Question

A 6.70 -microC 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 1.24 T points in the positive z direction. If the net force acting on the particle is 6.25×10^?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.

vx, vy, vz =

Answer 1

First, you write your variables:
Q = 6.7 x 10^-6 C
E = 1250 N/C --->+x direction
B = 1.24 T ------>+z direction
Fnet = 6.25 x 10^-3 N ----->+x direction

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

v = (1/1.24)(1250-(6.25*10^-3/6.7*10^-6))

v = 255.78 m/s

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, -255.78 m/s , 0
make sure you put (-) because v is going - y direction