---
title: "Drift speed of electrons, 20 A"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:drift-speed-copper-twenty-amp"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/drift-speed-copper-twenty-amp
site: BCT Engineering
---

# Drift speed of electrons, 20 A

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2072 Chaitra, 5 marks
- Video: [Drift velocity (concept & intuition) | Electricity | Physics | Khan Academy](https://www.youtube.com/watch?v=v-En7qY5vzY) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/drift-speed-copper-twenty-amp
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

Calculate the drift speed of electrons when 20 A current is supplied through a copper wire of cross-sectional area 1 mm² and electron density 10²⁸ m⁻³.

## Short answer

Drift speed v_d = I/(nAe) = 20/(10²⁸ × 10⁻⁶ × 1.6×10⁻¹⁹) ≈ 1.25×10⁻² m/s (12.5 mm/s).

## Answer

Drift speed comes from I = n A e v_d, so **v_d = I/(nAe)**.

**Given:** I = 20 A, A = 1 mm² = 1×10⁻⁶ m², n = 10²⁸ m⁻³, e = 1.602×10⁻¹⁹ C.

```text
v_d = 20 / (10²⁸ × 1×10⁻⁶ × 1.602×10⁻¹⁹)
    = 20 / (1.602×10³)
    = 1.25×10⁻² m/s
```

**Answer: v_d ≈ 1.25×10⁻² m/s = 12.5 mm/s**

The drift speed is very small compared with the random thermal speed (~10⁶ m/s). The bulb lights at once because the electric field travels through the wire at nearly the speed of light, not because electrons rush from the switch.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Drift speed of carriers in a wire is found from:
   A. v_d = nAe/I
   B. v_d = IA/(ne)
   C. v_d = I n/(A e)
   D. v_d = I/(nAe)
   Answer: D. From I = nAe v_d. (Key point: Formula)

2. A cross-section of 1 mm² must be used in the formula as:
   A. 1×10⁻⁶ m²
   B. 1×10⁻³ m²
   C. 1×10⁻⁹ m²
   D. 1 m²
   Answer: A. 1 mm = 10⁻³ m, so 1 mm² = 10⁻⁶ m². (Key point: Unit conversion)

3. For I = 20 A, n = 10²⁸ m⁻³, A = 1 mm², the drift speed is about:
   A. 1.25×10⁶ m/s
   B. 1.25×10⁻⁵ m/s
   C. 1.25×10⁻² m/s (12.5 mm/s)
   D. 20 m/s
   Answer: C. v_d = 20/(1.602×10³) ≈ 1.25×10⁻² m/s. (Key point: Result)

4. Why does a bulb light at once although the drift speed is tiny?
   A. Electrons rush from the switch to the bulb at thermal speed
   B. The electric field travels through the wire at nearly the speed of light and moves all electrons together
   C. The drift speed rises to 10⁶ m/s when the switch closes
   D. Only the electrons near the battery carry the current
   Answer: B. The field is set up almost instantly along the wire, so electrons everywhere start drifting at once. (Key point: Bulb lights instantly)

## Close variants set in other exams

- [Drift speed of electrons, 10 A](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-ten-amp) — Engineering Physics, chapter 8, asked 2×

## Related in Engineering Physics

- [Drift velocity of electrons in copper](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-velocity-copper-two-hundred-amp) — Engineering Physics, chapter 8
- [Drift speed in copper wire, 17 mA](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-seventeen-ma) — Engineering Physics, chapter 8
- [Current in outer portion of wire](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acurrent-density-wire-outer-portion) — Engineering Physics, chapter 8

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