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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 a- particle to describe a circle of same radius in the sam

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 a- particle to describe a circle of same radius in the same field?

Option: 1

2mV


Option: 2

1MeV


Option: 3

0.5MeV


Option: 4

4MeV


Answers (1)

best_answer

According to the principle of circular motion in a magnetic field

\begin{aligned} & F_C=F_m \Rightarrow \frac{m v^2}{R}=q V B \\ & \Rightarrow R=\frac{m v}{q B}=\frac{P}{q B}=\frac{\sqrt{2 m \cdot K}}{q B} \\ & R_a=\frac{\sqrt{2(4 m) K^{\prime}}}{2q B} \\ & \frac{R}{R a}=\sqrt{\frac{K}{K^{\prime}}} \\ & \text { but } R=R_a \text { (given) } \\ & \text { Thus } K=K^{\prime}=1 \mathrm{MeV} . \end{aligned}

Posted by

Pankaj Sanodiya

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