A charged particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to $4$ times its initial value. What will be the ratio of new radius to the original radius of circular path of the charged particle

A uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3 R}{2}$ (region $2$ in the figure) pointing normally into the plane of the paper. A particle with charge $+Q$ and momentum $p$ directed along $x$-axis enters region $2$ from region $1$ at point $P_1(y=-R)$. Which of the following option(s) is/are correct?

$[A$ For $B>\frac{2}{3} \frac{p}{QR}$, the particle will re-enter region $1$

$[B]$ For $B=\frac{8}{13} \frac{\mathrm{p}}{QR}$, the particle will enter region $3$ through the point $P_2$ on $\mathrm{x}$-axis

$[C]$ When the particle re-enters region 1 through the longest possible path in region $2$ , the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from $y$-axis is $p / \sqrt{2}$

$[D]$ For a fixed $B$, particles of same charge $Q$ and same velocity $v$, the distance between the point $P_1$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle

At $t$ = $0$, a positively charged particle of mass $m$ is projected from the origin with velocity $u_0$ at an angle $37^o $ from the $x-$axis as shown in the figure. A constant magnetic field ${\vec B_0} = {B_0}\hat j$  is present in space. After a time interval $t_0$ velocity of particle may be:-

A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is $R$. When energy of particle is doubled, then new radius will be

Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :…….$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )