
Physics II course sequence · Interactive lesson
The Ampere-Maxwell law
Ampere's law works beautifully until a wire stops in mid-air at a capacitor plate. This lesson shows where it breaks, what charge conservation says is missing, and how to use the repaired law to compute a real magnetic field in a region where no charge is moving at all.
UPII-12.04 · Lesson 4 of 8 in this unit. Position is not a completion score.Build the model
A changing electric field is a source of magnetic field.
The key idea
The Ampere-Maxwell law says the circulation of B around a closed loop equals μ₀ times the sum of the conduction current through any surface bounded by that loop and ε₀ times the rate of change of electric flux through that same surface.
Example. Circular plates of radius 5.0 cm charge at I = 2.0 A. Nothing crosses the gap, yet ε₀ dΦE/dt = 2.0 A there, so a loop of radius 3.0 cm drawn between the plates carries B = 4.8 μT.
The complete physical model and assumptions
Ampere's law, ∮B⋅dl = μ₀Ienc, carries a hidden assumption. The loop on the left bounds infinitely many surfaces, and the law makes sense only if every one of them reports the same enclosed current. For a steady circuit it does. Push the loop up against a charging capacitor and it does not: a flat disc is pierced by the lead and returns I, while a bag-shaped surface that slips between the plates is pierced by nothing and returns zero. Charge conservation names the missing piece. Current stopping on the plate makes q grow, the field in the gap grows with it, and the electric flux through the bag climbs at exactly the rate needed to make ε₀ dΦE/dt = dq/dt = I. Maxwell added that term, calling it displacement current, and the law became ∮B⋅dl = μ₀(Ienc + ε₀ dΦE/dt) — surface-independent for every circuit, steady or not. The new term is not a flow of charge; nothing crosses the gap. It says that a changing electric field wraps magnetic field around itself just as a current does, which mirrors Faraday's law and is the reason fields can sustain one another. What the law never promises is that B is easy to extract: like Ampere's law before it, the integral form yields B only when symmetry makes B uniform along the chosen loop.
1One loop, many surfaces, two different answers
Ampere's law ties the circulation of B around a closed loop to the current through a surface bounded by that loop — and it never says which surface, because for steady currents the choice cannot matter. Charge accumulates nowhere, so whatever current enters a closed region leaves it, and every surface spanning the loop is pierced the same number of times. Now build a circuit whose current stops: a parallel-plate capacitor with circular plates of radius 5.0 cm, charging at I = 2.0 A. Take a circular loop of radius 3.0 cm around the lead. Span it with a flat disc and the lead pierces it, giving ∮B⋅dl = μ₀ × 2.0 A = 2.51 × 10⁻⁶ T m. Span the same loop with a bag-shaped surface that bulges past the plate and closes inside the gap, and no charge crosses it at all, giving zero. Both surfaces are legal. The law as Ampere wrote it contradicts itself the moment charge piles up somewhere.
2Charge conservation names the missing term
Follow the charge. Current arriving at the plate has nowhere to go, so dq/dt = I. Between the plates that charge sets up a field E = σ/ε₀ = q/(ε₀A), uniform if the gap is small, so the electric flux through the part of the bag lying inside the gap is ΦE = EA = q/ε₀. Differentiate: ε₀ dΦE/dt = dq/dt = I. The bag was never empty after all — the conduction current it misses is replaced, ampere for ampere, by a growing electric flux. Adding μ₀ε₀ dΦE/dt to the right-hand side therefore makes the two surfaces agree: the disc collects μ₀I from the lead and nothing from flux, the bag collects nothing from current and μ₀I from flux. The same argument works for any loop and any circuit, because it is only charge conservation written twice. For this capacitor, with A = 7.85 × 10⁻³ m², the field in the gap is climbing at dE/dt = I/(ε₀A) = 2.9 × 10¹³ V m⁻¹ s⁻¹.
3Using the law where symmetry allows
The repaired law earns its keep as a computing tool exactly where Ampere's law did: when symmetry forces B to be constant in magnitude and tangential along the loop. Between circular plates the charging field lies along the axis and is uniform, so the arrangement is symmetric about that axis and B must circle it with constant magnitude. Choose a circle of radius r centred on the axis, midway between the plates. Then ∮B⋅dl = 2πrB. Inside the rim the loop's surface catches only the flux passing through area πr², so the enclosed displacement current is I(r/R)². With I = 2.0 A and R = 5.0 cm, a loop at r = 3.0 cm encloses 0.72 A, and 2πrB = μ₀(0.72 A) gives B = 4.8 μT. Push r past the rim and the whole 2.0 A is enclosed, so B = μ₀I/(2πr), the field of the wire. The two expressions meet at r = R, where B reaches its largest value, 8.0 μT.
4The sense of the loop fixes the sign of both terms
Signs here are a convention you must apply consistently, not a physical extra. Pick a direction to walk the loop when evaluating ∮B⋅dl. Your right hand, fingers curling that way, points the thumb along the normal n̂ of the spanning surface. That single n̂ then grades both source terms: a conduction current counts as positive when it flows along n̂, and the flux is ΦE = ∫E⋅n̂ dA, so dΦE/dt is positive when E grows along n̂. Reverse the walking direction and all three quantities change sign together, leaving the physics untouched. The payoff is a real prediction. While the capacitor charges, E in the gap grows from the positive plate towards the negative one, and B circles that direction by the right-hand rule, in the same sense as around the lead. Let the capacitor discharge and E shrinks, dΦE/dt turns negative, and B in the gap reverses — even though no charge has ever crossed the gap.
5Why nobody noticed for a century
The two source terms are rarely comparable. For a sinusoidal field of amplitude E₀ and angular frequency ω in a material of conductivity σ and permittivity ε, the conduction current density is σE₀ and the displacement current density is εωE₀, so their ratio is ωε/σ. Inside copper, σ = 5.96 × 10⁷ S m⁻¹ and ε ≈ ε₀, so at 60 Hz that ratio is 5.6 × 10⁻¹⁷: the displacement term in household wiring sits seventeen orders of magnitude below the conduction term, which is why Ampere's law survived unamended for decades. The ordering reverses in poor conductors and at high frequency, the two terms matching at f = σ/(2πε). In seawater, σ ≈ 4 S m⁻¹ and ε ≈ 81ε₀, putting that crossover near 8.9 × 10⁸ Hz. In a vacuum gap σ is zero, so the displacement term is not merely dominant, it is the only term there is.
6The law is deliberately lopsided
Set the two circulation laws side by side. Faraday gives ∮E⋅dl = −dΦB/dt; the Ampere-Maxwell law gives ∮B⋅dl = μ₀Ienc + μ₀ε₀ dΦE/dt. The flux terms mirror each other, and that reciprocity is what the rest of this unit is built on. Two asymmetries survive on purpose. There is a conduction term here and none in Faraday, because isolated electric charge exists while no magnetic charge has ever been found, so there is no magnetic current to put in. And Faraday carries a minus sign where this law does not, which is Lenz's law: an induced effect opposes the change that made it, while a growing electric flux drives B in the positive sense. The constant on the new term repays a second look too. μ₀ε₀ = 1.113 × 10⁻¹⁷ s² m⁻², which is 1/c² — a speed assembled from two constants measured with capacitors and coils.
Key equations
Read the units, assumptions and limits before substituting.
The Ampere-Maxwell law
∮B⋅dl = μ₀(Ienc + ε₀ dΦE/dt)
Magnetic circulation counts two sources: charge in motion, and electric flux in change.
B in T, dl in m, Ienc in A, ΦE in V m, μ₀ = 4π × 10⁻⁷ T m A⁻¹Displacement current
Id = ε₀ dΦE/dt
Puts the flux term in amperes, so it adds straight onto the conduction current.
ε₀ = 8.854 × 10⁻¹² F m⁻¹, and ΦE in V m gives Id in AEvaluating Id across a uniform gap
Id = ε₀ A dE/dt
The working form whenever E is uniform and perpendicular to a flat surface.
A is the area the field crosses, in m²; dE/dt in V m⁻¹ s⁻¹Capacitor consistency
Id = dq/dt = IC
The two surfaces of one loop now agree: the bag reports what the disc reported.
holds when the surface covers a whole plate, so that ΦE = q/ε₀ exactlyField inside a charging circular gap
B = μ₀ I r / (2πR²), for r ≤ R
Only the fraction (r/R)² of the displacement current is enclosed, so B grows with r.
R is the plate radius in m, r the loop radius in m, B in TField beyond the plate rim
B = μ₀ I / (2πr), for r ≥ R
Outside the rim the whole displacement current is enclosed, so the gap mimics the wire.
same I as in the leads; B in T, peaking at r = R and falling as 1/rInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the loop out past the plate rim: the field stops growing and starts falling, with the peak always at r = R. Then shrink the plate radius at fixed current and watch the rim field climb, because the same displacement current now crosses a smaller disc.
Live results
- DISPLACEMENT CURRENT
- 1.20 A
- ENCLOSED BY THE LOOP
- 0.469 A
- B ON THE LOOP
- 3.75 μT
- PEAK B AT r = R
- 6.00 μT
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. Charging current I: 1.2 A; Plate radius R: 4.0 cm; Amperean loop radius r: 2.5 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to treat displacement current as a current: charge somehow bridging the gap, or leaking through the dielectric. Nothing crosses. Maxwell's name came from an ether model in which a medium was literally displaced, and only the name survived; the term is a changing electric field, and it is complete in a vacuum gap where there is nothing to displace. The practical damage is double counting. A student draws a flat surface cutting the lead, writes Ienc = I, then adds Id = I as well and reports 2I. On that surface the electric flux is barely changing, so Id ≈ 0. On the bag surface inside the gap there is no conduction current, so Ienc = 0. Each surface collects the whole source exactly once, from one term or the other — never from both.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A vacuum-gap capacitor has circular plates of radius 4.0 cm separated by 1.5 mm. The potential difference across it is ramped up at a steady 2.4 × 10⁵ V s⁻¹. Find the displacement current in the gap.
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.
- The gap field is uniform, E = V/d, so it grows at dE/dt = (dV/dt)/d = 2.4 × 10⁵ V s⁻¹ ÷ 1.5 × 10⁻³ m = 1.6 × 10⁸ V m⁻¹ s⁻¹.
- That field crosses the plate area A = πR² = π(0.040 m)² = 5.03 × 10⁻³ m².
- Displacement current: Id = ε₀ A dE/dt = (8.854 × 10⁻¹² F m⁻¹)(5.03 × 10⁻³ m²)(1.6 × 10⁸ V m⁻¹ s⁻¹) = 7.12 × 10⁻⁶ A.
- Check it against the circuit: C = ε₀A/d = 29.7 pF, so the leads carry I = C dV/dt = (29.7 × 10⁻¹² F)(2.4 × 10⁵ V s⁻¹) = 7.12 μA — the same number the gap reports.
Answer. Id = 7.1 μA, equal to the conduction current in the leads.
MediumExample 2
Circular plates of radius 5.0 cm, 2.0 mm apart in vacuum, are charged by a steady conduction current of 0.85 A. Midway between the plates, find B at r = 2.5 cm and at r = 8.0 cm from the axis, and say where B is largest.
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.
- No charge crosses the gap, so the only source there is displacement current, totalling Id = I = 0.85 A over the plate area A = π(0.050 m)² = 7.85 × 10⁻³ m². (The field is climbing at I/(ε₀A) = 1.2 × 10¹³ V m⁻¹ s⁻¹.)
- Symmetry about the axis makes B tangential and constant on any circle centred on it, so ∮B⋅dl = 2πrB.
- At r = 2.5 cm the loop lies inside the rim and catches the fraction (r/R)² = (2.5/5.0)² = 0.25 of the flux, so Id, enc = 0.213 A. Then B = μ₀(0.213 A)/(2π × 0.025 m) = 1.7 × 10⁻⁶ T.
- At r = 8.0 cm the loop is outside the rim and encloses all 0.85 A: B = μ₀(0.85 A)/(2π × 0.080 m) = 2.1 × 10⁻⁶ T.
- B rises as r inside the rim and falls as 1/r outside it, so it peaks on the rim itself: B(R) = μ₀(0.85 A)/(2π × 0.050 m) = 3.4 μT.
Answer. B = 1.7 μT at r = 2.5 cm and 2.1 μT at r = 8.0 cm; the maximum is 3.4 μT, at r = R = 5.0 cm.
HardExample 3
A capacitor has circular plates of radius 6.0 cm separated by 1.2 mm of a weakly conducting filling of conductivity σ = 2.0 × 10⁻⁷ S m⁻¹ and permittivity ε₀. It is charged to 40 V and its leads are then cut, so it discharges through its own filling. Find the conduction and displacement currents in the filling at the instant of cutting, and the magnetic field they produce.
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.
- At that instant E = V/d = 40 V ÷ 1.2 × 10⁻³ m = 3.33 × 10⁴ V m⁻¹, so the conduction current density is Jc = σE = 6.67 × 10⁻³ A m⁻². Over A = π(0.060 m)² = 1.131 × 10⁻² m² that gives Ic = 7.54 × 10⁻⁵ A, directed along E.
- The charge leaving the plate feeds that current: dq/dt = −σAE = −(σ/ε₀)q, since E = q/(ε₀A). So q and E decay as e(−t/τ) with τ = ε₀/σ = 8.854 × 10⁻¹² ÷ 2.0 × 10⁻⁷ = 4.43 × 10⁻⁵ s.
- Hence dE/dt = −E/τ = −7.53 × 10⁸ V m⁻¹ s⁻¹, and Id = ε₀ A dE/dt = −7.54 × 10⁻⁵ A: the same size as Ic and opposite in sign.
- The cancellation is exact at every instant, not only at t = 0: Id = ε₀A(−E/τ) = ε₀A(−σE/ε₀) = −σAE = −Ic, whatever E has decayed to.
- Apply the law to a circle of radius r about the axis inside the filling: 2πrB = μ₀(Ic, enc + Id, enc) = μ₀ × 0 = 0.
- So B = 0 for every r and every t. The relaxing charge produces no magnetic field at all, because the conduction current it drives and the collapsing electric flux cancel term for term.
Answer. Ic = +75 μA and Id = −75 μA at the instant of cutting, decaying together with τ = 44 μs; their sum is exactly zero, so B = 0 everywhere at every instant.
Quick check
Select an answer and explain it before checking.
Review & continue
The Ampere-Maxwell law says the circulation of B around a closed loop equals μ₀ times the sum of the conduction current through any surface bounded by that loop and ε₀ times the rate of change of electric flux through that same surface.
- 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.