In a certain region of space a uniform and constant electric field and a magnetic field parallel to each other are present. A proton is fired from a point `A` in the field with speed `v=4xx10^(4)m//s` at an angle of `alpha` with the field direction. The proton reaches a point `B` in the field where its velocity makes an angle `beta` with the field direction. If `(sinalpha)/(sinbeta)=sqrt(3)`. Find the electric potential difference between the points `A` and `B`. Take `m_(p)`(mass of proton) `=1.6xx10^(-27)kg` and `c`(magnitude of electronic charge)`=1.6xx10^(19)C`.

Options

A. `16 V`
B. `16//3 V`
C. `90 V`
D. `30 V`

Explanation

Work done by magnetic force is zero
so from work energy theorem
`(1)/(2)m_(p)v_(B)^(2)=(1)/(2)m_(p)v_(A)^(2)+qDeltaV`
and simultaneously there is no change of velocity component along the direction of perpendicular to electric and magnetic field.
`v_(A)sinalpha=v_(B)sinbeta`
After solving `Deltav=16` Volt