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(B) When an ion of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by: $E = qV = \frac{1}{2}mv^2

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$\frac{1}{R^{2}}$

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(B) When an ion of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by: $E = qV = \frac{1}{2}mv^2$. From this,the velocity $v$ is $v = \sqrt{\frac{2qV}{m}}$.When this ion enters a magnetic field $B$ perpendicular to its motion,it follows a circular path of radius $R$ due to the Lorentz force acting as the centripetal force: $qvB = \frac{mv^2}{R}$.Substituting the expression for $v$ into the force equation: $qvB = \frac{m}{R} \left(\frac{2qV}{m}\right) = \frac{2qV}{R}$.Simplifying for the charge-to-mass ratio: $qB = \frac{2qV}{vR} \implies B = \frac{mv}{qR} \implies R = \frac{mv}{qB}$.Substituting $v = \sqrt{\frac{2qV}{m}}$ into the radius equation: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Squaring both sides: $R^2 = \frac{2mV}{qB^2}$.Rearranging to find the ratio $\frac{q}{m}$: $\frac{q}{m} = \frac{2V}{R^2 B^2}$.Since $V$ and $B$ are constant,$\frac{q}{m} \propto \frac{1}{R^2}$.

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