Question

A proton and an $\alpha$-particle enter a uniform magnetic field perpendicular with the same speed. If proton takes $20\mu s$ to take $5$ revolutions, then the periodic time for the $\alpha$-particle would be-

• A. $5\mu s$
• B. $8\mu s$
• C. $10\mu s$
• D. $16\mu s$

Answer 1

The correct option is C $10\mu s$

Time taken for one revolution is $\frac{20}{5}=4\mu s$

Time period for one revolution of charged particle in a uniform magnetic field is $T=\frac{2\pi m}{Bq}$

For alpha and proton particle

$T_{\alpha }=\frac{2\pi m_{\alpha }}{B{q}_{\alpha }}{T}_p=\frac{2\pi m_p}{B{q}_p}\frac{T_{\alpha }}{T_p}=\frac{m_{\alpha }}{q_{\alpha }}\frac{m_p}{q_p}ATQ{q}_p=1unit{q}_{\alpha }=2unit{m}_{\alpha }=4unit{m}_p=1unit{T}_p=4\frac{T_{\alpha }}{4}=\frac{4\times 1}{2\times 1}=2T_{\alpha }=8\mu s$