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SCIENCE · INVESTIGATE AND EXPLAIN

Check inductance against a field-volume integral

A flux through one turn is not the same as total flux linkage. In the long-solenoid approximation, turns per length n, length ell and area A give total turns N=n ell and internal field B=μ0nI. This packet uses ell=2 m, A=0.01 m² and a vacuum interior; external field and end corrections are neglected.

Learning outcome: Compare n=100,200,400 at the same current, then set current to zero. Record flux per turn, linkage, inductance and both energy accounts. Explain why doubling N at fixed length differs from doubling length at fixed n.

Starting recap

Review flux integrals, linear differential equations, initial conditions, complex numbers and vector derivatives. Keep source constraints, RMS conventions and calibration roles explicit.

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Try

At n=200 per metre and I=2 A, N=400 turns, B≈0.000502655 T and L=0.00032π H≈0.001005310 H. Stored energy is 0.00064π J≈0.002010619 J. Integrating the internal field density over 0.02 m³ gives the same value.

Open the experiment

Predict

At fixed length, area and current, doubling turn density changes stored energy by:

Experiment

Compare n=100,200,400 at the same current, then set current to zero. Record flux per turn, linkage, inductance and both energy accounts. Explain why doubling N at fixed length differs from doubling length at fixed n.

Reconcile flux linkage with field energy

Worked teaching model · synthetic values · Long-solenoid approximation length2 m,area.01 m²,vacuum. Uniform internal field and negligible external field are stipulated approximations; total turns n times length. Rounded mu0. No finite-end, core saturation, real coil heating or handling model.

Reset keeps your prediction, captures and practice history.

Quadratic magnetic storage at fixed geometry

stored energy (J)current (A)040.0080424770

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

Equivalent numerical coordinates for Quadratic magnetic storage at fixed geometry

half L I squared

current (A)stored energy (J)
00
0.051.25664e-6
0.15.02655e-6
0.151.13097e-5
0.22.01062e-5
0.253.14159e-5
0.34.52389e-5
0.356.15752e-5
0.48.04248e-5
0.450.0001017876
0.50.0001256637
0.550.0001520531
0.60.0001809557
0.650.0002123717
0.70.0002463009
0.750.0002827433
0.80.0003216991
0.850.0003631681
0.90.0004071504
0.950.000453646
10.0005026548
1.050.0005541769
1.10.0006082123
1.150.000664761
1.20.0007238229
1.250.0007853982
1.30.0008494867
1.350.0009160884
1.40.0009852035
1.450.001056832
1.50.001130973
1.550.001207628
1.60.001286796
1.650.001368478
1.70.001452672
1.750.00153938
1.80.001628602
1.850.001720336
1.90.001814584
1.950.001911345
20.002010619
2.050.002112407
2.10.002216708
2.150.002323522
2.20.002432849
2.250.00254469
2.30.002659044
2.350.002775911
2.40.002895292
2.450.003017186
2.50.003141593
2.550.003268513
2.60.003397947
2.650.003529894
2.70.003664354
2.750.003801327
2.80.003940814
2.850.004082814
2.90.004227327
2.950.004374354
30.004523893
3.050.004675947
3.10.004830513
3.150.004987592
3.20.005147185
3.250.005309292
3.30.005473911
3.350.005641044
3.40.00581069
3.450.005982849
3.50.006157522
3.550.006334707
3.60.006514407
3.650.006696619
3.70.006881345
3.750.007068583
3.80.007258336
3.850.007450601
3.90.00764538
3.950.007842672
40.008042477

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
Total turns400count
Approximate internal magnetic field0.0005026548T
Flux per turn5.02655e-6Wb
Flux linkage0.002010619Wb turn
Self inductance0.00100531H
Circuit magnetic energy0.002010619J
Internal field energy density0.100531J/m³
Field-volume energy0.002010619J
Energy-account residual0J

Flux linkage determines inductance, and the same approximation must be used in the field-volume energy check.

The long-solenoid field is B=mu0 n I and N=n ell, so flux linkage N B A=mu0 n² A ell I. Hence L=mu0 n² A ell. Integrating B²/(2mu0) over the approximate internal volume A ell gives LI²/2. Agreement checks consistency of one ideal geometry; it does not validate neglected fringe fields or a nonlinear magnetic core.

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

Flux linkage determines inductance, and the same approximation must be used in the field-volume energy check.

The long-solenoid field is B=mu0 n I and N=n ell, so flux linkage N B A=mu0 n² A ell I. Hence L=mu0 n² A ell. Integrating B²/(2mu0) over the approximate internal volume A ell gives LI²/2. Agreement checks consistency of one ideal geometry; it does not validate neglected fringe fields or a nonlinear magnetic core.

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

Understand

State geometry, sources and initial data

A flux through one turn is not the same as total flux linkage. In the long-solenoid approximation, turns per length n, length ell and area A give total turns N=n ell and internal field B=μ0nI. This packet uses ell=2 m, A=0.01 m² and a vacuum interior; external field and end corrections are neglected.

Derive the field or circuit relation

Flux per turn is BA, so linkage NΦ=μ0n²A ell I and self-inductance is L=μ0n²A ell. Independently integrate magnetic energy density B²/(2μ0) over the approximate field volume A ell. The result equals LI²/2. Both derivations must use the same geometry and approximations to be meaningfully compared.

Audit energy, uncertainty and domain

Doubling turn density increases field linearly at fixed current but increases inductance and stored energy fourfold. Zero current gives zero field and energy without making geometric inductance zero. A nonlinear core or a finite coil with important fringe fields requires another constitutive or spatial model; an algebraic agreement does not certify a real coil.

Worked original case

At n=200 per metre and I=2 A, N=400 turns, B≈0.000502655 T and L=0.00032π H≈0.001005310 H. Stored energy is 0.00064π J≈0.002010619 J. Integrating the internal field density over 0.02 m³ gives the same value.

Investigation

Compare n=100,200,400 at the same current, then set current to zero. Record flux per turn, linkage, inductance and both energy accounts. Explain why doubling N at fixed length differs from doubling length at fixed n.

Changed-case practice

Try the changed case before feedback

At fixed geometry, current rises from 1 A to 3 A. Find final-to-initial stored-energy ratio.

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 evidence checked 31 October 2026 against the primary explanation. Constants are rounded teaching values and are mutually consistent where Maxwell relations require it. No live circuit, high-voltage, energized-inductor, magnet, beam or optical exposure activity is requested. Retain first work, criterion-linked critique, revision and newly assigned evidence; public numerical feedback is not practical or learner certification.

Apply

At fixed length, area and current, doubling turn density changes stored energy by:

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

Which source, initial-state, calibration or field-law premise makes the result conditional, and what unused evidence would challenge it?

Continue

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