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(C) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $E_k = \frac{1}{2}mv

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$(r_1 / r_2)^2$

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(C) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $E_k = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mE_k}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mE_k}}{qB}$.Given that the charge $q$,magnetic field $B$,and kinetic energy $E_k$ are the same for both particles,we have $r \propto \sqrt{m}$.Squaring both sides,we get $r^2 \propto m$.Therefore,the ratio of their masses is $\frac{m_x}{m_y} = \left( \frac{r_1}{r_2} \right)^2$.

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

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