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

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

An electron moves at right angles to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be ...... $cm$.

$A$ uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3R}{2}$ (region $2$ in the figure) pointing normally into the plane of the paper. $A$ particle with charge $+Q$ and momentum $p$ directed along the $x$-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 $B = \frac{8}{13} \frac{p}{QR}$,the particle will enter region $3$ through the point $P_2$ on the $x$-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 the $y$-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.

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)