
Physics II course sequence · Interactive lesson
Magnetic Dipoles & Materials
One vector stands in for a whole coil, and the same vector explains why matter responds to a magnet at all. Build the dipole from a loop, stack atomic dipoles into magnetization, then find out how far the linear model of a material can be trusted.
UPII-09.07 · Lesson 7 of 8 in this unit. Position is not a completion score.Build the model
A current loop is a magnet, and matter is built from current loops.
The key idea
A magnetic dipole moment μ is the single vector that stands in for a current loop — magnitude N I A, direction along the loop's normal by the right hand — and a material's magnetization M is the sum of the atomic moments in unit volume.
Example. A 200-turn coil of radius 1.5 cm carrying 0.40 A has μ = 200 × 0.40 × π(0.015 m)² = 5.65 × 10⁻² A m², so a 0.35 T field can twist it with at most τ = μB = 1.98 × 10⁻² N m.
The complete physical model and assumptions
Take a flat loop of current small compared with the distance you watch it from, and everything it does magnetically is carried by one vector: μ = N I A, along the loop's normal by the right hand. In a uniform field the loop feels no net force but a torque μ × B that turns it into line, so it carries an orientation energy U = −μ⋅B; where the field varies, the near and far sides are pulled unequally and a net force μz(dBz/dz) survives. The model earns its place because matter is built from such loops: an orbiting electron is a current loop, electron spin adds a moment of its own, and both are of order the Bohr magneton μB = 9.274 × 10⁻²⁴ A m². Sum those moments per unit volume and you have the magnetization M, which feeds back into the field as B = μ₀(H + M). Three responses follow from how the atomic dipoles answer an applied field: induced counter-moments give a weak negative susceptibility, permanent moments partly aligned against thermal agitation give a weak positive one that falls as 1/T, and exchange coupling in a few metals locks neighbouring spins parallel across whole domains. The linear shorthand B = μᵣ μ₀ H describes the first two well and the third only over a short piece of one curve.
1One vector replaces the loop
A loop carrying current I around an enclosed area A, wound N times, has dipole moment μ = N I A. Curl the right hand along the current and the thumb gives the direction of μ, perpendicular to the loop's plane. The claim worth making is that this single vector is enough: at distances large compared with the loop, the field depends on the current and the area only through their product, so a circular loop, a square loop and a 200-turn pancake of the same μ are indistinguishable. A 200-turn coil of radius 1.5 cm carrying 0.40 A encloses A = π(0.015 m)² = 7.07 × 10⁻⁴ m² and so has μ = 200 × 0.40 A × 7.07 × 10⁻⁴ m² = 5.65 × 10⁻² A m². Its far field falls as 1/r³ — faster than a straight wire's 1/r, because the two sides of the loop very nearly cancel.
2A uniform field turns a dipole but cannot pull it
Put that loop in a uniform field. Opposite sides carry current in opposite directions, so the forces on them are equal and opposite: the net force is zero and the loop's centre of mass does not move. What survives is a couple, τ = μ × B, of magnitude μB sin θ, greatest when μ lies across the field and zero when it lies along it. Integrating that torque through an angle gives the orientation energy U = −μ⋅B = −μB cos θ, a minimum when μ is aligned and a maximum when it is reversed, so turning an aligned dipole end for end costs 2μB. For the coil above in B = 0.35 T the maximum torque is 5.65 × 10⁻² A m² × 0.35 T = 1.98 × 10⁻² N m and the flip costs 3.96 × 10⁻² J. This is what a compass does, and what a moving-coil meter does against a hairspring.
3A force needs a gradient
Since U = −μz Bz along an axis, the force is Fz = −dU/dz = μz (dBz/dz) for a rigid moment: no gradient, no force. That single line explains the demonstration in which samples are hung near a pole piece. A paramagnetic sample develops μ parallel to B, so it is pulled toward the stronger field; a diamagnetic sample develops μ antiparallel, so it is pushed away. Numbers: a 1.0 cm³ sample with χₘ = 2.0 × 10⁻⁴ sitting in B = 1.2 T has M = χₘ B/μ₀ = 191 A m⁻¹, hence μ = M V = 1.91 × 10⁻⁴ A m², and in a gradient of 15 T m⁻¹ it feels F = 2.9 × 10⁻³ N — about a seventh of the weight of a 2 g sample. Water, with χₘ = −9.0 × 10⁻⁶, needs B(dB/dz) = μ₀ρg/|χₘ| = 1.37 × 10³ T² m⁻¹ to levitate, which is why the feat takes a 16 T magnet.
4Where the atomic moments come from
An electron circulating in an atom is a current loop of atomic size, and its moment comes in multiples of the Bohr magneton μB = eħ/(2mₑ) = 9.274 × 10⁻²⁴ A m². Electron spin contributes a moment of about the same size; nuclear moments are smaller by roughly the electron-to-proton mass ratio and can be ignored here. In a filled shell the moments pair off and cancel, which is why most substances carry no permanent moment at all. Add the survivors per unit volume and you have the magnetization M = (Σ μᵢ)/V in A m⁻¹. Iron packs n = 8.49 × 10²⁸ atoms m⁻³ and saturates at Mₛ = 1.71 × 10⁶ A m⁻¹, so each atom contributes 1.71 × 10⁶ ÷ 8.49 × 10²⁸ = 2.01 × 10⁻²³ A m², which is 2.17 μB — about two unpaired spins' worth, and well short of the four unpaired 3d electrons an isolated iron atom carries, because in the metal those electrons are shared through the crystal rather than owned by one atom.
5Two weak responses, set by induction and by temperature
Every material is diamagnetic. Switching on a field changes the flux through each atomic orbit, and by Lenz's law the induced change in that orbital motion opposes the change, so M points against B. The effect is tiny and temperature-independent: χₘ = −9.0 × 10⁻⁶ for water, −1.66 × 10⁻⁴ for bismuth. Where atoms keep a permanent moment a much larger alignment is available, but thermal agitation fights it. Compare the energies: at B = 1.0 T and T = 300 K, μB B/kBT = 9.274 × 10⁻²⁴ ÷ (1.381 × 10⁻²³ × 300) = 2.2 × 10⁻³. The alignment energy is smaller than the thermal energy by a factor of about 450, so only a whisper of net order survives — which is why aluminium manages no more than χₘ = +2.2 × 10⁻⁵, a separate measured number and not that ratio. Because the competition is with kBT, paramagnetic susceptibility obeys Curie's law, χₘ = C/T: cool the sample and the response grows.
6Ferromagnetism, domains, and where the linear model dies
In iron, cobalt, nickel and some alloys a quantum exchange interaction locks neighbouring spins parallel over regions from 10⁻² mm to a few mm — domains, each already magnetized close to saturation. An unmagnetized bar simply has its domains pointing every way. A modest applied field does not align single atoms; it moves the domain walls, letting favourably oriented domains grow, and the response is thousands of times larger than any paramagnet's. Three consequences break the linear model. The B-H curve bends over at saturation: once every domain points one way M can rise no further than Mₛ, and for iron μ₀Mₛ ≈ 2.15 T. It is not single-valued — B depends on the history of H, which is hysteresis, and what a permanent magnet lives on. And heating destroys the order: above the Curie temperature, 1043 K for iron and 627 K for nickel, the metal is an ordinary paramagnet.
Key equations
Read the units, assumptions and limits before substituting.
Dipole moment of a current loop
μ = N I A
Replaces the whole loop: at a distance, no other detail of its shape survives.
N turns, current I in A, enclosed area A in m²; μ in A m², along the right-hand normalTorque on a dipole
τ = μ × B, so τ = μB sin θ
The turning effect a motor, a galvanometer and a compass needle all run on.
θ is the angle from μ to B; τ in N m with μ in A m² and B in TOrientation energy
U = −μ⋅B = −μB cos θ
Flipping an aligned dipole costs 2μB — the energy that thermal agitation must beat.
U in J, taken as zero at θ = 90°; it is lowest when μ lies along BForce needs a field gradient
Fz = μz (dBz/dz)
Follows from Fz = −dU/dz, and is why a pole piece pulls a sample in while a uniform field cannot.
F in N with μz in A m² and the gradient in T m⁻¹; one component, one axisMagnetization of a sample
M = (Σ μᵢ)/V
Converts a count of aligned atoms into one field source for the whole specimen.
vector sum of the atomic moments inside volume V; M in A m⁻¹, the same unit as HField inside magnetized matter
B = μ₀(H + M)
Separates the part of the field you control from the part the material adds.
H in A m⁻¹ is set by the free current alone — H = n I inside a long solenoid; B in TInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Push T up towards 300 K, leaving the other two sliders wherever you like: x falls to {0.67154*m*b/300|3}, both dots collapse onto the origin where exact and linear agree, and that sliver near the origin is all room-temperature paramagnetism ever uses.
Live results
- x = μB / kBT
- 1.998
- ALIGNED FRACTION M/Mₛ
- 0.5369
- LINEAR LAW x/3
- 0.6659
- M FOR 3.0 × 10²⁷ IONS m⁻³
- 104571 A m⁻¹
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Ionic moment μ: 7.0 μB; Applied field B: 8.5 T; Temperature T: 20 K
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Relative permeability is not a material constant the way relative permittivity nearly is. A handbook quoting μᵣ = 4000 for soft iron means the initial slope of one B-H curve, at low field and after demagnetizing. Feed 3.0 A into a solenoid with 2000 turns per metre and a student applying B = μᵣ μ₀ n I predicts 4000 × 1.257 × 10⁻⁶ T m A⁻¹ × 6000 A m⁻¹ = 30 T — more than a research magnet. It cannot happen: H = n I = 6.0 × 10³ A m⁻¹ already drives the iron far past the knee, M has stalled at Mₛ = 1.71 × 10⁶ A m⁻¹, and B = μ₀(H + M) = 2.16 T. Worse, hysteresis means B is not even a function of H alone, so no single μᵣ can exist for iron. Keep μᵣ for dia- and paramagnets, and use a measured curve for anything ferromagnetic.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A flat rectangular coil of 120 turns measures 4.0 cm by 2.5 cm and carries 0.60 A. It sits in a uniform 0.18 T field with the plane of the coil parallel to the field. Find its magnetic dipole moment, the torque on it, and the energy released as it turns into alignment.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Enclosed area: A = 0.040 m × 0.025 m = 1.0 × 10⁻³ m².
- Dipole moment: μ = N I A = 120 × 0.60 A × 1.0 × 10⁻³ m² = 7.2 × 10⁻² A m², directed along the coil's normal.
- The coil's plane is parallel to B, so its normal is perpendicular to B: θ = 90° and sin θ = 1. Torque τ = μB sin θ = 7.2 × 10⁻² × 0.18 = 1.296 × 10⁻² N m.
- Energy: U = −μB cos θ, so U(90°) = 0 and U(0°) = −7.2 × 10⁻² × 0.18 = −1.296 × 10⁻² J. The drop in U is what the field releases as the coil swings into line.
Answer. μ = 7.2 × 10⁻² A m², τ = 1.3 × 10⁻² N m, and 1.3 × 10⁻² J is released on reaching alignment.
MediumExample 2
An iron rod 8.0 cm long and 6.0 mm in diameter is magnetized uniformly along its axis to 40% of saturation, where Mₛ = 1.71 × 10⁶ A m⁻¹. Iron contains 8.49 × 10²⁸ atoms m⁻³. Find the rod's magnetization, its total dipole moment, the equivalent surface current that would produce it, and the average moment per atom in Bohr magnetons.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Volume: V = πr²L = π(3.0 × 10⁻³ m)²(0.080 m) = 2.262 × 10⁻⁶ m³.
- Magnetization: M = 0.40 × 1.71 × 10⁶ = 6.84 × 10⁵ A m⁻¹.
- Total moment: μ = M V = 6.84 × 10⁵ × 2.262 × 10⁻⁶ = 1.547 A m².
- The magnetized rod is equivalent to a solenoid of surface current, and M is that current per unit length: I = M L = 6.84 × 10⁵ × 0.080 = 5.47 × 10⁴ A. Check it against the moment: 5.47 × 10⁴ × πr² = 5.47 × 10⁴ × 2.827 × 10⁻⁵ = 1.547 A m².
- Per atom: 6.84 × 10⁵ ÷ 8.49 × 10²⁸ = 8.06 × 10⁻²⁴ A m², and dividing by μB = 9.274 × 10⁻²⁴ A m² gives 0.87 μB.
Answer. M = 6.84 × 10⁵ A m⁻¹, μ = 1.55 A m², an equivalent surface current of 5.5 × 10⁴ A, and 0.87 μB per atom — 40% of the 2.17 μB iron reaches at saturation.
HardExample 3
A paramagnetic salt holds 3.0 × 10²⁷ magnetic ions per cubic metre, each of moment 7.0 μB, and sits at 300 K in B = 1.5 T. In the low-field limit its magnetization is M = nμ²B/(3kBT). Find the saturation magnetization, the actual magnetization, the fraction of saturation reached and the susceptibility; then find the temperature at which this formula would predict 10% of saturation, and say whether it can be trusted there.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Ionic moment: μ = 7.0 × 9.274 × 10⁻²⁴ = 6.49 × 10⁻²³ A m². Saturation: Mₛ = nμ = 3.0 × 10²⁷ × 6.49 × 10⁻²³ = 1.95 × 10⁵ A m⁻¹.
- The physics sits in one dimensionless ratio, alignment energy over thermal energy: x = μB/kBT = 6.49 × 10⁻²³ × 1.5 ÷ (1.381 × 10⁻²³ × 300) = 2.35 × 10⁻².
- The low-field law is then M = Mₛ x/3 = 1.95 × 10⁵ × 7.83 × 10⁻³ = 1.53 × 10³ A m⁻¹, so only 0.78% of the moments are aligned on average.
- Susceptibility: since χₘ ≪ 1, H ≈ B/μ₀ = 1.5 ÷ (1.257 × 10⁻⁶) = 1.194 × 10⁶ A m⁻¹, giving χₘ = M/H = 1.53 × 10³ ÷ (1.194 × 10⁶) = 1.28 × 10⁻³.
- Setting x/3 = 0.10 needs x = 0.30, so T = μB/(0.30 kB) = 9.74 × 10⁻²³ ÷ (0.30 × 1.381 × 10⁻²³) = 23.5 K.
- At x = 0.30 the exact Langevin result is 0.6% below x/3, so the linear law still holds. It fails further down: half saturation needs x ≈ 1.8, where x/3 overstates the alignment by about 20%.
Answer. Mₛ = 1.95 × 10⁵ A m⁻¹, M = 1.53 × 10³ A m⁻¹ (0.78% aligned), χₘ = 1.28 × 10⁻³, and 10% alignment needs T = 23.5 K — still inside the linear regime, but only just.
Quick check
Select an answer and explain it before checking.
Review & continue
A magnetic dipole moment μ is the single vector that stands in for a current loop — magnitude N I A, direction along the loop's normal by the right hand — and a material's magnetization M is the sum of the atomic moments in unit volume.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.