Wendler_K_152_lab5_

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Dec 6, 2023

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Karsyn Wendler Physics 152- Lab Section: Wednesday 2:00pm Lab #05- Magnetic Force September 27, 2023 INTRODUCTION My lab partner Samiksha and I built and tested an apparatus to investigate magnetic force on the electrons moving through a wire. PART ONE: THE APPARATUS We assembled the testing apparatus pictured above through the following process. We began by coiling enameled wire 10 times around a metal tube, then repeating this process but coiling the wire 40 times leaving some excess length on the both ends of each coiled wire. Sandpaper was used to remove the insulation off the ends of both wire coils. Then, we taped paper clips onto the center of a 3”x10” ½” thick piece of Styrofoam and the 10-coiled wire on the underside of the Styrofoam. The excess length was coiled around the ends of the paperclips and soldered to ensure a good connection. The apparatus was then placed across two angle brackets by these paperclip ends. A penny was taped on the opposite side of the Styrofoam to act as a counter weight. Finally, the 40- coil wire was placed in parallel to the 10-coil wire, but taped on the table below it. A small aluminum ball acted as an indicator of whether or not the styrofoam was resting on one side or properly balanced. Finally, the whole apparatus was connected via alligator clips and wires to a DC ammeter. This final set up is seen in the following picture:
Assembled experimental apparatus with DC current running through it PART TWO: DETERMINING THE PERMEABILITY OF FREE SPACE 2-A With all 3 aluminum weights on the far end of our balance, we turned on the current and slowly decreased it until the balance was once again centered. This current was then documented. We repeated this for 4 trials in total and then 4 more trials with 2 and then 1 weight(s) respectively. In retrospect, this may have been a source of error as the data for the current at the exact moment at which the balance began to tip would have been a more accurate current reading. Nonetheless, the data collected is summarized in the chart below: Number of Weights Current Reading 1 (A) Current Reading 2 (A) Current Reading 3 (A) Current Reading 4 (A) Average (A) Standard Deviation (A) 3 1.03 1.01 0.99 0.98 1.00 0.02217 2 0.78 0.76 0.80 0.81 0.79 0.02217 1 0.69 0.67 0.71 0.59 0.66 0.07411
2-B Using the data from 2-A, I plotted the square of the current vs. the number of weights: 2-C Based on the information given in lab, we know the per unit mass of aluminum to be: mass aluminum = 6.01 mg cm 2 And our aluminum weights’ size to be: area aluminum = 2 cm× 2 cm area aluminum = 4 cm 2 Thus, each aluminum weight is: mass aluminum = 24.04 mg ¿ 24 × 10 6 kg 2-D Using the previously calculated mass of each weight, we will calculate the gravitational force acting on each weight: F g = ( 24 × 10 6 kg ) × 9.81 m / s 2 F g = 2.35 × 10 4 N Previously, the lab guide led us through a collection of equations that resulted in one to determine the magnetic force acting on the apparatus. It is as follows: F = μ 0 n U n L i 2 R d
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Where μ 0 is the permeability of free space, n U is the turns of the upper coil, n L is the turns of the lower coil, i 2 is the current squared, R is the radius of the upper coil, and d is the distance between the upper and lower coils. With the newly calculated gravitational force on one weight, we can determine that when the apparatus is balanced with n weights, the equation can be set equal to F ( ¿¿ g ) n ¿ : ( 2.35 × 10 4 ) n = μ 0 n U n L i 2 R d 2-E Because the slope of the graph in 2-B is equivalent to i 2 n , we will rearrange the equation from Part 2-D, setting M equal to the rest of the equation parts. i 2 n = ( 2.35 × 10 4 ) d μ 0 n U n L R 2-F When i 2 n = 0.3274 , n U = 10 , n L = 40 , R≈ 3.5 cm and d ≈ 7 mm , we can use the 2-E equation to solve for μ 0 , the permeability of free space: 0.3274 2 = ( 2.35 × 10 4 )( .007 ) μ 0 ( 10 ) ( 40 ) ( 0.035 ) μ 0 = ( 2.35 × 10 4 ) ( .007 ) ( 0.3274 2 )( 10 )( 40 )( 0.035 ) μ 0 = 1.10 × 10 6 N / A 2 theoretical value from physics textbooks: μ 0 = 4 π× 10 7 N / A 2 1.26 × 10 6 N / A 2 While these values are not the same, they are similar, insinuating that we likely did this lab correctly. I also think we could have simply been lucky with the data inserted. My lab partner and I forgot to take exact measurements of the coil radius, R, and the distance between the coils, d, so their values are best estimates based on the ratio of the coil diameter to the styrofoam piece and the fact that we balanced our styrofoam with the weights and currents rather than just noting the instant it shifted. These values may have been a lucky guess of perfect ratios that help hide human error or they could be highly accurate, making our calculation very close to the original.
Review Questions 1. You can change the force between the two coils from attracting to repelling by doing the following except: D. turn the bottom coil upside down 2. The force between the coils increases if: A. The current through the coils gets larger 3. There should be no force between the coils if: D. alternating (AC) current is flowing through the coils instead of DC current. 4. If, as shown in the figure, the top coil is not directly above the bottom one, then the o that you compute from your data will be: B. smaller than the accepted textbook value. 5. In this lab you measured the permeability o of free space (of air, actually). You can also measure the permeability of other materials by inserting them in the space between the two coils. If a piece of steel or iron is placed between the two coils, the magnetic force between the coils will be greatly increased, implying that: A. >> o
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