Moving Charges and Magnetism - Result Question 5
5. 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 $\alpha$-particle to describe a circle of same radius in the same field?
[2012M]
(a) $2 MeV$
(b) $1 MeV$
(c) $0.5 MeV$
(d) $4 MeV$
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Answer:
Correct Answer: 5. (b)
Solution:
- (b) According to the principal 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 _{\alpha}=\frac{\sqrt{2(4 m) K^{\prime}}}{2 q B} \end{aligned} $
$\frac{R}{R _{\alpha}}=\sqrt{\frac{K}{K^{\prime}}}$
but $\quad R=R _{\alpha}$ (given)
Thus $K=K^{\prime}=1 MeV$
