Question

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of electric field $\overrightarrow{E}=E_0\hat{i}$ and $\overrightarrow{B}=B_0\hat{i}$ with velocity $\overrightarrow{V}=V_0\hat{j}$. The speed of particle will become $2V_0$ after a time:

• A. $t=\frac{2m{V}_0}{qE}$
• B. $t=\frac{2Bq}{m{V}_0}$
• C. $t=\frac{\sqrt{3}Bq}{m{V}_0}$
• D. $t=\frac{\sqrt{3}m{V}_0}{qE}$

Answer 1

The correct option is D $t=\frac{\sqrt{3}m{V}_0}{qE}$

The velocity of particle is,
$\overrightarrow{V}=V_0\hat{j}$

Magnetic field $\overrightarrow{B}=B_0\hat{i}$ means along positive $x-$axis.

Thus,$\overrightarrow{V}{\perp }_r\overrightarrow{B}$, hence magnetic force $F$ will tend to revolve the particle in a circular path.

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

Work done by magnetic force, $W_F=0$, so magnetic field will not contribute to the changing of speed of the particle.

But due to present of $\overrightarrow{E}=E_0\hat{i}$ in $x-$direction, particle will accelerate in $x$- direction.

$F_x=q{E}_0$

$\Rightarrow a_x=\frac{F_x}{m}=\frac{q{E}_0}{m}$

Let at time $t$, the speed becomes $2V_0$.

$\therefore 2V_0=\sqrt{V_x^2+V_y^2}$

$2V_0=\sqrt{V_x^2+V_0^2}(\because V_y=V_0=\text{const})$

$4V_0^2=V_x^2+V_0^2$

$V_x=\sqrt{3}{V}_0$

Using kinetic equation in $x-$ direction:

$v_x=u_x+a_{x}t$

$\sqrt{3}{V}_0=0+\left(\frac{q{E}_0}{m}\right)t$

$\therefore t=\frac{\sqrt{3}m{V}_0}{qE}$

Hence, option $\left(d\right)$ is correct.

Alternate Solution:

Using Work-energy theorem we can write,

$\frac{3}{2}m{V_0}^2=\left(qE\right)x=\left(qE\right)\frac{qE}{2m}{t}^2$

$t=\frac{\sqrt{3}m{V}_0}{qE}$