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

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catfisherman
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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!
 
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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.