When a charged particle enters a uniform magnetic field,its kinetic energy

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon:

The magnetic field vector of an electromagnetic wave is given by $\vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t)$,where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$-axes,respectively. At $t = 0 \, s$,two electric charges $q_1 = 4\pi \, C$ and $q_2 = 2\pi \, C$ are located at $(0, 0, \pi/k)$ and $(0, 0, 3\pi/k)$,respectively. Both charges have the same velocity $\vec{v} = 0.5c\hat{i}$,where $c$ is the speed of light. The ratio of the magnetic force acting on charge $q_1$ to that on $q_2$ is:

$A$ particle having the same charge as an electron moves in a circular path of radius $0.5 \, cm$ under the influence of a magnetic field of $0.5 \, T$. If an electric field of $100 \, V/m$ makes it move in a straight path,then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19} \, C$):

If $\alpha$ and $\beta$ particles are moving with equal velocity perpendicular to the magnetic flux density $B$,then the radii of their paths will be