An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

The magnetic force acting on a charged particle of charge $2\,\mu C$ in a magnetic field of $2\, T$ acting in the $y-$ direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$ is:

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

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ negatively charged particle projected towards east is deflected towards north by a magnetic field. The direction of the magnetic field is

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is: