A proton is moving in a circular path of radius R in a uniform magnetic field B. It carries a kinetic energy of 2 MeV. What should be the energy of an ∝-particle to move in a circle of same radius in the same field ?
Answer & explanation
Correct answer: option 2
$R = \frac{\sqrt{2mE}}{qB} \Rightarrow E = \frac{q^2B^2R^2}{2m}$
$\Rightarrow \frac{E_{\alpha}}{E_p} = \frac{q_{\alpha}^2}{q_p^2}\times \frac{m_p}{m_{\alpha}} = 4\times \frac{1}{4} = 1$
$\Rightarrow E_{\alpha} = E_p = 2MeV$