Question

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A)If the net force acting on the particle is 6.21×10−3 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.

Enter your answers numerically separated by commas.

Answer 1

First, you write your variables:
Q = 6.50 x 10^-6 C
E = 1300 N/C --->+x direction
B = 1.02 T ------>+z direction
Fnet = 6.25x 10^-3 N ----->+x direction

F_net = F_e - F_b
=qE - qvB
Solve for v so v = (1/B)(E - (F_net/q))

v=458.507 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: v_x, v_y,v_z = 0, -(458.507), 0