$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is:

$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: mass of electron $= 9.1 \times 10^{-31} \, kg$)

When an electron is accelerated by a potential difference of $V$ volts and enters a magnetic field,the force acting on it is $F$. What will be the force acting on the electron when it is accelerated by a potential difference of $5V$ volts and enters the same magnetic field?

The ratio of time periods of $\alpha$-particle and proton moving on circular path in a uniform magnetic field is . . . . . . .

An $\alpha$-particle (mass $4 \text{ amu}$) and a singly charged sulfur ion (mass $32 \text{ amu}$) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region,the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_s$,respectively. The ratio $(r_s / r_\alpha)$ is:

$A$ beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20 \ V m^{-1}$ and $0.5 \ T$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons in $m s^{-1}$?