In the previous question, if the pitch of the helical path is equal to the maximum distance of the particle from the x-axis

Options

A. `cos theta = (1)/(pi)`
B. `sin theta = (1)/(pi)`
C. `tan theta = (1)/(pi)`
D. `tan theta = pi`

Answer

D

Explanation

`p = ((2pi m)/(Bq))v cos theta`
`r = (mv sin theta)/(Bq)`
Maximum distance of the particle from `X`-axis `= 2r`
`p =2r`
`((2pi m)/(Bq)) v cos theta = (2mv sin theta)/(Bq)`
`tan theta = pi`