$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move

An electron enters a region where the electrostatic field is $20\,N/C$ and the magnetic field is $5\,T$. If the electron passes undeflected through the region,then the velocity of the electron will be.....$m\,s^{-1}$.

$A$ proton and an alpha particle of the same kinetic energy enter a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and the proton is ....

$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .

Assertion : $A$ proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow{B}$ in a direction perpendicular to $\overrightarrow{B}$,will describe circular trajectories of equal radii.