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 $1.6 \times 10^{-27} \;\mathrm{kg}$ )

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and $\vec B = {B_0}\hat i$ with velocity ${\rm{\vec v}} = {{\rm{v}}_0}\hat j$. The speed of the particle will become $2v_0$ after a time

A particle of mass $'m'$ and carrying a charge $'q'$ enters with a velocity $'v'$ perpendicular to a uniform magnetic field. The time period of rotation of the particle

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

The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to……. $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$