$A$ charged particle is moving in a uniform magnetic field in a circular path. The radius of the circular path is $R$. When the energy of the particle is doubled,the new radius will be:

Two charged particles of mass $m$ and charge $q$ each are projected from the origin simultaneously with the same speed $V$ in a transverse magnetic field $B$. If $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of the particles (with respect to the origin) at $t = \frac{\pi m}{qB}$, then the value of $\vec{r}_1 \cdot \vec{r}_2$ at that time is:

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.

$A$ 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

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

$A$ rectangular region $ABCD$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. $A$ narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $BC$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$,the magnetic field has to be