$A$ proton is projected with a uniform velocity $v$ along the axis of a current-carrying solenoid,then

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a uniform magnetic field,speed and energy remain the same for a moving charged particle.
Reason $(R)$: $A$ moving charged particle experiences a magnetic force perpendicular to its direction of motion.

When an electron beam passes through an electric field,it gains kinetic energy. If the same beam passes through a magnetic field,then:

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:

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

If a charge is moving parallel or antiparallel to a magnetic field,what is the magnetic force acting on it?