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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)`A. `2.5X 10^(-10)` newtonB. `8X 10^(-11)` newtonC. `2.5X 10^(-11)` newtonD. `8X 10^(-12)` newton

Answer» Correct Answer - 4
`F=Bqv,v=sqrt((2KE)/(m))`


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