# Derivation Hall Voltage due to the Hall Effect

## Video - Hall Effect and a SIMPLE Derivation of The Hall Voltage

Subscribe on YouTube In the lesson on the Hall effect, you learned how Hall voltage is generated. In the following we want to derive a formula for the **Hall voltage** \( U_{\text H} \), which depends only on the quantities we can determine in the experiment.

We consider a Hall plate of **width** \( h \), **thickness** \( d \) and **length** \( L \). It can be made of a metal or a semiconductor.

Then we send an electric current \( I \) through the Hall plate. Electric current \( I \) means that in the material - positive or negative charge carriers move in a certain direction with a **drift velocity** \( v \). These charge carriers are either negatively charged (**electrons** with the charge \( q = -e\)) or positively charged (so-called **holes** with the charge \( q = +e\)).

Perpendicular to the direction of electron motion (\(v \perp \class{violet}{B}\)), a magnetic field with constant **magnetic flux density** \( \class{violet}{B} \) penetrates the Hall sample.

## Electric and magnetic force inside the Hall sample

On the *moving* charge carriers always acts a magnetic force \( F_{\text m} \) (**Lorentz force**), which deflects the charge carriers perpendicular to the magnetic field and perpendicular to the direction of motion. In this case, the magnitude of the Lorentz force is:

**Formula for magnetic force**

**Note!** Lorentz force has a different direction, depending on whether you put for \( q \) negative elementary charge \( -e \) (for electrons) or positive elementary charge \( +e \) (for holes).

Due to the deflection of the charge carriers by the Lorentz force, there is a negative charge excess at one edge of the sample and a positive charge excess at the other edge. This charge difference generates an **electric field** \( E \), which exerts a **electric force** \( F_{\text e} \) on all subsequent charges. It acts against the magnetic force.

The top and bottom edge of the Hall sample can be considered as plates of a parallel plate capacitor. The electric force between the plates is given by:

**Formula for electric force in a plate capacitor**

The electric force is not only directed in the opposite direction to the magnetic force; it also becomes larger the more the charge carriers have been deflected by the magnetic force. After a short time a **equilibrium of forces** between the magnetic and the electric force is established, which is why you may equate the formulas 1

and 2

:

**Electric force equated with the magnetic force**

With the equilibrium of forces, the electric field stabilizes at a certain value. At the edges of the sample this E-field can be measured as Hall voltage \( U_{\text H} \). Electric field is related to the voltage across the **distance** \( h \) between the edges of the sample: \(E = \frac{U_{\text H}}{h} \). Substitute the equation into the formula 3

for the E field. Charge \( q \) cancels out:

**Formula for Hall voltage via drift velocity**

You cannot measure the average velocity \( v \) of the charge carriers directly, so substitute it using the formula for uniform motion. If a charge carrier travels the **distance** \( L \) (length of the sample) within the **time** \( t \), then we can write the velocity as follows:

**Velocity is distance per time**

The time \( t \) can then be determined using the formula for electric current:

**Current is charge per time**

Rearrange Eq. 6

with respect to time:

**Time equals charge per current**

Substitute equation 7

into equation 5

:

**Velocity equals distance times current divided by charge**

Here, the amount of total charge \( Q \) moving through the sample is the product of the number \( N \) of moving charge carriers and their individual charge \( q \) (depending on the type of charge, \(q\) can be positive or negative):

**Total charge is number of charge carriers miltiplied by single charge**

Substitute Eq. 9

into Eq. 8

to get the following formula for drift velocity:

**Speed is distance multiplied by current divided by charge carrier number multiplied by charge**

Substituting the velocity 10

into the Hall voltage formula 4

yields:

**Formula for Hall voltage via distance, current, thickness, magnetic field, number of charges and charge**

We can express the number of charge carriers \( N \) contributing to the current with the **charge carrier density** \( n \). Charge carrier density is defined as the number of charge carriers \( N \) per **volume** \( V \) of the conductor (here: Hall sample):

**Charge carrier density is number of charges per volume**

Therefore, replace charge carrier number \( N \) in equation 11

using Eq. 12

by rearranging 12

for \(N\):

**Formula for Hall voltage via distance, current, thickness, magnetic field, charge carrier density, volume and charge**

Now you can simplify the formula 13

a little bit. The volume \(V= h \, L \, d\) is the product of height \( h \), length \( L \) and thickness \( d \) of the Hall sample. Substitute it into Eq. 13

and then cancel \(L\) and \(h\):

**Formula for Hall voltage via charge carrier density**

~&=~ \frac{1}{n \, q} ~ \frac{I \, \class{violet}{B}}{d} \end{align} $$

The coefficient \( \frac{1}{n \, q} \) is called the **Hall constant** and is abbreviated as \( A_{\text H} \):

**Formula for Hall voltage via Hall constant**