An electron is projected with velocity $v_{0}$ in a uniform electric field $E$ perpendicular to the field. Again it is projected 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 :

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Answer 1

For electric field,

Refer image 1.

$q \to$ change of electron.

Four due to electric field $= q E$

$∴ \frac{m v _{0^{2}}}{R _{e}} = q E$

where $R_{e}$ is radius of curvature,

$∴ \frac{1}{R _{e}} = \frac{q E}{m v _{0^{2}}}$

$∴ = \frac{m v _{0^{2}}}{q E} = r_{1}$

In magnetic field,

Focus on electron $= q V_{0} B$

Refer image 2.

Balancing of centripetal and centrifugal focuses,

$\frac{m V _{0^{2}}}{R b} = q V_{0} B$

$∴ R_{b} = \frac{m V _{0}}{q B} = r_{2}$

$∴ \frac{r _{1}}{r _{2}} = \frac{R _{c}}{R _{b}} = \frac{\frac{m V _{0^{2}}}{q E}}{\frac{m V _{0}}{q B}}$

$∴ \frac{r _{1}}{r _{2}} = \frac{B V _{0}}{E}$

Answer is option $D$

Answer 2

For electric field,

Refer image 1.

q \to

change of electron.

Four due to electric field

= q E

∴ \frac{m v _{0^{2}}}{R _{e}} = q E

where

R_{e}

is radius of curvature,

∴ \frac{1}{R _{e}} = \frac{q E}{m v _{0^{2}}}

∴ = \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

∴ R_{b} = \frac{m V _{0}}{q B} = r_{2}

∴ \frac{r _{1}}{r _{2}} = \frac{R _{c}}{R _{b}} = \frac{\frac{m V _{0^{2}}}{q E}}{\frac{m V _{0}}{q B}}

∴ \frac{r _{1}}{r _{2}} = \frac{B V _{0}}{E}

Answer is option

D