Here the proton beam passes without deviation, hence
`F_b = F_e impliesBev=eE`
`:. V=E/B=(120xx10^3)/(50xx10^-3)=2.4xx10^6 ms^-1`
Number of protons reaching per second
`(nq)/t=0.8xx10^-3`
or `(nxx1.6xx10^-19)/1=0.8xx10^-3 or n=5xx10^15`
Total mass of proton reaching per second
`=(5xx10^15)xx(1.67xx10^-27)=8.35xx10^-12kg`
`Force= ("Change in momentum")/("time")`
`=[(8.35xx10^-12)(2.4xx10^6)]//1`
`20.04xx10^-6N=20.04muN`

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