---
title: "Hall effect, coefficient and mobility"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:hall-effect-coefficient-mobility-conductivity"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-effect-coefficient-mobility-conductivity
site: BCT Engineering
---

# Hall effect, coefficient and mobility

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2073 Shrawan, 5 marks
- Video: [Hall Effect | Origin & Experimental Observation || Expression of Hall Coefficient & Hall Mobility](https://www.youtube.com/watch?v=HmeJh30oFf0) — VMG Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-effect-coefficient-mobility-conductivity
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

What is Hall-effect? Derive an expression for the Hall coefficient and establish the relation between mobility of charge carrier and conductivity of material of wire.

## Short answer

A transverse Hall voltage develops when a conductor carrying current I is placed in a perpendicular magnetic field B. Force balance gives Hall coefficient R_H = E_H/(JB) = 1/(nq). Since conductivity σ = neμ, the carrier mobility is μ = σ|R_H|.

## Answer

#### Hall effect
When a conductor carrying current I is placed in a magnetic field B perpendicular to the current, a transverse electric field, and hence a potential difference (Hall voltage), appears across the conductor, perpendicular to both I and B.

#### Hall coefficient
Strip of width w, thickness t, carrying current I along x with field B along z. Carriers: charge q (magnitude e), density n, drift velocity v_d.

```text
   strip: width w, thickness t
   current I along x, field B along z (thickness)
   Hall voltage appears across the width w
```

1. Magnetic force deflects carriers sideways, charges accumulate on the edges, and a Hall field E_H builds up until the forces balance: eE_H = e v_d B → E_H = v_d B
2. Current density J = n e v_d, so v_d = J/(ne)
3. E_H = J B/(n e)
4. The **Hall coefficient** is defined as R_H = E_H/(J B), so

**R_H = 1/(n q)**

(R_H is negative for electrons and positive for holes. R_H = −1/(ne) for electrons.) In terms of the Hall voltage, V_H = E_H w = R_H I B / t.

#### Mobility and conductivity
1. Conductivity: σ = J/E = n e v_d/E.
2. Mobility is the drift velocity per unit electric field: μ = v_d/E.
3. So σ = n e μ.
4. Multiply by R_H: |R_H| σ = (1/ne)(neμ)

**μ = σ |R_H|**

Therefore, measuring R_H (from Hall voltage) and σ (from resistance) gives n and the carrier mobility of the wire material.

## Check yourself

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

1. Hall field E_H equals
   A. v_d B
   B. B/v_d
   C. v_d/B
   D. J B
   Answer: A. Forces balance: eE_H = e v_d B. (Key point: Balance)

2. The Hall coefficient is
   A. n q
   B. 1/(nq)
   C. nq/B
   D. B/(nq)
   Answer: B. R_H = E_H/(JB) = 1/(nq). (Key point: Coefficient)

3. Conductivity is
   A. σ = n e μ
   B. σ = n μ/e
   C. σ = e μ/n
   D. σ = μ/(ne)
   Answer: A. With μ = v_d/E, σ = J/E = n e μ. (Key point: Conductivity)

4. The mobility is
   A. μ = σ/|R_H|
   B. μ = σ|R_H|
   C. μ = |R_H|/σ
   D. μ = nσ
   Answer: B. |R_H|σ = (1/ne)(neμ) = μ. (Key point: Mobility)

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