Measurement of e/m for electrons question

This means that the $\frac {e}{m}$ ratio can be calculated as the square of the slope, and the Earth's magnetic field can be calculated from the intercept using equation (A).
  • #1
azntoon
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Homework Statement


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.

Homework 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

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.
 

Attachments

  • electron charge to mass ratio.xls
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  • #2
I am not sure if I am missing a step or if I linearized it wrong. I know the numbers I am entering are correct so I am confused as to why this is happening. Any help would be greatly appreciated. A:Once you have it in the form $αI = \sqrt {\frac {m}{e}} \frac {\sqrt {2V}}{r} + B_E$, you can plot $V$ on the $x$-axis and $αI$ on the $y$-axis. The slope of the plot is $\sqrt {\frac {m}{e}} \frac {1}{r}$ and the intercept is $B_E$.
 

FAQ: Measurement of e/m for electrons question

1. How is e/m for electrons measured?

The e/m ratio for electrons is typically measured using an experiment called the Thomson's tube experiment. This involves passing a beam of electrons through a perpendicular magnetic field and measuring the deflection of the electrons. By knowing the strength of the magnetic field and the voltage applied to the tube, the e/m ratio can be calculated.

2. What is the significance of measuring e/m for electrons?

The e/m ratio is a fundamental property of an electron and is crucial in understanding its behavior in electromagnetic fields. It also played a crucial role in the development of the electron and atomic theories.

3. How accurate is the measurement of e/m for electrons?

The measurement of e/m for electrons has been refined over the years and is now known to be accurate to 1 part in 10 billion. This level of precision has been achieved through advancements in technology and experimental techniques.

4. Can the e/m ratio for electrons vary?

No, the e/m ratio for electrons is a constant value and does not vary. This has been experimentally verified and is a fundamental constant in physics.

5. Are there any other methods for measuring e/m for electrons?

Yes, other methods such as the Millikan oil drop experiment and the J.J. Thomson's charge-to-mass ratio experiment have also been used to measure the e/m ratio for electrons. However, the Thomson's tube experiment is the most commonly used method due to its simplicity and accuracy.

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