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

An electron moves along vertical line and away from the observer, then pattern of concentric circular magnetic field lines which are produced due to its motion

A charged particle of specific charge $\alpha$ is released from origin at time $t = 0$ with velocity $\vec V = {V_o}\hat i + {V_o}\hat j$ in magnetic field $\vec B = {B_o}\hat i$ . The coordinates of the particle at time $t = \frac{\pi }{{{B_o}\alpha }}$ are (specific charge $\alpha = \,q/m$) 

A particle moving with velocity v having specific charge $(q/m)$ enters a region of  magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram……$^o$ 

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