Question

In a certain region there exist a uniform electric field $\overrightarrow{E}$ and a uniform magnetic field $\overrightarrow{B}$ both directed along z-axis. A particle of charge $Q$ and mass $m$ is projected at time $t=0$ with a velocity $v_0$ along $x$-axis. Find the velocity at any instant of time $t$.

• A. $v_0\cos \left(\frac{QBt}{m}\right)\hat{i}-v_0\sin \left(\frac{QBt}{m}\right)\hat{j}+\left(\frac{QEt}{m}\right)\hat{k}$
• B. $v_0\cos \left(\frac{QBt}{m}\right)\hat{i}+v_0\sin \left(\frac{QBt}{m}\right)\hat{j}+\left(\frac{QEt}{m}\right)\hat{k}$
• C. $v_0\sin \left(\frac{QBt}{m}\right)\hat{i}-v_0\cos \left(\frac{QBt}{m}\right)\hat{j}-\left(\frac{QEt}{m}\right)\hat{k}$
• D. $v_0\cos \left(\frac{QBt}{m}\right)\hat{i}-v_0\sin \left(\frac{QBt}{m}\right)\hat{j}-\left(\frac{QEt}{m}\right)\hat{k}$

Answer 1

The correct option is A $v_0\cos \left(\frac{QBt}{m}\right)\hat{i}-v_0\sin \left(\frac{QBt}{m}\right)\hat{j}+\left(\frac{QEt}{m}\right)\hat{k}$

The Lorentz's force on a charged particle is,

$\overrightarrow{F}=Q(\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B}\left)\right)$

where $\overrightarrow{E}=E\hat{k},\overrightarrow{B}=B\hat{k}$

Let $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$

$\overrightarrow{F}=Q[E\hat{k}+(v_x\hat{i}+v_y\hat{j}+v_z\hat{k})\times B\hat{k}]$

$m[\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k}]]=Q[v_{y}B\hat{i}-v_{x}B\hat{j}+E\hat{k}]$

First we consider $z$-component ; we get

$\frac{d{v}_z}{dt}=\frac{QE}{m}.....\left(1\right)$

Integrating both sides with proper limits we get,

${\int }_0^{v_z}d{v}_z={\int }_0^t\frac{QE}{m}dt$

Hence $v_z=\frac{qEt}{m}$ (as $v_z=0$ when $t=0$)

Now considering motion in the $x$ and $y$-directions. We have

$\frac{d{v}_x}{dt}=\frac{QB{v}_y}{m}.....\left(2\right)$

and $\frac{d{v}_y}{dt}=-\frac{QB{v}_x}{m}.....\left(3\right)$

This is a pair of coupled equations. It can be decoupled by differentiating one of them w.r.t. time and substituting the result into the other equation. We get,

$\frac{d^2{v}_x}{d{t}^2}=\frac{Q^2{B}^2}{m^2}{v}_x$

and $\frac{d^2{v}_y}{d{t}^2}=-\frac{Q^2{B}^2{v}_y}{m^2}$

These equations are similar to equation of S.H.M., with the general solutions ,

$v_x=a\cos \omega t+b\sin \omega t$

$v_y=c\cos \omega t+d\sin \omega t$

Where, $\omega =\frac{QB}{m}$

From the given initial condition, $t=0,v_x=v_0$ and $v_y=0$

Also $\frac{d{v}_x}{dt}=0$

From eq (3) we get,

and $\frac{d{v}_y}{dt}=-\frac{QB{v}_0}{m}=-\omega v_0$

thus $a=v_0,b=0$

$c=0,d=-v_0$

We can thus finally write the velocity at time $t$ as,

$v=v_0\cos \left(\frac{QBt}{m}\right)\hat{i}-v_0\sin \left(\frac{QBt}{m}\right)\hat{j}+\frac{QEt}{m}\hat{k}$

Hence, option (a) is the correct answer.