Q92 / 125 · BITSAT 2015 · Physics

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

  • Correct answer
    A
    $2 \%$
  • B
    $4 \%$
  • C
    $10 \%$
  • D
    $12 \%$

Solution

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

Hence, the correct answer is option: 'A'.

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