A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$ direction, and a magnetic field of magnitude 1.28 T points in the positive $z$ direction. If the net force acting on the particle is 6.40E-3 N in the positive $x$ direction, calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the $x$-$y$ plane.

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle,

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force,

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

So, the particle's velocity is 212.15 m/s.

answer; how far out of place which the object finished relative to the starting location;