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 $60°$ 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 =$1.69\times {10}^{-19}\mathrm{C}$)

• A. $4cm$
• B. $2cm$
• C. $12cm$
• D. $5cm$

Answer 1

The correct option is A

$4cm$

Answer:(A)

Explanation for correct option:

When a charged particle is projected at an angle $\mathrm{\theta }$ to magnetic field, the component of velocity parallel to the field is $\mathrm{vcos\theta }$ while perpendicular to the field is $\mathrm{vsin\theta }$, so the particle will move in a circle of radius,\

Step 1:

$$\begin{gathered} \mathrm{r}=\frac{\mathrm{m}\left(\mathrm{vsin\theta }\right)}{\mathrm{qB}} \\ \mathrm{Given}: \\ \mathrm{m}=1.67\times {10}^{-27}\mathrm{Kg} \\ \mathrm{vsin\theta }=4\times {10}^5{\mathrm{ms}}^{-1} \\ \mathrm{q}=1.69\times {10}^{-19}\mathrm{C} \\ \mathrm{B}=0.3\mathrm{T} \\ \frac{(1.67\times {10}^{-27})\times \left(4\times {10}^5\times {\displaystyle \frac{\sqrt{3}}{2}}\right)}{1.6\times {10}^{-19}\times 0.3}=\frac{2\times {10}^{-2}}{\sqrt{3}} \end{gathered}$$

Step 2:

$$\begin{gathered} \mathrm{Time}\mathrm{period}:\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{vsin\theta }}=\frac{2\mathrm{\pi }\times \frac{2\times {10}^{-2}}{\sqrt{3}}}{4\times {10}^5\times \sin 60°} \\ =\frac{2\mathrm{\pi }\times \frac{2\times {10}^{-2}}{\sqrt{3}}}{4\times {10}^5\times {\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{2\mathrm{\pi }}{3}}\times {10}^{-7}} \end{gathered}$$

Step 3:

$$\begin{gathered} \mathrm{Pitch}:\mathrm{P}=\mathrm{vcos\theta T}=(4\times {10}^5)\times \cos 60°\times \frac{2\mathrm{\pi }}{3}\times {10}^{-7} \\ =\frac{4\mathrm{\pi }}{3}\times {10}^{-2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{m}\mathbf{\approx }\mathbf{4}\mathbf{m} \end{gathered}$$

Hence the correct option is (A)