A negatively charged particle projected towards east is deflected towards north by a magnetic field. The field may be

When a charged particle moving with velocity $\vec V$ is subjected to a magnetic field of induction $\vec B$ , the force on it is non-zero. This implies the

A particle having charge of $10\,\mu C$ and $1\,\mu g$ mass moves along circular path of $10\, cm$ radius in the effect of uniform magnetic field of $0.1\, T$. When charge is at point $'P'$, a uniform electric field applied in the region so charge moves tangentially with constant speed. The value of electric field is......$V/m$

The magnetic field is uniform for $y>0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y<0$. A proton denoted by filled circle leaves $y=0$ in the $-y$-direction with some speed as shown below.Which of the following best denotes the trajectory of the proton?

An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.

(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )

A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be