Assume electron speed to be 3$\times$10$^{7}$ m/s.

Given:

$B = 0.23 \times 10^{-2} Wb/m^2$

$\theta = 45^0$

$v = 3 \times 10^7 m/s$

$R = ?$

$P = ?$

Solution:

$\begin{aligned} R &= \frac{mv \ sin\theta}{eB} \\ &= \frac{9.1 \times 10^{-31} \times 3 \times 10^7}{1.6 \times 10^{-17} \times 0.23 \times 10^{-2}} \times sin 45^0 \\ &= 52.46 \times 10^{-3}m \\ &= 5.246 cm \end{aligned}$

$\begin{aligned} \text{Pitch P} &= \frac{2 \pi m}{Be}(v cos \theta) \\ &= \frac{2 \times 3.14 \times 9.1 \times 10^{-31} \times 3 \times 10^7 \times cos 45^0}{0.23 \times 10^{-2} \times 1.6 \times 10^{-19}} \\ &= 3.9 mm \end{aligned}$