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.

Answered: Image /qna-images/answer/2875ad6a-1538-456d-8a26-1de9356aecf0.jpg

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

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

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

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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