Question

Question 22 1 pts Determine the magnitude and direction of the force on an electron (q=-1.610^-19 C) traveling 7.7510^5 m/s horizontally to the right in an upward magnetic field of strength 0.45 T. 5.610^-14 N Into the page 5.610^-14 N Out of the page 5.610^-15 N Into the page 5.610^-15 N Out of the page

          Question 22
1 pts
Determine the magnitude and direction of the force on an electron (q=-1.6*10^-19 C) traveling
7.75*10^5 m/s horizontally to the right in an upward magnetic field of strength 0.45 T.
5.6*10^-14 N Into the page
5.6*10^-14 N Out of the page
5.6*10^-15 N Into the page
5.6*10^-15 N Out of the page

Question 22
1 pts
Determine the magnitude and direction of the force on an electron (q=-1.6*10^-19 C) traveling
7.75*10^5 m/s horizontally to the right in an upward magnetic field of strength 0.45 T.
5.6*10^-14 N Into the page
5.6*10^-14 N Out of the page
5.6*10^-15 N Into the page
5.6*10^-15 N Out of the page

Added by Sandra S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vipin Kumawat

Step 1

The formula for the magnetic force on a charged particle moving in a magnetic field is given by: F = q * v * B * sin(theta) Where: F is the force on the particle q is the charge of the particle v is the velocity of the particle B is the magnetic field Show more…

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Question 22 Determine the magnitude and direction of the force on an electron (q = -1.6*10^-19 C) traveling 7.75*10^5 m/s horizontally to the right in an upward magnetic field of strength 0.45 T. Options: A. 5.6*10^-14 N into the page B. 5.6*10^-14 N out of the page C. 5.6*10^-15 N into the page D. 5.6*10^-15 N out of the page