A positive ion having charge `q= 3.2xx10^(-19)C,` enters a uniform magnetic field of induction ` 10^(-2)W//bm^(2),` in a direction perpendicular to the field. If the force exerted by the field on the ions is `1.6xx10^(-16)N,` then the speed with which the ion enters the field, is

Options

A. `5xx10^(3)m//s`
B. `5xx10^(4)m//s`
C. `8xx10^(5)m//s`
D. `6xx10^(4)m//s`

Answer

B

Explanation

`q=3.2 xx10^(-19) C, B=10^(-2) Wb//m^(2), theta =90^(@),`
`F =1.6 xx10^(-16) N`
`F =qvB sin theta`
`therefore v=(F)/(qB sin theta)=(1.6xx10^(-16))/(3.2xx10^(-19) xx10^(-02) xx sin 90^(@))`
`thereforev=1/2 xx10^(5) =5xx 10^(4) m//s`