An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3}\,T$. The radius of path of the circle will approximately be :
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$\because \mathrm{evB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$
$\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{\sqrt{2 \mathrm{mE}}}{\mathrm{eB}}$
or $\quad \mathrm{r}=\frac{\sqrt{2 \times 9.1 \times 10^{-31} \times 880 \times 1.6 \times 10^{-19}}}{1.6 \times 10^{-19} \times 2.5 \times 10^{-3}} \approx 0.04\, \mathrm{m}=4 \,\mathrm{cm}$
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