Question

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.

Answer 1

v = 248.8 m/s

Step-by-step explanation:

Given that,

Charge, q =
`6.47\ \mu C`

Electric field, E = 1300 N/C

Magnetic field, B =1.33 T

The force acting on the particle,
`F=6.27* 10^(-3)\ N`

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

`F=qE+qvB\\\\6.27* 10^(-3)=6.47* 10^(-6)* 1300+6.47* 10^(-6)* 1.33v\\\\6.27* 10^(-3)-6.47* 10^(-6)* 1300=6.47* 10^(-6)* 1.33v\\\\-0.002141=6.47* 10^(-6)* 1.33v\\\\v=(0.002141)/(6.47* 10^(-6)* 1.33)\\\\v=248.8\ m/s`

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