A rectangular region $A B C D$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. A narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $B C$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$, the magnetic field has to be

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 $m$ 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 

A positive charge $'q'$ of mass $'m'$ is moving along the $+ x$ axis. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that

An electron emitted by a heated cathode and accelerated through a potential difference of $ 2.0 \;kV$, enters a region with uniform magnetic field of $0.15\; T$. Determine the trajectory of the electron if the field

$(a)$ is transverse to its initial velocity,

$(b)$ makes an angle of $30^o$ with the initial velocity

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.