Question

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1250 N/C points in the positive x direction, and a magnetic field of magnitude 1.03 T points in the positive z direction.

.If the net force acting on the particle is 6.22×10⁻³ 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?

Answer 1

First, you write your variables:
Q = 6.70 x 10^-6 C
E = 1250 N/C --->+x direction
B = 1.03 T ------>+z direction
Fnet = 6.22x 10^-3 N ----->+x direction

Fnet = Fe - Fb
=qE - qvB
Solve for v so v = (1/B)(E - (Fnet/q))

v= 312.27 m/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, -312.27, 0