Question

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?

• A. 2 MeV
• B. 1 MeV
• C. 0.5 MeV
• D. 4 MeV

Answer 1

Answer: (B)

1 MeV

For a charged particle's motion in a magnetic field

$F_C=F_m\Rightarrow \frac{m{v}^2}{R}=qVB$

$\Rightarrow R_p=\frac{m_p{v}_p}{q_{p}B}=\frac{P_p}{q_{p}B}=\frac{\sqrt{m{K}_p}}{qB}$

$R_{\alpha }=\frac{\sqrt{2\left(4m\right)K_{\alpha }}}{2qB}$

$\frac{R_p}{R_{\alpha }}=\sqrt{\frac{K_p}{K_{\alpha }}}$

but $R_p=R_{\alpha }$ (given)

Thus $K_p=K_{\alpha }=1MeV$