An electron is moving along the positive $x$-axis. To make it move in an anticlockwise circular path in the $x-y$ plane,a magnetic field must be applied:

$A$ 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 speed of the particle is $v = 10^7 \, m/s$. 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 :-

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis, under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

$A$ particle is projected with a velocity of $10 \ m/s$ along the $y-$axis from the point $(2, 3)$. $A$ uniform magnetic field of $(3\hat{i} + 4\hat{j}) \ T$ exists in the space. What is its speed when the particle passes through the $y-$axis for the third time? (Neglect gravity)