If a charged particle enters perpendicularly in a uniform magnetic field,then:

An electron is travelling in the east direction and a magnetic field is applied in the upward direction. In which direction will the electron deflect?

$A$ proton of mass $m$ moving with a speed $v$ ($v << c$, where $c$ is the velocity of light in vacuum) completes a circular orbit in time $T$ in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}v$, what will be the time needed to complete the circular orbit?

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_0 \hat{j}$ and $\vec{B}=B_0 \hat{j}$. At time $t=0$,this charge has velocity $\vec{v}$ in the $x-y$ plane,making an angle $\theta$ with the $x$-axis. Which of the following option$(s)$ is(are) correct for time $t>0$?
$(A)$ If $\theta=0^{\circ}$,the charge moves in a circular path in the $x-z$ plane.
$(B)$ If $\theta=0^{\circ}$,the charge undergoes helical motion with constant pitch along the $y$-axis.
$(C)$ If $\theta=10^{\circ}$,the charge undergoes helical motion with its pitch increasing with time,along the $y$-axis.
$(D)$ If $\theta=90^{\circ}$,the charge undergoes linear but accelerated motion along the $y$-axis.