Calculating Hall Potential Difference in a Subway Third Rail

In summary: So, in summary, the question asks to calculate the Hall potential difference across the steel third rail in a subway system, using its dimensions and the Earth's magnetic field components. The formulas used are |Vh|vert = (I*Bhoriz)/(n*e*w) and |Vh|horiz = (I*Bvert)/(n*e*L). The calculated values for Vmax and Vmin may be incorrect due to ambiguity in the diagonal being referred to in the question.
  • #1
NikitaY
2
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Homework Statement



The electric third rail in a subway system is made of steel, measures h=0.1m high, w=0.02m wide, and carries a current I=110A. Calculate the Hall potential difference across the rail from corner to opposite corner due to the Earth's magnetic field with components Bhoriz=3e-05T and Bverti=2.7e-05T. Steel has 8e+28 conducting electrons per cubic meter, or a conducting charge density of 12800000000C/m3. You can relate this conducting charge density to the speed of the electrons and the current density through the rail.

Vmax = __ V
Vmin = __ V

Homework Equations



|Vh|vert = Evert*L = Vd*Bhoriz*L = (I/newL)*Bhoriz*L = (I*Bhoriz)/(n*e*w)

|Vh|horiz = Ehoriz*w = (I/(n*e*w*L))*Bvert*w = (I*Bvert)/(n*e*L)

The Attempt at a Solution



Plugging in the numbers into the formulas from above, I got the following:

|Vh|vert = (110A*3e-5T)/(8e28*1.6e-19C*.02m) = 1.29e-11

|Vh|horiz = (110A*2.7e-5T)/(8e28*1.6e-19C*.1m) = 2.32e-12

The problem is asking me to find the Hall potential difference, but what I really need to do is find the max and min voltage, which I thought I did - I am getting the answer marked as wrong, however. If anyone can shine any light on this problem, it would be greatly appreciated! Thank you.
 
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  • #2
NikitaY said:

Homework Statement



The electric third rail in a subway system is made of steel, measures h=0.1m high, w=0.02m wide, and carries a current I=110A. Calculate the Hall potential difference across the rail from corner to opposite corner due to the Earth's magnetic field with components Bhoriz=3e-05T and Bverti=2.7e-05T. Steel has 8e+28 conducting electrons per cubic meter, or a conducting charge density of 12800000000C/m3. You can relate this conducting charge density to the speed of the electrons and the current density through the rail.

Vmax = __ V
Vmin = __ V

Homework Equations



|Vh|vert = Evert*L = Vd*Bhoriz*L = (I/newL)*Bhoriz*L = (I*Bhoriz)/(n*e*w)

|Vh|horiz = Ehoriz*w = (I/(n*e*w*L))*Bvert*w = (I*Bvert)/(n*e*L)

The Attempt at a Solution



Plugging in the numbers into the formulas from above, I got the following:

|Vh|vert = (110A*3e-5T)/(8e28*1.6e-19C*.02m) = 1.29e-11

|Vh|horiz = (110A*2.7e-5T)/(8e28*1.6e-19C*.1m) = 2.32e-12

The problem is asking me to find the Hall potential difference, but what I really need to do is find the max and min voltage, which I thought I did - I am getting the answer marked as wrong, however. If anyone can shine any light on this problem, it would be greatly appreciated! Thank you.

When they say to find the potential "across the rail from corner to opposite corner", presumably this means a diagonal of the cross section? If so, which diagonal do they mean? A-->C, or B --D?

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  • #3
Vmax = Vvert + Vhoriz
Vmin = Vvert - Vhoriz
 

What is Hall potential difference?

Hall potential difference is an electric potential difference that occurs in a conductor when it is placed in a magnetic field that is perpendicular to the direction of current flow. It is a result of the Lorentz force acting on the moving charges in the conductor, causing them to accumulate on one side of the conductor, creating a potential difference.

How is Hall potential difference measured?

Hall potential difference is measured using a Hall probe, which is a device that consists of a thin strip of conducting material and a magnetic field sensor. When placed in a magnetic field, the Hall probe produces a voltage that is proportional to the strength of the magnetic field and the current flowing through the conductor. This voltage is the Hall potential difference.

What is the relationship between Hall potential difference and magnetic field strength?

The Hall potential difference is directly proportional to the strength of the magnetic field. This means that as the magnetic field strength increases, the Hall potential difference also increases. However, the direction of the Hall potential difference depends on the direction of the magnetic field and the direction of current flow.

What are the applications of Hall potential difference?

Hall potential difference has various applications in science and technology. It is commonly used in electronic devices such as Hall effect sensors, which are used to measure magnetic fields and detect the presence of magnetic materials. It is also used in the study of semiconductors and in the measurement of flow rates in fluids.

How can Hall potential difference be manipulated?

Hall potential difference can be manipulated by changing the strength or direction of the magnetic field, or by changing the current flowing through the conductor. It can also be affected by the properties of the conductor, such as its resistivity and thickness. The Hall potential difference can also be used to manipulate the magnetic field, as seen in the operation of some electric motors.

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