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

Procedure
The Procedure that we followed can be found here

Equipment
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}

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

The accepted value of e/m is

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

Error
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.

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.