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Question

# Two particles X and Y having equal charge, 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) (R1/R2)1/2 (b) R1/R2 (c) (R1/R2)2 (d) R1R2.

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Solution

## (c)$\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}$ Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 such that ${R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}$ Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy. $\therefore \frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{2}{{v}_{2}}^{2}\phantom{\rule{0ex}{0ex}}\because {R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}⇒{v}_{1}=\frac{{R}_{1}qB}{{m}_{1}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}⇒{v}_{2}=\frac{{R}_{2}qB}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\therefore {m}_{1}{\left(\frac{{R}_{1}qB}{{m}_{1}}\right)}^{2}={m}_{2}{\left(\frac{{R}_{2}qB}{{m}_{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{m}_{1}}{{m}_{2}}=\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}$

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