A proton with energy of 2 MeV enters a region of a uniform magnetic field of 2.5 T normally. The magnetic force on the proton is (Take mass of proton to be

Options

A. `3xx10^(-12)N`
B. `3xx10^(-8)N`
C. `8xx10^(-12)N`
D. `3xx10^(-10)N`

Answer

C

Explanation

Here E = 2 MeV=`2xx1.6xx10^(-13)J,`
`B=2.5T,m=1.6xx10^(-27)kg, theta=90^(@),q=1.6xx10^(-19)C`
As E `=(1)/(2)mv^(2)orv=sqrt((2e)/(m))`
Now F =qv`Bsintheta=qxxsqrt((2e)/(m))xxBxxsin90^(@)`
`=(1.6xx10^(-19))xxsqrt((2xx2xx1.6xx10^(-19))/(1.6xx10^(-27)))xx2.5xx1`
`= 8 xx 10^(-12)` N