User:John Callow/Notebook/Junior Lab 307/2009/09/30
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e/m Lab NotebookLink to Dr. Gould's manual Link to my summary Equipment ListSJK 14:17, 14 October 2009 (EDT) Excellent job with equipment, safety, and setup 2 BK precision digital multimeter model 2831B 1 Hewlett Packard low voltage high amp dc power supply model 6384a 1 Lichida e/m experimental apparatus model tg-13 1 Soar Corperation low voltage high amp DC power supply 7403 1 Gelman Instrument Company deluxe regulated power supply 1 black hood for the experimental apparatus. 1 blanket to block other light in the room when taking measurements Recommend having a flashlight or desk lamp for when the lights are out too. SafetyFor Experimenters Most of the power supplies are low voltage high amp, and so are very dangerous because of risk of shock. Be absolutely sure everything is off and preferably unplugged when connecting wires and that the set up is as neat as possible to prevent accidental unplugging/contact with exposed parts of the wires. Specifically with the equipment I used the wire ends in the jacks tended to still be somewhat exposed. The heat sinks on the power supplies may get hot so be cautious with those. For the Equipment Be sure the setup is correct before turning anything on, each part of the apparatus uses very different voltage and amp values so an improperly plugged power supply could damage the equipment. SJK 14:16, 14 October 2009 (EDT)Apparently the new bulb arrived today! My inkling is that groups should start off using the old bulb, and then switch to the new bulb once they've gotten used to things. At least until the old bulb finally dies. And very much agree with you that new bulb should be kept below 6.3 V. The manual suggests not setting the heater voltage above 6.3 V. If the bulb is fairly new it is probably best to stick to this recommendation or the bulbs life will be short. In my setup the bulb was already nearing the end of it's life and so required slightly more (around 7.4V) to be viewable. Setup Begin by connecting a 6-9 Volt dc power supply rated at 2 A to the coil jacks on the apparatus in series with an ammeter. In my case I used the Soar Corperation 7403 power supply which did not quite meet these specs (I believe it was max 6.3 volts at 1.5 A), but it still did the job. Next connect another low voltage high amp power supply to the heater jacks on the apparatus. The manual recommended 6.3 volts at 1.5 amps max but I used the Hewlett Packard 6384a power supply which allowed the voltage to be a little higher as the bulb used is nearing the end of its life. The last power supply to set up is a Dc high voltage low amp supply with manual recommendations of 150-300 Volts and 40 mA. I used the Gelman Instrument Company deluxe regulated power supply for this which has similar specs to that recommended. Now just connect a voltmeter to the voltmeter jacks on the apparatus and everything should now be set up. If there is a hood for the apparatus then now is a good time to put it on. Also if a flashlight or desk lamp is available it is probably ok to turn off the lights and use either of those from now on to see.  If using the same apparatus set the switch to e/m and turn the current adjust to zero. Turn on the power supply connected to the heater jacks and let it heat for about 2 minutes. After it has heated up turn on the high voltage power supply. There should now be a visible beam, if not the heater power supply may need to be turned up a little. Once the beam is viewable turn on the coil power supply and adjust the current so that the beam does a loop. If the loop does not loop back on itself turn the bulb till it does (as in turn until the beam doesn't spiral outwards). Solving for e/m formulaBefore taking data it is a good idea to check that the bulb is in the center. To do this line the reflection of the each side of the bulb with itself and take the measurement. If they are both equal the bulb is centered, if not you may have to adjust your scale to compensate. My bulb appeared to be centered properly. The method of lining up with reflections is also how to take the measurement for the radius.SJK 14:19, 14 October 2009 (EDT)This is a really good idea for centering the bulb. From the lab manual assuming the apparatus is the same (I hope it is) the magnetic field B acting on the electrons is found by [math]\displaystyle{ B = \frac{{\mu}*{R^2}*N*I}{({{R^2}+{x^2}})^{1.5}} }[/math] From the manual the number of wires in the coils N = 130, radius of coils R = .15 meters, axis of symmetry x = R/2, and the permeability of free space μ = 4*π*10^-7 weber per amp-meter. Now B becomes [math]\displaystyle{ B = 7.8*{10^{-4}}*I }[/math] weber per amp-meter. The velocity may be found from [math]\displaystyle{ e*V = \frac{m*{v^2}}{2} }[/math] With e being charge, V accelerating voltage, v velocity, and m the mass. solving for v produces [math]\displaystyle{ v = {(\frac{2*e*V}{m})^{.5}} }[/math] The centripetal force is found by [math]\displaystyle{ F = \frac{m*{v^2}}{r} }[/math] where m is the mass, v is velocity, and r is the radius And the magnetic force on the particle by [math]\displaystyle{ F = e*(B*v) }[/math] with e the charge, B magnetic field, and v velocity. This is not a cross product because the velocity is perpendicular to the magnetic field. Because both forces in the centripetal force and magnetic force equations are the same we may set them equal to each other obtaining [math]\displaystyle{ \frac{m*{v^2}}{r} = e*(B*v) }[/math] substituting the formula found for v and solving for e/m gives the formula [math]\displaystyle{ \frac{e}{m} = \frac{2*V}{{B^2}*{r^2}} }[/math] This will be the formula I use to calculate e/m. DataFor this data I just used random voltages and currents to take measurements. {{#widget:Google Spreadsheet |
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}} So from this data I was pretty far off from the expected value but did manage to be at least in the same order of magnitude. Data part 2For this data I kept the voltage constant when taking measurements. {{#widget:Google Spreadsheet |
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}}  The least square fitted formula and plot found with maple 12, r being in centimeters [math]\displaystyle{ \frac{1}{I} = 0.0803+.182*r }[/math] is what is plotted above, along with each of my data points. Substituting this value into the above e/m equation ignoring the .0803 term as r won't cancel if I don't and it is insignificant, I haveSJK 14:39, 14 October 2009 (EDT)You can actually do linear regression without allowing for the constant. In Excel or GoogleDocs, you set the [const] parameter = 0, which then fits to y = m*x, forcing fit through zero. For your data, this changes the fit from 0.18 to 0.20. As an experimentalist, the first question to ask is, should I expect an offset? That is, if I have infinite current, would I measure zero radius? That's a tough question to answer. Since the beam has a finite thickness, I sort of have an inkling that allowing for an offset is OK. [math]\displaystyle{ \frac{e}{m} = \frac{2*V}{{(7.8*{10^{-4})}^2}*{r^2}}*{(.182*100*r)^2} }[/math] which simplifies to [math]\displaystyle{ \frac{e}{m} = \frac{2*V}{{(7.8*{10^{-4})}^2}}*{(.182*100)^2} }[/math] and after plugging in 250 Volts that I used as my constant voltage I find [math]\displaystyle{ \frac{e}{m} = 2.72*{10^{11}} \frac{coul}{kg} }[/math] Which gives an error of about 55% when compared to the accepted value. Still at least of the right order of magnitude. I had to multiply my r value by 100 to match units in the above. SJK 14:44, 14 October 2009 (EDT)You can get the uncertainty using linear regression. Using LINEST in Excel, I get 0.18 +/- 0.02 for the slope.
Data part 3For these measurements the current was now held constant and I let the voltage vary. {{#widget:Google Spreadsheet |
key=tzh7KPYKs5WYUZnBtMf0FXQ | width=760 | height=200
}}  In the above I used my data to find a least squares fit of V vs. r^2 using maple again. The formula from the result is [math]\displaystyle{ V = 125+6.78*r^2 }[/math] where r is in centimeters. Plugging this into the formula from earlier I have
again multiplying by 100 so units match (I multiply by 100 so the least squares fit becomes V vs. r^2 in meters). The 125 term then becomes insignificant and so I'll ignore it to get [math]\displaystyle{ \frac{e}{m} = \frac{2*(6.78*{(100)^2})}{{B^2}} }[/math] Now plugging the constant value of I to find B into the equation [math]\displaystyle{ B = 7.8*{10^{-4}}*I }[/math] and then plugging B into the equation just found results in [math]\displaystyle{ \frac{e}{m} = 1.42*{10^{11}} \frac{coul}{kg} }[/math] SJK 14:47, 14 October 2009 (EDT)Linear regression produces an uncertainty in both fit values, as we briefly discussed on Monday (and we'll discuss further in coming weeks). In the following file, I show how to do this with your data in Excel. File:E over m John Callow.xlsx I would guess that Maple can do this too (for sure you can program in the formula yourself, but I'd guess it's built in too). Using LINEST in Excel, I get 6.8 +/- 0.3 for your constant current data slope. My error is only 19% here and I have no good reason to explain this. I do know that there were issues with the power supply for the coils being a bit unstable when the nobs where turned (the current would suddenly drop off and then the supply had to be reset) so perhaps being stuck on one current setting allowed it to stabilize a bit. I also don't know how to deal with standard error after the linear fit. I know think there is a way in maple to plot the error bars at each point, but I don't know how to find it.
It is great that you plotted the data that you were fitting. Of the two graphs, which of the two look most appropriate for a linear fit? Do either of them look distinctly non-linear? Another way of looking at this is to plot the residuals, which is the data minus the fit...and then judging whether residuals are normally distributed or show some other trend.
Qualitative Experiment QuestionsThe questions here are from the lab manual. What happens when the bulb is rotated?  Rotating the bulb will cause the electron beam to spiral. This is caused by the Lorentz vector as it is a cross product between the magnetic field and v. The cross product always points towards the center and so the beam spirals.
The beam bends the opposite direction, which is because changing the polarity in the coils changes the sign of the direction of B, so the cross product also changes sign.
The charged deflection plates attract/repel the electron causing the beam to head up or down depending on polarity.
Yes, if the top plate is positive then if the beam redirects up then the charge of the particles in the beam is negative. If it goes down it is positive. The opposite is true if the top plate is the negatively charged plate.
The beam bounces or deflects away for a short period of time from what I could see. This part of the qualitative experiment was difficult to see as the beam didn't want to bend that far.
Questions from the lab manual1. Why do we see the electron beam at all? A beam is visible because as the manual says there is helium in the bulb that the electrons from the electron gun interact with by exciting electrons of the helium atoms. We have electrons being launched from the electron as the heated filament in the bulb has freed electrons to pull off with the accelerating voltage.
From http://hypertextbook.com/facts/1999/DanielleCaruso.shtml the magnetic field of the earth is about 50 µT. From my data the field generated by the coils ranged from .8 to 1 milliwebers/meters^2 or .8 to 1 mT. So the earth's field is only about 5% of the value of the coil's. 3. Suppose that protons were emitted in the bulb instead of electrons. How would this effect the experiment. Protons are both much more massive and have an equal but opposite charge. This would mean with the same accelerating voltage the proton would be launched much slower and the field would accelerate the proton towards the center much slower. Just looking at [math]\displaystyle{ \frac{e}{m} = \frac{2*V}{{B^2}*{r^2}} }[/math] These two effects do not cancel each other, and the radius looks like it would be much much bigger to compensate for the changes. Also using the same magnetic field the particle would loop down instead of up, unless the method for accelerating stayed the same, then the proton would probably do a similar loop in the opposite direction. 4. Show that if the magnetic field is held constant, the time t required for an electron to make a complete circle in the bulb is independent of the accelerating voltage by deriving an expression for this time. The reciprocal f=1/t is called the electron's cyclotron frequency. The time it would take to go around the circle would be the circumference of the loop made divided by the velocity or [math]\displaystyle{ t = \frac{2*{\pi}*r}{v} }[/math] From before [math]\displaystyle{ \frac{m*{v^2}}{r} = e*(B*v) }[/math] and solving this for r results in [math]\displaystyle{ r = \frac{m*{v}}{e*B} }[/math] now substituting this in to the formula for t results in [math]\displaystyle{ t = \frac{2*{\pi}*m*v}{e*B*v} }[/math] and simplifies to [math]\displaystyle{ t = \frac{2*{\pi}*m}{e*B} }[/math] Which shows t to be independent of the accelerating voltage V. The cyclotron frequency is then [math]\displaystyle{ f = \frac{e*B}{2*{\pi}*m} }[/math]
Using the formula [math]\displaystyle{ v = {(\frac{2*e*V}{m})^{.5}} }[/math] and using V = 300 volts (higher than anything recorded in my lab) the velocity is found to be [math]\displaystyle{ v = 1.027*{10^{7}} \frac{m}{s} }[/math] which is less than 1% the speed of light. This is no where near fast enough for a relativistic correction to be relevant compared to the accuracy of the measurements I took. Questions about the labSJK 14:06, 14 October 2009 (EDT)These are great pictures! I really liked your interest and pursuit of this part of the lab. I personally don't understand it either yet. I do like your probabilistic argument, but don't have any more fundamental understanding of why those transitions would be more likely. It doesn't seem like resonance, because it's not all that sensitive to exact voltages (rather voltage thresholds). Still a fun puzzle! When viewing the electron beam the color would change from the teal color to violet as it slowed down or came to a stop. The electrons should definitely have enough enough energy to create any color when starting out based on the information found at http://hyperphysics.phy-astr.gsu.edu/HBASE/atomic but I could not find a clear reason why as the beam slowed down it would prefer to excite helium to the higher energy levels to produce violet light as the beam slowed down. My guess is there is some sort of probabilistic reason dealing with resonance but did not have enough time to figure it out in time for this report. At the very least though, here is some cool pictures.  I found it odd that even though the voltage and current for taking this picture (I believe it was 231 volts and 1.31 amps) are not much different than the values in my other measurements yet the radius is much smaller (about 1-2 cm). Also my professor Steve Koch had moved a magnet next to the bulb when the beam looked as it does above and it ended up changing back to a fully teal beam that had a radius similar to my other measurements and remained this way when pulling the magnet away again. This suggests that the beam might be in a position that is probably unstable.  Here the voltage was turned way down and it looks like at this point the whole beam became violet. This was around 50 volts with the coil current off. | |
