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A beam of protons moving with a velocity $1.6 \times 10^5 \mathrm{~ms}^{-1}$ enter a uniform magnetic field of $\frac{\pi}{10} \mathrm{~T}$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $=1.6 \times 10^{-27} \mathrm{~kg}$)
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The correct answer is:
$1.6 \times 10^{-2} \mathrm{~m}$
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