If a charged particle moves through a magnetic field with a velocity that has both perpendicular and parallel components,what is the path followed by the particle?

$A$ particle of charge $16 \times 10^{-16} \, C$ moving with velocity $10 \, ms^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \, Vm^{-1}$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $\vec{B}$ is:

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

Force acting on an electron moving with velocity $v$ in a magnetic field $B$ is ($e$ is the charge of electron).

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.
Reason $(R) :$ The magnetic field in that region is along the direction of velocity of the electron.
In the light of the above statements,choose the correct answer from the options given below.

$A$ proton (or charged particle) moving with velocity $v$ is acted upon by an electric field $E$ and a magnetic field $B$. The proton will move undeflected if