An $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is

In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

$A$ 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:

An electron and a proton enter a region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius $r_e$ and $r_p$ respectively. Then:

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field: