$A$ 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?

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

$A$ rectangular region of dimensions $(\omega \times l)$ where $\omega \ll l$ has a constant magnetic field $B$ directed into the plane of the paper as shown in the figure. On one side,the region is bounded by a screen. On the other side,positive ions of mass $m$ and charge $q$ are accelerated from rest by a parallel plate capacitor at a constant potential difference $V$ and enter the magnetic field region through a small hole. Which one of the following statements is correct regarding the charge $q$ on the ions that hit the screen?

$A$ particle having some charge is projected in the $x-y$ plane with a speed of $5\ m/s$ in a region having a uniform magnetic field along the $z-$ axis. Which of the following cannot be the possible value of velocity at any time?

The magnetic force acting on a charged particle of charge $-2\, \mu C$ in a magnetic field of $2\, T$ acting in the $y$-direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$,is:

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field: