An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

$A$ 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?

If a charged particle enters a uniform magnetic field normally with a certain velocity,then the time period of revolution of the particle

$A$ particle of mass $m$ and charge $q$ moving with velocity $v$ enters region-$b$ from region-$a$ along the normal to the boundary as shown in the figure. Region-$b$ has a uniform magnetic field $B$ perpendicular to the plane of the paper. Also, region-$b$ has length $L$. Choose the correct statements:

Assertion : $A$ proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow{B}$ in a direction perpendicular to $\overrightarrow{B}$,will describe circular trajectories of equal radii.

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be: