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

Force on a charged particle of charge, moving in an external magnetic field B with a velocity v is given by:
`F = qvB sin theta` ,
where `theta` is the angle between v and B .
Since the magnetic field is perpendicular to initial velocity of electron i.e. `theta = 90^(@)`. Therefore, the electron will move in a circular path.
` B = 0.15 T`
Potential difference through which electron is accelerated,` V = 2000 V`
Ife is the charge and m is the mass of electron, its kinetic energy is given by
`K = eV = 1/2 mv^(2)`
` :. " " v^(2) = (2eV)/m`
`rArr " " v = sqrt((2eV)/m) ` ....(i)
`rArr " " v = sqrt((2 xx 1.6 xx 10^(-19) xx 2000)/(9.1 xx 10^(-31))) `
` v = ( 8xx 10^(7))/3`
` v = 2.66 xx 10^(7) ms^(-1)`
The necessary centripetal force for circular motion of the electron will be provided by the magnetic force. Ifr is the radius of circular path of electron, then:
` Bev = (mv^(2))/r`
` rArr" " r = (mv)/(Bq) `
or ` r = (9.1 xx 10^(-31) xx 2.66 xx 10^(7))/(0.15 xx 1.6 xx 10^(-19))`
` = (91 xx 83.86 xx 10^(-5))/(15 xx 1.6) 10^(-3) m`
` = 1.008 xx 10^(-3) m`
` 1 mm`
(b) When magnetic field makes an angle 30 with the initial velocity, there will be two components of velocity. One along the field and one perpendicular to the field.
Component of velocity perpendicular to the field.
`v. = v sin theta = sqrt((2eV)/m) sin 30^(@) ["From" (i)]`
`rArr " " v. = 8/3 xx 10^(7) xx 1/2`
` = 4/3 xx 10^(7) m s^(-1)`
The electron will move in a circular path due to this component of velocity.
Component of velocity along the field:
` v.. = v cos theta`
`= sqrt((2eV)/m) cos 30^(@) ["From"(i)]`
` = 8/3 xx 10^(7) xx sqrt3/2`
` = (4sqrt3)/3 xx 10^(7) ms^(-1) = 2.3 xx 10^(7) ms^(-1)`
The electron will move in a straight line path due to this component as the magnetic field on the electron due to parallel component of velocity will be zero. So, the combination of circular motion and straight line motion will result in a helical path of electron in the magnetic field.
The radius of the helical path is:
` r = (mv.)/(Be)`
` = (9xx10^(-31) xx (4/3 xx 10^(7)))/(0.15 xx 1*6 xx 10^(-19)) `
` = 0.5` mm