An electron emmited by a heated cathode and accelerated through a potential difference of `2*0kV` enters a region with a uniform magnetic field of `0*15T`. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity (b) makes an angle of `30^@` with the initial velocity.

Explanation

Here, `V=2*0kV=2*0xx10^3V` , `B=0*15T`. Let e, m be the charge and mass of the electron. If v is the velocity gained by electron when accelerated under a potential difference V, then
`eV=1/2mv^2` or `v=((2eV)/(m))^(1//2)=((2xx1*6xx10^(-19)xx2*0xx10^3)/(9*0xx10^(-31)))^(1//2)=(8xx10^7)/(3)m//s`
(a) Force on electron due to transverse magnetic field is `=Bev`, which is perpendicular to `vecB` as well as `vecv`. It will provide the required centripetal force to the electron for its circular motion. Therefore, the trajectory of electron in magnetic field is circular. Its radius r can be given by
`Bev=(mv^2)/(r)` or `r=(mv)/(Be)=(9xx10^(-31)xx(8xx10^7//3))/(0*15xx1*6xx10^(-19))=10^-3m=1mm`
(b) When electron makes an angle `theta` with the direction of magnetic field, its component velocity perpendicular to the field, `v_1=v sin 30^@=8/3xx10^7xx1/2=4/3xx10^7m//s`. Due to this velocity, the force acting on the electron due to magnetic field will be providing the required centripetal force, hence the electron will describe a circular path of radius `r_1`. The component velocity along the magnetic field `=vcos30^@=8/3xx10^7xxsqrt3/2=(4sqrt3)/(3)xx10^7m//s`. Due to this velocity, no force acts on the electron in magnetic field, and electron moves without change in its speed along the magnetic field. Therefore, due to two perpendicular component velocities, the electron describes a helical path.
The radius of the helical path, `r=(mv_1)/(Be)=(9xx10^(-31)xx(4//3)xx10^7)/(0*15xx1*6xx10^(-19))=0*5xx10^-3m=0*5mm`