An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 Kv, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity , (b) makes an angle of `30^@` with the intial velocity .

Explanation

v=2.0 kV `= 2.0 xx 10^3 v, B = 0.15 T`
`KE = eV = 1/2 mv^2 " " therefore v = sqrt(2eV)/(m)`
`v= sqrt(2xx 1.6 xx 10^(-19) xx 2 xx 10^3)/(9 xx10^(-31)) = sqrt((2 xx 1.6 xx 2 xx 10^(15))= 8/3 xx 10^7 ms^(-1)`
`(mv^2)/(Bq)=(9 xx 10^(-31) xx 8 xx 10^7)/(3 xx 0.15 xx 1.6 xx 10^(-19)) = 1 ` mm (transverse)
b . r=0.5 mm inclined through `30^@` with B .
Here v `sin theta` component is `_|_ = 2.3 xx 10^7 ms^(-1)`