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\title{Final Draft Lab 3 (LL and NZ) Determination of Carrier Density through Hall Measurements and Determination of Transition Temperature ($Tc$) in a High-Tc Superconductor}
\author{Ning Zhu}
\affil{Affiliation not available}
\author{Lucy Liang}
\affil{Smith College}
\date{\today}
\maketitle
\section{Abstract}
An electromagnet was used to provide a magnetic field to 3 different conducting samples: n type Geranium (n-Ge), p type Geranium (p-Ge), and silver (Ag). A calibrated Hall probe was used to obtain the current ($\vec{I}_{mag}$) to magnetic field ($\vec{B}$) calibration of the iron-core electromagnet. The Hall voltages ($V_H$) produced by each of the three samples were plotted against $B$, and a linear line was produced, as expected. The slope ($\frac{\Delta V_H}{\Delta B}$) of each of the graphs were used to calculate the Hall coefficient for each sample, which we found to be $-4.99\cdot 10^{-3}\pm -0.0998 \cdot 10^{-3}\frac{\textrm{Vm}}{\textrm{AT}}$, $5.64 \cdot 10^{-3}\pm 0.11 \cdot 10^{-3} \frac{\textrm{Vm}}{\textrm{AT}}$, $-2.24 \cdot 10^{-10}\pm -0.04 \cdot 10^{-10} \frac{\textrm{Vm}}{\textrm{AT}}$ respectively. These do not really agree with given values of $-5.6\cdot 10^{-3}\frac{\textrm{Vm}}{\textrm{AT}}$ for n-Ge, $6.6\cdot 10^{-3}\frac{\textrm{Vm}}{\textrm{AT}}$ for p-Ge, and $-8.9\cdot 10^{-11}\frac{\textrm{Vm}}{\textrm{AT}}$ for silver by the manufacturer. Using the Hall coefficients, we found their carrier densities to be $-1.25\cdot 10^{21}\pm 0.025 \cdot 10^{21} \textrm{m}^{-3}$ for n-Ge, $1.11\cdot 10^{21}\pm 0.02 \cdot 10^{21} \textrm{m}^{-3}$ for p-Ge, $-2.79\cdot 10^{28}\pm 0.06\cdot 10^{28} \textrm{m}^{-3}$ for silver, which are all in the same order of magnitude as the given absolute values of $1.2 \cdot 10^{21} \textrm{m}^{-3}$, $1.1 \cdot 10^{21} \textrm{m}^{-3}$, $6.6 \cdot 10^{28} \textrm{m}^{-3}$.
A current was applied to the superconductor $Bi_2 Sr_2 Ca_2 Cu_3 O_{10}$ which was cooled in liquid nitrogen until it became superconducting, and was allowed to warm slowly. Its voltage and temperature were monitored in the warming process which we used to produce a graph of voltage against temperature. The graph showed a transition temperature of about $118\textrm{K}\pm 2\textrm{K}$, similar to the provided critical temperature of $108\textrm{K}$.
\section{Aims}
The Hall Effect is an important part of finding out about charge transport in a material.In this experiment, we aim to determine the charge density in Ge n-type, Ge p-type semiconductor samples, and in a Ag (silver) sample. We will also be determining the transition temperature ($Tc$) of the \href{https://en.wikipedia.org/wiki/Bismuth_strontium_calcium_copper_oxide}{high-Tc superconductor} $Bi_2 Sr_2 Ca_2 Cu_3 O_{10}$ (aka $Bi2223$).
\section{Introduction}
Hall Effect was discovered by Edwin Herbert Hall, an American physicists, in 1879. This phenomenon can be used to determine the sign of the charge carrier in electrical conductors such as semiconductors and superconductors. The fundamental idea of Hall Effect can be illustrated in Figure \ref{fig:HallIllustration}:\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Intro/Intro}
\caption{{\label{fig:HallIllustration}
Hall Effect Illustration: The magnetic field is pointing into the page. By convention, the direction of current is defined as the direction of the moving positive charge. Therefore, $\vec{v}$ is opposite for positive and negative charge carriers for a $\vec{I}$. And according to right hand rule, the magnetic force $F_M$ (Equation \ref{eq:righthand}) will be upward for both cases. As charges accumulate at one side of the plate, an electric field $E$ is created between the sides. The electric force $F_E$ on the current is opposite to $F_M$. The Hall voltage is the voltage across the plate when the current does not curve up or down anymore, when $F_M=F_E$.%
}}
\end{center}
\end{figure}
Hall effect is directly caused by a current passing through a conducting sample in a magnetic field $\vec{B}$. The conducting sample may have either positive or negative charge carriers (charges that are free to move in a conductor). There will be no potential difference in the direction perpendicular to the current when there is no magnetic field applied. However, if a magnetic field $\vec{B}$ is present in the direction perpendicular to the sample, a Lorentz force (see Eq.\ref{eq:righthand}) will be produced which will initially induce a movement of the current (charge carriers). In this way, carriers with opposite charges will accumulate on opposite sides of the sample by the Right Hand Rule. Once the electric force produced by the oppositely charged sides balance the magnetic force, the current will not curve anymore. The resulting voltage across the two sides when $F_M=F_E$ is the Hall voltage. The sign of the carrier decides whether the carrier will curve upward or downward. In other words, the sign of the carrier is directly related to the sign of the Hall voltage. %Therefore, by measuring the Hall voltage we will be able to figure out the sign of the carrier, and take a step further, to identify the type of the conducting sample as illustrated in Figure \ref{fig:HallIllustration}.
%For the two cases shown in Figure \ref{fig:HallIllustration}. The magnetic field is pointing into the page. By convention, the direction of current is defined as the direction of the moving positive charge. Therefore, according to right hand rule, \textbf{(I will finish this later)}
N-type Germanium and P-type Germanium semiconductors have been used in this lab. Carriers are mainly electrons ($q=-e$) in N-type semiconductors and mainly holes ($q=+e$) in P-type semiconductors. Electrons move opposite to the applied current while holes move along with it. According to Equation \ref{eq:righthand}:
\begin{equation}
\label{eq:righthand}
\vec{F}=q \vec{v}\times \vec{B}
\end{equation}
With the same $+\vec{I}$, the direction of the force will be the same for both positive and negative charge carriers. Therefore, carriers will accumulate on the same side for either type of carrier. The voltage difference resulted from this cumulation of charge on one side of the sample is the Hall voltage. Due to the different signs of the charge carriers ($\pm e$), the sign of the Hall voltage will be different.
%\textbf{Explain carrier density, and how to find it using data collected from Hall Effect. (Lots of equations)}
Carrier density \href{http://libproxy.smith.edu:2048/login?url=http://search.ebscohost.com/login.aspx?direct=true&db=cat00321a&AN=fivecol.000716202&site=eds-live}{(Melissinos 2003)} is an important concept in Hall Effect, since it determines the magnitude of $V_{H}$, and its sign also reflects the type of carrier present. It represents the number of carriers per unit volume, by definition, and mathematically can be calculated using Equation \ref{eq:Hallcoefficient} and Equation \ref{eq:density}:
\begin{equation}
\label{eq:Hallcoefficient}
R_{H}=\frac{V_{H}\cdot t}{I\cdot B}
\end{equation}
where $R_{H}$ is the Hall coefficient, $V_{H}$ is the measured Hall voltage, $t$ is the thickness of the sample, $I$ is the supplied current to the sample and $B$ is the magnetic field.
\begin{equation}
\label{eq:density}
n=\frac{1}{R_{H}e}
\end{equation}
where $n$ is the carrier density and $e$ is the elementary charge, which equals to $1.60\cdot10^{-19}$ coulombs.
Two other important values are resistance $R$ and resistivity $\rho$. The resistance describes how difficult it is for a current to go through, non-specific to the material. It depends both on the nature of the material and the shape of the material (may change with temperature). Resistivity, however, is an intrinsic property that describes how strongly a material opposes the flow of electric current (collisions within a material as current passes, related to carrier density), and thus, does not depend on its shape. It can be calculated both resistance $R$ with Equation \ref{eq:resistivity}.
\begin{equation}
\label{eq:resistivity}
\rho=\frac{RA}{L}
\end{equation}
Resistivity becomes an important part of this experiment in assessing our data and analyzing material property. The importance of resistance is that its dependence (or independence) of magnetic field ($\vec{B}$) during the experiment can help us determine whether we can isolate $V_H$ in our experimental data analysis.
\section{Calibrations}
In order to carry out our experiment in determining the carrier charge density of a conducting material through the Hall Effect, it is necessary to know the magnitude and direction of the magnetic field that is present for each measurement.
For our experiments on metallic silver (Ag) and semiconducting n-type and p-type doped Ge, we use the value of the magnetic field from our iron electromagnet (LEYBOLD 562 13) and measurements of the Hall voltage $V_H$ to determine the Hall coefficient $R_H$ and carrier density. Interestingly, we can reverse this process and use the Hall voltage from a specially calibrated Hall Effect probe (TEACHSPIN) to measure the strength and direction of magnetic fields.
\subsection{Hall Probe Calibration}
\subsubsection{Method}
\begin{equation}
\label{BiotSavart}
B= \beta\frac{NI}{R}
\end{equation}
A \href{http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/helmholtz.html}{Helmholtz coil} was built as a source of known magnetic field that can be used to figure out the proportionality constant between the output voltage and the magnetic field of the Hall probe HE1-A. The Helmholtz coil used in this calibration has 15 turns ($N$) and the radius ($R(\textrm{m})$) is $2.3 \cdot 10^{-2}$ meters. $\beta$ is a constant, which equals to $0.9\cdot 10^{-6} \frac{\textrm{Tm}}{\textrm{#of turns}\cdot(\textrm{A})} $. The magnetic field at the center of the Helmholtz coil can be calculated using Equation \ref{BiotSavart}, known as Biot-Savart Law.
Therefore:
\begin{equation}
B= 0.59\ I \textrm{ [mT/A]}= 0.59 \cdot 10^{-3}\ I \textrm{ [T/A]}
\end{equation}
We measured the Hall voltage as a function of magnetic field strength by varying the current through the Helmholtz coil as shown in Fig ~\ref{fig:ProbeCal_Setup}.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Probe/Probe}
\caption{{\label{fig:ProbeCal_Setup}
\textbf{(a)} Setup for radial Hall sensor calibration. Probe is set at "Radial $\times 10$". The Helmholtz coil is oriented in the way shown above that $\vec{B}$ is pointing radially toward/away from the probe. \textbf{(b)} Setup for transverse Hall sensor calibration. Probe setting is changed to "Transverse $\times 10$", and $\vec{B}$ is perpendicular to the probe. (the transverse sensor has a specific orientation that it only detects $\vec{B}$ pointing up/down)%
}}
\end{center}
\end{figure}
%The magnetic field at the center of our Helmholtz coils was calculated using Equation \ref{BiotSavart}:
%\begin{equation}
%\label{BiotSavart}
%B=0.9\cdot 10^{-6}\frac{NI}{R}
%\end{equation}
%where $B$ is the magnetic field in Tesla, $N$ is the number of turns per winding, which in our case is $15$, $R$ is the radius of the coil in meter, which in our case is $2.3 \cdot 10^{-2}$ meters and $I$ is the current supplied to the Helmholtz coil in amperes. Equation \ref{BiotSavart} is known as Biot-Savart Law.
%Therefore:
%\begin{equation}
%B= 0.59\ I \textrm{ [mT/A]} (Millitesla)= 0.59 \cdot 10^{-3}\ I \textrm{ [mT/A]} (Tesla)
%\end{equation}
%After converting the applied current to magnetic field, we obtained the relationship between the voltage and the magnetic field by applying linear line fit to our data points as shown in Figure \ref{fig:VBradial} and Figure \ref{fig:VBtransverse}. In this way, we get:\\
%For radial sensor:
%\begin{equation}
%V_{B}=(65.8) B-0.000769 \textrm{ WHAT ARE MY UNITS?}
%\end{equation}
%For transverse sensor:
%\begin{equation}
%\label{eq:TransCal}
%V_{B}=(66.6) B-0.001460 \textrm{ WHAT ARE MY UNITS?}
%\end{equation}
\subsubsection{Result}
we found that the Hall voltage is a linear function of magnetic field
\begin{equation}
\label{eq:MissingCalibrationEq}
V_{\textrm{Hall sensor}} = \alpha B = \alpha \beta \frac{N}{R} I_{coil}
\end{equation}
As shown in Figs.~\ref{fig:VBtransverse} and \ref{fig:VBradial} below, we found from the slope of output voltage vs. magnetic field that $\alpha_t = 66.6 \pm 1.3\textrm{ [Volts/Tesla]}$ for the transverse sensor and $\alpha_r = 65.8 \pm 1.3\textrm{ [Volts/Tesla]}$ for the radial (axial) sensor, hence
\begin{equation}
\label{eq:AnotherCalEquation}
%B \textrm{ [tesla, or perhaps mT would be more convenient]} = \gamma V_{\textrm{Hall sensor}}
B = \gamma V_{\textrm{Hall sensor}}
\end{equation}
where $\gamma_t = 1 / \alpha_t $ = $0.015 \pm 0.003\textrm{ [T/V]}$ and $\gamma_r = 1 / \alpha_r $ = $0.015 \pm 0.003\textrm{ [T/V]}$.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/V-versus-B-(transverse)-(1)/V-versus-B-(transverse)-(1)}
\caption{{\label{fig:VBtransverse}
Hall Effect Probe (HE1-A) transverse sensor calibration using a $15$ turn/winding Helmholtz Pair. X-axis represent the magnetic field imposed at the sensor, and Y-axis represents the voltage at the output terminals of the probe. A linear function has been used to fit the data points in order to get the proportionality constant between the voltage and the magnetic field.%
}}
\end{center}
\end{figure}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/V-versus-B-(radial)/V-versus-B-(radial)}
\caption{{\label{fig:VBradial}
Hall Effect Probe (HE1-A) radial sensor calibration using a $15$ turn/winding Helmholtz Pair. X-axis represents the magnetic field imposed at the sensor, and Y-axis represents the voltage at the output terminals of the probe. A linear function has been used to fit the data points in order to get the proportionality constant between the voltage and the magnetic field.%
}}
\end{center}
\end{figure}
We calculated the chi square of the fits shown in Figure \ref{fig:VBtransverse} and Figure \ref{fig:VBradial}, and we found that $\chi^2$ = $11.48$ for radial and $\chi^2$ = $9.27$. According to the Chi Square Distribution Table , $P(\chi^2;\nu)$ for radial sensor is in the range of $0.25$ and $0.5$, but it is closer to $0.5$. Since a reasonable fit means $P(\chi^2;\nu)$ is around $0.5$, we conclude that the fit for radial sensor is a reasonable fit. Although not perfect, it's acceptable and will not be questioned. $P(\chi^2;\nu)$ for transverse sensor is in the range of $0.50$ and $0.75$, but it is also much closer to $0.50$, so we conclude that our fit for transverse sensor is also a reasonable fit.
\subsection{Electromagnet Calibration}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Electromagnet/Electromagnet}
\caption{{\label{fig:MagCal}
Setup for electromagnet is mostly the same as the Hall probe calibration (Figure \ref{fig:ProbeCal_Setup}), except the Helmholtz coil is replaced by the electromagnet.%
}}
\end{center}
\end{figure}
\subsubsection{Method}
For calibrating the electromagnet, we used the calibrated Hall probe to find $\vec{B}$. As shown in Figure \ref{fig:MagCal}, we used the transverse sensor of the Hall probe, so we used the transverse calibration equation to calculate the corresponding magnetic field using the measured voltage $V$.\\
%For radial sensor
%\begin{equation}
%B=\frac{V}{658} \textrm[WHAT ARE MY UNITS?]
%\end{equation}
For transverse sensor
\begin{equation}
\label{eq:Transverse}
B=0.015\pm 0.003\textrm{[Tesla/Volts]}\cdot V
\end{equation}
In this way, we were able to obtain a graph of $B[\textrm{T}]$ versus $I[\textrm{A}]$ as shown in Figure \ref{fig:hysteresis}. It can be seen that the magnetic field does not exactly follow the original path when the current reverse in sweep direction. Therefore, the output has a hysteresis loop, meaning that the output will depend on both the magnitude and the sweep direction of the input current.
%When using the result from electromagnetic calibration,
To determine the absolute value of the magnetic field, we need to pay close attention to whether we are increasing or decreasing magnetic current as we take our measurements.
%strictly follow the \textbf{path?? a better word??}.
Figure \ref{fig:hysteresis} also shows that the relationship between $B$ and $I$ fits perfectly well with a linear line ($\chi^2$ and $P(\chi^2;v)$ calculations in Section \ref{GoodnessOfFit}) when $I$ falls in the region of $-2A$ to $2A$.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/Electromagnetresult/Electromagnetresult}
\caption{{\label{fig:hysteresis} This shows the hysteresis loop of the iron-core electromagnet. Hysteresis loop is a result of the difference between the up sweep and down sweep(reversal in direction of applied $\vec{B}$ field) magnetization of a ferromagnetic material. The red circles are for increasing $\vec{B}$, and the blue are for decreasing $\vec{B}$. The green dots are again for an up sweep in $\vec{B}$, showing the retraceability of the hysteresis loop.%
}}
\end{center}
\end{figure}
\section{Experiment on Resistance}
In this experiment we want to explore the relationship between resistance and resistivity, and we did this by calculating both of them. By definition, resistance is independent of voltage or current. It can be calculated using Ohm's Law, which states that:
\begin{equation}
R=\frac{V_{L}/I}{I}
\end{equation}
where $V$ is the recorded value of longitudinal voltage and $I$ is the supplied current in our case. In this way, we got $41.09 \pm 0.03 \Omega$ for n-Ge, $50.49 \pm 0.02 \Omega$ for n-Ge and $0.0662 \Omega$ for Silver. Th uncertainty is calculated from the resistance obtained at different magnetic fields.
It can be concluded from our results that resistance does not depend on the magnetic, which is different from the hall voltage. This is important to notice since in this way we can separate $V_{H}$ from $V_{L}$ in our later experiment. However, the resistance becomes different for different material. Therefore, there must be a property of the material that determines the resistance. This property is called resistivity ($\rho$), which will be discussed in the next section.
\section{Experiment on Resistivity}
We also measured the resistivity $\rho$ of the material, and like the Hall coefficient, we determined it from $I$ and and a voltage ($V_L$ instead of $V_H$). This too depends on $n$. The two measurements are related theoretically but also in a practical sense, since some fraction of $V_L$ tends to show up in our intended measurement of $V_H$. That is, we end up measuring $V_H(B) + \eta V_L$ instead of just $V_H$, and we can separate $V_{H}$ from $V_{L}$ since $V_{L}$ is not field dependent as discussed above.
By plotting, we will theoretically obtain a $V_{\perp}$
\begin{equation}
\label{eq:MeasuredHallVoltage}
V_{\perp} = V_H(B) + \eta V_L
\end{equation}
as a function of $B$, since $\eta V_L$ is constant,
\begin{equation}
\label{eq:HallSlope}
\frac{\Delta V_{\perp}}{\Delta B} = \frac{\Delta V_H}{\Delta B}
\end{equation}
The relationship holds because only $V_{H}$ is field dependent.
\subsection{Method}
As discussed above, resistivity ($\rho$) is a property of the material. It can be calculated using Equation \ref{eq:resistivity}, where $R = \frac{V_L}{I}$ is the resistance in $\Omega$, $L[\textrm{m}]$ is the length of the material and $A[\textrm{m}^2]$ is the cross sectional area where the current passes (width$\cdot$thickness) in $m^2$.
\begin{equation}
\rho=\frac{V_L/I}{A}{L}
\end{equation}
Therefore, the SI unit of resistivity is $\Omega \cdot m$.
It's important to keep in mind that although resistivity is calculated using the dimensions of the resistor, it is independent of the dimensions.
\subsection{Results}
By plugging numbers into Equation \ref{eq:resistivity}, we get $\rho_{n-Ge} = 0.02053 \pm 0.00005$ $\Omega \cdot m$ for n-Ge, $\rho_{p-Ge} = 0.02524 \pm 0.00001$ $\Omega \cdot m$ for p-Ge and $\rho_{Ag} = 1.117 \cdot 10^{-6}$ $\Omega \cdot m$ for Silver.The theoretical value of resistivity for Silver is $1.58\cdot 10^{-8}$ $\Omega \cdot m$, which is much smaller than our calculated result.
\subsection{Analysis and Discussion}
Both $\rho_{n-Ge}$ and $\rho_{p-Ge}$ are reasonable results. Resistance of the n-Ge sample calculated with $\rho_{n-Ge}$ is $41.05 \Omega$, and resistance of the p-Ge sample calculated with $\rho_{p-Ge}$ is $50.49 \Omega$. Both values are within $3\%$ difference with the measured resistances of $40.1$ and $52.4\Omega$ respectively. This ensures that we have indeed measured the correct current that will be used in calculating the Hall coefficients and carrier densities for n-Ge and p-Ge.
We calculated the resistance for silver with its resistivity as well. Its resistance turned out to be $0.0662\Omega$. This is a very small resistance, and since the calculated resistivity is about a factor of a hundred greater than the literature value, our expected resistance for silver is $0.00662$, a even smaller value. Looking at the differences between calculated and measured resistances for n-Ge and p-Ge, this calculated resistivity for our silver sample could be well dominated by contact resistance in the $V_L$ measurement.
To be sure that this is the cause of our inaccurate resistivity for silver, we propose to make a 4 terminal measurement instead of a 2 terminal measurement for current and resistance. A 4 terminal measurement allows the voltage to be measured across just the sample and not the contact where current runs through that may have a voltage drop because of its contact resistance.
\section{Experiment on the Hall Effect}
\subsection{Method}
We first tested both n-Ge and p-Ge. We calculated the carrier density of each type using Equation \ref{eq:Hallcoefficient} and Equation \ref{eq:density}.
$B$ in the equation represents the magnetic field, which can be calculated from the current sent through the electromagnet using the electromagnet calibration result for transverse sensor as shown in Equation \ref{eq:Transverse}. We then plotted $V_{H}$ vs $B$ as shown in Figure \ref{fig:n}, Figure \ref{fig:p} and Figure \ref{fig:silver}, and used this slope to calculate the Hall coefficient and carrier density.
We then tested a silver(Ag) sample. Eventhough the carrier type of silver is unknown, we expect it to be negative due to its conductive nature. Again, we calculated the carrier density using the same method as shown above for n-Ge and p-Ge.
\subsection{Result}
We found carrier densities and Hall coefficients as follows:
For n-Ge:%the number of carriers (which has no units) per cubic meter is given by
$$R_{H}=-4.99\cdot 10^{-3}\pm -0.0998 \cdot 10^{-3}\frac{\textrm{Vm}}{\textrm{AT}}$$
$$n=-1.25\cdot 10^{21}\pm 0.025 \cdot 10^{21} \textrm{m}^{-3}$$
where $n$ represents the carrier density in number of carriers (which has no units) per cubic meter and $R_{H}$ represents Hall coefficient in $\frac{m^3}{C}$.
For p-Ge:
$$R_{H}=6.6\cdot 10^{-3}\frac{\textrm{Vm}}{\textrm{AT}}$$
$$n=1.11\cdot 10^{21}\pm 0.02 \cdot 10^{21} \textrm{m}^{-3}$$
Since $V_H$ and $B$ in Equation \ref{eq:Hallcoefficient} have different signs, it makes sense that $n$ for n-type is negative, and $n$ for p-type is positive.
By using the same method, we found the carrier density and Hall coefficient of the silver sample, which are:
\\
$$R_{H}=-2.24 \cdot 10^{-10}\pm -0.04 \cdot 10^{-10} \frac{\textrm{Vm}}{\textrm{AT}}$$
$$n=-2.79\cdot 10^{28}\pm 0.06\cdot 10^{28} \textrm{m}^{-3}$$\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/n-Ge/n-Ge}
\caption{{\label{fig:n}
Hall voltage versus magnetic field for n-Ge, where hall voltage is the recorded value and magnetic field is calculated from the electromagnet calibration. Theoretically there should be a linear relationship between the hall voltage and the field. The blue line represents the linear fit.The slope is $-0.0499 \frac{V}{T}$.%
}}
\end{center}
\end{figure}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/p-Ge/p-Ge}
\caption{{\label{fig:p}
Hall voltage versus magnetic field for p-Ge, where hall voltage is the recorded value and magnetic field is calculated from the electromagnet calibration. Theoretically there should be a linear relationship between the hall voltage and the field. The blue line represents the linear fit. The slope is $0.0564\frac{V}{T}$.%
}}
\end{center}
\end{figure}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/Silver/Silver}
\caption{{\label{fig:silver}
Hall voltage versus magnetic field for silver, where hall voltage is the recorded value and magnetic field is calculated from the electromagnet calibration. Theoretically there should be a linear relationship between the hall voltage and the field. The blue line represents the linear fit. The slope is $-2.24\cdot10^{7} \frac{V}{T}$.%
}}
\end{center}
\end{figure}
\subsection{Analysis}
According to our calculation above, both n-Ge and Silver have negative carrier densities and p-Ge has positive carrier density. This is a direct result of the signs of $\frac{V_{H}}{B}$. Since the sign of carrier density of the Silver is consistent with it of n-Ge, we conclude that Silver is n-type. For both n-Ge and p-Ge, the carrier densities are consistent with the absolute theoretical values, which are $1.2 \cdot 10^{21} \textrm{m}^{-3}$ and $1.1 \cdot 10^{21} \textrm{m}^{-3}$ respectively. However, although our calculated Silver carrier density has the same order of magnitude as the theoretical value ($6.6 \cdot 10^{28} \textrm{m}^{-3}$), they are not consistent with each other. One possibility we can think of is that we only have three data points for each sample as shown in Figure \ref{fig:n}, Figure \ref{fig:p} and Figure \ref{fig:silver}. This is due to the fact that the linear relationship between the magnetic field and the magnet current only falls in the region of $-2A$ < $I$ <$2A$. Due to the time constraint, we are unable to go back and redo the experiment, but this should be something worth noting in the future experiments.
\section{High $T_{c}$ Superconductor}
\subsection{Method}
In this section of the experiment, we are trying to find the transition (to superconducting state) temperature($Tc$) of a high $Tc$ superconductor, $Bi_2 Sr_2 Ca_2 Cu_3 O_{10}$ (Bi2223).
We placed our superconductor in a jug, and filled the jug with small glass beads to cover the Bi2223 superconductor. This will allow the temperature to change slowly so we can obtain a smoother graph of $V[\textrm{V}]$ vs. $T[\textrm{K}]$ to determine the transition temperature.
We used a commercial superconductor where all the leads for current input, voltage and type T thermo-coupled are readily attached. The current input was an AC signal of $10\textrm{mA}$ at $f = 101.19\textrm{Hz}$, and connected the voltage leads to a lock-in amplifier (sensitivity=$200\mu V$) and then to a DMM (Digital Multimeter) and thermo-couple leads to another DMM. We recorded an offset of $10\textrm{K}$ in room temperature. The phase of the signal were recorded before the sample was cool. When the sample becomes superconducting, we expect the phase to shift by $180$ degrees, since resistance in the sample and thus the voltage will go to $0$.
Then, when all the connections are made, we added liquid nitrogen to the jug, little by little, until the temperature of the superconductor read about $30\textrm{K}$ below the expected transition temperature of $108\textrm{K}$. We did, indeed, observe a $180$ degrees shift in the phase of the voltage. As the sample starts to warm back up, we started collecting data for $V$ and $T$, until the sample is warmed back to almost room temperature, and that the sample has shifted back to a non-superconducting state.
The result of $V$ is plotted against $T$ as shown in Figure \ref{fig:TransTemp}.
\subsection{Result}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/Superconductor/Superconductor}
\caption{{\label{fig:TransTemp}
This is a graph showing the voltage change of Bi2223, a high temperature superconductor, as its temperature changes. When a material becomes superconducting, its resistance goes to 0, and so does the voltage. In this graph we can see that $Tc$ is about $118\textrm{K}$ for this material.
*The voltage on the y-axis is the voltage read from the multimeter. It was not converted to the actual voltage read by the lock-in amplifier since we are only interested in when $R$ went to $0$, and not the exact voltages for this particular experiment. The voltage is negative because of a reversal in the connection of the positive and negative $V$ leads, but it does not affect our conclusion for the transition temperature.%
}}
\end{center}
\end{figure}
\subsection{Analysis and Discussion}
The superconducting onset temperature is $130\pm 2\textrm{K}$, and the zero resistance transition temperature for the this sample is $118\pm 2\textrm{K}$ from our graph. A $10\textrm{K}$ that was recorded in the temperature before the cool down will correct these two values to $120\pm 2\textrm{K}$ and $108\pm 2\textrm{K}$ for the superconducting onset temperature and the zero resistance transition temperature respectively.
The value for our measured zero resistance transition temperature is very consistent with the \href{http://en.wikipedia.org/wiki/Bismuth_strontium_calcium_copper_oxide}{reported transition temperature} of $108\textrm{K}$.
\section{Conclusion}
In conclusion, the results (Hall coefficient, carrier density, resistance and resistivity) associated with both n-Ge and p-Ge are consistent with the theoretical values. However, the results of the Silver sample are not really satisfying. There can be multiple possibilities and we mentioned some above such as inadequacy of the number of data points and additional contacting resistance. Due to time constraints, at this point we are unable to go back to redo our experiments or try with more samples, so these should be considered in future experiments. One important lesson we learned from this lab is that we need to carefully perform all the preliminary calculations. In this way, we can have a sense of what do we expect so we can quickly identify any error appeared during the process and find a way to fix it. We will definitely keep this in mind in our future experiments.
\section{Appendix}
\label{GoodnessOfFit}
The detailed calculation of chi square can be found in \href{https://osf.io/yw62r/files/}{Open Science Framework}.
\section{Acknowledgement}
We would like to thank Prof. Nat Fortune for helping us throughout the experiment, and we also would like to thank our classmates for their support.
\selectlanguage{english}
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\end{document}