In an experiment,electrons are accelerated from rest by applying a voltage of $500 \, V$. Calculate the radius of the path if a magnetic field of $100 \, mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19} \, C$,Mass of the electron $= 9.1 \times 10^{-31} \, kg$]

${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ ions having same kinetic energy pass through a region of space filled with a uniform magnetic field $B$ directed perpendicular to the velocity of the ions. The masses of the ions ${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ are respectively in the ratio $1:4:16$. As a result:

The radius of curvature of the path of a charged particle moving in a static uniform magnetic field is

$A$ magnetic field $\overrightarrow{B} = B_0 \hat{j}$ exists in the region $a < x < 2a$ and $\overrightarrow{B} = -B_0 \hat{j}$ in the region $2a < x < 3a$,where $B_0$ is a positive constant. $A$ positive point charge moving with a velocity $\overrightarrow{v} = v_0 \hat{i}$,where $v_0$ is a positive constant,enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like,

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, Wb/m^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is

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 $:1$)