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(C) The radius of curvature $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula:$r = \frac{mv}{qB} = \frac{\sqrt{

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Reduced to $\frac{1}{\sqrt{2}}$ times of its initial value

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(C) The radius of curvature $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula:$r = \frac{mv}{qB} = \frac{\sqrt{2m(K.E.)}}{qB}$Since $m$,$q$,and $B$ are constant,we have $r \propto \sqrt{K.E.}$.Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = \frac{K_1}{2}$.The ratio of the radii is given by:$\frac{r_2}{r_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{K_1/2}{K_1}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,the radius of curvature is reduced to $\frac{1}{\sqrt{2}}$ times its initial value.

$A$ charged particle is moving in a uniform magnetic field. It penetrates a layer of lead and thereby loses half of its kinetic energy. What happens to the radius of curvature of its path?

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