A beam of electrons enters a uniform magnetic field of `0*3T` with a velocity of `4xx10^5ms^-1` at an angle of `60^@` to the field. Find the radius of the helical path taken by the beam. Also find the pitch of the helix (distance travelled by a proton parallel to the magnetic field during one period of rotation). Mass of proton is `1*67xx10^(-27)kg`.

Answer

`1*2cm`; `4*35cm`

Explanation

Let `v_1`, `v_2` be the rectangular components of proton's velocity perpendicular and parallel to the direction of magnetic field. Then
`v_1=v sin 60^@=(4xx10^5)xxsqrt3//2`
`=3*464xx10^5ms^-1`
`v_2=v cos 60^@=(4xx10^5)xx1//2`
`=2xx10^5ms^-1`
The component velocity `v_1` makes the proton to move along a circular path of radius r, given by
`qv_1B=(mv_1^2)/(r)`
or `r=(mv_1)/(qB)=((1*67xx10^(-27))xx(3*464xx10^5))/((1*6xx10^(-19))xx0*3)`
`=1*2xx10^-2m=1*2cm`
Time period of revolution of proton
`T=(2pir)/(v_1)=(2xx3*142xx1*2xx10^-2)/(3*464xx10^5)`
`=21*75xx10^-8s`
The component velocity `v_2` of proton, makes the proton to move along the direction of magnetic field.
Therefore, due to velocity components `v_1` and `v_2`, the proton moves on the helical path.
Pitch of the helix, `p=v_2xxT`
`=(2xx10^5)xx(21.75xx10^-8)`
`=4*35xx10^-2m=4835cm`