An electron and a proton have equal kinetic energies. They enter a magnetic field perpendicularly. Then:

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

If cathode rays are projected at right angles to a magnetic field,their trajectory is

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be