If a proton enters perpendicularly a magnetic field with velocity $v$, then time period of revolution is $T$. If proton enters with velocity $2 v$, then time period will be

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

$\alpha $ particle, proton and duetron enters in a uniform (transverse) magnetic field $'B'$ with same acceleration potential find ratio of radius of path followed by these particles.

Two very long, straight, parallel wires carry steady currents $I$ and $-I$ respectively. The distance  etween 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

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat k$ .

An electron beam passes through a magnetic field of $2 \times 10^{-3}\,Wb/m^2$ and an electric field of $1.0 \times 10^4\,V/m$ both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively