The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to

A charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a

A proton and an $\alpha - $particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25$ $\mu \, sec$ to make $5$ revolutions, then the periodic time for the $\alpha - $ particle would be........$\mu \, sec$

In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$

A charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$ . When electric field is switched off, the particle will move in a circular path of radius

A particle of specific charge $(q/m)$ is projected from the origin of coordinates with initial velocity $[ui - vj]$. Uniform electric magnetic fields exist in the region along the $+y$ direction, of magnitude $E$ and $B.$ The particle will definitely return to the origin once if