Question

A proton and an $\alpha -$particle enters a uniform magnetic field in a perpendicular direction with the same speed. If the proton takes $25\mu \text{s}$ to make $5\text{revolutions}$, then the time period for the $\alpha -$particle will be:

Answer 1

Proton is completing $5\text{revolutions}$ in $25\mu \text{s}$. Thus, its time period will be

$T_p=\frac{25}{5}=5\mu \text{s}$

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is given by

$T=\frac{2\pi m}{qB}$

$T\propto \frac{m}{q}(\because B\text{is same in both case})$

$\therefore \frac{T_{\alpha }}{T_p}=\frac{m_{\alpha }}{m_p}\times \frac{q_p}{q_{\alpha }}$

An $\alpha -$particle is actually a helium nucleus which consists of two protons and two neutrons, so its charge is twice the proton's charge and the mass is four times greater. Therefore

$\frac{T_{\alpha }}{T_P}=\frac{4m_p}{m_p}\times \frac{q_p}{2q_p}$

$T_{\alpha }=2T_p$

As, $T_p=5\mu \text{s}$

$\therefore T_{\alpha }=2\times 5=10\mu \text{s}$

Hence, option $\left(c\right)$ is the correct answer.