One proton beam enters a magnetic field of ${10^{ - 4}}$ $T$ normally, Specific charge = ${10^{11}}\,C/kg.$ velocity = ${10^7}\,m/s$. What is the radius of the circle described by it....$m$

An electron moving with a speed $u$ along the positive $x-$axis at $y = 0$ enters a region of uniform magnetic field $\overrightarrow B = - {B_0}\hat k$ which exists to the right of $y$-axis. The electron exits from the region after some time with the speed $v$ at co-ordinate $y$, then

A beam of well collimated cathode rays travelling with a speed of $5 \times {10^6}\,m{s^{ - 1}}$ enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If $| B |=0.02\; T$, the magnitude of the electric field is

A charged particle projected in a limited magnetic field according to figure. The charged  particle does not strike to the opposite plate provided

Statement $-1$ : Path of the charge particle may be straight line in uniform magnetic field.
Statement $-2$ : Path of the charge particle is decided by the angle between its velocity and the magnetic force working on it

Two charges of same magnitude move in two circles of radii $R_1=R$ and $R_2=2 R$ in a region of constant uniform magnetic field $B _0$. The work $W_1$ and $W_2$ done by the magnetic field in the two cases respectively, are such that