$A$ negatively charged particle projected towards east is deflected towards north by a magnetic field. The direction of the magnetic field is

$A$ charged particle of charge $q$ and mass $m$ is accelerated through a potential difference of $V$ volts. It enters a region of orthogonal magnetic field $B$. The radius of its circular path will be:

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

$A$ 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?

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j}) \ \text{T}$ has an acceleration $(4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$. The value of $x$ is . . . . . . .

$A$ neutron,a proton,an electron,and an $\alpha$-particle enter a region of uniform magnetic field with the same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labeled in the figure. Which track does the electron follow?