A charge $q$ moving with a velocity $r$ at an angle $\theta$ to the magnetic field $B$ experiences a force given by ${\overrightarrow{F}}_B=q\overrightarrow{v}\times \overrightarrow{B}$
The magnitude of the force is $F_B=q\times B\sin \theta (N)$ $B$ in tesla $(T)1T=\frac{1\text{ Newton }}{\text{ Ampere meter }}$
where $\theta$ is the angle between
$\overrightarrow{v}$ and $\overrightarrow{B}$ and the direction of this force is perpendicular to both the velocity $\overrightarrow{v}$ of the particle and the magnetic field $\overrightarrow{B}$. That is the magnetic force is perpendicular to the plane formed by $\overrightarrow{v}$ and $\overrightarrow{B}$.
The direction of this force is also suggested by Fleming's left hand rule which states.
If the thumb, the first finger and the middle finger of the left hand are held mutually perpendicular such that the first finger points in the direction of the field and the middle finger in the direction of a moving positive charge, the thumb points in the direction of the force on the charge. Now, following the left hand rule, we have
If however the charge $q$ is a positive charge, then the force will be in the opposite direction.urther note:ecause $\mathrm{\Sigma }F=F_B=ma$
$$\Rightarrow F_B=qVB=ma$$
for circular motion,
$$\begin{array}{rl}& a=\frac{v^2}{r}\\ & F_B=qvB=\frac{m{v}^2}{r}\end{array}$$
so that
$r=\frac{mv}{qB}$ (radius of circular path)
also $w=\frac{v}{r}=\frac{qrB}{mr}=\frac{qB}{m}$
$\omega =\frac{qB}{m}-\left(\begin{array}{l}\text{ angular speed of the }\\ \text{ particle }\end{array}\right)$
and also,
$$\begin{array}{rl}& T=\frac{2\pi }{\omega }=\frac{2\pi }{qB}\\ & T=\frac{2\pi m}{qB}-\left(\begin{array}{c}\text{ Period of motion }\\ \text{ of the particle }\end{array}\right)\end{array}$$
In summary, the motion of the electron will be in a circle (cyclotron) and in the direction of the force it experiences in the magnetic ficld.s stated earlier, the magnetic force will be perpendicular to the plane formed by both $\overline{v}$ and $\overline{B}$ i.e. $y-x$ or $x^{-1}$
plane, though rendering Option $A$ incorrect.