The Hall effect
Derive Hall voltage from magnetic deflection of charge carriers and use it to infer carrier sign and density.
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Guided steps
- Name the current direction, field direction and the two labelled voltage contacts before assigning signs. Width separates the Hall contacts, while thickness lies along the magnetic field. The current-carrying area contains both dimensions. A positive A-minus-B reading means contact A is at higher potential.
- Edge charge builds a Hall electric field until transverse electric and magnetic forces balance. In magnitude, Hall field is drift speed times perpendicular magnetic field. Voltage also includes the contact separation. At fixed current, increasing width lowers drift speed while increasing that separation, so width cancels; thickness remains in the denominator.
- At fixed conventional current, changing carrier sign reverses drift direction. Magnetic force pushes both carrier types towards the same edge, but that edge accumulates opposite charge signs. The figure keeps current, field and contacts fixed to expose the resulting reversed Hall field and voltage. Field direction is not electric-force direction for a negative carrier. The transverse net force is zero in the final steady state.
- For a magnitude calculation, use charge magnitude and thickness in metres. A large carrier density gives a small transverse voltage at the same current, field and geometry. Keep magnitude separate from the signed contact reading; this example does not ask you to infer a polarity from an unsigned number.
- Start from signed current density equal to carrier number density times signed charge times drift velocity. Use the transverse vector balance for the Hall field, then potential difference opposite to that field. Keep the same contact labels throughout; merely putting signed charge into an unsigned formula does not define a voltmeter convention.
- Find the signed drift component, Hall-field component and A-minus-B voltage in the requested units. Keep both material dimensions in the current-carrying area. Compare edge charge and the two force directions separately.
- Explain electron drift, the sign of charge at the lower edge, the balancing electric force and the defined voltage polarity. Compare your reasoning with the public model yourself; the three checked values do not mark that argument. Paper working is fine.
- A geometry change must name what stays fixed. Width changes both current density and contact separation; thickness changes area without changing separation. Treat each trial as a fresh change from the original strip. Current and field reversals are signed changes, whereas the requested geometry factors compare magnitudes.
- Give voltage-magnitude factors for the separate width and thickness changes, then the signed A-minus-B reading after reversing only the magnetic field. Each trial starts from the original state. Explain cancellation and the double reversal separately.
- Show why width cancels under fixed current and why thickness does not. Explain the sign of the field-reversal reading and what reversing both current and field would do. Compare your constraint and contact reasoning with the public solution yourself.
- Predict the Hall-voltage magnitude change when magnetic field magnitude doubles and the current, carriers and geometry remain fixed.
- Infer the positive carrier number density from the supplied voltage magnitude. Convert thickness and voltage to SI units; carrier charge sign does not make number density negative.
- Explain why the same conventional current and field can push either carrier type to the same edge while producing opposite voltage polarities. Compare your drift and accumulated-charge argument with the public answer yourself, within the single-carrier model.
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The Hall effect connects carrier motion to a transverse voltage. Its magnitude helps infer carrier density; its polarity can identify carrier sign in a single-carrier model. To use both, distinguish conventional current from carrier drift and define which contact’s potential is subtracted.
Use signed carrier flow and current density and the magnetic force on a particle. We use a uniform rectangular strip with one carrier species, steady transverse balance and negligible transverse current.
Define the geometry and the measured voltage
Let conventional current Iₓ be positive along +x, field component Bz positive along +z, width w > 0 along y and thickness t > 0 along z. The current-carrying section is A = wt. Width is the Hall-contact separation; thickness lies along the magnetic field.
Contact A is at the upper +y edge and contact B at the lower −y edge. Define the signed reading VH = V(A)-V(B). A positive reading means A is at higher potential. Keep these contact labels fixed when reversing current, field or carrier sign. In magnitude-only calculations, I, B and vd below denote nonnegative magnitudes.
Balance the transverse forces, then find the voltage
Magnetic deflection separates edge charge until its Hall electric field balances the transverse magnetic force: q(EH + vd × B) = 0, EH = -vd × B. This is a transverse balance; it does not set the longitudinal driving field to zero. For perpendicular drift and field, the magnitudes obey |EH| = vdB and |VH| = |EH|w. Using I = n|q|wtvd gives |VH| = BI/(n|q|t). Width cancels because a wider strip at fixed current has a smaller drift speed. It does not cancel if the drift speed is independently held fixed. Thickness does not supply the voltage-measuring separation.
The units provide a check: BI/(n|q|t) has units (T A)/(m⁻³ C m) = V. If current or perpendicular field is zero, this ideal Hall voltage is zero. At fixed current, field and geometry, a larger carrier density makes its magnitude smaller. The transverse-balance basis also appears in OpenStax’s Hall-effect treatment.
Keep carrier sign and contact polarity together
For our axes, x hat × z hat = -y hat. With Jₓ = Iₓ/(wt) = nqv_(d,x), E_(H,y) = v_(d,x)Bz = JₓBz/(nq). Potential decreases along the field, so the defined contact reading is VH = -E_(H,y)w = -IₓBz/(nqt). The minus sign follows from our A-minus-B contact convention. Merely replacing |q| by q in a magnitude formula does not define a voltmeter polarity.
At fixed conventional current and magnetic field, changing carrier sign reverses drift direction. The magnetic part of the force nevertheless points towards the same edge: F_(B,y) = -qv_(d,x)Bz = -JₓBz/n. What reverses is the sign of the charge accumulating there, and therefore the Hall field and contact polarity. At steady state the electric force balances this magnetic part; a downward magnetic-force arrow is not a continuing net downward acceleration.
| Carriers, with Iₓ > 0 and Bz > 0 | Drift | Lower edge | Hall field | V(A)-V(B) |
|---|---|---|---|---|
| Negative | −x | Negative | −y | Positive |
| Positive | +x | Positive | +y | Negative |
Polarity identifies carrier sign within this one-carrier model. A material with multiple contributing carriers or a more complex electronic structure requires a richer transport analysis; the elementary inversion is not a complete microscopic description.
Hall voltage in a metal sample
Worked example 1
A small transverse voltage
Problem
A sample carries I = 5.0 A in a perpendicular field B = 0.80 T. Its thickness along the field is t = 1.0 mm and its carrier density is n = 8.5 × 10²⁸ m⁻³. Each carrier has charge magnitude e = 1.60 × 10⁻¹⁹ C. Find the Hall-voltage magnitude in the single-carrier model.
Show full solution
Convert thickness and substitute
Method
Use t = 1.0 × 10⁻³ m in |VH| = BI/(n|q|t).Reason
This relation follows from the steady electric and magnetic force balance; SI thickness gives the voltage in volts.Working
Answer
|VH| = BI/(n|q|t); = (0.80)(5.0)/((8.5 × 10²⁸)(1.60 × 10⁻¹⁹)(1.0 × 10⁻³)); ≈ 2.94 × 10⁻⁷ V.State the magnitude
Working
Answer
The Hall-voltage magnitude is 2.9 × 10⁻⁷ V, or 0.29 μ V. The large carrier density makes it small.
Quick check
At the same nominal I, B and t, the single-carrier model with n = 10²² m⁻³ predicts a magnitude of about 2.5 V. This is a rescaling within the stated model.
Calculate a signed Hall state
Guided practice 1
Keep the current and contact labels fixed
Problem
A rectangular strip has electron carriers with n = 5.00 × 10²⁴ m⁻³ and q = -1.60 × 10⁻¹⁹ C. Its width is w = 8.00 mm along y and thickness t = 0.400 mm along z. Conventional current is Iₓ = +2.00 A and Bz = +0.500 T. Use the steady, uniform single-carrier model. Contact A is at the upper +y edge and contact B at the lower −y edge.
Find the signed electron drift component v_(d,x), Hall field component E_(H,y) and VH = V(A)-V(B) in mV. Explain the edge charge and why the Hall electric field direction alone is not the electric-force direction for an electron.
State the directions before calculating
Hints
Hint 1
Keep the carrier sign
Start with Jₓ = Iₓ/(wt) = nqv_(d,x), retaining q < 0.Hint 2
Convert field to voltage
For these axes E_(H,y) = v_(d,x)Bz. Then V(A)-V(B) = -E_(H,y)w.Show solution step by step
Find the signed drift
Method
Convert both dimensions and use the full current-carrying area.Reason
Conventional current is fixed along +x while negative carriers drift oppositely.Working
Answer
wt = (8.00 × 10⁻³)(0.400 × 10⁻³) = 3.20 × 10⁻⁶ m². Thus Jₓ = 6.25 × 10⁵ A/m² and v_(d,x) = Jₓ/(nq) = -0.781 m/s.
Balance transverse forces
Method
Use EH = -vd × B.Reason
The Hall field builds until the transverse electric and magnetic forces balance.Working
Answer
E_(H,y) = v_(d,x)Bz = -0.391 V/m. The magnetic force on an electron is downward, while its electric force is upward because q < 0. With guard digits their components are -6.25 × 10⁻²⁰ N and + 6.25 × 10⁻²⁰ N.
Read the defined contacts
Method
Integrate potential opposite to the field direction.Reason
A is above B, and potential rises towards A when the Hall field points down.Working
Answer
VH = -E_(H,y)w = +3.13 mV using the unrounded field. Electrons accumulated at the lower edge, leaving A at higher potential. Check the edge-charge and force argument yourself; the checker verifies the three signed values only.
Quick check
Changing the contact labels would reverse the reported voltage. It would not change the physical edge charges or the Hall field.
Change geometry and infer the carriers
Check your understanding 1
Separate contact spacing, thickness and polarity Each trial starts from the original electron-strip state in the signed example. In W only the width doubles; in T only the thickness doubles; in R only the magnetic field reverses. Current, carrier density and all other stated parameters remain fixed in each trial.
Give the final/original Hall-voltage magnitude factors for W and T, and the signed V(A)-V(B) in mV for R. Explain why width cancels from the fixed-current voltage formula even though it is the contact separation. Would reversing both current and field restore the original polarity?
Show hint
Show answer
W gives factor 1: doubling width halves drift speed and the Hall field, while doubling the distance across which voltage is measured. T gives factor 1/2: drift speed and field halve, but contact separation stays fixed. R gives -3.13 mV, with the original current and contacts retained.
The signed result is proportional to the product IₓBz. Reversing either one reverses polarity; reversing both restores it. Width cancels here because current, rather than drift speed or current density, is held fixed. Compare the constraint and polarity reasoning yourself; the checker verifies the three numerical results only.
Check your understanding 2
If the magnetic field magnitude doubles while current, carrier density and sample dimensions stay fixed, how does the Hall-voltage magnitude change?
Show hint
Identify the dependence on B in |VH| = BI/(n|q|t).
Show answer
It doubles because |VH| is directly proportional to B when the other quantities are fixed.
Check your understanding 3
A sample with electron carriers has Hall-voltage magnitude 4.0 × 10⁻⁶ V at B = 1.0 T and I = 2.0 A. Its thickness along the field is 0.50 mm. Find the carrier density using e = 1.60 × 10⁻¹⁹ C.
Show hint
Rearrange the magnitude relation for n and convert millimetres to metres.
Show answer
n = BI/(e t |VH|); = (1.0)(2.0)/((1.60 × 10⁻¹⁹)(0.50 × 10⁻³)(4.0 × 10⁻⁶)); = 6.25 × 10²⁷ m⁻³. To two significant figures, n = 6.3 × 10²⁷ m⁻³.
Check your understanding 4
For known conventional current and field directions, why does Hall-voltage polarity distinguish positive from negative carriers in the single-carrier model?
Show hint
Negative carriers drift opposite to conventional current. Track both their drift direction and the sign of the charge accumulating at an edge.
Show answer
With conventional current fixed, changing the carrier sign also reverses its drift velocity. The force q vector vd × vector B therefore pushes either carrier type towards the same edge, but that edge accumulates opposite charge signs. The Hall field and voltage polarity consequently reverse.
State the constraints before interpreting a Hall reading
Use the magnitude formula for a density or size calculation, and the signed vector balance with defined contacts for polarity. At fixed current, width cancels while thickness remains in the denominator. Reversing current or field alone reverses polarity; reversing both restores it. Keep the single-carrier and steady-state assumptions explicit when interpreting the result.
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