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A charged particle moving in a uniform magnetic field and losses $4 \%$ of its kinetic energy. The radius of curvature of its path changes by
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$2 \%$
As we know $\mathrm{F}=\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}$
And $\mathrm{KE}=\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$
$\therefore \mathrm{mv}=\sqrt{2 \mathrm{~km}}$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~km}}}{\mathrm{qB}}$
$\Rightarrow r \propto \sqrt{k}$ or $r=c^{1 / 2}(c$ is a constant $)$ $\frac{\mathrm{dr}}{\mathrm{dr}}=\mathrm{c} \frac{\mathrm{dk}^{1 / 2}}{\mathrm{dr}}$ or $\frac{\mathrm{c} \Delta \mathrm{k}}{\Delta \mathrm{r}}=2 \sqrt{\mathrm{k}}$
$\frac{\Delta r}{r}=\frac{\mathrm{c} \Delta \mathrm{k}}{2 \sqrt{\mathrm{k}} \mathrm{c} \sqrt{\mathrm{k}}}=\frac{\Delta \mathrm{k}}{2 \mathrm{k}}$
Therefore percentage changes in radius of path,
$\frac{\Delta r}{r} \times 100=\frac{\Delta k}{2 k} \times 100=2 \%$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}$
And $\mathrm{KE}=\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$
$\therefore \mathrm{mv}=\sqrt{2 \mathrm{~km}}$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~km}}}{\mathrm{qB}}$
$\Rightarrow r \propto \sqrt{k}$ or $r=c^{1 / 2}(c$ is a constant $)$ $\frac{\mathrm{dr}}{\mathrm{dr}}=\mathrm{c} \frac{\mathrm{dk}^{1 / 2}}{\mathrm{dr}}$ or $\frac{\mathrm{c} \Delta \mathrm{k}}{\Delta \mathrm{r}}=2 \sqrt{\mathrm{k}}$
$\frac{\Delta r}{r}=\frac{\mathrm{c} \Delta \mathrm{k}}{2 \sqrt{\mathrm{k}} \mathrm{c} \sqrt{\mathrm{k}}}=\frac{\Delta \mathrm{k}}{2 \mathrm{k}}$
Therefore percentage changes in radius of path,
$\frac{\Delta r}{r} \times 100=\frac{\Delta k}{2 k} \times 100=2 \%$
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