A beam of protons with a velocity of $4 \times 10^{5} m/s$ enters a uniform magnetic field of 0.3 T. The velocity makes an angle of $60°$ with the magnetic field. The radius of the helical path taken by the proton beam and the pitch of the helix are (nearly)

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2.1 cm, 4.9 cm

12 cm, 44 cm

1.2 cm, 4.4 cm

answer is B.

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Detailed Solution

$v = 4 \times 10^{5} m/s, B = 0.3 T, θ =60° m_{p} = 1.6 \times 10^{- 27} kg q_{p} = 1.6 \times 10^{- 19} kg r = ?, pitch = ? r = \frac{{mv}_{⊥}}{Bq} = \frac{1.6 \times 10^{- 27} \times 4 \times 10^{5} \times \sin 60 °}{0.3 \times 1.6 \times 10^{- 19}} = \frac{4 \times 10^{5} \times 10^{- 7}}{3} \times \sqrt{\frac{3}{2}} = \frac{2}{\sqrt{3}} \times 10^{- 2} = 1.1547 cm Pitch, h = \frac{v_{∥} \times 2 π r}{v_{⊥}} = \frac{2 π rv \cos θ}{v \sin θ} = 2 π r \cot θ = 2 \times \frac{22}{7} \times 1.15 \times \cot 60 ° = \frac{44}{7} \times 1.1547 \times \frac{1}{\sqrt{3}} = 4.2 cm$