$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

$A$ particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform electric and magnetic fields, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

$A$ charge moves in a circular path perpendicular to a magnetic field. The time period of the revolution is independent of

The magnetic force acting on a charged particle of charge $-2\, \mu C$ in a magnetic field of $2\, T$ acting in the $y$-direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$,is:

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.

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .