When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that

$A$ stream of electrons is projected horizontally to the right. $A$ straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right,then what will be the effect on the electron stream?

$A$ particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then

Two ions have equal masses but one is singly ionized and the second is doubly ionized. They are projected from the same place in a uniform transverse magnetic field with the same velocity. Then:
$(a)$ Both ions will move along circles of equal radii.
$(b)$ The radius of the circle described by the singly ionized charge is double the radius of the circle described by the doubly ionized charge.
$(c)$ Both circles do not touch each other.
$(d)$ Both circles touch each other.

$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ particle of mass $m$ and charge $q$ moving with velocity $v$ enters region-$b$ from region-$a$ along the normal to the boundary as shown in the figure. Region-$b$ has a uniform magnetic field $B$ perpendicular to the plane of the paper. Also,region-$b$ has length $L$. Choose the correct statements: