$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

$A$ particle having charge of $1 \, C$,mass $1 \, kg$ and speed $1 \, m/s$ enters a uniform magnetic field,having magnetic induction of $1 \, T$,at an angle $\theta = 30^\circ$ between the velocity vector and the magnetic induction. The pitch of its helical path is (in meters):

Two very long,straight,and parallel wires carry steady currents $I$ and $I$ respectively in opposite directions. The distance between the wires is $d$. At a certain instant of time,a point charge $q$ is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

$OABC$ is a current-carrying square loop. An electron is projected from the center of the loop along its diagonal $AC$ as shown. The unit vector in the direction of initial acceleration will be

$A$ charged particle is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. For the magnetic force on it to be maximum,which of the following is correct?

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be: