A 6.47 micro-coloumb particle moves through a region of space where an electric field of magnitude 1300 N/C points in the positive x direction, and a magnetic field of magnitude 1.33 T points in the positive z direction. If the net force acting on the particle is 6.27E^-3 N in the positive x direction> Required:
Calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the xy plane.

v = 248.8 m/s

Explanation:

Given that,

Charge, q =

Electric field, E = 1300 N/C

Magnetic field, B =1.33 T

The force acting on the particle,

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

So, the magnitude of the particle's velocity is 248.8 m/s.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west