$A$ deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ beam of electrons,initially at rest,is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating potential is increased to $V^{\prime}$ and the force experienced by the electrons in the same magnetic field becomes $2F$. The ratio $\frac{V}{V^{\prime}}$ is:

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

An electron is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. The angle $\theta$ between $\vec{v}$ and $\vec{B}$ lies between $0^\circ$ and $\frac{\pi}{2}$. Its velocity vector $\vec{v}$ returns to its initial value in a time interval of:

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.