User:Ryan P. Long/Notebook/Physics 307L/2009/11/09: Difference between revisions

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with my slope and constant current e/m can be calculated:
with my slope and constant current e/m can be calculated:


<math>\frac{e}{m}=\frac{1}{slope}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}</math>
<math>\frac{e}{m}=\frac{1}{slope}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}=</math>


<math>\frac{e}{m}=\frac{1}{.00000814}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}</math>
<math>\frac{e}{m}=\frac{1}{.00000814}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}</math>

Revision as of 20:37, 21 November 2009

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e/m Lab

Equipment

  • BK Precision Multimeter 2831B (x2)
  • HP 6384A DC Power Supply
  • Uchida Yoko e/m experimental apparatus model TG-13
  • Soar Corp. 7403 DC power supply
  • Gelman Instrument Company 500V Deluxe Regulated power supply
  • BNC Cables

Setup

First we connected the HP power supply to the Helmholtz coils on the Uchida apparatus with one of the BK multimeters in series. Next the other BK multimeter was connected to the voltmeter jacks on the Uchida apparatus. The Soar power supply was connected to the heater jacks on the Uchida. Finally, the Gelman was connected to the electrodes on the Uchida. We followed the procedure in Professor Gold's manual, which included letting the heater warm up for 2 minutes before applying voltage to the electrodes on the apparatus.

To take data, we let either let the voltage of the electrodes or the current running through the coils be constant, while varying the other parameter. We measured the radius of the rings on the right and the left side using the built in ruler.


Data

Note: At a current of -1.35A, maximum accelerating voltage is 320V, the minimum is 165V. At -1.05A, the maximum voltage is 200V and the minimum is 100V.

11/16/09 Voltage applied to heater jacks: 6.24 V

Raw Data:

{{#widget:Google Spreadsheet

key=0Ajaf_99vseiXdEIyXzJxWGJhLUQ2SE9qNGo4UURmMUE width=760 height=300

}}

Analysis

Given that: (From Alexandra Andrego's Lab Notebook):

[math]\displaystyle{ B=(7.8\times10^{-4}\frac{weber}{amp-meter})\times I\,\! }[/math]
[math]\displaystyle{ {e}{V}=\frac{1}{2}{m}{v}^{2}\,\! }[/math]
[math]\displaystyle{ {F}_{B}={q}{v}{B}\,\! }[/math]
[math]\displaystyle{ \frac{e}{m}=\frac{{2}{V}}{{r}^{2}{B}^{2}}=\frac{{2}{V}}{{r}^{2}{({{7.8}\times10}^{-4}{I})}^{2}}\,\! }[/math]

We can calculate [math]\displaystyle{ \frac{e}{m}\,\! }[/math] based on the slope of a [math]\displaystyle{ \frac{V}{{r}^{2}}\,\! }[/math] line holding current constant, or the slope of a [math]\displaystyle{ \frac{r}{\frac{1}{I}}\,\! }[/math] line holding voltage constant.

Part 1

I calculated the slope and slope uncertainty of the line from my plot using the linest array function in excel.

Using the equation:

[math]\displaystyle{ \frac{e}{m}=\frac{{2}{V}}{{r}^{2}{B}^{2}}=\frac{{2}{V}}{{r}^{2}{({{7.8}\times10}^{-4}{I})}^{2}}\,\! }[/math]

with my slope and constant current e/m can be calculated:

[math]\displaystyle{ \frac{e}{m}=\frac{1}{slope}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}= }[/math]

[math]\displaystyle{ \frac{e}{m}=\frac{1}{.00000814}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}} }[/math]

Part 2


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