\input{labpreamble}
\begin{document}

\subsubsection*{PHSX 222L\hfill Charge to Mass Ratio of Electrons\hfill Name:\underline{\hspace*{2in}}}

\underline{Object:}  Determine the charge to mass ratio of the 
electron by observing its motion in a magnetic field. 

\underline{Theory:} In experimental work involving laboratory-produced
magnetic fields, it has been found convenient to use an arrangement of
windings known as a \underline{Helmholtz coil} to generate the needed
magnetic field.  The Helmholtz set up consists of two coils of equal
radius $R$ placed parallel to each other and separated by a distance
equal to the radius of the coils.  The magnetic field produced at the
center of such a pair of coils is very uniform over a significant
region of space in which an experiment can be performed (both $\dif
B/\dif x$ and $\dif^{2}B/\dif x^{2}$ vanish at the center).  The
magnitude of this field can be derived from the Biot-Savart Law and is
given by
\begin{equation}\label{B}
B={8\mu_{o}NI\over\sqrt{125}R}
\end{equation}
where $N$ is the number of turns per 
coil, and $I$ is the current in the 
coils.

If electrons of mass $m$ and charge $e$ are accelerated through a 
potential difference $V$ and injected into a magnetic field 
$\boldsymbol{B}$ at 
right angles to it, they will move in a circular orbit having radius
\begin{equation}\label{R}
r = {\sqrt{2mV\over e}\over B}.
\end{equation}
Equation \ref{R} can be solved for the charge to mass ratio:
\begin{equation}\label{e/m}
{e\over m} = {2V\over r^{2}B^{2}}.
\end{equation}

% From the official apparatus instruction sheet:
% 
% "The cathode 'gun' is located about 0.254 cm below the anode plate 
% opening.  This causes the center of curvature of the emerging beam 
% toie just below the anode plate.  When conditions are adjusted so that 
% the beam strikes a circle on the anode disk, the true diameter of the 
% beam circle is larger than the recorded value.  This results in a 
% recorded value of e/m that is larger than the actual value.  To 
% account for the displacement of the cathode below the anode plate, 
% Equation (4) becomes:
% 
% e/m = 2V/[B^2 (r^2 + (0.254 x 10^(-2))^2)]."

The electrons will be ``boiled off'' a heated filament (cathode) in a
vacuum tube (see photograph on page 908 in Serway) and then
accelerated by an anode plate (at a higher potential) which has a small
hole in it.  The cathode ``gun'' is located a small distance $d$ below
the anode plate opening.  This causes the center of curvature of the
emerging beam to lie just below the anode plate.  When conditions are
adjusted so that the beam strikes a circle on the anode disk, the true
diameter of the beam circle is larger than the recorded value.  This
results in a recorded value of $e/m$ that is larger than the actual
value.  To account for the displacement of the cathode below the anode
plate, replace $r^{2}$ with ${r'}^{2}$ where ${r'}^{2}= r^{2}+d^{2}$and
$d$ is 0.254 cm in our case.

If the electron beam is not exactly perpendicular to the magnetic
field the electrons will follow a helical trajectory rather than a
circular one.  (This is not what you want in this experiment.)

\underline{Apparatus:}  Carefully draw a detailed diagram of the 
apparatus.  Include the wiring.


\pagebreak
\underline{Procedure:} 
\vspace{-.25in}
\begin{enumerate}

\item Derive equations \ref{B} and \ref{R}.

\item Determine the radius and the number of windings in each loop of 
the pair of Helmholtz coils.  This data, along with the current, will 
be used to calculate the magnetic field at the center of the coil 
arrangement.

\item While the effect of the Earth's magnetic field is not large 
compared to the Helmholtz field, you may want to consider it and make 
a correction for it.  You should also insure that there aren't other 
stray magnetic fields nearby (use a compass or gaussmeter).

\item First set the electron tube filament current to be 0.6--0.8 A at
5.0--7.0 V (usually 6.3 V).  After a 1-min filament warm-up, the plate
potential (accelerating voltage) may be turned on (to about 100 V). 
You should see a beam of electrons going straight up.  Reduce the
filament current to the minimum level that still yields a visible beam
(higher settings can prematurely burn out the tube which is very
costly).  Use the grid voltage to focus the beam.

\item Adjust the current in the coils (0--4 A) and the accelerating
voltage (80--250 V) so as to produce an electron beam which curves into
a visible orbit inside the apparatus tube.  Measure the radius of the
orbit for each coil current and accelerating voltage setting.  The
flourescent circles painted on the anode plate have radii 0.50 cm, 1.0
cm, 1.5 cm, and 2.0 cm.  (These radii are the diameters of the electron
orbits, taking $d$ into account.)

\end{enumerate}

\goodbreak
\underline{Results:} 
\vspace{-.25in}
\begin{enumerate}

	\item Substitute equation \ref{B} into equation \ref{e/m} (modified 
	for ${r'}^{2}$) and then 
	solve for $V$ as a function of ${r'}^{2} I^{2}$.
% \item From the measurements of $V$, $I$, and $R_{\rm orb}$, calculate 
% $e/m$ for each set of values.

	\item A plot of $V$ vs.~$(r' I)^{2}$ should produce a straight 
	line whose slope involves $e/m$.
	
	\item Determine the least squares slope of your line and solve it
	for $e/m$.  Then compare your result with the accepted value of
	$e/m$ ($1.75882\times 10^{11}$ C/kg), and discuss possible sources
	of uncertainty.

\end{enumerate}


\underline{Questions:} 
\vspace{-.25in}
\begin{enumerate}

\item The Millikan oil drop experiment gives $e$ 
(current best value is $1.602\,176\,462 \times 10^{-19}$~C), so now compute the 
mass of the electron.

\item Compute the speed $v$ of the electrons in this experiment and 
then determine if a relativistic correction to the mass is needed.  
Special relativity says $m = m_{o}/\sqrt{1-v^{2}/c^{2}}$, where $m_{o}$ 
is the rest mass.

\item What would be the mass of an electron traveling at half the 
speed of light?  At 99.5\% of $c$?

\item What potential difference would an electron have to fall from 
rest through to acquire a speed of $3.00\times 10^{6}$~m/s ({\it 
i.e.,} $0.10 c$)?

\item How could you determine the size of the error contributed by the 
earth's magnetic field in this experiment?  How could you account for 
it or counter-act it?

\end{enumerate}


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