$A$ 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

An electron of mass $m$ and charge $q$ is moving with a speed $v$ in a circular path of radius $r$ perpendicular to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved,what will be the radius of the resulting path?

An electron enters the space between the plates of a charged capacitor as shown. The surface charge density on the plates is $\sigma$. The electric field intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in the space,perpendicular to the direction of $E$. The electron moves perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$ without any change in direction. The time taken by the electron to travel a distance $l$ in the space is:

$A$ 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 $\times$)

${O^{++}}, {C^+}, {He^{++}}$ and ${H^+}$ ions are projected on a photographic plate with the same velocity in a mass spectrograph. Which one will strike the farthest?

$A$ charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when