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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

`2.5X 10^(-10)` newton

B

`8X 10^(-11)` newton

C

`2.5X 10^(-11)` newton

D

`8X 10^(-12)` newton

Text Solution

Verified by Experts

The correct Answer is:
4

`F=Bqv,v=sqrt((2KE)/(m))`
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