The distance between two plates is $1 \ cm$ and the potential difference is $1000 \ V$. $A$ magnetic field $B = 1 \ T$ is applied. If an electron passes through without any deflection,what will be its velocity?

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ ions having same kinetic energy pass through a region of space filled with a uniform magnetic field $B$ directed perpendicular to the velocity of the ions. The masses of the ions ${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ are respectively in the ratio $1:4:16$. As a result:

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?

$A$ particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent,as shown. The particle after entering the region collides elastically with a rigid wall. The time after which the velocity of the particle becomes anti-parallel to its initial velocity is

$A$ proton of mass $m$ moving with a speed $v$ ($v << c$, where $c$ is the velocity of light in vacuum) completes a circular orbit in time $T$ in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}v$, what will be the time needed to complete the circular orbit?