(adsbygoogle = window.adsbygoogle || []).push({}); 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 B_{E}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.

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# Measurement of e/m for electrons question

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