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:

$A$ particle with charge $-Q$ and mass $m$ enters a magnetic field of magnitude $B$, which exists only to the 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:

$A$ charged particle with charge $q$ enters a region of constant,uniform,and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B}$,and comes out without any change in the magnitude or direction of $\vec{v}$. Then:

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is:

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

Two charged particles $A$ and $B$ of masses $m$ and $2m$, charges $2q$ and $3q$ respectively, moving with the same velocity, enter a uniform magnetic field such that both particles make the same angle $( < 90^{\circ} )$ with the direction of the magnetic field. Then the ratio of the pitches of the helical paths of the particles $A$ and $B$ is