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

Answer:

v = -227.785 m/s

Explanation:

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

F = qE

$F = 6.65 \times 10^{-6} \times 1230$

$F = 0.0081795 \ N$

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

F = qvB

$F = 6.65\times 10^{-6} \times v \times 1.32$

$F = 8.778 \times 10^{-6} \times v$

Total force acting is:

F = qvB + qE

$6.18 \times 10^{-3} = 0.0081795 + 8.778 \times 10^{-6} \times v$

$v = \dfrac{6.18 \times 10^{-3} -0.0081795 }{8.778 \times 10^{-6}}$

v = -227.785 m/s