water |
| Paramagnetic | small, positive | weakly attracted | aluminium |
| Ferromagnetic | large, positive | strongly attracted | iron |
Practice questions
- 1What is the sign of χ for a diamagnetic substance?
- 2Name one ferromagnetic material.
- 3What is the Curie temperature?
- 4A paramagnetic sample has χ = 2.0 × 10⁻⁵ at 300 K. Find χ at 150 K.
- 5Which material is better for the core of an electromagnet, soft iron or steel?
- 6Which material is better for a permanent magnet?
- 7What happens to a ferromagnet above its Curie temperature?
- 8Why is soft iron used for transformer cores?
Show answers (try them yourself first)
- 1Negative.
- 2Iron (or nickel, or cobalt).
- 3The temperature above which a ferromagnet turns paramagnetic.
- 4χ ∝ 1 ÷ T, so χ = 4.0 × 10⁻⁵.
- 5Soft iron (low coercivity, high permeability).
- 6Steel (high retentivity and coercivity).
- 7It becomes paramagnetic.
- 8Its narrow hysteresis loop means little energy is lost as heat in each AC cycle.