A electron which is accelerated by a potential difference of 2.0 kV will have a kinetic energy gained 2000 eV.
E = 1/2mev2 = 2000 × 1.6 × 10−19
v = √

4×1.6×10−16

9×10−31

= 2.66 × 107 m s−1
(a) When the electron enters in the uniform magnetic field which is normal to the velocity of electron the electron follows a circular path of radius.
r =

mv

Bq

=

9.1×10−31×2.66×107

0.15×1.6×10−19

= 99.75 × 10−5 = 1 mm
(b) When the magnetic field makes an angle 30° with the initial velocity, the trajectory of the electron becomes helical. radius of the helical path is

r =

mvsinθ

Bq

r = 99.75 × 10−5×

1

2

r = 49. 875 × 10–5 m ≈ 0.5 mm
v = v cosθ = 2.66 × 107 cos 30 = 2.3 × 107ms–1
Pitch of the helical path is
Pitch = T × v cos θ =

2πm

Bq

× v cos θ =

2π×9×10−31×2.66×107

0.15×1.6×10−19

×

√3

2

= 542.5 × 10−5 m
so, pitch = 5.42 mm.