User:Garrett E. McMath/Notebook/Junior Lab/2008/09/15
E/M Ratio Lab  Main project page Previous entry Next entry 
Lab 2: The Ratio e/m for ElectronsNotes By: Garrett McMath and Paul Klimov ^{SJK 00:22, 4 October 2008 (EDT)}Purpose and ObjectivesThe purpose of this laboratory is to measure the e/m ratio of electrons. This will be done by accelerating electrons into a strong magnetic field, which will cause the electrons to rotate in a circle on a plane perpendicular to the field. TheoryThe Lorentz force, in the absence of an Electric field gives us: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \vec{F}=e(\vec{v} \times \vec{B}) = m \frac{\vec{v}^{2}}{R}} this implies: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\frac{\vec{v}}{R\vec{B}}} The electrons are accelerated through a potential V, implying: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{1}{2}mv^{2}=eV} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle v=\sqrt{\frac{2eV}{m}}} Now, we use this velocity in the above equation for the ratio e/m: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\sqrt{\frac{e}{m}}\frac{\sqrt{2V}}{RB}} This, then, boils down to give e/m ratio in terms of V, r, and B, all of which are variables that we can find: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{e}{m}=\frac{2V}{(RB)^{2}} } From here, we must find the strength of the magnetic field which is produced by the Helmholtz coils: We use the BiotSavart Law, to find the field due to one coil, and then add the field due to two coils. A single coil, with N loops, of radius R, stands with its opening aligned with the yaxis. The point of interest will be at y=a. Due to symmetry, we know that there will only be a field in the positive y direction (where current flows counterclockwise, as seen looking from positive y). Theta will rotate in the xz plane, which will allow us to define a small current element. BiotSavart Law: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle d\vec{B}=\frac{\mu_{0}i}{4\pi}\frac{ \vec{dl}\times\vec{r}}{r^{3}}} we know that dl is perpendicular to r always, and so we can turn this into a scalar equation for the y coordinate. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle dl = R d\theta } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{4\pi}\int_{0\leq\theta\leq2\pi}\frac{Rd\theta}{(a^{2}+R^{2})}\frac{R}{\sqrt{a^{2}+R^{2}}}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{4\pi}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}\int_{0\leq\theta\leq2\pi}d\theta } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}=\frac{N\mu_{o}i}{2}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}} } Now, because we have 2 coils, which make up the Helmholtz coils, we simply multiply the above expression to find the field as a function of the current: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B_{y}(i)=\frac{N\mu_{o}iR^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}} } with this information, we should be able to complete the lab. Experimental Setup
Heater connected to SOAR PS3630. Accelerating Electrodes connected in series with AMPROBE multimeter and then to the Gilman power supply. Voltmeter jacks connected to BK Precision multimeter Helmholtz coils connected in series with FLUKE multimeter to SOAR 7403 DC Power Supply.
Connected in series with an ammeter/voltmeter and connected to the heater.
Connected in series with an ammeter/voltmeter and connected to the Helmholtz coils.
Connected directly to the voltmeter jacks on the e/m experimental apparatus.
Connected in series with an ammeter/voltmeter and connected to the electron gun.
Connected in series to the SOAR 7403 and e/m experimental apparatus.
Connected in series between e/m apparatus and Gelman power supply.
Procedure1.) Turn on SOAR PS3630 PSU for heater to:
Nothing connected to ground. This starts the heating of the filament. It starts radiating as it heats up. 2.) Turn on Gilman power supply for electron gun
3.) Settings on e/m experimental apparatus:
Experimental Procedure
DataSource of error: Our voltmeters accuracy doesn't go into the decimals. Each data point has an uncertainty of at least ±.1 cm because this is roughly the width of the beam Constant Current
1.)200V, 4.1±.1cm 2.)210V, 4.35±.1cm 3.)220V, 4.45±.1cm 4.)230V, 4.5±.1cm 5.)240V, 4.55±.1cm 6.)250V, 4.60±.1cm 7.)260V, 4.65±.1cm 8.)270V, 4.70±.1cm 9.)280V, 4.75±.2cm 10.) 290V, 4.80±.3cm 11.) 300V, 4.90±.3cm At the end, the beam seemed to get a bit thicker. This introduced more uncertainty into our measurements. Constant VoltageOne source of experimental error: The beam seems like it is somewhat elliptical. The radius on the right is slightly smaller than on the left. We have decided to use only the left radius. This will definitely introduce uncertainty into our calculations. However, after rotating the bulb, we have a spiral forming. With the spiral, we cannot take any measurements because there is simply too much going on. V=280V 1.)i=1.000A, 4.75±.2cm 2.)i=1.050A, 4.75±.2cm 3.)i=1.100A, 4.60±.1cm 4.)i=1.151A, 4.55±.2cm 5.)i=1.202A, 4.47±.1cm 6.)i=1.253A, 4.35±.1cm 7.)i=1.297A, 4.25±.2cm 8.)i=1.351A, 4.15±.2cm 9.)i=1.399A, 4.05±.2cm 10.)i=1.449A, 4.00±.2cm ANOTHER ROUND OF DATA FOR CONSTANT VOLTAGES:
V=299.8 1.)i=1.245A, R=3.90±.1cm 2.)i=0.989A, R=4.25±.2cm
V=249.6V 1.)i=.997A, R=3.95±.1cm 2.)i=1.252A, R=3.70±.2cm LAB SUMMARY^{SJK 00:34, 4 October 2008 (EDT)}Lab Manual Questions1. Why do we see the electron beam at all? The visual beam that we see is merely the ionization of the helium gas caused by the electrons losing energy to the helium causing the helium to emit photons of such wavelength in the visible light spectrum. 2. We ignored the Earth’s magnetic ﬁeld in our procedure. How much error does this introduce into this experiment? Using the website http://www.ngdc.noaa.gov/geomagmodels/IGRFWMM.jsp, you can get the magnectic field and its components of your location. The total magnetic field in Albuquerque is 50,292.7 nT, with components of 23,027.7 nT North, 3858.3 nT East, 23,348.7 nT Horizontal, and 44,544.3 nT Upward. Our apperatus was more or less aligned with the east component of the earths Bfield so plugging this value in using 300 V will give a reasonable approximation of the error the earths field could have caused. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle R = \sqrt{\frac{2Vm}{e}}\frac{1}{B} = 15.15m } This value is actually pretty large using MatLab paul analyzed this result which you can see in the data analysis section.
Using the same formula as in question 2: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{R_{p}}{R_{e}}=42.81 } So other than the difference in polarity the major difference would be a much stronger magnetic field would be needed to bend the beam. 4. Show that if the magnetic ﬁeld is held constant, the time t required for an electron to make a complete circle in your e/m tube and return to the anode is independent of the accelerating voltage by deriving an expression for this time. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle T=\frac{2\pi R}{v} } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle R = \sqrt{\frac{2Vm}{e}}\frac{1}{B} } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle v = \sqrt{\frac{2eV}{m}} } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle T = \frac{2\pi m}{e B} } Data Analysis^{SJK 00:06, 4 October 2008 (EDT)}
File:ConstantI2.tif File:ConstantIDiffSlope.tif File:ConstantV.tif File:ConstantVDiffSlope.tif File:Empoints2.tif File:Emvsradius.tif ProgramsUsing data analysis programs written in MATLAB by Paul klimov we calculated least squares regressions for both sets of data, constant voltage and constant current. The program then calculated an e/m ratio for both sets of data using the formula that is shown in theory section. Margin of ErrorAs will be discussed later in sources of error we had uncertainty in all of our measurements. The most obvious reason being the width of the beam caused a minimum uncertainty of one millimeter, though we often had larger uncertainty either due to thicker beams caused by higher voltage, or simply a large discrepency between Paul's and my measurements. Constant CurrentUsing the MatLab programs, the following information was obtained:
Constant Voltage
Slope Modification: Figure 2b
Final Stats^{SJK 23:46, 3 October 2008 (EDT)}
Sources of Error^{SJK 00:01, 4 October 2008 (EDT)}This lab we were told has one of the worst percentage errors of any of the labs we will be doing this semester. With that in mind we watched very closely for sources of this while doing the lab.
Improvements
