E-6  Measurement of Charge to Mass Ratio (e/m) for Electrons

OBJECTIVES:

To observe magnetic deflection of electrons, at fixed energy, in a uniform magnetic field and then use this information to obtain e/m, the electron charge/mass ratio.

APPARATUS:

Figure 1: Foreground: dip-needle magnetic-field sensor sitting on power supplies with built-in digital meters; Background: Sargent-Welch e/m equipment and black cardboard.

Figure 1

INTRODUCTION

Your e/m vacuum tube contains a number of features for producing and visualizing a thin uniform electron beam. Refer to Figs. 2 & 3 for schematic detais. Passing current through a wire filament (F) causes it to become hot and if the hot filament is in close proximity to a higher potential electrical element (the anode C) some of the electrons will accelerate through the vacuum towards the anode. To allow a narrow beam of electron to leave the vacinity of the anode a thin slit, S, has been cut in the anode cylinder.

Normally the electron beam is invisible to your eye. However the e/m tube also contains a saturated vapor pressure of Hg. When electrons, with energies in excess of 10.4 eV collide with Hg atoms, some atoms become ionized and some become excited. (These electrons are, of course, now permenently lost from the beam). Most of the Hg atoms/ions quickly recombine and/or de-excite to emit a bluish light before moving appreciably from the collision center. Hence the bluish light marks the path taken by the electron beam.

The potential difference $V$ between the filament and anode (Fig. 2 and 3) accelerates electrons thermionically emitted by the filament. Those electrons travelling toward the slit S emerge with a velocity $v$ given by

$$Ve = \frac{1}{2}mv^2$$ (1)

provided that the thermal energy at emission is small compared to $Ve$.

Preliminary Questions:

1. What is purpose of the Helmholtz coil (read through the appropriate section first)?

2. What aspect of the electron trajectory will indicate that it is undergoing uniform circular motion?

Figure 2: The e/m tube viewed along the earth's magnetic field (i.e., $B_{\mbox{\tiny earth}}$ is perpendicular to the page).

Figure 3: Side view of Figure 2.

If the tube is properly oriented, the velocity of the emerging electrons is perpendicular to the magnetic field. Hence the magnetic force vector $\vec{F}$, which is described as a cross product $\vec{F}=e (\vec{v}\times\vec{B})$, supplies the centripetal force $\frac{mv^{2}}{r}$ for a circular path of radius $r$. Since $\vec{v}$ is perpendicular to $\vec{B}$

$$evB = \frac{mv^{2}}{r}$$ (2)

Eliminating $v$ between (1) and (2) gives

$$e/m = \frac{2V}{B^2r^2}$$ (3)

Our digital voltmeter measures accurately the accelerating voltage $V$. The radius of curvature, $r$ or $D/2$, is half the distance between the filament $F$ and one of the cross bars attached to rod $A$. The cross bar positions are as follows:

Crossbar No.	Distance to Filament	Radius of Beam Path
1	0.065 meter	0.0325 m
2	0.078 meter	0.039  m
3	0.090 meter	0.045  m
4	0.103 meter	0.0515 m
5	0.115 meter	0.0575 m

To find $e/m$ we still need to know the magnetic field strength $B$. Instead of measuring $B$, we calculate it from the dimensions of the Helmholtz coils and the measured coil current, $I$, needed to bend the beam so that it hits a given cross bar.

Helmholtz coils consist of two identical coaxial coils which are separated as in Fig. 4 by a distance $R$ equal to the radius of either coil. They are useful because near the center ($X = R/2$) the field is nearly uniform over a large volume. (Most textbooks assign this proof as a problem. Hint: Find how $dB/dX$ changes with $X$ for a single coil at $X = R/2$, etc.)

Figure 4: Helmholtz coil geometry.

The field at $X = R/2$ is in fact: ²

$$B = \frac{8}{\sqrt{125}} \left( \frac{N\mu _0I}{R}\right) \mbox {teslas}$$ (4)

as one easily sees by adding the axial fields $B(x)$ from each of the single coils:

$$B(X) = \frac{N\mu _0IR^2}{2\left[ R^2 + X^2\right] ^{3/2}}.$$

Thus when X = R/2, then 

$$B = 2B(X) = \frac{N\mu _0I^2R^2}{\left[ R^2 + R^2/4\right] ^{... ...{3/2}} = \frac{8}{\sqrt{125}} \left(\frac{N\mu _0I}{R}\right) .$$

In Eqn. 4,

$N =$

number of turns on each coil (72 for these coils)

$I =$

current through each coil in amperes

$R =$

mean radius of the coils in meters (approximately 0.33 m but varies slightly from unit to unit)

$\mu _0 =$

$4\pi \times  10^{-7}$ tesla meter/ampere.

Finally if we substitute (4) into (3) we obtain

$$e/m = \left( 2.47 \times  10^{12} \frac{R^2}{N^2}\right) \frac{V}{I^2r^2}   \mbox {coulombs/kg .}$$ (5)

CIRCUIT DIAGRAM: Fig. 5 should help you hook up the components. The current must have the same direction in both sets of Helmholtz coils.

Figure 5: Basic wiring layout.

SUGGESTIONED PROCEDURE:

1.

Set the axis of the Helmholtz coils along the local direction of the earth's field, $B_{e}$, as determined with a compass and dip needle. The axis should point towards the geographic north but at an angle about 60 degrees from the horizontal. Thus the axis should point deep under Canada. Put the long axis of the e/m tube in the north-south direction and with the cross bars up.

CAUTION: Nearby ferromagnetic material (e.g. steel in the power supply, the table and in the walls) can alter the local direction of $B_{e}$. As long as the field direction does not change during the experiment, there should be no problem.

2.

Set the filament supply knob to zero before turning the power on. A red light comes on when excessive filament current or a too rapid increase trips a protective relay. Reset by putting the filament control to zero and slowly turning it up.

3.

Since $e/m$ depends on $V/I^2$, one needs high quality meters for $V$ and $I$. Our digital meters have an accuracy of $\pm$ one digit in the last displayed digit.

4.

Start with $\sim 22 V$ between filament and anode. With the room dark, gradually turn up the filament control until the beam is visible. To make the beam more visible use black cloth and black cardboard to block out stray light: place the black cardboard inside the Helmholtz coils and view from the top. If the protection circuit trips and a red light comes on, reset by zeroing the filament control; then turn it up slowly. When the beam appears, adjust the filament control to give 5-10 mA of anode current. Do not exceed 15 mA! With no current through the Helmholtz coils, the earth's magnetic field should slightly curve the beam (like the dotted line in Fig. 2.)

5.

Correction for the earth's magnetic field: Increase $I$ through the Helmholtz coils until the beam deflects. If the sense of deflection is the same as that from the earth's field (i.e. when $I=0$), reverse the leads to the Helmholtz coils; the curvature should then be like B in Fig. 2. The field from the coils then opposes the earth's field; hence adjust the coil current until the beam path is straight. Since light travels in straight lines, the beam should then hit the glass at the center of the area illuminated by light from the filament.

This field current which just cancels the earth's field, we will call $I_{e}$. Clearly the $I$ for Eqn. 5 must have $I_{e}$ subtracted: Let the correct field current for crossbar $n$ be $I_{n}$, and let $I_{n}'$ be the measured current to bend the beam around so that the outside sharp edge of the beam hits the center of the $n^{\mbox{th}}$ crossbar. Hence $I_{n} = I_{n}'$ - I$_{e}$ is the correct current to bend the beam to the appropriate radius, $r_{n}$.

MORE ACCURATE METHODS TO ELIMINATE THE EFFECT OF THE EARTH'S FIELD:

Uncertainty in $I_{e}$ can dominate the uncertainty in $e/m$, so consider using the following more accurate methods to eliminate the effect of the earth's field:

i.

Take two readings of coil currents at $n=3$, $I_3'$ and $I_3''$: one with the earth's field opposing that of the Helmholtz coils, the other with it aiding. To accomplish this, first (with the cross bars up) measure $I_3'$ (as described above); then rotate just the e/m tube 180$^o$ about its long axis.

CAUTION: Always rotate the tube in the sense to REDUCE the TWIST in the filament and anode leads. Otherwise one can twist the leads off.

Next reverse the current in the Helmholtz coil and adjust it so that the beam again hits the same cross bar. Call this reading $I_3''$. Then

$$I_3'  -  I_3''  =  2I_{e}.$$

ii.

Or better yet, for each cross bar record an $I_{n}'$ and $I_{n}''$ reading. The average

$$\frac{I_{n}'  +  I_{n}''}{2}  =  I_{n}$$

is the current required if the earth's field were absent. You may find it easier to measure $I_n'$ for $n=1$ to 5 and then rotate the e/m tube just once to measure $I_n''$.

MEASUREMENTS AND ANALYSIS: (If time is short, measure only two cross bars at each voltage.)

1. Make two determinations of $e/m$ for each cross bar.
2. Repeat for an accelerating voltage of $\sim$ 44 volts. (Note $I_{e}$ should remain the same and one can measure it more precisely at the lower accelerating voltage.)

   SUGGESTED DATA TABULATION: {If you use the alternative method (ii. above), replace $I_{e}$ by $I_{n}''$ and $I_{n} = I_{n}'-I_{e}$ by $I_{n}=\frac{1}{2} (I_{n}'+ I_{n}''$).}
   

   Volts	bar	radius	$I_{e}$	$I_{n}'$	$I_{n} = I_{n}'-I_{e}$	$e/m$

3. Estimate the reliability of your value of $e/m$ and compare to $1.76 \times 10^{11}$ C/kg.

OPTIONAL:

Note that the electron beam expands in a fan shape after leaving the slit. Interchange the filament leads and note how the fan shape flips to the other side of the beam.

Explanation: The fan shape deflection arises from the filament current's magnetic field acting on the beam. Our filament supply purposely uses unfiltered half wave rectification of 60 Hz AC so that for half of the cycle no current flows thru the filament but which is still hot enough to emit sufficient electrons. The electrons emitted in this half cycle constitute the sharp beam edge which we use for measurement purposes. This trick not only avoids deflection effects for this part of the beam, but also avoids the uncertainty in electron energy arising from electrons being emitted from points of different potential along the filament. (During this half cycle of no filament current the filament is $\sim$ an equipotential.) The fan shape deflection relates to electrons emitted during the other half cycle when filament current flows. [See: F.C. Peterson, Am. J. Phys. 51, 320, (1983)]