User:Justin Roth Muehlmeyer/Notebook/307L Notebook/EoverM
Introduction and Theory
By analyzing the motion of a beam of electrons in a magnetc and electric field, Thomson found that there is a ratio independent of all other experiemental factors: the charge-to-mass ratio. This of course proved his hypothesis that there is a constituent in this beam that is common to all matter: the electron. In this lab, we will re-live Thomson's experiment (or atleast a version of it), by acaccelerating electrons into a strong magnetic field created by Helmholtz coils surrounding the vacuum bulb. The electrons, experiencing a magnetic field force, will rotate in a circle on a plane perpendicular to the field. By oberving the radius of this circle, and by knowing the magnitude of the fileds, we will be able to determine the e/m ratio. We will do this numerous times for both constant V and changing V situations. Our hope is to approximate the accepted value of
e/m= 1.76e-11 C/kg
By heating a filament, particles are accelerated from the filament by a potential difference between two anodes directly n fron of the filament. The particles are given a kinetic energy of [math]\displaystyle{ \frac{1}{2}mv^{2}=eV }[/math] which can can be solved for its velocity [math]\displaystyle{ v=\sqrt{\frac{2eV}{m}} }[/math]. This particle then is thrown into the magnetic field.
The particle will then experience the Lorentz force for a particle moving in a magnetic field [math]\displaystyle{ \vec{F}=e(\vec{v} \times \vec{B}) }[/math]. We see that this puts the beam of particles into circular motion, and using the centripetal acceleration from Newton's Second Law where [math]\displaystyle{ \vec{F}=m \frac{\vec{v}^{2}}{R} }[/math] we can solve for the e/m ratio, using the velocity attained from the potential difference, to get:
[math]\displaystyle{ \frac{e}{m}=\frac{2V}{R^{2}B^{2}} }[/math]
Which is an expression for the ratio with parameters we can adjust a measure directly.
Set Up
We Followed procedure listed in Gold's manual tightly, with meters between every source of power/current and the experiment itself for accuracy (the power sources seem old, we do not completely trust the accuracy of their dials).
Power Supply for heater: SOAR DC POWER SUPPLY, model PS-3630
Connected voltmeter in parallel to measure the V for the heating coils PRECISELY at 6.200 V. This is a great improvement over the gross dial meter on the actual power supply.
Now we won't touch the knobs...
Rearranged, measured current from same supply: .727 A
Now for V on electrodes: 200.0 V
Current to coils: 1.199 A
Gold's manual is WRONG and the electrode voltage needs to be ~ 300 V in order to get a measurable circle radius.
Data
We decided, with the help of our TA Aram, to use high resolution photography to measure the radius of our circles. Placing the camera directly in front of the bulb and without moving it between trials, we figured that this will be a better way to measure the radius, rather than trying to align our heads with the mirror. We used a shutter speed of 30 seconds and received descent images of our circle. We will have to digitally measure the radius of the circles, and our hopes is that this will be better than measuring by eye, however this might not be the case and it might be of interest that we compare our values to the labs of other who did it by eye.
SJK 02:49, 23 October 2008 (EDT)
This is a clever method, and I think probably is better than doing it by eye. Whether the time is worth it, though???? Probably not, given the huge systematic error in this lab.
Sadly, I did not have the digital photo program capabilities I thought I had, so I had to resort to Microsoft Paint as my was of finding the radius. Paint has no tools for measuring pixels, so what I had to do be creative with my radius measuring (although maybe still better than eye?). I drew a line along the diameter of the circles, making sure that the diameter was in fact going through the middle of the circle and reaching to its farthest diameter (for the circles that had a slight eliptical nature to them, this means my diameter is a measure of their major axis). Then I cut this line accurately from the image and re-aligned it with the ruler. This gave me the length of the diameter, and dividing by 2 obviously will give me the radius.
The radii are recorded below. The error of ± .1 cm that I record is based on the following sources of error:
- Due to the fact that the ruler has the minimum increments of .1 cm. Usually it is standard procedure to estimate error one incremnent beyond the smallest division mark, however, in consideration of the below I decided that I will disregard that procedure to account for the other possible sources of error that certainly have an effect.
- Due to the misallignment of the ruler with the diameters of the circles (I could not rotate my diamter cutting to match the slight slope of the ruler relative to my line).
- Due to the width of the beam (which did not appear to exceed .1 cm).
- Due to the incoherency of the beam.
The e/m ration was determined as explained in the introduction below in an excel spreadsheet.SJK 02:50, 23 October 2008 (EDT)
Where is the spreadsheet?
Averages for the e/m ratio for the two experiements are found below.
SJK 02:50, 23 October 2008 (EDT)
These photos are great!
| Constant V Trials | Corresponding Pictures |
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1) DSC_0006
2) DSC_007
3) DSC_008
4) DSC_009
5) DSC_010
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| Changing V and I Trials | Corresponding Pictures |
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1) DSC_011
2) DSC_012
3) DSC_013
4) DSC_014
5) DSC_015
6) DSC_016
7) DSC_017
8) DSC_018
9) DSC_019
10) DSC_020
|
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Calculating the e/m Ratio
SJK 02:43, 23 October 2008 (EDT)
I presume for these answers you are using the mean +/- SEM of all the numbers above? The way you're writing it implies a mean +/- SEM, of course, but it would help to explain which data you're using from where, etc. Also, what software are you susing?
[math]\displaystyle{ \frac{e}{m}=\frac{2V}{R^{2}B^{2}} }[/math]
Changing V and I
[math]\displaystyle{ \frac{e}{m}=4.51731 \pm .128121 \cdot 10^{12}\frac{C}{kg} }[/math]
Constant V
[math]\displaystyle{ \frac{e}{m}=4.72128 \pm .48565645561 \cdot 10^{11}\frac{C}{kg} }[/math]
Other Analysis
SJK 02:51, 23 October 2008 (EDT)
As I mentioned above, I can't find your spreadsheet, so I don't know what you did with this graph. I'll mention that you'd want to set the linear fit to force it through zero (y=m*x instead of y=m*x + b). It doesn't look like you used this slope anywhere to calculate a value?
Plotting Radius vs inverse current with a least squares linear fit line:
Qualitative Set Up and Discussion
Heater 6.2 V
V= 371 V
I= 1.175 A
Helix
Filament discussion. Why does the radius shrink when we turn down the heater voltage? If the energy of the electrons leaving the filament is negligible (the work function consumes most of the energy as the electons leave) compared to the accelerating voltage, then why does the radius change?
When we switch the direction of the current in the helmholtz coil we see now that the force is sending the electron radius in the opposite direction (downward into the bottom of the bulb). This is because the helmholtz force has now been negated, and the circle forms in the opposite direction.
When we apply a potential difference to the deflection plates we create an electric field for the electrons to pass through. As they pass through this electric field they shoot upwards (there is no current in the helmholtz coils so they do not go into circular motion). They shoot upwards because they are negatively charged and thus are repelled by the negative plate (or by the force form the electric field) and thus shoot upwards toward the positive plate on top. When we add more accelerating voltage, and thus more potential deffierence (they are connected in parralel) between the deflection plates, the beam goes higher up. This is surprising, we thought that the increase of the elctric field between the plates would cause the beam to bend. This does not occur.
reversing the polarity on the deflection plates reverses the direction of the beam
