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

The electron will move along the direction of the field with velocity `v'= v sin 30^(@) ` and radius r with the magnetic field as the axis.
`r'=(mv)/(Be)=(sin 30^(@))/(B) sqrt((2mV)/(e))`
`=((1)/(2))/(0.15) sqrt((2xx9xx10^(-31)xx2xx10^(3))/(1.6xx10^(-19)))` `=0.5xx10^(-3)m`
The electron will follow a helical path. Its pitch is,
`2 pi r' cot theta=2xx3.14xx10(0.5xx10^(-3))xx sqrt(3)`
`=5.44xx10^(-3)m=5.44 mm`