Electricity & Magnetism · 20.1
Fields Driving Current
Flip a switch and the lamp lights at once — yet each electron crawls along at a fraction of a millimetre per second. An electric field inside the wire pushes every charge at the same moment, and I = nAvq counts the result.
Build the model
Connect the measurement to the mechanism.
A cell maintains a potential difference, and that p.d. keeps an electric field alive inside the wire. The field pushes on every free electron at once, so the whole circuit responds together even though each electron only drifts. The current is I = nAvq: carrier density, times cross-sectional area, times drift speed, times the charge on each carrier.
- Simple definition
- Electric current is the rate at which charge passes a point; it flows because a potential difference maintains an electric field inside the conductor that pushes the free charges along.
- Example
- A 20 A household supply moves 20 coulombs past the meter every second — about 1.3 × 10²⁰ electrons.
Stand at one point of the circuit and count the charge crossing each second — that count is the current.
One ampere is one coulomb per second
The potential difference is the energy handed to each coulomb — it is what keeps the field, and the push, alive.
Volts are joules per coulomb
Current is how many carriers there are, times how much room they have, times how fast they drift, times the charge each one carries.
n = carriers per m³, A = cross-section
A field inside a conductor
An isolated conductor has no internal field. Connect a cell and the p.d. across the wire maintains one — every free electron feels the push almost simultaneously, which is why the lamp lights the instant you close the switch.
Drift, not sprint
Free electrons already fly about at enormous random speeds, colliding constantly with the lattice. The field adds only a tiny net crawl toward the positive terminal — the drift velocity, typically well under a millimetre per second.
Conventional current vs electron flow
Franklin guessed the moving charge was positive, so circuit diagrams show conventional current from + to −. The electrons actually drift the other way. Both descriptions give identical physics — only the arrow labels differ.
Change one variable at a time
Make the relationship visible.
I = nAvq rearranged gives v = I/(nAq). A fatter wire or a denser metal carries the same current with a slower crawl.
Drift speed v = I/(nAq)0.15 mm/s
Time to drift 1 m1.9 h
Charge past a point each second2 C
Catch the common trap
Explain before calculating.
A wire is swapped for one of twice the cross-sectional area, carrying the same current with the same carrier density. What happens to the drift velocity?
Choose an answer to test the model.
Worked examples
State the rule, substitute, then check units.
EasyA phone charger delivers 0.50 A for 120 s. How much charge flows, and how many electrons is that?
- q = It = 0.50 × 120 = 60 C.
- Number of electrons = 60 ÷ 1.60 × 10⁻¹⁹ ≈ 3.8 × 10²⁰.
Answer60 C ≈ 3.8 × 10²⁰ electrons
MediumA 2.5 mm diameter wire carries 1.5 A. The free-electron density is 5.0 × 10²⁶ m⁻³. Find the drift velocity.
- A = πd²/4 = π(2.5 × 10⁻³)²/4 = 4.9 × 10⁻⁶ m².
- v = I/(nAq) = 1.5 ÷ (5.0 × 10²⁶ × 4.9 × 10⁻⁶ × 1.6 × 10⁻¹⁹).
- v ≈ 3.8 × 10⁻³ m/s — about 4 mm every second.
Answerv ≈ 3.8 mm/s
HardCopper has n = 8.5 × 10²⁸ m⁻³. A 1.0 mm² lamp cord carries 2.0 A. Find the drift velocity, and how long one electron takes to travel the 3.4 m to the lamp.
- v = I/(nAq) = 2.0 ÷ (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹) ≈ 1.5 × 10⁻⁴ m/s.
- t = d/v = 3.4 ÷ 1.5 × 10⁻⁴ ≈ 2.3 × 10⁴ s ≈ 6.4 hours.
- Yet the lamp lights instantly — the field starts every electron in the cord moving at once.
Answerv ≈ 0.15 mm/s; ≈ 6.4 h — but the lamp is instant
ChallengingAn electron is accelerated from rest through a p.d. of 25.0 V. Find the work done in electronvolts and joules, and the electron’s final speed.
- By definition W = qV = 25.0 eV — the electronvolt makes this step trivial.
- In joules: W = 25.0 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻¹⁸ J.
- ½mv² = W → v = √(2 × 4.0 × 10⁻¹⁸ ÷ 9.11 × 10⁻³¹) ≈ 3.0 × 10⁶ m/s — about 1% of the speed of light.
Answer25 eV = 4.0 × 10⁻¹⁸ J; v ≈ 3.0 × 10⁶ m/s