Answer:
[tex]9.6\cdot 10^{-15} N[/tex]
Explanation:
The magnetic force acting on a charged particle moving perpendicularly to the field is given by:
[tex]F=qvB[/tex]
where
q is the charge
v is the speed of the particle
B is the magnetic field strength
In this problem, we have:
[tex]q=-1.6\cdot 10^{-19} C[/tex] is the charge
[tex]v=3.0\cdot 10^6 m/s[/tex] is the speed
[tex]B=0.020 T[/tex]
so, the magnetic force is
[tex]F=(-1.6\cdot 10^{-19} C)(3.0\cdot 10^6 m/s)(0.020 T)=9.6\cdot 10^{-15} N[/tex]
