$A$ 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 $\left(g=10 \,ms^{-2}\right)$.

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.

$A$ charged particle of charge '$q$' is accelerated by a potential difference '$V$' and enters a region of uniform magnetic field '$B$' at right angles to the direction of the field. The charged particle completes a semicircle of radius '$r$' inside the magnetic field. The mass of the charged particle is

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

An electron (mass $m$) is accelerated through a potential difference of $V$ and then it enters a magnetic field of induction $B$ normal to the field lines. The radius of the circular path is ($e$ = electronic charge).

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).