Charge-to-Mass Ratio (e/m) of the Electron
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).Hoveroverthese!
- 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
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, \(F_c=\frac{mv^2}{r}\).1
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:
$$\frac{mv^2}{r}=evB_\perp\label{FEq}$$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:
$$eV=\frac{1}{2}mv^2\label{EnEq}$$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.
In general, we know that magnetic field is proportional to the current that generates it:
$$B=\alpha I$$The constant of proportionality \(\alpha\) in general depends on the geometry of the wires making the magnetic field and the point at which you measure. If we measure at the center of a Helmholtz coil of radius \(a\) with \(N\) turns, the constant \(\alpha\) is:
$$\alpha=\frac{8}{5\sqrt{5}}\frac{\mu_0 N}{a}\label{BHH}$$This means that our final formula for \(e/m\) will be:
$$\frac{e}{m}=\frac{2}{\alpha^2}\frac{V}{(Ir)^2}$$Or, rewritten in the form that we will plot:
$$V=\frac{\alpha^2}{2}\left(\frac{e}{m}\right)(Ir)^2$$Adding the B Field of the Earth
Now, let's go back to equation \eqref{em} 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:
$$B_\perp=\frac{1}{r}\sqrt{\frac{2V}{e/m}}\label{bperp}$$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 \(B_\text{E}\). Then, let's define the angle between \(B_\text{HH}\) and \(B_\text{E}\) as \(\theta\). We then find:1
$$B_\perp=B_\text{HH}+B_\text{E}\cos(\theta)\label{btot}$$Therefore, we can rewrite our above equation as:
$$\sqrt{\frac{2V}{(e/m)r^2}}=B_\text{E}\cos(\theta)+B_\text{HH}$$Thus, we can measure \(B_\text{E}\) as the slope of a \(B_\perp\) vs. \(\cos(\theta)\) plot, where \(B_\perp\) is calculated with formula \eqref{bperp}.
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.1
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 diameter, 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). You can stop when you get sufficiently many data points.
Part II: Observing the Magnetic Field of the Earth
Begin by determining what direction the magnetic field is directed in your apparatus. The questions on the data sheet will guide you. You may assume that the electron beam is travelling downwards when it is emitted (by the little device in the glass chamber in your apparatus). You'll need to know this information to do this part correctly.
Turn your current down low (but at least 1A) 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; the electron beam won't move much, but it should move!
Record the current and diameter 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 (whose direction you previously determined) points the same direction as the (expected) direction of Earth's magnetic field
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. (Your TA will help you orient yourself.)
Then, turn your voltage knob until the innermost edge of the electron beam lines up with the largest distance marker again. Record this voltage.
Part I: Measuring \(e/m\)
First, from coil diameter, calculate coil radius, \(a\). Then, calculate \(\alpha\) and its uncertainty.2
For each data point, calculate \(r\) from beam diameter, then calculate \((Ir)^2\). Propagate uncertainties.
Finally, 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
For your voltage value, calculate \(B_\perp\) using equation \eqref{bperp}, and propagate uncertainty. Use the theoretical value of 1.75882009\(\times\)10\(^{11}\)C/kg as your value for \(e/m\) in this part, not your measured \(e/m\) from part I.
Using \(B_\perp\) and \(B_\text{HH}\), with \(\cos(\theta) = 1\), use Equation 10 to extract a measurement for the horizontal component of the magnetic field of the Earth.
Consider whether your answer is reasonable, compared to the total magnetic field of Earth of ~50μT.
Your TA will ask you to answer some of the following questions (they will tell you which ones to answer):
Experimental Questions:
- Look up what the horizontal component of Earth's magnetic field is (e.g., in Wolfram|Alpha). How does your measurement compare? (Does it agree? Did you at least get the right order of magnitude?)
- How could we (with only a bit more equipment, but potentially more risk to our apparatus) get the total magnetic field strength of the Earth from an experiment very similar to Part II of this one?
- Why is it far easier to measure \(e/m\) experimentally than either \(e\) or \(m\) individually?
Theoretical Questions:
- Justify equations \eqref{FEq} and \eqref{EnEq} in your own words.
- From equations \eqref{FEq} and \eqref{EnEq}, show that equation \eqref{em} holds.
- Using the expression for the magnetic field from a loop of wire, show that equation \eqref{BHH} holds.
For Further Thought:
- 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?
Hovering over these bubbles will make a footnote pop up. Gray footnotes are citations and links to outside references.
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For a review of magnetic forces on free charges, see KJF ch. 24.5. For a review of circular motion, see KJF ch. 6.
Technically, this equation is not exactly correct: it assumes that the plane of motion of the electron is always perpendicular to \(B_{HH}\). While this is not exactly true, under an assumption that \(B_E\ll B_{HH}\), it is approximately true.
Protip: using relative errors here saves you a fair amount of re-typing the prefactor, if you know how to do it.
This is really the same measurement we did in part I, except in part I we took \(e/m\) as unknown and \(B_{HH}\) as known. Here, we take \(B_{HH}\) as unknown and \(e/m\) as known, but use the same relationship.
Sometimes it can appear more visibly if you look through a smartphone camera.
Excel, Google Sheets, etc. have built-in trigonometric functions, but they all assume that the argument is entered in radians, thus why we need that conversion.