An electron gun ` G` emits electrons of energy `2ke V` travelling in the positive ` x- direction` . The electrons are required to hit the spot `S` where `GS = 0.1m`, and the line `GS` makes an angle of `60^(@)` with the ` x-axis` as shown in the fig. A uniform magnetic field `vec(B)` parallel to `GS` exists in the region outside the electron gun. Find the `GS` exists in the region outside the electron gun. Find the minimum value of `B` needle to make the electron hit `S`.

Answer

B, C

Explanation

Let us resolvethe velocity into two rectangular components `v_(1) ( = vcos 60^(@)) and v_(2) (= v sin 60^(@)). V_(1)` components of velocity is responsible to move the charge particle in the direction of the magnetic field whereas ` v_(2)` component is responsible for revolving the charged particle in circular motion . The overall path is helical. The condition for the charged particle to strike ` S` with minimum value of `B` is that Pitch of Helix `= GS`
` Txxv_(1) = GS rArr (2 pim)/(q^(B))xx v cos 60^(@) = 0.1`
` B = ( 2 pi mv cos 60^(@))/( q xx 0.1)`
But ` (1)/(2) mv^(2) = E rArr v = (sqrt(2E))/( m)`
:. ` B = ( 2 pi m)/( qxx0.1)xx( sqrt(2E))/(m) xx cos 60^(@)`
` = ( 2 pi)/( qxx0.1)xx sqrt(2mE)xx cos 60^(@) = ( 2xx 3.14)/( 1.6xx10^(-19)xx0.1)`
` sqrt ( 2xx9.1xx10^(-31)xx2xx10^(3)xx1.6xx10^(-19))xx1/2`
` (149.8)/(10^(-19))xx0.316xx10^(-23) = 47.37xx10^(-4)`
` = 4.737xx10^(-3) T`