$A$ charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other. The particle will move in a:

The radius of the circular path of an electron when subjected to a perpendicular magnetic field is:

$A$ proton of mass $1.67 \times 10^{-27} \, kg$ and charge $1.6 \times 10^{-19} \, C$ is projected with a speed of $2 \times 10^6 \, m/s$ at an angle of $60^\circ$ to the $X$-axis. If a uniform magnetic field of $0.104 \, T$ is applied along the $Y$-axis,the path of the proton is:

$A$ charged particle is moving in a magnetic field of strength $B$ perpendicular to the direction of the field. If $q$ and $m$ denote the charge and mass of the particle respectively,then the frequency of rotation of the particle is

Three ions $H^+$,$He^+$,and $O^{2+}$ having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. Then:

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: