What is the unique drift speed for a charged particle in crossed E and B fields?

[num1cutiey]

Homework Statement

I am working on a problem and don't know if I am going about it right because I am stuck.

A charged particle of mass m and positive charge q moves in uniform electric and magnetic fields, E pointing in the y direction and B in the z direction (an arrangement called "crossed E and B fields"). Suppose the particle is initially at the origin and is given a kick at time t=0 along the x axis with v$$_{}x$$=v$$_{x0}$$ (positive or negative). (those are supposed to be subscripts but I can't get it to work)

(a) Write down the equation of motion for the particle and resolve it into it's three components. (done) Show that the motion remains in the plane z=0 (done)

(b) Prove that there is a unique value of v$$_{x0}$$, called the drift speed v$$_{dr}$$,for which the particle moves undeflected through the fields. (This is what I can't get)

Homework Equations

F=ma
F=q(E+vXB)

The Attempt at a Solution

I wrote down the equation of motion and when separated I got

v$$_{y}$$*B$$_{z}$$=m*dv$$_{x}$$/dt
v$$_{x}$$*B$$_{z}$$=m*dv$$_{y}$$/dt
0=m*dv$$_{z}$$/dt

I proved the whole z=o plane thing.

Now I get to b

Iknow that x(t)=f some arbitrary function and y(t)=0. (We already know z(t)=0). So I used the second equation above and solve it to get y(t)=(v$$_{x}$$*B$$_{z}$$*t^2)/2m=0. Ok so what do I do from here. I am lost I can divide everything and get v$$_{x}$$=0 but I know that is wrong. Did I go about this the wrong way??

 

[Dick]

E is in the y direction. So is vxB, since v is along x and B is along z. You just have to pick an absolute value of v so the total of all the forces, E+vxB=0.

 

[num1cutiey]

shoot I missed the E in the y direction in my equations for a)! Thanks!