Is the Solution for Finding Hall Voltage Correct?

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In summary: I don't understand where you disagree with me. We both say that the current is in the x-axis and that the hall voltage is in the y-axis. The way the OP did the calculation implies that the hall voltage is along the x-axis.Perhaps it is a nomenclature difference.
  • #1
PhysicsTest
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Homework Statement
a. Find the magnitude of the Hall voltage VH in an n-type germanium bar used in fig, having majority-carrier concentration ND = 10^17 /cm^3. Assume Bz=0.1 Wb/m^2, d=3 mm, Ex=5V/cm.
b. What happens to VH if an identical p-type germanium bar having NA=10^17/cm^3 is used in part a?
Relevant Equations
##V_H = Ed=Bvd = \frac{BJd} {\rho w} ##
The question is to find the Hall voltage
1610202843009.png

The magnetic field is in the +ve Z-direction, the electric field is in the +ve X-direction, the current will be in -ve Y direction.
There are many equations to find the hall voltages ##V_H = Ed=Bvd = \frac{BJd} {\rho w} ##. But i find the equation ##V_H = Ed## to solve and other parameters are not much of use.
a. n-type 4 will be more negative than 3 and ##V_H = 5*0.3V = 1.5V##
b. p-type 4 will be more positive than 3 and ##V_H = 5*0.3V = 1.5V##
Is the solution correct, or missing something?
 
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  • #2
No I don't think it is correct. The E-field in the x direction ##E_x=5V/cm## is responsible for creating the current density ##J=\sigma E_x## according to Ohm's law.
It is the E-field in the y-direction ##E_y=Bv## that creates the hall voltage.
 
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  • #3
Delta2 said:
No I don't think it is correct. The E-field in the x direction ##E_x=5V/cm## is responsible for creating the current density ##J=\sigma E_x## according to Ohm's law.
It is the E-field in the y-direction ##E_y=Bv## that creates the hall voltage.
I disagree with the disagreement.

The interaction of the current flow and B field creates a Hall voltage that is perpendicular to both, think Lorentz force. The problem statement specifies the direction of both current direction, X-axis, and B-field direction, Z axis. This leave only the Y axis for the Hall voltage. Note that the Terminals "1" and "2" (sense terminals) are in the Y dimension. (But that graph sure shows an unusual axes orientation.)

As for the numerical solution, I have no idea! That must be left to those that have studied this in more detail.
 
  • #4
Tom.G said:
I disagree with the disagreement.

The interaction of the current flow and B field creates a Hall voltage that is perpendicular to both, think Lorentz force. The problem statement specifies the direction of both current direction, X-axis, and B-field direction, Z axis. This leave only the Y axis for the Hall voltage. Note that the Terminals "1" and "2" (sense terminals) are in the Y dimension. (But that graph sure shows an unusual axes orientation.)

As for the numerical solution, I have no idea! That must be left to those that have studied this in more detail.
I don't understand where you disagree with me. We both say that the current is in the x-axis and that the hall voltage is in the y-axis. The way the OP did the calculation implies that the hall voltage is along the x-axis.
 
  • #5
Perhaps it is a nomenclature difference.

Re-reading your two posts I see that we agree on the orientation of the Hall voltage.

I was thrown off by the use of 'E-field' being the cause of the Hall voltage. I was thinking of 'fields' being externally applied things, rather than being electron paths.

Sorry for any unnecessary confusion.

Tom
 
  • #6
Tom.G said:
I was thrown off by the use of 'E-field' being the cause of the Hall voltage
I said that this E-field is in the y-direction. Of course the mechanism of hall voltage is known to me, it is not the electric field the real cause. It is the Lorentz force ##(\mathbf{v}\times\mathbf{B})q## that creates charge separation and a coulomb electric field ##\mathbf{E_H}## such that $$\mathbf{E_H}q=(\mathbf{v}\times\mathbf{B})q$$ and the hall voltage is of course $$V_H=E_Hd$$
 
  • #7
Still not able to solve, not sure if some of the data is missing or to assume something. I understood the above discussion
##V_H = \frac{BJd} {\rho w} ## ->eq1
##J = \sigma E_x## ->eq2
substitute eq2 in eq1
##V_H = \frac{B\sigma E_x} {\rho w}## ->eq3
If conduction primarily due to charges of one sign, the conductivity ##\sigma## is related to mobility ##\mu## by
##\sigma = \mu \rho## -> eq4
substitute in eq3 to give
##V_H = \frac{B\mu E_x} {w} ## -> eq5
Now i do not know the value of ##\mu##. Is the above derivation correct? Can i assume its value at 300K which is a standard value i assume.
 
  • #8
PhysicsTest said:
Still not able to solve, not sure if some of the data is missing or to assume something. I understood the above discussion
##V_H = \frac{BJd} {\rho w} ## ->eq1
##J = \sigma E_x## ->eq2
substitute eq2 in eq1
##V_H = \frac{B\sigma E_x} {\rho w}## ->eq3
If conduction primarily due to charges of one sign, the conductivity ##\sigma## is related to mobility ##\mu## by
##\sigma = \mu \rho## -> eq4
substitute in eq3 to give
##V_H = \frac{B\mu E_x} {w} ## -> eq5
Now i do not know the value of ##\mu##. Is the above derivation correct? Can i assume its value at 300K which is a standard value i assume.
In going from Eq1 to Eq3 (using Eq2) you have forgotten ##d##.
What is ##\rho##? The density of charge carriers?
 
  • #9
Delta2 said:
In going from Eq1 to Eq3 (using Eq2) you have forgotten ##d##.
Sorry my mistake the eq(3) is
##V_H = \frac{B\sigma E_xd} { \rho w}## ->eq(3)
Delta2 said:
What is ##\rho##? The density of charge carriers?
Yes ##\rho## is the density of charge carriers
 
  • #10
I feel there is something wrong regarding the ##w## in eq(3). As far as I know $$V_H=Bvd, J=\rho v\Rightarrow V_H=\frac{BJd}{\rho}$$ so I really don't see how you get that ##w## there. Is this equation $$V_H=\frac{BJd}{\rho w}$$ taken straight from the book?
 
  • #11
I am really confused with all these equations
1610429340806.png

Yes it is the current I not the J the current density.
 
  • #12
PhysicsTest said:
I am really confused with all these equations
View attachment 276051
Yes it is the current I not the J the current density.
Ah this settles it, yes the book is right (and I was right too in post #10). It is $$I=JS=Jdw$$ that's how the last equality is explained. (##S## is the cross section of the semiconductor ##S=dw## according to the figure provided).
So to sum it up:
The correct (eq3) is $$V_H=\frac{B\sigma E_xd}{\rho}$$ and IF $$\sigma=\mu\rho$$ is correct then (eq3) leads to $$V_H=B\mu E_xd$$. I suppose from this last equation you can find the hall voltage, if you know ##\mu## as everything else is given.

Something else is the density of the charge carriers ##N_D## (or ##N_A##) given in ##cm^{-3}## or ##Cb\cdot cm^{-3}## because the ##\rho## appearing in the above equations has units of the latter case (##Cb\cdot cm^{-3}##).
 
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  • #13
Delta2 said:
I suppose from this last equation you can find the hall voltage, if you know ##\mu## as everything else is given
But that is not given, can i assume a value?
Delta2 said:
Something else is the density of the charge carriers ##N_D## (or ##N_A##) given in ##cm^{-3}## or ##Cb\cdot cm^{-3}## because the ##\rho## appearing in the above equations has units of the latter case (##Cb\cdot cm^{-3}##).
I really did not understand this.
 
  • #14
PhysicsTest said:
But that is not given, can i assume a value?
Not sure, what is given is ##N_D## or ##N_A##. I think we must use those somehow (but in the correct units, see below). Look in book for table with the ##\sigma## of germanium at various temperatures. I am very unsure about the equation ##\sigma=\mu\rho##.
I really did not understand this.
I am asking in what units are the ##N_D## or ##N_A## given. In the OP it seems that the units are ##\frac{1}{cm^3}## or simply ##cm^{-3}##.
 
  • #15
i am not sure if the original question is wrong, i have taken from the book as it is

1610443864516.png

1610443997183.png

1610444087380.png

This is how i derived the equations. Generally in some of the previous questions i used to see ##N_D## units as ##atoms/cm^3##
 
  • #16
Ok i see. The only problem i see is whether the ##\rho## in this equation ##\sigma=\rho\mu## is the same quantity as the ##\rho## in this equation ##V_H=\frac{B\sigma E_xd}{\rho}##
Also what is q that appears in the equation for ##\sigma## in the image above. Is it the charge of the electron?

It seems that we need ##\mu## to calculate the hall voltage...
 
  • #17
Yes charge of the electron is "q" in the equation.
 
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  • #18
It seems to me that ##\rho=N_Dq## or ##\rho=-N_Aq## where q the charge of electron
 
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  • #19
One question i wanted to ask is that an external voltage (electric field ##E_e##) is used to drive the current I. Because of magnetic field a force is applied on the charged electrons and holes and hence the Hall voltage (Hall electric field ##E_H##) is set up. Will there be a net electric field of ##E_e \text{ and } E_H## ?
 
  • #20
Yes there will be a net electric field from those two E-fields, but the effect of the field ##E_H## on the charge carriers is "neutralized" by the force ##(\mathbf{v}\times \mathbf{B})q## which is equal and opposite of the force ##E_Hq##. So ##E_H## is like it doesn't exist for the charge carriers, it doesn't affect their motion, it doesn't make a new current if that's what you were thinking of.
 
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Related to Is the Solution for Finding Hall Voltage Correct?

1. What is the Hall effect and how does it relate to finding Hall voltage?

The Hall effect is a phenomenon in physics where a magnetic field applied perpendicular to an electric current causes a voltage difference perpendicular to both the magnetic field and the current. This voltage difference is known as the Hall voltage, and it can be used to measure the strength of a magnetic field or the properties of a material. Therefore, the Hall voltage is crucial in finding the Hall effect and understanding its applications.

2. How is the Hall voltage calculated?

The Hall voltage is calculated by multiplying the magnetic field strength, the electric current, and the Hall coefficient of the material being tested. The Hall coefficient is a material-specific constant that relates the strength of the magnetic field to the generated Hall voltage. The formula for calculating Hall voltage is VH = B * I * RH, where B is the magnetic field strength, I is the electric current, and RH is the Hall coefficient.

3. What factors can affect the accuracy of finding Hall voltage?

Several factors can affect the accuracy of finding Hall voltage, such as the quality of the equipment used, the uniformity of the magnetic field, and the temperature of the material being tested. The Hall coefficient can also vary with temperature, so it is essential to maintain a constant temperature during the experiment to ensure accurate results.

4. How can the accuracy of finding Hall voltage be improved?

To improve the accuracy of finding Hall voltage, it is crucial to use high-quality equipment and calibrate it properly before conducting the experiment. Additionally, the magnetic field should be as uniform as possible, and the temperature of the material should be carefully controlled. Using multiple measurements and taking the average can also help improve accuracy.

5. What are the applications of finding Hall voltage?

The Hall effect and the resulting Hall voltage have several practical applications, such as measuring the strength of a magnetic field, detecting the presence of a magnetic field, and determining the properties of a material, such as its charge carrier density and mobility. It is also used in various electronic devices, such as Hall sensors, which are used in speedometers, tachometers, and other devices that require precise measurements of magnetic fields.

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