1.

A proton is moving perpendicular to a uniform magnetic field of 2.5 tesla with 2MeV kinetic energy. The force on the proton is ______ .

Answer»

`8XX10^(-11)`
`3XX10^(-10)`
`3xx10^(-11)`
`8xx10^(-12)`

Solution :`k=1/2mv^(2)`
`thereforev=SQRT((2k)/m)`
Now, F = Bqv
= `Bexxsqrt((2k)/m)`
= `2.5xx1.6xx10^(-19)sqrt((2xx2xx10^(6)xx1.6xx10^(-19))/(1.6xx10^(-27)))`
= `4xx10^(-19)xxsqrt(4xx10^(6)xx10^(8))`
= `4xx10^(-19)xx2xx10^(7)`
`thereforeF=8xx10^(-12)N`


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