$A$ 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

An electron is orbiting in a circular orbit under the influence of a constant transverse magnetic field of strength $B$. Assuming that Bohr's postulate regarding the quantisation of angular momentum holds good for this electron,find the kinetic energy of the electron in the $n^{th}$ orbit. ($m_e =$ mass of electron,$e =$ magnitude of charge of electron)

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: mass of electron $m_e = 9 \times 10^{-31} \ kg$)

$A$ particle with charge $q$,moving with a momentum $p$,enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$,where $d < \frac{p}{Bq}$. The particle is deflected by an angle $\theta$ in crossing the field.

$A$ charged particle of charge $q$ and mass $m$ is accelerated through a potential difference of $V$ volts. It enters a region of orthogonal magnetic field $B$. The radius of its circular path will be:

$A$ charged particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,mT$ at a right angle to the direction of the field. The charged particle completes a semicircle of radius $3\,cm$ inside the magnetic field. The mass of the charged particle is $........\times 10^{-18}\,kg$.