$A$ proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

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 the:

Particles having positive charges occasionally come with high velocity from the sky towards the earth. On account of the magnetic field of the earth,they would be deflected towards the

$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.

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.