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  • A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude    <img alt="{\frac{\pi}{2}}\times 10^{-13}T" src="https://entrancecorner.oncodecogs.co

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^{-13}T .  The angle between the direction of magnetic field and velocity of proton is 60o

. The pitch of the helical path taken by the proton is ______ cm.

(Take, mass of proton = 1.6 × 10-27 kg and Charge on proton = 1.6\times10^{19} C).

 

Option: 1

40


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{B}=\frac{\pi}{2} \times 10^{-3} \\ & \mathrm{~K} . \mathrm{E} .=\frac{1}{2} \mathrm{mV}^2 \\ & \Rightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{KE}}{\mathrm{m}}} \end{aligned}

PItch = v cos 60° × time period of one rotation

\begin{aligned} & =v \cos 60^{\circ} \times \frac{2 \pi \mathrm{m}}{\mathrm{eB}} \\ & =\sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-9}}{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}} \end{aligned}

\begin{aligned} & =2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5} \\ & =4 \times 10^{-1} \mathrm{~m}=40 \mathrm{~cm} \end{aligned}

 

 

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