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. $4\mathrm{cm}$
• B. $2\mathrm{cm}$
• C. $12\mathrm{cm}$
• D. $5\mathrm{cm}$

Answer 1

The correct option is A

$4\mathrm{cm}$

The explanation for correct option:

Option (A):

When a charged particle is projected at an angle $\mathrm{\theta }$ to a 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: Given

Speed of proton $\mathrm{v}$: $4\times {10}^5{\mathrm{ms}}^{-1}$

magnetic field $\mathrm{B}$= $0.3\mathrm{T}$

Mass of the proton $\mathrm{m}$ = $1.67\times {10}^{-27}\mathrm{kg}$

charge of the proton$\mathrm{q}$ =$1.69\times {10}^{-19}\mathrm{C}$

Radius of circle = $\mathrm{r}$

$\mathrm{\theta }=\mathrm{angle}\mathrm{between}\mathrm{magnetic}\mathrm{feild}\mathrm{and}\mathrm{velocity}$

Step 2: To calculate the radius

The radius of circle of particle in a cyclotron is given by

$$\begin{gathered} \mathrm{r}=\frac{\mathrm{m}\left(\mathrm{vsin\theta }\right)}{\mathrm{qB}} \\ =\frac{\left(1.67\times {10}^{-27}\right)\times \left(4\times {10}^5\times \sin {60}^{°}\right)}{1.6\times {10}^{-19}\times 0.3} \\ =\frac{\left(1.67\times {10}^{-27}\right)\times \left(4\times {10}^5\times {\displaystyle \frac{\sqrt{3}}{2}}\right)}{1.6\times {10}^{-19}\times 0.3} \\ =\frac{2\times {10}^{-2}}{\sqrt{3}}\mathrm{m} \end{gathered}$$

Step 2: To calculate the time period

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

Step 3: To calculate the pitch

$$\begin{gathered} \mathrm{Pitch}:\mathrm{P}=\mathrm{vcos\theta T} \\ =(4\times {10}^5)\times \cos 60°\times \frac{2\mathrm{\pi }}{3}\times {10}^{-7} \\ =\frac{4\mathrm{\pi }}{3}\times {10}^{-2}=4.35\times {10}^{-2}\mathrm{m} \end{gathered}$$

Hence, option (A) is the correct answer.