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

Kinetic energy of electron, `K=1/2mv^2=2keV`
`:.` Speed of electron, `v=sqrt((2k)/m)=sqrt((2xx2xx1.6xx10^-16)/(9.1xx10^-31)) ms^-1`
`=2.65xx10^7ms^-1`
Since the velocity `(vecv)` of the electron makes an angle of `theta=60^@` with the magnetic field `vecB`, the path will be a helix.
So, the particle will hit S if GS=nP, Here, n=1,2,3,....
p=pitch of helix `=(2pim)/(qB) v cos theta`
`implies GS=(n 2pi mvcos theta)/(qB)`
`implies B=(n2pimvcos theta)/(q(GS))`

But for B to be minimum, `n=1`
`B_(min)=(2pimvcos theta)/(q(GS))`
Subtituting the values, we have
`B_(min)=((2pi)(9.1xx10^-31)(2.65xx10^7)(1/2))/((1.6xx10^-19)(0.1))`
or `B_(min)=4.73xx10^-3T`