\section{E-4~~Electron~Charge~to~Mass~Ratio}
\noindent OBJECTIVES:
\begin{enumerate} 
\item[] \latex{\vspace*{-0.05in}}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.
\end{enumerate} 

\noindent APPARATUS:
\begin{enumerate} 
\item[] \latex{\vspace*{-0.05in}}
\begin{figure}[h] \centering
\latex{\hspace*{.0in}\includegraphics[height=3in]{figs/e6-01.eps}}
\html{ {\bf Figure 1} \\
      \includegraphics[height=3in]{figs/e6-01.eps}}
\latex{\vspace*{-.1in}}
\caption{Foreground: dip-needle magnetic-field sensor sitting on power supplies with built-in
digital meters; Background: Sargent-Welch e/m equipment and black cardboard.}
\end{figure}
\end{enumerate}

\vspace*{-.15in}
\noindent INTRODUCTION
\begin{enumerate} 
\item[] \latex{\vspace*{-0.05in}}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. \medskip  

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. \medskip   

\noindent 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 {\html \\}
\begin{equation}
Ve = \frac{1}{2}mv^2
\end{equation}
\noindent provided that the thermal energy at emission is small compared to 
$Ve$. 
\end{enumerate} 

{\bf{\em{\noindent Preliminary Questions:}}}
\begin{description}
\item \latex{\vspace*{-0.05in}} 1. What is purpose of the Helmholtz coil (read through the appropriate section first)?
\item \latex{\vspace*{-0.05in}} 2. What aspect of the electron trajectory will indicate that it is undergoing uniform
circular motion?
\end{description}
\setcounter{figure}{4}
\begin{htmlonly}
\begin{center}
{\bf Figure 2:} The e/m tube viewed along the earth's magnetic field (i.e.,
$B_{\mbox{\tiny earth}}$ is perpendicular to the page).
\includegraphics[width=4.8in]{figs/e6-02.eps}
\end{center}
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\begin{latexonly}
\begin{figure}[!h] \centering 
\includegraphics[height=2.4in]{figs/e6-02.eps}
\end{figure}
\vspace*{-.25in}
\hspace*{1.7in}
\parbox{4.2in}{Figure 2: The e/m tube viewed along the earth's magnetic field (i.e.,
$B_{\mbox{\tiny earth}}$ is perpendicular to the page).\\ \\}
\end{latexonly}
\begin{htmlonly}
\begin{center}
{\bf Figure 3:} Side view of Figure 2. \\ 
\includegraphics[width=4.8in]{figs/e6-03.eps}
\end{center}
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\begin{latexonly}
\begin{figure}[!h] \centering 
\latex{\includegraphics[height=2.4in]{figs/e6-03.eps}}
\end{figure}
\vspace*{-.35in} \hspace*{1.0in}\parbox{4.2in}{Figure 3: Side view of Figure 2.}\\
\end{latexonly}
\begin{enumerate} 
\item[] \latex{\vspace*{-0.15in}}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 $\vec{B}$  \html{\\}
\begin{equation}
evB = \frac{mv^{2}}{r}
\end{equation}
\noindent Eliminating $v$ between (1) and (2) gives \html{\\} 
\begin{equation}
e/m = \frac{2V}{B^2r^2}
\end{equation}
\noindent 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:
\begin{center}
\begin{tabular}{|c|c|c|} \hline 
{Crossbar~No.} & {Distance~to~Filament} & {Radius~of~Beam~Path}  \\
 \hline   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 \\
    \hline          
\end{tabular}\end{center}

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.

\html{ \begin{tabular}{p{4.5in}l} }
\latex{\parbox{0.50\linewidth}}
{\noindent 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.) }
\html{ & }
\latex{\parbox{0.50\linewidth}}{
\latex{\hspace*{.1in}\includegraphics[height=1.6in]{figs/e6-04.eps}}
\html{\includegraphics[height=1.5in]{figs/e6-04.eps}}
\begin{center} \hspace*{-.3in}Figure~4: Helmholtz coil geometry.\\
\end{center} \latex{\vspace*{-.1in}} }
\html{ \cr \end{tabular}  }

The field at $X=R/2$ is in fact:
\footnote{The actual $B$ at the maximum electron orbit is only $\sim $~0.5\% less.  See 
Price, ``Electron trajectory in an $e/m$ experiment'', 
{\it Am.~J.~Phys.}~{\bf 55},~18,~(1987)}


\begin{equation}
B = \frac{8}{\sqrt{125}}
\left( \frac{N\mu _0I}{R}\right)  \mbox {teslas} 
\end{equation}

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~ 
\[B~=~2B(X)~=~\frac{N\mu _0I^2R^2}{\left[ R^2~+~R^2/4\right] ^{3/2}}
~=~\frac{N\mu _0I}{R~\left[ \frac{
5}{4}\right] ^{3/2}}~=~\frac{8}{\sqrt{125}}
\left(\frac{N\mu _0I}{R}\right) .\]
In Eqn.~4,
\begin{enumerate} \item[]
\begin{enumerate} 
  \item[$N~=$] \latex{\vspace*{-.04in}}number of turns on each coil (72 for these coils)
  \item[$I~=$] \latex{\vspace*{-.04in}}current through each coil in amperes
  \item[$R~=$] \latex{\vspace*{-.04in}}mean radius of the coils in meters (approximately
               0.33~m but varies slightly from unit to unit)
  \item[$\mu _0~=$] \latex{\vspace*{-.04in}}$4\pi \times ~10^{-7}$ tesla meter/ampere.
\end{enumerate}
\end{enumerate}
Finally if we substitute (4) into (3) we obtain
\begin{equation}
e/m~=~\left( 2.47~\times ~10^{12}~\frac{R^2}{N^2}\right) 
\frac{V}{I^2r^2} ~~\mbox {coulombs/kg~.}
\end{equation}
\end{enumerate}

\noindent{CIRCUIT~DIAGRAM:}  Fig.~5 should help you hook up the 
components.  The current must have the same direction in both sets of Helmholtz coils.
\begin{figure}[!h] \centering 
\latex{\includegraphics[height=2.2in]{figs/e6-05.eps}
}
\html{\includegraphics[width=6.0in]{figs/e6-05.eps}}
\caption {Basic wiring layout.}
\end{figure}
\latex{\vspace*{.03in}}

\noindent SUGGESTIONED PROCEDURE: 
\begin{enumerate}
\item \latex{\vspace*{-.07in}}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. 

\html{
\begin{rawhtml}  
<IMG SRC="../gif/caution.gif" ALIGN=center>  <br> 
\end{rawhtml}}
\begin{center}
\begin{tabular}{|p{4.5in}|} 
\hline
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. \\ \hline
\end{tabular}
\end{center}
\html{\begin{rawhtml} <IMG SRC="../gif/caution.gif" ALIGN=center>  <br> \end{rawhtml}}
\item 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. 
\item 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.
\item Start with $\sim 22~V$ between filament and 
anode.  With the room dark, {\em 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 {\bf 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 
{\bf slowly}.  When the beam appears, adjust the 
filament control to give 5-10~mA of anode current. {\bf 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.)
\item {\em 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.  

\item[] 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, $r_{n}$.
\end{enumerate}

\noindent MORE ACCURATE METHODS TO ELIMINATE THE EFFECT 
OF THE EARTH'S FIELD:
\begin{enumerate}
\item[] \latex{\vspace*{-.07in}}
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: 
\begin{enumerate} 
\item[i.] \latex{\vspace*{-.07in}}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$^\circ $ about its {\em long} axis.

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\begin{center}
\begin{tabular}{|p{\latex{4.5in}\html{7.5in}}|} 
\hline
CAUTION: 
Always rotate the tube in the sense to {\em REDUCE} the {\em TWIST} in the 
filament and anode leads.
{\em Otherwise one can twist the leads off.}  \\ \hline
\end{tabular}
\end{center}
\begin{rawhtml}  
<IMG SRC="../gif/caution.gif" ALIGN=center>  <br> 
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\noindent 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
\[I_3'~~- ~I_3''~~=~~2I_{e}.\]
\item[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 $I_n''$.
\end{enumerate}
\end{enumerate}

\noindent MEASUREMENTS~AND~ANALYSIS: (If time is short, measure only two cross bars 
at each voltage.)
\begin{enumerate}
\item Make two determinations of $e/m$ for each cross bar.  
\item 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.)

\noindent 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 $I_{n}=\frac{1}{2} (I_{n}'+
I_{n}''$).\}\\

\begin{center}
\begin{tabular}{|p{12mm}| p{12mm}| p{12mm}| p{17mm}| p{17mm}| p{19mm}| p{17mm}| } \hline
        Volts& bar & radius & $I_{e}$ & $I_{n}'$& $I_{n}=
I_{n}'-I_{e}$ ~ & ~ $e/m$ \\ \hline
 ~ & ~ & ~ & ~ & ~ & ~ & ~ \\ \hline 
 ~ & ~ & ~ & ~ & ~ & ~ & ~ \\ \hline 
 ~ & ~ & ~ & ~ & ~ & ~ & ~ \\ \hline 
 ~ & ~ & ~ & ~ & ~ & ~ & ~ \\ \hline 
\end{tabular}
\end{center}
\item Estimate the reliability of your value of $e/m$ and compare to 
$1.76 \times 10^{11}$~C/kg.  
\end{enumerate}
\noindent OPTIONAL:
\begin{enumerate}
\item[] \latex{\vspace*{-.07in}}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. 
\end{enumerate}

\begin{description} \item{\em 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, {\it Am.~J.~Phys.} {\bf 51}, 320, (1983)]
\end{description}
