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Hall Effect

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data

    OeP0E2B.png

    2. Relevant equations



    3. The attempt at a solution

    $$Δ{ V }_{ H }=\frac { IB }{ nqt } \\ 9.6*{ 10 }^{ -6 }=\frac { 50*1.3 }{ n*e*3.3*{ 10 }^{ -4 } } \\ n=1.28*{ 10 }^{ 29 }\quad electrons/{ m }^{ 3 }$$
     
  2. jcsd
  3. Sep 30, 2013 #2

    mfb

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    Staff: Mentor

    You can use WolframAlpha to check calculations like that.
     
  4. Oct 1, 2013 #3
    I thought the method was wrong because it gives us d=1mm but we don't use it in our calculations.
     
  5. Oct 1, 2013 #4

    mfb

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    I'm not sure where the Hall voltage is measured, either 0.330mm or 1mm is not necessary.
     
  6. Oct 1, 2013 #5

    gneill

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    Staff: Mentor

    Take a look at the Hyperphysics web page on the Hall Effect.

    @Turion: Your calculation looks good.
     
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