The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

$A$ 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 $B = 1 \, T$ along the direction shown in the figure. The speed of the particle is $v = 10^7 \, m/s$. The magnetic field is directed along the inward normal to the plane of the paper. The particle enters the field at $C$ and leaves at $D$. Then the angle $\theta$ must be :-

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat k$. $A$ particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{mv}{2qB}$,the acceleration of the charged particle at the point of its emergence at the other side is

$A$ tightly wound long solenoid has $n$ turns per unit length,a radius $r$ and carries a current $I$. $A$ particle having charge $q$ and mass $m$ is projected from a point on the axis in a direction perpendicular to the axis. The maximum speed of the particle for which the particle does not strike the solenoid is

An electron and a proton have equal kinetic energies. They enter a magnetic field perpendicularly. Then:

$A$ 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......$MeV$