Question:
A beam of protons with speed $4 \times 10^{5} \mathrm{~ms}^{-1}$ enters a uniform magnetic field of $0.3 \mathrm{~T}$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.69 \times 10^{-19} \mathrm{C}$
Correct Option: , 2
Solution:
Pitch $=\frac{2 \pi \mathrm{m}}{\mathrm{qB}} \mathrm{v} \cos \theta$
Pitch $=\frac{2(3.14)\left(1.67 \times 10^{-27}\right) \times 4 \times 10^{5} \times \cos 60}{\left(1.69 \times 10^{-19}\right)(0.3)}$
Pitch $=0.04 \mathrm{~m}=4 \mathrm{~cm}$