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

An electron having a charge $e$ moves with a velocity $\vec{v}$ in the positive $x$-direction. $A$ magnetic field $\vec{B}$ acts on it in the positive $y$-direction. The force on the electron acts in (where the outward direction is taken as the positive $z$-axis):

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

The radius of curvature of the path of a charged particle moving in a static uniform magnetic field is

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:

$A$ proton (or charged particle) moving with velocity $v$ is acted upon by an electric field $E$ and a magnetic field $B$. The proton will move undeflected if