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An alpha particles is projected with a v...

An `alpha` particles is projected with a velocity of `10^(7)` m/s, in a uniform magnetic field of induction, `1.2xx10^(-6)Wb//m^(2),` in a direction perpendicular to the field. If the lateral force action on the particle is `3.84xx10^(-18)N,` what is the change on the a particle ?

A

`1.6xx10^(-19)C`

B

`2.4xx10^(-19)C`

C

`3.2xx10^(-19) C`

D

`4xx10^(-19)C`

Text Solution

Verified by Experts

`F= qvB sin theta but theta =90^(@)`
`therefore q = (F)/(Bv) =(3.84xx10^(-18))/(1.2xx10^(-6) xx10^(7))`
`q =3.2 xx10^(-19) C`
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