An electron emitted by a heated cathode and accelerted through a potential difference of `2.Kv,` enters a region of uniform magnetic field of `0.15T`. Determine the trajectory of the electron if the field makes an angle `30^(@)` with the initial velocity.

Explanation

When magnetic field makes an anlge `30^(@)` with the initial velocity i.e. `theta=30^(@)`
Then `V^(1)=V sin theta =sqrt((2ev)/(m))=sin 30^(@)`
`V^(1)=(8)/(3)xx10^(7)xx(1)/(2)=(4)/(3)xx10^(7) m//s`
The radius of the helical path is
`r=(mV^(1))/(Be)`
`=(9xx10^(-31)xx((4)/(3)xx10^(7)))/(0.15xx1.6xx10^(-19))=0.5 mm`