Question

The charge on a particle Y is double the charge on particle X. These two particles X and Y after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is

• A. ${{(2{{R}_{1}}/{{R}_{2}})}^{2}}$
• B. ${{({{R}_{1}}/2{{R}_{2}})}^{2}}$
• C. $R_{1}^{2}/2R_{2}^{2}$
• D. $2{{R}_{1}}/{{R}_{2}}$

Answer 1

Answer: (C)

$R_{1}^{2}/2R_{2}^{2}$

Answer 2

Given, ${{q}_{y}}=2{{q}_{x}}$ Radius of circular path in a magnetic field is given by $r=\frac{mv}{Bq}$ $\therefore$ $v=\frac{Brq}{m}$ ${{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{{{m}^{2}}}$ or $m{{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{m}$ $\therefore$ $KE=\frac{1}{2}m{{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{2m}$ ? (i) When charge particle is accelerated by potential $V$ , then its kinetic energy $KE=Vq$ ... (ii) From Eqs. (i) and (ii) $Vq=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{2m}$ $m=\frac{{{B}^{2}}{{r}^{2}}q}{2V}$ $\therefore$ $m\propto {{r}^{2}}q$ $\therefore$ $\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{r_{1}^{2}{{q}_{1}}}{r_{2}^{2}{{q}_{2}}}$ $=\frac{R_{1}^{2}}{R_{2}^{2}}\times \frac{q}{2q}=\frac{R_{1}^{2}}{2R_{2}^{2}}$