$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

An electron having mass $9.1 \times 10^{-31} \ kg$,charge $1.6 \times 10^{-19} \ C$ and moving with velocity of $10^6 \ ms^{-1}$ enters a region where a magnetic field exists. If it describes a circle of radius $0.2 \ m$,then the intensity of the magnetic field must be . . . . . . $\times 10^{-5} \ T$.

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. . . . . .

$A$ charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other. The particle will move in a:

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a uniform magnetic field,speed and energy remain the same for a moving charged particle.
Reason $(R)$: $A$ moving charged particle experiences a magnetic force perpendicular to its direction of motion.

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be: