# 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

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#### Solution Preview

The solution is written in the attached file 93338.pdf

COMMENT FROM STUDENT:

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