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Hall potential difference

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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)

    3. 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.
  2. jcsd
  3. Apr 20, 2012 #2


    User Avatar

    Staff: Mentor

    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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  4. Apr 20, 2012 #3
    Vmax = Vvert + Vhoriz
    Vmin = Vvert - Vhoriz
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