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A transverse Hall voltage develops when a longitudinal current flows through a sample in a perpendicular magnetic field. With $R_H=E_y/(j_x{B}_z)$, a uniform slab of thickness $t$ along the magnetic field has a signed transverse voltage magnitude set by $R_{H}I{B}_z/t$, with the final sign determined by the chosen voltage leads.

The Hall-plateau voltmeter calibration is obtained as follows. Let $D_p$ be a plateau height in displayed units after any voltage offset is removed, with identified integer plateau index $p$. The true Hall voltage is $I_x{R}_K/p$, so the calibration factor would beThe nominal current is $15\mathrm{nA}$. If several plateau heights are available, their inverse-integer ratios help identify indices and a fit of $D_p$ against $1/p$ can separate a constant offset from the gain. Reversing magnetic field and antisymmetrizing the voltage also removes an even offset. A unique absolute index should not be asserted solely from an unspecified sequence of steps.

For an established exact integer $p$ and independent current and height errors,The current alone contributes $2\%$ relative uncertainty; it is a common calibration uncertainty and does not average away by fitting more plateaux. Voltage-offset uncertainty and deviations from quantization must also be included where relevant. The $25\pm 2\mu \mathrm{m}$ width cancels because the Hall voltage divided by current is a sheet Hall resistance; it would enter the transverse electric field, not this calibration.

TODO — missing source data: the held PDF's last-page stream refers to an absent /Im6 graphic, and its recovery log reports an unknown XObject. The supplied TeX records the same source gap. No plateau heights, labels or axis values survive, so neither a numerical $D_p$ nor a unique $p$ can be read. The numerical voltmeter calibration and its complete graph-reading error cannot be supplied without that graph; the formula and known $2\%$ current contribution above are determined.

A perfectly clean two-dimensional system has sharply degenerate Landau levels and integer filling only at isolated fields for fixed density. Real disorder broadens each level into a band, with states localized in the tails and extended states near its centre. When the chemical potential moves through a localized region, changing field can change the occupation of those localized states without changing the number of conducting extended channels. The Hall conductance then remains $p{e}^2/h$ and the Hall voltage at fixed current remains $I_{x}h/(p{e}^2)$ over a finite field interval.

In the same regime, separated edge channels and localized bulk states suppress dissipative backscattering, so ${\rho }_{xx}$ is very small. At a transition the chemical potential reaches the next band of extended states; longitudinal conduction rises and the Hall conductance changes to another integer. This is a mobility-gap explanation: a plateau does not require the density of states itself to vanish, nor does it require the carrier density to grow linearly with field. Low temperature, sufficient mobility and suitable contacts are needed to resolve it.

This physical explanation is determined by the surviving question text. Particular measured plateau heights and widths cannot be quoted because the source graph is absent, as recorded in part (b).

Take a two-dimensional conducting sheet in the $xy$ plane with magnetic flux density $\mathbf{B}=B\hat{\mathbf{z}}$ and electric current density along $x$. Moving carriers feel the transverse Lorentz force $q{\mathbf{v}}_d\times \mathbf{B}$. They accumulate at opposite edges, producing a transverse electric field until its force balances the magnetic deflection:For the transverse-voltage convention $E_y/j_x$, this gives $E_y=v_{x}B$ andThe 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 $E_y/j_x$ ${\rho }_{yx}$ rather than ${\rho }_{xy}$; 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 $|V_H|=IB/(n|q|t)$; for a sheet, use the sheet number density $n_s$ and omit $t$.

Take the electron's signed charge to be $e=-q$, where $q>0$ is the elementary-charge magnitude. Conventional electric current flows along $+x$, while the mean electron drift velocity is negative. With $\mathbf{B}=B_z\hat{\mathbf{z}}$, the magnetic Lorentz force separates charge across the width until a transverse electric field balances it. The accumulated edge charge then sustains a Hall voltage without a transverse steady current. In a one-carrier model, the Hall coefficient identifies the sign and number density of carriers; real multiband materials require a more careful interpretation.

The acceleration term contains the effective mass describing the local band response. The relaxation term removes drift momentum on the mean momentum-relaxation time $\tau$. The remaining two terms are the electric and magnetic Lorentz force on signed charge $e$; the magnetic force is perpendicular to velocity and does no instantaneous work. The relaxation-time approximation replaces individual scattering events by an averaged drift equation.

Let $n$ be the electron number density, distinct from the negative charge $e$. In transverse open circuit $⟨\mathbf{v}⟩=(v_x,0,0)$ and $j_x=ne{v}_x=I_x/(wt)$. The steady component equations givehenceConsequentlyThere is no sign ambiguity in this field-defined $R_H$ once axes and carrier charge are fixed.

A measured voltage ratio additionally needs probe polarities, which the geometric bar diagram does not label. To reproduce the positive ratio specified for electrons, define the longitudinal voltage drop $V_x=\phi (0)-\phi (l)=E_{x}l$ and the transverse rise $V_y=\phi (w)-\phi (0)=-E_{y}w$. ThenReversing either voltage's probes reverses the measured ratio, without changing the physical carrier density. With both voltages instead defined as drops along their positive axes, the algebraic ratio would have the opposite sign. The figure records the convention used here.

For the density, geometry gives $\rho =Rwt/l=Rt/(l/w)$ and $|R_H|=t|V_y|/(I_x|B_z|)=tR|V_y/V_x|/|B_z|$. ThusandThe mobility is $\mu =|R_H|/\rho =0.0130{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. With $m^{\ast }=m_e$,

A monovalent copper model assigns roughly one delocalized conduction electron per atom to a broad, approximately free-electron-like band. Using $m_e$ is therefore a sensible first estimate for its transport scale. Monovalence alone does not prove that every band curvature or orbit has exactly the free mass: periodic-potential mixing, the nonspherical Fermi surface and direction-dependent scattering limit the model. The calculated $\tau$ is conditional on the stipulated effective-mass approximation.

Experimentally, reverse both current and magnetic field and extract the part of the transverse voltage odd in each. This rejects fixed contact offsets and much longitudinal pickup; use four-terminal longitudinal sensing, measure dimensions independently and check heating before assigning material parameters. These controls are implemented in a reusable Hall/conductivity teaching experiment, although that experiment uses germanium rather than the copper sample here.^[ref]

Use a Hall effect measurement. Pass a known current $I$ through a thin rectangular specimen, apply a perpendicular magnetic field $B$, and measure the transverse Hall voltage $V_H$. For a uniform single-carrier sample of thickness $t$,The Hall-voltage sign identifies the carrier sign. Reverse $B$ and the current to separate the field-odd Hall signal from voltage-probe misalignment and offsets, and measure the specimen thickness accurately. The extraction assumes a one-carrier Hall factor near one. In intrinsic or mixed-carrier material, the two-carrier Hall coefficient is mobility weighted, so Hall voltage and conductivity need a two-carrier interpretation rather than this simple inverse formula.

The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental electrons.

For a single carrier type with unit Hall factor, the magnitude of the Hall voltage is $V_H=IB/(net)$, where $t$ is the slab thickness parallel to the magnetic field. ThereforeThe longitudinal electrical resistivity is $\rho =Rwt/L=1000(2\times 10^{-3})(0.5\times 10^{-3})/(5\times 10^{-3})=0.20\Omega \mathrm{m}$. Using $1/\rho =ne\mu$, the electron mobility magnitude isWithout the Hall polarity these data do not identify the carrier charge sign. The inference assumes one dominant carrier population; electron-hole compensation would require a multicarrier analysis.

In a Hall effect measurement, a current flows along $x$ and a magnetic field is applied along $z$. The transverse Lorentz force initially deflects the mobile charges, building a transverse electric field until the net transverse current vanishes. Defining $R_H=E_y/(j_x{B}_z)$, the one-carrier Drude model gives $R_H=1/(nq)$. Thus electron-like carriers give $R_H=-1/(ne)$ and hole-like carriers give $R_H=+1/(pe)$.

A nearly filled band can conduct through holes: its missing electrons behave as positive mobile charges. Metals can have both electron-like and hole-like Fermi surface sheets. In the weak-field two-carrier approximation the Hall coefficient isThe sign depends on the carrier densities and squared mobilities; it need not identify the most numerous carrier. Experimentally the Hall voltage changes sign when the field or current reverses, allowing subtraction of unwanted longitudinal-voltage pickup. Anisotropic bands and different scattering rates complicate a literal conversion of $R_H$ into a carrier count.

A current-carrying sample in a transverse magnetic field develops a transverse electric field because the Lorentz force deflects its charge carriers. In steady state this Hall voltage balances the transverse deflection. With Hall coefficient $R_H=E_y/(j_x{B}_z)$, one isotropic electron band gives $R_H=-1/(ne)$ and one hole band gives $R_H=+1/(pe)$.

When electrons and electron holes both conduct, the weak-field two-carrier result isThe two-carrier Hall coefficient weights each species by mobility squared. Its sign changes whenTemperature changes dopant ionization, intrinsic electron-hole generation, and the two charge carrier mobilities, so it can move the system across this condition. For example, a p-type semiconductor can be hole-dominated in its extrinsic regime but become negative in its intrinsic regime if $n\simeq p$ and ${\mu }_e>{\mu }_h$. A sign change is possible, not mandatory: in many n-type semiconductors electrons dominate at both temperatures. The density majority alone does not determine the Hall sign in a two-carrier material.