THE HALL EFFECT

A magnetic field pushes a current sideways, and the voltage it builds reveals the carriers' number and even their sign. Drive current I through a strip of thickness t, cross it with field B, and the Lorentz force deflects the moving charges until a transverse field balances it: the Hall voltage VH = I·B / (n·q·t). Rendered, not quoted.

source E. H. Hall, On a New Action of the Magnet on Electric Currents (1879), Am. J. Math. 2, 287–292.

Blue Team · builds & defendsthe model, its lineage, its live witness

3 The Model

In an idealized conductor the carriers drift at velocity vd under field E. A transverse B exerts the Lorentz force qvd×B, pushing carriers to one edge. Charge piles up until the transverse electric field cancels the push:

q·EH = q·vd·B  ⇒  EH = vdB

With vd = I/(n q t w) and VH = EH·w, the width w cancels:

VH = I·B / (n·q·t)  ·  RH = 1/(n·q)

Linear in I and B, inverse in carrier density n, and signed by q. That single sign was the shock of 1879.

5 The Lineage

Current pushed sideways — Hall 1879. The transverse voltage VH = IB/(nqt) reads both carrier density and sign.

Upstream: the-lorentz-force acting inside a conductor, counting the carriers. Downstream: the the-semiconductor-doping set, where the measured sign of RH separates n-type from p-type material.

the-lorentz-force → the-hall-effect → the-semiconductor-doping

7 The Witness

Live re-check of the load-bearing law: the Hall voltage of a conductor with 2n carriers must be exactly half that of one with n. Re-run continuously; flips red the instant window 6 tampers.

…

asserts VH(2n) = ½·VH(n) — the inverse-in-n signature

The Machinedata in → the panel → data out

4 Data In in ↓

A strip of chosen material carrying current I, in a field B, of thickness t. Choose the carrier sign.

↓ LORENTZ DEFLECTION ↓

0 The Panel lit

Transverse Hall voltage
—
drift velocity vd
—
Hall field EH
—
Hall coeff RH
—
n recovered from RH
—
Sign read-out: —
↓ PROVEN RESULT ↓

8 Data Out out ↓

The voltmeter across the strip reads VH; its magnitude gives n = 1/(RHq) and its polarity gives the carrier sign — a full carrier census from two wires.

…
Red Team · attacks & breaksthe adversary, the graveyard, the tamper

1 The Adversary wall

"Voltage across a wire? Ohm's law already gives V = IR — the field does nothing new." Wrong axis. The Hall voltage is transverse, perpendicular to both I and B, and vanishes at B = 0. It is not a resistive drop along the current; it is the standing transverse field that balances the magnetic deflection. No B, no VH.

"Then it just measures resistance." No — it measures the product nq. Two metals of equal resistivity but different carrier density give different VH. Resistance alone cannot separate n from mobility; the Hall voltage can.

2 The Graveyard

  • "Charge carriers must be negative electrons, so RH is always negative."Some conductors (Zn, Cd, many p-type semiconductors) show positive RH — conduction by holes. The Hall sign is the direct evidence.
  • "VH depends on the strip's width w."w cancels: VH = IB/(nqt). Only thickness t (along B) survives.
  • "The free-electron / Drude single-carrier picture is exact."amberIt is a classical idealization. With two carrier types or energy-dependent scattering, RH becomes a weighted average and can even change sign with field — the simple 1/(nq) is the clean limit, not the whole story.

6 The Tamper

Plant the disclosed void: drop the carrier density from the law, so VH = IB/(qt), independent of n. Now a denser conductor wrongly reports the same voltage — the measurement can no longer read n, and RH = 1/(nq) is broken. The Witness (7) catches it live.

law intact