\experiment{The Hall Effect}

In 1879, American physicist Edwin Herbert Hall (1855--1938) observed a
small potential difference across a conducting sample through which a
current flowed perpendicular to an applied magnetic field.
% The electric field that produces this potential difference was
% perpendicular to both the magnetic field and the direction of current
%flow.
In this experiment you will study this phenomenon, called the Hall
effect, in a semiconductor.  Measurements of the Hall potential will
yield the sign and density of the charge carriers in the
semiconductor.

\begin{comment}
\emph{Do we want to revise this?}
In preparation for this experiment you should review Halliday \&
Resnick, Section 33-5 (Hall Effect) and the discussion in Section 37-6
(Ferromagnetism) on the magnetization curve and hysteresis. Also see
the discussion of the Hall effect in, e.g., Eisberg and Resnick,
\emph{Quantum Physics} and \emph{Modern Physics} by Robert Sproull. In
connection with the following theoretical discussion you should
determine the sign of the Hall potential $(V_{a} -V_{b}) = \Delta V$
in \fref{Hall1} for positive and negative charge carriers. Also show
how the Hall potential depends on $B$, $I$, $n$, $w$ and $d$ as
defined below.
\end{comment}


% \begin{wrapfigure}[nlines]{<rl>}[overhang into margin]{width or 0pt}
\begin{wrapfigure}{r}{0pt}
	\igw[2.25in]{Hall1}
	\caption{The Hall effect.}
	\label{fig:Hall1}
\end{wrapfigure}

\section{Theoretical Background } 

The charge carriers that conduct electricity in metals are
electrons.  If a metal strip is placed in a magnetic field and a
current is established in the strip, then a small transverse electric
field is set up across the strip.  The resulting difference of
potential is the Hall potential.  Note that this potential is
perpendicular to both the current flow and the magnetic field (see
\fref{Hall1}).  When semiconducting materials are used in place of the
metal, the Hall potential is generally much larger and may be of
opposite sign.  The change in sign implies that, in such cases, the
charge carriers are positive and that a different conduction process
is occurring than in metals.

Two quantities are of interest here.  First, the sign of the Hall
potential, which depends on the sign of the charge carriers.  Second, the
magnitude of the potential, from which may be deduced the density of charge
carriers.  This deduction is briefly the following: Let $E_{H}$ be the
transverse electric field generated in the strip carrying a current of
density $j$ in a field $B$ in the geometry shown above.  The quantity
\begin{equation*}
	R = \frac{E_{H}}{j B} 
\end{equation*}
is called the Hall coefficient. In equilibrium, the transverse force
of $q E_{H}$ acting on the charge carriers of charge magnitude $q$ must
just compensate the Lorentz force of $q \vv B$ acting upon these charges
moving with drift velocity $\vv$. Since also $j =qn\vv$, where $n$ is
the density of charge carriers, we have
\begin{equation*}
	R = \frac{E_{H}}{j B} = \frac{\vv B}{q n \vv B} = \frac{1}{q n}  
\end{equation*}

For both metals and semiconductors, it turns out that $|q| =e$, the
magnitude of the charge of an electron. Thus $n$ may be calculated
from measured values of $R$.

For a strip of width $w$ and thickness $d$, carrying a uniform current
density $j$, the conduction current is $I =jwd$.

The Hall potential is $\Delta V = E_{H}w$. Thus,
\begin{equation}
	\label{eq:Hall2}
	\boxed{ R = \frac{\Delta V}{w}  \frac{w d}{I} \frac{1}{B} 
      = \frac{\Delta V}{I B} d
      }
\end{equation}


\section{Measurement of the Hall Potential}

\begin{figure}[b!]
  \centering
  \subfigure[Electromagnet circuit]{\igw[4.25in]{Hall2a}\label{fig:Hall2a}}
    \subfigure[]{\igw[4.25in]{Hall2b}\label{fig:Hall2b}}
  \caption{Circuits}
  \label{fig:Hall2}
\end{figure}

\begin{description}
\item[Circuits] Figs.~\ref{fig:Hall2a} and \ref{fig:Hall2b} show the
  circuits used in obtaining the Hall potential. \fref{Hall2a} shows the
  circuit used to produce the required magnetic field and \fref{Hall2b}
  shows the Hall effect circuit itself.

Concerning these figures, note:
\begin{itemize}
\item The Hall effect requires both a current $I$ passing through the
  sample and a magnetic field. The magnetic field is not shown in Fig.
  2b, but the sample is oriented in the magnet to give the geometry
  shown in \fref{Hall1}.

\item Two reversing switches (RS1 and RS2) are present in the
  circuits, one to reverse the current in the magnet, the other to
  reverse the direction of current flow $I$ in the Hall effect device.
  The reason for RS2 is to compensate for thermal emf's which can lead
  to small zero offsets. 

\item The reversing switch RS1 is included for a rather subtle reason:
  In an ideal Hall effect device $\Delta V = 0$ if there is no
  magnetic field. But you will find that $\Delta V \ne 0$ even when $B = 0$. This
  occurs because the sample has a finite resistance and therefore a
  voltage drop occurs along the direction of current flow; the
  potential leads are soldered onto the sample at points which are not
  quite on the same equipotential line (in zero magnetic field). This
  potential difference between the two leads, which we'll call the $IR$
  effect, adds onto the Hall potential $\Delta V$. In order to
  eliminate the $IR$ effect, the magnetic field direction is reversed by using RS1.
  This changes the sign of $\Delta V$, but since $I$ is still flowing
  in the same direction, the IR effect will not change sign. The
  results for both magnetic field directions are averaged to get
  $\Delta V$.

\caution{Ramp the current through the electromagnet down to zero\\
  before reversing RS1.}

\end{itemize}

% \item[The Electromagnet] Calibration data for the magnet will be supplied.
%   These data were taken along the magnetization curve of the magnet
%   and apply only when the magnet is so used. Due to hysteresis (see
%   H\&R 37-6), a given current through the coils produces a unique field
%   only if the history in establishing this current is identical from
%   one time to the next. The best procedure to use is first to set up
%   the maximum current in the coils. Then reverse this current by means
%   of the reversing switch. Reduce the current by perhaps one half and
%   again reverse the current. Proceed by reducing and reversing the
%   current until the current is almost zero. In this way the field at
%   zero current is reduced arbitrarily closely to zero and a unique
%   starting point is established. The magnet calibration corresponds to
%   a continually increasing coil current from the zero field, zero
%   current point. Once a given current is established, it may not be
%   decreased without destroying the calibration. Each time reduction of
%   the field is necessary, the demagnetization procedure must be
%   repeated.

\item[Gaussmeter] You will measure the magnetic field using the
  LakeShore gaussmeter. It may be used in either SI mode (in
  milliteslas) or Gaussian mode (in gauss). Note that $\val{1}{T} =
  \val{10^{4}}{G}$.

\item[Measurements] Establish a field of some 1000 to 1700
  gauss and determine its polarity by means of a compass. \scream{The
    current in the magnet circuit must not exceed 1.9 amps!!} With a
  current of some 20 mA to 300 mA in the Hall probe, measure the potential
  difference across the Hall-effect sample. Vary both the current and
  magnetic field to establish the constancy of $R$ and obtain a best
  value from your data. Evaluate the sign and density of charge
  carriers in the sample of indium arsenide.

  \scream{Note:} Consider the physical significance of your value for
  $n$.
\end{description}

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