Given magnetic field, particle charge, and force (vector) Calculate velocity?

[Goomba]

A particle with charge -5 nC is moving in a uniform magnetic field B = -(1.2 T)k. The magnetic force on the particle is measured to be F = -(3.6*10^(-7) N)i + (7.6*10^-7)j. Calculate the x and y components of the particle's velocity.

F = q(V X B)
(force equals charge multiplied by the cross product of V and B)

Calculate the scalar product (dot product) v*F
vx*Fx + vy*Fy + vz*Fz?

What is the angle between v and F (in degrees)?
F = qvB sin theta
theta = arcsin [F/(qvB)]?

 

[eigenglue]

Fz is zero, right? So do you need to worry about Vz, or the angle between V and B? Or do you just need to worry about the component of V that is perpendicular to B? (Which would be in the xy plane only).

 

[Goomba]

I believe that I just need to worry about the component of V that is perpendicular to B (x and y components of V) since force exists in the x and y directions only.

 

[Goomba]

I found equations for each of the x and y components:

V_x = (-F_y)/(q*B)
V_x = -(-7.6E-7)/(-5E-6*-1.20)
V_x = -0.127 m/s

V_y = (-F_x)/(q*B)
V_y = -(-3.6E-7)/(-5E-6*-1.2)
V_y = 0.06 m/s

But those answers aren't right... What am I doing wrong? Are the equations not right? Am I using incorrect values in correct equations? I'm so confused!

 

[Goomba]

Kay, nevermind...

 

[neutrino]

Just calculate the cross product q(vxB), assuming a certain variable for each component of the velocity, and equate the result to the force.

Btw, vxF is zero and it doesn't help much.

 

[Goomba]

Thanks. I found out my problem. I'm so used to converting micro 10^-6 that when I came across "nano," I used micro's conversion value.

 

[mj21181]

v*F being zero tells you that the angle between them is 90 degrees, so it does actually help