# User:Richard T. Meyers/Notebook/Phys307l/E/m Lab

## Procedure

SJK 22:22, 13 October 2010 (EDT)
22:22, 13 October 2010 (EDT)
This is a good start on the primary notebook, but some things are missing. It's good that you include the photos and your embedded spreadsheets for the raw data. Your calculations are good and I can tell what you did for the analysis, which is the point. (I don't agree with some analysis, but that's separate from me being able to reproduce what you did.) On the other hand, it is a bit lacking in recording what you did while you were taking data. There are many things that you probably did which are either not described in the manual, or are different. Remember that the point of your primary notebook page is for you or someone else to be able to reproduce your work. It should also include failed attempts, bad data, etc.

The Procedure that we followed can be found here

The Apparatus
The Apparatus working
No arc

## Equipment

SJK 22:17, 13 October 2010 (EDT)
22:17, 13 October 2010 (EDT)
Good use of photo for setup.

6-9V Power supply rated at 2A (Gelman Deluxe Regulated Power Supply)

Power supply of 6.0V max rated at 1.5A (Soar DC Power Supply model PS 3630)

150-300V power supply rated at 40mA (Hewlett Packard 6236B)

3 Voltmeters

e/m Experimental Apparatus

## Calculations

At first we use the equation for the magnetic field

$B=\frac{\mu R^2NI}{(R^2+x^2)^{3/2}}$

So from this we know that:

${N=130}\,\!$

${R=0.15m}\,\!$

${x=\frac{R}{2}}$

${\mu=4\pi x10^{-7}}\,\!$

so

${B=7.79x10^{-4}I}\,\!$

Then we can say that since $F_c=F_B\,\!$

So

$F_c=eV=\frac{mv^2}{2}$ and

$F_B=evB\,\!$

so we get

$v=\frac{eBr}{m}$

Also from the data of the plot of $r^2 versus V\,\!$

$slope=0.002954x10^{-5}m^2/V\,\!$

this is just the inverse of the slope of second chart converted from cm to m, and the average current is $I=1.42875A\,\!$

$\frac{e}{m}=\frac{2}{((7.79x10^{-4})(1.42875))^2 slope}=6.747x10^8\,\!$

The second approximation of e/m is as follows:

$\frac{e}{m}=\frac{2}{({7.79x10^{-4}I})^2{0.2954x10^{-5}}}$

from the plot of radius versus inverse current we get:

$\frac{1}{r}=\sqrt{\frac{(7.79*10^{-4})^2eI}{2Vm}}\,\!$

So we can use the slope of the first graph to say

$\frac{1}{slope}=\sqrt{\frac{(7.79*10^{-4})^2e}{2Vm}}\,\!$

And $slope=0.04898 \frac{1}{mA}$

$\frac{e}{m}=\frac{2V}{{7.79x10^{-4}}^2 slope^2}=$\frac{2x199.55}{{7.79x10^{-4}}^2slope^2}[/itex]

$\frac{e}{m}=2.741x10^{11}$

The accepted value of e/m is

$\frac{e}{m}=1.758x10^{11} \frac{C}{kg}\,\!$

## Error

SJK 22:20, 13 October 2010 (EDT)
22:20, 13 October 2010 (EDT)

With our accepted value at:

$\frac{e}{m}=1.758x10^{11} \frac{C}{kg}\,\!$

and the two calculated values at:

$\frac{e}{m}=6.747x10^8\,\!$

$\frac{e}{m}=2.741x10^{11}$

we can obviously see that there is a large discrepancy with the data.

One set is large and the other is small so we assume that the correct number is somewhere between the two.

So the average of the two gets:

$\frac{e}{m}=1.374x10^{11}\,\!$

Compare this with the actual number and we get and error of:

$error=\frac{1.758x10^{11}-1.374x10^{11}}{1.758x10^{11}}x100=21.84%$

This error for the amount of data points taken is not unreasonable.

SJK 22:19, 13 October 2010 (EDT)
22:19, 13 October 2010 (EDT)
You do link to the procedure in Gold's manual, but you don't appear to note any significant changes. In your primary notebook, this is what you should do. For example, did you do the qualitative experiments? Did you adjust the stated procedure at all? Any other notes as you went along?

## Citation

1) I found the accepted value of e/m here

## Thanks

1)Nathan for being my excellent lab partner and for taking down the equipment list when I had failed to.

2)Emran for setting up the lab previously and for a general outline of my report.

3)Randy also for setting up the lab previously.