Physics · Electricity and magnetism

Hall effect and Hall voltage

Q: What do the Hall effect and Hall voltage describe physically?

Hall voltage arises from transverse charge separation: magnetic force on drifting carriers is balanced by a transverse electric field in steady state. For positive conventional current along x and positive field along z, we define voltage as lower potential minus upper potential. Electrons then give negative voltage, positive carriers positive voltage. The result uses thickness parallel to the field, not transverse contact separation. The simple relation connects this geometry with positive volume density and signed carrier charge. It assumes one dominant carrier type and does not describe arbitrary complex material response.

Q: How can I check a calculation of Hall voltage?

Record current direction, field direction, probe polarity and active thickness before interpreting a sign. For $I=1.0\,\mathrm A$, $B_z=0.50\,\mathrm T$, $n=8.5\cdot10^{28}\,\mathrm{m^{-3}}$, $|q|=e$ and $t=1.0\,\mathrm{mm}$, the magnitude is 36.7 nanovolts. This equals 0.0367 microvolts, not 36.7 microvolts; that would differ by a factor of one thousand. Electrons give negative voltage under our polarity definition. Also check scaling: doubling current or field doubles magnitude, doubling thickness halves it. A tiny signal requires checking contact misalignment and offsets before extracting density. Reversing probe polarity changes the voltage sign alone.

Q: What assumptions and limits apply to the Hall-voltage formula?

The classical equation requires a homogeneous sample with one dominant carrier type, known active thickness and negligible transverse measurement current. A steady Hall field must balance the sideways force. With multiple species, mobilities also affect the coefficient, so the signal does not simply yield their total density. Nonuniform layers or large, misplaced contacts may violate the simple geometry. Temperature, magnetisation and nonlinear field response can require extended models. Field reversal helps remove even-in-field offsets but not every disturbance. An experimental density should therefore always be reported together with the assumptions used to infer it.

Q: How do I check units and scale for Hall voltage?

Enter t in metres and n as carriers per cubic metre. Sheet density instead has inverse-square-metre units and must not be multiplied by the same thickness again. In SI, $\frac{\mathrm{A\,T}}{\mathrm{m^{-3}\,C\,m}}$ reduces to volts. Dimensions alone cannot distinguish width and thickness because both are lengths; that needs the derivation’s geometry. For scale, carefully distinguish nano-, micro- and millivolts: the existing metal-like inputs produce only nanovolts. A negative particle density always signals incorrect magnitude or sign handling; the carrier sign belongs to q.

Q: When should I combine Hall voltage with another relation?

Current density $J_x=nqv_{{\rm d},x}$ connects Hall measurement to signed drift. Conversely, known carrier density permits magnetic-field determination. With $n_s=nt$, infer sheet density of a homogeneous thin layer even without knowing active thickness. A resistance measurement may add mobility information under a suitable model, but does not measure the same quantity as Hall voltage. Field reversal splits measured voltage into odd and even components. Only when conditions remain fixed and Hall response is antisymmetric can the odd term be directly identified with a corrected Hall signal; it does not remove every possible measurement artifact.

The Hall voltage appears when a magnetic field deflects moving charge carriers sideways and builds a transverse charge separation.

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Formula

Calculate

U_{H} = \frac{I B _{z}}{n qt}

I in A; Bz in T; n in m⁻³; q in C; t in m; UH in V.

Variables & units

Variables & units
Symbol	Meaning	Unit
$U_{\mathrm{H}}$	Signed voltage V(y−)−V(y+) transverse to current and magnetic field	V
$I$	Current through the sample	A
$B_{\mathrm{z}}$	Signed z component of magnetic field perpendicular to x-directed current	T
$n$	Density of the dominant carrier type	m⁻³
$q$	Signed carrier charge	C
$t$	Sample thickness along the field	m

LaTeX

U_{\rm H}=\frac{IB_z}{nqt}

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Hall plate with +y upward, +x rightward and field dots out of the plane. Upper edge positive and lower edge negative. Electron drift points left, magnetic force down, electric force up. Positive probe terminal below gives negative voltage. — Original electron convention: conventional I along +x, B along +z out of the plane. Electron drift along −x; negative charge below. UH=V(lower)−V(upper)≈−3.121 mV for I=5 mA, B=0.40 T, n=2·10²² m⁻³, t=0.20 mm. Thickness t lies along z, width w along y; not drawn to scale.

Hall voltage calculator

Find the Hall voltage, current, magnetic field or carrier density of a strip with one type of charge carrier, with a fixed probe polarity.

Calculation model

Find

Current I (positive along +x)

Current along the strip, positive in the +x direction, e.g. 5 mA.

Magnetic field Bz (positive along +z)

Magnetic field perpendicular to the strip, positive along +z, e.g. 0.4 T.

Signed charge q of one carrier

Charge of one carrier with its sign: −1 e for electrons, +1 e for holes.

Carrier density n

Charge carriers per cubic metre, e.g. 2·10²² m⁻³ in a doped semiconductor, about 8.5·10²⁸ m⁻³ in copper.

= 2 · 10²²

Thickness t along the field

Thickness of the strip along the field, not its width, e.g. 0.2 mm.

Filled with example values. Type your own and the result updates straight away.

Result: Hall voltage: lower minus upper UH

UH = -3.121 mV

With I = 5 mA, Bz = 0.4 T, q = -1 e, n = 2 · 10²² m⁻³ and t = 0.2 mm, UH = -3.121 mV.

Working · rearrange for UH, substitute, evaluate

1. Rearrange for UH
$U_{\rm H}=\frac{IB_z}{nqt}$
2. Substitute values (in SI units)
- I = 5 mA = 0.005 A
- q = -1 e = -1.60218 · 10⁻¹⁹ C
- t = 0.2 mm = 0.0002 m
$\frac{0.005\cdot 0.4}{2\times10^{22}\cdot (-1.60218\times10^{-19})\cdot 0.0002}=-0.00312075$
3. Result
UH = -3.121 mV

Quick tips

The width of the strip cancels out: UH = IB/(nqt) only contains the thickness t along the field.
The thinner the strip and the smaller n, the larger the magnitude of UH. That is why probes use semiconductors: in metals UH often stays at microvolts.

Assumptions and limits

Homogeneous classical one-carrier plate: conventional +x current reference, +z field reference, thickness along z, transverse width along y, UH=V(y−)−V(y+). n>0,t>0, signed q≠0; electrons have q=−e. Open transverse branch and stationary mean force balance. Uniform field and suitable contacts. Not a multicarrier, quantum-Hall or unverified sensor-calibration model.

Derivation & background

Homogeneous classical one-carrier model, steady transverse force balance, negligible transverse load current and known t parallel to the field. Polarity UH=V(y−)−V(y+) with current along x and field along z. Multicarrier transport, nonuniform active layers, large contacts and nonlinear/material-dependent Hall effects need extended models.

Exam blueprint

Validity range

Derivation steps

Transverse force balance sets the Hall field; current density connects signed drift and carrier density.

1Define I along +x, Bz along z, width w along y and thickness t along z.
2Use polarity $U_{\rm H}=V(y_-)-V(y_+)$ ; q carries the sign of the dominant carrier type.
3The steady open transverse branch satisfies $\mathbf E_{\rm H}=-\mathbf v_{\rm d}\times\mathbf B$ .
4The cross product gives $E_y=v_{{\rm d},x}B_z$ ; obtain drift from $I=nqv_{{\rm d},x}wt$ .
5Since $\frac{\partial V}{\partial y}=-E_y$ , the defined potential difference is $U_{\rm H}=E_yw$ .
6Width cancels: $U_{\rm H}=\frac{IB_z}{nqt}$ . Negative q gives negative voltage at positive I and Bz under this polarity.

Rearrangements

Volume density from magnitude

n=\frac{|IB_z|}{|q|t|U_{\rm H}|}

n is positive. Requires nonzero signal, active thickness and one-carrier model; check probe polarity separately.

Task variant

A homogeneous one-carrier electron model has $I=1.0\,\mathrm A$ along +x, $B_z=+0.50\,\mathrm T$ , $n=8.5\cdot10^{28}\,\mathrm{m^{-3}}$ , $q=-e$ , $t=1.0\,\mathrm{mm}$ along z. Find $U_{\rm H}=V(y_-)-V(y_+)$ in the open transverse branch.

Approach: $U_{\rm H}=\frac{IB_z}{nqt}$ .
With $t=0.001\,\mathrm m$ and $q=-1.602176634\cdot10^{-19}\,\mathrm C$ , $U_{\rm H}\approx-3.6715\cdot10^{-8}\,\mathrm V=-36.715\,\mathrm{nV}$ .
Lower potential is below upper potential; magnitude is about 36.7 nV.
This uses a steady homogeneous one-carrier model with the specified probe polarity.

Common mistakes

Use sample width as thickness or interpret voltage sign without defining measurement polarity.

Define current, field and voltage probes. Active thickness runs along the magnetic field; carrier charge is signed.

Exam context

Find carrier type, volume or sheet density, and offset-corrected Hall voltage from field reversal.

Formula cluster

Electricity and magnetism

Lorentz force and the transverse electric field explain the Hall effect; current density connects drift to measured carrier density.

F = B|I|ℓ sinθ

Magnetic force on a current-carrying wire

F = q(v × B)

Lorentz Force

R = ρ·l/A

Resistivity (Conductor Resistance)

Worked example

For $I=1.0\,\mathrm A$ , $B_z=0.50\,\mathrm T$ , $n=8.5\cdot10^{28}\,\mathrm{m^{-3}}$ , $|q|=e$ and $t=1.0\,\mathrm{mm}$ , $|U_{\rm H}|\approx3.67\cdot10^{-8}\,\mathrm V=36.7\,\mathrm{nV}$ . For electrons with the specified polarity UH is negative.

Applications

Measure magnetic fields and estimate carrier type and density in conductors and semiconductors.

How do you calculate with Hall Effect and Hall Voltage?

Here is how to work through a typical Hall Effect and Hall Voltage (UH = IB/(nqt)) task step by step:

1
Task
A homogeneous one-carrier electron model has $I=1.0\,\mathrm A$ along +x, $B_z=+0.50\,\mathrm T$ , $n=8.5\cdot10^{28}\,\mathrm{m^{-3}}$ , $q=-e$ , $t=1.0\,\mathrm{mm}$ along z. Find $U_{\rm H}=V(y_-)-V(y_+)$ in the open transverse branch.
Solution path
1. Approach: $U_{\rm H}=\frac{IB_z}{nqt}$ .
2. With $t=0.001\,\mathrm m$ and $q=-1.602176634\cdot10^{-19}\,\mathrm C$ , $U_{\rm H}\approx-3.6715\cdot10^{-8}\,\mathrm V=-36.715\,\mathrm{nV}$ .
3. Lower potential is below upper potential; magnitude is about 36.7 nV.
4. This uses a steady homogeneous one-carrier model with the specified probe polarity.

Understand and apply the equation

Hall voltage makes transverse charge separation measurable. Its magnitude connects current, magnetic field and carrier density; its sign becomes meaningful only with a complete axis and probe convention. The simple equation is a one-carrier model, not a universal material diagnosis.

A small transverse voltage reveals carrier information

Current flows lengthwise through a thin plate. A perpendicular magnetic field deflects moving carriers sideways. Edge charge builds an opposing electric field. Once transverse forces balance on average, a voltage can be measured between the sides. Inferring carrier type from its sign first requires current direction, field direction and the order of voltage probes.[1][2]

Prior knowledge: for negative carriers, conventional current opposes drift motion.
Width and thickness play different roles. Transverse width is probe separation; active thickness runs parallel to the magnetic field.
Use positive number density n and signed carrier charge q. Negative density would not be a physical answer.

[1][2]

Which probe polarity and thickness belong in the Hall-voltage equation?

Consider a homogeneous rectangular plate. Conventional current $I>0$ flows along +x and $B_z>0$ points along +z. Width w lies along y, thickness t along z. Define Hall voltage as the lower y− edge potential minus the upper y+ edge potential. This fixes probe polarity completely. Carrier charge q is signed; electrons have $q=-e$ . Density n counts carriers per volume and is positive.[1]

\begin{aligned}U_{\rm H}&=V(y_-)-V(y_+)\\J_x&=\frac I{wt}=nqv_{{\rm d},x}\\U_{\rm H}&=\frac{IB_z}{nqt}\end{aligned}

This signed form uses the axes and contacts specified here.[1]

Reversing voltage leads changes the measured sign without changing carrier type. Reversing current or field also changes the sign in this model. Thus “negative Hall voltage means electrons” is incomplete without current direction, field direction and positive probe terminal. A different voltage definition writes the same physics with an extra minus sign.[1]

Same probes, reversed current

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Probe polarity remains lower minus upper. Curves use n=2×10²² m⁻³, t=0.20 mm and I=±5 mA as in the first example. At Bz=+0.40 T, voltages are −3.121 mV for +I and +3.121 mV for −I. Lines show ideal corrected signal without offset.[1]

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From Lorentz force to transverse voltage

Magnetic force deflects drifting carriers sideways. Edge charge builds a Hall field until transverse electric and magnetic forces balance in steady state. This concerns average drift, not each carrier’s thermal motion. The open transverse branch carries no steady net transverse current. The measuring instrument should therefore not disturb the voltage with a substantial transverse load current. The Lorentz force must therefore be checked using the specified drift and field directions before interpreting the probe voltage.[1]

Force balance first determines transverse field, not voltage. Current then determines mean drift speed. At fixed current, a wider plate has slower drift but a longer transverse distance between probes. These width effects cancel when converting field to voltage. Thickness remains in the result.[1]

\begin{aligned}\mathbf E_{\rm H}&=-\mathbf v_{\rm d}\times\mathbf B\\E_y&=v_{{\rm d},x}B_z\\v_{{\rm d},x}&=\frac I{nqwt}\\U_{\rm H}&=E_yw=\frac{IB_z}{nqt}\end{aligned}

Since $\frac{\partial V}{\partial y}=-E_y$ , $V(y_-)-V(y_+)=E_yw$ . Width cancels.[1]

At equal conventional current, negative carriers drift against the current arrow. With electrons under our convention magnetic force points downward and the Hall field also points downward, while electric force on electrons points upward. Negative charge accumulates below and positive charge above. Positive carriers at fixed current and field still experience downward magnetic force, but edge charge and Hall-field signs reverse.[1]

Explore further: Hall coefficient and sheet density

\begin{aligned}R_{\rm H}&=\frac{E_y}{J_xB_z}=\frac1{nq}\\U_{\rm H}&=R_{\rm H}\frac{IB_z}{t}\\n&=\frac{|IB_z|}{|q|t|U_{\rm H}|}\\\left[R_{\rm H}\right]&=\mathrm{m^3\,C^{-1}}\\\left[U_{\rm H}\right]&=\mathrm V\end{aligned}

One-carrier model; density inversion needs a nonzero signal and known geometry.[1][2]

Thickness t is measured parallel to the field, not between transverse contacts. Density n is volumetric. A thin homogeneous conducting layer can use sheet density $n_s=nt$ , eliminating t from the voltage formula. That redefines density; it does not establish quantum Hall behaviour. The following calculations use the exact SI elementary charge e=1.602176634·10⁻¹⁹ coulomb.[2][3]

Example: electron drift and Hall polarity

Connect all directions in one calculation

Given: One-carrier electron model: $n=2.0\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=5.0\,\mathrm{mA}$ along +x, $B_z=0.40\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ , $w=3.0\,\mathrm{mm}$ .
Find: Find drift component, Hall field and lower-minus-upper voltage.

Model: use signed charge in current density and force balance.
$\begin{aligned}v_{{\rm d},x}&=\frac I{nqwt}\\E_y&=v_{{\rm d},x}B_z\\U_{\rm H}&=E_yw\end{aligned}$
Substitute $q=-e$ ; separately convert width and thickness to metres.
$\begin{aligned}v_{{\rm d},x}&=\frac{0.005}{(2.0\cdot10^{22})(-1.602176634\cdot10^{-19})(0.003)(0.0002)}\,\mathrm{m\,s^{-1}}\\&\approx-2.60063\,\mathrm{m\,s^{-1}}\\E_y&\approx-1.04025\,\mathrm{V\,m^{-1}}\\U_{\rm H}&\approx-3.12075\,\mathrm{mV}\end{aligned}$

Result: Electrons drift against +x. The field points toward −y; the lower potential is 3.121 mV below the upper potential.

Check: Check: negative charge times negative drift gives positive current. Direct voltage substitution gives the same −3.121 mV. Upward electric and downward magnetic forces balance.

What does field reversal separate from the Hall signal, and what still needs checking?

Misaligned contacts can pick up longitudinal voltage; thermoelectric and instrument offsets may contribute too. Field reversal separates an odd-in-field Hall component from an even component if current, temperature and geometry remain fixed and Hall response is antisymmetric at the two field magnitudes. Not every disturbance is even in field, so reversal does not replace contact checks or a suitable material model. Electric field strength and measured probe voltage remain different quantities; relating them also requires transverse separation and a suitable field model.[2]

\begin{aligned}U_{\rm odd}(B)&=\frac{U_{\rm meas}(B)-U_{\rm meas}(-B)}2\\U_{\rm even}(B)&=\frac{U_{\rm meas}(B)+U_{\rm meas}(-B)}2\end{aligned}

Algebraic odd/even decomposition; assigning the odd term to Hall voltage requires the stated measurement assumptions.[2]

Example: field reversal and density extraction

Remove a constant offset

Given: $I=10\,\mathrm{mA}$ , $t=0.50\,\mathrm{mm}$ ; measured voltage is $+4.6\,\mathrm{mV}$ at $B_z=+0.50\,\mathrm T$ , and $-3.4\,\mathrm{mV}$ at $-0.50\,\mathrm T$ . One-carrier model with $|q|=e$ , constant even offset and fixed probe polarity.
Find: Find Hall component, offset, carrier sign and density.

Model: the difference removes offset and the sum removes Hall signal. A positive Hall coefficient here identifies positive carriers.
$\begin{aligned}U_{\rm H}&=\frac{U_+-U_-}2\\U_0&=\frac{U_++U_-}2\\n&=\frac{|IB_z|}{et|U_{\rm H}|}\end{aligned}$
Substitute the corrected 4.0 mV into the density equation.
$\begin{aligned}U_{\rm H}&=\frac{4.6-(-3.4)}2\,\mathrm{mV}=4.0\,\mathrm{mV}\\U_0&=0.6\,\mathrm{mV}\\R_{\rm H}&=\frac{(0.004)(0.0005)}{(0.010)(0.50)}\,\mathrm{m^3\,C^{-1}}\\&=4.0\cdot10^{-4}\,\mathrm{m^3\,C^{-1}}\\n&\approx1.56038\cdot10^{22}\,\mathrm{m^{-3}}\end{aligned}$

Result: The offset is +0.6 mV. Under the specified one-carrier model the data indicate positive carriers with n≈1.56·10²² per cubic metre.

Check: Check: $(+4.0+0.6)\,\mathrm{mV}=+4.6\,\mathrm{mV}$ and $(-4.0+0.6)\,\mathrm{mV}=-3.4\,\mathrm{mV}$ reproduce both measurements. Density is positive; the coefficient carries the carrier sign. An uncorrected reading would imply a different apparent density.

Four readings separate three contributions

Given: At equal current and field magnitudes, readings are $U_{++}=-1.50\,\mathrm{mV}$ , $U_{+-}=+2.50\,\mathrm{mV}$ , $U_{-+}=+1.90\,\mathrm{mV}$ and $U_{--}=-2.10\,\mathrm{mV}$ . Signs denote current and field. Model: Hall term H is odd in both, contact term L is odd only in current, and offset O is constant.
Find: Find H, L and O for positive current and field and check all four readings.

Approach: separate readings by their behaviour under current and field reversal.
$\begin{aligned}U(s_I,s_B)&=H s_Is_B+L s_I+O\\H&=\frac{U_{++}-U_{+-}-U_{-+}+U_{--}}4\\L&=\frac{U_{++}+U_{+-}-U_{-+}-U_{--}}4\\O&=\frac{U_{++}+U_{+-}+U_{-+}+U_{--}}4\end{aligned}$
Substitute all readings in millivolts.
$\begin{aligned}H&=\frac{-1.50-2.50-1.90-2.10}4\,\mathrm{mV}=-2.00\,\mathrm{mV}\\L&=\frac{-1.50+2.50-1.90+2.10}4\,\mathrm{mV}=+0.300\,\mathrm{mV}\\O&=\frac{-1.50+2.50+1.90-2.10}4\,\mathrm{mV}=+0.200\,\mathrm{mV}\end{aligned}$

Result: Under the stated model, Hall signal at positive current and field is −2.00 mV. The current-odd contact contribution is +0.300 mV and constant offset +0.200 mV.

Check: Back-substitution: −2+0.3+0.2=−1.5; +2+0.3+0.2=+2.5; +2−0.3+0.2=+1.9; −2−0.3+0.2=−2.1 mV. Another disturbance odd in both current and field would not be distinguishable from Hall signal by this procedure.

When Hall voltage is not a simple carrier count

The relation assumes a homogeneous classical one-carrier model, known active thickness and suitable contacts. Multiple carrier species cannot simply be combined into a total count with one q: their mobilities influence the measured Hall coefficient. Anisotropic transport, magnetisation effects and nonlinear field response may also invalidate a simple density interpretation. Outside this model a positive signal does not automatically establish a single positive carrier population.[1][2]

Reliable measurements record temperature, contact size, uniformity, field, current and active layer thickness. NIST discusses small ohmic contacts, uniform samples and separate Hall/resistance measurements. The cited page is explicitly archived, not a continuously updated measurement standard. The equation explains the principle; calibrating a real sensor requires its actual measurement conditions.[2]

Five exercises on calculation, polarity and measurement models

Calculate · Foundation

Positive carriers with $q=+e$ , $n=5\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=20\,\mathrm{mA}$ , $B_z=0.20\,\mathrm T$ , $t=0.10\,\mathrm{mm}$ . Find U_H under the article convention.

Show hint

Convert thickness to metres; use lower minus upper potential.

Learning goal: Calculate a signed Hall voltage.

Show solution and calculation

Model and approach: One dominant carrier type and open transverse branch; polarity lower minus upper.
$U_{\mathrm H}=\frac{IB_z}{nqt}$ Use the one-carrier voltage equation with positive q.
Substitution gives +4.9932 mV; the lower edge is at higher potential.
Reversing current at unchanged field would give −4.9932 mV.

U_{\rm H}=\frac{(0.020)(0.20)}{(5\cdot10^{22})(1.602176634\cdot10^{-19})(0.0001)}\,\mathrm V\approx4.9932\,\mathrm{mV}

[1]

Rearrange · Application

An electron model gives $|U_{\rm H}|=1.0\,\mathrm{mV}$ at $I=5.0\,\mathrm{mA}$ , $B_z=0.30\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ . Find n.

Show hint

Use magnitudes; a particle density is not negative.

Learning goal: Extract a positive volume density from magnitude.

Show solution and calculation

Model and approach: Electron model; n is positive and charge sign is checked separately.
$n=\frac{|IB_z|}{et|U_{\mathrm H}|}$ Isolate n from the magnitude equation.
The result is n≈4.6811·10²² per cubic metre.
Using $q=-e$ in the signed equation gives −1.0 mV, matching the measured magnitude.

n=\frac{(0.005)(0.30)}{(1.602176634\cdot10^{-19})(0.0002)(0.001)}\,\mathrm{m^{-3}}\approx4.6811\cdot10^{22}\,\mathrm{m^{-3}}

[1]

Interpret · Foundation

I points along +x and B along +z. Positive edge charge accumulates above and negative charge below. What drift, field and voltage directions follow in the one-carrier model?

Show hint

The Hall field points from positive to negative edge charge.

Learning goal: Interpret carrier sign only with full measurement convention.

Show solution and calculation

The field points along −y, so $U_{\mathrm H}=V(y_-)-V(y_+)$ is negative.
Positive I and B imply $q<0$ .
Negative drift times negative charge gives positive conventional current.
Swapping probe leads changes the reading but neither drift nor carrier type.

E_y<0,\quad U_{\rm H}<0,\quad q<0,\quad v_{{\rm d},x}<0

[1]

Error analysis · Application

With $w=3.0\,\mathrm{mm}$ and $t=0.20\,\mathrm{mm}$ , a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?

Show hint

Width appears in both current density and potential difference.

Learning goal: Distinguish magnetic-field thickness from contact width.

Show solution and calculation

The correct model uses t parallel to the field.
The wrong result is $\frac tw=\frac1{15}$ of the correct magnitude: 3.121 mV would become about 0.208 mV.
Width cancels because a wider plate at fixed current has smaller drift speed but a longer transverse distance.

\frac{|U_{\rm wrong}|}{|U_{\rm correct}|}=\frac tw=\frac{0.20}{3.0}=\frac1{15}

[1]

Transfer · Challenge

A thin homogeneous layer gives $|U_{\rm H}|=2.0\,\mathrm{mV}$ at $I=10\,\mathrm{mA}$ , $|B_z|=0.20\,\mathrm T$ . In a one-carrier model with $|q|=e$ , find sheet density n_s without knowing thickness. Can n also be obtained?

Show hint

Combine n and t into n_s.

Learning goal: Distinguish volume density and sheet density.

Show solution and calculation

Use $n_s=nt$ in the magnitude relation.
The measurement gives n_s≈6.2415·10¹⁸ per square metre.
Volume density also needs active thickness: $n=\frac{n_s}{t}$ .
The signal alone cannot separate n and t.
The inverse-square-metre unit confirms that a volume density has not been claimed.

n_s=\frac{|IB_z|}{e|U_{\rm H}|}\approx6.2415\cdot10^{18}\,\mathrm{m^{-2}}

[2]

What to take away

Voltage sign is meaningful only with current, field and probe conventions. Swapping probes does not change carrier type.
Transverse width cancels; active thickness parallel to the magnetic field remains.
Reversal measurements separate specific symmetry components. They do not automatically remove every artifact or fix an unsuitable multicarrier model.

[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed Hall polarity, signed drift and offset removal; corrected the existing metal example to 36.7 nV; added two examples and five exercises.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

Cite this reference page

Quanta. Hall Effect and Hall Voltage. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/hall-effekt-spannung

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Hall Effect and Hall Voltage

What do the Hall effect and Hall voltage describe physically?+

Hall voltage arises from transverse charge separation: magnetic force on drifting carriers is balanced by a transverse electric field in steady state. For positive conventional current along x and positive field along z, we define voltage as lower potential minus upper potential. Electrons then give negative voltage, positive carriers positive voltage. The result uses thickness parallel to the field, not transverse contact separation. The simple relation connects this geometry with positive volume density and signed carrier charge. It assumes one dominant carrier type and does not describe arbitrary complex material response.

How can I check a calculation of Hall voltage?+

Record current direction, field direction, probe polarity and active thickness before interpreting a sign. For $I=1.0\,\mathrm A$ , $B_z=0.50\,\mathrm T$ , $n=8.5\cdot10^{28}\,\mathrm{m^{-3}}$ , $|q|=e$ and $t=1.0\,\mathrm{mm}$ , the magnitude is 36.7 nanovolts. This equals 0.0367 microvolts, not 36.7 microvolts; that would differ by a factor of one thousand. Electrons give negative voltage under our polarity definition. Also check scaling: doubling current or field doubles magnitude, doubling thickness halves it. A tiny signal requires checking contact misalignment and offsets before extracting density. Reversing probe polarity changes the voltage sign alone.

What assumptions and limits apply to the Hall-voltage formula?+

The classical equation requires a homogeneous sample with one dominant carrier type, known active thickness and negligible transverse measurement current. A steady Hall field must balance the sideways force. With multiple species, mobilities also affect the coefficient, so the signal does not simply yield their total density. Nonuniform layers or large, misplaced contacts may violate the simple geometry. Temperature, magnetisation and nonlinear field response can require extended models. Field reversal helps remove even-in-field offsets but not every disturbance. An experimental density should therefore always be reported together with the assumptions used to infer it.

How do I check units and scale for Hall voltage?+

Enter t in metres and n as carriers per cubic metre. Sheet density instead has inverse-square-metre units and must not be multiplied by the same thickness again. In SI, $\frac{\mathrm{A\,T}}{\mathrm{m^{-3}\,C\,m}}$ reduces to volts. Dimensions alone cannot distinguish width and thickness because both are lengths; that needs the derivation’s geometry. For scale, carefully distinguish nano-, micro- and millivolts: the existing metal-like inputs produce only nanovolts. A negative particle density always signals incorrect magnitude or sign handling; the carrier sign belongs to q.

When should I combine Hall voltage with another relation?+

Current density $J_x=nqv_{{\rm d},x}$ connects Hall measurement to signed drift. Conversely, known carrier density permits magnetic-field determination. With $n_s=nt$ , infer sheet density of a homogeneous thin layer even without knowing active thickness. A resistance measurement may add mobility information under a suitable model, but does not measure the same quantity as Hall voltage. Field reversal splits measured voltage into odd and even components. Only when conditions remain fixed and Hall response is antisymmetric can the odd term be directly identified with a corrected Hall signal; it does not remove every possible measurement artifact.

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More formulas: Electricity and magnetism

F = k_e \cdot \frac{|q_1 q_2|}{r^2}

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\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

Faraday’s law of induction

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

U = R \cdot I

Ohm's Law

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Hall effect and Hall voltage

The Hall voltage appears when a magnetic field deflects moving charge carriers sideways and builds a transverse charge separation.

UniversityBachelor foundationsSubject specialization

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Variables & units

Symbol

Meaning

Unit

U_{\mathrm{H}}

Signed voltage V(y−)−V(y+) transverse to current and magnetic field

I

Current through the sample

B_{\mathrm{z}}

Signed z component of magnetic field perpendicular to x-directed current

n

Density of the dominant carrier type

m⁻³

q

Signed carrier charge

t

Sample thickness along the field

Derivation & background

Exam blueprint

Validity range

Derivation steps

Transverse force balance sets the Hall field; current density connects signed drift and carrier density.

1Define I along +x, Bz along z, width w along y and thickness t along z.
2Use polarity $U_{\rm H}=V(y_-)-V(y_+)$ ; q carries the sign of the dominant carrier type.
3The steady open transverse branch satisfies $\mathbf E_{\rm H}=-\mathbf v_{\rm d}\times\mathbf B$ .
4The cross product gives $E_y=v_{{\rm d},x}B_z$ ; obtain drift from $I=nqv_{{\rm d},x}wt$ .
5Since $\frac{\partial V}{\partial y}=-E_y$ , the defined potential difference is $U_{\rm H}=E_yw$ .
6Width cancels: $U_{\rm H}=\frac{IB_z}{nqt}$ . Negative q gives negative voltage at positive I and Bz under this polarity.

How do you calculate with Hall Effect and Hall Voltage?

Here is how to work through a typical Hall Effect and Hall Voltage (UH = IB/(nqt)) task step by step:

Task

Solution path

Approach: $U_{\rm H}=\frac{IB_z}{nqt}$ .
With $t=0.001\,\mathrm m$ and $q=-1.602176634\cdot10^{-19}\,\mathrm C$ , $U_{\rm H}\approx-3.6715\cdot10^{-8}\,\mathrm V=-36.715\,\mathrm{nV}$ .
Lower potential is below upper potential; magnitude is about 36.7 nV.
This uses a steady homogeneous one-carrier model with the specified probe polarity.

A small transverse voltage reveals carrier information

Prior knowledge: for negative carriers, conventional current opposes drift motion.
Width and thickness play different roles. Transverse width is probe separation; active thickness runs parallel to the magnetic field.
Use positive number density n and signed carrier charge q. Negative density would not be a physical answer.

[1][2]

Which probe polarity and thickness belong in the Hall-voltage equation?

\begin{aligned}U_{\rm H}&=V(y_-)-V(y_+)\\J_x&=\frac I{wt}=nqv_{{\rm d},x}\\U_{\rm H}&=\frac{IB_z}{nqt}\end{aligned}

This signed form uses the axes and contacts specified here.[1]

Same probes, reversed current

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From Lorentz force to transverse voltage

\begin{aligned}\mathbf E_{\rm H}&=-\mathbf v_{\rm d}\times\mathbf B\\E_y&=v_{{\rm d},x}B_z\\v_{{\rm d},x}&=\frac I{nqwt}\\U_{\rm H}&=E_yw=\frac{IB_z}{nqt}\end{aligned}

Since $\frac{\partial V}{\partial y}=-E_y$ , $V(y_-)-V(y_+)=E_yw$ . Width cancels.[1]

Explore further: Hall coefficient and sheet density

\begin{aligned}R_{\rm H}&=\frac{E_y}{J_xB_z}=\frac1{nq}\\U_{\rm H}&=R_{\rm H}\frac{IB_z}{t}\\n&=\frac{|IB_z|}{|q|t|U_{\rm H}|}\\\left[R_{\rm H}\right]&=\mathrm{m^3\,C^{-1}}\\\left[U_{\rm H}\right]&=\mathrm V\end{aligned}

One-carrier model; density inversion needs a nonzero signal and known geometry.[1][2]

Example: electron drift and Hall polarity

Connect all directions in one calculation

Given: One-carrier electron model: $n=2.0\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=5.0\,\mathrm{mA}$ along +x, $B_z=0.40\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ , $w=3.0\,\mathrm{mm}$ .
Find: Find drift component, Hall field and lower-minus-upper voltage.

Model: use signed charge in current density and force balance.
$\begin{aligned}v_{{\rm d},x}&=\frac I{nqwt}\\E_y&=v_{{\rm d},x}B_z\\U_{\rm H}&=E_yw\end{aligned}$
Substitute $q=-e$ ; separately convert width and thickness to metres.
$\begin{aligned}v_{{\rm d},x}&=\frac{0.005}{(2.0\cdot10^{22})(-1.602176634\cdot10^{-19})(0.003)(0.0002)}\,\mathrm{m\,s^{-1}}\\&\approx-2.60063\,\mathrm{m\,s^{-1}}\\E_y&\approx-1.04025\,\mathrm{V\,m^{-1}}\\U_{\rm H}&\approx-3.12075\,\mathrm{mV}\end{aligned}$

Result: Electrons drift against +x. The field points toward −y; the lower potential is 3.121 mV below the upper potential.

Check: Check: negative charge times negative drift gives positive current. Direct voltage substitution gives the same −3.121 mV. Upward electric and downward magnetic forces balance.

What does field reversal separate from the Hall signal, and what still needs checking?

\begin{aligned}U_{\rm odd}(B)&=\frac{U_{\rm meas}(B)-U_{\rm meas}(-B)}2\\U_{\rm even}(B)&=\frac{U_{\rm meas}(B)+U_{\rm meas}(-B)}2\end{aligned}

Algebraic odd/even decomposition; assigning the odd term to Hall voltage requires the stated measurement assumptions.[2]

Example: field reversal and density extraction

Remove a constant offset

Given: $I=10\,\mathrm{mA}$ , $t=0.50\,\mathrm{mm}$ ; measured voltage is $+4.6\,\mathrm{mV}$ at $B_z=+0.50\,\mathrm T$ , and $-3.4\,\mathrm{mV}$ at $-0.50\,\mathrm T$ . One-carrier model with $|q|=e$ , constant even offset and fixed probe polarity.
Find: Find Hall component, offset, carrier sign and density.

Model: the difference removes offset and the sum removes Hall signal. A positive Hall coefficient here identifies positive carriers.
$\begin{aligned}U_{\rm H}&=\frac{U_+-U_-}2\\U_0&=\frac{U_++U_-}2\\n&=\frac{|IB_z|}{et|U_{\rm H}|}\end{aligned}$
Substitute the corrected 4.0 mV into the density equation.
$\begin{aligned}U_{\rm H}&=\frac{4.6-(-3.4)}2\,\mathrm{mV}=4.0\,\mathrm{mV}\\U_0&=0.6\,\mathrm{mV}\\R_{\rm H}&=\frac{(0.004)(0.0005)}{(0.010)(0.50)}\,\mathrm{m^3\,C^{-1}}\\&=4.0\cdot10^{-4}\,\mathrm{m^3\,C^{-1}}\\n&\approx1.56038\cdot10^{22}\,\mathrm{m^{-3}}\end{aligned}$

Result: The offset is +0.6 mV. Under the specified one-carrier model the data indicate positive carriers with n≈1.56·10²² per cubic metre.

Four readings separate three contributions

Given: At equal current and field magnitudes, readings are $U_{++}=-1.50\,\mathrm{mV}$ , $U_{+-}=+2.50\,\mathrm{mV}$ , $U_{-+}=+1.90\,\mathrm{mV}$ and $U_{--}=-2.10\,\mathrm{mV}$ . Signs denote current and field. Model: Hall term H is odd in both, contact term L is odd only in current, and offset O is constant.
Find: Find H, L and O for positive current and field and check all four readings.

Approach: separate readings by their behaviour under current and field reversal.
$\begin{aligned}U(s_I,s_B)&=H s_Is_B+L s_I+O\\H&=\frac{U_{++}-U_{+-}-U_{-+}+U_{--}}4\\L&=\frac{U_{++}+U_{+-}-U_{-+}-U_{--}}4\\O&=\frac{U_{++}+U_{+-}+U_{-+}+U_{--}}4\end{aligned}$
Substitute all readings in millivolts.
$\begin{aligned}H&=\frac{-1.50-2.50-1.90-2.10}4\,\mathrm{mV}=-2.00\,\mathrm{mV}\\L&=\frac{-1.50+2.50-1.90+2.10}4\,\mathrm{mV}=+0.300\,\mathrm{mV}\\O&=\frac{-1.50+2.50+1.90-2.10}4\,\mathrm{mV}=+0.200\,\mathrm{mV}\end{aligned}$

Result: Under the stated model, Hall signal at positive current and field is −2.00 mV. The current-odd contact contribution is +0.300 mV and constant offset +0.200 mV.

When Hall voltage is not a simple carrier count

Five exercises on calculation, polarity and measurement models

Calculate · Foundation

Positive carriers with $q=+e$ , $n=5\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=20\,\mathrm{mA}$ , $B_z=0.20\,\mathrm T$ , $t=0.10\,\mathrm{mm}$ . Find U_H under the article convention.

Show hint

Convert thickness to metres; use lower minus upper potential.

Learning goal: Calculate a signed Hall voltage.

Show solution and calculation

Model and approach: One dominant carrier type and open transverse branch; polarity lower minus upper.
$U_{\mathrm H}=\frac{IB_z}{nqt}$ Use the one-carrier voltage equation with positive q.
Substitution gives +4.9932 mV; the lower edge is at higher potential.
Reversing current at unchanged field would give −4.9932 mV.

U_{\rm H}=\frac{(0.020)(0.20)}{(5\cdot10^{22})(1.602176634\cdot10^{-19})(0.0001)}\,\mathrm V\approx4.9932\,\mathrm{mV}

[1]

Rearrange · Application

An electron model gives $|U_{\rm H}|=1.0\,\mathrm{mV}$ at $I=5.0\,\mathrm{mA}$ , $B_z=0.30\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ . Find n.

Show hint

Use magnitudes; a particle density is not negative.

Learning goal: Extract a positive volume density from magnitude.

Show solution and calculation

Model and approach: Electron model; n is positive and charge sign is checked separately.
$n=\frac{|IB_z|}{et|U_{\mathrm H}|}$ Isolate n from the magnitude equation.
The result is n≈4.6811·10²² per cubic metre.
Using $q=-e$ in the signed equation gives −1.0 mV, matching the measured magnitude.

n=\frac{(0.005)(0.30)}{(1.602176634\cdot10^{-19})(0.0002)(0.001)}\,\mathrm{m^{-3}}\approx4.6811\cdot10^{22}\,\mathrm{m^{-3}}

[1]

Interpret · Foundation

I points along +x and B along +z. Positive edge charge accumulates above and negative charge below. What drift, field and voltage directions follow in the one-carrier model?

Show hint

The Hall field points from positive to negative edge charge.

Learning goal: Interpret carrier sign only with full measurement convention.

Show solution and calculation

The field points along −y, so $U_{\mathrm H}=V(y_-)-V(y_+)$ is negative.
Positive I and B imply $q<0$ .
Negative drift times negative charge gives positive conventional current.
Swapping probe leads changes the reading but neither drift nor carrier type.

E_y<0,\quad U_{\rm H}<0,\quad q<0,\quad v_{{\rm d},x}<0

[1]

Error analysis · Application

With $w=3.0\,\mathrm{mm}$ and $t=0.20\,\mathrm{mm}$ , a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?

Show hint

Width appears in both current density and potential difference.

Learning goal: Distinguish magnetic-field thickness from contact width.

Show solution and calculation

The correct model uses t parallel to the field.
The wrong result is $\frac tw=\frac1{15}$ of the correct magnitude: 3.121 mV would become about 0.208 mV.
Width cancels because a wider plate at fixed current has smaller drift speed but a longer transverse distance.

\frac{|U_{\rm wrong}|}{|U_{\rm correct}|}=\frac tw=\frac{0.20}{3.0}=\frac1{15}

[1]

Transfer · Challenge

A thin homogeneous layer gives $|U_{\rm H}|=2.0\,\mathrm{mV}$ at $I=10\,\mathrm{mA}$ , $|B_z|=0.20\,\mathrm T$ . In a one-carrier model with $|q|=e$ , find sheet density n_s without knowing thickness. Can n also be obtained?

Show hint

Combine n and t into n_s.

Learning goal: Distinguish volume density and sheet density.

Show solution and calculation

Use $n_s=nt$ in the magnitude relation.
The measurement gives n_s≈6.2415·10¹⁸ per square metre.
Volume density also needs active thickness: $n=\frac{n_s}{t}$ .
The signal alone cannot separate n and t.
The inverse-square-metre unit confirms that a volume density has not been claimed.

n_s=\frac{|IB_z|}{e|U_{\rm H}|}\approx6.2415\cdot10^{18}\,\mathrm{m^{-2}}

[2]

What to take away

Voltage sign is meaningful only with current, field and probe conventions. Swapping probes does not change carrier type.
Transverse width cancels; active thickness parallel to the magnetic field remains.
Reversal measurements separate specific symmetry components. They do not automatically remove every artifact or fix an unsuitable multicarrier model.

[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed Hall polarity, signed drift and offset removal; corrected the existing metal example to 36.7 nV; added two examples and five exercises.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

Cite this reference page

Quanta. Hall Effect and Hall Voltage. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/hall-effekt-spannung

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Frequently asked questions about Hall Effect and Hall Voltage

What do the Hall effect and Hall voltage describe physically?+

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What assumptions and limits apply to the Hall-voltage formula?+

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Hall effect and Hall voltage

Variables & units

Hall voltage calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electricity and magnetism

Worked example

Applications

How do you calculate with Hall Effect and Hall Voltage?

A small transverse voltage reveals carrier information

Which probe polarity and thickness belong in the Hall-voltage equation?

Same probes, reversed current

From Lorentz force to transverse voltage

Explore further: Hall coefficient and sheet density

Example: electron drift and Hall polarity

Connect all directions in one calculation

What does field reversal separate from the Hall signal, and what still needs checking?

Example: field reversal and density extraction

Remove a constant offset

Four readings separate three contributions

When Hall voltage is not a simple carrier count

Five exercises on calculation, polarity and measurement models

Positive carriers with q=+eq=+eq=+e, n=5⋅1022 m−3n=5\cdot10^{22}\,\mathrm{m^{-3}}n=5⋅1022m−3, I=20 mAI=20\,\mathrm{mA}I=20mA, Bz=0.20 TB_z=0.20\,\mathrm TBz​=0.20T, t=0.10 mmt=0.10\,\mathrm{mm}t=0.10mm. Find UH under the article convention.

An electron model gives ∣UH∣=1.0 mV|U_{\rm H}|=1.0\,\mathrm{mV}∣UH​∣=1.0mV at I=5.0 mAI=5.0\,\mathrm{mA}I=5.0mA, Bz=0.30 TB_z=0.30\,\mathrm TBz​=0.30T, t=0.20 mmt=0.20\,\mathrm{mm}t=0.20mm. Find n.

I points along +x and B along +z. Positive edge charge accumulates above and negative charge below. What drift, field and voltage directions follow in the one-carrier model?

With w=3.0 mmw=3.0\,\mathrm{mm}w=3.0mm and t=0.20 mmt=0.20\,\mathrm{mm}t=0.20mm, a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?

What to take away

Sources for this explanation

Frequently asked questions about Hall Effect and Hall Voltage

Subject references and further reading

More formulas: Electricity and magnetism

Retain Hall Effect and Hall Voltage for exams

Hall effect and Hall voltage

Variables & units

Hall voltage calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electricity and magnetism

Worked example

Applications

How do you calculate with Hall Effect and Hall Voltage?

A small transverse voltage reveals carrier information

Which probe polarity and thickness belong in the Hall-voltage equation?

Same probes, reversed current

From Lorentz force to transverse voltage

Explore further: Hall coefficient and sheet density

Example: electron drift and Hall polarity

Connect all directions in one calculation

What does field reversal separate from the Hall signal, and what still needs checking?

Example: field reversal and density extraction

Remove a constant offset

Four readings separate three contributions

When Hall voltage is not a simple carrier count

Five exercises on calculation, polarity and measurement models

Positive carriers with q=+eq=+eq=+e, n=5⋅1022 m−3n=5\cdot10^{22}\,\mathrm{m^{-3}}n=5⋅1022m−3, I=20 mAI=20\,\mathrm{mA}I=20mA, Bz=0.20 TB_z=0.20\,\mathrm TBz​=0.20T, t=0.10 mmt=0.10\,\mathrm{mm}t=0.10mm. Find UH under the article convention.

An electron model gives ∣UH∣=1.0 mV|U_{\rm H}|=1.0\,\mathrm{mV}∣UH​∣=1.0mV at I=5.0 mAI=5.0\,\mathrm{mA}I=5.0mA, Bz=0.30 TB_z=0.30\,\mathrm TBz​=0.30T, t=0.20 mmt=0.20\,\mathrm{mm}t=0.20mm. Find n.

I points along +x and B along +z. Positive edge charge accumulates above and negative charge below. What drift, field and voltage directions follow in the one-carrier model?

With w=3.0 mmw=3.0\,\mathrm{mm}w=3.0mm and t=0.20 mmt=0.20\,\mathrm{mm}t=0.20mm, a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?

What to take away

Sources for this explanation

Frequently asked questions about Hall Effect and Hall Voltage

Subject references and further reading

More formulas: Electricity and magnetism

Retain Hall Effect and Hall Voltage for exams

Positive carriers with $q=+e$ , $n=5\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=20\,\mathrm{mA}$ , $B_z=0.20\,\mathrm T$ , $t=0.10\,\mathrm{mm}$ . Find U_H under the article convention.

An electron model gives $|U_{\rm H}|=1.0\,\mathrm{mV}$ at $I=5.0\,\mathrm{mA}$ , $B_z=0.30\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ . Find n.

With $w=3.0\,\mathrm{mm}$ and $t=0.20\,\mathrm{mm}$ , a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?

Positive carriers with $q=+e$ , $n=5\cdot10^{22}\,\mathrm{m^{-3}}$ , $I=20\,\mathrm{mA}$ , $B_z=0.20\,\mathrm T$ , $t=0.10\,\mathrm{mm}$ . Find U_H under the article convention.

An electron model gives $|U_{\rm H}|=1.0\,\mathrm{mV}$ at $I=5.0\,\mathrm{mA}$ , $B_z=0.30\,\mathrm T$ , $t=0.20\,\mathrm{mm}$ . Find n.

With $w=3.0\,\mathrm{mm}$ and $t=0.20\,\mathrm{mm}$ , a solution uses w instead of t in the Hall equation. By what factor is its magnitude wrong?