In this lab, an electron beam will be accelerated through a known voltage and then ejected into a magnetic field of known strength. Because of the magnetic field, it will then travel in a circle. We will use the radius of this circle to determine the charge-to-mass ratio (\(e/m\)) of the electron.

Then, we will change the vary the orientation of our device, and use how the radius of the beam changes to measure the strength of the magnetic field generated by the Earth (or rather, more precisely, the horizontal component of it).

• 1 e/m apparatus consisting of:
  • 1 set of Helmholtz coils (wires & mount)
  • 2 DC power supplies (one for current in coils, one for accelerating voltage)
  • 1 vacuum tube (with some low-pressure Helium gas)
  • 1 electron gun
• 1 light-blocking hood
• Record data in this Google Sheets data table

See KFJ Ch. 24.4 for the physics behind magnetic forces.

Calculating \(e/m\)

In this lab, we'll be using the Lorentz force law, \(\vec{F}=q\vec{v}\times\vec{B}\), and the centripetal force law from classical mechanics, \(F_c=\frac{mv^2}{r}\).

Let \(B_\perp\) denote the component of the magnetic field perpendicular to the plane of motion of the electron.

Since the component of \(\vec{F}\) in the inwards direction is made by \(B_\perp\) (check this with the right-hand rule!), we can ignore the other components of magnetic field.

Since the inward magnetic force is the centripetal force and \(|q|=e\), we must have:

$$\frac{mv^2}{r}=evB_\perp$$

This would be nice if we knew the velocity. Fortunately, we know the voltage through which the electrons are accelerated. Setting the change in electrical potential energy equal to the final kinetic energy of the electrons, we find:

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

From this point, some algebra lets us eliminate the velocity variable entirely, giving us:

$$\frac{e}{m}=\frac{2V}{(B_\perp r)^2}\label{em}$$

Helmholtz Coils

We'll be generating our magnetic field using a Helmholtz coil. This coil consists of two circles of wires (each \(N=130\) turns) specially configured to have their radius \(a\) equal to the separation distance between them. This makes the magnetic field in the center of the pair of coils as close to constant as possible.

The magnetic field at the center of a Helmholtz coil of radius \(a\) with \(N\) turns is:

$$B=\frac{8}{5\sqrt{5}}\frac{\mu_0 N I}{a}$$

This means that our final formula for \(e/m\) will be:

$$\frac{e}{m}=\frac{125a^2}{32(\mu_0 N)^2}\frac{V}{(Ir)^2}$$

Or, written in other words:

$$V=\alpha\left(\frac{e}{m}\right)(Ir)^2$$

where we have defined the convenient constant:

$$\alpha=\frac{32(\mu_0 N)^2}{125a^2}$$

Adding the B Field of the Earth

Now, let's go back to our equation for \(\frac{e}{m}\) and figure out what happens when we add the magnetic field of the Earth.

Let's first re-write it in a more convenient form for this purpose:

$$B_\perp^2=\frac{2V}{(e/m)r^2}$$

Let's denote the magnetic field made by the Helmholtz coil as \(B_\text{HH}\) and the horizontal component of the magnetic field of the Earth as \(B_\text{E}\).

Let's define the angle between \(B_\text{HH}\) and \(B_\text{E}\) as \(\theta\). Then, we find:

$$B_\perp=B_\text{HH}+B_\text{E}\cos(\theta)$$

Therefore, we can rewrite our above equation as:

$$\sqrt{\frac{2V}{(e/m)r^2}}=B_\text{E}\cos(\theta)+B_\text{HH}$$

For convenience, we will call the left-hand side of that equation \(\beta\).

Thus, we can measure \(B_\text{E}\) as the slope of a \(\beta\) vs. \(\cos(\theta)\) plot.

Part I: Measuring \(e/m\)

First, grab a meter stick and measure the diameter, \(2a\), of the Helmholtz coil. Estimate an uncertainty on this value. Calculate the radius and propagate uncertainty.

Next, plug in and turn on your machine. Wait the 30 seconds it takes to boot up. Drape the hood over the top of the machine, if it is not there already. Align your machine so that the plane of the coils lies in the (magnetic) North-South direction, for optimal precision.

Turn your voltage and current knobs until you see a green circle of light illustrating the electron beam. This circle may be dim, so look closely! It is easiest to see where it crosses the measuring rod in the center.

Once you have a beam, it is time to take measurements. Choose a value for current, and adjust your voltage until the beam is centered on the smallest marker. The marker tells you the diameter, in cm; record the radius, the current, and the voltage (with associated uncertainties).

Then, increase the voltage until you hit the next marker, and again record the same quantities. Repeat until you hit the largest marker.

Then, choose another value for current, and repeat the process (again, for a variety of radii).

Part II: Measuring the Magnetic Field of the Earth

Turn your current down low (but at least \(1\)A) and set your voltage so that the innermost edge of the electron beam lines up with the largest distance marker. For this part of the experiment, you'll need to be especially precise.

Record the current and radius that you are working at. These quantities will remain the same through all of the following trials.

Orient your machine so that the magnetic field of the Helmholtz coil is directed the same direction as the (expected) direction of Earth's magnetic field (i.e., \(\theta=0^\circ\)).

Here at Stony Brook, the magnetic field is about 13 degrees west of (true) north. The north-south direction is (to a good enough approximation) across the classroom, and east-west is from the front to back of the classroom. (Whether the front of the room is east or west depends on which specific lab room you are in.)

Then, turn your voltage knob until the innermost edge of the electron beam lines up with the largest distance marker again. Record this voltage.

Repeat the process (reorient, tweak voltage until it lines up in the same way, record voltage) for \(\theta=45^\circ,\ 90^\circ,\ 135^\circ,\) and \(180^\circ\).

Part I: Measuring \(e/m\)

For each data point, calculate \((Ir)^2\) and the uncertainty. Also calculate \(\alpha\) and its uncertainty.

Then, make a plot of \(V\) vs. \((Ir)^2\). From the slope and your calculation of \(\alpha\), determine a measurement of \(e/m\).

Part II: Measuring the Magnetic Field of the Earth

If you enter the voltages (and uncertainties), the Google Sheet will take care of the calculations of \(\beta\) and \(\Delta\beta\) for you.

Plot \(\beta\) against \(\cos(\theta)\), with y error bars. The slope will be your estimate of the horizontal component of the magnetic field of the Earth.

Analysis questions:

1. Does your measurement of \(e/m\) from part I match up with the "correct" value of \(1.75882009\times 10^{11}\)C/kg?
2. Is your measurement of magnetic field strength roughly accurate to what you expect (approximately \(20\)μT)?
3. Does the direction of magnetic field of the Earth (North/South) match what you expect?

Theoretical questions:

4. Work through the algebra to derive the eq. 3 for \(e/m\) from the equations for magnetic force, centripetal force, and kinetic energy of the electron.

For further thought:

5. At the edges of the coil, the deviations of the magnetic field from the value at the center is approximately 7%. How does this compare to the other uncertainties you have in this lab?

Guide to Uncertainty Propagation & Error Analysis

Plotting Tool

Helmholtz Coils