First slide

Motion of a charged particle

Question

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

Moderate

A

12cm
B

4cm
C

2cm
D

5cm

Solution

$\begin{array}{l} p i t c h = V \cos θ . T = V \cos θ \times \frac{2 π m}{q B} \\ \Rightarrow p i t c h = \frac{4 \times 10^{5} \times \cos 60 º \times 2 π \times 1.67 \times 10^{- 27}}{1.69 \times 10^{- 19} \times 0.3} \\ \Rightarrow p i t c h = \frac{1.67 \times 4 π}{1.69 \times 0.3} \times 10^{- 3} m ≃ 4 c m \end{array}$

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