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

# Hall voltage and mobility, wide copper strip

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2068 Shrawan, 5 marks
- 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-two-cm
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A copper strip 2 cm wide and 10mm thick is placed in a magnetic field of 1.5T. If a current of 200A is set up in the strip, calculate (i) Hall voltage (ii) Hall mobility if the no. of electrons per unit volume is 8.4×10²⁸/m³ and resistivity is 1.72×10⁻⁸ Ohm-m.

## Short answer

V_H = IB/(n e t) = (200×1.5)/(8.4×10^28 × 1.60×10^-19 × 0.01) ≈ 2.2×10^-6 V. Hall mobility μ_H = 1/(n e ρ) = 1/(8.4×10^28 × 1.60×10^-19 × 1.72×10^-8) ≈ 4.3×10^-3 m^2 V^-1 s^-1.

## Answer

**Formulas:** V_H = IB/(n e t) with t the thickness along B; Hall mobility μ_H = |R_H|/ρ = 1/(neρ).

**Given:** w = 2 cm, t = 10 mm = 1×10⁻² m (as printed), B = 1.5 T, I = 200 A, n = 8.4×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m.

#### (i) Hall voltage
```text
n e t = 8.4×10²⁸ × 1.602×10⁻¹⁹ × 1×10⁻² = 1.3457×10⁸
V_H   = (200 × 1.5) / 1.3457×10⁸
      = 300 / 1.3457×10⁸
      = 2.23×10⁻⁶ V
```

#### (ii) Hall mobility
```text
R_H = 1/(ne) = 1/(8.4×10²⁸ × 1.602×10⁻¹⁹) = 7.43×10⁻¹¹ m³/C
μ_H = R_H/ρ = 7.43×10⁻¹¹ / 1.72×10⁻⁸ = 4.32×10⁻³ m²/(V·s)
```

**Answer: V_H ≈ 2.2 μV; 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. Hall voltage for a strip is:
   A. V_H = IB/(net)
   B. V_H = IB/(new)
   C. V_H = ne tB/I
   D. V_H = IBt/(ne)
   Answer: A. t is the thickness along B. (Key point: Formula)

2. For I = 200 A, B = 1.5 T, t = 10 mm, n = 8.4×10²⁸ m⁻³, V_H is about:
   A. 2.2 mV
   B. 2.2 μV
   C. 22 μV
   D. 0.22 μV
   Answer: B. 300/1.3457×10⁸ ≈ 2.23×10⁻⁶ V. (Key point: Hall voltage)

3. The Hall mobility is:
   A. R_H ρ ≈ 1.3×10⁻¹⁸
   B. ρ/R_H ≈ 230
   C. R_H/ρ = 1/(neρ) ≈ 4.3×10⁻³ m²/(V·s)
   D. n e ρ ≈ 232
   Answer: C. 7.43×10⁻¹¹/1.72×10⁻⁸ ≈ 4.32×10⁻³. (Key point: Mobility)

## Close variants set in other exams

- [Hall voltage and mobility, copper strip](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-numerical-copper-strip-three-cm) — Engineering Physics, chapter 8, asked 2×

## Related in Engineering Physics

- [Hall voltage, coefficient and mobility](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-numerical-copper-strip-thin) — Engineering Physics, chapter 8
- [Hall voltage and mobility in silver](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-numerical-silver-foil) — Engineering Physics, chapter 8