When a charged particle moves perpendicular to a uniform magnetic field,then . . . . . . .

$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

$A$ charge $Q$ moves parallel to a very long straight wire carrying a current $I$ as shown. The force on the charge is

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

Two very long,straight,parallel wires carry steady currents $I$ and $-I$ respectively. 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 the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is