Slow carrier drift can produce a substantial current
You will be able to: Connect microscopic carrier motion with current density and electric field.
Why does a thicker wire need less drift speed for the same current?
A wire contains a huge number of mobile electrons. Even a slow average drift can move considerable charge through its cross-section each second. That drift speed is not the speed at which a circuit’s electromagnetic response propagates.
A useful starting point: Current is the rate at which charge crosses a surface →
Words and symbols before equations
- Carrier density n
- Number of mobile carriers per cubic metre.
- Drift speed v_d
- Magnitude of the average carrier velocity, not its random thermal motion.
- Current density J
- Current per perpendicular area for uniform flow, in A/m².
- Resistivity ρ
- Material coefficient relating electric field to current density, in Ω·m.
What this picture assumes
Uniform electron flow; e = 1.60×10⁻¹⁹ C and fixed ρ = 2×10⁻⁸ Ω·m. Arrows show direction only. Drift speed is not signal speed.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- J = 1600000 A/m²; electron drift speed = 1 mm/s; field magnitude = 0.032 V/m at ρ = 2×10⁻⁸ Ω·m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For one carrier species with uniform density and drift, the volume crossing area A in time dt is Av_d dt. Multiplying by n and charge magnitude e gives I = neAv_d.
Uniform current density is J = I/A = nev_d. In vector form J = nqv_d, so negative electron charge makes conventional J opposite electron drift velocity.
A resistive wire carrying steady current is not in electrostatic equilibrium: it has a nonzero driving field. For an isotropic ohmic material E = ρJ. This differs from E = 0 inside settled metal with no current.
A worked example, step by step
Let I = 1.6 A, n = 1.0×10²⁸ m⁻³ and A = 1.0 mm². Use e = 1.60×10⁻¹⁹ C to find drift speed.
- Convert A = 1.0×10⁻⁶ m².
- Solve v_d = I/(neA).
- v_d = 1.6/(1.0×10²⁸ × 1.60×10⁻¹⁹ × 10⁻⁶) = 0.001 m/s.
- This is 1 mm/s; the electron drift points opposite conventional current.
Do not confuse slow drift with circuit signal speed or assume a current-carrying resistive wire has zero E.
If area doubles at fixed I and n, what happens to v_d?
Compare with an explanation
It halves, because more carriers cross in parallel.
Predict. Change one thing. Explain.
Increase cross-sectional area at fixed current and carrier density. Predict the changes in drift speed and J; compare the opposing electron and current arrows.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
J = 1600000 A/m²; electron drift speed = 1 mm/s; field magnitude = 0.032 V/m at ρ = 2×10⁻⁸ Ω·m.
Uniform electron flow; e = 1.60×10⁻¹⁹ C and fixed ρ = 2×10⁻⁸ Ω·m. Arrows show direction only. Drift speed is not signal speed.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, circuit topology, charge or energy conservation, or RC relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA uniform wire carries 3.2 A through 2 mm² with n = 10²⁸ m⁻³ and ρ = 2×10⁻⁸ Ω·m. Find J, v_d and E.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: J = 3.2/(2×10⁻⁶) = 1.6×10⁶ A/m².
- 1 point: v_d = J/(ne) = 0.001 m/s.
- 1 point: E = ρJ = 0.032 V/m.
- 1 point: Conventional J and E align; electron drift opposes them.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why can small drift create large I?
There are many mobile carriers in each cubic metre.
RECALL 2What does J describe?
Local current flow per perpendicular area.
RECALL 3When is E = 0 in ideal metal?
At electrostatic equilibrium, not generally during resistive current flow.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Slow carrier drift can produce a substantial current
- I = neAv_d for uniform single-carrier flow.
- J = I/A.
- E = ρJ for an isotropic ohmic material.
Remember: Do not confuse slow drift with circuit signal speed or assume a current-carrying resistive wire has zero E.
Conditions: Uniform electron flow; e = 1.60×10⁻¹⁹ C and fixed ρ = 2×10⁻⁸ Ω·m. Arrows show direction only. Drift speed is not signal speed.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.1.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.1, objectives 11.1.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 4: Electric Circuits, numbered Unit 11 in the official combined Physics C sequence. Topics 11.1–11.8 retain their official identifiers. Models include signed charge flow, prescribed current-density and resistivity integrals, DC resistor networks, nonideal batteries and meters, capacitor combinations and finite-resistance RC transients. Circuit schematics use conventional-current references and explicit node connectivity. Initial capacitor voltage and positive time constants are stated. Unequal ideal sources are never directly wired in parallel. The optional spatial wire view supplements a complete 2D current-density explanation. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about Slow carrier drift can produce a substantial current. Your explanation and answers remain free to access.