$A$ proton and an $\alpha$-particle enter a uniform magnetic field perpendicularly with the same speed. If a proton takes $25 \ \mu s$ to make $5$ revolutions,then the periodic time for the $\alpha$-particle would be ........ $\mu s$.

$A$ charged particle moving along a straight line path enters a uniform magnetic field of $4 \ mT$ at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7 \ C \ kg^{-1}$,the angular velocity of the particle in the magnetic field is

An electron with kinetic energy $5 \ eV$ enters a region of uniform magnetic field of $3 \ \mu T$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$,so that the electron moves along the same path,is . . . . . $N C^{-1}$.
(Given: mass of electron $= 9 \times 10^{-31} \ kg$,electric charge $= 1.6 \times 10^{-19} \ C$)

When a charged particle moves perpendicular to a uniform magnetic field,then . . . . . . .

The figure shows three situations when an electron moves with velocity $\vec v$ through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

As shown in the figure,the uniform magnetic field between the two identical plates is $B$. There is a hole in the bottom plate. If a particle of charge $q$,mass $m$,and energy $E$ enters this magnetic field through this hole,then the particle will not collide with the upper plate provided: