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

# Hall voltage and mobility, copper strip

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2072 Chaitra, 5 marks
- Video: [HALL EFFECT EXPERIMENT WITH CALCULATIONS | M.Sc. | B.Sc. | PHYSICS |](https://www.youtube.com/watch?v=_bI2IDrQwfo) — Professor Bagga
- 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-three-cm
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A copper strip 3.0 cm wide and 2.0 mm thick is placed in a magnetic field 1.75T. If a current of 150 A is setup in the strip, calculate (i) Hall voltage and (ii) Hall mobility if the number of electrons per unit volume is 8.4×10²⁸ m⁻³ and resistivity is 1.72×10⁻⁸ ohm-m.

## Short answer

Hall voltage V_H = IB/(net) ≈ 9.75 μV; Hall mobility μ_H = 1/(neρ) ≈ 4.32×10⁻³ m²V⁻¹s⁻¹.

## Answer

**Formulas:** the Hall field balances the magnetic force, eE_H = e v_d B. For a strip of width w and thickness t, v_d = I/(n e w t), so

- **V_H = IB/(n e t)** (t is the thickness along B)
- Hall coefficient R_H = 1/(ne) (magnitude)
- Hall mobility μ_H = |R_H|/ρ = 1/(neρ)

**Given:** w = 3.0 cm, t = 2.0 mm, B = 1.75 T, I = 150 A, n = 8.4×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m, e = 1.602×10⁻¹⁹ C.

#### (i) Hall voltage
```text
n e t = 8.4×10²⁸ × 1.602×10⁻¹⁹ × 2×10⁻³ = 2.691×10⁷
V_H   = (150 × 1.75) / 2.691×10⁷
      = 262.5 / 2.691×10⁷
      = 9.75×10⁻⁶ 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 ≈ 9.75 μV; μ_H ≈ 4.32×10⁻³ m²V⁻¹s⁻¹**

## Check yourself

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

1. For a strip of thickness t (along B), the Hall voltage is:
   A. V_H = IB/(n e w)
   B. V_H = n e t/(IB)
   C. V_H = IB/(n e t)
   D. V_H = I/(n e t B)
   Answer: C. From E_H = v_d B and v_d = I/(n e w t), V_H = E_H w = IB/(net). (Key point: Hall voltage formula)

2. With I = 150 A, B = 1.75 T, n = 8.4×10²⁸ m⁻³, t = 2 mm, V_H is about:
   A. 9.75 μV
   B. 9.75 mV
   C. 9.75×10⁻³ μV
   D. 97.5 μV
   Answer: A. V_H = 262.5/2.691×10⁷ ≈ 9.75×10⁻⁶ V. (Key point: Hall voltage value)

3. Hall mobility is obtained from:
   A. μ_H = R_H ρ
   B. μ_H = |R_H|/ρ = 1/(neρ)
   C. μ_H = ρ/(ne)
   D. μ_H = ne/ρ
   Answer: B. Mobility is R_H times conductivity, i.e. R_H/ρ. (Key point: Hall mobility)

4. Using R_H = 7.43×10⁻¹¹ m³/C and ρ = 1.72×10⁻⁸ Ω·m, the mobility is about:
   A. 4.32×10⁻⁶ m²/(V·s)
   B. 1.28×10⁻¹⁸ m²/(V·s)
   C. 232 m²/(V·s)
   D. 4.32×10⁻³ m²/(V·s)
   Answer: D. 7.43×10⁻¹¹/1.72×10⁻⁸ ≈ 4.32×10⁻³. (Key point: Mobility value)

## Close variants set in other exams

- [Hall voltage and mobility, wide copper strip](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-numerical-copper-strip-two-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