Question

A 6.80 $\mu C$ particle moves through a region of space where an electric field of magnitude 1230 N/C points in the positive $x$ direction, and a magnetic field of magnitude 1.32 T points in the positive $z$ direction. If the net force acting on the particle is 6.18E-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

v = -227.785 m/s

Step-by-step explanation:

The electric field exerts the following force on the electric particle:

F = qE

`F = 6.65 * 10^(-6) * 1230`

`F = 0.0081795 \ N`

The magnetic field exerts the following force on the particle::

F = qvB

`F = 6.65* 10^(-6) * v * 1.32`

`F = 8.778 * 10^(-6) * v`

Total force acting is:

F = qvB + qE

`6.18 * 10^(-3) = 0.0081795 + 8.778 * 10^(-6) * v`

`v = (6.18 * 10^(-3) -0.0081795 )/(8.778 * 10^(-6))`

v = -227.785 m/s