A particle with positive charge q moving in the vertical direction with speed v enters a region where both a magnetic field and an electric field is present. The magnetic field has magnitude Band it is directed at an ang degrees with the vertical axis. Find the magnitude and the direction of the electric field if the particle continues to move on a straight line with the same speed. (sin(45) = cos(45) = (1/V2) O a. E= , to the left O b. E= Y down
A particle with positive charge q moving in the vertical direction with speed v enters a region where both a magnetic field and an electric field is present. The magnetic field has magnitude Band it is directed at an ang degrees with the vertical axis. Find the magnitude and the direction of the electric field if the particle continues to move on a straight line with the same speed. (sin(45) = cos(45) = (1/V2) O a. E= , to the left O b. E= Y down
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter22: Magnetic Forces And Magnetic Fields
Section: Chapter Questions
Problem 63P
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![A particle with positive charge q moving in the vertical direction with speed v enters a region where both a magnetic field and an electric field is present. The magnetic field has magnitude Band it is directed at an angle of 45°
degrees with the vertical axis. Find the magnitude and the direction of the electric field if the particle continues to move on a straight line with the same speed. (sin(45) = cos(45) = (1//2))
B 4
O a. E= to the left
O b. E=
down
2
O . E =
to the right
O d. E=
out of the page
О е. Е —
up
2
O f. E=
into the the page](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff94f2f9-80ab-4bc0-b156-c84512ad3445%2F92ff762d-c8dc-4c49-b080-7d6d5b704101%2Fzfvbnnl_processed.png&w=3840&q=75)
Transcribed Image Text:A particle with positive charge q moving in the vertical direction with speed v enters a region where both a magnetic field and an electric field is present. The magnetic field has magnitude Band it is directed at an angle of 45°
degrees with the vertical axis. Find the magnitude and the direction of the electric field if the particle continues to move on a straight line with the same speed. (sin(45) = cos(45) = (1//2))
B 4
O a. E= to the left
O b. E=
down
2
O . E =
to the right
O d. E=
out of the page
О е. Е —
up
2
O f. E=
into the the page
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