$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

An electron enters with a velocity $\vec{v} = v_0 \hat{i}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the $xy$-plane. Suggest a configuration of fields $\vec{E}$ and $\vec{B}$ that can lead to this.

$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the 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)

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ proton and a helium nucleus are shot into a magnetic field at right angles to the field with the same kinetic energy. The ratio of their radii is: