Answer to Question #111472 in Electricity and Magnetism for Yayra

Question

A particle with charge 6.80 micro coloumbsis moving with velocity

Answer 1

Consider the magnetic field $B$ perpendicular to the velocity $v$ and the charge on the particle $q=6.80\mu C$

So, the force acting upon the charged particle upon entering the magnetic field$=q(\vec v\times \vec B)$