Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are $\overrightarrow{v}=3\hat{i}+4\hat{j}$ and $\overrightarrow{a}=2\hat{i}+x\hat{j}$. Select the correct options.

Under the influence of magnetic field only, speed and hence, kinetic energy remains constant.

As $\overrightarrow{F}\perp \overrightarrow{v}$ 

So $\overrightarrow{F}\cdot \overrightarrow{v}=0\Rightarrow m\vec{a}\cdot \overrightarrow{v}=0$

$\Rightarrow m(2\hat{i}+x\hat{j})\cdot (3\hat{i}+4\hat{j})=0$

$\Rightarrow 6+4x=0\Rightarrow x=-1.5$

Let the magnetic field is $\overrightarrow{B}=a\hat{i}+b\hat{j}+c\hat{k}$, then from $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$\Rightarrow m(2\hat{i}+x\hat{j})=q(3\hat{i}+4\hat{j})\times (a\hat{i}+b\hat{j}+c\hat{k})$

$\Rightarrow 2m\hat{i}-1.5m\hat{j}=q(3b\hat{k}-3c\hat{j}-4a\hat{k}+4c\hat{i})$

$4c=2m\Rightarrow c\ne 0$

Hence, magnetic field has a component along z-direction.