---
title: "Drift speed in copper wire, 17 mA"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:drift-speed-copper-seventeen-ma"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/drift-speed-copper-seventeen-ma
site: BCT Engineering
---

# Drift speed in copper wire, 17 mA

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2067 Asar, 5 marks
- Video: [How to Calculate Drift Velocity in Copper Wires](https://www.youtube.com/watch?v=m377fnPQJw0) — Physics Ninja
- 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-seventeen-ma
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

What is the drift speed of the conduction electrons in a copper wire (molecular mass = 63.54 g/mol, density 8.96 g/cm³) with radius 900 μm when it has a uniform current 17mA flowing in the wire?

## Short answer

Electron density n = (density/M) N_A ≈ 8.49×10^28 m^-3. Cross-section A = π(9×10^-4)^2 ≈ 2.55×10^-6 m^2. Drift speed v_d = I/(nAe) ≈ 4.9×10^-7 m/s.

## Answer

The free-electron density is found from the density and molar mass, taking one free electron per copper atom. Then v_d = I/(nAe).

**Given:** M = 63.54 g/mol, density = 8.96 g/cm³ = 8.96×10³ kg/m³, r = 900 μm = 9×10⁻⁴ m, I = 17 mA = 1.7×10⁻² A, N_A = 6.022×10²³ mol⁻¹.

```text
n = (density / M) × N_A
  = (8.96×10⁶ g/m³ / 63.54 g/mol) × 6.022×10²³
  = 1.410×10⁵ mol/m³ × 6.022×10²³
  = 8.49×10²⁸ m⁻³

A = π r² = π (9×10⁻⁴)² = 2.545×10⁻⁶ m²

v_d = I / (n A e)
    = 1.7×10⁻² / (8.49×10²⁸ × 2.545×10⁻⁶ × 1.602×10⁻¹⁹)
    = 1.7×10⁻² / 3.462×10⁴
    = 4.91×10⁻⁷ m/s
```

**Answer: v_d ≈ 4.9×10⁻⁷ m/s** (about 0.5 μm per second).

## Check yourself

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

1. The free-electron density of copper is found from:
   A. n = density × M / N_A
   B. n = M N_A / density
   C. n = (density/M) × N_A, with one electron per atom
   D. n = density × N_A × M
   Answer: C. Moles per m³ times atoms per mole. (Key point: Finding n)

2. The area for r = 900 μm is:
   A. πr² = 2.545×10⁻⁶ m²
   B. 2πr = 5.7×10⁻³ m²
   C. πr = 2.8×10⁻³ m²
   D. r² = 8.1×10⁻⁷ m²
   Answer: A. π(9×10⁻⁴)². (Key point: Area)

3. The drift speed for 17 mA is about:
   A. 4.9×10⁻³ m/s
   B. 4.9×10⁻⁷ m/s
   C. 4.9×10⁻¹⁰ m/s
   D. 4.9 m/s
   Answer: B. 1.7×10⁻²/3.462×10⁴. (Key point: Result)

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