A electron which is accelerated by a potential difference of 2.0 kV will have a kinetic energy gained 2000 eV. E = 1/2$m_{ev}^2$ = 2000 × 1.6 × $10^{-19}$ v = $\sqrt{\frac{4 \times 1.6 \times 10^{-16}}{9 \times 10^{-31}}}$ = 2.66 × $10^7$ 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 = $\frac{mv}{Bq}$ = $\frac{9.1 \times 10^{-31} \times 2.66 \times 10^7}{0.15 \times 1.6 \times 10^{-19}}$ = 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 = $\frac{mv \sin \theta}{Bq}$ r = 99.75 × $10^{-5} \times \frac{1}{2}$ r = 49. 875 × $10^{-5}$ m ≈ 0.5 mm v = v cosθ = 2.66 × $10^7$ cos 30 = 2.3 × $10^7 \, \mathrm{m} \, \mathrm{s}^{-1}$ Pitch of the helical path is Pitch = T × v cos θ = $\frac{2 \pi m}{Bq}$ × v cos θ = $\frac{2 \pi \times 9 \times 10^{-31} \times 2.66 \times 10^7}{0.15 \times 1.6 \times 10^{-19}}$ × $\frac{\sqrt{3}}{2}$ = 542.5 × $10^{-5}$ m so, pitch = 5.42 mm.