A proton of energy `2 MeV` is moving perpendicular to uniform magnetic field of `2.5T`. The form on the proton is `(Mp=1.6xx10^(-27)Kg` and `q=e=1.6xx10^(-19)C)`

Options

A. `3 xx 10^(-10)` N
B. `8 xx 10^(11)` N
C. `3 xx 10^(-11)` N
D. `8 xx 10^(-12)` N

Answer

D

Explanation

F = qvB
`E = K.E = (1)/(2)mv^(2)`
`sqrt((2E)/(m)) = v`
`F = q xx sqrt((2E)/(m)) xx B`
`F = 1.6 xx 10^(-19) xx sqrt((2E)/(m)) xx 2.5`
`F = 4 xx 10^(-19) xx sqrt((2 xx 2 xx 1.6 xx 10^(-19) xx 10)/(1.66 xx 10^(-27)))`
`F = 4 xx 10^(-19) xx 2 xx sqrt(10^(14))`
`F = 8 xx 10^(-19) xx 10^(7)`
`F = 8 xx 10^(-12)`N