$A$ proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?

$A$ 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 particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

$A$ 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 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:

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

$A$ beam of electrons,initially at rest,is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating potential is increased to $V^{\prime}$ and the force experienced by the electrons in the same magnetic field becomes $2F$. The ratio $\frac{V}{V^{\prime}}$ is: