An electron is projected with velocity v_0 in | vedclass

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Question

For electric field,

Refer image $1.$

$q ightarrow$ change of electron.

Four due to electric field $=q E$

$	herefore \frac{m v_{0} {^2}}

Vedclass · Accepted Answer

$\frac{{B{v_0}}}{E}$

Vedclass · Answer

For electric field,

Refer image $1.$

$q ightarrow$ change of electron.

Four due to electric field $=q E$

$	herefore \frac{m v_{0} {^2}}{R_{e}}=q E$

where $R_{e}$ is radius of curvature,

$	herefore \frac{1}{R_{e}}=\frac{q E}{m v_{0} {^{2}}}$

$	herefore=\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,

$\frac{m V_{0} {^{2}}}{R b}=q V_{0} B$

$	herefore R_{b}=\frac{m V_{0}}{q B}=r_{2}$

$	herefore \frac{r_{1}}{r_{2}}=\frac{R_{c}}{R_{b}}=\frac{\frac{m V_{0^{2}}}{q E}}{\frac{m V_{0}}{q B}}$

$	herefore \frac{r_{1}}{r_{2}}=\frac{B V_{0}}{E}$

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