$A$ particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction,extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

$A$ proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

An electron is moving along the positive $x$-axis. If a uniform magnetic field is applied parallel to the negative $z$-axis,then:
$A.$ The electron will experience a magnetic force along the positive $y$-axis.
$B.$ The electron will experience a magnetic force along the negative $y$-axis.
$C.$ The electron will not experience any force in the magnetic field.
$D.$ The electron will continue to move along the positive $x$-axis.
$E.$ The electron will move along a circular path in the magnetic field.
Choose the correct answer from the options given below:

$A$ charge of $1\,C$ is moving in a magnetic field of $0.5\,T$ with a velocity of $10\,m/s$ perpendicular to the field. The force experienced is.....$N$.

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$