---
title: "Drift velocity from Hall experiment"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:hall-numerical-conductor-drift-velocity"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-numerical-conductor-drift-velocity
site: BCT Engineering
---

# Drift velocity from Hall experiment

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2070 Chaitra, 5 marks
- Video: [Resistivity, Drift Velocity, and the Hall Effect](https://www.youtube.com/watch?v=u_4xOxoKu1I) — Dr. Pierce's Physics & Math
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-numerical-conductor-drift-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

In a Hall Effect experiment, a current of 3.2A lengthwise in a conductor 1.2 cm wide, 4.0 cm long and 9.5μm thick produces a transverse Hall voltage (across the width) of 40μV when a magnetic field of 1.4T is passed perpendicularly through the thin conductor. From this data, find (a) the drift velocity of the charge carriers and (b) the number density of charge carriers.

## Short answer

Drift velocity v_d = V_H/(B w) = 2.4×10⁻³ m/s. Carrier density n = I/(e v_d w t) ≈ 7.4×10²⁸ m⁻³.

## Answer

**Formulas:** the Hall field balances the magnetic force, E_H = v_d B, and V_H = E_H w. The carrier density follows from I = n e v_d (w t).

**Given:** I = 3.2 A, w = 1.2 cm = 0.012 m, t = 9.5 μm = 9.5×10⁻⁶ m, V_H = 40 μV = 4×10⁻⁵ V, B = 1.4 T.

#### (a) Drift velocity
```text
v_d = V_H / (B w) = 4×10⁻⁵ / (1.4 × 0.012)
    = 4×10⁻⁵ / 0.0168
    = 2.38×10⁻³ m/s
```

#### (b) Number density
```text
n = I / (e v_d w t)
  = 3.2 / (1.602×10⁻¹⁹ × 2.38×10⁻³ × 0.012 × 9.5×10⁻⁶)
  = 3.2 / (4.35×10⁻²⁹)
  = 7.36×10²⁸ m⁻³

(check with n = IB/(e t V_H) = 3.2×1.4 / (1.602×10⁻¹⁹ × 9.5×10⁻⁶ × 4×10⁻⁵) = 7.36×10²⁸ m⁻³)
```

**Answer: (a) v_d ≈ 2.4×10⁻³ m/s; (b) n ≈ 7.4×10²⁸ m⁻³** (the length 4.0 cm is not needed).

## Check yourself

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

1. From the Hall data, drift velocity is:
   A. v_d = V_H B w
   B. v_d = V_H/(B t)
   C. v_d = V_H/(Bw)
   D. v_d = B w/V_H
   Answer: C. E_H = v_dB and V_H = E_H w. (Key point: Drift velocity)

2. With V_H = 40 μV, B = 1.4 T, w = 1.2 cm, v_d is about:
   A. 2.4×10⁻³ m/s
   B. 2.4×10⁻⁶ m/s
   C. 2.4 m/s
   D. 2.4×10⁻⁵ m/s
   Answer: A. 4×10⁻⁵/0.0168 ≈ 2.38×10⁻³ m/s. (Key point: Value)

3. The number density is found from:
   A. n = e v_d w t/I
   B. n = I/(e v_d w t)
   C. n = I/(e v_d L)
   D. n = I w/(e v_d)
   Answer: B. I = n e v_d (w t). (Key point: Carrier density)

4. The result is n ≈ 7.4×10²⁸ m⁻³. Which given value is not needed?
   A. The 9.5 μm thickness
   B. The 1.2 cm width
   C. The current 3.2 A
   D. The 4.0 cm length
   Answer: D. Length plays no role in Hall measurements. (Key point: Result and unused data)

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