$A$ neutron,a proton,an electron,and an $\alpha$-particle enter a region of uniform magnetic field with the same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labeled in the figure. Which track does the electron follow?

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

If an electron and a proton having same momenta enter perpendicular to a magnetic field,then

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 is moving in the north direction. It experiences a force in the vertically upward direction. The magnetic field at the position of the electron is in the direction of

An electron having kinetic energy of $100 eV$ circulates in a path of radius $10 cm$ in a magnetic field. The magnitude of magnetic field $|B|$ is approximately [Mass of electron $= 0.5 MeV/c^2$,where $c$ is the velocity of light].