How Does Free Electron Density Affect Hall Voltage in a Semiconductor?

[catfisherman]

The question is...

A Hall probe is constructed from a semi-conducting material of thickness 0.4 millimeters. When 5 amps of current passes through the material a Hall voltage of .6 volts is produced when the probe is placed in a uniform magnetic field of 0.01 Tesla. Assuming the current is due to "free electrons" in motion, determine the following:
a. The "free electron density" of the semi-conducting material

b. The Hall voltage produced by the Hall probe when it is placed in a magnetic field of 0.3 Tesla


My opinion- I am mostly concerned with part a because I am lost. I am thinking you would use the equation for Coloumb's law Fe=ke*(q1*q2)/r^2 and solving for ke. Does it sound like I'm on the right track?

For part b I was thinking I would use DeltaV=E*d=vdBd

One of my main problems is just knowing what to plug in where. If anyone can help please do. Your assistance will be greatly appreciated! Thanks!

 

[Nate810]

For part a, the equation you need to use is VH = (ρe/B)j, where ρe is the free electron density of the semi-conducting material and B is the magnetic field strength. You can rearrange this equation to obtain ρe = (VH*B)/j, where VH is the Hall voltage produced by the Hall probe when it is placed in a uniform magnetic field of 0.01 Tesla and j is the current flowing through the material. Plugging in the values given, the free electron density of the semi-conducting material is 0.6 x 0.01 / 5 = 1.2 x 10^-4 m^-3.For part b, you can use the same equation, VH = (ρe/B)j, but with the new magnetic field strength of 0.3 Tesla. Plugging in the values given, the Hall voltage produced by the Hall probe when it is placed in a magnetic field of 0.3 Tesla is 1.2 x 10^-4 x 0.3 / 5 = 7.2 x 10^-6 V.

 

[Moetasim]

I can provide a response to the content and help clarify the concepts involved. Firstly, the term "free electron density" refers to the number of free electrons per unit volume in a material. In a semi-conducting material, the free electrons are responsible for conducting electricity.

To determine the free electron density in this scenario, we need to use the formula for the Hall voltage (VH) in a Hall probe, which is VH = (IB)/(nq), where I is the current passing through the material, B is the magnetic field strength, n is the free electron density, and q is the charge of an electron.

Using the given values of I = 5 amps, VH = 0.6 volts, and B = 0.01 Tesla, we can rearrange the formula to solve for n. This gives us n = (IB)/(qVH) = (5*0.4*10^-3)/(1.6*10^-19*0.6) = 6.25 x 10^22 electrons/m^3.

For part b, we can use the same formula but with the new value of B = 0.3 Tesla. This gives us VH = (5*0.4*10^-3)/(1.6*10^-19*0.3) = 5.56 volts.

It is important to note that the Hall voltage is directly proportional to the magnetic field strength, so a higher magnetic field will result in a higher Hall voltage. Also, the thickness of the material (0.4 millimeters) plays a role in the calculation, as it affects the number of free electrons per unit volume.

In summary, the free electron density in the semi-conducting material is 6.25 x 10^22 electrons/m^3, and the Hall voltage produced by the Hall probe in a magnetic field of 0.3 Tesla is 5.56 volts. I hope this helps to clarify the concepts involved.