$A$ particle of mass $m$ and charge $q$ moving with velocity $\vec v$ enters a region of uniform magnetic field $\vec B.$ Then:

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.

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .

$A$ proton with a kinetic energy of $2.0 \, eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \, T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $.......... \, cm$. (Take,mass of proton $= 1.6 \times 10^{-27} \, kg$ and charge on proton $= 1.6 \times 10^{-19} \, C$)

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be