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 the $y$-axis. The electron exits from the region after some time with the speed $v$ at coordinate $y$,then

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

$A$ uniform magnetic field acts at right angles to the direction of motion of electrons. As a result,the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled,then the radius of the circular path will be.....$cm$.

In a region,steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. $A$ charged particle is released from rest in this region. The path of the particle will be a

An electron,a proton,and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e, r_p$,and $r_{\alpha}$ respectively in a uniform magnetic field $B$. The relation between $r_e, r_p$,and $r_{\alpha}$ is:

$A$ 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?