A proton enters in a uniform electric and magnetic fields $\overrightarrow{E}and\overrightarrow{B}$ respectively. Velocity of proton is $\overrightarrow{v}$. All the three vectors are mutually perpendicular. The proton is deflected along positive x-axis when either of the fields or both are switched on simultaneously. Which of the following statement(s) is/are correct?
1)  $\overrightarrow{v}$  may be along positive y-axis
2)  $\overrightarrow{E}$  is along positive x-axis
3)  $\overrightarrow{B}$ may be along positive z-axis
4)  $\overrightarrow{B}$ may be along negative y-axis
 

$\text{ Since, }\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$

So, for the proton to suffer deflection along +x axis, the force acting on it must be along +x axis. For that to happen, we must/ may have

$\text{ Case }\mathrm{I}:\overrightarrow{\mathrm{E}}\text{ along }\mathrm{x}-\text{ axis }(\overrightarrow{\mathrm{B}}\text{ is switched off) }$

$\text{ Case II: }\overrightarrow{v}\text{ along }+y\text{ axis and }\overrightarrow{B}\text{ along }+\text{ zaxis, }$

$\text{ so that }\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}\text{ is along }+x\text{ axis ( }\overrightarrow{\mathrm{E}}\text{ switched off) }$

$\text{ Case III: }\overrightarrow{E}\text{ along }x\text{-axis, }\overrightarrow{B}\text{ along }+y\text{ axis }\overrightarrow{v}$

$=\pm v_{\mathrm{y}}\hat{\mathrm{j}}-{\mathrm{v}}_{\mathrm{z}}\hat{\mathrm{k}}\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}\text{ is along }+\mathrm{x}\text{ axis }$

$\text{ Case IV: }\overrightarrow{E}\text{ along }+x\text{ axis }\overrightarrow{B}\text{ along }-y\text{ axis }$

$\overrightarrow{v}=\pm v_y\hat{j}+v_2\hat{k}$

$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}\text{ is along }+\mathrm{x}\text{ axis }$