An electron is moving along the positive $x$-axis. $A$ uniform electric field exists towards the negative $y$-axis. What should be the direction of a magnetic field of suitable magnitude so that the net force on the electron is zero?

If an electron is moving in the direction of the magnetic field $\overrightarrow{B}$ with a velocity $\overrightarrow{v}$,then the force on the electron is:

The direction of the magnetic force on the electron as shown in the diagram is along:

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

The figure shows the face of a cathode-ray oscilloscope tube,as viewed from the front. The electron beam is coming out normally from the plane of the paper. The electron beam passes through a region where there are electric and magnetic fields directed as shown. The deflections of the spot from the center of the screen produced by the electric field $E$ and the magnetic field $B$ separately are equal in magnitude. Which one of the diagrams below shows a possible position of the spot on the screen when both fields are operating?

$A$ charge '$q$' moves with a velocity $2 \ m/s$ along the $x$-axis in a uniform magnetic field $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$. The charge will experience a force: