Physics - Magnetism Question with Solution | TestHub

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, $\frac{\Delta r}{r}\times 100=\frac{\Delta k}{2k}\times 100=2\%$