An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

$A$ capacitor with surface charge density $\sigma$ has a magnetic field $\overrightarrow{B}$ between its two plates. If an electron passes through without any deflection,how long will it take to exit the plates of length $l$?

The radius of curvature of the path of a charged particle moving in a static uniform magnetic field is

Two identical charged particles enter a uniform magnetic field with the same speed but at angles $30^{\circ}$ and $60^{\circ}$ with the field. Let $a, b,$ and $c$ be the ratios of their time periods,radii,and pitches of the helical paths,respectively. Then:

$(a)$ $A$ monoenergetic electron beam with electron speed of $5.20 \times 10^{6} \;m s^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} \;T$ normal to the beam velocity. What is the radius of the circle traced by the beam,given $e/m$ for electron equals $1.76 \times 10^{11} \;C \;kg^{-1}$?
$(b)$ Is the formula you employ in $(a)$ valid for calculating the radius of the path of a $20 \;MeV$ electron beam? If not,in what way is it modified?

Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $\theta$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then,they will meet again for the first time at a point whose distance from the point of projection is