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 \(=1.69 \times 10^{-19}\,C\))

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Answer & Solution

Correct Answer

(B) \(4\)

Detailed explanation

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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