A proton with a kinetic energy of \(2.0\,eV\) moves into a region of uniform magnetic field of magnitude \(\frac{\pi}{2} \times 10^{-3}\,T\). The angle between the direction of magnetic field and velocity of proton is \(60^{\circ}\). The pitch of the helical path taken by the proton is \(..........cm\) (Take, mass of proton \(=1.6 \times 10^{-27}\,kg\) and Charge on proton \(=1.6 \times 10^{-19}\,kg)\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(C) \(40\)

Step-by-step Solution

Detailed explanation

\(B=\frac{\pi}{2} \times 10^{-3}\)

\(K E=\frac{1}{2} m V^2\)

\(\Rightarrow V=\sqrt{\frac{2 K E}{m}}\)

\(=v \cos 60^{\circ} \times \frac{2 \pi m}{e B}\)

\(=\sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}} \times \cos 60^{\circ} \times \frac{2 \pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \frac{\pi}{2} \times 10^{-3}}\)

\(=2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5}\)

\(=4 \times 10^{-1} m =40\,cm\)

Same subject

Match List-\(I\) with List-\(II\) :-

List - \(I\) (EM-Wave)	List - \(II\) (Wavelength) (Range)
\((A)\) infra-red	\((i)\) \(<10^{-3} \mathrm{~nm}\)
\((B)\) Ultraviolet	\((ii)\) \(400 \mathrm{~nm}\) to \(1 \mathrm{~nm}\)
\((C)\) X-rays	\((iii)\) \(1 \mathrm{~mm}\) to \(700 \mathrm{~nm}\)
\((D)\) Gamma rays	\((iv)\) \(1 \mathrm{~nm}\) to \(10^{-3} \mathrm{~nm}\)

Choose the correct answer from the options given below:

JEE Mains 2024 Hard

Explore more questions on app

From JEE Mains

Explore more questions on app