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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 (Take, mass of the proton $= 1.67 \times 10^{- 27} kg$ and charge of the proton $= 1.6 \times 10^{- 19} C$ )

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2 cm

4 cm

5 cm

12 cm

answer is B.

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Detailed Solution

Pitch of helical path $= \frac{2 π m}{Bq} \cdot vcos θ = \frac{2 \times 3.14 \times 1.67 \times 10^{- 27} \times 4 \times 10^{5} \times \cos 60^{0}}{(1.6 \times 10^{- 19}) (0.3)}$
= 0.04 m = 4 cm

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