If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \, m/s$. It enters a region of uniform electric field of $300 \, V/cm$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:

An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

$A$ 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 $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

$A$ small block of mass $20 \,g$ and charge $4 \,mC$ is released on a long smooth inclined plane of inclination angle $45^{\circ}$. $A$ uniform horizontal magnetic field of $1 \,T$ is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is (in $\,s$)

${O^{++}}, {C^+}, {He^{++}}$ and ${H^+}$ ions are projected on a photographic plate with the same velocity in a mass spectrograph. Which one will strike the farthest?