$A$ beam of well-collimated cathode rays travelling with a speed of $5 \times 10^6 \, m/s$ enters a region of mutually perpendicular electric and magnetic fields and emerges undeviated from this region. If $|B| = 0.02 \, T$,the magnitude of the electric field is:

$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is:

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

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:

Two particles of masses $m_{a}$ and $m_{b}$ and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius $r_{a}$ and $r_{b}$ such that $r_{a} > r_{b}$. Then which is true?