An electron moves in a force field due to a uniform electric field E and a uniform magnetic field B that is at right angles to E. Let E = jE and B = kB. Take the initial position of the electron at the origin with initial velocity vo = ivo in the x direction. Find the resulting motion of the particle. Show that the path of motion is a cycloid.
x = a sin wt + bt
y = a (1 - cos wt)
z = 0
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Start out with writing the Lorenz force on the electron.
F = -e [E + v x B]
Let v = (vx, vy , vz) and F = (Fx, Fy, Fz)
F = -e [E j + (vx,vy,vz) x B k]
(Fx, Fy, Fz) = -e [E j + - vx j + vy i]
Equating the components,
Fx = - e vy B
Fy = -e(E - B vx)
Fz = 0
Apply Newton's second law. F = m a
Fx = m d^2x/dt^2 = - e vy B
Fy = m d^2y/dt^2 = ...
I have found the path of the electron when it is placed in a mutually perpendicular magnetic and electric fields provided that the electron is in motion. This problem and solution set will be a great practice set for a student in an Electromagnetic Theory course.
Magnitude of velocity of a electron
An electron e= 1.6*10^-19C, m= 9.1*10^-31kg is accelerated through a potential difference of 2kV. It then passes into a magnetic field perpendicular to its path, where it moves in a circular arc of diameter 0.36m.
1.) What is the magnitude of the velocity of the electron in a magnetic field?
2.) What is the magnitude of the magnetic field?
3.) What is the frequency of the electron's motion in the magnetic field?View Full Posting Details