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Measurement of e/m for electrons question

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Using Equations (A) and (B) explain how to obtain for e/m from the slope of an I vs. [tex] \sqrt{V}/{r} [/tex] graph. Show how to calculate a value for the Earth's magnetic field BE using the intercept of the graph.


    2. Relevant equations

    (A) [tex] B_H = \frac{8NI \mu_0}{R \sqrt{125}} [/tex] (B) [tex] \frac{e}{m} = \frac{2V}{r^2 (B_H -B_E)^2} [/tex]

    where N=72 turns R=0.33m [tex]\mu_0=4\pi*10^-7 \frac{Tm}{A}[/tex] for helmholtz coil and I measured both current and Voltage

    3. The attempt at a solution

    I first move the magnetic field to the other side
    [tex]\frac{e}{m}(B_H -B_E)^2 = \frac{2V}{r^2}[/tex]

    Then I square both sides
    [tex]\sqrt {\frac {e}{m}}(B_H -B_E)= \frac {\sqrt {2V}}{r} [/tex]

    I then move the qm ratio to the other side and insert equation (A) into it to find
    [tex] (\frac{8NI \mu_0}{R \sqrt{125}} -B_E)= \sqrt {\frac {m}{e}} \frac {\sqrt {2V}}{r}[/tex]

    let [tex]α= \frac{8N \mu_0}{R \sqrt{125}} [/tex] so then me equation becomes

    [tex]αI = \sqrt {\frac {m}{e}} \frac {\sqrt {2V}}{r} + B_E[/tex]

    I now have it in a y=mx+b form but when i enter my numbers and graph the equation on excel(ATTACHMENT) I find that my [tex]\frac {e}{m} and B_E[/tex] are off by a factor of 2
    I was wondering if I linearized the equation wrong or did I forget to put a another factor into finding the [tex]\frac {e}{m}[/tex] ratio. I tired using both the linest function and least square fit to find the slope and intercept and both are also off by a factor of 2.
     

    Attached Files:

    Last edited: Oct 1, 2011
  2. jcsd
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