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 fields $B$. If $r_1$ is initial radius of curvature just after entering into the electric field and $r_2$ is initial radius of curvature just after entering into the magnetic field then the ratio $r_1/r_2$ is equal to (Assume gravity free space)

• 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 $\frac{B{v}_0}{E}$


Electron, $e^{-}$ is projected with velocity ${\overrightarrow{v}}_0$ perpendicular to the uniform electric field $\overrightarrow{E}$.

Electric force on the $e^{-}$

$\overrightarrow{F_E}=q\overrightarrow{E}$

$\Rightarrow |\overrightarrow{F_E}|=eE$

This force is responsible for initial radius of currature ($r_1$)

$\therefore eE=\frac{m{v}_0^2}{r_1}$

$r_1=\frac{m{v}_0^2}{eE}.......\left(1\right)$

Again the same $e^{-}$ is projected with velocity ${\overrightarrow{v}}_0\perp$ to uniform magnetic field $\overrightarrow{B}$.

Magnetic force on the $e^{-}$ is,

$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\Rightarrow |{\overrightarrow{F}}_B|=e{v}_{0}B$

Now this force is responsible for initial radius of currature $\left(r_2\right)$

$\therefore e{v}_{0}B=\frac{m{v}_0^2}{r_2}$

$\Rightarrow r_2=\frac{m{v}_0}{eB}......\left(2\right)$

On dividing eqn $\left(1\right)$ by $\left(2\right)$ we get,

$\frac{r_1}{r_2}=\frac{m{v}_0^2/eE}{m{v}_0/eB}$

$\frac{r_1}{r_2}=\frac{B{v}_0}{E}$

Hence, option (d) is the correct answer.

Why this quetion ? To understand parameters on which radius of curvature depends under influence of different forces.