The figure shows a region of length 'l' with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity $$4 \times 10^5$$ m s$$^{-1}$$ making an angle 60$$°$$ with the field. If proton completes 10 revolution by the time it cross the region shown, 'l' is close to (mass of proton $$= 1.67 \times 10^{-27}$$ kg, charge of the proton $$= 1.6 \times 10^{-19}$$ C):

Time period of one revolution: $$T = \frac{2\pi m}{qB}$$

Pitch of the helical path: $$p = v \cos\theta \cdot T = \frac{2\pi m v \cos\theta}{qB}$$

Total length traveled along the field: $$l = N \cdot p = \frac{2\pi m v N \cos\theta}{qB}$$

$$l = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^5 \times 10 \times \cos 60^\circ}{1.6 \times 10^{-19} \times 0.3}$$

$$l = \frac{2 \times 3.14 \times 1.67 \times 4 \times 10 \times 0.5 \times 10^{-22}}{0.48 \times 10^{-19}} = \frac{41.95 \times 10^{-21}}{4.8 \times 10^{-20}} \approx 0.44\ \text{m}$$