Question

A particle of specific charge (charge/mass) $\alpha$ starts moving from the origin from rest under the action of an electric field $\overrightarrow{E}=E_0\hat{i}$ and magnetic field $\overrightarrow{B}=B_0\hat{k}$. Its velocity at $(x_0,y_0,0)$ is $(4\hat{i}-3\hat{j})$. The value of $x_0$ is :
(All parameters are in SI unit.)

• A. $\frac{13\alpha E_0}{2B_0}$
• B. $\frac{16\alpha B_0}{E_0}$
• C. $\frac{25}{2\alpha E_0}$
• D. $\frac{5\alpha }{2B_0}$

Answer 1

The correct option is C $\frac{25}{2\alpha E_0}$

Given:
$\overrightarrow{E}=E_0\hat{i}$ and $\overrightarrow{B}=B_0\hat{k}$

Initially the charged particle is at rest. Hence, at $t=0$, it will start to accelerate along $(+ve)x-$direction due to presence of electric field. So initial velocity of charge will be along $\text{+ve}x-$ direction.

Given,

Final velocity, ${\overrightarrow{V}}_f=4\hat{i}-3\hat{j}$

$\therefore |{\overrightarrow{V}}_f|=V_f=\sqrt{4^2+3^2}=5\text{m/s}$

Now using work energy theorem;

$W_{elect}+W_{mag}=\Delta K.E$

As we know that work done by magnetic force $\left(W_{mag}\right)=0$.

$W_{elec}+0=\frac{1}{2}\times m\left(5^2\right)-0$

$W_{elec}=\frac{25m}{2}........\left(1\right)$

The position of particle is $(x_0,y_0,0)$
Here, $\Delta x=x_0-0=x_0$

Workdone by electric force:

$W_{elec}=\left(q{E}_0\right)\Delta x=\left(q{E}_0\right)x_0.....\left(2\right)$

From Eqs $\left(1\right)$ & $\left(2\right)$ we get;

$\frac{25m}{2}=q{E}_0{x}_0$

According to problem, $\alpha =q/m$

$\therefore x_0=\frac{25m}{2q{E}_0}=\frac{25{\alpha }^{-1}}{2E_0}=\frac{25}{2\alpha E_0}$

Hence, option $\left(c\right)$ is the correct answer.