When a proton of KE = 1.0 MeV moving towards North enters a magnetic field (directed along East), it accelerate with an acceleration, `a=10^(12)m//s^(2)`. The magnitude of the magnetic field is

Options

A. 0.71 mT
B. 7.1 mT
C. 71 mT
D. 710 mT

Answer

A

Explanation

Kinetic energy of proton = `1/2mv^(2)`
`1xx10^(6)xx10^(-19)=1/2xx1.6xx10^(-27)xxv^(2)`
`thereforev^(2)=(2xx10^(-13))/10^(-27)`
`thereforev^(2)=2xx10^(14)`
`thereforev=sqrt2xx10^(7)m//s`
Now, force acting on proton moving in magnetic field,
F = Bqv
`thereforema=Bqv`
`thereforeB=(ma)/(qv)=(1.6xx10^(-27)xx10^(12))/(1.6xx10^(-19)xx1.414xx10^(7))`
`thereforeB=0.7072xx10^(-3)T`
`thereforeB~~0.71mT`