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A particle having charge of $10\,\mu C$ and $1\,\mu g$ mass moves along circular path of $10\, cm$ radius in the effect of uniform magnetic field of $0.1\, T$. When charge is at point $'P'$, a uniform electric field applied in the region so charge moves tangentially with constant speed. The value of electric field is......$V/m$
$0.1$
$1$
$10$
$100$
Solution
$\left|\overrightarrow{\mathrm{F}}_{\mathrm{m}}\right|=\left|\overrightarrow{\mathrm{F}}_{\mathrm{e}}\right|$
$\mathrm{qvB}=\mathrm{q} \mathrm{E} \Rightarrow \mathrm{E}=\mathrm{vB}$
from $\quad \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}} \quad \Rightarrow \quad \mathrm{v}=\frac{\mathrm{qBr}}{\mathrm{m}}$
Then $E=\left(\frac{\mathrm{qBr}}{\mathrm{m}}\right) \mathrm{B}=\frac{\mathrm{qB}^{2} \mathrm{r}}{\mathrm{m}}$
$\mathrm{E}=\frac{10 \times 10^{-6} \times 0.1 \times 0.1 \times 10^{-2}}{1 \times 10^{-6} \times 10^{-3}}=10\, \mathrm{V} / \mathrm{m}$
