# PHY 124 Lab 8 - Measurement of e/m for the electron

The purpose of this laboratory is the measurement of the charge (e) over mass (m) ratio e/m for the electron and to study qualitatively the motion of charged particles in a magnetic field.

## Video

NOTE: The video was made when two similar, but non-identical, apparatuses were being used for this lab. We now have enough of one of them (called Apparatus A in the video) to enable you all to use the same type for this experiment. Therefore, don't worry about the difference in details between Apparatus A and Apparatus B in the video. The principles of operation are the same for both types. The written text of this Lab manual has been edited to include only the details that have to do with the apparatus (A) you'll be using to do this lab.

## Equipment

• e/m aparatus (Daedalon Corporation) pdf of manual here
• meter stick
• black cloth

We are fortunate to have enough setups for each of you to use the Daedalon Corporation Model EP-20 e/m of the Electron Apparatus for this lab. It has internal power supplies that provide the voltage used to accelerate electrons in a beam and the current used to drive Helmholtz coils that produce the magnetic field B that deflects the electrons.

## Background to the experiment

The principle of the e/m measurement is as follows (see the figure below): electrons “boiled” off a heated metal cathode are accelerated by an accelerating voltage difference V applied between the negative cathode and an anode kept near ground potential. The anode has a small hole in it that allows a small fraction of the electrons to pass through, thereby creating an electron beam with energy kinetic energy $KE = eV$ inside the glass tube mounted on top of the power supply box. (Recall that $e$ is the elementary unit of charge: $1.602 \times 10^{-19}$ Coulomb.) A small amount of helium gas is present inside the tube, and when forced onto a circular path by the magnetic field supplied by the two “Helmholtz coils” shown in the figure, some of the electrons excite helium atoms in collisions. The excited helium atoms emit light that we see as the blue-green glow in the color pictures in this lab manual. We also see light emitted from marked areas along a glass “crossbar” hit by the electrons; these markers are delineated along a centimeter scale and allow us to determine – albeit with non-negligible experimental uncertainty – the diameter of the circular path of the electron beam. This diameter depends on the energy of the electron beam and on the strength of the magnetic field B inside the spherical vacuum tube. The B field is approximately perpendicular to the plane of the Helmholtz coils, and, when the tube is properly oriented, the circular path of the electrons lies in the midplane between the coils. The figure below will help you to use the right-hand-rule (remember the electrons have negative charge!) to visualize the directions of the electron velocity $v$, the magnetic field $B$, and the (centripetal) force $F$.

In the MapleTA lab pretest, you are going to derive the e/m relation you will use in your analysis for this lab. The kinetic energy KE, and hence the velocity v of the electrons, are determined by the accelerating voltage $V$ they have traversed (SV8, Chap. 16). The radius r of the circular path can be related to the electron velocity and magnetic field (SV8, Chap. 19). By combining these equations you will end up with a formula that relates e/m to the quantities you measure in this lab.

Using the equations from, e.g., SV8, Examples 16.1 and 16.2, and the formula $KE=\frac{1}{2}mv^{2}$, and then solving for the velocity gives the relation between the electron velocity $v$ and the accelerating voltage $V$. This relation and equation [19.10] in SV8 for the radius r of the electron circular path leads to an equation that relates e/m to the accelerating voltage V, the magnetic field B, and the radius $r$ of the electron-beam orbit. You should get the relation

$\Large \frac{e}{m}=\frac{2V}{B^{2}r^{2}}$
(8.1)

Equation [19.15] in SV8 gives a formula for the magnetic field in the center of the plane of a current loop with $N$ turns of wire. In this lab we will use the symbol $a$ for the radius of the coils carrying the current because we don't want to confuse this radius with the radius of the electron beam inside the glass cell. Make sure you refer to the coil radius with the letter $a$.

We don't use one coil (current loop) with $N$ turns of wire because this does not create a uniform magnetic field over an appreciable region of space. However, two identical sets of coils, each having $N$ turns of wire and identical radius $a$, can be used to create a reasonably uniform $B$ field over an appreciable region of space if the following conditions are met: (i) the planes of the two coils are parallel and separated by a distance $a$ equal to the radius of each coil; (2) the same current $I$ flows in the same sense (CW or CCW)through each of the coils; (3) the experimental cell is centered on the midplane between the two coils because it is in that region that the $B$ field is most uniform. This configuration is known everywhere as “Helmholtz coils” after its inventor.

The figure below (credit: GNU Free Documentation License) shows magnetic field lines for a pair of Helmholtz coils. The current is coming out of the page at the “top” of both sets of coils (at the black dots) and into the page at the “bottom” of both sets of coils (at the black crosses). The separation between the coils (dot-to-dot distance and cross-to-cross distance) is equal to the radius, i.e., one-half the diameter (dot-to-cross distance), of each coil, which satisfies the Helmholtz condition.) Notice that both the strength (density of field lines) and the direction of the $B$ field are quite uniform in the center of the midplane between the coils.

The equation for the magnetic field produced in this region of Helmholtz coils, which is where the electron beam circulates in our experiment, is given by the equation (all units are SI)

$B = \left(\frac{4}{5} \right)^{3/2} \frac{\mu_0NI}{a}$
(8.2)

where 4
N = the number of turns in each Helmholtz coil = 130
a = the radius of the Helmholtz coils (see the figure above)
$\mu_0 = 4 \pi \times 10^{-7}$ $\frac{Vs}{Am}$ (the universal magnetic constant)
$I$ = the current through the Helmholtz coils.

Substituting equation (8.2) for the magnetic field, B, in equation (8.1) gives, for e/m

$\Large \frac{e}{m}= \left( \frac{125}{32} \right) \left(\frac{a^2}{\mu_0^2N^2} \right) \left( \frac{V}{I^2r^2} \right)$
(8.3)

You will use this expression (8.3) in the quantitative Part II.

NOTE: It bears repeating that in this lab you will be measuring two radii, the radius $r$ of the circular path of the electron and the radius $a$ of the Helmholtz coils. Make sure you don't confuse the two radii!

## Part I: Qualitative Exploration

NOTE: The figure below was made when this lab manual needed to describe two similar but non-identical setups used for the e/m measurement. For this experiment, you need refer only to the apparatus on the left-hand side of the picture.

Before you use the equipment to determine e/m, you will first investigate the relation between the voltage and current controls of the apparatus and the radius of the path of the electron. In this part you explore how the accelerating voltage V and the magnet current I influence the radius r of the electron orbit. The accelerating voltage, V, sets the electron beam energy and the magnet current, I, sets the magnetic field of the Helmholtz coils, which deflects the electron beam.

### Procedure

Plug in the apparatus and turn the power switch (lower left hand corner) to the on position. You must then wait for 30 seconds while the power supply runs a self-test. The power supply has two knobs: on the left, for the accelerating voltage V; on the right, for the magnet current I. Set V to ~150 V and I to ~1.2 A.

You should see the electron circular path as a “blue-green” glow. As mentioned earlier, this glow comes from the electron-impact excitation of a small amount of helium gas left on purpose in the vacuum tube. Note that you will probably have to either cover the coil with a dark cloth or dim the room lights to see the glow.

NOTE: Looking at your tube from directly above, if you see that the electron beam is not staying in a plane containing the glass “crossbar”, i.e., the electron beam is making a spiral, then the position of your tube will have to be changed a little. Since the tube is made of glass and, therefore, breakable, do not adjust the position yourself. Ask the TA to do it.

Keeping V constant at 150 V, increase I in several steps and observe what this does to the radius r of the circular path followed by the electron beam. Explain your observation in terms of the functional relationships between the experimental variables.

Keeping I constant at 1.2 A, increase V in several steps and observe r. Again, explain your observation in terms of the functional relationships between the experimental variables.

## Part II: Measurement of e/m

To get an accurate value for coil radius, a, you will need to make several measurements and use their average for future calculations. Notice that each Helmholtz coil is made of 130 turns of wire, and because the wire has finite thickness, each “bundle” of wires doesn't have a unique radius, and the two bundles are not separated by a unique distance. A reasonable approximation is to assume that the effective location of each bundle is its center. Therefore, Measure the vertical (from top to bottom) and the horizontal (from left to right) diameter (from the middle of the bundle of turns making up the coil) for both coils and enter them on your worksheet . Notice that these values are referred to as 2a because they are (effective) diameters. Take the average of these 4 measurements to calculate the effective radius for the coils that you will used in the analysis:

$a=\frac{1}{2}(2a_{v1}+2a_{v2}+2a_{h1}+2a_{h2})/4$

Estimate the measurement uncertainty for one of your effective-diameter measurements and assume it is the same for all 4 measurements. Calculate the error of the average value of a using equation (E.5b) and (E.5a) in Uncertainty, Error and Graphs.

### Procedure

To measure the electron beam radius, r, use the marks etched as the scale on the glass crossbar in the vacuum tube. Be sure to notice that these marks are in cm and that they give the diameter – not the radius – of the beam path! You will take your data by keeping I fixed and varying V. Keep I fixed at ~2 A and start with V ~ 150 V. Increase V in ~ 6 steps such that at each step the electron beam hits the cm marks etched on the glass scale in the center of the vacuum tube. It is more important that the beam hits these marks than to have equal V steps. For each step record V, I, r and your estimate of the error $\Delta$r in the table on your worksheet. When estimating $\Delta$r for your measurements, consider things like the fuzziness or the brightness of blue-green blow of the electron beam and where it hits the crossbar. Note that $\Delta$r does not have to be – nor, indeed, should it be – the same for each measurement.

Assume that the errors of V and I are negligible. Solve equation (8.3) for V and write it such that it has the form $V=constant\times \frac{e}{m}\times (Ir)^2$ where “constant” stands for a factor containing only quantities that are constant during the experiment, including the radius a of the Helmholtz coils. Calculate the value of “constant” and enter it on your worksheet. Notice it includes the radius a of the Helmholtz coils, which has an uncertainty so you must propagate the error of a into the error of the constant using expressions (E.1) and (E.8) in Uncertainty, Error and Graphs.

Enter your recorded data into the form below.

V1  V   I1  A   r1  m   Δr1  m
V2  V   I2  A   r2  m   Δr2  m
V3  V   I3  A   r3  m   Δr3  m
V4  V   I4  A   r4  m   Δr4  m
V5  V   I5  A   r5  m   Δr5  m
V6  V   I6  A   r6  m   Δr6  m

When you click submit the computer calculates the quantity $(Ir)^2$ for each data point and then plots $V$ vs $(Ir)^2$.

Why do we plot $V$ vs $(Ir)^2$ and not $V$ vs $(Ir)$? Since you are considering the errors of V and I to be negligible, all the error bars in your plot will come from the measurement uncertainties in r, so you will only have horizontal error bars. The computer obtains the values for the error in $(Ir)^2$ by propagating the error of r into the error of the quantity $(Ir)^2$ using expression (E.1) and (E.8) in Uncertainty, Error and Graphs. Write the slope of the graph and its error on your worksheet.

Next, set up an equation that relates the slope, s, of the graph to the “constant” and to e/m. Solve the equation for the ratio e/m. Using your errors for s and the “constant”, calculate the error of e/m using expression (E.7) in Uncertainty, Error and Graphs. Make sure you simplify units as much as possible. Compare your value to the established value that you can obtain by dividing the charge on the electron, $e=1.602\times10^{-19} C$ by its mass $m_{e}=9.109\times10^{-31} kg$. 