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  • A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in th

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?

Option: 1

2 \mathrm{MeV}


Option: 2

1 \mathrm{MeV}


Option: 3

0.5 \mathrm{MeV}


Option: 4

4 \mathrm{MeV}


Answers (1)

best_answer

According to the principal of circular motion in a magnetic field

\mathrm{ F_{\mathrm{c}}=F_{\mathrm{m}} \Rightarrow \frac{\mathrm{mv}^2}{\mathrm{R}}=\mathrm{qVB} \\ }

\mathrm{ \Rightarrow \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m} \cdot \mathrm{k}}}{\mathrm{qB}} \\ }

\mathrm{ \mathrm{R}_\alpha=\frac{\sqrt{2(4 \mathrm{~m}) \mathrm{K}^{\prime}}}{2 \mathrm{qB}} \\ }
\mathrm{ \frac{\mathrm{R}}{\mathrm{R}_\alpha}=\sqrt{\frac{\mathrm{K}}{\mathrm{K}^{\prime}}} \\ }
\mathrm{ \text { but } R=R_\alpha \text { (given) } \\ }

\mathrm{ \text { Thus } K=K^{\prime}=1 \mathrm{MeV} }.
 

Posted by

Kuldeep Maurya

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