Hall Effect: A Simple Explanation of How Hall Voltage is Induced
Important Formula
What do the formula symbols mean?
Hall voltage
$$ U_{\text H} $$ Unit $$ \mathrm{V} $$Hall constant
$$ A_{\text H} $$ Unit $$ \frac{\mathrm{m}^3}{\mathrm{As}} $$Electric current
$$ \class{red}{\boldsymbol I} $$ Unit $$ \mathrm{A} $$Magnetic flux density (Bfield)
$$ \class{violet}{B} $$ Unit $$ \mathrm{T} $$Thickness
$$ d $$ Unit $$ \mathrm{m} $$Table of contents
Hall effect is a very important physical effect which is used to measure magnetic fields, charge carrier densities or Hall coefficients of different materials.
To demonstrate the Hall effect, you need four things:

A Hall sample (metal foil, plate, etc.)

External magnetic field

Voltmeter

Ammeter
Select Hall sample: Metal or semiconductor
For the Hall effect you take a conductive plate of thickness \( d \) and width \( h \).
The material of the Hall plate is also important. Most often it is made of...

a metal (e.g. Aluminum, Cuprium, Argentium, etc.)

or a semiconductor (e.g. Silicon, Germanium).
If the Hall plate is a metal, then the electrons are responsible for electrical conduction as soon as you apply a voltage to the metal plate. Aluminum, for example, is an exception. In aluminum, it is not electrons but socalled holes that are responsible for electrical conduction.
A hole is in contrast to the electron positively charged (with positive elementary charge +e) and means: the absence of an electron! This hole conduction can be realized in semiconductors, if you contaminate the semiconductor with elements from the 3rd main group in the periodic table. Then you have a pdoped semiconductor, where the hole conduction dominates.
Send current through the Hall plate
After you have chosen a suitable Hall plate, you apply a voltage to both ends of the plate (right and left). This creates an electric current \( I \) which flows from one end of the plate to the other. If you want to measure this current, you must of course install an ammeter in your circuit.
Depending on which material you have chosen, the electric current is dominated either by negatively charged electrons or by positively charged holes.
Usually, the movement of the charges in the plate is disordered (chaotic). However, by applying an external voltage, you force the moving charges to flow in a certain direction (i.e. to the opposite pole). The velocity at which they flow in a fixed direction is called drift velocity \( v \).
These electric charges move to the left or to the right with a certain drift velocity. If electrons (at the set voltage polarity) move to the left, then the holes (at the same voltage polarity) would move in the opposite direction  i.e. to the right in this case.
Place currentcarrying Hall sample into magnetic field
Now place the plate, through which a current flows, into a magnetic field \( B \). Align the plate in such a way that the magnetic field lines penetrate the plate perpendicularly, that is at a 90° angle. This makes it easier to study the Hall effect.
So the magnetic field points either into the plane (screen) or out of the plane.
What is the Hall effect and how is Hall voltage generated?
Let's take electrons to explain the Hall effect (the explanation works analogously with the holes). Let the electrons move from the left to the right part of the plate. Also, let's consider a magnetic field \(B\) pointing INTO the plane.
We have an electric current \(I\) consisting of negatively charged electrons moving through a magnetic field \(B\). Moving electrons experience a magnetic force \(F_{\text m} \) (Lorentz force) in the magnetic field, which deflects the electrons into the upper or lower part of the plate.
Whether the electrons are deflected downward or upward depends on the selected direction of the magnetic field and the direction of the current. The left hand rule then tells you the direction of the magnetic force. In our case:

Thumb points to the left because the electrons move to the left.

Index finger points into the screen because the magnetic field points into the screen.

The middle finger points in the direction of the magnetic force. The electrons are thus deflected into the upper part of the plate.
The electrons gradually accumulate in the upper part of the plate. So there are more electrons in the upper region than in the lower region. So at the top you have an electron excess and at the bottom you have an electron deficiency.
This difference in charge causes an electric field \( E \), which by definition points from the positive pole (lower part) to the negative pole (upper part). All other electrons traveling through the plate now experience not only a magnetic but also an electric force \( F_{\text e} \). This force acts downward on the electrons because they are attracted to the positive pole.
This deflection of the electrons happens until a equilibrium of forces is established.
The upper and lower parts of the plate correspond to the oppositely charged electrodes of a plate capacitor, which are placed at distance \( h \) from each other. Here \(h\) is also the width of the Hall plate.
The homogeneous Efield of a plate capacitor is related to the voltage \(U_{\text H}\) between the plates as follows:
The voltage \( U_{\text H} \) is called Hall voltage. In contrast to the Efield inside the plate, you can easily measure the Hall voltage directly with a voltmeter.
After a few simple steps, we can derive the formula for the Hall voltage:
Here \(d\) is the thickness of the plate and \(A_{\text H}\) is a socalled Hall constant (or Hall coefficient).
What is the Hall constant and how do I determine it?
Hall constant \( A_{\text H} \) is a constant that depends on the used material.
The Hall constant depends on the charge carrier density \( n \). In our case, it depends on the number of conduction electrons per volume of the plate. But it also depends on the charge \( q \) that a single particle carries. In our case \( q \) is the negative elementary charge \(q =  e \). Consequently, the Hall constant is negative. However, it can also be positive if holes, rather than electrons, are responsible for conduction.
With the help of the derived formula 3
you can determine the Hall constant in an experiment. You can measure the current with an ammeter, the Hall voltage with a voltmeter, set the magnetic field yourself and measure the thickness of the plate with a ruler:
Once you have determined the Hall constant, you can use it to find out the charge carrier density of the plate according to Eq. 4
, and from the sign of the Hall constant you know whether the current through the plate is dominated by electrons or holes.
Material  Hall constant \(A_{\text H}\) 

Cuprium (Copper)  5.3 · 10^{11} m^{3}/C 
Aluminum  +9.9 · 10^{11} m^{3}/C 
Argentium (Silver)  8.9 · 10^{11} m^{3}/C 
Also note that the charge carrier density depends on the temperature of the plate and so does the Hall constant.
Exercises with Solutions
Use this formula eBook if you have problems with physics problems.Exercise #1: Calculate Magnetic Field in Horseshoe Magnet using a Hall Probe
You want to determine the strength of the magnetic field inside a horseshoe magnet. To do this, you need to determine the magnetic flux density \( B \).
To measure this, you insert a Hall probe into the horseshoe magnet. The Hall probe has a Hall coefficient \( A_{\text H} \) = 5·10^{7} m^{3}/As and an electric current \( I \) = 5A flowing through it. Additionally, the Hall probe indicates a Hall voltage of \( U_{\text H} \) = 25mV. The dimensions of the Hall probe are: \( l \) = 3.5cm, \( d \) = 0.05mm, and \( b \) = 3.5cm.
Solution to Exercise #1
A Hall probe utilizes the Hall effect to measure the strength of the magnetic field. You are given the Hall voltage, so you can use the following formula:
1 \[ U_{\text H} ~=~ A_{\text H} \, \frac{I \, B}{d} \]Rearrange equation 1
for the magnetic flux density \( B \):
Now, substitute the given values into equation 2
to find the strength of the magnetic field:
Exercise #2: Magnetic Field inside Horseshoe Magnet using a Hall Probe
A silicon wafer with length \( l = 1 \, \mathrm{cm}\), width \(b = 1 \, \mathrm{mm}\), and thickness \(d =1 \, \mathrm{mm}\) is traversed by a current \(I = 3.36 \, \mathrm{A} \) longitudinally, perpendicular to a magnetic field \(B = 1 \, \mathrm{T}\). The wafer has a density of \(n_{\text p} = 42 \cdot 10^{20} \, 1/\mathrm{m}^3 \) defect electrons.
 Calculate the drift velocity \(v\) of the defect electrons.
 Calculate the "force" \( F_{\text L} \) on a positive defect electron.
 Calculate the Hall voltage \( U_{\text H} \).
Solution to Exercise #2.1
Drift velocity is the velocity \( v \) at which the defect electrons move along the wafer. They cover a certain distance, namely the wafer length \( l \), within a certain time \( t \), which precisely corresponds to the definition of velocity: 1 \[ v ~=~ \frac{l}{t} \]
All defect electrons together form a current flow with a certain current strength \( I \), defined as charge per time: 2 \[ I ~=~ \frac{Q}{t} \]
Although you don't know how much charge \( Q \) flows within a time interval, what you do know is that the total charge \( Q \) can be written as the number \( N \) of charges multiplied by the elementary charge \( e \): 3 \[ Q ~=~ N \, e \]
Unfortunately, you don't know the number \( N \), but you do know the volume of the wafer \( V ~=~ l \, b \, d \) and the defect electron density \( n_{\text p} \). To calculate the number \( N \) of defect electrons in this volume, you calculate: 4 \[ \begin{align}N ~&=~ n_{\text p} \, V \\\\ ~&=~ n_{\text p} \, l \, b \, d \end{align} \]
Substitute equation 4
into 3
and then 3
into 2
, and rearrange for time:
5
\[ t ~=~ \frac{ n_{\text p} \, l \, b \, d \, e}{I} \]
Now you can plug the time from 5
into the velocity formula 1
(where length l cancels out):
6
\[ v ~=~ \frac{ I }{ n_{\text p } \, b \, d \, e} \]
If you now substitute the given values, you'll get the velocity of the electrons: 7 \[ \begin{align} v ~&=~ \frac{ 3.36 \, \text{A} }{ 42*10^{20} \, \frac{1}{\text{m}^3} ~\cdot~ 0.001 \, \text{m} ~\cdot~ 0.001 \, \text{m} ~\cdot~ 1.6 \cdot 10^{19} \, \text{C} } \\\\ ~&=~ 5000 \, \frac{\text m}{\text s} \end{align} \]
Solution to Exercise #2.2
The electric current \( I \), more precisely the flowing charge carriers, are affected by the Lorentz force \( F_{\text L} \) in the magnetic field \( B \), which deflects the charge carriers of the current to the edge of the wafer. Each one experiences a force of: 8 \[ F_{\text L} ~=~ e \, v \, B \]
Since these are defect electrons, each of them carries the elementary charge \( +e \). Substituting the given values (including the drift velocity determined in Exercise 2.1) yields the following Lorentz force: 9 \[ \begin{align} F_{\text L} ~&=~ 1.6 \cdot 10^{19} \, \text{C} ~\cdot~ 5000 \, \frac{\text m}{\text s} ~\cdot~ 1 \, \text{T} \\\\ ~&=~ 10^{16} \, \text{N} \end{align}\]
Solution to Exercise #2.3
Hall voltage \( U_{\text H} \) occurs precisely when the Lorentz force \( F_{\text L} \) is completely neutralized by the attractive force of opposite charges (electric force) \( F_{\text e} ~=~ Q \, E \). So, a force equilibrium occurs at the value of the Hall voltage: 10 \[ Q \, v \, B ~=~ Q \, E \]
Since the charge carriers collect at the upper and lower ends of the wafer, this can be interpreted as a kind of plate capacitor. In this case, the electric field \( E \) is the voltage \( U_{\text H} \) (here, Hall voltage) per distance \( b \) (which corresponds to the wafer width): 11 \[ E ~=~ \frac{U}{b} \]
Now substitute 11
into 10
and rearrange for the Hall voltage:
12
\[ U_{\text H} ~=~ b \, v \, B \]
Just plug in all the values (including the drift velocity from Exercise 2.1) and you'll get the desired Hall voltage: 13 \[ U_{\text H} ~=~ 0.001 \, \text{m} ~\cdot~ 5000 \, \frac{\text m}{\text s} ~\cdot~ 1 \, \text{T} ~=~ 5 \, \text{V} \]