An electron emitted by a heated cathode and accelerated through a p.d. of 2.0 kV enters a region with 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

(a) `V=2 times 10^(-3)V, theta=90^(@), B=0.15T`
`" "e/m=1.76 times 10^(11)Ckg^(-1)`
The path of the electron will be circular and of radius `r=(mv)/(eB)`
i.e., `" "r=v/((e/m)B) " and " v=sqrt((2eV)/m)`
`" "e/m=v^(2)/(2V)`
i.e., `" "r=v/B times (2V)/V^(2)=(2v)/(BV)`
and `" "r=(2.64 times 10^(7))/(1.76 times 10^(11) times 0.15)`
`" "r=1 times 10^(-3)m`
`" "=1mm`
(b) If the field makes an angle of `30^(@)`, then `v_("hor")=vcostheta=2.64 times 10^(7) times cos30^(@)`
i.e., `" "v_(H)=2.64 times 0.866 times 10^(7)=2.286 times 10^(7)ms^(-1)`
in the direction of B and the path will be helical.