A magnetic field of `0.40T` is applied on a proton moving with a velocity of `5xx10^(6) m*s^(-1)`. The magnetic field acts at a angle `30^(@)` with the direction of velocity of the proton. What will be the acceleration of the proton ? ( mass of proton `=1.6xx10^(27)kg)`.

Explanation

Force acting on the proton `F=qvB sin theta`.
`:.` Acceleration of the proton `=("force acting")/("mass of proton")=(qvB sin theta)/(m)`
Here, `q=1.6xx10^(-19)C, v=5xx10^(6) m*s^(-1)`
`B=0.40T, theta=30^(@) and m=1.6xx10^(-27) kg`.
`:.` Acceleration of the proton
`=(1.6xx10^(-19)xx5xx10^(6)xx0.40xx sin 30^(@))/(1.6xx10^(-27))`
`=10^(-14)m*s^(-2)`.