A beam of protons with speed $4 \times 10^{5}\, ms ^{-1}$ enters a uniform magnetic field of $0.3\, T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $=1.67 \times 10^{-27}\, kg$, charge of the proton $=1.69 \times 10^{-19}\,C$)
JEE MAIN 2020, Medium
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Pitch $=\frac{2 \pi m }{ qB } v \cos \theta$
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 $=0.04 m =4 cm$
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