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Charged Particle in Constant Electromagnetic Field, Lorentz

The motion of a charged particle in an electromagnetic field can be obtained by the Lorentz Equation for the force on a particle in such a field. If the electric field vector is E and the magnetic field vector is B, the force on a particle of mass m that carries a charge q and has a velocity v is given by:

F= qE + qv (cross product) B

Where we assume v<< c (the speed of light)

a.) If there is no electric field and the particle enters the magnetic field in a direction perpendicular to the lines of flux show that the trajectory is a circle with radius:

r = (mv)/(qB)

b.) Choose thr z axis to lie in the direction of B and let the plane containing E and B be the xy plane.

Then: B = Bk, E= Ey j + Ez k

Show that z(t) = zo = zo't + (qEz)/(2m) t2

Where z(0) = zo and z'(0) = zo'

c.) Continue the calculation and find x'(t) and y'(t); show that the time averages of these velocities are:
(x') = Ey/B, (y') = 0

d.) Integrate the velocitiy equations above (c) with the initial conditions:

x(0) = -(Am)/( qB); x'(0) = Ey/B; y(0) = 0y'(0) = A

Note I cannot label vectors but hopefully it is clear from the context which are vectors

Solution Preview

The solution is written in the attached file 93338.pdf

Could you do the calculations using the unit vectors(i.j.k) and vector ...