TI-Nspire™ Activities: Physics
February 19, 2008 through February 26, 2008
A Magnetic Attraction
| Part 1 – Using a regression to quantify the field strength-distance relationship | |
| Step 1: Students should open the file PhysWeek04_magstrength.tns and read the first two pages. Page 1.3 contains the data students will use for their analyses. Before moving on to the next step, discuss the trend in the data with the students. They should recognize that field strength decreases as distance increases and that the relationship is not linear. | 
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| Step 2: Next, instruct the students to create a scatter plot of the data on page 1.4. They should use distance for the x-values and intensity for the y-values. They should use the Zoom-Data or Window Settings function to resize the graph window so all the data are visible, but there is no extra space around the edge of the graph. Discuss the shape of the graph with the students. Some students will probably assume that this is an inverse-square relationship. Ask them how they could determine whether an inverse-square function fits the data. They should reason that a power regression can provide information about the curve. You may need to guide them to this conclusion. |
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| Step 3: Next, students will use a power regression to find an equation that fits the data points. Instruct them to move back to page 1.3 and highlight columns A and B. They should then carry out a power regression (Menu > Statistics > Stat Calculations > Power Regression) on columns A and B. They should use column A for the X List and column B for the Y List. They should store the values in column C. |
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| Discuss the calculated value of variable b with students. They should conclude that the relationship is not, in fact, an inverse-square relationship. Instead, it is an inverse-cube relationship—that is, intensity decreases in direct proportion to distance cubed. |
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| Step 4: Have students graph the regression equation on the plot of the data on page 1.4. They will need to change the graph type to Function, then select f1(x) in the function bar. When they press •, the function should be plotted on the graph. Have the students record the equation on a separate sheet of paper. They should round the variables to three significant figures. | 
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| Part 2 – Building a model for the relationship | |
Step 1: Next, students will attempt to find the best-fit equation of the form  to model the data. They should hide the graph of f1(x) using the Show/Hide tool. Then, they should use the Trace tool to locate and mark the coordinates of the leftmost point, as shown. Then, they should solve the equation above for k, using the coordinates of the leftmost point for intensity and d. The data set provided yields 9.97 × 10–5 for the value of k. |
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| Step 2: Now, students should place a text box on the graph to contain the value of k. This will make it easier for them to vary k and observe the resulting changes in the graph. After placing the text box on the graph, they should type 0.0000997 and press •. Then, they should click once on the text box to highlight it and click h. They should select Store Var and type k to assign the variable k to this value. |
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| Step 3: Next, students should plot the function they have estimated by clicking on the function bar at the bottom of the screen. They should enter the expression k/x3 in the f2(x) function bar. When they press •, the graph should update with the curve they have entered, as shown. |
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| Step 4: Have students vary the value of k by editing the text box. They should attempt to find a value of k that produces the best fit to the data. Note that it will be impossible for students to find the absolute best-fit value of k. They should simply focus on finding a slightly better fit than the original value. Have students record the value of k they find on a separate sheet of paper. | 
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| Part 3 – Determining the magnetic moment | |
Next, students will calculate the magnetic moment of the field. The equation relating magnetic field strength (B) to distance (d) is given below (μ0 is the magnetic permeability constant, and μ is the magnetic moment of the field):  In the previous steps, students have approximated this equation with the equation  and then calculated the value of k for their data. They will now use this value of k to calculate the magnetic moment of the field using the following equation:  The magnetic permeability constant, μ0, is equal to 4  × 10–4. |
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| Step 1: Guide students through the derivation of the equation for μ. Then, have them move to page 1.5, which contains a blank Calculator application. They should use the Calculator to define the variable mu0 (which will represent the magnetic permeability constant), as shown. Note: Make sure students use the := notation when entering the variable definition. | 
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| Step 2: Next, students will use the nSolve function to calculate the value of the magnetic moment by solving the equation above for μ, as shown. Note that students should not define the variable mu before entering it into the nSolve function. Students should record the value of μ that they calculate. The values they come up with will vary based on the best-fit values of k they found. Finally, students should answer the questions on pages 1.6 through 1.9. They may answer in the Notes application, inserting new pages as necessary, or on a blank piece of paper. The questions and their answers (in italics) are given below. Q1. How well do the results of the power regression support the “ideal” relationship between field strength and distance given below?  (In this equation, μ0 and μ are constants.) Explain your answer. A. The results of the power regression agree with the “ideal” relationship. The ideal relationship indicates that magnetic field intensity should decrease in inverse proportion to distance cubed, and the power regression yields an exponent of nearly –3. Q2. What is the value of k that you calculated from the data? A. Students' values of k will vary. Q3. What is the value of μ that you calculated from the data? A. Students' values of μ will vary. Q4. Electric currents produce magnetic fields. For a single loop of wire, the magnetic moment of the field is related to current by the following equation, in which μ is magnetic moment, I is current, and A is the area of the loop through which the current flows: μ = IA The magnetic field in this activity was produced by a round magnet with a radius of 29 mm. What current would be required to produce an equivalent magnetic field in a loop of wire with the same radius as the magnet? (Note: The units of μ that you calculated are m2•A.) A. If the radius of the wire loop is 29 mm (0.029 m) and the magnetic moment of the field is 0.4995 m2•A, the current in the loop can be calculated as follows:  |
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