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An electron gun G emits electons of ener...

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`.

Text Solution

Verified by Experts

Kinetic energy of electron
`K=1/2mv^2=2keV`
`:.` Speed of electron `v=sqrt((2K)/m)`
`v=sqrt((2xx2x1.6xx10^-16)/(9.1xx10^-31))m/s`

`=2.65xx10^7m/s`
Since the velocity v of the electron makes an angle of `theta=60^@` with ethe magnetic field `B`, 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) vcostheta`
But for to be minimum `n=1`
Hence, `GS=P=(2pim)/(qB) vcostheta`
`B=B_(min)=(2pimvcostheta)/(q(GS)`
Substituting the values we have
`B_(min)=((2pi)(9.1xx10^-31)(2.65xx10^7)(1/2))/((1.6xx10^-19)90.1)`
or `B_(min)=(4.73xx10^-3T)`
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