If an electron is moving in the direction of the magnetic field $\overrightarrow{B}$ with a velocity $\overrightarrow{v}$,then the force on the electron is:

$A$ velocity selector is to be constructed to select ions with a velocity of $6 \,km \,s^{-1}$. If the electric field used is $400 \,V \,m^{-1}$, then the magnetic field to be used is

$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field will be:

Two charged particles,having the same kinetic energy,are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of the radii of their circular paths is $6:5$ and their respective masses ratio is $9:4$,then the ratio of their charges will be.

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

$A$ 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. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.