Question

# A 6.70 −μC particle moves through a region of space where an electric field of magnitude...

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1200 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction. If the net force acting on the particle is 6.24×10−3 N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane Find all three components and enter in Vx, Vy, Vz format.

Q = 6.70 x 10^-6 C
E = 1200 N/C --->+x direction
B = 1.25T ------>+z direction
Fnet = 6.24x 10^-3 N ----->+x direction

Fnet = Fe - Fb
=qE - qvB
Solve for v

so v = (1/B)(E - (Fnet/q))

= (1/1.25)(1200-(6.24*10^-3/6.70*10^-6)) = 214.925m/s

once you get the value for velocity. You need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for Fnet.
Do the right hand rule where:
your right thumb goes toward the Fnet, then your index finger points to B (z direction) Then curl your middle,ring, and pink 90 angle. This shows where v is going which is -y direction.

Answer: vx, vy,vz = 0, -(number you got for v), 0
make sure you put (-) because v is going - y direction