A particle with ${10^{ - 11}}\,coulomb$ of charge and ${10^{ - 7}}\,kg$ mass is moving with a velocity of ${10^8}\,m/s$ along the $y$-axis. A uniform static magnetic field $B = 0.5\,Tesla$ is acting along the $x$-direction. The force on the particle is

A proton enters a magnetic field of flux density $1.5\,weber/{m^2}$ with a velocity of $2 \times {10^7}\,m/\sec $ at an angle of $30^\circ $ with the field. The force on the proton will be

A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. The speed of the particle is $10^7\, m/s.$ The magnetic field is directed along the inward normal to the plane of the paper. The particle enters the field at $C$ and leaves at $D.$ Then the angle $\theta$ must be :-.........$^o$

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let $r_p, r_e$ and $r_{He}$ be their respective radii, then

A charged particle moves through a magnetic field perpendicular to its direction. Then

An electron is moving along positive $x$-axis. To get it moving on an anticlockwise circular path in $x-y$ plane, a magnetic filed is applied