An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

Ionized hydrogen atoms and $\alpha$-particles with same momenta enter perpendicular to a constant magnetic field $B$. The ratio of the radii of their paths $r_{H}: r_{\alpha}$ will be

$A$ moving electron produces

$A$ particle having a mass of $10^{-2} \, kg$ carries a charge of $5 \times 10^{-8} \, C$. The particle is given an initial horizontal velocity of $10^5 \, m/s$ in the presence of an electric field $\vec{E}$ and a magnetic field $\vec{B}$. To keep the particle moving in a horizontal direction,it is necessary that:
$(1)$ $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity.
$(2)$ Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity.
$(3)$ Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.
$(4)$ $\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity.
Which one of the following pairs of statements is possible?

$A$ capacitor with surface charge density $\sigma$ has a magnetic field $\overrightarrow{B}$ between its two plates. If an electron passes through without any deflection,how long will it take to exit the plates of length $l$?

An electron (mass $= 9.1 \times 10^{-31} \, kg$; charge $= -1.6 \times 10^{-19} \, C$) experiences no deflection if subjected to an electric field of $3.2 \times 10^5 \, V/m$ and a magnetic field of $2.0 \times 10^{-3} \, Wb/m^2$. Both the fields are normal to the path of the electron and to each other. If the electric field is removed,then the electron will revolve in an orbit of radius.....$m$: