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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2023/paper-2/section-a-ii/15/i/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 2 / Section A(ii) / 15 / i / Solution by

Codex  0 2026-10-20

Take a two-dimensional conducting sheet in the

x y

plane with magnetic flux density

B = B \hat{z}

and electric current density along

x

. Moving carriers feel the transverse Lorentz force

q v_{d} \times B

. They accumulate at opposite edges, producing a transverse electric field until its force balances the magnetic deflection:

q (E_{H} + v_{d} \times B) = 0.

For the transverse-voltage convention

E_{y} / j_{x}

, this gives

E_{y} = v_{x} B

and

\frac{E _{y}}{j _{x}} = \frac{B}{n q} .

The Hall effect therefore measures carrier sign as well as density in the single-carrier approximation: electrons give the opposite transverse-voltage sign from positive carriers. Some tensor conventions call

E_{y} / j_{x}

ρ_{y x}

rather than

ρ_{x y}

; the physical sign is fixed by the stated lead and field directions.

The Hall voltage is the integral of this transverse electric field across the sample width. In a three-dimensional slab of thickness

t

, its magnitude is

∣ V_{H} ∣ = I B / (n ∣ q ∣ t)

; for a sheet, use the sheet number density

n_{s}

and omit

t

.

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