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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 600 to the magnetic field. The pitch of the resulting helical path of protons is close to: (in cm) (Mass\; \:o\! f \; \:the \ :proton=1.67 \times 10^{-27}\:kg, charge \: \; o\! f \:proton=1.69\times 10^{-19}C
Option: 1 4
Option: 2 5
Option: 3 12
Option: 4 2

Answers (1)


\begin{array}{l} \text { Pitch }=\frac{2 \pi \mathrm{m}}{\mathrm{qB}} \mathrm{v} \cos \theta \\ \\ \text { 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)} \\\\ \text { Pitch }=0.04 \mathrm{~m}=4 \mathrm{~cm} \end{array}

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