If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field, then

Give expression for the force on a current carrying conductor in a magnetic field.

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

Two charged particle $A$ and $B$ each of charge $+e$ and masses $12$ $amu$ and $13$ $amu$ respectively follow a circular trajectory in chamber $X$ after the velocity selector as shown in the figure. Both particles enter the velocity selector with speed $1.5 \times 10^6 \,ms^{-1}.$ A uniform magnetic field of strength $1.0$ $T$ is maintained within the chamber $X$ and in the velocity selector.

If two protons are moving with speed $v=4.5 \times 10^{5} \,m / s$ parallel to each other then the ratio of electrostatic and magnetic force between them