Hall effect

The Hall effect

Procedure

Constant magnetic field

In this part of the experiment, the magnetic field will be held constant and the current through the semi-conductor will be varied.

1. Set the Excitation Current $(I_M)$ 0-1000 mA to a desired value (e.g., 500 mA) so the magnetic field strength will be constant.
2. Make sure the Hall Current $I_s$ 0-10mA is zero.
3. Click Preview and adjust the Hall Current to 0.5 mA. Press Keep to record the voltage and current. Then increase the Hall Current by increments of 0.5 mA, pressing Keep for each, until the Hall Current is 5.5 mA. Then press Stop. Return the Hall Current to zero.
4. Move the Hall Probe out of the magnet. Press the tare button on the side of the Magnetic Field Sensor. Move the end of the Magnetic Field Sensor into the center of the magnet. Click Preview and Keep to record the magnetic field strength. Then press Stop.
5. Set the current 0-1000mA to another value (e.g., 800 mA), and then record data again, repeating Steps 2 through 4.
6. Hall Voltage Compensation: Because the leads that measure the Hall Voltage may not be exactly opposite each other across the semi-conductor, a voltage may appear that is due to the potential difference along the direction of the current. To measure this and compensate for it, slide the Hall Probe completely out of the magnet, set the magnet current to zero, and perform Step 3 to record the Hall Voltage without any magnetic field. Apply a linear fit to the V vs. I data. The slope of this line will be subtracted from the slopes of the other lines to compensate for the offset of the Hall Voltage due to misalignment of the leads.

Constant Current through the Semi-Conductor

In this part of the experiment, the magnetic field will be varied by varying the current through the electromagnet while the current through the semi-conductor will be held constant. Rather than measuring the magnetic field for each data point, we will first characterize the magnetic field as a function of the current.

1. To discover the relationship between the magnetic field strength and the current through the magnet coils, create a table of Magnetic Field Strength (Perpendicular). In the second column, create a User-Entered data set called Magnet Current (with units of mA). Pre-fill the column with values from 50 to 900 mA in steps of 50 mA. Create a graph of Magnetic Field Strength (Perpendicular) vs. Magnet Current (A).
2. With the Hall Current set to zero, press the tare button on the side of the Magnetic Field Sensor. Move the Magnetic Field Sensor into the center of the magnet.
3. Click Preview and adjust the “Excitation Current” (the current through the electromagnet) to 50 mA as read on the digital readout on the Hall Effect Apparatus. Then click Keep.
4. Adjust the “Excitation Current” to each value in the table and click Keep for each value. Then click Stop. Return the Excitation Current to zero to prevent the magnet from getting too hot.
5. On the graph, apply a cubic curve fit.
6. Set the Hall current $I_s$ (0-10 mA) to a desired value (e.g., 5 mA).
7. Make sure the Excitation Current $(I_M)$ 0-1000 mA is zero.
8. Set the sampling rate to 10 Hz.
9. Move the Magnetic Field Sensor out of the magnet and move the Hall Effect Probe into the center of the magnet.
10. Click Preview and increase the Excitation Current to 50 mA and click Keep. Continue to increase the Excitation Current by steps of 50 mA up to 900 mA, clicking Keep for each current setting. Then click Stop. With the fit done in step 5, you can deduce the magnetic field produce at every value of the current.
11. Set the Hall Current 0-10 mA to another value (e.g., 8 mA), then record data again,

repeating Steps 7 through 10.

Results & analysis

Magnetic field as a function of the current

The results of the measurement are displayed in the figure below.

Magnetic field as a function of the current passing through the coils.

There, we can make a cubic fit (as recommended by the manufacturer) or a linear fit. Which one is better? The parameters of the fit with $B=a_0 +a_1 I + a_2 I^2 + a_3 I^3$ are the following:

Fit	$a_0\, (\mathrm{mT})$	$a_1\, (\mathrm{mT}/\mathrm{mA})$	$a_2\, (\mathrm{mT}/\mathrm{mA}^2)$	$a_3\, (\mathrm{mT}/\mathrm{mA}^3)$
Linear	-(1.17 $\pm$ 0.87)$\times 10^{-1}$	-(7.70 $\pm$ 0.01)$\times 10^{-2}$	-	-
Cubic	-(5.88 $\pm$ 0.28)$\times 10^{-1}$	-(7.48 $\pm$ 0.03)$\times 10^{-2}$	(2.01 $\pm$ 0.64)$\times 10^{-7}$	-(3.05 $\pm$ 0.45)$\times 10^{-9}$

Hall voltage at fixed magnetic field

The results of the measurement are shown in the figure below (note that the error bars are within the size of the dots!).

Voltage induced across the semiconductor for various magnetic fields (whose magnitude depends on the current passing through the coils).

The data can be linearly fitted, i.e., with an expression as $V_H = a_0 + a_1 I$, with the following parameters:

Current (mA)	$a_0\, (\mathrm{mV})$	$a_1\, (\mathrm{mV}/\mathrm{mA})$
0	(0.156 $\pm$ 0.010)	-(0.993 $\pm$ 0.003)
500	(0.019 $\pm$ 0.023)	-(8.529 $\pm$ 0.007)
800	(0.114 $\pm$ 0.033)	-(13.220 $\pm$ 0.011)

From the measurement of the magnetic field as a function of the current in the coils, done in the previous section, we know that the magnetic fields corresponding to 500 mA and 800 mA are $-(38.59 \pm 0.13)\,\mathrm{mT}$ and $-(61.68 \pm 0.14)\,\mathrm{mT}$, respectively. The Hall voltage is related to the current $I$ passing through the semiconductor with the follwoing expression \begin{equation} V_H = \left(\frac{B}{nde} \right) I = \left (\frac{R_H B}{d} \right) I\,, \end{equation} where $B$ is the magnetic field, $n$ is the density of charge carriers in the semiconductor, $d$ is the depth of the semiconductor sample and $e$ is the charge of the electron. We have also defined $R_H = 1/(ne)$, known as the Hall coefficient, which depends on the material and the temperature, but is independent of the geometry of the sample. With the data collected and having that $d=1.2\,\mathrm{mm}$, we find that the Hall coefficient of the semiconductor of the setup is $R_H = (0.2361 \pm 0.0005)\,\mathrm{m}^3/C$, and therefore the density of charge carriers on the sample is $n=(2.6438 \pm 0.0006)\times 10^{19}\,\mathrm{m}^{-3}$.