Question

Three particles, an electron $\left(e\right),$ a proton $\left(p\right)$ and a helium atom $\left(He\right)$ 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 $He$ 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

Answer: (B)

$t_{He}>t_p>t_e$

Time period $T=\frac{2\pi m}{qB}=\frac{2\pi }{B}\frac{m}{q}$
$\frac{m_{He}}{q_{He}}=\frac{4}{2}=2$
$\frac{m_p}{q_p}=\frac{1}{1}=1$
$\frac{m_e}{q_e}=\frac{m_e}{1}=m_e=\frac{1}{1836}$
Since $\frac{m_{He}}{q_{He}}>\frac{m_p}{q_p}>\frac{m_e}{q_e}$, $T_{He}>T_p>T_e$