Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to the magnetic field?

$A$ charge $+Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

$A$ 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?

When a charged particle enters a uniform magnetic field,its kinetic energy

$A$ proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

$A$ charged particle of charge $q$ and mass $m$ gets deflected through an angle $\theta$ upon passing through a square region of side $a$,which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side,what is the speed of the particle?