The magnetic field in a certain region of space is given by `B=8.35 xx10^(-2) hati T.`A proton is shot into the field with velocity `v=(2xx10^(5)hati+4xx10^(5)hatj)m//s.` the proton follows a helical path in the field. The distance moved by proton in the x-direction during the period of one revolution in the yz-plane will be ( mass of proton`=1.67 xx 10^(-27) kg)`

Options

A. 0.053 m
B. 0.136 m
C. 0.157 m
D. 0.236 m

Answer

C

Explanation

`vecv = (2 xx 10^(5) hati + 4 xx 10^(5) hatj) m//s`
`vecB = 8.35 xx 10^(-2) hati T`
Pitch = `v_(i) T = (2pi m V_(1))/(qB) " "` [As `T = (2pi m)/(qB)]`
`= (2 xx 3.14 xx 1.67 xx 10^(-27) xx 2 xx 10^(5) )/(1.6 xx 10^(-19) xx 8.35 xx 10^(-2)) = 0.157` m