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