Asked by Mara
An electron is accelerated by a 4.7 kV potential difference. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 8.0 cm?
Answers
Answered by
drwls
Use conservation of energy to compute the velocity, v.
(1/2) m v^2 = e V
v = sqrt (2V*e/m)
V is the 4700 volt accelerating potential. e amd m are the electron's charge and mass.
If R is the radius of the circular path,
m v^2/R = e v B
Solve for B
(1/2) m v^2 = e V
v = sqrt (2V*e/m)
V is the 4700 volt accelerating potential. e amd m are the electron's charge and mass.
If R is the radius of the circular path,
m v^2/R = e v B
Solve for B
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