C Search Textbook soluti x D hw week 4.pdf C Secure https://learning semo edu/pl
ID: 1535343 • Letter: C
Question
C Search Textbook soluti x D hw week 4.pdf C Secure https://learning semo edu/pluginfile.php/1614562/mod resource/content/5/hw.week 4pdf Apps Bookmarks Kent Library ISout fin learning-semo edu D SPOT Authentication hea 1. A particle with a mass of 6.64 x 10 27 kg and a charge of 3.20 x 10h BC is accelerated from rest through a potential difference of 2.45 x 10 V. The particle then enters a uniform 1.60-T magnetic field. If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle? (+3points) What is the radius of the particles trajectory? 2 points) (Note: FC is the equation for the centripetal force) NamExplanation / Answer
Given
m = 6.64 * 10^-27 kg
q = 3.2 * 10^-19 C
potential difference V = 2.45 * 10^6 V
from work energy theorem
W = change in kinetic energy
q * V = 1/2 * m * v^2
3.2 * 10^-19 * 2.45 * 10^6 = 1/2 * 6.64 * 10^-27 * v^2
speed v = 1.5 * 10^7 m/s
the magnetic force
F = q * v * B
F = 3.2 * 10^-19 * 1.5 * 10^7 * 1.6
F = 7.68 * 10^-12 N
to calculate the radius
F = m * v^2 / r
7.68 * 10^-12 = 6.64 * 10^-27 * (1.5 * 10^7)^2 / r
r = 0.19 m
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