In a crossed field, the magnetic field induction is $2.0 \,T$ and electric field intensity is $20 \times 10^3 \,V/m$. At which velocity will the electron travel in a straight line without being deflected by the electric and magnetic fields?

Force acting on an electron moving with velocity $v$ in a magnetic field $B$ is ($e$ is the charge of electron).

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ particle having charge of $1 \, C$,mass $1 \, kg$ and speed $1 \, m/s$ enters a uniform magnetic field,having magnetic induction of $1 \, T$,at an angle $\theta = 30^\circ$ between the velocity vector and the magnetic induction. The pitch of its helical path is (in meters):

Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is: