**Problem 14: Magnetic Force on an Electron**An electron moving at a speed of \(4.8 \times 10^5 \, \text{m/s}\) enters a \(0.60 \, \text{T}\) magnetic field from the left, as depicted in the diagram below. Determine the direction and magnitude of the magnetic force on the electron. (Charge of electron \( e = 1.60 \times 10^{-19} \, \text{C} \))**Diagram:**- The diagram consists of a grid with 'x' marks representing the magnetic field (\(B\)) directed into the page.- The electron is shown entering the field from the left with velocity \(\mathbf{v}\) directed to the right.**Options:**A. \(4.6 \times 10^{-14} \, \text{N} \) Downward B. \(2.3 \times 10^{-13} \, \text{N} \) Upward C. \(9.0 \times 10^{-14} \, \text{N} \) Downward D. \(0.6 \times 10^{-14} \, \text{N} \) Upwards **Explanation of Diagram:**- The grid of 'x' marks indicates the magnetic field is uniform and directed into the plane of the page.- The velocity vector \(\mathbf{v}\) is shown as an arrow pointing to the right, indicating the initial direction of motion of the electron.**Solution Approach:**To solve this problem, use the formula for the magnetic force on a moving charge:\[ F = qvB \sin \theta \]Where:- \( F \) is the magnetic force,- \( q \) is the charge of the electron (\(1.60 \times 10^{-19} \, \text{C}\)),- \( v \) is the speed of the electron (\(4.8 \times 10^5 \, \text{m/s}\)),- \( B \) is the magnetic field strength (\(0.60 \, \text{T}\)),- \(\theta\) is the angle between the velocity and the magnetic field direction.In this scenario, \(\theta = 90^\circ\) as the velocity is perpendicular to the magnetic field, making \(\sin

Given, The speed of the electron, v=4.8×105 m/s The magnetic field, B=0.60 T The charge on the electron, e=1.60×10-19 CThe magnitude of the force on the particle in presence of the magnetic field is given by, F=qv→×B→=1.60×10-19×4.8×105×0.6=4.608×10-14 =4.6×10-14 N The direction of the magnetic force will be, F^=-x^-z^=-y^ So the direction of the magnetic field will be downwards. Hence the magnitude of force will be 4.6×10-14 N in a downward direction. So in this question option A is correct.

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14. An electron moving at a speed of 4.8 x 105 m/s enters a 0.60 T magnetic field from the left as shown below. Find the direction and the magnitude of the Magnetic Force on the electron. (e= 1.60 × 10-19 C) V x xB ххххх Xx X X хххххххххх хххххххххх хххххх xx хххххххххх A. 4.6x10-14N Downward в. с.9.0х10-14N Downward B. 2.3x10-13N Upward D.0.6x10-14N Upwards

14. An electron moving at a speed of 4.8 x 105 m/s enters a 0.60 T magnetic field from the left as shown below. Find the direction and the magnitude of the Magnetic Force on the electron. (e= 1.60 × 10-19 C) V x xB ххххх Xx X X хххххххххх хххххххххх хххххх xx хххххххххх A. 4.6x10-14N Downward в. с.9.0х10-14N Downward B. 2.3x10-13N Upward D.0.6x10-14N Upwards

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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