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

$A$ proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2 \text{ mT}$ perpendicular to the direction of the field. After some time,it is found that the velocity of the proton has changed in direction by $90^{\circ}$. Then,at this time,the angle between the velocity vectors of the proton and the $\alpha$-particle is (in $^{\circ}$)

$A$ particle having a charge of $10\,\mu C$ and $1\,\mu g$ mass moves along a circular path of $10\, cm$ radius in the presence of a uniform magnetic field of $0.1\, T$. When the charge is at point $P$,a uniform electric field is applied in the region such that the charge moves tangentially with a constant speed. The value of the electric field is......$V/m$

An electron,a proton,and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e, r_p$,and $r_{\alpha}$ respectively in a uniform magnetic field $B$. The relation between $r_e, r_p$,and $r_{\alpha}$ is:

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is:

If a charged particle moves through a magnetic field with a velocity that has both perpendicular and parallel components,what is the path followed by the particle?