Question

A proton and an $\alpha -$particle enter in a uniform magnetic field moving with the same speed. If the proton takes $25\mu s$ to make $5$ revolutions, then the periodic time for the $\alpha -$particle would be

• A. $50\mu s$
• B. $25\mu s$
• C. $10\mu s$
• D. $5\mu s$

Answer 1

Answer: (C)

$10\mu s$

Time period of revolution for the charge $q$ and mass $m$ in a circular path: $T=\frac{2\pi m}{qB}$
$T_{\alpha }=\frac{2\pi m_{\alpha }}{q_{\alpha }B}$
$T_p=\frac{2\pi m_p}{q_{p}B}$ where $p$ in the subscript is for proton.
$\frac{T_{\alpha }}{T_p}=\frac{\frac{2\pi m_{\alpha }}{q_{\alpha }B}}{\frac{2\pi m_p}{q_{p}B}}$
$T_{\alpha }=T_p\times \frac{m_{\alpha }}{m_p}\times \frac{q_p}{q_{\alpha }}=5\times \frac{4}{1}\times \frac{1}{2}=10\mu s$ since $m_{\alpha }=4m_p$ and $q_{\alpha }=2q_p$