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A proton of energy 2 MeV is moving perpe...

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

`2.5xx10^(-16)N`

B

`8xx10^(-11)N`

C

`2.5xx10^(-11)N`

D

`8xx10^(-12)N`

Text Solution

Verified by Experts

The correct Answer is:
4
`F=Bqv=Bqsqrt((2KE)/(m))( :'(1)/(2)mv^(2)=KE)`
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