An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

Two electrons,$e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the kinetic energy of $e_1$ is double than that of $e_2$. The relation of their frequencies of rotation,$f_1$ and $f_2$ is

An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

The magnetic field vector of an electromagnetic wave is given by $\vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t)$,where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$-axes,respectively. At $t = 0 \, s$,two electric charges $q_1 = 4\pi \, C$ and $q_2 = 2\pi \, C$ are located at $(0, 0, \pi/k)$ and $(0, 0, 3\pi/k)$,respectively. Both charges have the same velocity $\vec{v} = 0.5c\hat{i}$,where $c$ is the speed of light. The ratio of the magnetic force acting on charge $q_1$ to that on $q_2$ is:

Ionized hydrogen atoms and $\alpha$-particles with same momenta enter perpendicular to a constant magnetic field $B$. The ratio of the radii of their paths $r_{H}: r_{\alpha}$ will be

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?