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A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loses half of its kinetic energy, then the radius of curvature of its path is
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Reduced to $\frac{1}{\sqrt{2}}$ times of its initial values
We know that
$r=\frac{m V}{q B}=\frac{\sqrt{2 m(K . E)}}{q B}$
$\begin{aligned} & \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{K} . \mathrm{E}} \\ & \Rightarrow \frac{\mathrm{r}_2}{\mathrm{r}_1}=\sqrt{\frac{\mathrm{K} \cdot \mathrm{E}_2}{\mathrm{~K} \cdot \mathrm{E}_1}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{r}_2=\frac{\mathrm{r}_1}{\sqrt{2}}\end{aligned}$
$r=\frac{m V}{q B}=\frac{\sqrt{2 m(K . E)}}{q B}$
$\begin{aligned} & \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{K} . \mathrm{E}} \\ & \Rightarrow \frac{\mathrm{r}_2}{\mathrm{r}_1}=\sqrt{\frac{\mathrm{K} \cdot \mathrm{E}_2}{\mathrm{~K} \cdot \mathrm{E}_1}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{r}_2=\frac{\mathrm{r}_1}{\sqrt{2}}\end{aligned}$
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