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An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again it is projetced with velocity $v_0$ perpendicular to a uniform magnetic field $B/$ If $r_1$ is initial radius of curvature just after entering in the electric field and $r_2$ is initial radius of curvature just after entering in magnetic field then the ratio $r_1:r_2$ is equal to
$\frac{{Bv_0^2}}{E}$
$\frac{B}{E}$
$\frac{{E{v_0}}}{B}$
$\frac{{B{v_0}}}{E}$
Solution
For electric field,
Refer image $1.$
$q \rightarrow$ change of electron.
Four due to electric field $=q E$
$\therefore \frac{m v_{0} {^2}}{R_{e}}=q E$
where $R_{e}$ is radius of curvature,
$\therefore \frac{1}{R_{e}}=\frac{q E}{m v_{0} {^{2}}}$
$\therefore=\frac{m v_{0} {^{2}}}{q E}=r_{1}$
In magnetic field,
Focus on electron $=q V_{0} B$
Refer image $2 .$
Balancing of centripetal and centrifugal focuses,
$\frac{m V_{0} {^{2}}}{R b}=q V_{0} B$
$\therefore R_{b}=\frac{m V_{0}}{q B}=r_{2}$
$\therefore \frac{r_{1}}{r_{2}}=\frac{R_{c}}{R_{b}}=\frac{\frac{m V_{0^{2}}}{q E}}{\frac{m V_{0}}{q B}}$
$\therefore \frac{r_{1}}{r_{2}}=\frac{B V_{0}}{E}$
