Physics307L:People/Knockel/Notebook/070926

Electron's e/m Ratio

Experimentalists: Nik and I

Goal

If a beam of electrons moving at a certain speed is bent by a magnetic field, the entire beam will curve into a circle of fixed radius. This result means that there is a charge to mass ratio for the electron, and it can be calculated! The accepted value for the charge to mass ratio of an electron is

[math]\displaystyle{ \frac{e}{m}=1.759\times10^{11} \frac{C}{kg} }[/math].

Theory

I will find two expressions for velocity of an electron beam moving in a circle and then set them equal to each other. I do this in hopes of finding a useful expression for e/m.

First expression for velocity

If the beam is moving in a circle, the velocity, [math]\displaystyle{ v }[/math], 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

[math]\displaystyle{ F_r=-\frac{m v^2}{r} }[/math].

The force (radial) from the perpendicular magnetic field is

[math]\displaystyle{ F_r=-e v B \, }[/math].

Solving for [math]\displaystyle{ v }[/math] in these two equations by first setting [math]\displaystyle{ F_r=F_r }[/math] gives

[math]\displaystyle{ v=\frac{reB}{m} }[/math].

Second expression for velocity

To create the electron beam, the electrons will be accelerated between two plates. The voltage between these plates, [math]\displaystyle{ V }[/math], 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 ([math]\displaystyle{ \Delta }[/math]U) between the two plates, which equals charge (e) times the voltage between the plates (V):

[math]\displaystyle{ K=\frac{1}{2}m v^2=-\Delta U=Ve }[/math].

Solving for [math]\displaystyle{ v }[/math] gives

[math]\displaystyle{ v=\sqrt{\frac{2Ve}{m}} }[/math].

Why find these expressions?

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

[math]\displaystyle{ \frac{e}{m}=\frac{2V}{B^2 r^2} }[/math].

This equation is nice since one must only measure [math]\displaystyle{ r }[/math], [math]\displaystyle{ V }[/math], and [math]\displaystyle{ B }[/math].

An alteration to our result...

Instead of measuring [math]\displaystyle{ B }[/math] which can be difficult, one needs only to know the relationship between current (I) and magnetic field and then measure the current. Since I will be using a Helmholtz coil to generate [math]\displaystyle{ B }[/math], the following derivations are appropriate.

Whenever there are [math]\displaystyle{ n }[/math] circular loops of current of radius [math]\displaystyle{ R }[/math] that are in basically the same spot, the magnetic field along the axis perpendicular to and in the center of the loops is given by

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

where [math]\displaystyle{ \mu }[/math] is magnetic permeability and [math]\displaystyle{ x }[/math] 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 I am doing the experiment half way between both coils, [math]\displaystyle{ x=R/2 }[/math]. Also, since there are two coils, [math]\displaystyle{ n=2N }[/math], where N is the number of loops in one of the two coils, we can manipulate the above equation into

[math]\displaystyle{ B=\left(\frac{4}{5}\right)^{\frac{3}{2}}\frac{\mu N I}{R} }[/math].

Since N=130, R=0.15m and [math]\displaystyle{ \mu=\mu_0=4\pi }[/math]x10^-7 Wb/(A*m), the Helmholtz coil I am using approximately abides by the following equation at the point on the axis exactly between both coils:

[math]\displaystyle{ B=(7.8\times10^{-4} \frac{Wb}{A\cdot m^2})\cdot I }[/math].

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 [math]\displaystyle{ B }[/math] can still be used in the expression for [math]\displaystyle{ e/m }[/math] 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!

[math]\displaystyle{ \frac{e}{m}=3.29\times10^6\frac{V}{B^2 r^2} }[/math].

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 Device! (Uchida Model TG-13)
  • Helmholtz coil
  • Vacuum tube with helium gas and electron gun inside
  • Mirror with ticks for measuring length
  • Black cloth hood to block out light

Setup

To plug everything in I...

• acquired The Device!
• positioned The Device! 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 into the wall and the two multimeters
• connected one of the weak power supplies into the heater for the electron gun with banana plugs and turned it on not exceeding 6.3V
• 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 Device! to measure voltage across electrodes

To adjust the electron beam I...

1. turned off lights
2. waited for heater to heat up (after about two minutes it is at equilibrium)
3. adjusted voltage to electrodes and current to Helmholtz coils so that the beam made a helix
4. rotated the vacuum tube so the helix became a circle
5. moved the power supplies until they no longer affected the size of the circle
6. adjusted the focus on The Device! until the circle size was maximized (the focus aims the electron beam from the anode to the hole in the cathode)

My Procedure

Once everything was set up, the procedure was very easy. We simply took many radius measurements of the electron beam loop at various voltages at a constant current, and then we took more radius measurements at various currents at a constant voltage.

Measuring the radius was slightly tricky. One thing we kept in mind was that the helium would make the circle smaller due to drag, so we measured the outermost radius of the circle (the beam width is large enough to give a substantial distribution of circle radii). Another difficulty was our inability to put a ruler into the vacuum tube. The mirror in the back of the device had a ruler on it. So, to measure the radius, we simply had to line up the real beam with its image on the mirror and take the reading off of the mirror where they line up (it's hard to explain how we did this unless The Device! is in front of me to show you, but it's really not that difficult). Since the zero mark on the mirror is between the two edges of the circle, we took measurements from the left and right edges and then averaged them together. Our measurements from the right edge were longer because the mirror was not centered.

Data

A datum is good but data are better.

Calculations and Results

e/m=[math]\displaystyle{ \pi }[/math]

Where to mail the Nobel Prize money

Bradley Knockel
Bahama Street
Bahamas