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`.

Explanation

`R = (mv sin theta)/(Bq), T = (2pi m)/(Bq)`
`p = v cos theta xx T = v cos theta.(2pi m)/(Bq)`
The particle will hit `S` if
`GS = np = nv cos theta.(2pi m)/(Bq)`, where `n = 1, 2,…`
`B = (nv cos theta. 2pi m)/(q(GS)), B` is minimum if `n = 1`
`B_(min) = (2pi cos theta m v)/(e(GS))`
`= (2pi cos theta sqrt(2mK))/(e(GS))`
`= (2pi xx cos 60^(@) sqrt(2 xx 9.1 xx 10^(-31) xx 2 xx 1.6 xx 10^(-16)))/(1.6 xx 10^(-19) xx 0.1)`
`=(2pi xx (1)/(2) xx 3 xx 8 xx 10^(-24))/(1.6 xx 10^(-20)) = 15 pi xx 10^(-4) T`