Q 12-04-191JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
A region of length $l$ contains a uniform magnetic field of $0.3\ \text{T}$ directed along its length. A proton enters the region with velocity $4 \times 10^{5}\ \text{m s}^{-1}$ making an angle $60^\circ$ with the field. If the proton completes $10$ revolutions by the time it crosses the region, $l$ is close to (mass of proton $= 1.67 \times 10^{-27}\ \text{kg}$, charge of the proton $= 1.6 \times 10^{-19}\ \text{C}$)
Answer: (C) $0.44\ \text{m}$
The proton moves on a helix whose axis is along $B$. Pitch:
$$p = v\cos60^\circ\cdot\frac{2\pi m}{qB} = \frac{2\pi\times1.67\times10^{-27}\times2\times10^{5}}{1.6\times10^{-19}\times0.3} \approx 0.0437\ \text{m}$$
Ten revolutions: $l = 10p \approx 0.44$ m.
Solution by Sreeraj P, M.Sc Physics