# Physics307L F07:People/Joseph/Notebook/071003

## The e/m Ratio

SJK 00:00, 11 October 2007 (CDT)
00:00, 11 October 2007 (CDT)
Cool, thanks for citing that. Check out his notebook for any comments I made. Also, it would be helpful to specify exactly what was copied. Also should link to it.

The first half of this report is borrowed (with permission) from Brad's notebook, because I could not make it to the first lab.

## Goal

When a stream of electrons, moving at a certain velocity, is influenced by a magnetic field, the stream will curve into a circle of fixed radius. This implies that there is a charge to mass ratio for the electron, which can be determined empirically. The accepted value for the charge to mass ratio of an electron is

$\displaystyle \frac{e}{m}=1.759\times10^{11} \frac{C}{kg}$ .

## Theory

Finding two expressions for the velocity of the electrons moving in the circular stream and setting them equal to one another will hopefully unveil a useful expression for e/m.

### First expression for velocity

If the beam is moving in a circle, the velocity, $\displaystyle v$ , can be discovered in terms of magnetic field strength (B), charge (e), radius of circle (r), and mass of electron (m). The force (radial) required to keep anything in circular motion is

$\displaystyle F_r=-\frac{m v^2}{r}$ .

The force (radial) from the perpendicular magnetic field is

$\displaystyle F_r=-e v B \,$ .

Solving for $\displaystyle v$ in these two equations by first setting $\displaystyle F_r=F_r$ gives

$\displaystyle v=\frac{reB}{m}$ .

### Second expression for velocity

To create the electron stream, the electrons will be accelerated by a potential between two plates. The voltage between these plates, $\displaystyle V$ , is the only significant means for the electron to speed up since the electric field practically goes to zero on the other side of the plate and since gravity is negligible with the speeds we are dealing with. Therefor, kinetic energy (K) equals the negative change in potential energy ($\displaystyle \Delta$ U) between the two plates, which equals charge (e) times the voltage between the plates (V):

$\displaystyle K=\frac{1}{2}m v^2=-\Delta U=Ve$ .

Solving for $\displaystyle v$ gives

$\displaystyle v=\sqrt{\frac{2Ve}{m}}$ .

### Why find these expressions?

Setting both expression for velocity equal to each other and solving for e/m gives

$\displaystyle \frac{e}{m}=\frac{2V}{B^2 r^2}$ .

This equation allows us to determine the $\displaystyle \frac{e}{m}$ simply, considering only the parameters $\displaystyle r$ , $\displaystyle V$ , and $\displaystyle B$ .

### An alteration to our result

Instead of directly measuring the magnetic field ($\displaystyle B$ , we derive what the $\displaystyle B$ field should be in our experiment. If we have $\displaystyle n$ circular loops of current of radius $\displaystyle R$ that are wound parallel to one another, the magnetic field along the axis perpendicular to and in the center of the loops is given by

$\displaystyle B=\frac{\mu I R^2 n}{2\left(x^2+R^2\right)^{\frac{3}{2}}}$ ,

where $\displaystyle \mu$ is magnetic permeability and $\displaystyle x$ is the distance along the axis from the plane containing the loop.

For any well-made Helmholtz coil, the distance between both coils is R, so, since the experiment is to be done half way between both coils, $\displaystyle x=R/2$ . Also, since there are two coils, ${\displaystyle n=2N}$, where N is the number of loops in one of the two coils, the above equation becomes

${\displaystyle B=\left({\frac {4}{5}}\right)^{\frac {3}{2}}{\frac {\mu NI}{R}}}$.

Using N=130, R=0.15m, and ${\displaystyle \mu =\mu _{0}=4\pi }$x10-7 Wb/(A*m), the Helmholtz coil being used abides (approximately) by the following equation at the point on the axis exactly between both coils:

${\displaystyle B=(7.8\times 10^{-4}{\frac {Wb}{A\cdot m^{2}}})\cdot I}$.

To derive this I have assumed that the electron beam will be on the axis, when, in fact, it will make a loop around the axis. However, this expression for ${\displaystyle B}$ can still be used in the expression for ${\displaystyle e/m}$ since the magnetic field weakens only slightly as one deviates from the axis as long as the deviation is small (not larger than R/2).

And now, here's The Final Equation!

${\displaystyle {\frac {e}{m}}=(3.29\times 10^{6}{\frac {A\cdot m^{2}}{Wb}}){\frac {V}{B^{2}r^{2}}}}$.

## Equipment

All the equipment.
• 2 weak direct current power supplies (10V and 2A are the max outputs needed)
• 1 strong direct current power supply (500V needed)
• 2 multimeters
• Banana plugs
• The apparatus (Uchida Model TG-13)
• Helmholtz coil
• Vacuum tube with helium gas and electron gun inside
• Mirrored measuring stick, for measuring length
• Dark cloth hood to block out light so electron and helium collisions can be seen

## Setup

### To plug everything in we

• acquired
• positioned the apparatus so that its axis lie in the East/West direction to minimize interference from the Earth's magnetic field
• plugged in the three power supplies and the two multimeters into the wall
• connected one of the weak power supplies into the heater for the electron gun with banana plugs
• plugged in the other weak power supply into the Helmholtz coil using banana plugs, but intercepted the circuit to connect a multimeter (in series) to measure current, and then turned on the power to the Helmholtz coil and multimeter so there were about 2 amps flowing
• plugged in the large power supply to power the electrodes of the electron gun using banana plugs, and turned it on so that about 300V are applied
• connected a multimeter to the apparatus to measure voltage across electrodes

### To adjust the electron beam we

1. Turned off lights and dimmed nearby monitors.
2. Turned on the heater keeping the current ${\displaystyle \leq }$ 6.3 ${\displaystyle V}$.
3. Let heater warm up (equilibrates after about two minutes).
4. Adjusted voltage to electrodes and current to Helmholtz coils so that the beam made a helix
5. Rotated the vacuum tube so the helix became a circle
6. Moved the power supplies until their magnetic fields no longer affected the size of the circle (see error sources)
7. Adjusted the focus on the apparatus until the radius of the electron stream was maximized (the focus aims the electron beam from the anode to the hole in the cathode)

## Procedure

Once the experiment was set we initially played around with everything once or twice to be sure we knew what was controlling what. Our target was to measure the radius of the electron stream under various conditions. With everything in place we began taking data. We first set the current to constant and varied the voltage, then we set the voltage to constant and varied the current. We kept the heating current at ${\displaystyle /approx}$6.3${\displaystyle V}$ the entire time.

The radius that we actually measured was the electron stream, which was anything but simply. The stream appeared as an ethereal blue ring, emanating from the electron gun and looping back (not necessarily always) onto itself. There were no clear "edges" to the ring, it looked more like a flashlight shown on a wall a few feet away. The mirrored ruler that was on the far side of the helium-filled bulb was how we actually measured the radius. The goal was to align the eye with the stream, and line the stream up to its own reflection. This sounds complicated but its actually straight forward, just lining up all images. We took the measurements from the left and right edges of the ring and averaged them together. The blue light was due to the electrons ionizing the helium in the apparatus. The helium itself caused a drag force against the electrons and we felt this would influence our data. We measured from the outermost "edge" of the ring, but there was always some amount of uncertainty in what we measured.

## Data

This data was taken at Current=1.5 amps

Trial # Voltage, ${\displaystyle V}$ Left Radius (cm) Right Radius (cm)
1 230 2.6 3.8
2 250 2.8 4.0
3 270 2.9 4.2
4 290 3.0 4.3
5 310 3.0 4.5
6 330 3.1 4.6
7 350 3.2 4.7
8 370 3.2 4.8
9 390 3.2 4.8

We are disregarding trial #9 because at this radius the drag from the helium is too great.

=Data taken @ Constant Voltage=275 ${\displaystyle V}$

Trial # Current ${\displaystyle A}$ Left Radius (cm) Right Radius (cm)
1 1.8 2.2 3.4
2 1.7 2.4 3.6
3 1.6 2.6 3.9
4 1.5 2.8 4.1
5 1.4 3.1 4.4
6 1.3 3.2 4.7
7 1.2 3.4 4.9
8 1.1 3.5 5.1
9 1.0 3.4 5.3

We took several more measurements but the overall radius changed shape in a inexplicable way.

## Calculations

In determining the radius (${\displaystyle R}$) we just took the average of the two points, i.e. ${\displaystyle R={\frac {R_{left}+R_{right}}{2}}}$,

also using the equation for ${\displaystyle e/m}$ we have:

We are disregarding trial #9 because at this radius the drag from the helium is too great. =Data taken @ Constant Voltage=275 ${\displaystyle V}$
Trial # Current ${\displaystyle A}$ Left Radius (cm) Right Radius (cm)
1 1.8 2.2 3.4
2 1.7 2.4 3.6
3 1.6 2.6 3.9
4 1.5 2.8 4.1
5 1.4 3.1 4.4
6 1.3 3.2 4.7
7 1.2 3.4 4.9
8 1.1 3.5 5.1
9 1.0 3.4 5.3

We took several more measurements but the overall radius changed shape in a inexplicable way.

## Calculations

In determining the radius (${\displaystyle R}$) we just took the average of the two points, i.e. ${\displaystyle R={\frac {R_{left}+R_{right}}{2}}}$,

also using the equation for ${\displaystyle e/m}$ we have|}

We are disregarding trial #9 because at this radius the drag from the helium is too great.

=Data taken @ Constant Voltage=275 ${\displaystyle V}$

Trial # Current $\displaystyle A$ Left Radius (cm) Right Radius (cm)
1 1.8 2.2 3.4
2 1.7 2.4 3.6
3 1.6 2.6 3.9
4 1.5 2.8 4.1
5 1.4 3.1 4.4
6 1.3 3.2 4.7
7 1.2 3.4 4.9
8 1.1 3.5 5.1
9 1.0 3.4 5.3

We took several more measurements but the overall radius changed shape in a inexplicable way.

## Calculations

In determining the radius ($\displaystyle R$ ) we just took the average of the two points, i.e. ${\displaystyle R={\frac {R_{left}+R_{right}}{2}}}$,

also using the equation for ${\displaystyle e/m}$ we have the following tables.

With constant current:

Trial # Radius (m) e/m (x10^11 C/kg) Percent Error
1 .032 3.28 86.5
2 .034 3.16 79.6
3 .036 3.13 77.9
4 .037 3.18 80.8
5 .038 3.22 83.1
6 .039 3.26 85.3
7 .040 3.28 86.5
8 .040 3.38 92.2

With constant voltage:

Trial # Radius (m) e/m (x10^11 C/kg) Percent Error
1 .028 3.56 102.4
2 .030 3.48 97.8
3 .033 3.34 89.9
4 .035 3.38 92.2
5 .038 3.28 86.5
6 .040 3.43 95
7 .042 3.65 107.5
8 .043 4.04 129.7

## Error Analysis

Bradley and I agreed that there are so many possible sources of (systematic) error, as well as so much fluctuation in our data that based on the data one would not be able to accurately determine the ${\displaystyle e/m}$ ratio. Here are some of the possible sources of error we determined:

• The work function: the work required to strip electrons from the heating element.
• The focus aperture: Seems to effect the density of the stream of electrons that flow from the heating element, into the accelerating voltage. We don't really understand the mechanics of what is happening with the focus, and this is causing a lot of trouble. It is uncertain if the focusing apparatus is taking work from the electron beam.
• The field of the earth.
• That willey tail of electrons that no one can explain
• Non uniform ${\displaystyle B}$, its close but not perfect. Uneven E-field as well between the plates
• Drag force from the helium (large effect)
• Light bends as it travels through glass bulb
• The electrons are semi-relativistic
• External components, such as the power sources, marginally influence ${\displaystyle B}$ field

Superhuman Error

Also, our data shows that as the radii increase the ${\displaystyle e/m}$ we determined increases. Typically the percent error grows as well. This says that if we create eventually a very small radii stream, we could eventually determine a (nearly) true value. Knowing what we do, it would be interesting to perform this experiment again and acut out as many of these identified error sources as possible. Using the relationship I just talked about, perhaps one could even determine, by experiment, the accepted value of ${\displaystyle e/m}$.