Respuesta :
The force on a charged particle on a magnetic field is given by:
[tex]F=Bqv\sin\theta[/tex]where B is the magnitude of the field, q is the charge of the particle, v is its velocity and θ is the angle between the field and the velocity of the particle.
In this case we have that:
• The magnitude of the field is 0.09 T.
,• The charge of the particle is 3.2 x 10 –19C .
,• The velocity is 3.0 × 10 4 m/s
,• The angle between the field and the velocity is 90°, since they are perpendicular.
Plugging these we have:
[tex]\begin{gathered} F=(0.09)(3.2\times10^{-19})(3\times10^4)\sin90 \\ F=8.64\times10^{-16}\text{ N} \end{gathered}[/tex]Therefore, the force on the charge is:
[tex]F=8.64\times10^{-16}\text{N}[/tex]