JEE Main 2019 — Magnetic Effects of Current Question with Solution

Radius of the path (r) is given by $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$ \begin{aligned} \mathrm{r} &=\frac{\sqrt{2 \mathrm{mk}}}{\mathrm{eB}} \quad(\because \mathrm{p}=\mathrm{mv}=\sqrt{2 \mathrm{mk}}) \\ &=\frac{\sqrt{2 \mathrm{meV}}}{\mathrm{eB}} \quad(\because \mathrm{k}=\mathrm{eV}) \\ \mathrm{r} &=\frac{\sqrt{\frac{2 \mathrm{~m}}{\mathrm{e}} \mathrm{V}}}{\mathrm{B}}=\frac{\sqrt{\frac{2 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}}(500)}}{100 \times 10^{-3}} \end{aligned} $r=\frac{\sqrt{\frac{9.1}{0.16}\times 10^{-10}}}{10^{-1}}=\frac{3}{4}\times 10^{-4}$ $=7.5\times 10^{-4}$