$A$ homogeneous electric field $\vec{E}$ and a uniform magnetic field $\vec{B}$ are pointing in the same direction. $A$ proton is projected with its velocity parallel to $\vec{E}$. It will

Two charged particles of specific charges in the ratio $2:1$ and masses in the ratio $1:4$ moving with same kinetic energy enter a uniform magnetic field at right angles to the direction of the field. The ratio of the radii of the circular paths in which the particles move under the influence of the magnetic field is (in $:1$)

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

$A$ particle of charge $q = -16 \times 10^{-18} \, C$ moving with velocity $v = 10 \, m/s$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis,and an electric field of magnitude $E = 10^4 \, V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $B$ is:

$A$ charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$. When the electric field is switched off,the particle will move in a circular path of radius:

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)