# Measurement of e/m for electrons question

1. Oct 1, 2011

### azntoon

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. $$\sqrt{V}/{r}$$ graph. Show how to calculate a value for the Earth's magnetic field BE using the intercept of the graph.

2. Relevant equations

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

where N=72 turns R=0.33m $$\mu_0=4\pi*10^-7 \frac{Tm}{A}$$ 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
$$\frac{e}{m}(B_H -B_E)^2 = \frac{2V}{r^2}$$

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

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

let $$α= \frac{8N \mu_0}{R \sqrt{125}}$$ so then me equation becomes

$$αI = \sqrt {\frac {m}{e}} \frac {\sqrt {2V}}{r} + B_E$$

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 $$\frac {e}{m} and B_E$$ 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 $$\frac {e}{m}$$ 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:

• ###### electron charge to mass ratio.xls
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Last edited: Oct 1, 2011