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?

Answer 1

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