$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.

Two particles, $A$ and $B$, having equal charges, after being accelerated through the same potential difference, enter into a region of uniform magnetic field and the particles describe circular paths of radii $R_{1}$ and $R_{2}$ respectively. The ratio of the masses of $A$ and $B$ is

An electron gun with its collector at a potential of $100 \; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \; mm$ of $Hg)$. $A$ magnetic field of $2.83 \times 10^{-4} \; T$ curves the path of the electrons in a circular orbit of radius $12.0 \; cm$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons,and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e/m$ from the data.

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

$A$ deuteron of kinetic energy $50 \, keV$ is describing a circular orbit of radius $0.5 \, m$ in a plane perpendicular to a magnetic field $\vec{B}$. The kinetic energy of a proton that describes a circular orbit of radius $0.5 \, m$ in the same plane with the same $\vec{B}$ is ........ $keV$.

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.