Question

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.

Answer 1

The particle's velocity is 212.15 m/s.

Step-by-step explanation:

Given that,

Charge of particle,
`q=6.41\ \mu C=6.41* 10^(-6)\ C`

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

The magnitude of magnetic field, B = 1.28 T

Net force,
`F=6.4* 10^(-3)\ N`

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

`F=qE+qvB\\\\v=(F-qE)/(qB)\\\\v=(6.4* 10^(-3)-6.41* 10^(-6)* 1270)/(6.41* 10^(-6)* 1.28)\\\\v=-212.15\ m/s`

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