A proton moves with a speed of 2.0 × 10^5 m/s along the +x-axis. It enters a region where there is a magnetic field of 4.5 T, directed in the y-axis. Calculate the magnitude of the force on the proton. (q = 1.60 × 10-19 C) O 1.44× 10-13 N 1.3 × 10-10 N 1.3 × 10-18N 3.0 × 10-18 N ΟΟΝ

College Physics
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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In this exercise, a proton is moving with a speed of \(2.0 \times 10^5 \, \text{m/s}\) along the positive x-axis. The proton then enters a region with a magnetic field of \(4.5 \, \text{T}\) directed along the y-axis. The task is to calculate the magnitude of the force acting on the proton. The charge of the proton is given as \(q = 1.60 \times 10^{-19} \, \text{C}\).

The following multiple-choice options are provided for the magnitude of the force:

- \(1.44 \times 10^{-13} \, \text{N}\)
- \(1.3 \times 10^{-10} \, \text{N}\)
- \(1.3 \times 10^{-18} \, \text{N}\)
- \(3.0 \times 10^{-18} \, \text{N}\)
- \(0 \, \text{N}\)

To solve this, you can use the formula for the magnetic force on a charged particle: 
\[ F = qvB\sin\theta \]
where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. In this case, \( \theta = 90^\circ \) because the velocity is along the x-axis and the magnetic field is along the y-axis. Thus, \(\sin\theta = 1\).

By substituting the given values, calculate the magnitude of the force.
Transcribed Image Text:In this exercise, a proton is moving with a speed of \(2.0 \times 10^5 \, \text{m/s}\) along the positive x-axis. The proton then enters a region with a magnetic field of \(4.5 \, \text{T}\) directed along the y-axis. The task is to calculate the magnitude of the force acting on the proton. The charge of the proton is given as \(q = 1.60 \times 10^{-19} \, \text{C}\). The following multiple-choice options are provided for the magnitude of the force: - \(1.44 \times 10^{-13} \, \text{N}\) - \(1.3 \times 10^{-10} \, \text{N}\) - \(1.3 \times 10^{-18} \, \text{N}\) - \(3.0 \times 10^{-18} \, \text{N}\) - \(0 \, \text{N}\) To solve this, you can use the formula for the magnetic force on a charged particle: \[ F = qvB\sin\theta \] where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. In this case, \( \theta = 90^\circ \) because the velocity is along the x-axis and the magnetic field is along the y-axis. Thus, \(\sin\theta = 1\). By substituting the given values, calculate the magnitude of the force.
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