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

Magnetic field strength, `B = 0.15 T`
Charge on the electron, `e = 1.6 xx 10^(-19)C`
Mass of the electron, `m = 9.1 xx 10^(-31)kg`
Potential differece, `V = 2.0 kV = 2 xx 10^(3)V`
Thus, kinetic energy of the electron `=eV`
`rArr eV = (1)/(2)mv^(2)`
`v = sqrt((2eV)/(m))` ...(1)
where,
`v =` velocity of the electron
(a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron trances a circular path of radius r.
Magnetic force on the electron is given by the relation, B ev
Centripetal force `= (mv^(2))/(r)`
`:. Bev = (mv^(2))/(r)`
`r = (mv)/(Be)` ...(2)
From equations (1) and (2), we get
`r =(m)/(Be) [(2eV)/(m)]^(1//2)`
`= (9.1 xx 10^(-31))/(0.15 xx 1.6 xx 10^(-19)) xx ((2xx 1.6 xx 10^(-19) xx 2 xx 10^(3))/(9.1 xx 10^(-31)))^(1//2)`
`=100.55 xx 10^(-5)`
`=1.01 xx 10^(-3)m`
`1 mm`
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
(b) When the field makes an angle `theta` of `30^(@)` with initial velocity, the initial velocity will be,
`v_(1) = v sin theta`
From equation (2), we can write the expression for new radius as:
`r_(1) = (mv_(1))/(Be)`
`=(mv sin theta)/(Be)`
`=(9.1 xx 10^(-31))/(0.15 xx 1.6 xx 10^(-19)) xx ((2xx 1.6 xx 10^(-19) xx 2 xx 10^(3))/(9.1 xx 10^(-31)))^(1//2)xx sin 30^(@)`
`=0.5 xx 10^(-3)m`
`=0.5mm`
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.