My name is Alexander FufaeV and here I will explain the following topic:

Hall Voltage due to the Hall Effect

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.

Rectangular Hall sample and its dimensions.

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

A moving electron in the Hall sample experiences a magnetic and electric force.

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:

$$ \begin{align} F_{\text m} ~=~ q \, v \, \class{violet}{B} \end{align} $$

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:

$$ \begin{align} F_{\text e} ~=~ q \, E \end{align} $$

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:

$$ \begin{align} q \, v \, \class{violet}{B} ~=~ q \, E \end{align} $$

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:

$$ \begin{align} U_{\text H} ~=~ h \, v \, \class{violet}{B} \end{align} $$

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:

$$ \begin{align} v ~=~ \frac{L}{t} \end{align} $$

The time $ t $ can then be determined using the formula for electric current:

$$ \begin{align} I ~=~ \frac{Q}{t} \end{align} $$

Rearrange Eq. 6 with respect to time:

$$ \begin{align} t ~=~ \frac{Q}{I} \end{align} $$

Substitute equation 7 into equation 5:

$$ \begin{align} v ~=~ \frac{L \, I}{Q} \end{align} $$

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):

$$ \begin{align} Q ~=~ N \, q \end{align} $$

Substitute Eq. 9 into Eq. 8 to get the following formula for drift velocity:

$$ \begin{align} v ~=~ \frac{L \, I}{N \, q} \end{align} $$

Substituting the velocity 10 into the Hall voltage formula 4 yields:

$$ \begin{align} U_{\text H} ~=~ \frac{L\, I \, h \, \class{violet}{B}}{N \, q} \end{align} $$

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):

$$ \begin{align} n ~=~ \frac{N}{V} \end{align} $$

Therefore, replace charge carrier number $ N $ in equation 11 using Eq. 12 by rearranging 12 for $N$:

$$ \begin{align} U_{\text H} ~=~ \frac{L\, I \, h \, \class{violet}{B}}{n \, V \, q} \end{align} $$

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 $h$:

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

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

$$ \begin{align} U_{\text H} ~=~ A_{\text H} ~ \frac{I \, \class{violet}{B}}{d} \end{align} $$