Question

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 :

• A. $\frac{B{v}_0^2}{E}$
• B. $\frac{B}{E}$
• C. $\frac{E{v}_0}{B}$
• D. $\frac{B{v}_0}{E}$

Answer 1

The correct option is D $\frac{B{v}_0}{E}$

For electric field,

Refer image 1.

$q\to$ change of electron.

Four due to electric field $=qE$

$\therefore \frac{m{v}_{0^2}}{R_e}=qE$

where $R_e$ is radius of curvature,

$\therefore \frac{1}{R_e}=\frac{qE}{m{v}_{0^2}}$

$\therefore =\frac{m{v}_{0^2}}{qE}=r_1$

In magnetic field,

Focus on electron $=q{V}_{0}B$

Refer image 2.

Balancing of centripetal and centrifugal focuses,

$\frac{m{V}_{0^2}}{Rb}=q{V}_{0}B$

$\therefore R_b=\frac{m{V}_0}{qB}=r_2$

$\therefore \frac{r_1}{r_2}=\frac{R_c}{R_b}=\frac{\frac{m{V}_{0^2}}{qE}}{\frac{m{V}_0}{qB}}$

$\therefore \frac{r_1}{r_2}=\frac{B{V}_0}{E}$

Answer is option $D$