Here, `B=6*5xx10^-4T`, `v=4*8xx10^6m//s`, `e=1*6xx10^(-19)C`, `theta=90^@`, `m=9*1xx10^(-31)kg`, `r=?`
(i) Force on the moving electron due to magnetic field will be, `F=evBsin theta`.
The direction of this force is perpendicular to `vecv` as well as `vecB` therefore this force will only change the direction of motion of the electron without affecting its velocity i.e. this force will provide the centripetal force to the moving electron and hence, the electron will move on the circular path. If r is the radius of circular path traced by electron, then
`evBsin90^@=mv^2//r` or `r=(mv)/(Be)=((9*1xx10^(-31))xx(4*8xx10^6))/((6*5xx10^-4)xx(1*6xx10^(-19)))=4*2xx10^-2m=4*2cm`