# Hall effect

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u23-hall-effect/

This sub-topic explains how the Hall effect arises from the Lorentz force on moving charge carriers, derives the Hall voltage equation, and covers key applications including measuring magnetic fields and identifying semiconductor type.

**Prerequisites:** [Lorentz force on moving charges](https://www.owlsprep.com/study/cie-9702-u23-lorentz-force/); [Electric current as flow of charge carriers](https://www.owlsprep.com/study/cie-9702-u17-electric-current/)

## Learning objectives

- Explain the origin of the Hall effect from the Lorentz force
- Derive the equation for Hall voltage
- Calculate Hall voltage for given conductor and magnetic parameters
- Use the sign of Hall voltage to identify semiconductor type
- Describe applications of the Hall effect

## Origin of the Hall Effect

When a current-carrying conductor is placed in a magnetic field perpendicular to the direction of current flow, moving charge carriers experience a Lorentz force that deflects them toward one side of the conductor.

**Hall Effect** — The generation of a transverse potential difference (Hall voltage) across a current-carrying conductor placed in a perpendicular magnetic field, due to charge separation caused by the Lorentz force on moving charge carriers.

*Example:* A thin copper strip carrying current placed between the poles of a permanent magnet develops a small voltage across its width.

Equilibrium is reached when the magnetic Lorentz force on the charge carriers is balanced by the electric force from the separated charge: $F_B = F_E$

$$Bqv = \frac{qV_H}{w}$$

Where $B$ = magnetic flux density, $q$ = charge of one carrier, $v$ = drift velocity of carriers, $V_H$ = Hall voltage, $w$ = width of the conductor across which the voltage develops.

**Worked example:** A copper strip of width 2.0 cm has electrons moving with drift velocity $3.2 \times 10^{-4} \text{ m s}^{-1}$ in a 0.5 T magnetic field. What is the electric field strength across the strip at equilibrium?

1. At equilibrium, magnetic force equals electric force, so cancel the charge term $q$ from both sides:
2. $$E = Bv$$
3. Substitute the given values to get the electric field strength:
4. $$E = (0.5)(3.2 \times 10^{-4}) = 1.6 \times 10^{-4} \text{ V m}^{-1}$$

## Derivation of the Hall Voltage Equation

**Derivation:** Derive an expression for Hall voltage $V_H$ in terms of measurable quantities $B$, $I$, $n$, $q$, and $t$.

*Starting from:* Equilibrium condition $Bqv = qV_H/w$ and current $I = nAvq$

1. Rearrange the equilibrium condition to get: $V_H = Bvw$
2. Cross-sectional area $A = wt$, where $t$ is the thickness of the conductor parallel to the magnetic field. Substitute into the current equation:
3. $I = n(wt)vq$, rearrange to solve for $v$:
4. $$v = \frac{I}{nwtq}$$
5. Substitute $v$ into the expression for $V_H$:
6. $$V_H = B \left(\frac{I}{nwtq}\right) w$$

*Conclusion:* The width term $w$ cancels out, giving the final Hall voltage formula:

$$V_H = \frac{BI}{nqt}$$

**Worked example:** A Hall probe made from n-type semiconductor has charge carrier density $n = 1.0 \times 10^{20} \text{ m}^{-3}$ and thickness $t = 0.50 \text{ mm}$. A current of 10 mA flows through the probe. Calculate the Hall voltage in a 0.2 T magnetic field ($e = 1.6 \times 10^{-19} \text{ C}$).

1. Convert all quantities to SI units:
2. $$t = 0.50 \text{ mm} = 0.50 \times 10^{-3} \text{ m}, \quad I = 10 \text{ mA} = 0.010 \text{ A}$$
3. Substitute into the Hall voltage formula:
4. $$V_H = \frac{BI}{net}$$
5. Plug in the values:
6. $$V_H = \frac{(0.2)(0.01)}{(1.0 \times 10^{20})(1.6 \times 10^{-19})(0.5 \times 10^{-3})} = \frac{0.002}{8 \times 10^{-3}} = 0.25 \text{ V}$$

## Applications and Key Properties

The Hall effect has many practical applications, thanks to the linear relationship between Hall voltage and magnetic flux density. Two of the most important are:

- **Hall probes for magnetic measurement**: If $n$, $q$, $t$, and $I$ are fixed for the probe, $V_H \propto B$, so measuring $V_H$ directly gives the magnetic flux density.
- **Identifying semiconductor type**: The sign of the Hall voltage corresponds to the sign of the majority charge carriers, distinguishing n-type (electrons) from p-type (holes) semiconductors.

> **info**
>
> Semiconductors have much lower charge carrier density $n$ than metals, so $V_H$ is much larger (since $V_H \propto 1/n$). This makes semiconductors ideal for Hall probes.

**Worked example:** Current flows left to right along a semiconductor strip, with magnetic field directed into the plane of the strip. The top edge of the strip becomes positively charged. Is the semiconductor n-type or p-type?

1. For p-type semiconductors, majority charge carriers are positive holes that move in the same direction as conventional current (left to right).
2. Use Fleming's Left Hand Rule for force on positive charge: First finger (field) into page, second finger (current) left to right, thumb points upwards.
3. Positive holes are deflected upwards, so the top edge accumulates positive charge, matching the observation.

*Conclusion:* The semiconductor is p-type.

## Common pitfalls

- **Wrong:** Confusing thickness $t$ and width $w$ in the Hall voltage formula
  - Why it fails: The formula uses thickness along the magnetic field direction, not the width across which Hall voltage is measured
  - Correct: Remember $V_H = BI/(nqt)$, where $t$ is the dimension parallel to the magnetic field, not transverse.
- **Wrong:** Treating holes as electrons moving opposite when finding deflection direction
  - Why it fails: This leads to the wrong sign of Hall voltage because holes are positive charge carriers, not negative electrons
  - Correct: Apply Fleming's Left Hand Rule directly to the charge of the majority carrier, not electron flow opposite to current.
- **Wrong:** Forgetting to convert prefixed units to SI units before calculation
  - Why it fails: The Hall voltage formula is derived for SI units, so leaving mm or mA un-converted gives wrong orders of magnitude
  - Correct: Always convert all quantities to metres, amperes, and tesla before calculating $V_H$.
- **Wrong:** Assuming Hall effect only occurs in metals
  - Why it fails: The Hall effect occurs in any material with moving charge carriers, but is much weaker in metals than semiconductors
  - Correct: Recognize that low carrier density in semiconductors produces large measurable Hall voltages, so semiconductors are used for Hall probes.

## Cheatsheet

| Quantity/Rule | Symbol | Relation |
| --- | --- | --- |
| Equilibrium condition | $F_B = F_E$ | $Bqv = \frac{qV_H}{w}$ |
| Hall voltage formula | $V_H$ | $V_H = \frac{BI}{nqt}$ |
| Proportionality for probes | $V_H \propto B$ | Linear for constant current $I$ |
| Positive Hall voltage sign | - | Indicates p-type semiconductor |

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