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10 1.2 nm O + 1.6 nm Proton X 1.92 x 103 T. out of the page 1.25 x 10-3 T, out of the page 2.33 x 103 T, out of the page 8.33 x 103 T, out of the page 2.33 x 103 T, into the page 1.25 x 10-3 T. into the page 1.54 x 10-3 T, out of the page 8.33 x 103 T, into the page 1.92 x 10-3 T. into the page 1.54 x 10-3 T. into the page Electron An electron and a proton are each moving at 4.80 x 105 m/s in perpendicular paths as shown in the figure. At the instant when they are at the positions shown in the figure, find the magnitude and direction of total magnetic field proton and electron produce at the origin. (e = 1.60 x 10-¹9 C)

10 1.2 nm O + 1.6 nm Proton X 1.92 x 103 T. out of the page 1.25 x 10-3 T, out of the page 2.33 x 103 T, out of the page 8.33 x 103 T, out of the page 2.33 x 103 T, into the page 1.25 x 10-3 T. into the page 1.54 x 10-3 T, out of the page 8.33 x 103 T, into the page 1.92 x 10-3 T. into the page 1.54 x 10-3 T. into the page Electron An electron and a proton are each moving at 4.80 x 105 m/s in perpendicular paths as shown in the figure. At the instant when they are at the positions shown in the figure, find the magnitude and direction of total magnetic field proton and electron produce at the origin. (e = 1.60 x 10-¹9 C)

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### Problem Description An electron and a proton are each moving at \(4.80 \times 10^5 \, \text{m/s}\) in perpendicular paths as shown in the figure. At the instant when they are at the positions shown in the figure, find the magnitude and direction of the total magnetic field the proton and electron produce at the origin. (e = \(1.60 \times 10^{-19} \, \text{C}\)) ### Diagram Explanation - The figure includes an \(xy\)-coordinate plane. - A proton, represented by a red sphere with a "plus" sign, is located 1.2 nm below the origin on the y-axis, moving horizontally to the right (along the positive x-direction). - An electron, represented by a blue sphere with a "minus" sign, is located 1.6 nm to the right of the origin on the x-axis, moving vertically upwards (along the positive y-direction). ### Answer Options Choose the correct magnitude and direction of the total magnetic field at the origin: - \(1.92 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(1.25 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(2.33 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(8.33 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(2.33 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.25 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.54 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(8.33 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.92 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.54 \times 10^{-3} \, \text{T}, \text{ into the page}\)
Transcribed Image Text:### Problem Description An electron and a proton are each moving at \(4.80 \times 10^5 \, \text{m/s}\) in perpendicular paths as shown in the figure. At the instant when they are at the positions shown in the figure, find the magnitude and direction of the total magnetic field the proton and electron produce at the origin. (e = \(1.60 \times 10^{-19} \, \text{C}\)) ### Diagram Explanation - The figure includes an \(xy\)-coordinate plane. - A proton, represented by a red sphere with a "plus" sign, is located 1.2 nm below the origin on the y-axis, moving horizontally to the right (along the positive x-direction). - An electron, represented by a blue sphere with a "minus" sign, is located 1.6 nm to the right of the origin on the x-axis, moving vertically upwards (along the positive y-direction). ### Answer Options Choose the correct magnitude and direction of the total magnetic field at the origin: - \(1.92 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(1.25 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(2.33 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(8.33 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(2.33 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.25 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.54 \times 10^{-3} \, \text{T}, \text{ out of the page}\) - \(8.33 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.92 \times 10^{-3} \, \text{T}, \text{ into the page}\) - \(1.54 \times 10^{-3} \, \text{T}, \text{ into the page}\)
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