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Percent correction for Magnetic Field vs. Electric Field!

  1. Jun 10, 2014 #1
    Hi guys,
    In the photo is the problem. We set Fb =Fe to show for E then out E into the Voltage equation to get V=vlB. However, if we won;t ignore the gravity force, what is the percent correction? (Note: q =Electron)
     

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  2. jcsd
  3. Jun 10, 2014 #2

    Simon Bridge

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    Welcome to PF;
    What is the definition of "percent correction" - that will tell you how to work it out.
    In the presence of gravity - draw a free-body diagram for the electron.
     
  4. Jun 10, 2014 #3
    i did a little work, but idk if it is right?
     

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  5. Jun 10, 2014 #4

    Simon Bridge

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    You missed out defining what "percentage correction" means.
    What are you correcting and why?

    The fbd and calculation seems OK.

    qvlB = qV + mgl

    So V= vlB-gl(m/q) with gravity

    Without gravity: V0=vlB

    If you used B=V0/vl with V0=V then you have introduced a systematic error into your equation for determining B.
     
  6. Jun 19, 2014 #5

    berkeman

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    Please check your PMs...
     
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