Question

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

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

Answer 1

The correct option is C $10\mu sec$.

Time period of proton $T_p=\frac{25}{5}=5\mu sec$
By using $T=\frac{2\pi m}{qB}\Rightarrow \frac{T_{\alpha }}{T_p}=\frac{m_{\alpha }}{m_p}\times \frac{q_p}{q_{\alpha }}=\frac{4m_p}{m_p}\times \frac{q_p}{2q_p}\Rightarrow T_{\alpha }=2T_p=10\mu s$.