An electron beam passes through a magnetic field of $2 \times 10^{-3}\,Wb/m^2$ and an electric field of $1.0 \times 10^4\,V/m$ both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively
AIIMS 2011, Medium
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$\mathrm{B}=2 \times 10^{-3}\, \mathrm{Wb} / \mathrm{m}^{2}$
$\mathrm{E}=1 \times 10^{4}\, \mathrm{V} / \mathrm{m}^{2}$
Since the path of electron remains undeviated, $\mathrm{qvB}=\mathrm{q} \mathrm{E}$ or
$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}=\frac{1 \times 10^{14}}{2 \times 10^{-3}}=0.5 \times 10^{7}$
$=5 \times 10^{6}\, \mathrm{m} / \mathrm{s}$
If the electricfield is removed, the path of the charged particle is circular and magnetic field provides the necessary centripetal force.
i.e., $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bev} \Rightarrow \quad \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}$
$=\frac{9.1 \times 10^{-31} \times 5 \times 10^{6}}{2 \times 10^{-3} \times 1.6 \times 10^{-19}}$
$=14.3 \times 10^{-3}\, \mathrm{m}=1.43\, \mathrm{cm}$
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