An electron and a proton of equal linear momentum enter in a direction perpendicular to a uniform magnetic field. If the radii of their circular paths are $r_e$ and $r_p$ respectively,then $\frac{r_e}{r_p}$ is equal to - (mass of electron $= m_e$,mass of proton $= m_p$)

$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

An $\alpha$-particle travels in a circular path of radius $0.45\,m$ in a magnetic field $B = 1.2\,Wb/m^2$ with a speed of $2.6 \times 10^7\,m/s$. The period of revolution of the $\alpha$-particle is:

An electron (mass $= 9.1 \times 10^{-31} \, kg$; charge $= -1.6 \times 10^{-19} \, C$) experiences no deflection if subjected to an electric field of $3.2 \times 10^5 \, V/m$ and a magnetic field of $2.0 \times 10^{-3} \, Wb/m^2$. Both the fields are normal to the path of the electron and to each other. If the electric field is removed,then the electron will revolve in an orbit of radius.....$m$:

$A$ proton and an alpha particle moving with equal speeds enter normally into a uniform magnetic field. The ratio of times taken by the proton and the alpha particle to make one complete revolution in the magnetic field is

Bohr model is applied to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a transverse magnetic field $B$. The energy of the charged particle in the $n^{\text{th}}$ level will be $[h = \text{Planck's constant}]$