Group3_8_ch21

flat

Sam Fihma, Steven Thorwarth, and Phil Litmanov
= = =Lab: Capacitors in Parallel and Series Circuits=


 * Objective:** To compare the voltage of capacitors in series to those in parallel.


 * Hypothesis:** In series, the voltage to the capacitors is split and when added together equal the voltage of the power source (V B =V C1 +V C2 ). In parallel, both capacitors get the same voltage as the power source (V B =V C1 =V C2 ).


 * Materials:**
 * 1 voltmeter (Multimeter)
 * 2 Capacitors
 * 2 Resistors
 * Variable power supply (can change the voltage)
 * Metal wires (5)


 * Procedure:**


 * 1) Before turning on the power supply or setting up a circuit, use a metal wire to short circuit each capacitor by touching each plate with the ends of the same wire. This will insure that the capacitor is discharged.
 * 2) Use the wires to create one of the desired circuits, preferably starting with a basic one such as the series circuit.
 * 3) Turn on the power supply and set the Voltage to the desired amount. Make sure the amounts you use are within the limits of the capacitors.
 * 4) Turn on the voltmeter and place it within the circuit in parallel to the capacitor you would like to test. The absolute value of the number you get is the amount of voltage running through the tested capacitor.
 * 5) Complete this step for both capacitors, recording the results.
 * 6) Change the voltage to another amount that you would like to test. (NOTE: if you decrease the voltage, you need to discharge the capacitors first, because they will still have the charge from earlier and could alter the results. If you INCREASE the voltage then there is no need and the capacitor will simply receive more charge)
 * 7) Repeat these steps for multiple voltages (we did 5) and then disconnect the circuit to make the next circuit.
 * 8) When you have connected the circuit, make sure you discharge the capacitors before testing to remove potential error.
 * 9) Repeat steps 3-7 for each circuit, testing the same voltages every time.

Circuit A



Circuit B



Circuit C

Circuit D




 * Data:**






 * Sample Calculations:**

Circuit A



Circuit B



Circuit C



Circuit D




 * Analysis:**

Circuit A:

Because the capacitors are in series, Q is the same for each. Because of this, we are able to find the theoretical Q for each capacitor and then find the theoretical V for each capacitor through the equation Q=CV.

Circuit B:

The capacitors are in parallel which means each should fill to the voltage of the power supply, regardless of capacitance. In this case, theoretical V is the voltage the power supply is set to.

Circuit C:

Because the capacitor is in parallel with the power supply, it will fill to the voltage of the power source. This is the capacitor's theoretical V. The resistors are in series with each other, so by finding the equivalent resistance we can find the current throughout that branch. The current will not split at the junction because when the capacitor is done charge, current will no longer flow through that branch. We use this calculated current to find the voltage through each resistor via Ohm's law equation, V=IR, where R is the respective resistance of each resistor.

Circuit D:

The capacitors are in parallel, so they will fill to the voltage of the power source. This is also each capacitors theoretical V. When charging is done, current will no longer flow and the voltage of the resistor will be 0V. This is why the theoretical voltage of the resistor is 0 and the experimental values are close to 0.


 * Discussion Questions:**

1. How does the voltage on the individual capacitors in series compare to the voltage when they are in parallel? When individual capacitors are in series, the voltage is split up. The potential difference of all the capacitors summed up will be equal to the total voltage supplied by the power source. The amount of voltage each capacitor has is dependent upon their own capacitance. When they are in parallel however, each capacitor on a difference branch has the same voltage as one another. The potential difference of each capacitor will be equal to the voltage supplied by the power source.

2. What is the effect of the resistor on the voltage of the capacitors? If a resistor is added in parallel to the capacitor, the voltage of the capacitor should not change. Voltage is evenly distributed among parallel branches, so no matter what conductor you add in parallel to the capacitor, the voltage will remain the same. However, if the resistor is added in series to a capacitor it has an effect on the voltage. The resistor will take some of the voltage away from the capacitor. This is seen in Circuits B and D. In Circuit B, C1 does not have a resistor and has a larger voltage than C1 in Circuit D, which does have a resistor.

3. How does the potential difference of the capacitors in series compare to the voltage of the source? What about when they are in parallel? The answer to this question was touched upon in question #1, but here are the two equations. For series, V(source) = V1 + V2 + ..., meaning the voltage of the source is equal to the sum of the voltage of the capacitors. For parallel, V(source) = V1 = V2 = ..., meaning the voltage of the source is equal to the voltage of the capacitor in each parallel branch. These equations can continue for however many capacitors are added.

4. How is the amount of voltage on the individual capacitors related to the known capacitance? C=Q/V There is an inverse relationship between capacitance and voltage. In parallel, voltage will remain the same for each capacitor, but charge will be the variable and split among the capacitors. It is different for series though, where charge remains the same throughout the circuit. Voltage is now the variable and changes depending on the capacitance. The higher the capacitance, the lower the voltage and vice versa.

5. Discuss the effect of switching out your bigger capacitor for one that is 10 times as big. V=Q/C In parallel, this would have no effect on voltage because each branch receives the same amount no matter the capacitance. But charge of the capacitor will change. There would be more charge on the bigger capacitor and now less charge on the smaller one. In series, voltage is changed and charge remains constant. Voltage will now be one tenth of its original size and for the smaller capacitor, voltage will be ten times as large as it once was.

Our Hypothesis was correct, and through this experiment we see that capacitors within series split the voltage of the circuit (V eq =V 1 +V 2 ), and that capacitors that are parallel both have the same voltage equal to the voltage of the power supply (V eq =V 1 =V 2 ). Through our data for the four different circuits, we notice that the relationship between capacitors in series and parallel are very different. Similar to resistors, capacitors that are parallel to one another will receive the same amount of voltage from the power source. In circuit B we calculated that both of the capacitors should have the same voltage as the power supply, since V eq =V 1 =V 2, and each capacitor had a voltage reading that was very similar to the voltage on the power supply. For the Capacitors that were in series we noticed that the sum of the voltages of the two capacitors sum to equal the voltage of the power supply. This can be most clearly seen in Circuit A, where when we tested the circuit with 6 Volts of power Capacitor 1 gave a reading of 2.41V and capacitor 2 gave a reading of 3.56V. 2.41+3.56=5.97V, very close to the 6V from the power supply. Circuits C and D were used to see how capacitors reacted in a circuit that had resistors in series and parallel to the capacitor. In Circuit C it was designed for a resistor to be in series with a set of parallel branches, one with a capacitor and one with a resistor. Through our knowledge of resistors and capacitors we are able to find the voltage running through each resistor when the capacitor is fully charged, and use this to find the voltage expected to run through the capacitor. Our data proved that when in series with a resistor the voltage of a capacitor and the voltage of the resistor sum to the voltage of the power supply. This also proves that the voltages of a capacitor and a resistor that are parallel are equal. The last Circuit, Circuit D, was the same as Circuit C, except that the resistor parallel to the capacitor was replaced with another capacitor. The data from the circuit showed that the capacitors filled to the voltage of the power supply, showing that capacitors in parallel receive the voltage of the power supply. The voltage of the resistor, however, was very small, and this is because when the capacitors are both charged, there is no longer a flow of current. The resistor is no longer able to resist and the capacitors are fully charged.
 * Conclusion:**

=Lab: Magnetic Field=


 * Objective** - To find the relationship between magnetic field strength and distance from the source (magnet).


 * Hypothesis** - As the distance between the magnet and the sensor increases, the magnetic force of the magnet on the sensor will decrease. We used this hypothesis because it was logical to believe that the farther away an object is from a source, the less power that source has over the object; similar to the relationship of Electric force and distance.


 * Materials** - magnetic field sensor, data studio and science workshop interface, wooden stick, meterstick, neodymium magnet

1. Tape the measuring tape or meter stick to the table, and tape the Magnetic Field Sensor to a convenient location. The sensor should be perpendicular to the stick, with the white spot inside the rod facing along the meter stick in the direction of increasing distance. Carefully measure the location of the sensor on the meter stick. This will be your origin for all distance measurements.
 * Procedure**:

2. As a convenient way to measure the center of the magnet, and to ease the handling of the small magnets, allow the two magnets to attract one another through a wooden stick, about .5 cm from either edge near the corner. The magnets should stay in place on the stick. The stick itself will serve to mark the center of the magnetic pair.

3. Connect the Magnetic Field Sensor to Channel A of the interface. Set the switch on the sensor to 1x.

4. Open Data Studio and choose "Create Activity." Click on "Setup" and add a Magnetic Force Sensor to the icon of the interface. For a display, click on "314 Digits," which will show the magnetic field strength in Gauss.

5. Zero the sensor when the magnets are far away from the sensor in order to remove the effect of the Earth's magnetic field and any local magnetism. The sensor will be zeroed only for this location, so instead of moving the sensor in later steps, you will move the magnet. 1. Move the magnets far away from the sensor. 2. When the reading in the meter window is stable, click "Tare" on the sensor.

6. Now you are ready to collect magnetic field data as a function of distance. 1. Click "Start" to begin data collection. 2. Place the stick with the magnets against the meter stick, 2 cm from the sensor, so the stick is perpendicular to the meter stick. Measure from the stick to the center of the sensor. 3. The current magnetic field measurement is shown in the meter window. If necessary, reverse the magnets so the reading is positive, and reposition the stick 2 cm from the sensor. 4. Carefully measure the distance of the card to the sensor. 5. Record your data in a table.

7. Continue taking readings every 0.5 cm until you get no more charge in Magnetic Field Strength.


 * Data:**


 * Graph:**


 * Sample Calculations:**



We calculated the percent error of the exponent of the distance. In theory, the distance should be cubed, but since this is an experiment, it was a bit off at 3.298. Our percent error was 9.93% which was pretty small compared to other groups.
 * Analysis:**

The equation for magnetic field strength and our equation of our graph is shown above. The y value correlates to B, while the x value correlates to d. The rest of the symbols in the equation are kept constant throughout. We know the value for all the constant symbols in that equation except for the magnetic moment of our magnet, mew. Using the slope of our equation, we calculated what the magnetic moment of our magnet was equal to and we got a value of 0.4 A//m^//2.


 * Discussion Questions:**

1. On Excel, create a graph of magnetic field //vs.// the distance from the magnet. Produce a best fit line using a “Power” function. Shown above. 2. Compare your data to the ideal inverse-cube model:
 * What value do you get for the constant, //A//, or [(//m//0 2 //m// ) / (4p)]? How well does this agree with the value that the rest of the class measured?
 * 8x10^-8.
 * What exponent do you get for d? How well does this agree with the ideal expression?
 * 3.298. In a perfect world, it would be 3. Our result had a percent error of 9.93% which is pretty close to the ideal expression.
 * From the above comparison, does your magnet show the magnetic field pattern of a dipole?
 * Yes because the magnetic field strength very nearly followed an inverse cube relationship with distance.

3. Use your value of // A //  to determine the magnetic moment //m// of your magnet. m = 0.4 A//m^//2 Work is shown above in "Analysis."

4. The units of //m// may suggest a relationship of a magnetic moment to an electrical current. In fact, a current flowing in a closed loop is a magnetic dipole. A current //I// flowing around a loop of area (pi)//r//^2 has a magnetic moment //m// = //I// (p//i)r^//2. If a single current loop had the same radius as your permanent magnet, what current would be required to create the same magnetic field?

Radius of magnet = ?

m = I* π*r^2 .4 = I* π*r^2 .4/(πr^2) = I I =.4/(πr^2) Amps

5. Discuss the precision of your data, referencing the correlation coefficient to support your conclusion.

<span style="font-family: Arial,Helvetica,sans-serif;">Our data was very precise as our R^2 value was very near 1 (.992) meaning all of our data points fit the line of best fit very well.


 * Conclusion** -

Based on our results in the data we know that our hypothesis is true. As the magnet is placed farther away, it has less magnetic force exerted on the sensor. This can be seen in our data table or in the graph itself since it always has a negative slope. As we look closer at the graph we can see that it follows the path of a polynomial graph, specifically a cubic graph. This makes sense in understanding the relationship as we look at the equation for magnetic force and distance. In this equation, there should be a cubic, indirect relationship between distance, d, and magnetic force, //B//, in our resulting graph where B gets smaller as d gets larger. We had some error and were very close to getting this in our trend line, which had an equation with x 3.298. The error most probably came from the sensor itself, which was receiving results even when the magnet was impossibly far away. These results could have come from various places. It could have been error in the sensor, or it could have been the magnetic field of the Earth. Regardless, it was error that we could not account for in our results.

=Lab: Magnetic Force on a Wire=

__Objective:__ For a conductor placed in a magnetic field, find the relationships between the magnetic force, magnetic field strength, length of the conductor, current, and angle between the field and the current.

__Hypothesis:__ Magnetic force has a direct relationship with the magnetic field strength, current, wire length, and the sine of the angle between the field and the current. My rationale for this is the equation F(magnetic)=BILsin(theta), where B is the magnetic field strength, I is the current, L is the length of the conductor, and theta is the angle.

__Materials:__
 * Lab stand
 * Main unit
 * Current loop
 * Magnet assembly
 * 0.01 gram Balance
 * SF-8808 Accessory Unit

__Procedure:__

For theta as variable -

3. Turn the electronic balance on with everything in place WITHOUT current running through the circuit. This is most easily done by turning off the power supply. 4. Once the balance has a measurement for mass of the magnet and setup, set the balance to zero. There should be a set/zero button to the left of the screen on the balance. 5. After setting the balance to zero, connect the circuit by turning on the power supply. NOTE: With the magnets and wires at an angle of 0º the balance should still read a mass of 0, as there should be no magnetic force. 6. In small intervals, we chose 15º, measure the magnetic force downward on the balance with each respective angle. Record the data to create a graph and measure from 0º up to 90º. 7. Create a graph to find the relationship between sin ø and the magnetic force.
 * 1) Set up the lab As seen in the diagram to the right, [[image:Untitled2.png]].[[image:Screen_shot_2011-12-01_at_2.24.04_PM.png]]
 * 2) Make sure that once everything is set up as in the picture above that the magnet is parallel to the metal wires, and that the measurement on top of the SF-8088 reads a 0º measurement.

For current as variable -

1. Set up the main circuit seen here. 2. Repeat steps 2-5 as seen above. 3. Vary the current using the power source and record the weight on the balance.

For length as variable - 1. Set up the main circuit as seen above. 2. Repeat steps 2-5 as seen above. 3. Vary the loops that complete the circuit and record the weight for each loop.

For number of magnets as variable - 1. Set up the main circuit as seen above. 2. Repeat steps 2-5 as seen above. 3. Vary the number of magnets that are near the loop and record the weight for each trial.


 * __Data and Graphs__**




 * Sin(theta)**



This graph turned out to be a linear fit which gives evidence for the direct relationship theory. In this graph, y is equal to the force while x is equal to sin(theta). The slope is the magnetic field multiplied by the current and the length (BIL).


 * Length**



This graph turned out to be a linear fit which gives evidence for the direct relationship theory. In this graph, y is equal to the force while x is equal to length. The slope is the magnetic field multiplied by the current and sine of the angle (BIsin(theta)).


 * Current**



This graph turned out to be a linear fit which gives evidence for the direct relationship theory. In this graph, y is equal to the force while x is equal to current. The slope is the magnetic field multiplied by the length and sine of the angle (BLsin(theta)).


 * Number of Magnets**



This graph turned out to be a linear fit which gives evidence for the direct relationship theory. In this graph, y is equal to the force while x is equal to the number of magnets. The slope is the magnetic field multiplied by the length, current, and sine of the angle (BILsin(theta)).


 * __Calculations__**


 * Force**




 * Sin(Theta)**




 * Length**




 * Current**




 * Number of Magnets**




 * Sample Percent Difference**



__Conclusion and Analysis:__

1. Using the equation of the trendline from the graph of Force vs. Current, find the magnitude of the magnetic field. Show your work.
 * The graph of the trendline from the graph of Force vs. Current was y = 0.0063 * x, with Y= Force and x=Current. When we plug this into the equation F B =B*i*l*sinø and substitute F with y and i with x, we are left with B*l*sinø, which must be equal to the 0.0063 slope of the graph, since it is always constant in this experiment. To find this equation, we found the experimental results for magnetic force when current changed, but magnetic field, the angle, and the length remained the same. Therefore, according to the equation F B =B*i*l*sinø, we know that Magnetic Field Strength times the Length times sinø (ß*i*l) must be equal to the 0.0063 slope for the resulting equation. We know that the length = 0.0232m ,and the angle=90º, so we can say that ß * (.0232) * sin(90º) = 0.0063. Therefore we find the experimental magnitude to be ß=0.212T.

2. Discuss the relationship of the quantities shown in the graphs. How do they agree with the theoretical relationships?
 * In each of the four graphs, F B v. i, F B v. l, F B v. sinø, and F B v. n (number of magnets), we noticed an almost identical direct relationship between F B and each of the other variables. Because of this similarity, we find that each group has a very small % difference when solving for ß in the slope of each graph.

3. Do the experimental relationships shown in the 4 graphs validate the theoretical relationships? Explain your reasoning using specific evidence from the lab to support your answer.
 * Yes, the experimental relationships shown in each of the four graphs prove that Current, Length, and sinø all have the same direct relationship with the Magnetic Force. We also know that the slope of this relationship is determined by the other variables involved in the calculations.

4. Is it reasonable to assume that the strength of the magnetic field is directly proportional to the number of magnets? Why or why not?
 * No this is not a reasonable assumption. As the number of magnets increases, the Magnetic **Force** increases due to the direct relationship per the equation F = NBILsin(theta). However, Magnetic Field strength and the number of magnets has an inverse relationship. As the number of magnets increases, B decreases, and vice-versa.

Our hypothesis was correct that magnetic force has a direct relationship with the magnetic field strength, current, wire length, and the angle between the field and the current. We were able to prove these by testing the magnetic force when each of these values are changed, and created a graph that would show us the relationship. On each of the graphs we saw that there was a positive slope, indicating that as the current, magnetic field, and angle increased, the magnetic field strength increased. There was definitely error in the many parts of this experiment. The first source of error was when we tested the angles of the magnet and the coil. In order to have an angle of 0º we needed to begin measuring with the magnet parallel to the coil, and this was done by the human eye. As a result, the angle itself may or may not have been 0º. Therefore each angle that we tested was slightly more than what we were testing, and therefore altered our results. The second source of error is the many potential inconsistencies in the data recorded. There were four lab groups collecting data, and some details of the set up could have been altered. The scale could have been zeroed differently, and would cause a change in results amongst the groups. The best way to correct this lab would be to have each group record all of the data on their own, with their own lab set up. This would prevent inconsistencies and would allow the group to keep everything the same. This lab information is useful when building motors or generators. These devices have to be as efficient as possible in converting electrical energy to mechanical and vice versa. Therefore it is important to know what is necessary to get the maximum magnetic force from this type of set up between a magnet and a coil.

=<span style="font-family: Arial,Helvetica,sans-serif;">Lab: Magnetic Field Strength of a Solenoid =


 * Prelab Questions:**



<span style="font-family: Arial,Helvetica,sans-serif;">**Objective**:

<span style="font-family: Arial,Helvetica,sans-serif;">To find the magnetic field strength in a solenoid.

<span style="font-family: Arial,Helvetica,sans-serif;">**Hypothesis**:

<span style="font-family: Arial,Helvetica,sans-serif;">The magnetic field strength of a solenoid will be proportional to the distance from the center of the solenoid. Magnetic field strength will be the greatest in the center of the solenoid and will become weaker as the distance from the center increases.

Rationale: The magnetic field strength can be found per this equation:



Where N is the number of loops, L is the length of the solenoid, I is the current and μ0 is the permeability constant.

<span style="font-family: Arial,Helvetica,sans-serif;">**Materials** - magnetic field sensor, power supply, meter stick, patch chord, and solenoid

<span style="font-family: Arial,Helvetica,sans-serif;">**Procedure:**

1. Using two wires, connect the ends of a solenoid to the ends of a variable power supply. 2. Place a meter stick parallel and next to the solenoid with the meter stick facing out from the end of the solenoid. 3. Plug the variable power supply into a power outlet and turn on the power. 4. Set the variable power supply to a specific Current (A) that is within a small voltage. (Ours was set to a current of 1A). 5. Plug in a magnetic field sensor to a computer, setting the screen to a grapher (EZScreen) for recording data. 6. Press play on the grapher to record results, and hold the sensor at the center of one end of the solenoid. 7. Record the data for this at distance (0). 8. Move the sensor a measured distance into the solenoid (we did every .01m), and record the magnetic field strength at this new point using an excel spreadsheet. 9. Repeat this step #8 for many points until the sensor is as far into the solenoid as it can be. 10. If the sensor is shorter than the solenoid, repeat steps 6-9 for the opposite end of the solenoid.



<span style="font-family: Arial,Helvetica,sans-serif;">**Data:** <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">


 * Graphs:**




 * Calculations:**

Max Theoretical Magnetic Field Strength:

Percent Error



We took measurements that ranged from the very outside of the solenoid to the middle of it. The best measurement would be the middle of the solenoid. This is because there is a lot of interference near the outer edges for the magnetic field measurements to be considered accurate.The equation we used to find the theoretical magnetic field strength should match up to our best measurement, the one taken in the middle of the solenoid (.08m). At .08m, we got a value of 0.004364 Teslas compared to the theoretical value of 0.004054 Teslas, which produced a percent error of 7.66%.
 * Analysis:**

We had two different graphs because the two sides of the solenoid (if one were to separate the solenoid in half down the middle) produced very different results. The first graph and those values had better results, so we decided to use that to compare it to the theoretical magnetic field strength. The x values (distance measurements) that are high and low on the graphs gave us magnetic field strengths that are very low. This was mentioned above and was due to the interference near the outer edges of the solenoid. As you get closer, to the center of the solenoid, the measurements only keep getting better.

The parabolic shape of the graphs can be seen above and makes sense for our data. Near the outer parts of the solenoid, the strength was very low. As the sensor moved closer and closer to the middle, the strength kept exponentially increasing until it neared the middle. At this part, it didn't increase nearly as much and just hit a maximum. After this maximum, the graph begins to decrease and the rest can be seen on the second graph. The trendline provides evidence for our parabolic fit, showing that the graph does indeed increase, hit a maximum, and then decrease once again.

<span style="font-family: Arial,Helvetica,sans-serif;">**Discussion Questions**

<span style="font-family: Arial,Helvetica,sans-serif;">1. Did the axial reading change when the sensor was moved radially outward from the center toward the windings on the coil?

<span style="font-family: Arial,Helvetica,sans-serif;">Yes. The reading was highest when the sensor was in the direct sensor. When it was moved radially toward the windings, the reading decreases a little bit.

<span style="font-family: Arial,Helvetica,sans-serif;">2. Was the axial reading different from the reading in the middle of the coil when the sensor was inside but near the ends of the coil? Why?

The reading was greatest when the sensor was inside the coil, at center of the solenoid and it was weakest at the ends of the coil. This is because the magnetic field of a solenoid converges in the center, where it is greatest, and it weakest at the ends.

<span style="font-family: Arial,Helvetica,sans-serif;">3. By comparing the axial and radial readings, what can you conclude about the direction of the magnetic field lines inside of a solenoid?

<span style="font-family: Arial,Helvetica,sans-serif;">The magnetic field lines wrap around the solenoid going from one end to the other, and through the center of the solenoid where the amount of field lines is the greatest. Here the magnetic field strength will be greatest.

<span style="font-family: Arial,Helvetica,sans-serif;">4. At what position in the solenoid should you get the greatest magnetic field strength?

<span style="font-family: Arial,Helvetica,sans-serif;">At the direct center of the solenoid.

<span style="font-family: Arial,Helvetica,sans-serif;">5. How does the theoretical value compare to the value at this position?

The value we measured was very close to the theoretical value of the magnetic field strength at the center. Our percent error was just over 7%.


 * Conclusion**:

Based on the results of our experiment, we can conclude successfully that the magnetic force of a solenoid is directly proportional to the distance from the center of the solenoid and it's magnetic field. Our data proves something that we didn't know from the equation of B = µ o * I * n that is true for solenoids with magnetic field strength, as we found when current, length, and number of coils remained the same. As you can see from our data, the closer we place the sensor to the center of the solenoid, the stronger the magnetic field strength that is measured. This is because the magnetic field is stronger in the middle of the solenoid and weaker as it gets farther away. We can also see when we measure the magnetic field strength from both sides, that the magnetic field strength is not exactly centered with the solenoid. As we measured we found two different graphs for each end of the solenoid, and the only possible reason is that the area of greatest strength is not centered, just near the middle of the solenoid. We experienced some error with the sensor because it was difficult to get a constant, steady value at each position. This was most likely because it was difficult to hold in the solenoid by hand. The results would have been more steady if we had a way of using a lab stand to hold the sensor in place. Another source of error came from the sensor, which received results even when it was far away from the solenoid. This magnetic field strength could be coming from the Earth's magnetic field or from a nearby lab groups set up and had a reading of no less than 4 Gauss, but it caused our sensor to get results even when there was no power running through the solenoid and caused variation in our data. This problem could have been prevented if the sensors could be zeroed, similar to how the scales could be in our previous lab. Therefore, the readings from the sensor would only be from our solenoid. This lab is relevant when trying to use magnetic fields in devices to generate a force on an object, such as with the starters for cars and fuel injectors.