Calculating Hall Potential Difference in a Subway Third Rail

[NikitaY]

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.

 

[gneill]

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?

 

[dancingyeti]

Vmax = Vvert + Vhoriz
Vmin = Vvert - Vhoriz