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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
$10\times 10^6\,m/s,$ $2.43\,cm$
$2.5\times 10^6\,m/s,$ $0.43\,cm$
$5\times 10^6\,m/s,$ $1.43\,cm$
none of these
Solution
$\mathrm{B}=2 \times 10^{-3}\, \mathrm{Wb} / \mathrm{m}^{2}$
$\mathrm{E}=1 \times 10^{4}\, \mathrm{V} / \mathrm{m}^{2}$
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., $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bev} \Rightarrow \quad \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}$
$=\frac{9.1 \times 10^{-31} \times 5 \times 10^{6}}{2 \times 10^{-3} \times 1.6 \times 10^{-19}}$
$=14.3 \times 10^{-3}\, \mathrm{m}=1.43\, \mathrm{cm}$
