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SCIENCE · INVESTIGATE AND EXPLAIN

Enclose a nonuniform current and its return path

Ampère’s law becomes a local-field formula only after the symmetry is justified. The ideal source here is infinitely long and coaxial: an inner conductor of radius 1 m carries +4 A with J=J0[1+α(r/R)²], with R=1 m, while a uniform annulus from radius 2 to 3 m carries −4 A. Vacuum permeability and no end effects are stipulated.

Learning outcome: Change α while comparing one radius inside the inner conductor, one in the empty gap and one outside the return. Predict which observations depend on the inner profile and which depend only on total current. Retain the symmetry statement before extracting B from circulation.

Starting recap

Review cross products, source integration measures, line integrals, differential initial states and signed work. Distinguish symmetry premises from integral laws.

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Try

For α=2 at r=0.5 m, enclosed current is 4×(1/4+1/16)/2=0.625 A and field is 0.25 µT. At r=2 m the field is 0.4 µT. At r=2.5 m the enclosed current is 2.2 A and field 0.176 µT. At r=3 m the return cancels the total exactly.

Open the experiment

Predict

Outside the complete return annulus, what fixes the zero field in this model?

Experiment

Change α while comparing one radius inside the inner conductor, one in the empty gap and one outside the return. Predict which observations depend on the inner profile and which depend only on total current. Retain the symmetry statement before extracting B from circulation.

Normalize enclosed current before using symmetry

Worked teaching model · synthetic values · Infinitely long coaxial model: inner radius 1 m carries +4 A with J=J0[1+alpha(r/R)²], R=1 m; annular return from radius 2 to 3 m carries uniform total −4 A. Empty region between. Cylindrical symmetry and vacuum permeability, no ends or transient displacement current. Radius zero uses its exact limit.

Reset keeps your prediction, captures and practice history.

Piecewise field including the return current

azimuthal field (T)radius (m)048.00000e-70

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: normalized enclosed current (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Piecewise field including the return current

normalized enclosed current

radius (m)azimuthal field (T)
00
0.052.00500e-8
0.14.04000e-8
0.156.13500e-8
0.28.32000e-8
0.251.06250e-7
0.31.30800e-7
0.351.57150e-7
0.41.85600e-7
0.452.16450e-7
0.52.50000e-7
0.552.86550e-7
0.63.26400e-7
0.653.69850e-7
0.74.17200e-7
0.754.68750e-7
0.85.24800e-7
0.855.85650e-7
0.96.51600e-7
0.957.22950e-7
18.00000e-7
1.057.61905e-7
1.17.27273e-7
1.156.95652e-7
1.26.66667e-7
1.256.40000e-7
1.36.15385e-7
1.355.92593e-7
1.45.71429e-7
1.455.51724e-7
1.55.33333e-7
1.555.16129e-7
1.65.00000e-7
1.654.84848e-7
1.74.70588e-7
1.754.57143e-7
1.84.44444e-7
1.854.32432e-7
1.94.21053e-7
1.954.10256e-7
24.00000e-7
2.053.74439e-7
2.13.49714e-7
2.153.25767e-7
2.23.02545e-7
2.252.80000e-7
2.32.58087e-7
2.352.36766e-7
2.42.16000e-7
2.451.95755e-7
2.51.76000e-7
2.551.56706e-7
2.61.37846e-7
2.651.19396e-7
2.71.01333e-7
2.758.36364e-8
2.86.62857e-8
2.854.92632e-8
2.93.25517e-8
2.951.61356e-8
30
3.050
3.10
3.150
3.20
3.250
3.30
3.350
3.40
3.450
3.50
3.550
3.60
3.650
3.70
3.750
3.80
3.850
3.90
3.950
40

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Enclosed signed current0.625A
Azimuthal field2.50000e-7T
Inner central current density0.6366198A/m²
Circular field circulation7.85398e-7T m
Total current outside return0A

Ampere symmetry uses the current actually enclosed at each radius, not the full current everywhere.

Inside the central conductor, integrate J0[1+alpha(r/R)²] over annular area, with R=1 m and displayed radial polynomial arguments in units of R, to obtain the normalized enclosed fraction. Between radii 1 and 2 all +4 A is enclosed. The uniform return annulus subtracts 4(r²−4)/5 A until radius 3, beyond which the net is exactly zero. Cylindrical symmetry justifies B times 2pi r; zero circulation alone without that symmetry would not force zero local field.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Ampere symmetry uses the current actually enclosed at each radius, not the full current everywhere.

Inside the central conductor, integrate J0[1+alpha(r/R)²] over annular area, with R=1 m and displayed radial polynomial arguments in units of R, to obtain the normalized enclosed fraction. Between radii 1 and 2 all +4 A is enclosed. The uniform return annulus subtracts 4(r²−4)/5 A until radius 3, beyond which the net is exactly zero. Cylindrical symmetry justifies B times 2pi r; zero circulation alone without that symmetry would not force zero local field.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Specify sources, directions and independent premises

Ampère’s law becomes a local-field formula only after the symmetry is justified. The ideal source here is infinitely long and coaxial: an inner conductor of radius 1 m carries +4 A with J=J0[1+α(r/R)²], with R=1 m, while a uniform annulus from radius 2 to 3 m carries −4 A. Vacuum permeability and no end effects are stipulated.

Derive the integral or vector evolution

Normalize the inner current by integrating 2πrJ dr. Writing u=r/R, the enclosed inner fraction is (u²+αu⁴/2)/(1+α/2). The later radius-polynomial numbers use radius in metres with the stipulated R=1 m. Between radii 1 and 2 the full +4 A is enclosed. Within the return annulus subtract 4(r²−4)/5 A. At and beyond radius 3 the enclosed current is exactly zero.

Check energy, symmetry and evidence limits

Cylindrical symmetry makes the field azimuthal and constant around each centered circle, giving B=μ0Ienclosed/(2πr). At r=0 use the exact zero limit, not division by zero. The field is continuous across the stated finite-density interfaces even though its radial slope can change. Without this symmetry, zero circulation would not imply zero local field.

Worked original case

For α=2 at r=0.5 m, enclosed current is 4×(1/4+1/16)/2=0.625 A and field is 0.25 µT. At r=2 m the field is 0.4 µT. At r=2.5 m the enclosed current is 2.2 A and field 0.176 µT. At r=3 m the return cancels the total exactly.

Investigation

Change α while comparing one radius inside the inner conductor, one in the empty gap and one outside the return. Predict which observations depend on the inner profile and which depend only on total current. Retain the symmetry statement before extracting B from circulation.

Changed-case practice

Try the changed case before feedback

At radius 2.5 m, find enclosed signed current in amperes for any allowed inner profile.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic numerical evidence checked 31 October 2026 against the primary explanation. The rounded teaching permeability is not asserted to be an exact modern SI constant; NIST CODATA 2022 lists its measured uncertainty. No live electrical construction, beam, strong-magnet handling, medical or physical certification claim is requested. Public feedback remains separate from retained first work, critique, revision and fresh assigned evidence.

Apply

Outside the complete return annulus, what fixes the zero field in this model?

Public practice checks a specific claim. It does not establish independent mastery.

Which source geometry, symmetry, initial state or calibration premise is necessary, and what independent observation could reject it?

Continue

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