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

`B_(min) = 4.7 xx 10^(-3) T`

Explanation

`GS =V cos theta xx (nT) = V cos theta xx(n2p m)/(qB)`
`For B_(min) n = 1 so GS = (2pi)/(qB) mV cos 60^(@) = (pmV)/(qB_(min))`
` = (p)/(qB_(min)) sqrt(2mE)`
`B = (pi)/(q(GS)) sqrt(2mE) = (3.14)/(1.6 xx 10^(-19) xx 0.1`
`sqrt(2 .9 .1 xx 10^(-31) xx 2 x 1.6 xx 10^(-19) xx 10^(3))`
`B = 31.4sqrt((4 xx 9 xx 10^(-28))/(1.6 xx 10^(-19))) = 31.4 xx 1.5 xx 10^(-4)`
` = 4. 71 xx 10^(-3) T`

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