Question

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction, and a magnetic field of magnitude 1.15 T points in the positive direction. If the net force acting on the particle is 6.13E-3 N in the positive direction, calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the - plane.

Answer 1

Answer:

                    $v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

     $F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

                    $F_{xnet}- F_{x1 } = F_{x2 }$

                    $(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

                    $v = (F_{xnet} - F_{x1}) / (F_{x2} )$

                        $=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

                       $= (- 104.25) m/s$

                      $v_{y} = -104 m/s$