$A$ proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1 \, MeV,$ the energy acquired by the alpha particle will be......$MeV$

An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2 B_0 \hat{j} \ T$. At a particular instant of time,the velocity of the electron is $\vec{v} = 3 \hat{i} + 5 \hat{j} \ m/s$. If the magnetic force acting on the electron is $\vec{F} = 5e \hat{k} \ N$,where $e$ is the magnitude of the charge of an electron,then the value of $B_0$ is . . . . . . $T$.

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

Consider the mass spectrometer as shown in the figure. The electric field between the plates is $E \ V/m$,and the magnetic field in both the velocity selector and in the deflection chamber has magnitude $B$. Find the radius $r$ for a singly charged ion of mass $m$ in the deflection chamber.

An electron gun with its collector at a potential of $100 \; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \; mm$ of $Hg)$. $A$ magnetic field of $2.83 \times 10^{-4} \; T$ curves the path of the electrons in a circular orbit of radius $12.0 \; cm$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons,and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e/m$ from the data.

$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.