---
title: "Drift velocity of electrons in copper"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:drift-velocity-copper-two-hundred-amp"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/drift-velocity-copper-two-hundred-amp
site: BCT Engineering
---

# Drift velocity of electrons in copper

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2067 Asoj, 5 marks
- Video: [Derivation of drift velocity formula:  electron drift velocity in a conductor + example with copper.](https://www.youtube.com/watch?v=v_JfuD_GCMg) — Zak's Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/drift-velocity-copper-two-hundred-amp
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

Assuming that each atom of copper contributes one free electron, calculate the drift velocity of free electrons in a copper conductor of cross sectional area 10⁻⁴ m² carrying a current of 200A. Given: Atomic weight of copper = 63.5 g/mol, Density of copper = 8.94×10³ kg/m³, Charge of an electron = 1.6×10⁻¹⁹ C

## Short answer

Electron density n = (ρ_m/M)N_A ≈ 8.48×10²⁸ m⁻³. Drift velocity v_d = I/(nAe) ≈ 1.47×10⁻⁴ m/s ≈ 0.15 mm/s.

## Answer

With one free electron per atom, n = (ρ_m/M) N_A, and v_d = I/(nAe).

**Given:** A = 10⁻⁴ m², I = 200 A, M = 63.5 g/mol = 63.5×10⁻³ kg/mol, density = 8.94×10³ kg/m³, e = 1.6×10⁻¹⁹ C, N_A = 6.022×10²³ mol⁻¹.

```text
n = (8.94×10³ / 63.5×10⁻³) × 6.022×10²³
  = 1.408×10⁵ mol/m³ × 6.022×10²³
  = 8.48×10²⁸ m⁻³

v_d = I / (n A e)
    = 200 / (8.48×10²⁸ × 10⁻⁴ × 1.6×10⁻¹⁹)
    = 200 / (1.357×10⁶)
    = 1.47×10⁻⁴ m/s
```

**Answer: v_d ≈ 1.5×10⁻⁴ m/s = 0.15 mm/s**

## Check yourself

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

1. n is found from density and atomic weight as:
   A. n = (ρ_m/M) N_A
   B. n = ρ_m M/N_A
   C. n = M N_A/ρ_m
   D. n = ρ_m/(M N_A)
   Answer: A. One electron per atom. (Key point: Electron density)

2. The molar mass 63.5 g/mol must be used with density in kg/m³ as:
   A. 63.5 kg/mol
   B. 63.5×10⁻³ kg/mol
   C. 6.35×10⁻⁵ kg/mol
   D. 0.635 kg/mol
   Answer: B. Units must match. (Key point: Unit care)

3. The electron density is about:
   A. 8.48×10²² m⁻³
   B. 8.48×10³⁴ m⁻³
   C. 8.48×10²⁸ m⁻³
   D. 1.41×10⁵ m⁻³
   Answer: C. 1.408×10⁵ mol/m³ × 6.022×10²³. (Key point: Value of n)

4. The drift velocity is about:
   A. 1.5×10⁻¹ m/s
   B. 1.5×10⁻⁷ m/s
   C. 1.5×10⁴ m/s
   D. 1.5×10⁻⁴ m/s
   Answer: D. 200/(1.357×10⁶). (Key point: Drift velocity)

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