Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :…….$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )

Which one of the following options represents the magnetic field $\vec{B}$ at $O$ due to the current flowing in the given wire segments lying on the $x y$ plane?

A proton of mass $1.67 \times {10^{ – 27}}\,kg$ and charge $1.6 \times {10^{ – 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X – $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y – $ axis, the path of proton is

Two ions of masses $4 \,{amu}$ and $16\, amu$ have charges $+2 {e}$ and $+3 {e}$ respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :

A charged particle is moving in a circular orbit of radius $6\, cm$ with a uniform speed of $3 \times 10^6\, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4}\, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is