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

One proton beam enters a magnetic field of ${10^{ - 4}}$ $T$ normally, Specific charge = ${10^{11}}\,C/kg.$ velocity = ${10^7}\,m/s$. What is the radius of the circle described by it....$m$

A particle of charge $ - 16 \times {10^{ - 18}}$ $coulomb$ moving with velocity $10\,\,m{s^{ - 1}}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis, and an electric field of magnitude ${10^4}\,\,V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis, the magnitude of $B$ is

Mark the correct statement

Two particles $A$ and $B$ of masses ${m_A}$ and ${m_B}$ respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are ${v_A}$ and ${v_B}$ respectively, and the trajectories are as shown in the figure. Then

The electrostatic force $\left(\vec{F}_1\right)$ and magnetic force $\left(\vec{F}_2\right)$ acting on a charge $q$ moving with velocity $v$ can be written :