An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels:

$A$ uniform magnetic field acts at right angles to the direction of motion of electrons. As a result,the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled,then the radius of the circular path will be.....$cm$.

$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then:

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be

In a crossed field, the magnetic field induction is $2.0 \,T$ and electric field intensity is $20 \times 10^3 \,V/m$. At which velocity will the electron travel in a straight line without being deflected by the electric and magnetic fields?

Consider a long thin conducting wire carrying a uniform current $I$. $A$ particle having mass $M$ and charge $q$ is released at a distance $a$ from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [$\mu_0$ is vacuum permeability]