An electron gun `G` emits electons of energy `2 keV` travelling in the positive x-direction. The electons are required to hit the spot `S` where `GS=0.1 m`, and the line `GS` makes an angle of `60^@` with the x-axis as shown in figure. A uniform magnetic field `B` parallel to `GS` exists in the region outside the electron gun.

find the minimum value of `B` needed to make the electrons hit `S`.

Answer

3

Explanation

`(a)` Let us revolve the velocity two rectangular components `V_(1)(V cos theta ) ` and `V_(2)(=V sin 60^(@))`, V component of velocity is responsible to move the charge particle in the direction of the magnetic field whereas `V_(2)` component is responsible for rotating the charged particle in circular motion. The overall path is helical, the condition for the charged particle to strike S with minimum vlaue of B is pitch of Helix `=GS`
`TxxV_(1)=GSimplies(2pim)/(qB)xxv cos 60^(@)=0.1(1)/(mv^(2))=E`
`=(2pi)/(qxx0.1)xxsqrt(2mE)xxcos 60^(@)=(2xx3.14)/(1.6xx10^(-19)xx0.1)`
`sqrt(2xx9.1xx10^(-31)xx2xx10^(3)xx1.6xx10^(-19))xx(1)/(2)`
`=(149.8)/(10^(-19))xx0.316xx10^(-23)=47.37xx10^(-4)`
`=4.737xx10^(-3)T`