An electron having a charge e moves with a velocity $v$ in positive $x$ direction. A magnetic field acts on it in positive $y$ direction. The force on the electron acts in (where outward direction is taken as positive $z$-axis).

An electron enters a region where magnetic $(B)$ and electric $(E)$ fields are mutually perpendicular to one another, then

An $\alpha -$ particle of $1\,MeV$ energy moves on circular path in uniform magnetic field. Then kinetic energy of proton in same magnetic field for circular path of double radius is......$MeV$

A charged particle projected in a limited magnetic field according to figure. The charged  particle does not strike to the opposite plate provided

A particle moving with velocity v having specific charge $(q/m)$ enters a region of  magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram......$^o$ 

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. They will move along circular paths.