A particle of charge $q$, mass $m$ and kinetic energy E enters in magnetic field perpendicular to its…
Solution

$\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB}$
$\mathrm{mv}=\mathrm{qBr}$
$\mathrm{E}=\frac{1}{2} \mathrm{mv}^2$
$\mathrm{E}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{q}^2 \mathrm{~B}^2 r^2}{\mathrm{~m}^2}\right)=\frac{\mathrm{q}^2 \mathrm{~B}^2 \mathrm{r}^2}{2 \mathrm{~m}}$
$\mathrm{E}=\left(\frac{\mathrm{q}^2 \mathrm{~B}^2}{2 \mathrm{~m}}\right) \mathrm{r}^2$
$r^2 \propto E$

Asked in: JEE Main 2025 (07 Apr Shift 1)
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