# Physics307L F09:People/Allen/Eoverm

## Objective

SJK 17:40, 15 November 2009 (EST)
17:40, 15 November 2009 (EST)
excellent primary lab notebook, up until the point of producing e/m values, where your descriptions did not contain enough information and / or supporting excel sheet to illustrate what you were doing.

This lab is undertaken to determine the ratio of the electron's charge to its mass by observing the curvature of an electron beam under the influence of a magnetic field. A vacuum tube filled with helium and containing an electron gun is placed at the center of a Helmholtz coil. The current of the coil that generates the magnetic field and the voltage of the electron gun that accelerates the electrons are varied to create different radii of the electron beam. The measurements of corresponding current, voltage and radius allow one to calculate the ratio of e/m for the electron.

## Notes

David Weiss and I worked together on this lab. We started by addressing safety concerns with Professor Koch, and checking the set up of the equipment with Pranav Rhathi. We tried using the 6 volt power supply for the electron gun so as to keep below the required limit, but found that it did not supply enough power to create a visible electron beam, so we replaced it with the 36V supply. We also misconnected the multimeter to measure the current, which was not properly set up to be in series.

We set the heater supply to 6.1 volts, and then raised it to 6.3V and let the heater gun filament heat up until it glowed. Then we applied 250 V potential to the electrodes and were able to see the blue light of the electron beam. When the coil current was turned on, we could adjust the curvature of the beam to circle in on itself by rotating the vacuum tube in its socket and by adjusting the relationship between the coil current and the velocity of the electrons, due to the voltage of the accelerator. The circle of light would pop in and out between two diameters, and it took some time to develop familiarity with the adjustments to get a clear path image, and to understand the process of aligning the path image with its reflection to determine an accurate reading.

We had time for only five readings of the radius measurements from the left and the right, at five different settings of the voltage of the electrodes potential and the coil current.

On the second day, we replaced the vacuum tube on Professor Koch's suggestion. We adjusted the focus knob, which worked better with the new tube, to give a narrow electron beam. This was limited by the need to maintain a workable brightness. Professor Koch mentioned another aspect of set-up, used by other students, that helped with the radius readings. They adjusted the vacuum tube so as to center both the electron beam and its reflection across the ruler, which I proceeded to do. This gave closer radius measurements between the left and right sides.

We repeated the radius measurements with different settings of both the voltage and current with the new tube. We also took measurements of the radius first with a constant current of 1.35 Amps and varied voltage, then with a constant voltage of 143 volts and varied current, for ten different readings each.

We repeated measurements of the radius for the settings of 143V and 1.1A, and for 183V and 1.09A to find a sample SEM for our radius measurements.

My thanks to David Weiss for working with me on this lab.

## Safety

• Electrical: Voltage to 300 volts, 500 volts to the electron gun can cause injury to the heart, so take care that all cords are safe and not worn, no uninsulated connections. High current to 2 Amps, a greater concern with AC than with DC current.
• Breakage: Take care with handling the glass vacuum tube housing the electron gun and the fragile filaments. It is also important to keep the voltage that passes through the electron gun no higher than 6.3 volts.

## Equipment and Set-up

Digital Voltmeter BK Precision Model 2831B

Deluxe Regulated Power Supply Gelman Interment CO. Model 38520 115V

DC Power Supply Soar Corporation 7400 Series 117V

Hewlett Packard dc Power Supply Model 6384A

e/m Experimental Apparatus Model TG-13 Uchida Yoko

These we set up according to the lab manual; see Dr Golds Lab Manual. The horizontal bar seen in the second photograph inside the Helmholtz coil and behind the vacuum tube is the ruler used for measurement of the radius of the electron beam.

## Data

This is the data from both day one and day two.

These are charts that we worked on together, and David entered into an Excel worksheet. Unfortunately, I am unable to go back and adjust the labels and formats.

Slope = 0.097351 Standard error of the slope = 0.003094

These came from David's calculations. My calculations follow below in the data analysis.

SJK 16:42, 15 November 2009 (EST)
16:42, 15 November 2009 (EST)
Nice charts, showing good data. I notice some features: The first graph shows more than random deviation from a linear relationship (the residuals would show a pattern), whereas the second doesn't. Would this give you any hints as to your systematic error discussions below? Also, do your error bars seem to under- or over-estimate the random error in your radius measurements?

## Theoretical Basis

This I am borrowing from Paul Klimov (Paul V Klimov's Lab Notebook) to understand the data analysis.

Theory

The Lorentz force, in the absence of an Electric field gives us:

$\vec{F}=e(\vec{v} \times \vec{B}) = m \frac{\vec{v}^{2}}{R}$

this implies:

$\frac{e}{m}=\frac{|\vec{v}|}{R|\vec{B}|}$

The electrons are accelerated through a potential V, implying:

$\frac{1}{2}mv^{2}=eV$

$v=\sqrt{\frac{2eV}{m}}$

Now, we use this velocity in the above equation for the ratio e/m:

$\frac{e}{m}=\sqrt{\frac{e}{m}}\frac{\sqrt{2V}}{RB}$

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:

$\frac{e}{m}=\frac{2V}{(RB)^{2}}$

From here, we must find the strength of the magnetic field which is produced by the Helmholtz coils: We use the Biot-Savart 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 y-axis. 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 counter-clockwise, as seen looking from positive y). Theta will rotate in the xz plane, which will allow us to define a small current element. Biot-Savart Law:

$d\vec{B}=\frac{\mu_{0}i}{4\pi}\frac{ \vec{dl}\times\vec{r}}{r^{3}}$

we know that dl is perpendicular to r always, and so we can turn this into a scalar equation for the y coordinate.

dl = Rdθ

$B_{y}=\frac{N\mu_{o}i}{4\pi}\int_{0\leq\theta\leq2\pi}\frac{Rd\theta}{(a^{2}+R^{2})}\frac{R}{\sqrt{a^{2}+R^{2}}}$

$B_{y}=\frac{N\mu_{o}i}{4\pi}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}\int_{0\leq\theta\leq2\pi}d\theta$

$B_{y}=\frac{N\mu_{o}i}{2}\frac{R^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}$

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:

$B_{y}(i)=\frac{N\mu_{o}iR^{2}}{(a^{2}+R^{2})^{\frac{3}{2}}}$

with this information, we should be able to complete the lab.

My thanks to Paul Klimov.

## Data Analysis

In a Helmholtz coil, two circular coils are placed a distance apart equal to the diameter of the coils, in order to have both the first and second derivatives of the magnetic field with respect to distance from the center equal to zero at the center, which gives a relatively uniform field between the coils. In the arrangement of this equipment , each coil has 65 turns, so N = 130. The radius of the coils is 0.15m. Therefore,

B = (7.8 × 10^−4 weber)/(amp − meter^2) × I.

By Lorentz, F = q(v x B) = -evB.r^ = -mv^2/R r^. This gives, evB = mv^2/R which gives e/m = v/BR.

The velocity can be found in terms of the voltage from the fact that the electrons acquire their velocity by being accelerated through a potential V.

Kinetic energy = charge times voltage, which gives 1/2 mv^2 = eV .
Solving for velocity in terms of voltage gives v = (2eV/m)^1/2 = (e/m)^1/2 * (2V)^1/2 .
Now solving for e/m gives e/m = 2V/(B^2*R^2)

Substituting the above value for the magnetic field into the equation for e/m gives

e/m = (2/((7.8x10^-4)^2)) * (V/(I^2*R^2))

This equation is used for the ten measurements of the electron beam radius with random voltage and current settings.

Solving for the radius squared as a function of voltage with constant current gives

R^2(V) = 2/((7.8 x 10^-4)^2*I^2) * (m/e)V or R^2(v) = (m/e)2/(B^2) * V

This equation is used for the ten measurements of the electron beam radius with constant current and varied voltage, where

e/m = (2/(B^2*I^2))/slope

Solving for the radius as a function of the inverse of the current gives

R(I) = (m/e)^1/2 * (2V/B^2)^1/2 * I^-1

This equation is used for the ten measurements of the electron beam radius with constant voltage and varied current, where

e/m = (2V/B^2)/(slope^2)

From the first equation for e/m using the ten readings with varied voltage and current, I calculated an average value for e/m of 2.384 x 10 ^11 +/- 0.0445 and a percent error of 35.5%.

From the second equation for e/m using the ten readings with constant current and varied voltage, I calculated a value for the slope of .0651 +/- 3.479, giving a value for e/m of 2.771 x 10^11 with a percent error of 57.4%. SJK 17:08, 15 November 2009 (EST)
17:08, 15 November 2009 (EST)
I cannot figure out where you got this slope of 0.0651 +/- 3.479. It must be some kind of typo, given that your error in the linear fit is 50 times your value. You also don't really say how you calculated the slope? Using LINEST I presume, but it's not shown anywhere that I can see.
From the third equation for e/m using the ten readings with constant voltage and varied current, I calculated a value of 4.121 for the slope with a standard error of the slope (SES) to be 0.129, giving a value for e/m of 2.768 x 10^11 with a percent error of 57.3%. SJK 17:11, 15 November 2009 (EST)
17:11, 15 November 2009 (EST)
I also cannot figure out where this slope and SES comes from, so my guess is that you're doing something wrong?

The slopes I get match that shown on your graphs above.

We took multiple readings of the radius of the electron beam at two different given settings of voltage and current and found the standard deviation = 0.096 and the SEM = 0.034.

## Topics From the Lab Manual

• The electron beam is visible due to collisions of the electrons with helium atoms of the helium gas contained in the vacuum tube that houses the electron gun. Energy is transferred from the electrons to the helium atoms in these collisions that is then emitted as radiation when the helium atoms return to a lower energy state. This light is seen along the path of the electrons as an electron beam.
• From reading other students' work on this question, I found that the Earth's magnetic field does not introduce much error into this experiment. Further data on this can be found on my lab partner's report (David's E/M Ratio Notes).
• If protons were emitted in the vacuum tube instead of electrons, the magnetic field would need to be much larger, for the same velocity, to draw the beam into a circle that would fit inside the dimensions of the vacuum tube. From the equation for R(I), it is shown that the radius is proportional to the mass of the electron, giving

Rp/Re ~ ((mp)^1/2)/((me)^1/2)

With protons, the radius of the proton beam would be 42.8 times the radius of the electron beam for the same acceleration speed and magnetic field strength. Lowering the acceleration speed would affect the amount of energy transferred and therefore the visibility of the beam.
It would also be necessary to turn the vacuum tube 180 degrees so that the positively charged particles could circle clockwise in the magnetic field and the beam still be contained within the expanded portion of the vacuum tube.

• The time for an electron to circle around the path is equal to the circumference /velocity.

t = 2pi*R/v where as shown above, R = (2vm/(B^2*e))^1/2 and v = (2ve/m)^1/2, giving

t = 2pi*((2vm^2)/(vB^2*e^2))^1/2 = (2pi/B)*(m/e)

Therefore, the time required for an electron to make a complete circle in the e/m tube and return to the anode is independent of the accelerating voltage. This allows for electrons to be accelerated through a voltage difference at a given location in a cyclotron.

• A relativistic correction for the electron's momentum would not be appreciable for the present experiment because our highest voltage was only 274 volts resulting in gamma very close to 1.

momentum p = gamma*m*v, gamma = 1/((1 - (v/c)^2)^1/2), v = (2eV/m)^1/2, (v^2)/(c^2) = 2eV/mc^2 ~ approximately 10^-4, such that gamma is approximately 1.0005 .

## Concerns and Sources of Error

• Systematic Error: The velocity of the electrons is lowered by collisiions with the helium atoms in the vacuum tube that are present to make the electron beam visible. This causes the radius of the electron beam to be less than it would be in a true vacuum. There is also a diminishing effect on the radius due to the relationship between the accelerating field and the anode. These should lead to a measurement of e/m higher than the accepted value for a given voltage, which we found to be so in our data.
• Random Error: There is error due to aligning the electron beam with its reflection on the ruler which was made more difficult by reflections of other lights, even though dim, in the room.

I kept the electron beam radius well below that of the vacuum tube in order to minimize distortion of the radius readings from the curvature of the glass. As a result, I was not able to keep the voltage as high as reccommended in the lab manual at the current levels that we were able to apply to the coil.

During the collection of data, I did not understand the reason to measure the outer radius of the electron beam as measuring the electrons less slowed by collisions. At first it seemed simply an arbitrary choice that would make the calculations come out closer to the accepted value. Without understanding the actual reason, I was reluctant to make this shift in my measurements for this data.

If I were to work further on this lab, I would like to calculate the effects of the known sources of systematic error to allow for them in the analysis of data. I would also like to compare this with the original experiment by Thomson.

Since there is a horizontal component to the gravitational field, it might be interesting to see whether its affect could be measured by turning the orientation of the equipment from one set of readings to another.

I liked this lab in that there was a more visible component and not so much of the calculations hidden in the technology of the electronic boxes.