An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

An electron enters a chamber in which a uniform magnetic field is present as shown below. An electric field of appropriate magnitude is also applied,so that the electron travels undeviated without any change in its speed through the chamber. We are ignoring gravity. Then,the direction of the electric field is

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

$A$ 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:

$A$ charge $+Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards