Question

Three particles, an electron $\left(e\right)$, a proton $\left(p\right)$ and a helium $\left(H{e}^{2+}\right)$ are moving in circular paths with constant speeds in the $x-y$ plane in a region where a uniform magnetic field $B$ exists along $z-$axis. The times taken by $e,p$ and $H{e}^{2+}$ inside the field to complete one revolution are $t_e,t_p$ and $t_{He}$ respectively. Then,

• A. $t_{He}>t_p=t_e$
• B. $t_{He}>t_p>t_e$
• C. $t_{He}=t_p=t_e$
• D. None of these

Answer 1

The correct option is B $t_{He}>t_p>t_e$

Here, the centripetal force is balanced by the magnetic force which is experienced by the charge particle.
$\frac{m{v}^2}{r}=qvB$
$\frac{r}{v}=\frac{m}{qB}......\left(1\right)$
Now the time period of revolution is given by
$T=\frac{2\pi r}{v}.........\left(2\right)$
From equation (1) and (2),
$T=\frac{2\pi m}{qB}$
$t_e:t_p:t_{He}=\frac{2\pi m_e}{q_{e}B}:\frac{2\pi m_p}{q_{p}B}:\frac{2\pi m_{He}}{q_{He}B}$
$=\frac{2\pi m_e}{q_{e}B}:\frac{2\pi m_p}{q_{e}B}:\frac{2\pi \times 4m_p}{2q_{e}B}=m_e:m_p:2m_p$
$t_{He}>t_p>t_e$