A proton is moving perpendicular to a uniform magnetic field of 2.5 tesla with 2MeV kinetic energy. The force on the proton is ______ .

Options

A. `3xx10^(-10)N`
B. `8xx10^(-11)N`
C. `3xx10^(-11)N`
D. `8xx10^(-12)N`

Answer

D

Explanation

`v=sqrt((2K)/m)`
k = 2 Mev
= `2xx1.6xx10^(-13)J`
`m=1.66xx10^(-27)kg`
`B=2.5Tq=1.6xx10^(-19)C`
`q=1.6xx10^(-19)C`
F = qvB
`thereforeF=qBsqrt((2k)/m)`
`thereforeF=2.5xx1.6xx10^(-19)sqrt((2xx2xx1.6xx10^(-13))/(1.66xx10^(-27)))`
= `4xx10^(-19)xx1.96xx10^(7)`
= `7.84xx10^(-12)~=8xx10^(-12)N`