A beam of protons with speed
Question:

A beam of protons with speed $4 \times 10^{5} \mathrm{~ms}^{-1}$ enters a uniform magnetic field of $0.3 \mathrm{~T}$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.69 \times 10^{-19} \mathrm{C}$ )

  1. (1) $2 \mathrm{~cm}$

  2. (2) $5 \mathrm{~cm}$

  3. (3) $12 \mathrm{~cm}$

  4. (4) $4 \mathrm{~cm}$


Correct Option: , 4

Solution:

(4) Pitch $=(v \cos \theta) T$ and $T=\frac{2 \pi m}{q B}$

$\therefore$ Pitch $=(V \cos \theta) \frac{2 \pi m}{q B}$

$=\left(4 \times 10^{5} \cos 60^{\circ}\right) \frac{2 \pi}{0.3}\left(\frac{1.67 \times 10^{-27}}{1.69 \times 10^{-19}}\right)=4 \mathrm{~cm}$

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