31 Magnetic Force

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St. John's University *

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Physics

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Apr 3, 2024

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April 2008 Rev 2/23 Experiment 31: Magnetic Forces Purpose (1) To study a magnetic force on a wire with a current in it. We will explore the relationship between the magnitude of the current in the wire, the length of the wire, and the strength of the magnetic field affect the force. (2) To find the magnetic field strength of the Magnet Assembly. Apparatus Basic Current Balance (Main Unit, 6 Current loop PC boards, Magnet Assembly with 6 magnets), DC power supply, ammeter, balance, lab stand, hook-up wires with banana plug connectors Theory A current-carrying wire experiences a force in a magnetic field. The magnitude and direction of this force depends upon the the strength of the magnet’s magnetic field (B), the length (L) of the wire inside the magnet’s magnetic field, the magnitude of the current (I), and the angle between the direction of the current and the direction of the magnetic field ( θ ). In this experiment, the direction of the magnetic field is horizontal (see Figure 1 next page). The direction of the current within the magnetic field is horizontal but perpendicular to the magnetic field. Using the right hand rule, we find that this creates a component of the magnetic force on the wire which will be either up or down depending on the relative directions of the current and the magnetic field. As we vary the three relevant quantities, namely: I, L, and B, we will measure the magnetic force by measuring changes in the apparent mass of the magnetic assembly on the balance as the magnetic force either pushes down or pulls up on the magnet assembly. We can easily calculate the force on a wire in a magnetic field. The equation for this force F * is: F = I d L x B = BLI sin θ where θ is the angle between the magnetic field at the wire and the direction of the current in the wire, B is the magnetic field of the permanent magnets, L is the length of the wire in the magnetic force, I is the magnitude of the current. In our case, the angle θ is 90º which gives us Equation 1: F = BLI 1 We will use this equation to calculate the magnetic field of the permanent magnets by setting the measured force equal to BLI. * Bold capital letters indicate vectors . 160

Experiment 31 Procedure Part I. Force versus Current [Note: In this experiment, the force is determined from a balance reading in grams. This times the acceleration due to gravity (g = 980 cm/s 2 ) yields the actual magnetic force, but we will use grams as a proxy for the force. ] 1. Set up the Main Unit as shown in Figures 1, 2, and 3. Some setups will have an ammeter in the power supply. Ask your instructor if you are not sure how to set this up. 2. Copy Table 1 onto your data sheet. Record the Current Loop you are using (use SF 42); ; the mass in grams (to 0.01 g); and the difference in mass, ΔMass, proxy for the magnetic force in grams. Table 1 3. Plug Current Loop 42 into the ends of the arms of the Main Unit, with the foil extending down as shown in Figure 1. 4. Check the colors of the magnets in the Magnet Assembly. All reds should be on one side. Ask your instructor for help if they are not. Place the Magnet Assembly on the balance. Position the lab stand so the horizontal portion of the conductive foil on the current loop passes through the pole region of the magnets. Make sure the current loop doesn't touch the magnets.. Measure the mass of the Magnet Assembly with no current flowing to 0.01g. Current Loop: Current (amps) Mass (grams) ΔMass (grams) 0.0 -------- 1.0 ... 161

Experiment 31 5. Set the current to 1.0 Amp – first with the coarse adjustment knob, then with the fine adjustment knob. Determine the new mass of the magnet assembly. Record this value under Mass in your data table. Subtract the mass value with the current flowing from the value with no current flowing. Record this difference as ΔMass. 6. Increase the current in 1.0 Amp increments to a maximum of 5.0 Amp, each time recording the current, Mass and calculating the proxy for the force, ΔMass. Procedure Part II. Force versus Length of Wire You can vary the wire length by using one of the six different current loops. The lengths for each current loop are given in the following table: Changing Current Loops : 1. Copy Table 2 onto your data sheet with six rows. Copy the mass of the Magnet Assembly with no current. Table 2 I = ______ mass no current = ________ Current Loop Length (cm) Mass (grams) ΔMass (grams) 2. Start with current loop SF40. Determine the length of the conductive foil on the current loop by referring to the table above. Record this value under "Length" in Table 2. Current Loop Length SF 40 1.2 cm SF 37 2.2 cm SF 39 3.2 cm SF 38 4.2 cm SF 41 6.4 cm SF 42 8.4 cm 162

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Experiment 31 3. Set the current to 4.0 amps. Determine the new mass of the Magnet Assembly. Record this value under "Mass" in Table 2. Subtract the mass that you measured with no current flowing from the mass that you measured with the current flowing. Record this difference as ΔMass. 4. Turn the current off. Swing the arm of the Main Unit up, to raise the present current loop out of the magnetic field gap. Pull the current loop gently from the arms of the base unit. Replace it with a new current loop. 5. Repeat steps 2-4 for each of the five remaining current loops. Procedure Part III. Force versus Magnetic Field The magnetic field is varied by changing the number of magnets that are mounted on the Magnet Assembly. The magnetic field strength may not be exactly proportional to the number of magnets, but it is reasonably close. 1. Copy Table 3 onto your data sheet. Set up the apparatus using the longest length current loop (SF42). Table 3 Current Loop: Number of Magnets Mass (grams) ΔMass (grams) I = 0.0 A I = 4.0 A 1 2 …. 2. Mount a single magnet in the center of the holder. 3. With no current flowing, determine the mass of the Magnet Assembly. Record this value in the I = 0 A column under "Mass" in Table 3. 4. Set the current to 4.0 amps. Determine the new mass of the Magnet Assembly. Record this value in the I = 4.0 A column under "Mass" in Table 3. 163

Experiment 31 5. Subtract the mass you measured when there was no current flowing from the value you measured with current flowing. Record this difference as ΔMass. 6. Add additional magnets, one at a time. (Make sure the red sides of the magnets are all on the same side in the Magnet Assembly.) Each time you add a magnet, repeat steps 3-5. Lab Report (we are using Δmass = Δgrams as a proxy for force) Part I 1. Plot a graph of ΔMass (grams, vertical axis) versus Current (Amperes, horizontal axis). Find the slope of the best fit line and include its units. 2. What is the nature of the relationship between these two variables? What does this tell us about how changes in the current will affect the force acting on a wire that is inside a magnetic field? 3. Calculate B from the slope: Your slope represents force /current. Since F = BLI, F/I = BLI/I = BL = the slope We must be very careful with the units. The units of B are Tesla if F is in Newtons, I is in Amperes and L is in meters. Your slope’s units are g/A. To get your slope into the correct units (1) multiply the slope times the acceleration of gravity g = 9.8 m/s 2 (2) multiply by the conversion factor from grams to kilograms 1 kg/1000 g. B Part I (in Tesla) • L(in meters) = Slope • 9.8 m/s 2 • 0.001 kg/g Check the table for the length of the loop that you used in Part I, convert it to meters , and calculate B Part I . Part II 4. Plot a graph of ΔMass (grams, vertical axis) versus Length (cm, horizontal axis). Find the slope of the best fit line including units. 5. What is the nature of the relationship between these two variables? What does this tell us about how changes in the length of a current-carrying wire will affect the force that it feels when it is in a magnetic field. 6. Calculate B from the slope: Your slope represents force/length. Since F = BLI, F/L= BLI/L= BI. = the slope. 164

Experiment 31 . Your slope’s units are g/cm. To get your slope into the correct units (1) multiply the slope times the acceleration of gravity g = 9.8 m/s 2 (2) multiply by the conversion factor from grams to kilograms 1 kg/1000 g (3) multiply by the conversion factor from centimeters to meters 100 cm/1m: B Part II (in Tesla) • I = 9.8 m/s 2 • 0.001 kg/g •100 cm/m ·Slope Check Table 2 for the value of the current I that you used in Part II and calculate B Part II . Part III 7. Plot a graph of ΔMass (grams, vertical axis) versus Number of Magnets (n,horizontal axis). Find the slope of the best fit line including units (Number of Magnets has no units) 8. What is the relationship between these two variables? How does the number of magnets affect the force between a current-carrying wire and a magnetic field? Is it reasonable to assume that the strength of the magnetic field is directly proportional to the number of magnets? 9. Calculate B from the slope: The slope of this line represents the force/magnet. In Equation 1, F = BLI. B is the field of the set of six magnets. B/6 is the field due to a single magnet. F = n·(B/6) (I·L) where n is the number of magnets. From this, the slope = F/n = (B/6) (I·L) . Your slope’s units are grams. To get your slope into the correct units, (1) multiply the slope times the acceleration of gravity g = 9.8 m/s 2 (2) multiply by the conversion factor from grams to kilograms 1 kg/1000 g. B Part III (in Tesla) · I · L(in meters) / 6 = Slope • 9.8 m/s 2 • 0.001 kg/g Check Table 3 for the value of the current I that you used in Part III in amperes. From the number of the loop you used, find the length of the loop, convert it to meters and calculate B Part III . 165

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Experiment 31 Part IV 10. Copy the following table to your report and fill in your results: Table 4 B measured from Slope (Tesla) Average Value of B measured (Tesla) % Deviation Part I 0.0361 0.0336 7.44 % Part II 0.0312 7.14 % Part III 0.0337 0.29 % 11. Find the % deviation between each B measured and B avg by using the following: % Deviation = (B avg - B measured ) / B avg • 100% 12. Display the percent discrepancy between your values for the magnetic field of the permanent magnets obtained in Part I and Part II by using the following equation: % discrepancy = B(Part I) – B(Part II) x 100% .5 * [B(Part I) + B(Part II)] 166

31 Magnetic Force

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