Q 12-04-190JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A beam of protons with speed $4 \times 10^{5}\ \text{m s}^{-1}$ enters a uniform magnetic field of $0.3\ \text{T}$ at an angle $60^\circ$ to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton $= 1.67 \times 10^{-27}\ \text{kg}$, charge of the proton $= 1.69 \times 10^{-19}\ \text{C}$)
Answer: (D) $4\ \text{cm}$
Pitch = (velocity component along $B$) $\times$ (time period):
$$p = v\cos60^\circ\cdot\frac{2\pi m}{qB} = \frac{2\pi\times1.67\times10^{-27}\times4\times10^{5}\times0.5}{1.69\times10^{-19}\times0.3}$$
$$p \approx 0.041\ \text{m} \approx 4\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics