Integrate current in space and charge in time
Current is dq/dt through a selected cross-section.
Learning outcome: Compare beta = 0, 1 and 2 at fixed radius. Predict the ratio of mean to edge density and retain the annular integral. Reverse the selected normal without changing physical flow. Then integrate the supplied time ramp and distinguish total charge from mean and endpoint current.
Starting recap
Integrate current or resistance, preserve node topology and derive transient state with a signed energy account.
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At R = 1 mm, beta = 1 and J0 = 2×10⁶ A/m², I = 3pi A ≈ 9.42478 A. Mean J is 3×10⁶ A/m². With n = 8×10²⁸ per m³ and electron charge −e, mean drift is approximately −2.34057×10⁻⁴ m/s. The separate 2 s ramp transfers approximately 37.6991 C.
Open the experimentPredict
For positive conventional current carried by electrons, how is electron drift directed?
Experiment
Compare beta = 0, 1 and 2 at fixed radius. Predict the ratio of mean to edge density and retain the annular integral. Reverse the selected normal without changing physical flow. Then integrate the supplied time ramp and distinguish total charge from mean and endpoint current.
Integrate a nonuniform current density
Worked teaching model · synthetic values · Jz(r) = J0[1 + beta(r/R)²], with J0 = 2e6 A/m². Electron carrier density 8e28 per m³ and charge −e. Assigned steady profile; the carrier-density inference assumes one mobile species. Current is signed through the selected positive-z normal.
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Assigned radial current density
Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.
Assigned radial current density: numerical coordinates
| radius (mm) | Jz (A/m²) |
|---|---|
| 0 | 2.00000e+6 |
| 0.0166667 | 2.00056e+6 |
| 0.0333333 | 2.00222e+6 |
| 0.05 | 2.00500e+6 |
| 0.0666667 | 2.00889e+6 |
| 0.0833333 | 2.01389e+6 |
| 0.1 | 2.02000e+6 |
| 0.116667 | 2.02722e+6 |
| 0.133333 | 2.03556e+6 |
| 0.15 | 2.04500e+6 |
| 0.166667 | 2.05556e+6 |
| 0.183333 | 2.06722e+6 |
| 0.2 | 2.08000e+6 |
| 0.216667 | 2.09389e+6 |
| 0.233333 | 2.10889e+6 |
| 0.25 | 2.12500e+6 |
| 0.266667 | 2.14222e+6 |
| 0.283333 | 2.16056e+6 |
| 0.3 | 2.18000e+6 |
| 0.316667 | 2.20056e+6 |
| 0.333333 | 2.22222e+6 |
| 0.35 | 2.24500e+6 |
| 0.366667 | 2.26889e+6 |
| 0.383333 | 2.29389e+6 |
| 0.4 | 2.32000e+6 |
| 0.416667 | 2.34722e+6 |
| 0.433333 | 2.37556e+6 |
| 0.45 | 2.40500e+6 |
| 0.466667 | 2.43556e+6 |
| 0.483333 | 2.46722e+6 |
| 0.5 | 2.50000e+6 |
| 0.516667 | 2.53389e+6 |
| 0.533333 | 2.56889e+6 |
| 0.55 | 2.60500e+6 |
| 0.566667 | 2.64222e+6 |
| 0.583333 | 2.68056e+6 |
| 0.6 | 2.72000e+6 |
| 0.616667 | 2.76056e+6 |
| 0.633333 | 2.80222e+6 |
| 0.65 | 2.84500e+6 |
| 0.666667 | 2.88889e+6 |
| 0.683333 | 2.93389e+6 |
| 0.7 | 2.98000e+6 |
| 0.716667 | 3.02722e+6 |
| 0.733333 | 3.07556e+6 |
| 0.75 | 3.12500e+6 |
| 0.766667 | 3.17556e+6 |
| 0.783333 | 3.22722e+6 |
| 0.8 | 3.28000e+6 |
| 0.816667 | 3.33389e+6 |
| 0.833333 | 3.38889e+6 |
| 0.85 | 3.44500e+6 |
| 0.866667 | 3.50222e+6 |
| 0.883333 | 3.56056e+6 |
| 0.9 | 3.62000e+6 |
| 0.916667 | 3.68056e+6 |
| 0.933333 | 3.74222e+6 |
| 0.95 | 3.80500e+6 |
| 0.966667 | 3.86889e+6 |
| 0.983333 | 3.93389e+6 |
| 1 | 4.00000e+6 |
| Quantity | Value | Unit |
|---|---|---|
| Cross-sectional area | 3.14159 | mm² |
| Total conventional current | 9.42478 | A |
| Area-average current density | 3.00000e+6 | A/m² |
| Average electron drift component | -0.000234057 | m/s |
| Charge crossing in a supplied 2 s ramp I(t)=I0(1+t/1s) | 37.6991 | C |
| Mean current in that supplied ramp | 18.8496 | A |
Integrate density over annular area; the edge value is not the area average.
I = integral Jz 2pi r dr = pi R² J0(1 + beta/2). The electron drift is opposite conventional current because carrier charge is negative. The separate time packet integrates I0(1 + t/1s) from 0 to 2s, giving 4 I0 seconds. Zero average drift would not imply zero random carrier velocity.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
Integrate density over annular area; the edge value is not the area average.
I = integral Jz 2pi r dr = pi R² J0(1 + beta/2). The electron drift is opposite conventional current because carrier charge is negative. The separate time packet integrates I0(1 + t/1s) from 0 to 2s, giving 4 I0 seconds. Zero average drift would not imply zero random carrier velocity.
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