$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

$A$ proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of the radii of their circular paths 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:

Give an example of a situation in which an applied force does not result in a change in kinetic energy.