Current, drift velocity and current density
Relate current, drift velocity, and current density with clear sign conventions for charge carriers and SI-unit checks.
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Guided steps
- Current is signed charge transport through a chosen surface; current density is a local vector. Dividing a total current by area gives only a mean normal component. The simple product needs uniform normal flow, while a nonuniform profile needs the surface integral.
- Use signed carrier charge in current density equals number density times charge times drift velocity. Negative electron charge makes conventional current oppose electron drift. Random motion cancels in the mean and is distinct from the small directed drift and from signal propagation.
- Compare individual carriers with the group mean in this paused schematic. Some move against the drift even while the mean shifts right. The arrow tracks carrier drift, so electron conventional current points left. The speed and distance scales are illustrative. Playback is optional; use the static figure and captions if you prefer.
- In a time interval, mean drift transports a volume equal to area times drift speed times elapsed time. Multiply by number density for carrier count and charge magnitude for transported charge magnitude. Divide by time for current magnitude. Keep this speed relation distinct from the signed vector law.
- Square the millimetre-to-metre conversion for an area. In steady flow without side branches or accumulation, a smaller complete cross-section carries the same total current at a larger mean normal density and drift speed. That does not require local density to be uniform across the section.
- The copper example asks for speed, so use charge magnitude and rearrange the transport relation. The result is a small positive speed. Describe the electron direction separately; do not turn a signed velocity component into a negative speed.
- Find signed current density and electron velocity after converting the area. Keep the chosen positive direction fixed.
- Explain the negative electron component and what reverses with the current. Compare your reasoning with the public model yourself; the numerical checker does not mark it.
- A thin annulus has area two pi times radius times thickness. Multiply each local normal density by that area, then add across the section. The centre value alone cannot replace the area mean. Different local profiles can have the same total current.
- Find total current and mean normal density for the supplied profile. The integral and its interpretation remain a separate comparison.
- Show the area-weighted integration, explain the central-density overestimate and the limits of reconstructing a local profile from its total. Compare your paper or written work with the public model.
- Find a nonnegative current magnitude from the stated carrier data.
- Give the new-to-original drift-speed factor at fixed carrier density and geometry.
- Explain why that speed factor follows from the fixed quantities. Compare with the public answer yourself.
- Convert the perpendicular area and calculate the current magnitude for the uniform normal density.
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A wire carries charge through a cross-section, while its carriers move through the material. Current counts the net charge crossing per second; drift velocity describes the average carrier motion; current density describes that flow locally. Keeping those jobs separate lets you handle carrier signs, changing area and a non-uniform flow without confusing a total with a local value.
Use charge, vectors and area conversion as your starting tools. For the integral example, recall integration techniques. Magnetic-force calculations are a later use of the model, not a prerequisite for defining current.
Distinguish total current from local flow
Choose a surface normal n hat to define a positive crossing direction. The signed current is
where Q through is the net signed charge transported through that oriented surface. Current has unit A, equivalent to C s⁻¹; it is not the amount of charge sitting in the wire.
Current density J is a vector field with unit A m⁻². For a surface S,
For uniform density making angle θ with the normal, I = JA cos θ. The product I = JA applies to uniform flow normal to the section. A total I divided by area gives the mean normal component, not a general definition of the local vector J.
Connect carrier motion to conventional current
For one carrier species, let n be number density in m⁻³, q the signed charge per carrier in C, and vd the mean drift velocity in m s⁻¹. Then
Positive carriers contribute current along their drift velocity. For electrons, q = -e, so their contribution points opposite to their drift. If several species carry charge, add their vector contributions rather than assuming one common drift speed.
Random thermal motion does not by itself give a net current: opposite velocity contributions cancel in the mean. Drift is the small directed mean superimposed on that motion. It is also distinct from propagation of an electrical signal, which involves the electromagnetic response of the circuit.
- Start with representative carriers distributed through the wire. Individual displacements will differ from the group mean.
- At this intermediate state, some carriers have moved against the drift while the group's mean position has moved right.
- Over the full interval, random displacements cancel in this schematic ensemble and its mean moves right. For electrons, conventional current points left.
The optional motion separates individual movement from the mean; the particle count, speeds and distances are illustrative, not to scale. The rightward arrow tracks carrier drift, not conventional current. With electrons the conventional current points left. The signed example below independently chooses + x along its conventional current, so its electron component is negative. The static figure and all three captions remain available without playback.
Derive the uniform-wire magnitude relation
Consider uniform carrier density and drift normal to a wire section of area A. In time Δ t, the mean drift sweeps a volume AvdΔ t through the section. Here vd denotes speed. The carrier count is nAvdΔ t, and its transported charge magnitude is n|q|AvdΔ t. Thus
This derivation counts net drift transport, not every random crossing. For a signed component use Jₓ = nqv_(d,x); do not attach the negative electron charge to a speed and then call the resulting negative number a current magnitude.
Check the units and the orientation
The units of nqvd are C m⁻² s⁻¹, equivalent to A m⁻². Multiplying a normal current density by area gives A. Also, 1 mm² = 10⁻⁶ m² because the length conversion is squared.
In a steady series wire with no side branches or charge accumulation, the current is the same at each complete cross-section. If area decreases while n and |q| remain fixed, both mean normal density and drift speed increase. That continuity statement does not require density to be uniform inside each section.
Electron drift in copper
Worked example 1
Drift speed from a measured current
Problem
A copper wire of area 1.0 × 10⁻⁶ m² carries 2.0 A. Its electron number density is 8.5 × 10²⁸ m⁻³. Find the electron drift speed using e = 1.60 × 10⁻¹⁹ C.
Show full solution
Relate carrier flow to current
Method
Rearrange the magnitude relation I = neAvd.
Reason
Speed is non-negative, so use the electron charge magnitude. Its negative sign determines direction separately.
Working
Answer
The charge-density magnitude is ne = 1.36 × 10¹⁰ C m⁻³. Thus neA = 1.36 × 10⁴ C m⁻¹, and
vd = I/(neA) ≈ 1.47 × 10⁻⁴ m s⁻¹.Report the drift speed
Working
Answer
vd ≈ 1.5 × 10⁻⁴ m s⁻¹, opposite to conventional current. This slow carrier motion is distinct from rapid signal propagation.
Quick check
A mean drift displacement of 1 m corresponds to about 2 hours at this speed. This is not the electrical signal travel time.
Guided practice 1
Find the carrier velocity component
Problem
A uniform wire carries a conventional current of 4.80 A along + x. Its perpendicular area is 1.50 mm² and its electron density is 8.00 × 10²⁸ m⁻³. Use q = -1.60 × 10⁻¹⁹ C. Find Jₓ and the signed electron drift component v_(d,x). Explain which directed quantities change sign if the current is reversed with all material and geometry data fixed.
Choose the model and complete the calculation
Hints
Hint 1
Choose the oriented quantity
The prompt asks for an x-component, not just a speed. Fix the sign of conventional current before using the carrier charge.
Hint 2
Keep geometry and carrier signs separate
Convert the area first. Then use Jₓ = nqv_(d,x) with q = -e, rather than using a negative speed in a magnitude formula.
Show solution step by step
Convert the section and find density
Method
Use the perpendicular area in Jₓ = I/A.
Reason
The density is uniform and normal to this section, so the integral reduces to that product.
Working
Answer
A = 1.50 × 10⁻⁶ m². Dividing 4.80 A by this area gives Jₓ = +3.20 × 10⁶ A m⁻².
Retain the electron charge sign
Method
Divide Jₓ by nq.
Reason
The negative carrier charge reverses the velocity direction relative to conventional current.
Working
Answer
nq = -1.28 × 10¹⁰ C m⁻³. Hence v_(d,x) = -2.50 × 10⁻⁴ m s⁻¹. Substitution into nqv_(d,x) gives the positive density again.
Check reversal and interpretation
Method
Separate the signed component from its magnitude.
Reason
Reversing current changes the directed transport, not the fixed area or carrier density.
Working
Answer
The electron drift speed is 2.50 × 10⁻⁴ m s⁻¹, while its x-component is negative. Reversing the current reverses both Jₓ and v_(d,x), with their magnitudes unchanged under the stated model. Self-review: explain the role of the negative charge; saying that electrons carry a negative speed confuses a vector component with a speed.
Quick check
Recalculate the drift speed directly: |vd| = J/(n|q|) = 3.20 × 10⁶/(8.00 × 10²⁸ × 1.60 × 10⁻¹⁹) = 2.50 × 10⁻⁴ m/s, a quarter of a millimetre per second.
Use the surface integral when density varies
A circular section can be divided into thin annuli. At distance s from its axis, an annulus has area dA = 2π s ds. For axial density Jz(s) and normal + z, add the signed contributions:
where a is the outer radius. Using the centre value multiplied by the whole area would assume that every annulus has that centre density. The integral reduces to Jπ a² when J is uniform.
Check your understanding 1 Integrate a radial current-density profile
A circular section of radius a = 1.00 mm has axial current density Jz(s) = J₀[1-(s/a)²], with J₀ = 2.00 × 10⁶ A m⁻² and normal + z. Find total current and the mean normal density. A student uses I = J₀π a²: explain the error. Would the total current and its area mean alone determine the density at every point?
Self-review criteria: add area-weighted signed contributions; distinguish centre density from its mean; explain the limitation of reconstructing a local field from a total.
Show answer
The contribution decreases towards the outer edge, so the centre value overestimates the whole-area average.
With a = 1.00 × 10⁻³ m, I = 3.14 A along the chosen normal. The mean normal density is I/(π a²) = J₀/2 = 1.00 × 10⁶ A m⁻². Using J₀ everywhere gives twice the correct current. For a known area, the total and its mean are equivalent information. They do not uniquely reconstruct the local field: different profiles can have the same integral. Here the supplied profile, not the total alone, provides the local values.
Short checks
Check your understanding 2
A wire has electron density n = 5.0 × 10²⁸ m⁻³, area A = 2.0 × 10⁻⁶ m² and drift speed vd = 1.0 × 10⁻⁴ m s⁻¹. Find current magnitude using e = 1.60 × 10⁻¹⁹ C.
Show hint
Count charge crossing the section per second with I = neAvd.
Show answer
Use e, not -e, because a magnitude is requested.
Check your understanding 3
Current doubles in a wire while carrier density, charge and cross-sectional area stay fixed. How does drift speed change? Enter the factor v_(d,new)/v_(d,old), then explain the comparison.
Show hint
Compare vd = I/(n|q|A) before and after.
Show answer
It doubles: the denominator is fixed, so vd ∝ I. The slow net carrier motion must transport twice as much charge per second.
Check your understanding 4
A uniform current density 3.0 × 10⁶ A m⁻² is perpendicular to a section of area 0.50 mm². Find current magnitude.
Show hint
Square the millimetre-to-metre conversion when converting area.
Show answer
A = 0.50 × 10⁻⁶ m², so I = JA = (3.0 × 10⁶)(0.50 × 10⁻⁶) = 1.5 A. The simple product applies because the density is uniform and normal to the section.
Before you move on
Recover the three jobs: I counts signed charge transport through an oriented surface, J describes local directed flow, and vd describes mean carrier motion. Use the signed charge for components, its magnitude for a speed relation, and the surface integral when uniform normal flow is not established.
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