Question

A positive charge particle is projected in a region containing uniform electric and magnetic fields with a velocity $v_0$ as shown in figure.


The acceleration of the particle at time $t$ can be expressed as:
(The symbols used have usual meaning)

• A. $\frac{2qE}{m}\hat{i}-{\omega }^{2}R\cos \omega t\hat{k}$
• B. $\frac{qE}{m}\hat{i}+{\omega }^{2}R[-\sin \omega t\hat{j}+\cos \omega t\hat{k}]$
• C. $\frac{qE}{m}\hat{i}$
• D. $\frac{qE}{2m}\hat{i}+{\omega }^{2}R[\sin \omega t\hat{j}-\cos \omega t\hat{k}]$

Answer 1

The correct option is B $\frac{qE}{m}\hat{i}+{\omega }^{2}R[-\sin \omega t\hat{j}+\cos \omega t\hat{k}]$

The velocity of particle can be resolved in two component $v_{||}\&v_{\perp r}$ ; the component $v_{\perp r}$ will lead it to move in a circular path of radius $R$ and angular frequency $\omega$.


The circular path will be described in $y-z$ plane.
After time $t$ let the angle covered by particle is $\beta$.


The centripetal acceleration can be shown as;


$\overrightarrow{a_c}=a\cos \beta \hat{k}-a\sin \beta \hat{j}$

Here $a={\omega }^{2}R$

$\Rightarrow {\overrightarrow{a}}_c={\omega }^{2}R[-\sin \beta \hat{j}+\cos \beta \hat{k}]$

Also, $\beta =\omega t$

$\overrightarrow{a_c}={\omega }^{2}R[-\sin \omega t\hat{j}+\cos \omega t\hat{k}]----\left(i\right)$

The acceleration in $x-$direction due to electric force is,

${\overrightarrow{a}}_x=\frac{q{E}_0}{m}\hat{i}$

Therefore, net acceleration of particle at time ${}^{\prime }{t}^{\prime }$ is,

${\overrightarrow{a}}_{net}=\overrightarrow{a_c}+{\overrightarrow{a}}_x$

${\overrightarrow{a}}_{net}=\frac{q{E}_0}{m}\hat{i}+{\omega }^{2}R[-\sin \omega t\hat{i}+\cos \omega t\hat{k}]$

Hence, option $\left(\text{B}\right)$ is correct.