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

# Hall voltage, coefficient and mobility

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2068 Chaitra, 5 marks
- Video: [Solved problem related to Hall Coefficient](https://www.youtube.com/watch?v=cxrBr_2BGic) — Instant Engineering
- 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-copper-strip-thin
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A copper strip 150μm thick is placed in a magnetic field of strength 0.65T perpendicular to the plane of the strip and current of 23Amp is set up in the strip. Calculate: (i) the Hall voltage (ii) Hall coefficient and (iii) Hall mobility, if the number of electrons per unit volume is 8.5×10²⁸/m³ and resistivity is 1.72×10⁻⁸ Ohm-m.

## Short answer

V_H = IB/(n e t) = (23×0.65)/(8.5×10^28 × 1.60×10^-19 × 1.5×10^-4) ≈ 7.3 μV. Hall coefficient R_H = 1/(n e) ≈ −7.3×10^-11 m^3/C. Mobility μ_H = |R_H|/ρ ≈ 4.3×10^-3 m^2 V^-1 s^-1.

## Answer

**Formulas:** V_H = IB/(n e t); R_H = 1/(ne) (negative for electrons, magnitude used); μ_H = |R_H|/ρ.

**Given:** t = 150 μm = 1.5×10⁻⁴ m, B = 0.65 T, I = 23 A, n = 8.5×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m, e = 1.602×10⁻¹⁹ C.

#### (i) Hall voltage
```text
n e t = 8.5×10²⁸ × 1.602×10⁻¹⁹ × 1.5×10⁻⁴ = 2.0426×10⁶
V_H   = (23 × 0.65) / 2.0426×10⁶
      = 14.95 / 2.0426×10⁶
      = 7.32×10⁻⁶ V
```

#### (ii) Hall coefficient
```text
R_H = 1/(ne) = 1/(8.5×10²⁸ × 1.602×10⁻¹⁹)
    = 7.34×10⁻¹¹ m³/C   (negative sign for electrons)
```

#### (iii) Hall mobility
```text
μ_H = |R_H|/ρ = 7.34×10⁻¹¹ / 1.72×10⁻⁸
    = 4.27×10⁻³ m²/(V·s)
```

**Answer: (i) V_H ≈ 7.3 μV; (ii) R_H ≈ −7.3×10⁻¹¹ m³/C; (iii) μ_H ≈ 4.3×10⁻³ m²V⁻¹s⁻¹**

## Check yourself

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

1. For I = 23 A, B = 0.65 T, t = 150 μm, n = 8.5×10²⁸ m⁻³, V_H is about:
   A. 7.3 mV
   B. 0.73 μV
   C. 73 μV
   D. 7.3 μV
   Answer: D. 14.95/2.0426×10⁶ ≈ 7.32×10⁻⁶ V. (Key point: Hall voltage)

2. The Hall coefficient of a metal with electron carriers is:
   A. R_H = −1/(ne) ≈ −7.3×10⁻¹¹ m³/C
   B. R_H = +1/(ne)
   C. R_H = ne ≈ 1.4×10¹⁰
   D. R_H = −ne/B
   Answer: A. Negative sign for electrons. (Key point: Hall coefficient)

3. Hall mobility equals:
   A. |R_H|ρ ≈ 1.3×10⁻¹⁸
   B. |R_H|/ρ ≈ 4.3×10⁻³ m²/(V·s)
   C. ρ/|R_H| ≈ 230
   D. |R_H| B ≈ 4.8×10⁻¹¹
   Answer: B. 7.34×10⁻¹¹/1.72×10⁻⁸. (Key point: Hall mobility)

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